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http://ugrad.stat.ubc.ca/R/library/base/html/scale.html | 1,553,098,354,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202433.77/warc/CC-MAIN-20190320150106-20190320172106-00369.warc.gz | 205,608,405 | 1,868 | scale {base} R Documentation
## Scaling and Centering of Matrix-like Objects
### Description
`scale` is generic function whose default method centers and/or scales the columns of a numeric matrix.
### Usage
```scale(x, center = TRUE, scale = TRUE)
```
### Arguments
`x` a numeric matrix(like object). `center` either a logical value or a numeric vector of length equal to the number of columns of `x`. `scale` either a logical value or a numeric vector of length equal to the number of columns of `x`.
### Details
The value of `center` determines how column centering is performed. If `center` is a numeric vector with length equal to the number of columns of `x`, then each column of `x` has the corresponding value from `center` subtracted from it. If `center` is `TRUE` then centering is done by subtracting the column means (omitting `NA`s) of `x` from their corresponding columns, and if `center` is `FALSE`, no centering is done.
The value of `scale` determines how column scaling is performed (after centering). If `scale` is a numeric vector with length equal to the number of columns of `x`, then each column of `x` is divided by the corresponding value from `scale`. If `scale` is `TRUE` then scaling is done by dividing the (centered) columns of `x` by their root-mean-square, and if `scale` is `FALSE`, no scaling is done.
The root-mean-square for a column is obtained by computing the square-root of the sum-of-squares of the non-missing values in the column divided by the number of non-missing values minus one.
### Value
For `scale.default`, the centered, scaled matrix. The numeric centering and scalings used (if any) are returned as attributes `"scaled:center"` and `"scaled:scale"`
### References
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.
`sweep` which allows centering (and scaling) with arbitrary statistics.
For working with the scale of a plot, see `par`.
### Examples
```require(stats)
x <- matrix(1:10, nc=2)
(centered.x <- scale(x, scale=FALSE))
cov(centered.scaled.x <- scale(x))# all 1
```
[Package base version 2.2.1 Index] | 531 | 2,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-13 | latest | en | 0.76022 |
https://www.instasolv.com/question/ol-and-on-the-3e8-229-323-3-you-note-that-has-its-zexes-are-de-2-mint-as-0f5n7p | 1,610,945,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00522.warc.gz | 857,699,247 | 10,922 | ol and on the 3e8 229 - 323- 3 you ...
Question
# ol and on the 3e8 229 - 323- 3 you note that has its zexes are de 2 MINT : As a Ba 200 g ple) men (824) is a polynomial
11th - 12th Class
Maths
Solution
86
4.0 (1 ratings)
( 2 x^{4}-3 x^{3}-3 x^{2}+6 x-2=0 ) If two zero are ( frac{1}{2} & 1 ) Then ( (x-1)left(x-frac{1}{2}right)left(a x^{2}+b x+cright)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2 ) ( left[x^{2}-frac{3}{2} x+frac{1}{2}right]left[a^{x^{2}}+b x+cright]=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2 ) ( 9 x^{4}+left(5-frac{3 a}{9}right) x^{3}+left(3+frac{a}{2}-frac{3}{8}right) ) ( =3 x^{3}-3 x ) ( a x^{4}+left(-frac{3 a}{2}+bright) x^{3}+left(c+frac{a}{2}-frac{3 b}{2}right) x^{2}+left(-frac{3 c}{2}+frac{b}{2}right) x+frac{1}{2} ) ( =2 x^{4}-3 n^{3}-3 x^{2}+6 x-2+frac{20}{2} ) By Combaring ( frac{c}{8}=-2 Rightarrow(c=-4) ) ( -frac{3 c}{g}+frac{b}{2}=6 Rightarrow frac{c}{c}+frac{b}{g}=6 Rightarrow b=0 ) ( -frac{39}{8}+b=-3 Rightarrow-frac{39}{8}=-3 Rightarrow(29) ) Then ( 2 x^{2}-4=0 ) ( x^{2}=2 Rightarrow(x=pm sqrt{2}) ) | 513 | 1,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-04 | latest | en | 0.305148 |
https://sites.google.com/site/btrgrad/curriculum-planning/content-specific/mathematics/go-to-websites/graph-of-the-week | 1,653,397,747,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00039.warc.gz | 587,354,508 | 8,871 | Planning > Content-Specific > Mathematics > Go-To Websites >
## This weekly routine will help your students become critical thinkers, engage them in current or historical issues, increase their ability to read graphs, and allow you to insert more social justice into your curriculum all in one straightforward activity. Wow!
Visit Kelly Turner’s website, Graph of the Week, to access all her resources! There you will find …
• graphs for every week
• a prompt for students you could use or modify
• a list of resources for where to find graphs
Turner uses the same prompt every week with her students:
The Student Writing Prompt:
Analyze the graph below and write a reflection on what you think the graph is communicating to you. To guide you with your response, start with some observations.
• What is the topic of the graph?
• What do the x-axis and y-axis represent?
• What are some observations you can make based on the graph?
• What do you foresee happening about 5 years from now?
• Is there an upward or downward trend?
• Are there any sudden spikes in the graph?
• What is being compared in the graph?
• What prediction can I make for the future?
• What inferences can I make about the graph?
An example from a BTR Grad algebra teacher (with student work)
### Suggestions for use:
1. Create a clear “graph of the week” (GOW) routine. The teacher above used the GOW in the Do Now. Students were responsible for finishing their analysis by Friday at the end of the Do Now. You might use GOW as your entire warm-up every Monday. Or as HW every Friday. Whatever the routine is, teach it to students, practice it and give them feedback on it.
2. Decide how you will assess these. Suggested criteria include:
• Can correctly identify the topic of the graph
• Can correctly identify what the x-axis and y-axis represent
• Makes accurate and detailed observations
• Can accurately identify trends
• Makes a prediction that is supported by evidence from the graph | 431 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-21 | latest | en | 0.940055 |
https://www.jiskha.com/display.cgi?id=1237929871 | 1,503,135,750,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00254.warc.gz | 940,894,694 | 3,693 | # English
posted by .
1. I am Bora, your student last year.
2. I am Bora, your student from last year.
3. I am Bora, your student who is from last year.
(Are all correct? In #2, what does 'from last year' modify? What is the part fo speech of 'from last year'?)
• English -
Only #2 is correct and smoothly worded. The prepositional phrase "from last year" serves as an adjective, modifying "student."
from = preposition
year = noun, object of the preposition "from"
• English -
(from last year) preposition phrase
## Similar Questions
1. ### English
1. Do you know what? 2. What do you know?
2. ### English
1. I am in the first grade in London Middle School. 1-2. I am in the first grade at London Middle School. (Are both prepositions correct?
3. ### English
1. I am your student from last year. 2. I was your student from last year. 3. I am your student last year. 4. I was your student last year. 5. I am your student of last year. (Which ones are correct?
4. ### English
1. Who wrote this letter? Who is this letter for?
5. ### English
1. My school begins at 8:30 a.m. like Bora's school. 2. Our school begins at 8:30 a.m. like Bors'a school. (Which one is correct?
6. ### Math
Suppose your community has 4512 students this year. The student population is growing 2.5% each year. Write an equation to model the student population. (y=a*b^x)
7. ### Math
Suppose your community has 4512 students this year. The student population is growing 2.5% each year. Write an equation to model the student population. (y=a*b^x)
8. ### English
1. I am Bora, your student from last year. 2. I am Bora, your student last year. (Which one is correct?
9. ### English
Who is Ms. Kim? 1. She was Bora's teacher last year. 2. She is Bora's old teacher from last year. 3. She is Bora's teacher in elementary school from last year. 4. She taught Bora in elementary school last year. (Bora learned from Ms.
10. ### English
1. What did Bora want to do? - She wanted to do volunteer work. 2. Where do her friends usually volunteer?
More Similar Questions | 572 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-34 | longest | en | 0.970163 |
https://number.academy/1008059 | 1,725,859,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00569.warc.gz | 406,434,383 | 10,957 | Number 1008059 facts
The odd number 1,008,059 is spelled 🔊, and written in words: one million, eight thousand and fifty-nine, approximately 1.0 million. The ordinal number 1008059th is said 🔊 and written as: one million, eight thousand and fifty-ninth. The meaning of the number 1008059 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1008059. What is 1008059 in computer science, numerology, codes and images, writing and naming in other languages
What is 1,008,059 in other units
The decimal (Arabic) number 1008059 converted to a Roman number is (M)(V)MMMLIX. Roman and decimal number conversions.
Time conversion
(hours, minutes, seconds, days, weeks)
1008059 seconds equals to 1 week, 4 days, 16 hours, 59 seconds
1008059 minutes equals to 2 years, 1 month, 59 minutes
Codes and images of the number 1008059
Number 1008059 morse code: .---- ----- ----- ---.. ----- ..... ----.
Sign language for number 1008059:
Number 1008059 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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Is Prime?
The number 1008059 is not a prime number. The closest prime numbers are 1008043, 1008101.
Factorization and factors (dividers)
The prime factors of 1008059 are 13 * 77543
The factors of 1008059 are 1, 13, 77543, 1008059.
Total factors 4.
Sum of factors 1085616 (77557).
Powers
The second power of 10080592 is 1.016.182.947.481.
The third power of 10080593 is 1.024.372.365.854.749.440.
Roots
The square root √1008059 is 1004,021414.
The cube root of 31008059 is 100,267915.
Logarithms
The natural logarithm of No. ln 1008059 = loge 1008059 = 13,823537.
The logarithm to base 10 of No. log10 1008059 = 6,003486.
The Napierian logarithm of No. log1/e 1008059 = -13,823537.
Trigonometric functions
The cosine of 1008059 is -0,897257.
The sine of 1008059 is -0,441509.
The tangent of 1008059 is 0,492065.
Number 1008059 in Computer Science
Code typeCode value
1008059 Number of bytes984.4KB
Unix timeUnix time 1008059 is equal to Monday Jan. 12, 1970, 4:59 p.m. GMT
IPv4, IPv6Number 1008059 internet address in dotted format v4 0.15.97.187, v6 ::f:61bb
1008059 Decimal = 11110110000110111011 Binary
1008059 Decimal = 1220012210112 Ternary
1008059 Decimal = 3660673 Octal
1008059 Decimal = F61BB Hexadecimal (0xf61bb hex)
1008059 BASE64MTAwODA1OQ==
1008059 MD5cd413f75550f0af3b2efb46b71626988
1008059 SHA158bc102380932c3d256d1f464118072b42afa21b
1008059 SHA224bbddb9aeae0351469483de10e109ff2f627747ab53293ffe075f4261
1008059 SHA256c483be192e4af22af2ded5e4b2f4e5b4c438d411e709fda51dc7be5ef726386d
1008059 SHA384a7160280490f43022881f1df56de80c6d7f2ab5f3a0c1d149f6f5cbdec5887e27527d555105d2a9ec99ff7e8c6d702d3
More SHA codes related to the number 1008059 ...
If you know something interesting about the 1008059 number that you did not find on this page, do not hesitate to write us here.
Numerology 1008059
Character frequency in the number 1008059
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 3 8 1 5 1 9 1
Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1008059, the numbers 1+0+0+8+0+5+9 = 2+3 = 5 are added and the meaning of the number 5 is sought.
№ 1,008,059 in other languages
How to say or write the number one million, eight thousand and fifty-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.008.059) un millón ocho mil cincuenta y nueve German: 🔊 (Nummer 1.008.059) eine Million achttausendneunundfünfzig French: 🔊 (nombre 1 008 059) un million huit mille cinquante-neuf Portuguese: 🔊 (número 1 008 059) um milhão e oito mil e cinquenta e nove Hindi: 🔊 (संख्या 1 008 059) दस लाख, आठ हज़ार, उनसठ Chinese: 🔊 (数 1 008 059) 一百万八千零五十九 Arabian: 🔊 (عدد 1,008,059) مليون و ثمانية آلاف و تسعة و خمسون Czech: 🔊 (číslo 1 008 059) milion osm tisíc padesát devět Korean: 🔊 (번호 1,008,059) 백만 팔천오십구 Danish: 🔊 (nummer 1 008 059) en millioner ottetusinde og nioghalvtreds Hebrew: (מספר 1,008,059) מיליון ושמונת אלפים חמישים ותשע Dutch: 🔊 (nummer 1 008 059) een miljoen achtduizendnegenenvijftig Japanese: 🔊 (数 1,008,059) 百万八千五十九 Indonesian: 🔊 (jumlah 1.008.059) satu juta delapan ribu lima puluh sembilan Italian: 🔊 (numero 1 008 059) un milione e ottomilacinquantanove Norwegian: 🔊 (nummer 1 008 059) en million åtte tusen og femtini Polish: 🔊 (liczba 1 008 059) milion osiem tysięcy pięćdziesiąt dziewięć Russian: 🔊 (номер 1 008 059) один миллион восемь тысяч пятьдесят девять Turkish: 🔊 (numara 1,008,059) birmilyonsekizbinellidokuz Thai: 🔊 (จำนวน 1 008 059) หนึ่งล้านแปดพันห้าสิบเก้า Ukrainian: 🔊 (номер 1 008 059) один мільйон вісім тисяч п'ятдесят дев'ять Vietnamese: 🔊 (con số 1.008.059) một triệu tám nghìn lẻ năm mươi chín Other languages ...
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# «Strengthening the Nation through Diversity, Innovation & Leadership in STEM San Antonio,Texas · October 3-6, 2013 Get Connected! Connect with the ...»
-- [ Page 70 ] --
1 A graph G = (V,E) is a collection of vertices V and edges E where an edge is an unordered pair of vertices. An orientation G’ of a graph G is the same collection of vertices, but, if {i,j} is an edge in G, we can have directed edge (i,j) or (j,i) but not both, where (i,j) means there is an arc from vertex i to vertex j. Matrix A corresponds to oriented graph G’ if,for i and j distinct, the entry aij is nonzero exactly when (i,j) is a directed arc in G’. Note that the diagonal entries in A are free. The maximum nullity M(G’) is defined to be the largest possible nullity over all real matrices corresponding to G’. For an oriented graph G’, the zero forcing number Z(G’) is the minimum number of blue vertices needed to force all vertices in G’ blue according to the color change rule. The color change rule is, for oriented graph G’ with vertices initially colored blue or white, a blue vertex b forces a white vertex w blue if w is the only white outneighbor of b. We investigate M(G’) and Z(G’) for an oriented graph G’ and its underlying, unoriented graph G (the parameters M(G) and Z(G) have related definitions), including establishing that, for certain families of graphs, it is always possible to find an orientation G’ of G so that Z(G’) = Z(G) and M(G’) = M(G).
FRI-400
## TOWERS OF REPEATED EXPONENTIAL SEQUENCES
Caroline VanBlargan1, Justin Peters2.
St. Mary’s College Maryland, St. Mary’s City, MD, 2Iowa State University, Ames, IA.
1 Euler proved which numbers have convergent, iterated exponential towers. That is, given a real number x, he looked at the convergence of a sequence where the nth term in the sequence is x exponentiated by itself n times.
We have generalized this question from dealing with a constant number x to considering any sequence of positive real numbers. We provide results about regions of convergence, existence of certain types of sequences, and relationships between a sequence and its sequence of exponential towers.
SAT-406
## ON THE SCHUR POSITIVITY OF DIFFERENCES OF PRODUCTS OF SCHUR FUNCTIONS
Nadine Jansen1, Jeremy Meza2, Rosa Orellana3, Jeremiah Emidih4, Samuel Ivy5.
North Carolina Agricultural and Technical State University, Greensboro, NC, 2Carnegie Mellon University, Pittsburgh, 1 PA, 3Dartmouth College, Hanover, NH, 4University of California, Riverside, Riverside, CA, 5North Carolina State University, Raleigh, NC.
The Schur functions are a basis for the ring of symmetric functions indexed by partitions of nonnegative integers. A symmetric functions f is called Schur positive if, when expressed as a linear combination of Schur functions, f=Σcλsλ, each coefficient cλis nonnegative. We wish to investigate expressions of the form sλcsλ - sμcsμ (1) where λ partitions n and μ partitions n-1, and the complements sλc and sμc are taken over a sufficiently large m x m square. We give a necessary condition that if (1) is Schur positive, then μ is contained in λ. Furthermore, we show how conjugating partitions preserves Schur positivity. Lastly, we incorporate the Littlewood-Richardson rule to show that particular classes of λ and μ are never Schur positive.
FRI-396
## SPECTRAL PROPERTIES OF WEIGHTED CAYLEY DIGRAPHS
Christopher Cox1, Hannah Turner2, Gregory Michel3, Sung Song1, Katy Nowak1.
Iowa State University, Ames, IA, 2Ball State University, Muncie, IN, 3Carleton College, Northfield, MN.
1 From a subset S of a finite group G, we can define the Cayley digraph of G with connector set S to be the directed graph Cay(G,S) with vertex set G and arc set (x,xs) for some s in S. Naturally arising from this definition is the Cayley isomorphism (CI) property. A Cayley digraph Cay(G,S) is said to be a CI graph if, for every subset T in G such that Cay(G,T) is isomorphic to Cay(G,S), there exists a group automorphism of G that maps S to T. If no such group automorphism exists for any Cay(G,T) isomorphic to Cay(G,S), the graph is called a non-CI graph. Finally, if every Cayley graph of G is CI, the group is called a CI group. In this presentation, we provide a new view of Cayley
digraphs by studying the adjacency matrix of the weighted Cayley graph obtained by weighting the edges of the digraphs corresponding to group character representations of G. Given a non-CI group, we examine the spectra of the weighted adjacency matrices of two isomorphic non-CI graphs and investigate the relation between the CI property of Cayley graphs over G and the spectra of the weighted adjacency matrices of the graphs. We present our findings on the spectral properties of the weighted Cayley digraphs and show how they can be applied to better understand the CI property and the structures of finite groups in general. In particular, we show how to determine if each isomorphism class of Cayley graphs of G are CI classes or not.
SAT-399
## EGYPTIAN FRACTIONS, THE GREEDY ODD ALGORITHM, AND GROUPOIDS
Andres Vindas Melendez1, Julia Bergner2.
University of California, Berkeley, Berkeley, CA, 2University of California, Riverside, Riverside, CA.
1 There has been a vast amount of research performed on Egyptian fractions and groupoids separately. In the recent past, Baez and Dolan defined the notion of groupoid cardinality; it is a topic that can be used to see the connection between Egyptian fractions and groupoids. A recent paper by Bergner and Walker shows that any positive rational number occurs as the groupoid cardinality of some groupoid, and this problem can be reduced to the question of whether any positive rational number has an Egyptian fraction decomposition. This result, obtained through the use of the greedy algorithm, implies the fact that any positive real number is the cardinality of a groupoid with no 2 components having the same cardinality. However, if a different algorithm is used, the decompositions are not alike.
This research seeks to investigate the differing decompositions that are obtained when applying the greedy odd algorithm. For example, under this algorithm, some rational numbers might have infinite decompositions or repeated summands. These results can be used to explore groupoids that have a particular cardinality relating to the outcomes of the greedy odd algorithm.
FRI-405
– – –
West Point, NY, 4George Washington University, Washington, DC, 5Dartmouth College, Hanover, NH.
Given any simple graph, there is a corresponding symmetric function called the chromatic symmetric function (CSF).
Introduced by Richard Stanley in 1995, the CSF of a graph G = (V(G), E(G)) is defined as follows: ΧG = ∑κ∏v∈V(G)Xκ(v) where the sum is over all proper colorings κ of G. A proper coloring is a labeling of a graph such that no 2 adjacent vertices have the same label. In 2008, Scott presented several open problems in graph theory. In our poster, we investigate these open problems and generalize some of his results. Our goal is to find necessary conditions for any 2 graphs that will ensure that they have the same CSF. We first consider 2 special types of graphs: trees and unicycles and write a program in SAGE to compare the CSF for any simple graph and compile a library of graphs with a small number of vertices alongside their CSFs.
FRI-403
## MIDDLE SCHOOL STUDENTS’ OPINIONS REGARDING STEM BEFORE AND AFTER A SUMMER ROBOT CAMP
EXPERIENCE Leanne Cohn, Chrystal Johnson, Gregg Gold.
Humboldt State University, Arcata, CA.
Science, technology, engineering, and mathematics (STEM) are critically important fields. A number of studies have looked at various metrics, including how many students enroll in STEM courses, how many of them drop out, etc.
However, as far as we know, research has not been conducted looking at students’ potential attitude changes toward STEM in the context of attending a summer camp devoted to a highly STEM oriented subject (robots). Here, we administered a paper-based entrance and exit survey regarding STEM attitudes to middle school participants in a summer robot camp. Our data will be statistically analyzed to determine the extent to which robot camp had an effect on participants’ attitudes towards STEM subjects. Hopefully analysis will show us a positive correlation between STEM camp attendance and positive changes in participants’ attitudes towards STEM. We are hopeful this research will provide evidence that camps like this will motivate and encourage students to have more positive attitudes toward STEM.
– – –
SAT-407
## SUBGROUP STRUCTURES ON CAYLEY GRAPHS AND THE CAYLEY ISOMORPHISM PROPERTY
Hannah Turner1, Sung Song2, Christopher Cox2, Katy Nowak2, Gregory Michel3.
Ball State University, Muncie, IN, 2Iowa State University, Ames, IA, 3Carleton College, Northfield, MN.
1 For a finite group G and a symmetric subset S of G, the Cayley (undirected) graph Cay(G,S) is the graph whose vertex set is G and such that two vertices x and y are adjacent if y is xs for some s in S. A group G is said to have the Cayley-isomorphism (CI) property if for any 2 isomorphic Cayley Graphs Cay(G,S) and Cay(G,T), there exists an automorphism of the group that sends S to T. We know that if a group G has a subgroup H that does not show the CI property, then G also does not have the CI property. In this poster, we analyze CI and non-CI subgroups of a non-CI group G and we characterize certain Cayley graphs of G where the connector sets generate these subgroups. We study the relation between the particular non-CI isomorphism classes of non-CI groups and their non-CI subgroups, and we also characterize irreducibly non-CI groups, those non-CI groups for which every subgroup of the group is CI.
FRI-407
## CLASSIFICATION OF THE CAYLEY GRAPHS OF SYMMETRIC GROUPS
Gregory Michel1, Hannah Turner2, Christopher Cox3, Katy Nowak3, Sung Song3.
Carleton College, Northfield, MN, 2Ball State University, Muncie, IN, 3Iowa State University, Ames, IA.
1 In this poster, we completely classify the Cayley graphs of the symmetric group on 4 letters. We characterize all possible Cayley graphs that come from symmetric subsets and we identify the individual classes of isomorphic graphs for which the Cayley isomorphism (CI) property holds. In studying the CI property, we analyze whether or not certain graphs are connected, planar, bipartite, edge-transitive, and/or strongly regular. We further investigate whether some of the strongly regular graphs can be decomposed by a pair of directed regular graphs, especially into a pair of directed strongly regular graphs, doubly regular tournaments, or normally regular digraphs that are realized as directed Cayley graphs of the same group. By analyzing patterns among the isomorphism classes, we can generalize certain properties for arbitrarily large symmetric groups.
– – –
FRI-409
## STUDY OF THE HILL ESTIMATOR
Maria Correa1, Rebekah Starks2, McKenna Mettling3, Javier Rojo4.
St. Mary’s University, Midland, TX, 2University of Arizona, Tucson, AZ, 3Regis University, Denver, CO, 4Rice 1 University, Houston, TX.
Heavy-tailed distributions are used for modeling in many of the popular fields, for example telecommunications and finance. Thus, the development and study of methods to estimate the tail index for these distributions is highly significant. For our research, we studied the well-known Hill estimator developed in 1975 by Bruce M. Hill. We studied its accuracy and flaws, which include Hill horror plots, by using various heavy-tailed distributions and real datasets to create simulations in R studio.
SAT-409
## SURVIVAL ANALYSIS
Adrian Carballeira1, Noel Martinez2, Kourtney Howell3, Javier Rojo4.
University of Arizona, Tucson, AZ, 2University of Texas at El Paso, El Paso, TX, 3Xavier University of Louisiana, New 1 Orleans, LA, 4Rice University, Houston, TX.
In recent years, scientists have exponentially increased the amount of data that can be gathered from a given experiment. The “curse of dimensionality” along with computational restrictions can hinder attempts to extract meaningful information from the data. Principal components analysis is one common way to reduce dimensionality.
Alternately, the Johnson-Lindenstrauss theorem guarantees that, given the number of data points and a specified error tolerance, there exists a mapping, realized as a linear transformation, into a lower dimension with the property
that pairwise distances between points in the original data set are preserved up to a small error bound. We compare several suggested lower bounds and types of random projection matrices found in the literature and see that these bounds are quite conservative. We use the software R to generate different random projection matrices to determine which type of random projection matrix most frequently upholds the conclusions of the Johnson-Lindenstrauss theorem. Also using R, we use both random projections and PCA to reduce data in order to estimate survival curves for randomly-generated data and compute bias and mean squared error in order to compare these 2 methods.
Knowing which dimension reduction method yields more accurate results allows scientists to analyze summarized data without losing precision, a much more computationally efficient task.
SAT-408
## ESTIMATING TAIL INDICES: THE ACCURACY OF THE ROJO1 ESTIMATOR
McKenna Mettling1, Maria Correa2, Rebekah Starks3, Javier Rojo4.
Regis University, Denver, CO, 2St. Mary’s University, Texas, San Antonio, TX, 3University of Arizona, Tucson, AZ, 1 Rice University, Houston, TX.
4 The focus of our project is to investigate the accuracy of the current estimators used for finding the tail indices for heavy-tail distributions. After studying both the Hill and Pickands estimators and discovering that both had certain flaws, we noticed there is no true consensus on which of the classical estimators we should use for research.
Pages: | 1 | ... | 68 | 69 || 71 | 72 | ... | 112 |
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If you do not agree with the fact that your material is placed on this site, please, email us, we will within 1-2 business days delete him. | 4,851 | 20,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-13 | latest | en | 0.864227 |
https://brainly.ph/question/368985 | 1,487,646,970,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170624.14/warc/CC-MAIN-20170219104610-00018-ip-10-171-10-108.ec2.internal.warc.gz | 705,925,031 | 10,866 | # Look at each pair of expressions below. Identify the expression that is not a polynomial from each. Give reasons for your answers. A 1.) 2x+1 2.) x^-3+2x²-7 3.)2√x 4.) 2x³-3x^½+-4 5.) (x+5)(9x+1)²(x-4) B 2/x+1 x³+2x²-7 x√2 2x³-3x²+x-4 (x+5)(9x+1)^-2/(x-4) Till July 25 2016
2
by zazazxc
2016-07-16T09:56:29+08:00
1. No -Binomial
2.No -Irrational
3. No -Irrational
4.Yes.
5.Yes.
Binomial is a polynomial, therefore Item 1 is a polynomial; Item 4 is not a polynomial because it has a fractional exponent in the middle term.
• Brainly User
2016-07-16T10:21:47+08:00
PLEASE USE THIS GUIDE TO DIFFERENTIATE A POLYNOMIAL FRON NON - POLYNOMIAL EXPRESSIONS:
A polynomial has:
2) No fractional exponent
3) No negative exponent or no variable as denominator in any term in lowest or simplified form.
A.
1) Polynomial
2) Not a polynomial - it has a negative exponent in leading term (x⁻³)
3) Not a polynomial - it's a radical expression
4) Not a polynomial - it has a negative exponent in middle term
5) Polynomial - it's product (81x⁴ + 99x³ - 1601x² - 359x - 20) is a polynomial
B.
1) Not a polynomial - The leading term has a denominator x, meaning the term has a negative exponent (2/x is the same as 2x₋¹)
2) Polynomial
3) Not a polynomial - it's a radical expression
4) Polynomial
5) Not a polynomial - The second factor, (9x+1)⁻² , obviously has a negative exponent. | 487 | 1,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-09 | latest | en | 0.789429 |
http://mathhelpforum.com/pre-calculus/44016-powers-0-1-a-print.html | 1,527,473,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00338.warc.gz | 170,994,801 | 3,026 | # Powers of 0 and 1
• Jul 18th 2008, 02:55 PM
magentarita
Powers of 0 and 1
We all know that any number or variable raised to the zero power equals 1
and that any number or variable raised to the 1st power equals the number or variable.
SAMPLE A:
x^0 = 1 as well as 100^0 = 1
SAMPLE B:
x^1 = x as well as 100^1 = 100
Why is this the case?
Can anyone explain, in simple terms, the WHY concerning the above?
• Jul 18th 2008, 03:05 PM
arbolis
I'll do the why of
Quote:
x^0 = 1
. We have $\displaystyle x^0=x^{1-1}=x^1\cdot x^{-1}=\frac{x^1}{x^1}=1$. EDIT : Or even better, $\displaystyle x^0=e^{0\cdot\ln x}=e^0=exp(0)=1$ because $\displaystyle \ln(1)=0$ (From the definition of the logarithm function. If you define the exponential function as the inverse function of the logarithm one, then what I wrote holds.). Note that it holds $\displaystyle \forall x \neq 0$.
• Jul 18th 2008, 03:12 PM
Mathstud28
Quote:
Originally Posted by magentarita
We all know that any number or variable raised to the zero power equals 1
and that any number or variable raised to the 1st power equals the number or variable.
SAMPLE A:
x^0 = 1 as well as 100^0 = 1
SAMPLE B:
x^1 = x as well as 100^1 = 100
Why is this the case?
Can anyone explain, in simple terms, the WHY concerning the above?
Assuming you buy Arbolis's post then
$\displaystyle x^1=\frac{x^{a+1}}{x^a}$
Or more obvious
$\displaystyle \frac{x^2}{x}$
Assuming $\displaystyle x\ne{0}$
• Jul 19th 2008, 06:00 AM
magentarita
Thanks A Million!
I thank you both for that great mathematical insight. | 512 | 1,557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-22 | latest | en | 0.879211 |
https://www.jiskha.com/display.cgi?id=1119931503 | 1,516,497,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00544.warc.gz | 985,883,829 | 3,690 | # algebra anyone
posted by .
How to you compute the intercepts of a quadratic function?
For the x-intercept, set x to 0. For the y-intercept(s) (if any), set y to 0.
## Similar Questions
1. ### algebra
: In case of a quadratic function, why are there two x-intercepts and one y-intercept?
2. ### Math question?
How do you compute the x and y intercepts of a quadratic function?
5. Find the x and y intercepts of each linear equation. Be sure to write your answers as ordered pairs. a.y=3x-6 b.6y=-x+2 Set x = 0 and y = 0 to find the intercepts. will that get me 2 thingys for each.?
4. ### Mathematics
Determine the equation of a quadratic function that satisfies each set of conditions. a) x-intercepts 1 and -1, y-intercept 3 b) x-intercept 3, and passing through the point (1, -2) c) x-intercepts -1/2 and 2, y-intercept 4 Please …
5. ### College Algebra
I could not figure out the answer to this Problem: The quadratic function f(x)= -2x^2 =4x +3 can be used to solve the following inequality: -2x^2 -4x < -3 First use the Quadratic Formula to find the x-intercept of f. This problem …
6. ### Algebra, finding the intercept, help?
How do you find the x and y intercept of 3x+0.5y=6?
7. ### Algebra
The table below represents the function f(x). (-6, 12),(-5,-8),(-4,-4), (-3,0),(-2,4),(-1,8),(0,12). If g(x) is a linear function that has a slope of -3 and a y-intercept of -9, which statement is true?
8. ### Algebra 1--Please, help me!
1. What are the x- and y-intercepts of the line described by 6x - 2y = 4?
9. ### Algebra
What are the intercepts of the equation? Graph the equation. -3x + 4y = -12 A. x-intercept: (4, 0) y-intercept: (0, –3) B. x-intercept: (–3, 0) y-intercept: (0, 4) C. x-intercept: (4, 0) y-intercept: (0, –3) D. x-intercept: (–3,
10. ### Algebra
The equations for two functions are shown below. f(x) = x + 3 and g(x) = 2x - 5 Which intercept is farthest from the origin compared to the other three intercepts?
More Similar Questions | 619 | 1,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-05 | latest | en | 0.852645 |
http://salttheoats.blogspot.com/2011/09/modeling-unit-5.html | 1,369,142,025,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700014987/warc/CC-MAIN-20130516102654-00011-ip-10-60-113-184.ec2.internal.warc.gz | 237,793,260 | 20,438 | ## Wednesday, September 28, 2011
### Modeling Unit 5
Unit V: Constant Force Particle Model
"Atwood Machine" with vernier track
From there, we started the next unit. Here's what each end of the track looked like (the middle is just a track)
Jon changed the first question slightly:
What factors will effect the motion? (between letting go of the cart and hanging mass hitting ground)
What factor effects the cart’s acceleration?
Hanging mass
Mass of car
Friction: {adjust tilt of track until cart rolls at constant speed}
(pasco hanger approx. applies force to balance friction)
Mass of pulley
Mass of the earth/gravity
Angle of the track - ?
Starting speed -?
{Chris showed us a quicker way of working through the process by guiding us to eliminate the factors mentioned that cannot be adjusted (Mass of Earth) or that could be removed with creative lab design. The last two options we left open, that depending on your class, you may or may not want to divide and conquer.}
Purpose: What is the graphical & mathematical relationship that exist between the mass of the cart and the force that is accelerating it.
Before getting started, we talked about multiple variations to this experiment
• Keeping the mass of the hanger while adding mass to the car
• Using photogate(s) above the track instead of motion detector
• Variation of this option is to attach picket fence to cart the cart and use vernier program
• Using kinematic equations and measure total time with stopwatch for total distance measured
• Having the students predict the mass of the system from data, and then, after showing prediction to the teacher, measuring the mass and comparing results to predictions {I like this!}
Equipment
Attach right angle to cart
Pulley at end of track
Hanger
String
Motion detector
Standard masses
Next up were some demonstrations of Newton's 3rd Law:
Same track set up, with Force sensors attached to the top of each car. The twist for this demonstration is to use the magnets to apply to force between the two cars and not the direct contact. There are a couple of things to note for this demonstration. Have one car against the stopper and start with the second car "far away" from the first car. Zero both probes, and make sure you reverse the direction for one of them so they both have positive in the same direction of the track. Have the magnets inside the car so the same pole faces outward and thus repel the cars. Push the force probe of the second car (not the car itself)
From there, were picked up on the lab with which we finished yesterday. Before starting the experiment we briefly discussed the merit of breaking the lab groups into different types of investigations, in which we would have 3 different trials: "A" would look at keeping the mass of the cart constant, but adding mass to the hanger; "B" would keep the hanger constant, and add mass to the car; "C" would move mass from the car to the hanger, keeping the mass of the system constant. In an effort to save time, we have everyone do option "C" but I may or may not look at all the groups (maybe in my honors class?)
Also showing a way to move through the whiteboard process more quickly (as needed by time constraints or if class is not productive in meetings), Jon walked us through a "Circle the wagons" meeting. In this format, all the groups show their white boards, and the teacher leads the group to try to draw conclusions in looking at all the results at once.
{As we were getting started, Jon also mentioned that when you are "normal" whiteboard meeting after a lab, in subsequent labs, start with a different group each time and change the order you call the groups forward.}
During the meeting, we had a great discussion on whether you should explain/guide to the students before starting the lab that they will need to plot Force vs acceleration so that the slope is mass, or wait until the end.
{My thought is to wait until the end, have all the groups manipulate their graphs, as teacher does it on projected screen}
At this point, Jon showed us a quick follow up demo/lab (used vernier "Lab 9 – Newton’s 2nd Law")
Jon taped an accelerometer to the force probe (Jon uses Velcro tape at his school). Then you just click the record data button, and then push the cart back and forth. Viola, data showing \$F \propto a\$
From there, we started individual work on Unit V worksheet 1 (#'s 1-4) and worksheet 2 (#'s 1-3)
As we got started, we briefly discussed strategies for word problems (w/ forces). A summary of what we said was:
• Have students sketch what is happening and identify the system with dotted circle/box
• Get the words out of the word problem
• Create a Free Body Diagram
• Next to FBD, draw an arrow showing the direction of acceleration
• That will be "+" direction for the problem
• This convention will aid circular motion problems later in the year
Notes from whiteboard session:
• Wkst 2: #2 is a great problem since a given number isn’t use in the calculation, but rather for analysis at the end.
• Wkst 2: #3 mass not given, so students need to determine it from the Weight
• Chris- Make sure units are included in the calculation not just at the end
• Possibly change wording of problem since the normal force changes not Fw
Jon then went on to describe how he helps his students understand "elevator" problems. If you are standing on a bathroom scale, and you want to increase your "weight" you can pull on the bottom of the counter and squeeze the scale. This is the same effect as when the elevator is accelerating upwards. On the contrary, if you want to lose "weight," you can push on the top of the counter and push you body off the scale. This same effect occurs when the elevator accelerates downward.
From there, we Jon showed us some fun demos
1. Have student kneel w/elbows touching knees & hands “praying”. Put chapstick at tip if fingers. Then have student place hands behind his/her back. They then need to try to knock over the chapstick by touching their nose to it. Due to differences in center of mass, girls should be able to do this, while boys usually can't.
2. Have student stand facing the wall, with toes touching the base of the wall. Have student take 3 steps (toe to back of heel) away from the wall. Bend at the waist \$90^o\$ with their forehead touching the wall. Place a small chair (or other "small" mass) in their hands and tell them to stand up. Again, boys will struggle, girls will tend to be successful.
3. Have one student (biggest student) sit all the way back into a chair with his/her feet flat on the floor. Have a second student (smallest) stand in front of first student and push into the first student's forehead. Tell first student, without moving their feet, to stand up. At the same time, the second student pushes on the forehead of the first, preventing him/her from standing up.
4. Have student stand with right shoulder and outside of right foot touching the wall. Then tell the student to lift his/her left foot.
Friction Lab
We then moved on to a lab on Friction. We used a friction block and force probe. The basic procedure was to the block at constant speed with different masses resting on top of the block. We used the vernier file "Lab 12a Static Kinetic Fric." A sketch of the graph produced looked basically like this:
The max force represents the static friction force, and when the force is basically horizontal (red line) then the friction force equals the measured force.
To speed things up, each group given different normal force ("zero mass" was mass of block plus 250 g) and needed to get good data (slope of oscillating data was as close to horizontal as possible). Find the average value of the force using the statistics button.
Unit V Feedback
The Good:
• We felt we were becoming comfortable using computer based equipment
• Continuation of the sequence of showing 3rd law (adding non-contact interaction)
• Low tech demo’s/labs | 1,800 | 7,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2013-20 | longest | en | 0.92094 |
https://physics.stackexchange.com/questions/184646/are-magnetism-and-electricity-the-same-thing/184732 | 1,580,174,747,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00140.warc.gz | 593,836,461 | 28,609 | Are magnetism and electricity the same thing? [closed]
I have read at certain place electricity and magnetism are the same thing, bit in reality we see both have different properties.
• Hi Sohaib. At the moment your question is too broad to be usefully answered. Can you make it more precise? For example can you pick a particular example of the difference between the two that is puzzling you. – John Rennie May 18 '15 at 11:21
• They are not the same thing, but two inseparable aspects of one thing. Other than that, John Rennie is right. It takes a textbook's worth of material to talk about the details. – CuriousOne May 18 '15 at 14:09
• @JohnRennie What is to broad that could not be explained? I gave an answer and I'm interested, is the answer only a small part of the full aspect? – HolgerFiedler May 19 '15 at 4:51
Magnetic and electric fields can transform into each other under Lorentz transformations. One mans magnetic field can become other mans electric field. Like space-time coordinates transform under Lorentz transformations, so do the EM fields.
• the second after one man's "magnetic" should be "electric" – anna v May 18 '15 at 18:00
• Lol..of course... – Žarko Tomičić May 18 '15 at 19:14
They're not quite the same thing, but are very closely related (as described by Maxwell's equations).
The support of both fields are particles with electric charge and magnetic dipole moment, such like electron and proton. Also there is the neutron with magnetic dipole moment, but without electric charge. By the help of an external magnetic field one could align neutrons and this magnetic field will exist without electric field.
Both, the electron's electric field and magnetic dipole moment are existing permanently. To reinforce the electric field we have to separate electrons. To reinforce the magnetic field we have to align the electron's magnetic dipole moments. This we can do in two ways.
First we use a permanent or an electromagnet and align the electrons magnetic dipole moment. Second we move electrons and accelerate them non parallel to the movement direction like in a coil. This induces a magnetic field. How does this works?
The cause why we could use generators, electric drives and electromagnets is based on the connection between the magnetic dipole moment and the intrinsic spin. Both phenomena have a direction and the relation between this two directions is for all electrons equal. If this would be not made so in nature, we would not get macroscopic magnetic fields.
Magnetism and electricity are not the same thing in the sense that the magnetic field (for example the magnetic field of a coil) does not attract an electric charge nor a electric charged body will attract a magnet (if neglect polarisation effects).
• The magnetic field of a coil is not due to the spin of the electrons. A current of spinless ions also creates a magnetic field. Actually modern generators are often not using permanent magnets to create the magnetic fields (but rather electromagnets). Asynchronous three-phase electric motors use no permanent magnets (which are indeed based on parallel alignment of spin magnetic moments) either. – Sebastian Riese Oct 12 '15 at 17:46
• @SebastianRiese Please give a source about experiments with spineless ions and created magnetic field. – HolgerFiedler Oct 12 '15 at 20:03
• I do not know about experiments, but it is pretty clear from a special relativity point of view (this way the magnetic field of a current carrying wire can be deduced from Coulomb forces, no reference is made to the spin/magnetic dipole moment of the charge carriers). So if you doubt this, you refute special relativity. – Sebastian Riese Oct 13 '15 at 0:17 | 826 | 3,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-05 | latest | en | 0.925021 |
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# A point P lies inside the circle. So____tangents can be drawn from the point P to the circle.
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http://www.docstoc.com/docs/74986454/Uncertainty | 1,419,664,289,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447550643.22/warc/CC-MAIN-20141224185910-00065-ip-10-231-17-201.ec2.internal.warc.gz | 197,553,604 | 22,413 | # Uncertainty
Document Sample
``` Uncertainty
Michael Peters
June 3, 2005
1 Lotteries
In many problems in economics, people are forced to make decisions without
knowing exactly what the consequences will be. For example, when you buy a
lottery ticket, you don’t know whether or not you will win when you buy the
ticket. There are many important problems like this. If you buy car insurance,
you hope you won’t need it, but you aren’t sure. A politician who spends time
and money running for public office is not sure whether or not they will be
elected. A drug company that invests in developing a new drug is not sure
whether or not it will actually work. A corporate insider who sells shares using
inside information is never sure whether or not they will be caught.
One way to think about such problems is to use the concept of a lottery. A
lottery is a pair of objects. The first, X , is a list of possible consequences of a
decision. The second is a list p = {p1 , p2 , . . . pn } of probabilities with which you
think that each of the consequences will occur. The number of probabilities you
list, n, is exactly equal to the number of consequences in X . For example, if you
buy a lottery ticket, the set of consequences consists of two things, you either
1
win or lose. Each of these consequences occurs with probability 2 . If you sell
stock based on an insider’s tip, you either get away with it, or you don’t. More
generally, there could be many consequences. If you open a new restaurant you
might sell 1 or 2 or 3, or any number of meals per week. The set of consequences
could be very large.
In this definition, the set of consequences could be very general. In particu-
lar, each of the consequences could be a lottery. A lottery over lotteries is called
a compound lottery. Lotteries over anything else are called simple lotteries. As
an abstract example, consider the following game. I will flip a coin and if the
coin comes up heads right away, I will give you \$2. If it comes up tails, I will
flip the coin again. If it comes up heads on the second flip, I will pay you \$4. If
it comes up tails, we stop and you get nothing. This is a simple example of a
compound lottery. The first of the two lotteries has a single consequence, you
receive \$2 (of course you get this consequence with probability 1). The second
lottery has two possibly consequences - either you get \$4 or you get nothing. In
this second lottery, each consequence occurs with equal probability. The com-
1
pound lottery involves 2 chance that you will play the first lottery and get \$2
1
for sure. Then there is 1 chance that you will play the second lottery and get
2
either \$4 or \$0.
It should be clear to you that this compound lottery is pretty much the same
thing as a simple lottery where the consequences are that you receive either \$2,
\$4 or \$0 with probabilities 1 , 4 , and 4 . This type of lottery is sometimes
2
1 1
referred to as the reduced lottery associated with the original compound lottery.
1.1 Monty Hall
We normally assume that compound lotteries and the reduced lotteries associ-
ated with them can be used interchangeably. This isn’t always as straightfor-
ward as in the example above. In the following problem, it is very easy to make
a mistake calculating the reduced lottery. You are a contestant in a game show,
and are given the choice of three doors. Behind one door is \$1 million which
you will win if you happen to open this door. There is nothing behind the other
doors, and if you pick one of them you will win nothing. Once you choose one
of the doors, the host will open one of the remaining doors and show you that
it contains no prize. You are then given the option to change from the door
you picked in the first place, to the remaining door. The problem is to decide
whether or not to switch doors.
This is a compound lottery in which the prize is first placed randomly behind
a door, then after observing where the prize is, the host randomly opens one
of the remaining doors. It appears that the prize is equally likely to be behind
either of the three doors, so it can’t matter whether you switch or not after the
host opens the door. However, you will do better on average in this game if you
always switch doors. You might be able to see this from the following casual
reasoning - the prize could be behind any of the three doors. If it is behind
the door that you chose, then it would, of course, be a mistake to switch. In
either of the other alternatives, switching doors will win you the prize. To put
it another way, the host will actually tell you which door contains the prize in
two of the three situations you might face. Following his advice won’t always
work, but it will most of the time.
Another way to think through the problem is to try to compute the reduced
lottery that you actually face when you hold on to your original choice, and
the one you face when you switch. The first part of the lottery involves the
1
placement of the prize. With probability 3 it is placed behind either of the
three doors. The second part of the lottery is the host’s announcement about
the door that doesn’t contain the prize. We might as well assume that you pick
door A, since the thing works the same way no matter which door you pick. The
lottery you get when you stick with your original choice is depicted in Figure 1.
The important part to understanding the correct strategy in this game is to
observe that the outcome of the second lottery depends both on the outcome
of the first lottery (the door where the prize is placed) and on your choice. In
Figure 1, the first set of branches shows the various locations where the prize
can be placed. The second set of branches shows the doors that the host can
open. Notice that if your choice is A and the prize is actually there, then the
2
Door A,B or C
1 1
3 1 3
3
1 1
2 2 1 1
B, W in C, W in B, Lose C, Lose
Figure 1: You stick with your initial choice
host can choose completely randomly to open either door B or C. On the other
hand, if the prize is behind doors B or C, then the host doesn’t have any choice
and is forced to open a door that effectively reveals the location of the prize.
Figure 2 shows what the lottery looks like in the case where you switch
doors.
Door A,B or C
1 1
3 1 3
3
1 1
2 2 1 1
B, Lose C, Lose B, W in C, W in
Figure 2: You switch choices after the host opens a door
Then just glancing at the outcomes in these two figures, it should be clear
that you will win two thirds of the time if you switch, but only one third of the
time if you stick with your initial choice.
1.2 St. Petersburg
Here is a famous reduced lottery involving coins that actually provides most
of the motivation for the approach that we currently use in economics. This
resembles the previous coin flipping problem. As before, if the coin comes up
heads on the first flip, I give you \$2, and if it comes up tails on the first flip,
I flip again. If it comes up heads on the second flip I give you \$4, otherwise I
flip again. If it comes up heads on the third flip, I pay you \$8, otherwise I flip
again. We keep going until I flip a head, then if it takes me k flips to get the
3
head, I pay you \$2k . The set of consequences in this reduced lottery is the set
{1, 2, 3, . . . , k, . . . } .
It won’t take you too much thinking to see that the probabilities are
1 1 1
, ,..., k,...
2 4 2
I am going to ask you how much you would be willing to pay me to play
this game. You could refuse to play - then you would get nothing for sure. Or
you could offer to pay me \$2 or more (you can’t lose from this choice unless you
pay more than \$2). Both choices would involve different lotteries, though one
of them (not playing) is sort of degenerate.
You might try to decide whether or not to play this game by figuring out how
much you would win on average from the game. This calculation is straightfor-
1 1
ward. With probability 2 you win \$2 right away, with probability 4 you win \$4,
1 k
with probability. . . 2k you win \$2 . Averaging all these gives
∞ ∞
1 i
2 = 1=∞
i=1
2i i=1
You will never find anyone who is willing to pay an infinite amount of money
to play this game.
2 Choosing among lotteries
In each of the problems above, you need to choose among lotteries. There is
no obvious way to do this. However, as you actually make the choice, I am
probably safe in thinking that you will be able to express a preference over any
pair of lotteries, and that the preferences you express will be transitive (in other
words, if you say that you prefer lottery (X , p) to lottery (X , p ), and you say
you like lottery (X , p ) more than lottery (X , p ), then I should be sure that
you will prefer lottery (X , p) to lottery (X , p ).
Now, if there is some set of lotteries L from which a choice is to be made, I
can ask for pairwise comparisons between all the lotteries and eventually learn
all of your preferences. To make the notation a little simpler, let me suppose
that every lottery in my set of alternatives L has the same set X of consequences.
Then I can think of a lottery as a simple list of probabilities with which these
various outcomes occur. The outcomes don’t have to be amounts of money,
they can be anything imaginable, including lotteries as you have seen. Yet, as
with all choice problems I am probably not too far off the mark by assuming
you can express a preference between every pair of lotteries in L, and that
the preferences you express will be transitive (which means that if p p , and
p p then it must be that p p ). If your preferences are also continuous in
4
an appropriate was, I will be able to represent with a utility function u in the
sense that p p if and only if u (p) ≥ u (p ).
One of the more important discoveries in economics is that if your preferences
satisfy a third condition, referred to as the independence axiom 1 , then this utility
function will, in fact, be linear in probabilities. The independence axiom say
this: suppose that for any three lotteries p, p , and p , p p implies that for
any λ ∈ [0, 1]
λp + (1 − λ) p
λp + (1 − λ) p .
These last two objects are compound lotteries in which you are given lot-
tery (X , p) (or (X , p )) with probability λ and lottery (X , p ) with probability
(1 − λ). If you can rank two lotteries, you will rank them the same way if they
are mixed with a common third lottery.
For the utility function to be linear in probabilities, it means that we will be
able to find numbers ui , one for each of the n outcomes in the set X such that
n
u (X , p) = ui p i
i=1
Since there is one number ui for each of the n outcomes in X , it is convenient
to think of ui as the utility value associated with outcome xi . If the outcomes
happened to be expressed in dollars, then we could think of the utility numbers
as representing some underlying utility for wealth. The point to be emphasized
is that the existence of the utility for wealth function is not an assumption, but
an implication of a certain method of ranking lotteries.
This is such an important idea that it is worthwhile to see how it works.
To see it, suppose that there is a b ∈ L (the best lottery) such that b p
for all p ∈ L; a w ∈ L (the worst lottery) such that p w for all p ∈ L.
We can now try to mimic the proof of the existence of a utility function that
we did previously. Recall that we used monotonicity of preferences and con-
tinuity. So we will say preferences are monotonic if λb + (1 − λ) w λb+
(1 − λ ) w if and only if λ > λ .2 We will say that preferences are continuous
if the sets {λ ∈ [0, 1] : λb + (1 − λw) p} and {λ ∈ [0, 1] : p λb + (1 − λ) w}
are both closed intervals.
Now we can state the result.
Theorem 1 If preferences are complete, transitive, continuous, monotonic and
satisfy the independence axiom, then there is a utility function u representing
preferences over lotteries in L that is linear in probabilities.
1 This is a bit of a misnomer. It isn’t really an axiom, since it is far from self evident. It is
more like an assumption.
2 The statement p p means that p p but not p p. The notation p ∼ p means that
p p and p p.
5
Proof. The proof is constructive. Let’s start by creating the utility function.
This is a function that assigns a real number to each lottery p ∈ L. To do so
set
u (p) = {λ ∈ [0, 1] : λb + (1 − λ) w ∼ p} (1)
Notice that the λ that satisfies this relation (warning - it is not an equation)
always exists and is unique. To see why, observe that by completeness λ b +
(1 − λ ) w p or the reverse for every λ ∈ [0, 1]. Then
{λ ∈ [0, 1] : λb + (1 − λ) w p} ∪
{λ ∈ [0, 1] : p λb + (1 − λ) w}
is all of the interval [0, 1]. Since both these sets are closed by the continuity of
preferences, they must have at least one point in common. Since preferences
are monotonic they can’t have more than one point in common (Prove this by
Next, we should show that the function u (·) as constructed above, actually
represents preferences . This relies on the monotonicity of preferences and is
left as an exercise.
Finally, we come to the important step - showing that the utility function
defined in (1) above is linear in probabilities, i.e., that
u (λp + (1 − λ) p ) = λu (p) + (1 − λ) u (p )
for all λ, p, and p . I will not write down a series of relations - make sure that
you don’t mistake them for equations. First observe that
λp + (1 − λ) p ∼
λ [u (p) b + (1 − u (p)) w] + (1 − λ) p
This follows from the definition of the utility function u (p) and the independence
axiom (in the sense that we are mixing the lotteries p and u (p) b + (1 − u (p)) w
together with the common third lottery p ). Do the same thing again to get
λ [u (p) b + (1 − u (p)) w] + (1 − λ) p ∼
λ [u (p) b + (1 − u (p)) w] + (1 − λ) [u (p ) b + (1 − u (p )) w]
This is a fairly complicated compound lottery (first you mix lotteries b and w
together using u (p) and u (p ). Then you mix the result using λ.) The reduced
lottery associated with this is
[λu (p) + (1 − λ) u (p )] b + [1 − λu (p) − (1 − λ) u (p )] w
which, if you recall where we started, is then indifferent to λp + (1 − λ) p . If
you glance back at (1), you will see that we have just discovered
u (λp + (1 − λ) p ) = λu (p) + (1 − λ) u (p )
6
which is the linearity property we wanted.
Modern economic theory is concerned largely with problems where there is
some kind of risk or uncertainty about outcomes. In finance, this uncertainty
arises from the inherent unpredictability of asset returns. In mechanism design
(the theory of auctions and institutions), uncertainty arises because of the in-
ability to know the tastes of others. In game theory, uncertainty arises because
of the inability to predict exactly how another player will behave. Expected util-
ity is the cornerstone of the modern approach to uncertainty, so it is probably
one of the most useful ideas that you will encounter.
It has been challenged in a number of ways. The challenges reflect both the
strength of the theory and its weakness. The strength of the theory lies in the
fact that is lays out so precisely what can and cannot happen if the theory is
true. The ’can happen’ part is good, because theories are supposed to explain
things we see. The ’can’t happen’ part is also important since it shows what
kinds of behavior would allow us to reject the theory.
You might wonder why we need a theorem like the one above connecting
expected utility which is a model of the utility function, to the independence
axiom, which you might think of as a restriction on the way people behave. Why
couldn’t we just write down a specific utility function then somehow test that,
instead of worrying about behavioral properties like the independence axiom.
There are two reasons. The first is, that provided you buy completeness,
transitivity and continuity, the independence axiom and expected utility are
equivalent. So the independence axiom provides all the behavioral restrictions
that come from assuming the utility function is linear in probabilities. As-
sumptions about utility functions will typically make it easy to derive some
restrictions on behavior, but not all of them. Knowing all of the implications
makes it far easier to construct an effective empirical test.
The second reason is that it is quite possible to impose assumptions on utility
that have no implications at all for behavior. A trivial example might involve
assumptions about the utility value associated with the best or worst lotteries.
In any case, theorems like the one above lay out very clearly what the additional
implications of linearity are, and how they differ from the implications of other
assumptions.
3 Empirical tests
I’ll provide an example of a challenge to expected utility that involves experi-
mental tests (we already described how econometric tests could be used to test
the implications of completeness and transitivity). To describe the test, let me
simplify things a bit and suppose that the set of consequences consists of exactly
three things - i.e., X = {x1 , x2 , x3 }. The set of lotteries, L, is then just the set
of triples of probabilities q = {q1 , q2 , q3 }. Since the probabilities have to sum to
one, it is possible to depict L in a simple two dimensional diagram as in Figure
3
Every point in the triangle with sides of length 1 in the diagram above
7
q2
1
q
q2
λq + (1 − λ)q q
q1 1 q1
Figure 3: The set of lotteries
is a lottery in L. To see this, take a point like q. The coordinate on the
horizontal axis, q1 represents the probability with which consequence x1 occurs.
The coordinate on the vertical axis q2 represents the probability with which
consequence x2 occurs under lottery q. The probability of consequence x3 is
just the remainder 1 − q1 − q2 . This is given by the length of the horizontal line
segment from q over to the hypotenuse of the triangle (the dashed line with the
arrow at the end in the picture). Now the lottery q can be easily compared to
q. q lies down, and to the right of q, and is closer to the hypotenuse of the
triangle. So, it assigns lower probability to x1 , higher probability to x2 , and
lower probability to x3 .
Each point in the triangle represents a different simple lottery in L. Com-
pound lotteries are lotteries over lotteries. For example, one might form a lottery
with two consequences - consequence 1 is the lottery q while consequence 2 is the
lottery q . Let λ be the probability with which the first lottery q is played. Then
the reduced lottery associated with this compound lottery is λq+(1 − λ) q . This
lottery is just the simple lottery that lies λ/ (1 − λ) of the way along the line
segment between q and q . This point is illustrated in Figure 3.
Preferences over lotteries then consists of a family of indifference curves that
look exactly like the ones you are used to using to think about preferences
over commodity bundles. An indifference curve through the lottery q is set of
˜
lotteries q that have the same utility value as q. Formally, an indifference curve
is the set
q q
{˜ ∈ L : u (˜) = u (q)}
If we use the linearity property of preferences given in our theorem above, then
8
the equation that defines this indifference curve is
q1 u1 + q2 u2 + q3 u3 = u (q)
Since q3 = 1 − q1 − q2 , this becomes
q1 u1 + q2 u2 + (1 − q1 − q2 ) u3 = u (q)
or
u (q) − qu1 − (1 − q) u3
q2 =
u2 − u 3
This is a linear function of q1 , which means that indifference curves are straight
lines.
To see the argument another way, go back to the independence axiom, which
states that for any three lotteries q, q and q and any λ, the lotteries λq +
(1 − λ) q and λq + (1 − λ) q must be ranked the same way as q and q . Since
this must be true for any three lotteries, then it must be that we can use
q in place of q in the argument above to conclude that λq + (1 − λ) q and
λq + (1 − λ) q must be ranked the same way as q and q . Then reducing the
compound lottery, if q˜q then λq + (1 − λ) q ˜q ˜q. Which means that every
lottery on the line segment between q and q must be indifferent to q. That is
just another way of saying that the indifference curves are straight lines.
It is a bit hard to deal with indifference. If you offer me q and q in an
experiment I will choose one of the them. I might strictly prefer the one I
pick, or I might be indifferent between them. It would be hard to figure this
out in practice. Fortunately, the independence axiom provides a much stronger
condition that gets around this. You can see this condition in Figure 4.
In the Figure, q and q are two lotteries on the same indifference curve. By
the previous argument the indifference curve connecting them is a straight line.
Choose some other lottery like p which strictly preferred to q. We shall try to
determine what the indifference curve looks like through the lottery p.
Draw a line segment from q through p to some third lottery q as in the
figure. Suppose that λ is such that p = λq + (1 − λ) q . Now since q ∼ q , we
have by the independence axiom that
p = λq + (1 − λ) q ∼ λq + (1 − λq )
Now the line segment from p to λq + (1 − λq ) is evidently parallel to the line
segment from q to q , so the indifference curve through p must be parallel to
the indifference curve through q. In other words, all the indifference curves are
parallel to one another when the independence axiom holds.
Now this is something that can be tested with an experiment. Simply present
an experimental subject a choice between two lotteries and observe their choice.
Once you see what their choice is, present them another two lotteries whose
probabilities are scaled in such a way that knowing that indifference curves are
all straight and parallel to one another will allow you to predict their choice in
the second lottery.
9
q2
1
q
p
q
λq + (1 − λ)q = p
q
1 q1
Figure 4: Parallel Indifference Curves
This is the experiment that was suggested by Allais. The consequences are
monetary with x1 = \$100, x2 = \$50 and x3 = 0. The first pair of lotteries
offered to the experimental subject are
q = {0, 1, 0}
and
q = {.1, .89, .01}
In other words, you can have either a sure \$50, or take a chance on getting \$100
with a small chance that you will lose everything. Most people are inclined to
take the sure \$50 in this case.
The second pair of lotteries is
p = {0, .11, .89}
and
p = {.1, 0, .9}
In this case, you probably won’t win anything with either lottery. Lottery p
offers you a small chance to earn \$50. Lottery p gives you a slightly smaller
chance of earning \$100, but also increases the probability with which you won’t
win anything. Most people are inclined to take the chance in this case and opt
for lottery p - perhaps because they are so unlikely to win anything they feel
there is nothing to lose in going for the \$100.
You should plot each of these four lotteries in a Figure like the one above.
You will see that the line segment joining q and q is parallel to the line segment
joining p and p . If indifference curves are all straight lines, parallel to one
10
another, anyone who chooses q over q must also choose p over p (just draw in
the indifference curve that would induce them to choose q over q then shift it
down to see what they will do with p and p ).
11
```
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Dana Myers
On 3/14/2018 7:53 AM, Graham wrote:
You can even buy the end parts to make your own cables with the cable of
your choosing if you are particular.
Is this really practical, especially from a mechanical ruggedness perspective?
73,
Dana K6JQ
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
Dana Myers
On 3/14/2018 7:35 AM, David Ranch wrote:
For a 2-pack of 10' cables [Volutz], \$17 on eBay was a deal.
Looks like there's a better deal on Amazon - free same-day shipping
with Prime, \$12. Shop around. It's nice having a ridiculously long
cable in my shack, BTW.
Wow.. 10'? I'd assume there is some pretty serious voltage drop over that! Regardless, I'll have to give them a look. Thanks.
Let's do some math, because we're scientists :-). A graphic at the
Volutz website says the power wires in their USB cables is 22 AWG.
I believe we can safely ignore the data wires here; as long as the
cable isn't dropping frames, it's OK.
Consulting Wikipedia:
https://en.wikipedia.org/wiki/American_wire_gauge#Rules_of_thumb
We see that the approximate resistance of this wire is 16 ohms / 1000'.
So, 10' of the wire (power and return) has about ~ 10/1000 * 16, or
0.16ohms, resistance per leg, or a total of 0.32 ohms.
Assuming a 200mA load (I don't actually know the Airspy HF+ current
consumption), that means the +5V rail at the HF+ drops by 0.032V,
or 32mV, and the ground rail at the Airspy HF+ increases by 32mV. In
other words, the +5V supply to the Airspy HF+ drops by 64mV.
Given that I bet nothing inside the Airspy HF+ runs from higher than a 3.3V
rail, this is plenty of margin for an LDO (you can do worst case math with USB
host power rail of 4.5V and LDO minimum differential of 0.6V and still have
500mV of margin).
Assume I'm wrong and the HF+ uses 500mA (the max spec for a
USB port, right?). That's 160mV drop to the HF+; still plenty of margin
for a 3.3V LDO (worst-case above would be > 400mV).
The keen observer might have noticed I suggest the DC ground
in the HF+ is ~32mV higher than the USB host ground. Note that
this doesn't matter because the USB signalling is differential data,
not analog.
OK, maybe we're not scientists, we're ham radio operators. In which
case you can disregard the above and say "it works great" (like the way
J-Pole antennas are 3/4-wave verticals :-)).
73,
Dana K6JQ
HF+ Quickstart?
Roel
Is there a quickstart page just like https://airspy.com/quickstart/ but for the HF+?
I installed the WinUSB Compatibility Driver and I get a waterfall and lots of noise but e.g. no FM stations.
Are there any settings I have to do?
Thnx!
Re: Using SDR# and HF+ for FM modulator measurement
prog
On Wed, Mar 14, 2018 at 09:30 am, Moij2x wrote:
What can cause this
Same list discussed earlier in this thread.
how to find and fix?
Fix what? What are your expectations? How do you define normal operation?
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
PMM
One thing I noticed was with my printer, it not grounded through power supply so having a good usb lead was important to kill the noise even though it only in standby mode, in fact printer contributed 20db of noise to the sdr near computer. Plugging in a decent usb cable pulled noise floor down 20db verse unplugged and against lesser quality usb cable.
So sometimes it can be said a poorly grounded device/non grounded psu can benefit from a good grounding given by usb cable, this seems to validate some of the cable choices due to better shielding/grounding wires in there construction.
On Wed, 14 Mar 2018, 15:11 Graham, <planophore@...> wrote:
Another good document that some may find interesting on this subject:
https://www.ti.com/sc/docs/apps/msp/intrface/usb/emitest.pdf
cheers,
Re: Using SDR# and HF+ for FM modulator measurement
Moij2x
I made more test on different PC-s, even in way that SDR#+HF+ was on one PC, REW for measurement on other PC with EMU sound card on ASIO drivers, PC-s where connected from audio out to EMU input with good cable.
Measurements with digital sound transfer from SDR# to REW soft was made with Jack2 Audio and EMU ASIO driver in one Win 7 PC. When this digital transfer measurements where done, I disabled all Windows Audio services, so Windows audio was not involved only ASIO driver and Jack2 Audio.
In many measurements KT-6040 THD is lower on same modulator input levels than HF+ in higher than 500 Hz range. HF+ had lower noise on lower modulator input levels.
Probably it is caused by PC. I not find significant difference in SDR# result made with digital transfer or when made from one PC to other with physical cable.
Interesting thing I find is that THD peaks on SDR measurement are on different frequencies in different PC-s.
Below 2 images of THD measured with SDR. First where SDR was in one PC, measuring soft REW in other. Second where all was in on PC.
On first case SDR was running in Windows 10 PC 2,9 GHz CPU, where SDR# CPU usage was about 16% from total, on second faster Win 7 64bit 3,5 GHz PC, SDR# used about 7% from total CPU. In both cases all settings of SDR# where identical.
As can be seen generally lines and figures are very similar except THD peaks in higher region on lower levels are on different frequencies.
What can cause this, how to find and fix?
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
Graham
Another good document that some may find interesting on this subject:
https://www.ti.com/sc/docs/apps/msp/intrface/usb/emitest.pdf
cheers,
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
Graham
Some of the better manufacturers put some details on their web site but
at that is usually still somewhat sparse.
for USB micro and apple cables I prefer Anker which are quite well
reviewed and their web page does provide some technical details.
Also, Sabrent and Volutz.
Tyco / TE Connectivity makes some decent USB cables. In their USB A to B
cables they have versions which have 25 AWG for the data pair and 20
AWG for power.
I have destructably disassembled (i.e. cut apart) a number of cables
from cheap to moderately priced. There are differences and you do get
what you pay for.
You can even buy the end parts to make your own cables with the cable of
your choosing if you are particular.
something else to consider is the cable shields. The metal parts at both
end of the cables should not be connected to the ground pin of the
connectors but should be connected to the shield (if there is any) of
the cable. In fact, sometimes it is better to just have the shield
connected to the shell on the end that plugs into the computer and not
connected on the other end. This creates what is called a Farady shield
(or cage if you prefer) over the cable. Connecting both ends can
sometimes create ground loops in the shield. I have run out of fingers
and toes to count the number of times that I have seen where a cable
having the shield connected at one end only solves EMI/RFI issues where
another cable with the shield connected at both ends only made the
problem worse or was in fact the source of the problem to begin with.
Solving EMI/RFI issues is a mixed bag of science and sometimes a bit of
black magic.
This is a could introductory and practical text on the subject:
http://audiosystemsgroup.com/SAC0305Ferrites.pdf
cheers, Graham ve3gtc
On 3/14/2018, "David Ranch" <airspy-groupsio@trinnet.net> wrote:
For a 2-pack of 10' cables [Volutz], \$17 on eBay was a deal.
Wow.. 10'? I'd assume there is some pretty serious voltage drop over
that! Regardless, I'll have to give them a look. Thanks.
--David
KI6ZHD
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
David Ranch
For a 2-pack of 10' cables [Volutz], \$17 on eBay was a deal.
Wow.. 10'? I'd assume there is some pretty serious voltage drop over that! Regardless, I'll have to give them a look. Thanks.
--David
KI6ZHD
Re: Gain settings with Firmware 1.6.2
prog
On Wed, Mar 14, 2018 at 04:21 am, Chris Smolinski wrote:
I'd need to check my email history for the exact details, but I went through a series of tests, passing the information back to Youssef, when I first got the AirSpyHF+. From memory, the MW intermod/images/aliases were a function of the station frequency and the AirSpyHF+ center frequency, as well as a 5 kHz tuning offset (part of the library on the computer side). Again I don't recall the exact relationship, it has been several months. I can mostly work around them with careful selection of the center frequency. They're not present to any degree with other radios, SDRs or otherwise, using the same (or other) antennas. I mostly use the AirSpyHF+ on HF where it isn't as much of an issue. I believe I have occasionally noted some similar products on HF involving mixing of strong SWBC stations. It doesn't happen often, and Radio Habana Cuba (the land of inept radio engineering) is often involved, so draw your own conclusions :)
Chris Smolinski
Black Cat Systems
Westminster, MD USA
http://www.blackcatsystems.com
Chris, this is a completely different topic. We are discussing the impact of the latest firmware that affects the control of the analog front-end.
You can start a new thread for the IQ correction algorithm and how it behaves with different signal situations. This will avoid more confusion.
Re: Gain settings with Firmware 1.6.2
Chris Smolinski
I'd need to check my email history for the exact details, but I went through a series of tests, passing the information back to Youssef, when I first got the AirSpyHF+. From memory, the MW intermod/images/aliases were a function of the station frequency and the AirSpyHF+ center frequency, as well as a 5 kHz tuning offset (part of the library on the computer side). Again I don't recall the exact relationship, it has been several months. I can mostly work around them with careful selection of the center frequency. They're not present to any degree with other radios, SDRs or otherwise, using the same (or other) antennas. I mostly use the AirSpyHF+ on HF where it isn't as much of an issue. I believe I have occasionally noted some similar products on HF involving mixing of strong SWBC stations. It doesn't happen often, and Radio Habana Cuba (the land of inept radio engineering) is often involved, so draw your own conclusions :)
Chris Smolinski
Black Cat Systems
Westminster, MD USA
http://www.blackcatsystems.com
On Mar 13, 2018, at 11:03 PM, jdow <jdow@earthlink.net> wrote:
Check the FCC requirements for harmonic radiation imposed on broadcast stations. Once done you will have a suitable "Aha" moment, one hopes, when you figure out that X dB down from Y dB over S9 (er 50 uV at 50 ohms unless you are Collins in which case it is 100 uV at 50 ohms) and just how big a signal S9 really is. Of course you will see harmonics. And it's nothing to do with your receiver. Now you use the word "images". If the signals are not on frequency they are leakage through the front end filtering and that implies they would properly be called aliases. Images are a superheterodyne receiver phenomenon and occur on the opposite side of a conversion process's IF frequency from the tuned frequency. It will be a different frequency. And while the AirSpy HF+ SDR with your computer is "sort of" a superhet receiver the IF in question is possibly 0 Hz. So images are "interesting" to contemplate. (I've not tried to figure this out.) A good set of numbers taken with variable center frequency tuning can give enough numbers to determine all the variables that might be involved. That will help you define your local "Don't DO that!" conditions.
(And if the tech on the phone asks you to make sure your computer is turned on, fire back at him, "I didn't know computers had a sex life!")
{^_^}
On 20180313 06:42, Chris Smolinski wrote:
I notice the same levels of intermod on LW/LW with firmware version 1.6.2 vs previous versions, it does not seem any worse. This is using SdrConsole which AFAIK keeps the AirSpyHF+ AGC turned on.
While playing around with SDR# I was of course able to generate lots more intermod/etc by switching AGC off and turning the gains up too high and overloading things, as you'd expect.
I don't have any nearby strong MW stations, the nearest is about 10 miles away, and I think 5 kW. With AGC on I can occasionally see images from it on LW and MW, although careful tuning can eliminate them. (Setting the center frequency such that the station is outside the 768 kHz passband)
Any guesses when there will be a library update to further improve the AirSpyHF+ filtering?
Chris Smolinski
Black Cat Systems
Westminster, MD USA
http://www.blackcatsystems.com
On Mar 13, 2018, at 8:33 AM, Giovanni Carboni <iz5pqt@gmail.com> wrote:
Hello everyone,
I had to disinstall version 1.6.2 of the firmware (20180309) on my AirspyHF+ since it caused horrible
intermodulation below 1 MHz.
Less than 15 km from my QTH there is a 50 kW MW TX on 657 kHz (Rai Coltrano). The signal at the
radio antenna input is -21 dBm measured with an ELAD S2 and checked with a scope. The two measurements
agree within errors (20 mV peak on the scope). It is a quite large signal so I am clearly in an extreme
situation.
With all firmware versions until 1.6.1 the receiver did not show intermodulation problems (except of course
for a narrow band around 657 kHz). With 1.6.2 a terrible
intermodulation occurred making reception impossible on all the MW spectrum up to 1000 kHz. I did not check
carefully but it looks like reception below 500 kHz was not affected.
I had to go back to 1.6.1 and things became normal.
I do not have instrumentation to perform more accurate measurements but the test is
easily reproducible in a lab.
73 Giovanni IZ5PQT
Re: AirSpy HF+ flashing again
I L
Thanks Dave for answering my questions! I have now successfully updated the firmware from 1.00 to the very latest from 20180309 using the standard procedure.
Regards,
Ingemar
2018-03-13 14:21 GMT+01:00 D R via Groups.Io :
Hi Ingemar,
1. Look under Source:Airspy HF+ in SDR Sharp.
2. You can use any version whenever you like.
3. You'll get many answers to this question, as peoples' experiences appear to differ. In my case I need to use the recovery procedure for every update, but others can use the "standard" procedure. Neither option is anything to be afraid of.
4. It will probably die at some time, but hopefully not for several years, and after many updates
Regards,
Dave
From: I L <ingemar.lekteus@...>
To: main@airspy.groups.io
Sent: Tuesday, 13 March 2018, 12:48
Subject: [airspy] AirSpy HF+ flashing again #firmware
I am new to the group and a new HF+ owner. I have not been able to find answers to a couple of questions. In case they already have been discussed I hope you will apologize me for bringing them up again.
1. How do I find out what firmware version my HF+ uses?
2. When I want to upgrade, can I go directly to the latest version or do I need to follow a certain sequence?
3. There is a recovery procedure mentioned in README.txt. As I got the RX just recently I suppose the firmware version is not the first one so I hope I can use the standard procedure. But to be sure I need an answer to the first question and I also need to know the version number of the first firmware.
4. Can I be sure not to end up with a dead unit?
Best wishes,
Ingemar, SM0ECF
--
------------------------------------------------------
Ingemar Lekteus
Älgstigen 2
178 52 EKERÖ
E-post: ingemar.lekteus@...
Mobil: 0704-71 33 30
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
jdow
Screening is good. Breaking ground loops, AF or RF, is a good thing, too. Ferrite makes the ground loops lossy as it breaks them. This is sort of the best of both worlds. And there is something to be said for keeping RF that is outside of cables where it belongs, on the outside of the box. So making a good electrical bond from each of the box's connectors to the box itself may provide you with a quite worthwhile improvement in behavior. It should not make it any worse. (And done carefully it will not change the cosmetics of the box. Braided shield from small diameter coax is a nice tool for this. Clean the inside of the holes and the inside of the box for the 10 MHz ref (AirSpy only) and USB connector holes. Then when installing the cover mush some of the braid between the edges of the hole and the connector shield. For the RF clean off a space just to about half way to the outside of the lockwasher. Then when tightening it you will get a nice ground to the coax shields for the RF input connector(s).
Also, unless you are running a very low efficiency antenna at HF, test for the interference when the band is open. If you cannot see it then it doesn't matter even if you can see it with a 50 ohm dummy load and with a small test signal present.
{^_^}
On 20180313 21:32, Ian DXer wrote:
Hi Dana,
I was looking at the Volutz Facebook page & I note that Volutz have just released a new Apple Lighting Cable.
For noise immunity I like the metal shroud that links the foil shield of the cable & surrounds the connector & components
underneath. One also gets a better look at the dual cable shielding from the drawing on the Facebook page.
Are these SMD components part of the lightning spec or are they part of Volutz's attention to EMI/EMC or impedance
matching?
I see a similar shroud on their USB C cable from their Facebook page.
What's not clear to me is if Volutz use a metal shroud on their connectors for their USB 2.0 to micro USB cables or
any filtering SMD components (if that's what they are?) similar to their new Apple Lightning Cable.
The Volutz website is scant with close up pics & tech data.
A further improvement to their USB 2.0 to micro usb cables would be to have a foil shield between the data & power pairs.
I purchased a replacement USB cable (CableCreation/Xillie) from the Internet last week. It arrived on Monday. It is supposedly triple shielded, but I suspect these are 'just
words' from a Chinese supplier via Amazon. I compared the the noise performance between AIRSPYHF+ supplied USB cable with my own
added mix of ferrites attached, also a Samsung USB cable, the CableCreation USB & the CableCreation Lead with a mix of ferrites.
Test freqs were roughly every 1MHz between 2 & 30MHz. Conclusions: I didn't gain the results I expected from the CableCreation USB cable, results were
similar to the plain Samsung Cable. But what brought best results was having ferrites on either the existing AirSpyHF+ USB cable or the CableCreation USB cable (similar results).
That said, for my own electro smog environment results were typically only a 2dB improvement to noise floor for the USB cables with the ferrites attached.
But on the better end of the results scale 'on some' frequencies I noted a 6-10dB improvement (with ferrites).
I'm sure if I purchased a better screened USB cable (similar to the standard of my router CAT6 S/FTP patch cables) I should hopefully see better results.
Every dxer has a unique RFI environment with their own unique challenges, some requiring a huge investiture of time & effort (mine's only in the early stages).
Our new Samsung refrigerator is the latest to be added to the RFI hit list (I'll try to mute my grumble).
73
Ian
Re: Airspyhf+ flashing
Ian DXer
RE: >AirSpy HF+ Firmware Update Instructions
###### >Howard M. Harte, WZ2Q - February 22, 2018 Hi Howard,There's one omission from your write-up that I would strongly recommend adding or highlighting. That is to add a line WARNING owners that their calibration will be lost during the initial firmware update procedure after the RESET.That said, mine didn't vary too much & was still within spec. (But one has to then re-enter any stored notch entries within the IF processor plugin, if used)73Ian
Re: USB cable quality and QRM in a spyserver setup with an Airspy Hf+ and Raspberry PI3. 6 microUSB cable test.
Ian DXer
Edited
Hi Dana,
I was looking at the Volutz Facebook page & I note that Volutz have just released a new Apple Lighting Cable.
For noise immunity I like the metal shroud that links the foil shield of the cable & surrounds the connector & components
underneath. One also gets a better look at the dual cable shielding from the drawing on the Facebook page.
Are these SMD components part of the lightning spec or are they part of Volutz's attention to EMI/EMC or impedance
matching?
I see a similar shroud on their USB C cable from their Facebook page.
What's not clear to me is if Volutz use a metal shroud on their connectors for their USB 2.0 to micro USB cables or
any filtering SMD components (if that's what they are?) similar to their new Apple Lightning Cable.
The Volutz website is scant with close up pics & tech data.
A further improvement to their USB 2.0 to micro usb cables would be to have a foil shield between the data & power pairs.
I purchased a replacement USB cable (CableCreation/Xillie) from the Internet last week. It arrived on Monday. It is supposedly triple shielded, but I suspect these are 'just
words' from a Chinese supplier via Amazon. I compared the the noise performance between AIRSPYHF+ supplied USB cable with my own
added mix of ferrites attached, also a Samsung USB cable, the CableCreation USB & the CableCreation Lead with a mix of ferrites.
Test freqs were roughly every 1MHz between 2 & 30MHz. Conclusions: I didn't gain the results I expected from the CableCreation USB cable, results were
similar to the plain Samsung Cable. But what brought best results was having ferrites on either the existing AirSpyHF+ USB cable or the CableCreation USB cable (similar results).
That said, for my own electro smog environment results were typically only a 2dB improvement to noise floor for the USB cables with the ferrites attached.
But on the better end of the results scale 'on some' frequencies I noted a 6-10dB improvement (with ferrites).
I'm sure if I purchased a better screened USB cable (similar to the standard of my router CAT6 S/FTP patch cables) I should hopefully see better results.
Every dxer has a unique RFI environment with their own unique challenges, some requiring a huge investment of time & effort (mine's only in the early stages).
Our new Samsung refrigerator is the latest to be added to the RFI hit list (I'll try to mute my grumble).
73
Ian
Re: Gain settings with Firmware 1.6.2
jdow
And do remember that if you switch from a signal generator output with signal set at -70 dBm-ish (-90 dBm will also do) to a live antenna with some signals present and the noise increases by more than a little bit (1 or 2 dB) you can insert more attenuation at the receiver's input and not lose any weak signal ability. Obsessing about HF noise figure is one of the silliest obsessions I've seen in innately perverse animate homo sapiens, IPAHS. (Pronounce it.) And, I admit, before I found a good book describing noise figure and atmospheric noise I fell down that rabbit hole of frustration myself.
{o.o}
On 20180313 15:15, Roberto Zinelli wrote:
Giovanni, with 1.6.2 i put AIRSPY-HF ACG OFF and set attenuator -12 DB, the overral noise figure for me is better then 1.6.1
On Tue, Mar 13, 2018 at 11:03 PM, Giovanni Carboni <iz5pqt@gmail.com <mailto:iz5pqt@gmail.com>> wrote:
Here are three spectra around 738 kHz taken with a) FW 1.6.2; b) FW
1.6.2 plus an external 20 dB attenuator; c) FW 1.6.1.
As you see in case a) the reception is totally washed out. The only way to
get a reasonable spectrum with 1.6.2 is to turn off the AGC in SDR# and
insert a lot of attenuation.
73 Giovanni
IF Spectrum Tuning
Bingo TheClowno
Has the IF Spectrum tuning been disabled?
I can't fine tune in the IF Spectrum display anymore in the latest SDR#.
I prefer this way than zooming in all the way.
Is there an option I need to enable?
Re: Airspy noise figure
jdow
(She screws up her face thinking furiously before creebing that measuring RMS signal plus noise and noise within an FFT spectrum summing the power for every FFT bin within the specified bandwidth should potentially be VERY accurate when reduced to a noise figure when using an SDR. Thanks for provoking me into analyzing your statement when it "bothered me" a little. It will likely be most accurate when your bandwidth of interest includes 1000 or more FFT bins and you diddle the calculated values to account for the FFT weighting algorithm. And I bet Leif has already done this in the past.)
{^_-} Joanne
On 20180313 10:06, doug wrote:
There's another method of getting a noise figure number, if you have a
pulsed RF source: carefully adjust the source so that the detected rf
pulse is tangential. What this means is that
the average noise at the bottom of the detected pulse is equal to the
average noise of the baseband, where there is no pulse. The value is
equal to 8dB above signal equals noise. Now if you
know the bandwidth of the receiver, you can determine the noise figure,
since you know that at 1 Hz BW, a perfect receiver would have a
sensitivity of -174 dBm. For every increase in
bandwidth, the s/n ratio decreases by the value ratio. For example, at
1KHz bandwidth, a perfect receiver would have s=n at -144 dBm. This
method is not as precise as a calibrated noise source,
but it's close.
--doug. WA2SAY, retired RF engineer.
On 03/13/2018 05:46 AM, jdow wrote:
Given a choice I'd prefer to try liquid funnygas er Helium for the
cold source.
{^_-} Just being imaginative and silly again.
On 20180312 09:15, David Eckhardt wrote:
Yes, there is nothing to compete with the hot/cold 'noise' source.
kTB is very specific. However, knowing the information bandwidth
introduces errors for our amateur receivers unless we rigorously
measure that as well. I've always preferred the hot/cold noise
source, but don't keep LN2 around. However, these days, most larger
grocery stores carry dry ice.
Dave - WØLEV
On Mon, Mar 12, 2018 at 4:06 AM, Marcus D. Leech <mleech@ripnet.com
<mailto:mleech@ripnet.com>> wrote:
On 03/11/2018 09:00 PM, David Rowe wrote:
Thanks Dave - sounds like you have a great experimental set
up there for
NF measurements. For anyone interested in NF measurements,
I've blogged
on my humble experiments here:
http://www.rowetel.com/?p=5867
I mentioned your results in the post Dave, hope thats OK.
Cheers,
David
I'll point out that single-point noise-figure methods are rather
error-prone, which is why the industry has standardized on the
Y-factor
method, in which
you have TWO sources of known noise power distribution.
For noise figures above 3dB, you can usually use a high-quality
terminator
at room temperature -- whose PSD will be -174dBM/Hz, and then a
calibrated
noise source, usually in the 20-35dB ENR range. You take
averaged
measurements of both, and use that to determine the noise figure
of the DUT.
Once you start going to noise-figures below 2-3dB, it's better to
have a
"cold load" as one of the noise sources. Labs that care about
noise figure
measurements for "good" LNAs usually use an LN2-cooled load
for the
On 12/03/18 10:18, David Eckhardt wrote:
I have a calibrated noise source good to 18 GHz and several
precision attenuators (HP, both single value and
switched). Knowing
the noise output of the noise source, I attenuate until
the noise
just goes into the 'noise', tangential noise. The
difference
between the noise source known output and the
attenuation is my
noise figure.
Dave - WØLEV
On Sun, Mar 11, 2018 at 10:27 PM, David Rowe
<david@rowetel.com
<mailto:david@rowetel.com> <mailto:david@rowetel.com
<mailto:david@rowetel.com>>> wrote:
Thanks Dave,
Could you tell me a little more about how you
measured NF?
Thanks,
David
On 12/03/18 08:37, David Eckhardt wrote:
I've measured both the R1 and my present R2.
They both come in
around 4 dB. I believe 7 to 8 dB is a bit much
based on my
experience with them.
Dave - WØLEV
On Sun, Mar 11, 2018 at 9:34 PM, <david@rowetel.com
<mailto:david@rowetel.com>
<mailto:david@rowetel.com
<mailto:david@rowetel.com>>
<mailto:david@rowetel.com <mailto:david@rowetel.com>
<mailto:david@rowetel.com
<mailto:david@rowetel.com>>>> wrote:
Hi - I've been setting up a system to
measure noise
figures
of SDRs
in real time, an extension of some similat
tests we
performed a few
years ago:
https://www.rowetel.com/?p=5057
<https://www.rowetel.com/?p=5057>
<https://www.rowetel.com/?p=5057
<https://www.rowetel.com/?p=5057>>
<https://www.rowetel.com/?p=5057
<https://www.rowetel.com/?p=5057>
<https://www.rowetel.com/?p=5057
<https://www.rowetel.com/?p=5057>>>
I'm getting a NF of 7-8dB for a new Airspy
R2 and Airspy
mini. The
spec quotes 3.6dB. Has anyone else
measured the NF?
I note the 3.6dB figure is exactly the same
as the front
end tuner
R820 tuner. However a system noise figure
will always be
poorer
than the front end device.
Thanks,
David
-- *Dave - WØLEV
*
*/Just Let Darwin Work/*
-- *Dave - WØLEV
*
*/Just Let Darwin Work/*
--
*Dave - WØLEV
*
*/Just Let Darwin Work/*
Re: Gain control #HF+
jdow
Bingo - question 2 is very pertinent. You need to learn what "decimation" can do to apparent noise levels. And remember that with SDRSharp you are measuring dB relative to full scale and that AirSpy HF+ has nicely effective AGC options.
{^_^}
On 20180313 09:52, Pete Smith wrote:
Can't really quantify. This is something I encountered while doing a full HF survey for RFI. I ran into one emitter of clear signals every 8 KHz, and noticed that it jumped up in strength on the ham bands as compared to 1-MHz slices above and below the bands. What I was wondering is whether the Airspy HF+ has different preamp gains settable by band, or whether gain is supposed to be uniform across the HF range.
73, Pete N4ZR
Check out the Reverse Beacon Network
at<http://reversebeacon.net>, now
spotting RTTY activity worldwide.
"retail" DX cluster.
On 3/13/2018 9:41 AM, prog wrote:
On Tue, Mar 13, 2018 at 06:36 am, Pete Smith wrote:
I'm seeing what seem to be significant gain variations between the ham
bands and other frequencies.
How significant is "significant"? What are the numbers? What is your test setup? And what are your expectations in dB?
Re: Gain settings with Firmware 1.6.2
jdow
Check the FCC requirements for harmonic radiation imposed on broadcast stations. Once done you will have a suitable "Aha" moment, one hopes, when you figure out that X dB down from Y dB over S9 (er 50 uV at 50 ohms unless you are Collins in which case it is 100 uV at 50 ohms) and just how big a signal S9 really is. Of course you will see harmonics. And it's nothing to do with your receiver. Now you use the word "images". If the signals are not on frequency they are leakage through the front end filtering and that implies they would properly be called aliases. Images are a superheterodyne receiver phenomenon and occur on the opposite side of a conversion process's IF frequency from the tuned frequency. It will be a different frequency. And while the AirSpy HF+ SDR with your computer is "sort of" a superhet receiver the IF in question is possibly 0 Hz. So images are "interesting" to contemplate. (I've not tried to figure this out.) A good set of numbers taken with variable center frequency tuning can give enough numbers to determine all the variables that might be involved. That will help you define your local "Don't DO that!" conditions.
(And if the tech on the phone asks you to make sure your computer is turned on, fire back at him, "I didn't know computers had a sex life!")
{^_^}
On 20180313 06:42, Chris Smolinski wrote:
I notice the same levels of intermod on LW/LW with firmware version 1.6.2 vs previous versions, it does not seem any worse. This is using SdrConsole which AFAIK keeps the AirSpyHF+ AGC turned on.
While playing around with SDR# I was of course able to generate lots more intermod/etc by switching AGC off and turning the gains up too high and overloading things, as you'd expect.
I don't have any nearby strong MW stations, the nearest is about 10 miles away, and I think 5 kW. With AGC on I can occasionally see images from it on LW and MW, although careful tuning can eliminate them. (Setting the center frequency such that the station is outside the 768 kHz passband)
Any guesses when there will be a library update to further improve the AirSpyHF+ filtering?
Chris Smolinski
Black Cat Systems
Westminster, MD USA
http://www.blackcatsystems.com
On Mar 13, 2018, at 8:33 AM, Giovanni Carboni <iz5pqt@gmail.com> wrote:
Hello everyone,
I had to disinstall version 1.6.2 of the firmware (20180309) on my AirspyHF+ since it caused horrible
intermodulation below 1 MHz.
Less than 15 km from my QTH there is a 50 kW MW TX on 657 kHz (Rai Coltrano). The signal at the
radio antenna input is -21 dBm measured with an ELAD S2 and checked with a scope. The two measurements
agree within errors (20 mV peak on the scope). It is a quite large signal so I am clearly in an extreme
situation.
With all firmware versions until 1.6.1 the receiver did not show intermodulation problems (except of course
for a narrow band around 657 kHz). With 1.6.2 a terrible
intermodulation occurred making reception impossible on all the MW spectrum up to 1000 kHz. I did not check
carefully but it looks like reception below 500 kHz was not affected.
I had to go back to 1.6.1 and things became normal.
I do not have instrumentation to perform more accurate measurements but the test is
easily reproducible in a lab.
73 Giovanni IZ5PQT
11741 - 11760 of 40000 | 8,676 | 33,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-21 | latest | en | 0.916571 |
https://cstheory.stackexchange.com/questions/18630/a-theorem-regarding-statistically-hiding-commitment-schemes/18635#18635 | 1,638,646,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00169.warc.gz | 250,032,874 | 33,564 | # A theorem regarding statistically-hiding commitment schemes
Let $C_n$ be a non-interactive statistically-hiding commitment scheme, able to commit to an $n$-bit string. To commit to $m \in \{0,1\}^n$, the sender picks a random $r$ (of proper length), and sends $C_n(m ; r)$ to the receiver. To decommit, he simply reveals $m$ and $r$. We let $C_n(m)$ denote the random variable obtained by uniformly and independently picking $r$ (of proper length) and computing $C_n(m ; r)$.
Finally, let $U_n$, $U_n'$, and $U_n''$ be i.i.d. random variables with uniform distribution over $\{0,1\}^n$. We assume that repeated use of a random variable results in the same sample. For example, $\langle U_n, U_n+1 \rangle$ means a pair, where we first sample $U_n$ to obtain an $n$-bit string $m$, and then let the pair be $\langle m, m+1 \rangle$.
I want to prove that the following distribution ensembles are statistically indistinguishable: $$\mathcal {X} = \{\langle U_n, C_n(U_n) \rangle\}_{n \in \mathbb{N}} \enspace,$$ $$\mathcal {Y} = \{\langle U_n', C_n(U_n'') \rangle\}_{n \in \mathbb{N}} \enspace.$$
This certainly follows for $n=1$.
Fix any algorithm $A$ that attempts to distinguish, and let $p_{a,b} = \Pr[A(\langle a, C(b) \rangle)=1]$, where $a,b\in \{0,1\}$. Now by the assumption that $C$ is statistically hiding, we have $p_{0,0} \approx p_{0,1}$ (for otherwise you could prepend the fixed constant 0 to any commitment and then use $A$ to break the hiding property) and we have $p_{1,0} \approx p_{1,1}$ (for a similar reason).
Expanding out the definitions, we have $\Pr[A(X)=1]=(p_{0,0}+p_{1,1})/2$ and $\Pr[A(Y)=1]=(p_{0,0}+p_{0,1}+p_{1,0}+p_{1,1})/4$ (where $X,Y$ are samples from your distributions ${\cal X},{\cal Y}$). But from the facts above, we have $2p_{0,0} + 2p_{1,1} \approx p_{0,0} + p_{0,1} + p_{1,1} + p_{1,0}$, i.e., $\Pr[A(X)=1]\approx \Pr[A(Y)=1]$, and thus $X,Y$ are statistically indistinguishable.
I think from this you should be able to see how this generalizes to $n>1$ as well. Basically, you note that
$$\Pr[A(X)=1]=1/2^n\sum_b \Pr[A(\langle b,C(b)\rangle)=1]\approx 1/2^{2n}\sum_{a,b} \Pr[A(\langle a,C(b)\rangle)=1] = \Pr[A(Y)=1],$$
using the fact that $\Pr[A(\langle a,C(b))\rangle)=1] \approx \Pr[A(\langle a,C(b')\rangle)=1]$ for all fixed $a,b,b'$ (otherwise you could distinguish $C(b)$ from $C(b')$ by simply prepending $a$ and then running $A$). | 812 | 2,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.785483 |
http://mathoverflow.net/feeds/question/22390 | 1,368,989,540,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697974692/warc/CC-MAIN-20130516095254-00040-ip-10-60-113-184.ec2.internal.warc.gz | 164,766,390 | 1,697 | How far is the tangent bundle from projective space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:52:19Z http://mathoverflow.net/feeds/question/22390 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22390/how-far-is-the-tangent-bundle-from-projective-space How far is the tangent bundle from projective space? altgr 2010-04-23T20:19:54Z 2010-04-24T04:52:35Z <p>Is there a general theory of embeddings of the (total variety of) the tangent bundle on a (nonsingular) projective variety into projective space? I suppose what I really mean is (and to be more precise about the projective space), if $\mathbb{P}$ is the <em>projective completion</em> of $T_X\rightarrow X$, then what can be said about the relative dimension of "infinity": $D=\mathbb{P}\backslash T_X$?</p> <p>I apologize if this question is vague. Any thoughts or references will be greatly appreciated.</p> <p>Thanks!</p> http://mathoverflow.net/questions/22390/how-far-is-the-tangent-bundle-from-projective-space/22407#22407 Answer by Sasha for How far is the tangent bundle from projective space? Sasha 2010-04-24T04:52:35Z 2010-04-24T04:52:35Z <p>The simplest way to get a "projective completion" is to consider the projectivization on $X$ of $T_X \oplus L$ for some line bundle $L$ on $X$. In this case the complement will be the projectivization of $T_X$ and will have codimension 1. Sometimes you can contract this completion to get smaller complement, e.g. if $X$ is a curve (choose $L$ in such a way that $\omega_X\otimes L$ is very ample, then $P(T_X \oplus L) = P(T_X\otimes L^{-1} \oplus O)$ which is a blowup of the projective cone over $X$ in the embedding given by $\omega_X\otimes L$).</p> | 506 | 1,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2013-20 | latest | en | 0.756848 |
http://www.physicsforums.com/showthread.php?t=658226&page=2 | 1,411,055,526,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657128304.55/warc/CC-MAIN-20140914011208-00221-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 745,132,977 | 13,223 | # Photon is electromagnetic field, right?
by Barry_G
Tags: electromagnetic, field, photon
Mentor
P: 12,010
Quote by Barry_G Yes, they are synonyms. Electron has electric field, therefore it is electrically charged.
No, they are not synonyms. Charge is a property of a particle, a field is not. The field describes how particles will react to each other based on their charge, distance, and relative motion.
http://en.wikipedia.org/wiki/Field_%28physics%29
http://en.wikipedia.org/wiki/Electric_charge
I disagree. As all those diagrams show photon is not waves in some field, but rather waves of those fields. It's not photon that waves, it is magnetic and electric fields that move, and their combination is what photon is.
Please, that's like arguing that a water wave is a wave OF water instead of a wave IN water. It means the same thing.
P: 68
Quote by K^2 Yet you still don't realize that the charge density at every point in space in and around the photon is precisely zero. It never becomes non-zero. You made an assumption that EM field requires a charge density, and that's why you cling onto QED interpretation. But QED interpretation does not give you a charge. It gives you virtual particle production, but no charge.
So the charge of the electric field in a photon is zero? Then what is the magnitude of that electric field, say at peaks of its amplitude?
No. How many times do several people have to tell you that?
Insistence does not prove or solve anything. Can you point some reference that distinguishes electric field from electric charge?
Because photon is neutral. Neutral particles are not influenced by electric field. Photon also has zero magnetic moment, so it is not influenced by magnetic field either.
Is magnetic or electric field of a photon neutral?
It's the same thing. Photon is the field. Field is the photon. Waves in the field or waves of the field are exactly the same thing in this case.
It's not the same thing. Like gravitational waves are not the same thing as spatial motion of gravity field as whole.
Photon spin is measured experimentally by transferring the angular momentum from photon to a charged particle and measuring resonance in magnetic field.
So photon spin is has something do with orientation of its magnetic field?
P: 1,020
Quote by Barry_G So you mean "no", photon spin can not be experimentally measured?
no,it can be measured experimetally.Here is some old experiment on it
http://prola.aps.org/abstract/PR/v50/i2/p115_1
P: 2,470
Quote by Barry_G Can you point some reference that distinguishes electric field from electric charge?
Maxwell's. Equations. There are no other equations for electromagnetic field.
So the charge of the electric field in a photon is zero?
There is no such thing as charge of the electric field.
P: 68
Quote by Drakkith No, they are not synonyms. Charge is a property of a particle, a field is not. The field describes how particles will react to each other based on their charge, distance, and relative motion. http://en.wikipedia.org/wiki/Field_%28physics%29 http://en.wikipedia.org/wiki/Electric_charge
Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field.
The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.
Please, that's like arguing that a water wave is a wave OF water instead of a wave IN water. It means the same thing.
Water wave is effect caused by the motion OF water molecules. You have to distinguish cause and effect, what is moving and what is the product of that motion.
P: 68
Quote by andrien no,it can be measured experimetally.Here is some old experiment on it http://prola.aps.org/abstract/PR/v50/i2/p115_1
I'm afraid that's about polarization, not spin, unless those two are related, but I don't see how since polarization is about planes of oscillation of em fields and seems it can be arbitrary, while spin is supposed to be intrinsic and just 1 or -1.
P: 2,470
Quote by Barry_G I'm afraid that's about polarization, not spin, unless those two are related, but I don't see how since polarization is about planes of oscillation of em fields and seems it can be arbitrary, while spin is supposed to be intrinsic and just 1 or -1.
If you actually bothered to read the article on circular polarizaion that has been previously linked, you would know the answer to this question.
Mentor
P: 3,967
Quote by Barry_G Insistence does not prove or solve anything. Can you point some reference that distinguishes electric field from electric charge?
"Electricity and Magnetism" by Edward Purcell would be a good start. Google will find you a copy somewhere for cheap.
P: 68
Quote by K^2 Maxwell's. Equations. There are no other equations for electromagnetic field.
But there are charges, electric and magnetic fields outside of photon, so why are we having this silly semantic argument, why can not we find nice specific definition that would point out differences so we can distinguish what is field and what is charge?
Maxwell's equations do not explain how can photon have electric and magnetic field and yet we can not bend a beam of light by external electric or magnetic fields.
Actually polarization should answer this, if we can change polarization then that would be the way to influence these fields, right? Perhaps we can not bend a beam of light, but if we can change the plane of em fields oscillation, well that's pretty good too.
There is no such thing as charge of the electric field.
Depends on definition. We need a technical dictionary, or something that would make it clear how they relate and how they differ. Until then I think my definition is better than yours.
P: 68
Quote by K^2 If you actually bothered to read the article on circular polarizaion that has been previously linked, you would know the answer to this question.
It says almost nothing about it, just rises many questions. What am I to conclude, left and right spin corresponds to 1 and -1? Ok, so why then Wikipedia says spin of a photon is 1, why does it not say 1 or -1? And what would be the spin of non-spinning photon? Also, how does that compare to electron spin since it seems photon can be spinning at arbitrary speed?
P: 2,470
Quote by Barry_G Maxwell's equations do not explain how can photon have electric and magnetic field and yet we can not bend a beam of light by external electric or magnetic fields.
Yes they do. They do precisely that. This is absolutely obvious, but lets walk through it. Let $E_{\gamma}$ and $B_{\gamma}$ be electric and magnetic fields of some beam of light respectively. They, therefore, satisfy charge-free Maxwell's Equations.
$$\nabla \cdot E_{\gamma} = 0$$
$$\nabla \times E_{\gamma} = -\frac{\partial B_{\gamma}}{\partial t}$$
$$\nabla \cdot B_{\gamma} = 0$$
$$\nabla \times B_{\gamma} = \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t}$$
However, instead of sending that beam of light through the vacuum, we send it through some system of charges and currents that have already created some fields $E$ and $B$. These fields must satisfy Maxwell's Equations with charges and currents.
$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$
$$\nabla \times E = -\frac{\partial B}{\partial t}$$
$$\nabla \cdot B = 0$$
$$\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$$
Here, $\rho$ and $J$ are charge and current densities, respectively.
We wish to demonstrate that this does not alter the trajectory of the light beam. In other words, that the new electric field is simply $E + E_{\gamma}$ and new magnetic field is simply $B + B_{\gamma}$. We show that by demonstrating that these sums satisfy the Maxwell's equations with the same charge and current densities.
$$\nabla \cdot (E + E_{\gamma}) = \nabla \cdot E + \nabla \cdot E_{\gamma} = \frac{\rho}{\epsilon_0} + 0 = \frac{\rho}{\epsilon_0}$$
$$\nabla \times (E + E_{\gamma}) = \nabla \times E + \nabla \times E_{\gamma} = -\frac{\partial B}{\partial t} -\frac{\partial B_{\gamma}}{\partial t} = -\frac{\partial(B + B_{\gamma})}{\partial t}$$
$$\nabla \cdot (B + B_{\gamma}) = \nabla \cdot B + \nabla \cdot B_{\gamma} = 0 + 0 = 0$$
$$\nabla \times (B + B_{\gamma}) = \nabla \times B + \nabla \times B_{\gamma} = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t} + \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t} = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial (E + E_{\gamma})}{\partial t}$$
The equations are satisfied. Therefore, the electromagnetic field of a beam of light simply adds on top of electromagnetic fields already present without any kind of distortion. In other words, photons are not affected by electric or magnetic fields.
Are we finally done with your "charged photon" nonsense?
It says almost nothing about it, just rises many questions. What am I to conclude, left and right spin corresponds to 1 and -1? Ok, so why then Wikipedia says spin of a photon is 1, why does it not say 1 or -1? And what would be the spin of non-spinning photon? Also, how does that compare to electron spin since it seems photon can be spinning at arbitrary speed?
That's only because you have no idea what spin is. I strongly suggest reading an article on that. Pay attention to quantization axes and connection to angular momentum.
P: 68
Quote by Nugatory "Electricity and Magnetism" by Edward Purcell would be a good start. Google will find you a copy somewhere for cheap.
Do you answer all the question like that? It's not very helpful. I can not justify spending all that time or any money to just get definition of two terms which I already believe to know what they mean. Do you have that book, why not just write it down for everyone? Does it not say field is a description of geometrical distribution of charge, where charge is just a magnitude at some point of that field?
Sci Advisor P: 2,470 You aren't just confused on a couple of definitions. You have no clue what you are talkinga bout at all. You need to read a textbook. That was probably the most helpful suggestion you got.
PF Gold
P: 3,225
Barry, you seem to miss the point that K^2 tried so hard to show you.
There can be an electric field without any charge. An EM wave is just this.
For example the light coming from stars contain no charge at all, yet it is made of an electromagnetic field. All the light is like this.
You can take a region in space with an electric field and find no charge.
Quote by Barry_G Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field.
I'm not really understanding this.
Quote by Barry_G The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.
No, absolutely not as explained above.
P: 261
Quote by Barry_G Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field. The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.
After reading through this and the thread you hijacked in the Relativity section, I must be the fifth or sixth person to tell you what I'm about to tell you:
There does not need to be any charges for there to be an electric field. Maxwell's Equations tell us that a changing magnetic field will also produce an electric field. Similarly, you don't need any currents for there to be a magnetic field; a changing electric field will produce a magnetic field. Essentially what happens in an EM wave is that a changing electric field produces a changing magnetic field which produces a changing electric field which... This allows the electric and magnetic fields to propagate without any charges or currents. I'll even go through how this is derived, step by step:
As you've been shown numerous times, Maxwell's Equations in a vacuum are:
$$\nabla \cdot E=0$$
$$\nabla \times E = -\frac{\partial B}{\partial t}$$
$$\nabla \cdot B=0$$
$$\nabla \times B = \frac{1}{c^2} \frac{\partial E}{\partial t}$$
Now, using the following "curl of the curl" identity: $\nabla \times \nabla \times A=\nabla(\nabla\cdot A)-\nabla^2A$
$$\nabla \times \nabla \times E=\nabla(0)-\nabla^2E=-\nabla \times \frac{\partial B}{\partial t}=-\frac{\partial }{\partial t}[\nabla \times B]=-\frac{\partial }{\partial t}[\frac{1}{c^2} \frac{\partial E}{\partial t}]$$
Simplifying you get:
$$\nabla^2E=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}$$
You can apply the same identity to the curl of the magnetic field to get the following:
$$\nabla^2B=\frac{1}{c^2}\frac{\partial^2 B}{\partial t^2}$$
The solution to both of these differential equations is a sinusoidal wave which moves at a velocity c.
You accuse K^2 of being condescending, yet he's absolutely correct. You have no idea how basic electrodynamics works. What you need to do is stop pretending you know what you're talking about and go pick up a textbook. You're arguing about things you don't understand.
P: 68
Quote by K^2 The equations are satisfied. Therefore, the electromagnetic field of a beam of light simply adds on top of electromagnetic fields already present without any kind of distortion. In other words, photons are not affected by electric or magnetic fields. Are we finally done with your "charged photon" nonsense?
That's ridiculous. There are no any photons, nor external fields in those equations except for Faraday's law. Gauss's law is about electric flux through closed surface, it's analogous to Coulomb's law. Gauss's law for magnetism describes geometry of magnetic field. Faraday's law of induction is about generating voltage by external magnetic field in closed circuit loops. Ampère's circuital law is about magnetic field generated around close circuit loop.
That's only because you have no idea what spin is. I strongly suggest reading an article on that. Pay attention to quantization axes and connection to angular momentum.
Pay attention to what I said and if you have any idea what spin is you might understand.
You aren't just confused on a couple of definitions. You have no clue what you are talkinga bout at all. You need to read a textbook. That was probably the most helpful suggestion you got.
You aren't just confused about actual application of Maxwell's equations, but you also have no clue what you are talking about at all. You need to wake up.
P: 261
Quote by Barry_G That's ridiculous. There are no any photons, nor external fields in those equations except for Faraday's law. Gauss's law is about electric flux through closed surface, it's analogous to Coulomb's law. Gauss's law for magnetism describes geometry of magnetic field. Faraday's law of induction is about generating voltage by external magnetic field in closed circuit loops. Ampère's circuital law is about magnetic field generated around close circuit loop.
Your responses are quite painful to read. The "photon" in question consists of an electric and a magnetic field, which he denoted as $E_{\gamma}$ and $B_{\gamma}$. Maxwell's Equations aren't just for the narrow range of applications you list above. They completely describe how electric and magnetic fields behave!
Quote by Barry_G You aren't just confused about actual application of Maxwell's equations, but you also have no clue what you are talking about at all. You need to wake up.
I really don't see how you could possibly be under the impression that you're the one here with the most understanding of E&M. You understand practically nothing. In fact, I'm starting to suspect this entire thread is just a giant troll. I don't see how you could realistically be this delusional.
Mentor
P: 17,543
Quote by Barry_G Does electric field not imply electric charge?
No, electric field does not imply electric charge. See Maxwell's vacuum equations.
Quote by Barry_G All I am asking is if photon is electromagnetic field why can not be influenced by an external electric or magnetic field. What equation explains that?
Maxwell's equations. The key property of Maxwell's equations that lead to this is the linearity. The linearity of Maxwell's equations shows that EM follows the principle of superposition which in turn implies that an EM field won't be altered by passing through an external static electric or magnetic field.
http://en.wikipedia.org/wiki/Superposition_principle
Related Discussions Quantum Physics 11 Quantum Physics 2 Quantum Physics 13 Special & General Relativity 31 Quantum Physics 5 | 4,158 | 17,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-41 | latest | en | 0.9569 |
https://stats.stackexchange.com/questions/120802/glmer-conflicting-results-from-p-values-and-bootstrapped-confidence-intervals | 1,571,562,939,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00312.warc.gz | 727,529,068 | 33,163 | # GLMER: Conflicting results from p values and bootstrapped confidence intervals
Please don’t be shocked by the amount of code in this post, this problem is (hopefully) not as complex as it seems. This dataset (https://www.dropbox.com/sh/dfn6n9ipqocd1mh/AAD74pdCRw8RXVBjpwAl_Ajja?dl=0) contains 40 participants of the study I ran and I put it up, along with the code, to help people recreating my problem.
> head(dataset) #Ignore the column X
X Subject Dilemma Inevitable Personal SelfBeneficial Resp2
1 721 21 Bike Week 1 0 1 0
2 722 21 Shark Attack 1 0 0 0
3 723 21 Ebola 1 1 0 1
4 725 21 Nuke 0 1 0 1
5 727 21 Rowboat 0 1 1 1
6 729 21 Cinderblock 0 0 0 0
Description of study:
In my study I was interested in how different contextual factors (Inevitable, Personal & SelfBeneficial) in a moral dilemma influence the judgement of whether killing someone to save multiple others is morally acceptable (Resp2 = 1) or unacceptable (Resp2 = 0). Every participant had to answer to 24 dilemmas of which every dilemma had a personal (Personal = 0) and an impersonal (Personal = 1) variant, that means one sentence in the dilemma was different within the dilemma. This factor was varied between subjects, so when participant A saw the impersonal variant of the “Bike Week” dilemma then participant B saw the personal variant of the “Bike Week” dilemma. The Inevitable and SelfBeneficial factor were fixed to the dilemma; that means for example “Shark Week” was, for all participants, inevitable (Inevitable = 1) and not self beneficial (SelfBeneficial = 0). So you could say the factor Personal was varied within-dilemma whereas the factors Inevitable and SelfBeneficial were varied between-dilemmas. All factors were fully crossed with each other.
Analytical approach:
As I wanted to calculate the probabilities of stating that killing was morally acceptable I conducted a multilevel logistic regression, in other words a generalised mixed model with a logit function. For this I used the glmer function of the R package lme4 (version 1.1-7). Following recommendations by Barr et al. (2013) I wanted to include the maximal random effect structure to reduce the probability of Type I errors. The maximal random effect structure here would be a by-subject random intercept as well as by-subject random slopes for Inevitable, Personal and SelfBeneficial. Furthermore a by-dilemma random intercept and a by-dilemma random slope for Personal. We will omit the by-dilemma random slope for Personal for now though as it does not influence the problem.
The problem:
When running the model in glmer I get the following output:
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation)['glmerMod']
Family: binomial ( logit )
Formula: Resp2 ~ SelfBeneficial + Personal + Inevitable + (1 + SelfBeneficial + Personal + Inevitable | Subject) + (1 | Dilemma)
Data: dataset
Control: glmerControl(optCtrl = list(maxfun = 5e+05), optimizer = "bobyqa”)
> Random effects:
Groups Name Variance Std.Dev. Corr
Subject (Intercept) 2.79199 1.6709
SelfBeneficial 0.73805 0.8591 -0.50
Personal 0.07412 0.2723 0.43 0.53
Inevitable 0.43871 0.6624 -0.01 0.10 0.37
Dilemma (Intercept) 2.60176 1.6130
Number of obs: 960, groups: Subject, 40; Dilemma, 24
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.5726 0.6526 0.878 0.380
SelfBeneficial -1.0209 0.6993 -1.460 0.144
Personal 0.7037 0.1806 3.895 9.8e-05 ***
Inevitable -0.6884 0.6924 -0.994 0.320
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
As you can see the factors SelfBeneficial and Inevitable are not significant, however they should be as their significance has been shown in previous studies (however using anova instead of glmms). Most importantly: When I calculate bootstrapped confidence intervals (cutting of the 2.5 and 97.5 percentile with resampling only on the subject level) all of the three factors are significant as expected.
Result from bootstrapping (100 iterations):
> [1] "(Intercept)"
2.5% 97.5%
0.17 1.10
[1] "SelfBeneficial"
2.5% 97.5%
-1.58 -0.76
[1] "Personal"
2.5% 97.5%
0.46 1.07
[1] "Inevitable"
2.5% 97.5%
-1.13 -0.44
The questions:
Why are the results as indicated by the p-values of the glmer output and the bootstrapped confidence intervals so fundamentally different?
Am I over-fitting the model by adding too many random effects? A model without the by-dilemma random intercept yields the three significant factors but I want to include the maximal random effect structure since this has an effect on variables on the subject-level (not included in this example).
Do you spot any fundamental mistake I did in my experimental design or the bootstrapping?
Note that this is not a problem of too little bootstrap iterations (I tried a lot more), bias introduced via bootstrapping (bias is at between .1 or .2 in bootstrap samples with more iterations) or the omission of the by-dilemma random slope for Personal.
The code:
Here I added the code I used so you can reproduce/recreate my problem. Sorry that this is probably not the most elegant way of doing this.
dataset <- read.csv("datasetproblem.csv")
library(lme4)
#The model
mm <- glmer(Resp2 ~ SelfBeneficial+Personal+Inevitable + (1 + SelfBeneficial+Personal+Inevitable | Subject) + (1|Dilemma), family = binomial("logit"), data= dataset, control=glmerControl(optCtrl=list(maxfun=500000), optimizer = "bobyqa"))
summary(mm)
#Function needed in the bootstrap-function to exclude models that did not converge
check <- function(model){
em <- 0
tryCatch(model, warning=function(err) {
mes <- substr(err$message,1,19) em <<- ifelse(mes == "failure to converge","not converged","other warning") }) if(em == 0){return(model)} if(em != 0){return(0)} } sa <- unique(dataset$Subject)
#Actual bootstrapping function
bootcon <- function(Model,n.sim=500,n.iter=50000,verbose=F){
frameall <- data.frame()
for(j in 1:n.sim){
sam1 <- sample(sa,40,replace=T)
bdat <- data.frame()
for(i in sam1){
ret <- dataset[dataset$Subject == i,] bdat <- rbind(bdat,ret) } wi <-as.vector(as.character(summary(Model)$call)[2])
modl <- check(glmer(wi,family = binomial("logit"), data= bdat, control=glmerControl(optCtrl=list(maxfun=n.iter), optimizer = "bobyqa"), verbose=verbose))
if(class(modl)[1] == "glmerMod"){
q <- as.data.frame(summary(modl)$coefficients)[1:2] coefficient <- rownames(q) rownames(q) <- NULL Number <- rep(j,nrow(q)) frapart <- cbind(coefficient,q,Number) frameall <- rbind(frameall,frapart) print(paste("Iteration no. ",j,". System time: ",Sys.time()))} } return(frameall) } #Running the bootstrapping function output <- bootcon(mm,100) #Analyse bootstraps for(i in unique(output$coefficient)){
print(i)
#print(round(mean(output$Estimate[output$coefficient == i]),2 ))
print(round(quantile(output$Estimate[output$coefficient == i],probs = c(0.025,0.975)),2))
} | 2,002 | 7,203 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-43 | latest | en | 0.929704 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/121/2/c/e/ | 1,653,275,324,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662552994.41/warc/CC-MAIN-20220523011006-20220523041006-00647.warc.gz | 1,000,068,568 | 58,272 | # Properties
Label 121.2.c.e Level $121$ Weight $2$ Character orbit 121.c Analytic conductor $0.966$ Analytic rank $0$ Dimension $4$ CM no Inner twists $4$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$121 = 11^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 121.c (of order $$5$$, degree $$4$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.966189864457$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{10})$$ Defining polynomial: $$x^{4} - x^{3} + x^{2} - x + 1$$ x^4 - x^3 + x^2 - x + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 11) Sato-Tate group: $\mathrm{SU}(2)[C_{5}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{10}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - 2 \zeta_{10}^{3} + 2 \zeta_{10}^{2} - 2 \zeta_{10} + 2) q^{2} - \zeta_{10}^{2} q^{3} - 2 \zeta_{10}^{3} q^{4} - \zeta_{10} q^{5} - 2 \zeta_{10} q^{6} + 2 \zeta_{10}^{3} q^{7} + ( - 2 \zeta_{10}^{3} + 2 \zeta_{10}^{2} - 2 \zeta_{10} + 2) q^{9} +O(q^{10})$$ q + (-2*z^3 + 2*z^2 - 2*z + 2) * q^2 - z^2 * q^3 - 2*z^3 * q^4 - z * q^5 - 2*z * q^6 + 2*z^3 * q^7 + (-2*z^3 + 2*z^2 - 2*z + 2) * q^9 $$q + ( - 2 \zeta_{10}^{3} + 2 \zeta_{10}^{2} - 2 \zeta_{10} + 2) q^{2} - \zeta_{10}^{2} q^{3} - 2 \zeta_{10}^{3} q^{4} - \zeta_{10} q^{5} - 2 \zeta_{10} q^{6} + 2 \zeta_{10}^{3} q^{7} + ( - 2 \zeta_{10}^{3} + 2 \zeta_{10}^{2} - 2 \zeta_{10} + 2) q^{9} - 2 q^{10} - 2 q^{12} + (4 \zeta_{10}^{3} - 4 \zeta_{10}^{2} + 4 \zeta_{10} - 4) q^{13} + 4 \zeta_{10}^{2} q^{14} + \zeta_{10}^{3} q^{15} + 4 \zeta_{10} q^{16} + 2 \zeta_{10} q^{17} - 4 \zeta_{10}^{3} q^{18} + (2 \zeta_{10}^{3} - 2 \zeta_{10}^{2} + 2 \zeta_{10} - 2) q^{20} + 2 q^{21} - q^{23} - 4 \zeta_{10}^{2} q^{25} + 8 \zeta_{10}^{3} q^{26} - 5 \zeta_{10} q^{27} + 4 \zeta_{10} q^{28} + 2 \zeta_{10}^{2} q^{30} + (7 \zeta_{10}^{3} - 7 \zeta_{10}^{2} + 7 \zeta_{10} - 7) q^{31} + 8 q^{32} + 4 q^{34} + ( - 2 \zeta_{10}^{3} + 2 \zeta_{10}^{2} - 2 \zeta_{10} + 2) q^{35} - 4 \zeta_{10}^{2} q^{36} - 3 \zeta_{10}^{3} q^{37} + 4 \zeta_{10} q^{39} - 8 \zeta_{10}^{2} q^{41} + ( - 4 \zeta_{10}^{3} + 4 \zeta_{10}^{2} - 4 \zeta_{10} + 4) q^{42} - 6 q^{43} - 2 q^{45} + (2 \zeta_{10}^{3} - 2 \zeta_{10}^{2} + 2 \zeta_{10} - 2) q^{46} + 8 \zeta_{10}^{2} q^{47} - 4 \zeta_{10}^{3} q^{48} + 3 \zeta_{10} q^{49} - 8 \zeta_{10} q^{50} - 2 \zeta_{10}^{3} q^{51} + 8 \zeta_{10}^{2} q^{52} + ( - 6 \zeta_{10}^{3} + 6 \zeta_{10}^{2} - 6 \zeta_{10} + 6) q^{53} - 10 q^{54} - 5 \zeta_{10}^{3} q^{59} + 2 \zeta_{10} q^{60} - 12 \zeta_{10} q^{61} + 14 \zeta_{10}^{3} q^{62} + 4 \zeta_{10}^{2} q^{63} + ( - 8 \zeta_{10}^{3} + 8 \zeta_{10}^{2} - 8 \zeta_{10} + 8) q^{64} + 4 q^{65} - 7 q^{67} + ( - 4 \zeta_{10}^{3} + 4 \zeta_{10}^{2} - 4 \zeta_{10} + 4) q^{68} + \zeta_{10}^{2} q^{69} - 4 \zeta_{10}^{3} q^{70} + 3 \zeta_{10} q^{71} - 4 \zeta_{10}^{3} q^{73} - 6 \zeta_{10}^{2} q^{74} + (4 \zeta_{10}^{3} - 4 \zeta_{10}^{2} + 4 \zeta_{10} - 4) q^{75} + 8 q^{78} + ( - 10 \zeta_{10}^{3} + 10 \zeta_{10}^{2} - 10 \zeta_{10} + 10) q^{79} - 4 \zeta_{10}^{2} q^{80} - \zeta_{10}^{3} q^{81} - 16 \zeta_{10} q^{82} + 6 \zeta_{10} q^{83} - 4 \zeta_{10}^{3} q^{84} - 2 \zeta_{10}^{2} q^{85} + (12 \zeta_{10}^{3} - 12 \zeta_{10}^{2} + 12 \zeta_{10} - 12) q^{86} + 15 q^{89} + (4 \zeta_{10}^{3} - 4 \zeta_{10}^{2} + 4 \zeta_{10} - 4) q^{90} - 8 \zeta_{10}^{2} q^{91} + 2 \zeta_{10}^{3} q^{92} + 7 \zeta_{10} q^{93} + 16 \zeta_{10} q^{94} - 8 \zeta_{10}^{2} q^{96} + ( - 7 \zeta_{10}^{3} + 7 \zeta_{10}^{2} - 7 \zeta_{10} + 7) q^{97} + 6 q^{98} +O(q^{100})$$ q + (-2*z^3 + 2*z^2 - 2*z + 2) * q^2 - z^2 * q^3 - 2*z^3 * q^4 - z * q^5 - 2*z * q^6 + 2*z^3 * q^7 + (-2*z^3 + 2*z^2 - 2*z + 2) * q^9 - 2 * q^10 - 2 * q^12 + (4*z^3 - 4*z^2 + 4*z - 4) * q^13 + 4*z^2 * q^14 + z^3 * q^15 + 4*z * q^16 + 2*z * q^17 - 4*z^3 * q^18 + (2*z^3 - 2*z^2 + 2*z - 2) * q^20 + 2 * q^21 - q^23 - 4*z^2 * q^25 + 8*z^3 * q^26 - 5*z * q^27 + 4*z * q^28 + 2*z^2 * q^30 + (7*z^3 - 7*z^2 + 7*z - 7) * q^31 + 8 * q^32 + 4 * q^34 + (-2*z^3 + 2*z^2 - 2*z + 2) * q^35 - 4*z^2 * q^36 - 3*z^3 * q^37 + 4*z * q^39 - 8*z^2 * q^41 + (-4*z^3 + 4*z^2 - 4*z + 4) * q^42 - 6 * q^43 - 2 * q^45 + (2*z^3 - 2*z^2 + 2*z - 2) * q^46 + 8*z^2 * q^47 - 4*z^3 * q^48 + 3*z * q^49 - 8*z * q^50 - 2*z^3 * q^51 + 8*z^2 * q^52 + (-6*z^3 + 6*z^2 - 6*z + 6) * q^53 - 10 * q^54 - 5*z^3 * q^59 + 2*z * q^60 - 12*z * q^61 + 14*z^3 * q^62 + 4*z^2 * q^63 + (-8*z^3 + 8*z^2 - 8*z + 8) * q^64 + 4 * q^65 - 7 * q^67 + (-4*z^3 + 4*z^2 - 4*z + 4) * q^68 + z^2 * q^69 - 4*z^3 * q^70 + 3*z * q^71 - 4*z^3 * q^73 - 6*z^2 * q^74 + (4*z^3 - 4*z^2 + 4*z - 4) * q^75 + 8 * q^78 + (-10*z^3 + 10*z^2 - 10*z + 10) * q^79 - 4*z^2 * q^80 - z^3 * q^81 - 16*z * q^82 + 6*z * q^83 - 4*z^3 * q^84 - 2*z^2 * q^85 + (12*z^3 - 12*z^2 + 12*z - 12) * q^86 + 15 * q^89 + (4*z^3 - 4*z^2 + 4*z - 4) * q^90 - 8*z^2 * q^91 + 2*z^3 * q^92 + 7*z * q^93 + 16*z * q^94 - 8*z^2 * q^96 + (-7*z^3 + 7*z^2 - 7*z + 7) * q^97 + 6 * q^98 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q + 2 q^{2} + q^{3} - 2 q^{4} - q^{5} - 2 q^{6} + 2 q^{7} + 2 q^{9}+O(q^{10})$$ 4 * q + 2 * q^2 + q^3 - 2 * q^4 - q^5 - 2 * q^6 + 2 * q^7 + 2 * q^9 $$4 q + 2 q^{2} + q^{3} - 2 q^{4} - q^{5} - 2 q^{6} + 2 q^{7} + 2 q^{9} - 8 q^{10} - 8 q^{12} - 4 q^{13} - 4 q^{14} + q^{15} + 4 q^{16} + 2 q^{17} - 4 q^{18} - 2 q^{20} + 8 q^{21} - 4 q^{23} + 4 q^{25} + 8 q^{26} - 5 q^{27} + 4 q^{28} - 2 q^{30} - 7 q^{31} + 32 q^{32} + 16 q^{34} + 2 q^{35} + 4 q^{36} - 3 q^{37} + 4 q^{39} + 8 q^{41} + 4 q^{42} - 24 q^{43} - 8 q^{45} - 2 q^{46} - 8 q^{47} - 4 q^{48} + 3 q^{49} - 8 q^{50} - 2 q^{51} - 8 q^{52} + 6 q^{53} - 40 q^{54} - 5 q^{59} + 2 q^{60} - 12 q^{61} + 14 q^{62} - 4 q^{63} + 8 q^{64} + 16 q^{65} - 28 q^{67} + 4 q^{68} - q^{69} - 4 q^{70} + 3 q^{71} - 4 q^{73} + 6 q^{74} - 4 q^{75} + 32 q^{78} + 10 q^{79} + 4 q^{80} - q^{81} - 16 q^{82} + 6 q^{83} - 4 q^{84} + 2 q^{85} - 12 q^{86} + 60 q^{89} - 4 q^{90} + 8 q^{91} + 2 q^{92} + 7 q^{93} + 16 q^{94} + 8 q^{96} + 7 q^{97} + 24 q^{98}+O(q^{100})$$ 4 * q + 2 * q^2 + q^3 - 2 * q^4 - q^5 - 2 * q^6 + 2 * q^7 + 2 * q^9 - 8 * q^10 - 8 * q^12 - 4 * q^13 - 4 * q^14 + q^15 + 4 * q^16 + 2 * q^17 - 4 * q^18 - 2 * q^20 + 8 * q^21 - 4 * q^23 + 4 * q^25 + 8 * q^26 - 5 * q^27 + 4 * q^28 - 2 * q^30 - 7 * q^31 + 32 * q^32 + 16 * q^34 + 2 * q^35 + 4 * q^36 - 3 * q^37 + 4 * q^39 + 8 * q^41 + 4 * q^42 - 24 * q^43 - 8 * q^45 - 2 * q^46 - 8 * q^47 - 4 * q^48 + 3 * q^49 - 8 * q^50 - 2 * q^51 - 8 * q^52 + 6 * q^53 - 40 * q^54 - 5 * q^59 + 2 * q^60 - 12 * q^61 + 14 * q^62 - 4 * q^63 + 8 * q^64 + 16 * q^65 - 28 * q^67 + 4 * q^68 - q^69 - 4 * q^70 + 3 * q^71 - 4 * q^73 + 6 * q^74 - 4 * q^75 + 32 * q^78 + 10 * q^79 + 4 * q^80 - q^81 - 16 * q^82 + 6 * q^83 - 4 * q^84 + 2 * q^85 - 12 * q^86 + 60 * q^89 - 4 * q^90 + 8 * q^91 + 2 * q^92 + 7 * q^93 + 16 * q^94 + 8 * q^96 + 7 * q^97 + 24 * q^98
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/121\mathbb{Z}\right)^\times$$.
$$n$$ $$2$$ $$\chi(n)$$ $$-\zeta_{10}^{3}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
3.1
0.809017 + 0.587785i −0.309017 − 0.951057i −0.309017 + 0.951057i 0.809017 − 0.587785i
1.61803 1.17557i −0.309017 0.951057i 0.618034 1.90211i −0.809017 0.587785i −1.61803 1.17557i −0.618034 + 1.90211i 0 1.61803 1.17557i −2.00000
9.1 −0.618034 + 1.90211i 0.809017 0.587785i −1.61803 1.17557i 0.309017 + 0.951057i 0.618034 + 1.90211i 1.61803 + 1.17557i 0 −0.618034 + 1.90211i −2.00000
27.1 −0.618034 1.90211i 0.809017 + 0.587785i −1.61803 + 1.17557i 0.309017 0.951057i 0.618034 1.90211i 1.61803 1.17557i 0 −0.618034 1.90211i −2.00000
81.1 1.61803 + 1.17557i −0.309017 + 0.951057i 0.618034 + 1.90211i −0.809017 + 0.587785i −1.61803 + 1.17557i −0.618034 1.90211i 0 1.61803 + 1.17557i −2.00000
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
11.c even 5 3 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 121.2.c.e 4
11.b odd 2 1 121.2.c.a 4
11.c even 5 1 11.2.a.a 1
11.c even 5 3 inner 121.2.c.e 4
11.d odd 10 1 121.2.a.d 1
11.d odd 10 3 121.2.c.a 4
33.f even 10 1 1089.2.a.b 1
33.h odd 10 1 99.2.a.d 1
44.g even 10 1 1936.2.a.i 1
44.h odd 10 1 176.2.a.b 1
55.h odd 10 1 3025.2.a.a 1
55.j even 10 1 275.2.a.b 1
55.k odd 20 2 275.2.b.a 2
77.j odd 10 1 539.2.a.a 1
77.l even 10 1 5929.2.a.h 1
77.m even 15 2 539.2.e.h 2
77.p odd 30 2 539.2.e.g 2
88.k even 10 1 7744.2.a.k 1
88.l odd 10 1 704.2.a.c 1
88.o even 10 1 704.2.a.h 1
88.p odd 10 1 7744.2.a.x 1
99.m even 15 2 891.2.e.k 2
99.n odd 30 2 891.2.e.b 2
132.o even 10 1 1584.2.a.g 1
143.n even 10 1 1859.2.a.b 1
165.o odd 10 1 2475.2.a.a 1
165.v even 20 2 2475.2.c.a 2
176.v odd 20 2 2816.2.c.f 2
176.w even 20 2 2816.2.c.j 2
187.j even 10 1 3179.2.a.a 1
209.m odd 10 1 3971.2.a.b 1
220.n odd 10 1 4400.2.a.i 1
220.v even 20 2 4400.2.b.h 2
231.u even 10 1 4851.2.a.t 1
253.f odd 10 1 5819.2.a.a 1
264.t odd 10 1 6336.2.a.br 1
264.w even 10 1 6336.2.a.bu 1
308.t even 10 1 8624.2.a.j 1
319.k even 10 1 9251.2.a.d 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
11.2.a.a 1 11.c even 5 1
99.2.a.d 1 33.h odd 10 1
121.2.a.d 1 11.d odd 10 1
121.2.c.a 4 11.b odd 2 1
121.2.c.a 4 11.d odd 10 3
121.2.c.e 4 1.a even 1 1 trivial
121.2.c.e 4 11.c even 5 3 inner
176.2.a.b 1 44.h odd 10 1
275.2.a.b 1 55.j even 10 1
275.2.b.a 2 55.k odd 20 2
539.2.a.a 1 77.j odd 10 1
539.2.e.g 2 77.p odd 30 2
539.2.e.h 2 77.m even 15 2
704.2.a.c 1 88.l odd 10 1
704.2.a.h 1 88.o even 10 1
891.2.e.b 2 99.n odd 30 2
891.2.e.k 2 99.m even 15 2
1089.2.a.b 1 33.f even 10 1
1584.2.a.g 1 132.o even 10 1
1859.2.a.b 1 143.n even 10 1
1936.2.a.i 1 44.g even 10 1
2475.2.a.a 1 165.o odd 10 1
2475.2.c.a 2 165.v even 20 2
2816.2.c.f 2 176.v odd 20 2
2816.2.c.j 2 176.w even 20 2
3025.2.a.a 1 55.h odd 10 1
3179.2.a.a 1 187.j even 10 1
3971.2.a.b 1 209.m odd 10 1
4400.2.a.i 1 220.n odd 10 1
4400.2.b.h 2 220.v even 20 2
4851.2.a.t 1 231.u even 10 1
5819.2.a.a 1 253.f odd 10 1
5929.2.a.h 1 77.l even 10 1
6336.2.a.br 1 264.t odd 10 1
6336.2.a.bu 1 264.w even 10 1
7744.2.a.k 1 88.k even 10 1
7744.2.a.x 1 88.p odd 10 1
8624.2.a.j 1 308.t even 10 1
9251.2.a.d 1 319.k even 10 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{4} - 2T_{2}^{3} + 4T_{2}^{2} - 8T_{2} + 16$$ acting on $$S_{2}^{\mathrm{new}}(121, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4} - 2 T^{3} + 4 T^{2} - 8 T + 16$$
$3$ $$T^{4} - T^{3} + T^{2} - T + 1$$
$5$ $$T^{4} + T^{3} + T^{2} + T + 1$$
$7$ $$T^{4} - 2 T^{3} + 4 T^{2} - 8 T + 16$$
$11$ $$T^{4}$$
$13$ $$T^{4} + 4 T^{3} + 16 T^{2} + 64 T + 256$$
$17$ $$T^{4} - 2 T^{3} + 4 T^{2} - 8 T + 16$$
$19$ $$T^{4}$$
$23$ $$(T + 1)^{4}$$
$29$ $$T^{4}$$
$31$ $$T^{4} + 7 T^{3} + 49 T^{2} + \cdots + 2401$$
$37$ $$T^{4} + 3 T^{3} + 9 T^{2} + 27 T + 81$$
$41$ $$T^{4} - 8 T^{3} + 64 T^{2} + \cdots + 4096$$
$43$ $$(T + 6)^{4}$$
$47$ $$T^{4} + 8 T^{3} + 64 T^{2} + \cdots + 4096$$
$53$ $$T^{4} - 6 T^{3} + 36 T^{2} + \cdots + 1296$$
$59$ $$T^{4} + 5 T^{3} + 25 T^{2} + 125 T + 625$$
$61$ $$T^{4} + 12 T^{3} + 144 T^{2} + \cdots + 20736$$
$67$ $$(T + 7)^{4}$$
$71$ $$T^{4} - 3 T^{3} + 9 T^{2} - 27 T + 81$$
$73$ $$T^{4} + 4 T^{3} + 16 T^{2} + 64 T + 256$$
$79$ $$T^{4} - 10 T^{3} + 100 T^{2} + \cdots + 10000$$
$83$ $$T^{4} - 6 T^{3} + 36 T^{2} + \cdots + 1296$$
$89$ $$(T - 15)^{4}$$
$97$ $$T^{4} - 7 T^{3} + 49 T^{2} + \cdots + 2401$$ | 6,831 | 11,921 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-21 | latest | en | 0.387729 |
https://math.stackexchange.com/questions/1472887/existence-of-a-path-of-length-k-if-degu-degv-ge-k-for-every-pair-of-di | 1,726,431,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00711.warc.gz | 371,137,563 | 35,349 | # Existence of a path of length $k$ if $\deg(u)+\deg(v)\ge k$ for every pair of distinct vertices $u$ and $v$
The problem I am struggling to prove is the following:
Let $G$ be a connected simple graph with $n$ vertices and let $k<n$ be a positive integer. If $\deg(u)+\deg(v)\ge k$ for every pair of distinct vertices $u$ and $v$, then $G$ has a vertex disjoint path of length at least $k$, where, for a vertex $x$, deg$(x)$ stands for the degree of $x$.
The condition $\deg(u)+\deg(v)\ge k$, somehow, looks like the Ore condition in the discussion of Hamiltonian graphs.
Can someone show me the way to get into (or a counter example)? | 178 | 639 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.837319 |
https://forum.freecodecamp.org/t/only-one-condition-is-not-met/334415 | 1,675,714,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00199.warc.gz | 272,412,188 | 8,544 | # Only one condition is not met
so all conditions to pass this are met except one, and I cant tell why at the moment. If the drawer has only 50 cents worth or pennies (1 cent) and nothing else, the while statements work perfectly by continuing to subtract from the 50 cent change until its zero and the return array says [[“pennies”, .50]], but if the drawer has more than just pennies and the customer needs change that is all kinds of denominations, no subtractions and additions take place according to the 9 while loops that start at the top of the page with the largest denomination (100) and move down checking to see if the change is greater than the next largest denomination. I will put the code in here if it will let me.
CODE IS BELOW.
ABOVE IS EXPLANATION OF PROBLEM
THANKS
``````function checkCashRegister(price, cash, cid) {
price = price * 100;
cash = cash * 100;
for (let u = 0; u < cid.length; u++) {
cid[u][1] = cid[u][1] * 100;
};
// multiplying the price, cash provided and cid by 100 to avoid non-decimal calculations //
var cH = cash - price;
console.log(cH)
var cHA = [["PENNY", 0], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]];
var ind = [1, 5, 10, 25, 100, 500, 1000, 2000, 10000];
// line above has all denominations multiplied by 100 as to avoid subtraction and addition of decimal point numbers. The variable ind also acts as a reference, for example, if we have a few one hundreds in the drawer and the change required is above 100, we can say, subtract 10000 from the drawer (remembering that everything in the drawer was multiplied by 100 (or moved two decimal places forward), add 10000 to our array which has all nine denominations initailly at zero, and take 10000 from change (cH), then it can move forward by again asking if change is still above 10000 or not, if it isnt, the next 'while' loop will be checking to see if the change is now above the 20 dollar note and so on //
var drawTotal = 0;
//starting the while loops at the highest denomination. If the while loop is true, the empty array with all 9 denominations started at 0 will be added to accordingly, cid will be subtracted from accordingly and change will be subtracted from accordingly //
while (ind[8] <= cH && cid[8][1] >= cH) {
cHA[8][1] += ind[8];
cid[8][1] -= ind[8];
cH -= ind[8];
};
while (ind[7] <= cH && cid[7][1] >= cH) {
cHA[7][1] += ind[7];
cid[7][1] -= ind[7];
cH -= ind[7];
};
while (ind[6] <= cH && cid[6][1] >= cH) {
cHA[6][1] += ind[6];
cid[6][1] -= ind[6];
cH -= ind[6];
};
while (ind[5] <= cH && cid[5][1] >= cH) {
cHA[5][1] += ind[5];
cid[5][1] -= ind[5];
cH -= ind[5];
};
while (ind[4] <= cH && cid[4][1] >= cH) {
cHA[4][1] += ind[4];
cid[4][1] -= ind[4];
cH -= ind[4];
};
while (ind[3] <= cH && cid[3][1] >= cH) {
cHA[3][1] += ind[3];
cid[3][1] -= ind[3];
cH -= ind[3];
};
while (ind[2] <= cH && cid[2][1] >= cH) {
cHA[2][1] += ind[2];
cid[2][1] -= ind[2];
cH -= ind[2];
};
while (ind[1] <= cH && cid[1][1] >= cH) {
cHA[1][1] += ind[1];
cid[1][1] -= ind[1];
cH -= ind[1];
};
while (ind[0] <= cH && cid[0][1] >= cH) {
cHA[0][1] += ind[0];
cid[0][1] -= ind[0];
cH -= ind[0];
};
// ok, so the array (cHA) which initially holds all 9 denominations at value 0, should be either added to or not. The cid should show the 9 denominations with the values of some of them subtracted from or not, and the change (cH) should return either 0 or above, depending on whether the cid provided had enough total currency but lacked the currency to provide exact change, whether the cid provided had enough total currency and was able to provide exact change, or whether the cid provided didnt have enough total currency or the currency to provide exact change.//
console.log(cH)
for (let r = 0; r < cid.length; r++) {
drawTotal += cid[r][1];
};
// the above 'for loop' totals the money left in the till/cid/drawer so that if it equals exactly 0, the program can return "CLOSED" and the change needing to be given, or if it doesnt equal 0, the program can return "OPEN" and the change needing to be provided //
cH = cH / 100;
// line above brings the change back two decimal places as the change was originally amplfied by multiplication of 100 to avoid addition and subtraction of decimal point values //
for (let p = 0; p < cid.length; p++) {
cid[p][1] = cid[p][1] / 100;
cHA[p][1] = cHA[p][1] / 100;
}
// line above brings cid and cHA "back down to earth", likely into decimal point notation //
var rArr = [];
for (let e = 0; e < cHA.length; e++) {
if (cHA[e][1] !== 0) {
rArr.push(cHA[e]);
};
};
// for loop above pushes only non zero values of the complete 9 denomination change array into an array that starts completely empty (rArr) which was declared just above //
var retObj = {status: null, change: []};
if (cH !== 0.00) {
retObj.status = "INSUFFICIENT_FUNDS";
retObj.change = rArr;
return retObj;
} else if (cH === 0 && drawTotal === 0) {
retObj.status = "CLOSED";
retObj.change = cHA;
return retObj;
} else if (cH === 0 && drawTotal !== 0) {
retObj.status = "OPEN";
retObj.change = rArr;
return retObj;
};
};
checkCashRegister(3.26, 100, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]]);```
``````
WARNING
The challenge seed code and/or your solution exceeded the maximum length we can port over from the challenge.
You will need to take an additional step here so the code you wrote presents in an easy to read format.
Please copy/paste all the editor code showing in the challenge from where you just linked.
``````
Replace these two sentences with your copied code.
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``````
User Agent is: `Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/66.0.3359.181 Safari/537.36`.
Challenge: Cash Register
I’m not quite sure where the error is, but it definitely looks like something to do with the while loops. The cH variable is not returning 0. I would suggest seeing if you can condense those 9 while loops into one while loop. You could make a copy of the cH variable into a remainder variable that you could keep subtracting the denomination values from as long as the while loop returns true. That might be a start, but it looks like you are pretty close. I will try looking at it more tomorrow.
Hope this helps!
Thanks for the reply. It’s funny because when I put in 50 (price) and 100 (cash), it works well and gives two 20s and a ten. But when I put 30 (price) and 100 (cash), it gives three 20s and ten 1s. I changed the 9 while loops into an IF followed by 8 ELSE IFs, and then put that into a FOR loop that runs 200 times just to see if it was an order problem, but no, same response. I don’t see any way of condensing into one while loop. Anyway, yes, that would be helpful.
Works now. I made a FOR loop that goes through variable ‘ind’, starting at highest denomination, and then a ‘while’ loop inside of it that will keep running if change (cH) is greater than or equal to the current ind array index AND cid’s quantity of that denomination is greater than or equal to the current change value (cH). If you look at my old code, you can see that the if statement is the same but I put that into a while statement, and that while statement is inside a for loop). Also, everytime that statement ran true, after all subtractions and additions tok place, the while loop then rounded everything as some calculations caused rounding errors. Ill put the part of the code I am talking about if it saved.
``````for (let c = 8; c >= 0; c--) {
while (ind[c] <= cH && cid[c][1] >= ind[c]) {
cHA[c][1] += ind[c];
cid[c][1] -= ind[c];
cH -= ind[c];
cHA[c][1] = Math.round(cHA[c][1] * 1e2) / 1e2;
cid[c][1] = Math.round(cid[c][1] * 1e2) / 1e2;
cH = Math.round(cH * 1e2) / 1e2;
}
};
``````
1 Like | 2,392 | 7,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | latest | en | 0.870231 |
safepay.helpscoutdocs.com | 1,601,512,389,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00692.warc.gz | 529,276,940 | 16,074 | # Can I pass my transaction charges to my customers?
At the moment, Safepay does not allow merchants to automatically pass transaction charges to customers. Since we do not offer this feature, this is something you (the merchant) will have to achieve on your end.
There are two ways to achieve this:
• You can mark-up your prices, so you get close to the amount you want to receive.
So instead of charging Rs5,000 you can instead charge Rs5,200, which would come down to Rs4,998.40 at the standard 3.3% + Rs30 transaction fee.
• Alternatively, you can design your platform in such a way that a charge is applied right before the customer has to pay. This can be achieved using a simple formula:
For transactions of Rs 5000 for example:
({Payment Amount} + Rs30) / (1 – 0.033) = Final Amount
The "Final Amount" is the amount you should pass to your customers. Using the formula above, for a transaction that costs Rs5,000, your customer will pay Rs5,201.65. This will come down to exactly Rs5000 once we deduct transaction charge at 3.3% + Rs30. | 266 | 1,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-40 | longest | en | 0.873544 |
http://gdevtest.geeksforgeeks.org/tag/xor/ | 1,550,591,029,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247490225.49/warc/CC-MAIN-20190219142524-20190219164524-00298.warc.gz | 103,586,745 | 18,218 | ## Sum of XOR of all subarrays
Given an array containing N positive integers, the task is to find the sum of XOR of all sub-arrays of the array. Examples: Input :… Read More »
## Check whether XOR of all numbers in a given range is even or odd
Given a range [ L, R ], the task is to find if value of XOR of all natural numbers in range L to R… Read More »
## Composite XOR and Coprime AND
Given an array arr[], the task is to count the number of unordered pairs of indices (i, j) such the gcd(2, a[i]^a[j]) > 1 and… Read More »
## Find array using different XORs of elements in groups of size 4
Given an array q[] of XOR queries of size N (N is a multiple of 4) which describe an array of the same size as… Read More »
Prerequisite: XOR Linked List – A Memory Efficient Doubly Linked List | Set 1 XOR Linked List – A Memory Efficient Doubly Linked List |… Read More »
## Count of values of x <= n for which (n XOR x) = (n – x)
Given an integer n, the task is to find the number of possible values of 0 ≤ x ≤ n which satisfy n XOR x… Read More »
## Count no. of ordered subsets having a particular XOR value
Given an array arr[] of n elements and a number K, find the number of ordered subsets of arr[] having XOR of elements as K… Read More »
## Pairs from an array that satisfy the given condition
Given an array arr[], the task is to count all the valid pairs from the array. A pair (arr[i], arr[j]) is said to be valid… Read More »
## Find elements of array using XOR of consecutive elements
Given an array arr[] in which XOR of every 2 consecutive elements of the original array is given i.e if the total number of elements… Read More »
## All pairs whose xor gives unique prime
Given an array arr[], the task is to count all the pairs whose xor gives the unique prime, i.e. no two pairs should give the… Read More »
## Find position of left most dis-similar bit for two numbers
Given two numbers n1 and n2. The task is to find the position of first mismatching bit in the binary representation of the two numbers… Read More »
## Count pairs with Bitwise XOR as EVEN number
Given an array of N integers, the task is to find the number of pairs (i, j) such that A[i] ^ A[j] is even. Examples:… Read More »
## Count of cyclic permutations having XOR with other binary string as 0
Given two binary strings and . Let be set of all the cyclic permutations of string . The task is to find how many strings… Read More »
## Find a value whose XOR with given number is maximum
Given a value X, the task is to find the number Y which will give maximum value possible when XOR with X. (Assume X to… Read More »
## XOR of path between any two nodes in a Binary Tree
Given a binary tree with distinct nodes and a pair of two nodes. The task is to find the XOR of all of the nodes… Read More » | 683 | 2,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-09 | longest | en | 0.751678 |
http://www.ck12.org/book/CK-12-Math-Analysis-Concepts/r4/section/7.10/ | 1,493,006,889,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118963.4/warc/CC-MAIN-20170423031158-00590-ip-10-145-167-34.ec2.internal.warc.gz | 486,197,611 | 40,900 | 7.10: Sums of Infinite Geometric Series
Difficulty Level: At Grade Created by: CK-12
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Practice Sums of Infinite Geometric Series
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Sayber's mom told him to clean his room on Saturday morning.
"But, MOM! It's gonna take forever!" said Sayber.
"Oh, don't be overly dramatic," said mom.
"I am NOT being dramatic!" Sayber said.
"If I start right now, it is going to take me at LEAST an hour to clean this half alone, then it will take another half hour to clean half of the remainder, and 15 mins to clean half of THAT remainder... since I will always have half left, I will never be done!"
Do you agree with Sayber? Will Sayber be stuck with a vacuum in his hand forever?
Tune in next week...
Embedded Video:
Guidance
Let’s return to the situation in the introduction: Poor Sayber is stuck cleaning his room. He cleans half of the room in 60mins. Then he cleans half of what is left, 30 more minutes, half again for 15 more. If he keeps cleaning half of the remaining area, how will he ever finish the room?
We know that the pieces have to add up to some finite time period (no matter what it feels like, Sayber CAN get the room clean), but how is it possible for the sum of an infinite number of terms to be a finite number?
To find the sum of an infinite number of terms, we should consider some partial sums. Three partial sums, relatively early in the series, could be: \begin{align*}S_2 = 90, S_3 = 105,\end{align*} and \begin{align*}S_6 = 118.125\end{align*} or \begin{align*}118 \frac{1}{8}\end{align*}
Now let’s look at larger values of \begin{align*}n:\end{align*}
\begin{align*}S_7\end{align*} \begin{align*}= \frac{60 \left(1 - (\frac{1} {2})^7 \right)} {1 - \frac{1} {2}} \approx 119.06\end{align*} minutes
\begin{align*}S_8\end{align*} \begin{align*}= \frac{60 \left(1 - (\frac{1} {2})^8 \right)} {1 - \frac{1} {2}} \approx 119.5\end{align*} minutes
\begin{align*}S_{10}\end{align*} \begin{align*}= \frac{60 \left(1 - (\frac{1} {2})^{10} \right)} {1 - \frac{1} {2}} \approx 119.9\end{align*} minutes
As n approaches infinity, the value of Sn seems to approach 120 minutes. In terms of the actual sums, what is happening is this: as n increases, the nth term gets smaller and smaller, and so the nth term contributes less and less to the value of Sn . We say that the series converges, and we can write this with a limit:
\begin{align*}\lim_{n \rightarrow \infty} S_n\end{align*} \begin{align*}= \lim_{n \rightarrow \infty} \left(\frac{60 \left(1 - (\frac{1} {2})^{n} \right)} {1 - \frac{1} {2}} \right )\end{align*}
\begin{align*}= \lim_{n \rightarrow \infty} \left(\frac{60 \left(1 - (\frac{1} {2})^{n} \right)} {\frac{1} {2}} \right )\end{align*}
\begin{align*}= \lim_{n \rightarrow \infty} \left(120 \left(1 - \left(\frac{1} {2} \right)^{n} \right ) \right) \end{align*}
As n approaches infinity, the value of \begin{align*}\left(\frac{1} {2} \right)^{n}\end{align*} gets smaller and smaller. That is, the value of this expression approaches 0. Therefore the value of \begin{align*}1 - \left(\frac{1} {2} \right)^{n}\end{align*} approaches 1, and \begin{align*}120 \left(1 - \left(\frac{1} {2} \right)^n \right)\end{align*} approaches 120(1) = 120.
Therefore, no matter how long the process continues, Sayber will not spend more than 2hrs cleaning the room. Of course, it may SEEM like a lot more!
We can do the same analysis for the general case of a geometric series, as long as the terms are getting smaller and smaller. This means that the common ratio must be a number between -1 and 1: |r| < 1.
\begin{align*}\lim_{n \rightarrow \infty} S_n\end{align*} \begin{align*}= \lim_{n \rightarrow \infty} \left (\frac{a_1(1 - r^n)} {1 - r} \right)\end{align*}
\begin{align*}=\frac{a_1} {1 - r},\end{align*} as \begin{align*}(1 - r^n) \rightarrow 1\end{align*}
Therefore, we can find the sum of an infinite geometric series using the formula \begin{align*}S = \frac{a_1} {1 - r}\end{align*}.
When an infinite sum has a finite value, we say the sum converges. Otherwise, the sum diverges. A sum converges only when the terms get closer to 0 after each step, but that alone is not a sufficient criterion for convergence. For example, the sum \begin{align*}\sum_{n = 1}^\infty \frac{1} {n} = 1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ....\end{align*} does not converge.
Example A
Find the sum of the convergent series: \begin{align*}40 + -20 + 10 + -5 + ...\end{align*}
Solution
The common ratio is \begin{align*}\frac{-1} {2}\end{align*}. Therefore the sum converges to:
\begin{align*}\frac{40} {1 - \left (\frac{-1} {2} \right)} = \frac{40} {\frac{3} {2}} = 40 \left(\frac{2} {3} \right) = \frac{80} {3}\end{align*}
Example B
Determine if the series converges. If it converges, find the sum.
a) \begin{align*}1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27}+ ... \end{align*} b) \begin{align*}3 + -6 +12 + -24 + ... \end{align*}
Solution:
a) \begin{align*}1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27} + ... \end{align*} converges.
The common ratio is (1/3) . Therefore the sum converges to:
\begin{align*}\frac{1} {1 - \frac{1} {3}} = \frac{1} {\frac{2} {3}} = \frac{3} {2}\end{align*}
b) The series 3 + -6 + 12 + -24 + ... does not converge, as the common ratio is -2.
Remember that the idea of an infinite sum was introduced in the context of a realistic situation, albeit a paradoxical one. We can in fact use infinite geometric series to model other realistic situations. Here we will look at another example: the total vertical distance traveled by a bouncing ball.
Example C
A bouncing ball
A ball is dropped from a height of 20 feet. Each time it bounces, it reaches 50% of its previous height. What is the total vertical distance the ball travels?
Solution
We can think of the total distance as the distance the ball travels down + the distance the ball travels back up. The downward bounces form a geometric series:
20 + 10 + 5 +...
The upward bounces form the same series, except the first term is 10.
So the total distance is: \begin{align*}\sum_{n=1}^\infty 20 \left ( \frac{1}{2} \right )^{n-1}+\sum_{n=1}^\infty 10 \left ( \frac{1}{2} \right )^{n-1}\end{align*}.
Each sum converges, as the common ratio is (1/2). Therefore the total distance is:
\begin{align*}\frac {20}{1-\frac {1}{2}}+\frac {10}{1-\frac {1}{2}}=\frac {20}{\frac {1}{2}}+\frac {10}{\frac {1}{2}}=40+20=60\end{align*}
So the ball travels a total vertical distance of 60 feet.
Vocabulary
A geometric series is the sum of terms from a geometric sequence, which is a sequence of terms with a different number or value between each pair of terms.
If the limit of the partial sums of series does not exist or is infinite, then the series diverges.
If the limit of the partial sums of series exists and is finite, the series converges.
Guided Practice
Questions
For questions 1 – 3, determine if the series converges or diverges. If it converges, find the sum.
1) -3 + 6 + -12 + 24 + ...
2) 240 + 60 + 15 + ...
3) 9 + 6 + 4 + (8/3) + ...
4) In this lesson, we proved the formula for the sum of a geometric series, \begin{align*}S_n=\frac {a_1(1-r^n)}{1-r}\end{align*} using induction.
Prove this formula without induction:
a) Let Sn = a1 + a1r + a1r2 + ... + a1rn-1
b) Multiply Sn by r to obtain a second equation
c) Subtract the equations and solve for Sn.
5) A ball is dropped from a height of 40 feet, and each time it bounces, it reaches 25% of its previous height.
a) Find the total vertical distance the ball travels, using the method used in the lesson.
b) Find the total vertical distance the ball travels using a single series.
(Hint: write out several terms for each bounce. For example, the first bounce is: 40 feet down + 10 feet up = 50 feet traveled.)
6) Below are two infinite series that are not geometric. Use a graphing calculator to examine partial sums. Does either series converge?
a) \begin{align*}1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} + ...\end{align*}
b) \begin{align*}1 + \frac {1}{4} + \frac {1}{9} + \frac {1}{16} + ...\end{align*}
Solutions
1) The sum does not converge because r = -2.
2) The sum converges. S = 320.
3) The sum converges. S = 27.
4) Follow the steps below:
(1)\begin{align*} S_n = a_1 + a_1{r} + a_1{r^2} + ... + a_1{r^{n-1}}\end{align*}
(2) \begin{align*}r{S_n} = a_1{r} + a_1{r^2} + a_1{r^3} + ... + a_1{r^n}\end{align*}
(3) \begin{align*} S_n - rS_n = a_1 - a_1{r^n}\end{align*}
\begin{align*}\Rightarrow S_n(1 - r) = a(1 - r^n)\end{align*}
\begin{align*}\Rightarrow S_n = \frac {a(1 - r^n)}{(1 - r)}\end{align*}
5) a) \begin{align*}\sum_{n=1}^\infty 40 \left ( \frac{1}{4} \right )^{n-1}+\sum_{n=1}^\infty 20 \left ( \frac{1}{4} \right )^{n-1}=66\frac {2}{3}\end{align*}
b) \begin{align*}\sum_{n=1}^\infty 50 \left ( \frac{1}{4} \right )^{n-1}=66\frac {2}{3}\end{align*}
6) a) This series does not converge.
b) This series converges around 1.65. (The actual sum is \begin{align*}\frac {\pi^2}{6}\end{align*})
Practice
1. Find the sum of the first 10 terms of \begin{align*}\sum_{n = 1}^\infty \left (\frac{1} {5} \right)^n\end{align*} using a graphing calculator.
2. Find the sum of the first 20 terms of \begin{align*}\sum_{n = 1}^\infty \left (\frac{1} {5} \right)^n\end{align*} using a graphing calculator.
3. Conjecture on the possible convergence of the series in questions 1 and 2.
Evaluate the infinite sum of the following geometric series:
1. \begin{align*}-2 + 1 -\frac{1}{2}+...\end{align*}
2. \begin{align*} -6 + \frac{24}{5} -\frac{96}{25}+...\end{align*}
3. \begin{align*} 3 +\frac{3}{2} -\frac{3}{4}+...\end{align*}
4. \begin{align*} -6 + 4 -\frac{8}{3}+...\end{align*}
5. \begin{align*}1 + \frac{1}{2}+\frac{1}{4}+...\end{align*}
Evaluate the infinite sum of the following geometric series:
1. \begin{align*}\sum_{n=1}^{\infty} -3(\frac{1}{2})^{(n-1)}\end{align*}
2. \begin{align*}\sum_{n=1}^{\infty} -2(\frac{4}{7})^{(n-1)}\end{align*}
3. \begin{align*}\sum_{n=1}^{\infty} 7(\frac{-4}{5})^{(n-1)}\end{align*}
4. \begin{align*}\sum_{n=1}^{\infty} -9(\frac{-1}{5})^{(n-1)}\end{align*}
5. \begin{align*}\sum_{n=1}^{\infty} 5(\frac{-5}{7})^{(n-1)}\end{align*}
6. \begin{align*}\sum_{n=1}^{\infty} 6(\frac{1}{5})^{(n-1)}\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
TermDefinition
converge If a series has a limit, and the limit exists, the series converges.
convergent If a series has a limit, and the limit exists, the series is convergent.
divergent If a series does not have a limit, or the limit is infinity, then the series is divergent.
diverges If a series does not have a limit, or the limit is infinity, then the series diverges.
geometric sequence A geometric sequence is a sequence with a constant ratio between successive terms. Geometric sequences are also known as geometric progressions.
geometric series A geometric series is a geometric sequence written as an uncalculated sum of terms.
partial sums A partial sum is the sum of the first ''n'' terms in an infinite series, where ''n'' is some positive integer.
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http://www.perlmonks.org/?displaytype=displaycode;node_id=252219 | 1,521,824,791,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-13/segments/1521257648404.94/warc/CC-MAIN-20180323161421-20180323181421-00768.warc.gz | 437,031,595 | 940 | (\$latitude,\$longitude)=ll( MILES=>5, LATITUDE=>51+29/60, LONGITUDE=>0+0/60, DIRECTION=>315, ); if( \$latitude < 0 ){ \$latitude *= -1; \$ns='S'; }else{ \$ns='N'; } if( \$longitude < 0 ){ \$longitude *= -1; \$ew='W'; }else{ \$ew='E'; } print int \$latitude,"°",60*(\$latitude-int \$latitude),"' \$ns\n", int \$longitude,"°",60*(\$longitude-int \$longitude),"' \$ew\n"; use Math::Trig; sub ll{ my %l=@_; use constant PI=>4*atan2(1,1); my \$nm = \$l{MILES}*.87; my \$A=\$l{DIRECTION}*PI/180; my \$b=(90-\$l{LATITUDE})*PI/180; my \$c=\$nm/60/180*PI; my \$a=acos(cos(\$b)*cos(\$c)+sin(\$b)*sin(\$c)*cos(\$A)); my \$latitude=90-180/PI*\$a; my \$C=asin(sin(\$c)*sin(\$A)/sin(\$a)); my \$longitude=\$l{LONGITUDE}+180/PI*\$C; return \$latitude,\$longitude; } | 298 | 751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-13 | latest | en | 0.152615 |
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# Surveying How To
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## Surveyor Community Forum Discussions
### Grid to Ground Plus Laser Scanning
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### Basic Manual of Surveying
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Started by AxeMen Site Prep. Last reply by AxeMen Site Prep Aug 8, 2013.
## Surveying How-to Videos on Our Network (50)
### Rise and Fall, RL of point method in Surveying (Hindi/Urdu)
Rise and fall method in Surveying
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Leveling method by autolevel or Dumpy Level in Surveying
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### What are Geodetic Datums?
In this video we explain the basic concepts behind geodetic datums, where they are used, and why it is important to know about and use the correct datums.
For more information on geodetic datums, visit http://www.geodesy.noaa.gov/.
For more information and a gallery of reusable resources from this video see…
### Best Practices for Minimizing Errors during GNSS Data Collection
Aimed at surveyors and GIS professionals who use geodetic-quality GNSS equipment to determine positions for land planning, coastal monitoring and other purposes, this video covers best practices for reducing errors in the areas of: 1. location and environment, 2. equipment setup and 3. observation times and accuracy checks.
For…
### Traversing with Distance and Bearing in AutoCAD | Excel to AutoCAD | Surveying
In this video I have shown direct method of how to create a Traverse map when we have only distance and bearing with us.
I have showed a direct method by which you can draw the traverse line with just copy and paste. I used some easy formula in excel for this.
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### Welcome Surveyor! Please Sign in or register to continue.
Reply by Deddy Priatna on April 10, 2014 at 5:51pm
How about get Geo Tool Office software
Reply by Montano on April 13, 2012 at 7:26pm
Hello everyone. I have decided to pick up my surveying skills again. I have been doing a lot of GIS work and put my surveying on the back burner. I have AutoCAD Civil 3D but now need to learn how to use TDS Foresight DXM and Survey Pro. Anyone out there familiar with the two? I need your help just to get me started. Thank you.
Reply by ePalmetto on June 24, 2011 at 2:37pm
50 Surveying How-to Videos added to this group. for your photos and videos of 'how-to' scenarios to automatically show up in this group, simply tag them with either "how-to" or "tutorial"
Reply by Paul Quagge on May 2, 2011 at 1:09pm
Thanks again Kendall,i'll try what your doing and let you know, have a good day...paul
Reply by kendall payne on May 2, 2011 at 8:21am
i have never used autodesk field survey, what i know from autocad well is that when i import survey results into cad (points with their z,y,z ) and i do not need elevations , first step is to select all then right click , propperties and make z-value which is then in varies , change it to 0 the veruthing will be 0 elevations so you can inverse your disntances and have the without slope , in that way i normally do it
Reply by Paul Quagge on May 2, 2011 at 8:08am
Thanks kendall, we are using Autodesk Field Survey to process our field data...collecting in x,y and z but when downloading the office is only receiving slope distances...have a good day...Paul
Reply by kendall payne on May 2, 2011 at 7:44am
hi, depends on what software they use at the office for processing survey results, i think personally by sende x,y,z values is better and at the office depending your software you can always cancel the z-value and work further , if they use trimble geomatics office for processing that is the case ....foresight dxm i do not know, in survey pro version version 4.9 and newer you can make your survey 2d at start so no z-values will be recorded ..that can be an option
Reply by Paul Quagge on May 1, 2011 at 7:01pm
I'm using a Ranger data collector with Survey Pro software and Nikon total station.I am new to this company and this companys equipment although I have been using Survey Pro for awhile. My new supervisor mentioned that when data is downloaded they receive the distances in slope distance which they then convert to horizontal distances and asked me to look into this.They would like the data in horizontal right off. Could anyone tell me if there is a setting that effects this situation? Any advice would be appreciated. Thanks, Paul
Reply by MARK GREGORY HILL on May 10, 2010 at 10:13pm
I have a question rather than an answer. I was wondering if anyone other than me has experienced problems working under high power electical transmission lines. Running one static session with two sokkia recievers, the third on a known contol point, underneath some very poweerful high voltage electric lines and one unit collected good data while the other came in as floating. Swithced the units around the next day and still got poor results. Has anyone else experienced this situation and is there anything I can do to prevent it?
Reply by Ozan AGYUZ on December 17, 2009 at 7:58pm
looking for job
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# Union-Find Problem section 12.9.2 - PowerPoint PPT Presentation
Union-Find Problem section 12.9.2 Given a set {1, 2, …, n} of n elements. Initially each element is in a different set. {1}, {2}, …, {n} An intermixed sequence of union and find operations is performed. A union operation combines two sets into one.
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Union-Find Problemsection 12.9.2
• Given a set {1, 2, …, n} of n elements.
• Initially each element is in a different set.
• {1}, {2}, …, {n}
• An intermixed sequence of union and find operations is performed.
• A union operation combines two sets into one.
• Each of the n elements is in exactly one set at any time.
• A find operation returns the set that contains a particular element.
Using Arrays And Chains
• See Section 7.7 for applications as well as for solutions that use arrays and chains.
• Best time complexity obtained in Section 7.7 is O(n + u log u + f), where u and f are, respectively, the number of union and find operations that are done.
• Using a tree (not a binary tree) to represent a set, the time complexity becomes almost O(n +f) (assuming at least n/2 union operations).
Uses of Union/Find
A factory has a single machine that is to perform n tasks. Task i has an integer release time ri and an integer deadline di. The completion of each task requires one unit of time on this machine. A feasible schedule is an assignment of tasks to time slots on the machine such tat task I is assigned to a time slot between its release time and deadline and no slot has more than one task assigned to it.
Uses of Union/Find
Consider the following four tasks:
Task A B C D
Release time 0 0 1 2
Deadline 4 4 2 3
Tasks A and B are released at time 0, etc.
Uses of Union/Find
The following task-to-slot assignment is a feasible schedule:
0 1 2 3 4
A C D B
Task A B C D
Release time 0 0 1 2
Deadline 4 4 2 3
Uses of Union/Find
Intuitive algorithm:
1. Sort the tasks into nonincreasing order of release time.
2. Consider the tasks in this nonincreasing order. For each task determine the free slot nearest to, but not after, its deadline. If this free slot is before the task’s release time, fail. Otherwise, assign the task to this slot.
Uses of Union/Find
• Task scheduling. Can use Union/Find for part 2
• Let d denote the latest deadline of any task
• Usable time slots are j-1 to j where 1<=j <= d
• Refer to these slots as slots 1 through d
• For any slot a, define near(a) as the largest j such that j <= a and slot j is free.
• If no such j exists, then near(a)=near(0)=0
• Two slots a and b are in the same equivalence class iff near(a) = near(b)
Uses of Union/Find
• Task scheduling. Can use Union/Find for part 2
• Beginning: near(a) = a for all slots and each slot is in its own equivalence class.
• When slot a is assigned a task, near changes for all slots b with near(b) = a.
• The new value of near is near(a-1)
• Thus perform a union on the equivalence classes that currently contain slots a and a-1. F
• Find the equivalence class of a-1 by doing a find(a)
Uses of Union/Find
• Task scheduling. Can use Union/Find for part 2
• For each equivalence class e must retain in nearest[e] the value of near of its members
• Get near(a) by doing nearest[find(a)]
• Assume that the equivalence class name is taken to be whatever the find returns.
• Consider the following four tasks:
Task A B C D
Release time 0 0 1 2
Deadline 4 4 2 3
Putting in non-increasing order of release:
Task D C A B
Deadline 3 2 4 4
latest schedule 2 1 3 3
• Putting in non-increasing order of release:
Task D C A B
latest schedule 2 1 3 3
class
near 0 1 2 3
nearest: 0 1 2 3
0 1 2 3
0
1
2
3
• Schedule D in slot 2. All items in class 2 become class 1
Task D C A B
latest schedule 2 1 3 3
class
near 0 1 3
nearest: 0 1 2 3
0 1 1 3
0
1
3
2
2
• Schedule C in slot 1. All items in class 1 become class 0
Task D C A B
latest schedule 2 1 3 3
scheduled 2 1
class
near 0
nearest: 0 1 2 3
0 0 1 3
0
3
1
3
2
1
2
• Schedule A in slot 3. All items in class 3 become class 2
Task D C A B
latest schedule 2 1 3 3
scheduled 2 1 3
class
near 0
nearest: 0 1 2 3
0 0 1 0
0
3
1
2
3
1
2
• Schedule B in slot 3 which is in e-class 0 which has nearest value 0. All items in class 3 become class 2
Task D C A B
latest schedule 2 1 3 3
scheduled 2 1 3 0
class
near 0
nearest: 0 1 2 3
0 0 0 0
0
3
1
2
3
1
2
• Resulting task schedule.
The following task-to-slot assignment is a feasible schedule:
0 1 2 3 4
B C D A
Task A B C D
Release time 0 0 1 2
Deadline 4 4 2 3
5
4
13
2
9
11
30
5
13
4
11
13
4
5
2
9
9
11
30
2
30
A Set As A Tree
• S = {2, 4, 5, 9, 11, 13, 30}
• Some possible tree representations:
4
2
9
11
30
5
13
Result Of A Find Operation
• find(i) is to return the set that contains element i.
• In most applications of the union-find problem, the user does not provide set identifiers.
• The requirement is that find(i) and find(j) return the same value iff elements i and j are in the same set.
find(i) will return the element that is in the tree root.
13
4
5
9
11
30
2
Strategy For find(i)
• Start at the node that represents element i and climb up the tree until the root is reached.
• Return the element in the root.
• To climb the tree, each node must have a parent pointer.
7
13
4
5
8
3
22
6
9
11
30
10
2
1
20
16
14
12
Trees With Parent Pointers
Possible Node Structure
• Use nodes that have two fields: element and parent.
• Use an array table[] such that table[i] is a pointer to the node whose element is i.
• To do a find(i) operation, start at the node given by table[i] and follow parent fields until a node whose parent field is null is reached.
• Return element in this root node.
13
4
5
9
11
30
2
1
table[]
0
5
10
15
(Only some table entries are shown.)
Example
13
4
5
9
11
30
2
1
parent[]
0
5
10
15
Better Representation
• Use an integer array parent[] such that parent[i] is the element that is the parent of element i.
2
9
13
13
4
5
0
Union Operation
• union(i,j)
• i and j are the roots of two different trees, i != j.
• To unite the trees, make one tree a subtree of the other.
• parent[j] = i
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Union Example
• union(7,13)
### The Find Method
public int find(int theElement)
{
while (parent[theElement] != 0)
theElement = parent[theElement]; // move up
return theElement;
}
### The Union Method
public void union(int rootA, int rootB)
{parent[rootB] = rootA;}
Time Complexity Of union()
• O(1)
• Time for u unions: O(u)
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Time Complexity of find()
• Tree height may equal number of elements in tree.
• union(2,1), union(3,2), union(4,3), union(5,4)…
• So complexity of a single search is O(u).
Time Complexity of find()
• Worst case:
• do u unions resulting in one tree as on previous slide (height u)
• Do f finds on last element
• Time is O(uf)
Back to the drawing board.
u Unions and f Find Operations
• O(u + uf) = O(uf)
• Time to initialize parent[i] = 0 for all i is O(n).
• Total time is O(n + uf).
• Worse than solution of Section 7.7!
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Smart Union Strategies
• union(7,13)
• Which tree should become a subtree of the other?
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Height Rule
• Make tree with smaller height a subtree of the other tree.
• Break ties arbitrarily.
union(7,13)
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Weight Rule
• Make tree with fewer number of elements a subtree of the other tree.
• Break ties arbitrarily.
union(7,13)
Implementation
• Root of each tree must record either its height or the number of elements in the tree.
• When a union is done using the height rule, the height increases only when two trees of equal height are united.
• When the weight rule is used, the weight of the new tree is the sum of the weights of the trees that are united.
Height Of A Tree
• If we start with single element trees and perform unions using either the height or the weight rule. The height of a tree with p elements is at most floor (log2p) + 1.
• Proof is by induction on p. See next slide.
Height Of A Tree
• Proof is by induction on p.
• Trivial for p=1
• assume true for all trees with i nodes, i<=p-1.
• show for trees with p nodes.
• consider the last union operation union(k,j) used to create tree t which has p nodes.
• Let m be the number of nodes in tree j and let p-m be the number of nodes in tree k.
• assume that 1<=m<=p/2. So j is made a subtree of k.
Height Of A Tree
• Proof is by induction on p.
• assume that 1<=m<=p/2. So j is made a subtree of k.
• therefore height of t is either same as that of k or is one more that that of j.
• if former then, since num nodes in k is p-m and by induction, height of t is
Height Of A Tree
• Proof is by induction on p.
• if latter then height of t is by induction
since log2p/2 = log2p - log22 = log2p - 1
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Sprucing Up The Find Method
• find(1)
• Do additional work to make future finds easier.
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Path Compaction
• Make all nodes on find path point to tree root.
• find(1)
Makes two passes up the tree.
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Path Splitting
• Nodes on find path point to former grandparent.
• find(1)
Makes only one pass up the tree.
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Path Halving
• Parent pointer in every other node on find path is changed to former grandparent.
• find(1)
Changes half as many pointers.
Time Complexity
• Ackermann’s function.
• A(i,j) = 2j, i = 1 and j >= 1
• A(i,j) = A(i-1,2), i >= 2 and j = 1
• A(i,j) = A(i-1,A(i,j-1)), i, j >= 2
• Inverse of Ackermann’s function.
• alpha(p,q) = min{z>=1 | A(z, p/q) > log2q}, p >= q >= 1
• Means find the minimum z such that A(z,p/q) > log2q
Time Complexity
• Ackermann’s function grows very rapidly as p and q are increased.
• A(2,1) = A(1,2) = 22 = 4
• For j >= 2, A(2,j) = A(1,A(2,j-1)) = 2A(2,j-1).
• So A(2,2) = 2A(2,1) = 24 =16
• A(2,3) = 2A(2,2) = 216 =65,536
• A(2,4) = 265,536 .
• A(2,j) = 222…. Where there are j + 1 rows of twos.
• A(2,j) for j > 3 is very large
Time Complexity
• Ackermann’s function grows very rapidly as p and q are increased.
• A(2,4) = 265,536
• The inverse function grows very slowly.
• alpha(p,q) < 5 until q = 2A(4,1)
• A(4,1) = A(2,16) >>>> A(2,4)
• In the analysis of the union-find problem, q is the number, n, of elements; p = n + f; and u >= n/2.
• For all practical purposes, alpha(p,q) < 5.
Time Complexity
Theorem 12.2[Tarjan and Van Leeuwen]
Let T(f,u) be the time required to process any intermixed sequence of f finds and u unions. Assume that u >= n/2.
a*(n + f*alpha(f+n, n) <= T(f,u) <= b*(n + f*alpha(f+n, n)
where a and b are constants. These bounds apply when we start with singleton sets and use either the weight or height rule for unions and any one of the path compression methods for a find. | 3,873 | 12,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-47 | longest | en | 0.894838 |
http://www.bridgeworld.com/indexphp.php?page=/pages/learn/beginners/lesson3.html | 1,501,239,979,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549448198.69/warc/CC-MAIN-20170728103510-20170728123510-00007.warc.gz | 366,029,099 | 9,467 | LESSON 3: Introduction to Declarer's Play
### Sure Tricks at Notrump
The game of bridge revolves around the bidding for and the taking of tricks. Here we are not worried about the bidding, just the taking.
The most important single move that you must make before playing out a hand is to count your tricks. That seems easy enough, doesn't it?
Lets take a simple example:
DUMMY ♠ A 4 3 YOU ♠ K 5 2
Whenever you play a bridge hand as declarer, you get to see all of your partner's cards before you play. Your partner's hand is called the dummy, and that term has nothing to do with the way he may have bid his hand.
So what you do after the opponent on your left makes an opening lead is to look at one suit at a time; look, for example, at your spades and at dummy's spades and count the number of sure tricks you have in that suit. Then you go through the same process in each suit and come up with a figure. That is a very important figure. It tells you how many tricks you can take at a moment's notice. Remember that term, sure tricks, because we are going to work with it for a while.
Now let's go back to our example. In dummy we have the A 4 3 of spades, and in our own hand we have the K 5 2. The ace will take one trick and the king will take another, so we have the two sure spade tricks. This may seem elementary, but you will never learn to play a hand unless you do this.
Counting tricks has its hazards. Let's try this one:
DUMMY ♠ K Q YOU ♠ A 2
Now how many sure tricks do you have in spades? This answer is two, not three. You see, when you play a card from your hand, you must also take a card from the dummy. Let's say you play the ace; then the queen must be played from dummy. That leaves you with the two in your hand and the king in dummy. In other words, you have two tricks, not three.
The important thing to see is that you can never take more tricks in a suit than there are cards in the longer of the two hands. Look:
DUMMY ♠ A K Q YOU ♠ J 10
Between you and your dummy you have the ace, king, queen, jack, and ten. But you can only take three tricks. That is because the dummy, which is the longer hand in spades, has only three cards.
Practice counting sure tricks with these examples:
(a)
DUMMY ♠ K Q 3 YOU ♠ A 5 2
(b)
DUMMY ♠ A Q J 8 YOU ♠ K 7
(c)
DUMMY ♠ A J 3 YOU ♠ K Q 5 4
(d)
DUMMY ♠ Q J 10 5 4 YOU ♠ A K 3
Solutions
(a) Three tricks. You can take them in any order you like. You could play the king, then the queen, and then the three to your ace; or you could play the ace, and then a little one to the king, and then the queen. Or you could play the king, then the three to your ace, and then a little one back to your queen. You see, when you have the same number of cards in both hands (e.g., three cards on each side), you have quite a bit of flexibility. You would have to see all 26 cards before you knew which hand you wanted to end up in. I am merely showing you that you don't always have to play the ace first when taking tricks.
(b) Four tricks. Now this situation and the following ones are a little different because you do not have the same number of cards on both sides. In this case the dummy has four and you only have two. As a general rule, whenever you have a bunch of good tricks in a suit that is unevenly divided, you should play the high card(s) from the short side first. This means playing the king, which will take the eight from dummy, and then leading your seven over to the ace, queen, and jack in dummy. When cards are high it does not matter which one you play first. In this case, when you have played the king and are about to lead the seven over to the dummy, it doesn't matter if you play the jack, queen, or ace; they are all the same. In this little game we are playing, we are always assuming that the opponents have led some other suit and we have taken the trick. Now we are about to play our suit. Sometimes the trick we have taken will have been in dummy. Therefore, if the lead is in the dummy, we must play the eight of spades over to our king and then the seven back to the dummy. But in either case we are playing the high card from the short side first.
(c) Four tricks. If the lead is in the dummy (from the prior play), we should first play the ace, then the jack, and then the three over to our king and queen. Notice that we played the high cards from the short side first. Things would be exactly the same if the lead were in our hand. We would play the four over to the ace (or jack), then the jack, and then the three over to our king and queen. It is conceivable that the opponents might lead this suit themselves, in which case we would still play it the same way.
(d) Five tricks. This time we would play the king and ace (or the ace and king) from our hand and then lead the three over to the queen, jack, and ten in the dummy. Playing the high card or high cards from the short side first allows us to end up on the long side, where we can take the maximum number of tricks.
Now let's practice counting our sure tricks in an entire deal:
DUMMY ♠ A 4 3 ♥ K 4 ♦ 10 8 7 5 ♣ A K Q 3 YOU ♠ 7 5 2 ♥ A Q 3 ♦ A 4 3 2 ♣ J 4 2
Let's pretend the final contract was three notrump and West, your left-hand opponent, led the king of spades. How many sure tricks do you have altogether?
You should have come up with nine sure tricks. You have one in spades, three in hearts, one in diamonds, and four in clubs.
Sometimes counting tricks and taking them are two different things. But if you remember about the high card(s) from the short hand, you will not have any trouble. In clubs, you would play the jack first from your own hand and then play a little one over to the ace, king, and queen in dummy. In hearts, you would play the king first and then the four over to the ace and queen in your own hand.
Here are some additional practice deals. Count your sure tricks and see what you come up with:
(a)
DUMMY ♠ A 4 3 ♥ K Q ♦ A J 4 ♣ A J 7 6 5 YOU ♠ K Q 7 ♥ A 7 ♦ K Q 10 3 2 ♣ K 3 2
(b)
DUMMY ♠ K Q J ♥ Q J 10 9 ♦ J 10 9 ♣ K Q J YOU ♠ 10 9 8 ♥ K 8 7 6 ♦ K Q 8 7 ♣ 10 9
Solutions
(a) You should have come up with twelve tricks: three in spades, two in hearts (make sure you see why), five in diamonds (playing the ace and jack first), and two in clubs.
(b) You have zero sure tricks. That's right, not one. In order to take tricks in any one of these suits, you must first get rid of the opponent's ace. Until you get rid of that ace, you do not have a sure trick. The definition of a sure trick is a trick that you can take without giving up the lead. When you must give up the lead to take a trick, you are establishing tricks, which leads us to a new topic.
Summary: Key Pointers About Sure Tricks
(1) The first step in playing a bridge hand as declarer is to count your sure tricks.
(2) A sure trick is a trick that can be taken without giving up the lead.
(3) You can never take more tricks in a suit than there are cards in the longer hand. (If both you and your dummy have two cards in one suit, the most tricks you can take in that suit is two.)
(4) When taking sure tricks, play the high cards(s) from the short side first. This will allow you to end up on the long side, where you can cash the rest of the tricks in the suit.
### Establishing Tricks
In most of the deals that you play, you never seem to have enough sure tricks to make your contract. Let's say you are playing three notrump. You need nine tricks to fulfill your contract, and you usually count up only five or six sure tricks. What are you going to do? Well, there is another method of getting tricks, but it involves a little work. You have to establish, or make, tricks for yourself. Study this diagram:
DUMMY ♠ K Q J YOU ♠ 4 3 2
If this is your spade suit, you do not have a sure trick in spades. But if you were to play the king (or the queen or jack) from dummy and drive out the ace, you could establish two spade tricks for yourself. This method of establishing tricks, driving out the opponents' aces and kings, is the most common method of obtaining tricks in bridge.
You may be wondering what would happen if the opponents did not take their ace; actually, it would tum out the same. Let's say you lead the king and everyone plays low. Well, you've taken one trick already. Now you lead the queen. If everyone plays low again, you have taken two tricks in the suit, and that is all you ever had coming in the first place.
When it comes to establishing tricks, you follow the same general rule that you did when you were taking your sure tricks. Play the high card(s) from the short side first. For example:
DUMMY ♠ Q J 10 3 YOU ♠ K 2
Let's say you wish to establish some spade tricks for yourself. You should lead the king from your own hand. If the opponent takes it with the ace, you will still have the deuce, and the next time it is your lead you can take the queen, jack, and ten. In other words, you should get three tricks from this suit.
Now let's practice counting tricks in suits that we must establish. How many tricks can you establish in each of the following suits, and which card do you play first?
(a)
DUMMY ♠ K Q 7 YOU ♠ J 3
(b)
DUMMY ♠ Q 5 YOU ♠ K J 10 9 3
(c)
DUMMY ♠ 4 3 2 YOU ♠ Q J 10
(d)
DUMMY ♠ A 3 YOU ♠ Q J 10 9
Solutions
(a) Two tricks. You should play the jack first. If the lead is in the North hand, lead the seven to the jack.
(b) Four tricks. You should play the queen first. If the lead is in the South hand, you should lead the three to the queen.
(c) One trick. You can lead from either hand because you have the same number of cards on both sides. The queen will drive out the king, the jack will drive out the ace, and the ten will be an established trick.
(d) Three tricks. You should lead the ace and then the three. If you live right, the king might fall under the ace, and then you will get four tricks--but don't count on it. They have too many cards in the suit.
You are now ready to make a little progress. Your next step in planning the play of a contract is to count the sure tricks you have and see how many more tricks you can establish.
The important thing to remember is to keep the two counts separate until you have actually established some tricks. Once you establish some tricks, you can add the tricks you have established to your sure trick count. Take a look at this layout:
DUMMY ♠ A 4 3 ♥ K Q J 10 ♦ K 5 2 ♣ 9 8 7 YOU ♠ K 5 ♥ 5 4 3 2 ♦ A Q J 9 ♣ A K Q
Let's say you are playing a contract of six notrump. You must always ask yourself how many tricks you need to make your contract. In this case you need twelve (six plus your bid). The opponents lead the queen of spades. Now, after realizing how many tricks you need, which is really the first step, you must add up your sure tricks. So let's do that. You have two in spades, four in diamonds, and three in clubs. A total of nine. Notice that you did not count even one sure trick in hearts, simply because you cannot take a trick in that suit until you drive out the ace.
Well, you have nine sure tricks and you must establish at least three more tricks in hearts to make your contract. That's easy enough. You simply win the spade with your king and lead a heart. Let's assume that the opponents take it with their ace. Your sure trick count has just changed. You now have twelve sure tricks instead of nine, because you can add those extra three heart tricks to your total once the ace has been removed.
Now for the most important point in the whole lesson. When playing a bridge hand that does not have enough sure tricks, you must establish extra tricks. Establishing extra tricks should be the first thing you do. You establish the extra tricks you need before you take your sure tricks. Then, when you have established enough tricks to make your contract, you take all of your tricks at once.
Rules are not much good unless you know their reasons. So we are going to go back to our six notrump contract. For the first time we are going to look at all four hands.
DUMMY ♠ A 4 3 ♥ K Q J 10 ♦ K 5 2 ♣ 9 8 7 WEST ♠ Q J 10 ♥ A ♦ 10 8 7 6 4 ♣ J 4 3 2 EAST ♠ 9 8 7 6 2 ♥ 9 8 7 6 ♦ 3 ♣ 10 6 5 YOU ♠ K 5 ♥ 5 4 3 2 ♦ A Q J 9 ♣ A K Q
For the time being don't worry about why West led the queen of spades. Presently you are worried about taking twelve tricks.
Notice that after you take the first trick with the king of spades, you still have control (that is, you can take the next trick) in all suits except hearts, where you will soon be establishing your tricks.
What if you were to take your club tricks before knocking out the ace of hearts? Watch closely what would happen so that you never make this error--in fact, this is the most common error beginners make--of taking sure tricks too quickly.
If you were to take your three club tricks before playing hearts, West would still have the jack of clubs. It would be the only club left. Then, when you led a heart, West would take it with his ace and then would be able to take the next trick with his jack of clubs because you had surrendered control of the club suit by taking your sure tricks too quickly.
The same thing would happen in diamonds. If, after winning the first trick with the king of spades, you were to take four tricks in diamonds, West would still have one diamond. Then, when you played a heart, West would take that trick with the ace of hearts and the next trick with the ten of diamonds. In neither case would you make your contract, because you would have lost two tricks, while you can afford to lose only one in a contract of six.
Therefore, it is important that you see that by taking your sure tricks too quickly, you give up control in the suit, and--even worse--you establish tricks for your opponents.
Establish first: Take your sure tricks after you have established.
Now you are going to practice counting your sure tricks, seeing if you have tricks that can be established (and, if so, how many), and, finally, determining which suit you should play first.
(a)
DUMMY ♠ K Q 10 3 ♥ A 4 3 ♦ 7 6 5 ♣ K Q 2 YOU ♠ J 5 ♥ K 5 2 ♦ A 8 4 3 ♣ A J 10 9
Contract: Three notrump
(b)
DUMMY ♠ A K 5 ♥ 3 2 ♦ A 7 6 5 ♣ 5 4 3 2 YOU ♠ Q 7 ♥ Q J 10 9 ♦ K 4 3 2 ♣ A K 6
Contract: Three notrump
In each exercise: How many sure tricks do you count? How many more can you establish? Which suit should you play first? Which card should you play in that suit?
Solutions
(a) You have seven sure tricks and you can establish three more in spades. You should play spades first (after taking the first trick with the king of hearts) and you should lead the jack. If it takes the trick, you continue with spades until one of your opponents plays the ace. You will eventually wind up with ten tricks. Once you have driven out the ace of spades, you will have established enough tricks to make your contract. Then you can take all of your tricks at once.
(b) You have seven sure tricks and you can establish two more in hearts. Therefore, you should play hearts first. After taking the first trick in your hand (high card from the short side), you can begin by playing any heart For concealment, declarer usually plays his highest equal, or highest in a sequence, first. So you would first lead the queen of hearts. In this case you must give up the lead twice in hearts in order to establish two tricks of your own in the suit. Assume that the queen loses to the king or ace and that a spade is returned. You take this in the dummy and lead another heart, establishing your hearts before taking any of your sure tricks.
Summary: Key Pointers About Establishing Tricks
(1) When playing a hand as declarer, know how many tricks you must take to fulfill your contract.
(2) Count your sure tricks and, if you do not have enough, look for suits that can be established (usually suits that are missing the ace or the king). Once you lose a trick to the high card, the rest of your cards in that suit will be good.
(3) Do your establishing early. Establish first and then take your sure tricks.
(4) If you take your sure tricks too soon, you may find that when you start establishing, the opponents will by that time have good tricks established in the suits in which you hastily cashed your sure tricks.
(5) Don't be afraid to give up the lead. On most hands you must give up the lead two or three times.
(6) When playing equal cards (such as the jack, ten, and nine), declarer should usually play his highest equal first. This applies to both establishing and taking. By doing this, you make it harder for the opponents to know what is going on. If you have the ace, king, and queen of spades and you play the queen, naturally it will take the trick, but your opponents will know that you still have the king and ace. However, if you play the ace first, the opponents will not know who has the king and queen.
When playing equal cards from the dummy, it doesn't matter which one you play first, because the opponents can see the dummy. However, just to stay in practice, you should take the highest equal from dummy also. | 4,317 | 17,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-30 | latest | en | 0.976241 |
https://studylib.net/doc/25225604/macroeconomics-2018 | 1,566,667,879,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321351.87/warc/CC-MAIN-20190824172818-20190824194818-00484.warc.gz | 652,531,913 | 47,590 | Uploaded by Kevin Aguila
# Macroeconomics 2018
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```Macroeconomics 1500, Humber College, Winter 2018
Professor: Adela Nistor
Name: ________________________
Midterm (30%)
All questions are 1 point unless otherwise stated.
Answer all the questions for a total of 74 Points.
50 multiple choice: total 50 points
Short answer: 3 questions each 8 points, total 24 points
1. In a simple circular-flow diagram, how are total income and total expenditure interrelated?
a. They are seldom equal because of fluctuations in the business cycle that occur in an economy.
b. They are equal only when all goods and services produced are sold.
c. They are always equal because every transaction has a buyer and a seller.
d. They are always equal because of accounting rules.
2. What does the simple circular-flow diagram illustrate?
a. that expenditure generates income so that income and the value of production are equal
b. that the economy’s income exceeds its expenditure
c. that factors of production “flow” from firms to households
d. that money “flows” from households to firms
3. Suppose an apartment complex converts to a condominium (i.e., the renters become owners of their
apartments). What is included in GDP?
a. The rent was included in GDP; the purchases of the condominiums are not included.
b. The rent was included in GDP, and so is the purchase of the condominiums.
c. The rent was not included in GDP; the purchases of the condominiums are included.
d. The rent was not included in GDP; the purchases are not included either.
4. A professional gambler moves from a province where gambling is legal to a province where gambling
is illegal. What impact does this move have on Canada’s GDP?
a. It raises GDP.
b. It decreases GDP.
c. It does not change GDP because gambling is never included in GDP.
d. It does not change GDP because in either case his income is included.
5. The local Nissan dealership has an increase in inventory of 25 cars in 2015. How will the sale of all 25
cars in 2016 affect the GDP?
a. The value of the cars will be counted as part of GDP in 2015 but not in 2016.
b. The value of the cars will not affect the 2015 GDP, but will be included in 2016 GDP.
c. The value of the cars will be included in both 2015 and 2016 GDP.
d. The value of the cars will not be included in GDP in 2015 or 2016.
1
6. A company makes 500,000 smart phones in the second quarter. It sells 300,000 of them before the end
of the second quarter, and holds the others in its warehouse. How is the second quarter GDP affected?
a. The smart phones are included in second-quarter GDP.
b. The smart phones that are not purchased in the second quarter are not counted in the secondquarter GDP.
c. Since all the 500,000 smart phones will eventually be bought by consumers, they are included
as consumption in the third quarter.
d. The smart phones will be counted as a change in inventory in the second quarter, and when
sold in the third quarter will raise GDP.
7. Suppose an economy produces only wheat and rice. Last year, 20 bushels of wheat are sold at \$4 per
bushel, and 10 bushels of rice are sold at \$2 per bushel. If the price of wheat was \$2 per bushel and the
price of rice was \$1 per bushel in the base year, what can we conclude?
a. Nominal GDP is \$100, real GDP is \$50, and the GDP deflator is 50.
b. Nominal GDP is \$100, real GDP is \$50, and the GDP deflator is 200.
c. Nominal GDP is \$50, real GDP is \$100, and the GDP deflator is 200.
d. Nominal GDP is \$50, real GDP is \$100, and the GDP deflator is 50.
Table 5-4
Use the following table to answer the following questions.
Year Price of Burgers Quantity of Burgers Price of Magazines
2013 \$4.00
100
\$2.00
2014 \$5.00
120
\$2.50
2015 \$6.00
150
\$3.50
Quantity of Magazines
180
200
200
8. Refer to the Table 5-4. Using the GDP deflator to measure the average level of prices and using 2013
as the base year, what is the economy’s inflation rate?
a. 20 percent for 2014 and 12.5 percent for 2015
b. 20 percent for 2014 and 30 percent for 2015
c. 25 percent for 2014 and 28 percent for 2015
d. 44.7 percent for 2014 and 45.5 percent for 2015
9. How is a country’s wealth related to other measures of well-being such as life expectancy and literacy?
a. positively with both life expectancy and literacy
b. unrelated with life expectancy; positively with literacy
c. negatively with life expectancy; unrelated to literacy
d. negatively with both life expectancy and literacy
10. Which goods are supposed to be included in the CPI?
a. all goods and services produced in the economy
b. all goods and services that typical consumers buy
c. all goods and services in the consumption component of the GDP accounts
d. all the goods, but not the services, in the consumption component of the GDP accounts
Table 6-1
Year Peaches
2014 \$14 per bushel
2015 \$12 per bushel
Pecans
\$9 per bushel
\$14 per bushel
2
11. Refer to the Table 6-1. Suppose that the typical consumer basket consists of 5 bushels of peaches and
10 bushels of pecans and that the base year is 2014. What is the consumer price index for 2015?
a. 80
b. 100
c. 125
d. 200
12. Refer to the Table 6-1. What was the inflation rate in 2015?
a. 4 percent
b. 5 percent
c. 20 percent
d. 25 percent
Table 6-3
In the country of Shem, the CPI is calculated using a market basket consisting of 5 apples, 4 kg of
chicken, 3 shirts, and 2 litres of gasoline. The per-unit prices of these goods have been as follows:
Year Apples Chicken Shirts Gasoline
2012 \$1.00
\$2.00
\$10.00 \$1.00
2013 \$2.00
\$1.50
\$8.00
\$1.50
2014 \$2.00
\$2.00
\$11.00 \$2.00
2015 \$3.00
\$3.00
\$14.00 \$3.00
13. Refer to the Table 6-3. What was the inflation rate, as measured by the CPI, between 2013 and 2014?
Assume the base year is 2013.
a. 27.9 percent
b. 34.1 percent
c. 47.0 percent
d. 48.5 percent
14. What does “substitution bias” in the consumer price index refer to?
a. replacing old goods with new goods in consumer purchases
b. replacing low-quality goods with high-quality goods in consumer purchases
c. replacing expensive goods with cheaper goods in consumer purchases
d. replacing old technology goods with new technology goods in consumer purchases
15. Laura bought word processing software for \$50. A year later, Laura’s twin brother Laurence bought
an upgraded version of the same software for \$50. Which problem in the construction of the CPI does this
situation best represent?
a. substitution bias
b. unmeasured quality change
c. introduction of new goods
d. income bias
16. Suppose that Canadian mining companies purchase German-made ore trucks at a reduced price. By
itself, what will this do to the GDP deflator and the consumer price index?
a. The GDP deflator and the consumer price index will fall.
b. The GDP deflator and the consumer price index will be unaffected.
c. The GDP deflator will be unaffected, but the consumer price index will fall.
d. The GDP deflator will fall, but the consumer price index will be unaffected.
3
17. Babe Ruth’s 1931 salary was \$80,000. The price index for 1931 is 16 and the price index for the
current year is 210. What would Ruth’s 1931 salary be equivalent to in the current year?
a. \$536,000
b. \$635,000
c. \$828,000
d. \$1,050,000
18. Which statement best describes the relationship between inflation and interest rates?
a. There is no relationship between inflation and interest rates.
b. The interest rate is determined by the rate of inflation.
c. In order to fully understand inflation, we need to know how to correct for the effects of
interest rates.
d. In order to fully understand interest rates, we need to know how to correct for the effects of
inflation.
19. Which statement best explains economists’ understanding of the facts concerning the relationship
between natural resources and economic growth?
a. A country with few or no domestic natural resources is destined to remain undeveloped.
b. Differences in natural resources have virtually no role in explaining differences in standards of
living.
c. Some countries can be rich mostly because of their natural resources, and countries without
natural resources need not be poor, but can never have very high standards of living.
d. Abundant domestic natural resources may help make a country rich, but even countries with
few natural resources can have high standards of living.
20. What would increase productivity, everything else being the same?
a. an increase in immigration
b. an increase in the number of hours of work per week
c. an increase in prices
d. an increase in physical capital per worker
21. Other things equal, how do relatively poor countries tend to grow?
a. They grow slower than relatively rich countries; this is called the poverty trap.
b. They grow slower than relatively rich countries; this is called the Malthus effect.
c. They grow faster than relatively rich countries; this is called the catch-up effect.
d. They grow faster than relatively rich countries; this is called the constant-returns-to-scale
effect.
22. According to the traditional view of the production process, how does output per worker change when
capital per worker increases?
a. It increases. This increase is larger at larger values of capital per worker.
b. It increases. This increase is smaller at larger values of capital per worker.
c. It increases. This increase is the same at all values of capital per worker.
d. It decreases. This decrease is larger at larger values of capital per worker.
4
23. Which statement illustrates an implication of investment from abroad?
a. Foreign investment makes poor countries poorer.
b. Foreign investment promotes economic growth.
c. Foreign investment makes rich countries poorer.
d. Foreign investment is only beneficial to investors.
24. The economic development minister of a country has a list of things she thinks may explain her
country’s low growth of real GDP per person relative to other countries. She asks you to pick the one you
think most likely explains her country’s low growth. What contributes to low growth?
a. strong private property rights
b. tariffs and quotas
c. encouraging foreign investment
d. low population growth
25. Which statement illustrates an important fact about population growth?
a. There are no substantial differences between rates of growth in population among countries.
b. In developed countries, population tends to grow slower than in developing countries.
c. Higher rate of growth in population implies higher productivity.
d. A country that increases its population growth rate will increase its economic growth rate.
26. Economists differ in their views of the role of the government in promoting economic growth.
According to the text, at the very least, what should the government do?
a. lend support to the invisible hand by maintaining property rights and political stability
b. limit foreign investment to industries that don’t already exist
c. impose trade restrictions to protect the interests of domestic producers and consumers
d. subsidize key industries
27. Suppose that in a closed economy GDP is equal to 15,000, taxes are equal to 2500, consumption is
equal to 7500, and government expenditures are equal to 3100. What is public saving?
a. –600
b. –500
c. 500
d. 600
28. Suppose that consumption is 7500, taxes are 1800, and government expenditures are 1500. If national
saving is 1500 and the economy is closed, what is the GDP?
a. 9500
b. 10,000
c. 10,500
d. 11,200
29. In a closed economy, what is private saving?
a. the amount of income that households have left after paying for their taxes and consumption
b. the amount of income that businesses have left after paying for the factors of production
c. the amount of tax revenue that the government has left after paying for its spending
d. the amount of tax revenue that the government has left after paying for transfers
5
30. When is a budget surplus created?
a. when the government sells more bonds than it buys back
b. when the government spends more than it receives in tax revenue
c. when private savings are greater than zero
d. when the government spends less than the tax revenue
31. What would most likely happen in the market for loanable funds if the government were to decrease
the tax rate on interest income?
a. The supply of and demand for loanable funds would shift to the right.
b. The supply of and demand for loanable funds would shift to the left.
c. The supply of loanable funds would shift to the right, and the demand for loanable funds
would shift to the left.
d. The supply of loanable funds would shift to the right, and the demand for loanable funds will
remain unchanged.
32. Suppose that Parliament were to repeal (e.g., revoke) an investment tax credit. What would most
likely happen in the market for loanable funds?
a. The demand for and supply of loanable funds would shift right.
b. The demand for and supply of loanable funds would shift left.
c. The supply of loanable funds would shift right.
d. The demand for loanable funds would shift left.
33. When the government runs a budget deficit, what is most likely to happen?
a. Interest rates become lower than they would be otherwise.
b. National saving gets higher than it would be otherwise.
c. Investment gets lower than it would be otherwise.
d. Government debt decreases by the amount of the deficit.
34. Assuming that other things remain the same, what effect does a government budget deficit have on
saving?
a. It increases both private and national saving.
b. It increases public saving but reduces national saving.
c. It reduces both public and national saving.
d. It reduces private saving but increases national saving.
35. Who would NOT be included in the labour force?
a. Karen, who works most of the week in a steel factory
b. Beth, who is waiting for her new job at the bank to start
c. Dave, who does not have a job, but is looking for work
d. Jeremiah, who plans to travel for a few months before looking for work
36. A foreign governmental statistics agency recently reported that there were 47.6 million people over
age 25 who had at least a bachelor’s degree. Of this number, 38.0 million were in the labour force and
35.9 million were employed. What was the unemployment rate in this group?
a. about 2.3 percent
b. about 5.5 percent
c. about 22.8 percent
d. about 55.1 percent
6
37. Rick loses his job and immediately begins looking for another. Other things equal, what happens to
the unemployment rate?
a. The unemployment rate decreases because he is not in the labour force anymore.
b. The unemployment rate increases because he is still in the labour force.
c. The unemployment rate decreases because he is still in the labour force.
d. The unemployment rate increases because he is not in the labour force anymore.
Table 9-1
This table shows the 2013 data for males and females aged 15 and over in the country of Bolivar.
Not in the Labour Force
Unemployed
Employed
Male
Female
Male
Female
Male
Female
45 million
35 million
5 million
5 million
85 million
65 million
38. Refer to the Table 9-1. What is the adult population in Bolivar?
a. 150 million
b. 160 million
c. 210 million
d. 240 million
39. According to your text, what is NOT a reason that actual labour markets experience unemployment?
a. unions
b. job search
c. flexible wages
d. minimum-wage legislation
40. Nancy is searching for a job that suits her tastes about where to live and the people she works with.
Laura is looking for a job that makes best use of her skills. Which statement best describes the nature of
Nancy and Laura’s unemployment?
a. Nancy and Laura are both frictionally unemployed.
b. Nancy and Laura are both structurally unemployed.
c. Nancy is frictionally unemployed, and Laura is structurally unemployed.
d. Nancy is structurally unemployed, and Laura is frictionally unemployed.
41. Tom is looking for work after school, but everywhere he fills out an application he is told that so have
lots of others. Simon has a law degree. Several firms have made him offers, but he thinks he might be
able to find a firm where his talents could be put to better use. Which category of unemployment do Tom
and Simon belong to?
a. Tom and Simon are both frictionally unemployed.
b. Tom and Simon are both structurally unemployed.
c. Tom is frictionally unemployed, and Simon is structurally unemployed.
d. Tom is structurally unemployed, and Simon is frictionally unemployed.
7
42. Aaron is the owner of a firm that produces wind power in southern Alberta. There are many such
firms in the area. Aaron decides that if he pays his workers a wage higher than the going market wage, his
profits will increase. What is a likely explanation for his decision according to the efficiency wage
theory?
a. The higher the wage, the less often his workers will choose to leave his firm.
b. The higher the wage, the lower will be the cost of obtaining needed supplies.
c. The higher the wage, the more he can charge for his wind power.
d. The higher the wage, the less competition will be in the industry.
43. Suppose that the reserve ratio is 9 percent and that a bank has \$2000 in deposits. What are its required
reserves?
a. \$100
b. \$120
c. \$140
d. \$180
44. Suppose a bank has a 6 percent reserve ratio, \$4000 in deposits, and it loans out all it can, given the
reserve ratio. Which of the following describes the bank’s assets?
a. It has \$800 in reserves and \$16,000 in loans.
b. It has \$240 in reserves and \$3760 in loans.
c. It has \$240in reserves and \$4000 in loans.
d. It has \$800 in reserves and \$3200 in loans.
Table 10-2
Last Bank of Panorama Springs
Assets:
Liabilities:
Reserves \$25,000
Deposits
Loans
\$150,000
\$175,000
45. Refer to the Table 10-2. If the reserve requirement is 30 percent, what is this bank’s reserve position?
a. It has \$29,000 of excess reserves.
b. It needs \$10,000 more in reserves to meet its reserve requirement.
c. It needs \$27,500 more in reserves to meet its reserve requirement.
d. It just meets its reserve requirement.
46. Which list contains only actions that increase the money supply?
a. lowering the bank rate; raising the reserve requirement ratio
b. lowering the bank rate; lowing the reserve requirement ratio
c. raising the bank rate; raising the reserve requirement ratio
d. raising the bank rate; lowering the reserve requirement ratio
47. Which statement best describes the process of open-market sales conducted by the Bank of Canada?
a. The Bank of Canada sells Treasury bills, which increases the money supply.
b. The Bank of Canada sells Treasury bills, which decreases the money supply.
c. The Bank of Canada borrows from member banks, which increases the money supply.
d. The Bank of Canada lends money to member banks, which decreases the money supply.
8
48. Which statement best describes the outcome of a decrease in the bank rate?
a. Banks will borrow less from Bank of Canada, so reserves increase.
b. Banks will borrow less from Bank of Canada, so reserves decrease.
c. Banks will borrow more from Bank of Canada, so reserves increase.
d. Banks will borrow more from Bank of Canada, so reserves decrease.
49. How can the Bank of Canada directly protect a bank during a bank run?
a. by increasing reserve requirements
b. by selling government bonds to the bank
c. by lending reserves to the bank
d. by increasing the overnight rate
50. Suppose the reserve ratio is 20 percent and banks do not hold excess reserves. Suppose the Bank of
Canada buys \$10 million of government bonds. Which statement best describes the effects of this openmarket operation?
a. Bank reserves increase by \$10 million, and the money supply eventually decreases by \$50
million.
b. Bank reserves increase by \$10 million, and the money supply eventually increases by \$50
million.
c. Bank reserves decrease by \$10 million, and the money supply eventually increases by \$50
million.
d. Bank reserves decrease by \$10 million, and the money supply eventually decreases by \$50
million.
Bonus question (true/false)
True/false The CPI is computed by finding the price of a market basket of goods whose contents vary
each year.
a. True
b. False
Short answer: The following questions are short answer. Briefly explain your answer. Clarity will be
rewarded.
Question1 (7.5 points)
Describe three ways in which a government policymaker can try to raise the growth in living standards in
a society. Are there any drawbacks to these policies?
9
Question2 (6 points)
“Some economists worry that the aging populations of industrial countries are going to start
running down their savings just when the investment appetite of emerging economies is
growing” (The Economist, May 6, 1995). Illustrate the effect of this phenomenon on the world
market for loanable funds. What is the effect on the interest rate and the quantity of loanable funds?
(explain with graph + words).
10
Question 3 (5.5 points)
a. Suppose that the United States cracks down on illegal immigrants and returns millions of workers
to their home countries.
In the countries to which the immigrants return, explain how employment, the wage rate, and
potential GDP would change. (explain with graph + words)
11
Question 3 (5 points)
b. In Korea, real GDP per hour of labor is \$22, the wage rate is \$15 per hour, and people
work an average of 46 hours per week. In U.S., real GDP per hour of labor is \$60, the
wage rate is \$40 per hour, and people work an average of 34 hours per week.
Draw a graph of the production functions in Korea and the United States using the
information from the data in the text above. Explain the difference in the two production
functions.
Answers multiple choice:
1c
2a
3a
4b
5a
6a
7b
8c
9a
10b
11c
12d
13a
14c
15b
16b
17d
18d
19d
20d
21c
22b
23b
24b
25b
26a
27a
28c
29a
30d
31d
32d
33c
12
34c
35d
36b
37b
38d
39c
40a
41d
42a
43d
44b
45c
46b
47b
48c
49c
50b
Bonus question is false
13
14
15
```
Arab people
15 Cards
Radiobiology
39 Cards
Radioactivity
30 Cards
Nomads
17 Cards
Ethnology
14 Cards | 5,354 | 21,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | latest | en | 0.95807 |
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## Engineering
Civil Engineering Chemical Engineering Electrical & Electronics Mechanical Engineering Computer Engineering Engineering Mathematics MATLAB Other Engineering Digital Electronics Biochemical & Biotechnology
problem 1:
Demonstrate with sketches four dissimilar effects of shrinkage? How they can be eliminated?
problem 2:
prepare down the factors which limit the use of blind risers.
problem 3:
Illustrate out the following repair techniques of castings and their application:
a) Epoxy fillers
b) Straitening
c) Thermit welding.
problem 4:
Illustrate with the help of appropriate sketch of the tilting type oil-fired furnace and describe its operation.
problem 5:
Make a distinction between inspection and quality control.
problem 6:
Illustrate out the principle, limitations and applications of Magnetic particle testing.
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### Describe what you learned about the impact of economic
Describe what you learned about the impact of economic, social, and demographic trends affecting the US labor environmen | 1,084 | 4,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-13 | longest | en | 0.836635 |
https://zbmath.org/?q=an:0861.06001 | 1,723,711,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00003.warc.gz | 829,834,225 | 10,807 | ## Formal concept analysis. Mathematical foundations. (Formale Begriffsanalyse. Mathematische Grundlagen.)(German)Zbl 0861.06001
Berlin: Springer. x, 286 p. (1996).
The authors state in the introduction to this book that Formal Concept Analysis, based on a mathematical interpretation of concepts and hierarchies of concepts, is an area of applied mathematics. While Formal Concept Analysis played a special role in the development of mathematical logic in the nineteenth century, the area of study has not received much attention until its revitalization around 1980. Since then a large number of published articles have given the field considerable breadth, and it has become necessary to publish a systematic treatise on the subject.
This book provides a systematic treatise for the mathematical foundations of concept analysis, and it interprets these foundations in the context of lattice theory. The content of the book is mathematical and does not attempt to explain the mechanisms of human conceptual thought processes.
The material is organized into eight chapters, starting with a brief introduction of ordered sets, lattices, complete lattices, and Galois connections in Chapter 0. The goal of Chapter 1 is to introduce formal contexts, formal concepts of contexts and concept lattices.
A formal context $$K:=(G,M,I)$$ consists of two sets $$G$$ and $$M$$ and a relation $$I$$ between $$G$$ and $$M$$. The elements of $$G$$ are interpreted as subjects and the elements of $$M$$ as properties, and $$gIm$$ means that the subject $$g$$ has property $$m$$.
For a subset $$A$$ of $$G$$, $$A'=\{m\in M\mid gIm \text{ for all }g\in A\}$$, and for a subset $$B$$ of $$M$$, $$B'=\{g\in G\mid gIm$$ for all $$m\in B\}$$. These two derivation operators give rise to a Galois connection between the power sets of $$G$$ and $$M$$.
A partial ordering for formal concepts is defined as follows: $$(A,B)\leq (C,D)$$ if $$A$$ is a subset of $$C$$ (which is equivalent to $$D$$ being a subset of $$B$$). With this ordering, the set of all formal concepts is a complete lattice, the concept lattice of the context $$(G,M,I)$$.
While this may sound like standard lattice theory with a different verbal spin, the examples given in the text are very interesting. They range from a concept lattice describing a Hungarian video about organisms and water to a concept lattice interpreting the sites and temples of Rome/Italy according to the number of stars awarded to them by various travel guides (Baedecker, Les Guides Bleus, Michelin, Polyglott).
In Chapters 2-8 the mathematical theory of concept lattices is rigorously developed. Topics include decomposition (subdirect, atlas, tensor), constructions (gluing, doubling, tensor), distributivity and modularity, automorphisms and measures. Each chapter concludes with a section of references that are relevant to the topics of the chapter, and a very extensive bibliography at the end of the book gives a complete overview of the literature (221 titles).
Reviewer: M.Höft (Dearborn)
### MSC:
06-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to ordered structures 03-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to mathematical logic and foundations 06B23 Complete lattices, completions | 750 | 3,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.895465 |
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# Copper Tubes - Uninsulated Heat Losses
## Heat loss from uninsulated copper pipes - dimensions ranging 1/2 - 4 inches.
Heat loss at various temperature differences between pipe and ambient air for 1/2" to 4" copper tubes - or pipes - are indicated below:
Note that it is temperature difference that is used - and that 1 oC of temperature difference equals to 1.8 oF temperature difference.
Copper Tubes - Uninsulated Heat Losses
Nominal boreHeat loss for the fluid inside pipe
(W/m)
Heat loss for the fluid inside pipe
(Btu/hr ft)
Temperature difference (oC)Temperature difference (oF)
(mm)(inches)223855406899
15 1/2 21 32 45 22 34 47
22 3/4 28 43 60 29 45 64
28 1 34 53 76 36 56 79
35 1 1/4 41 64 89 43 67 93
42 1 1/2 47 74 104 49 77 108
54 2 59 93 131 62 97 136
67 2 1/2 71 111 156 74 116 162
76 3 83 129 181 87 135 189
108 4 107 165 232 111 172 241
### Example - Heat Loss from a 2" Pipe
A 2" pipe with temperature 75 oC (167 oF) is located in a room with temperature 20 oC (68 oF). The temperature difference between the pipe and the ambient room is 55 oC (99 oF).
The heat loss from the pipe to the room can be estimated from the table (or chart) above to 131 W/m.
## Related Topics
• ### Heat Loss and Insulation
Heat loss from pipes, tubes and tanks - with and without insulation. Use of materials lke foam, fiberglass, rockwool and more.
• ### Insulation
Calculate heat transfer and heat loss from buildings and technical applications. Heat transfer coefficients and insulation methods available for reduction of energy consumption.
• ### Water Systems
Design of hot and cold water service and utility systems with properties, capacities, sizing of pipe lines and more.
## Related Documents
• ### Arithmetic and Logarithmic Mean Temperature Difference
Arithmetic Mean Temperature Difference in Heat Exchangers - AMTD - and Logarithmic Mean Temperature Difference - LMTD - formulas with examples - Online Mean Temperature Calculator.
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Standard specifications for seamless copper tube used in Air Conditioning and Refrigeration Field Services.
• ### ASTM B302 - Threadless Copper Pipes - Dimensions
Dimensions of threadless copper pipes according ASTM B302.
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Dimensions of seamless copper pipes according ASTM B42.
• ### ASTM B43 - Seamless Red Brass Pipes - Dimensions
Standard sizes specification for seamless red brass pipes.
• ### ASTM B837 - Seamless Copper Tubes for Natural Gas and Liquified Petroleum - Dimensions
Dimensions of seamless copper tubes for natural gas and liquified petroleum.
• ### ASTM B88 - Seamless Copper Water Tubes - Dimensions
Water and gas copper tubes according ASTM B88 - type K, L and M - imperial units.
• ### ASTM B88M - Seamless Copper Water Tubes - Metric Dimensions
Water and gas copper tubes according ASTM B88M - type K, L and M - metric units.
• ### BS 2871 Copper Tubes Table X, Y and Z - Working Pressures vs. Size
Working pressures of metric sized copper tubes according the BS (British Standard) 2871.
• ### Copper Tubes - Heat Losses
Heat loss from uninsulated copper tubes vs. temperature differences between tube and air.
• ### Copper Tubes - Pressure Loss vs. Water Flow
Water flow and pressure loss (psi/ft) due to friction in copper tubes ASTM B88 Types K, L and M.
• ### Copper Tubes Supports
Maximum space between supports aand clips for copper tubes.
• ### Copper Tubes Type K - Dimensions and Physical Characteristics
Physical characteristics of copper tubes ASTM B 88 type K.
• ### Copper Tubes Type K, L and M - Working Pressures vs. Size
ASTM B88 seamless copper water tubes - working pressures.
• ### EN 1057 Copper Tubes for Water and Gas in Sanitary and Heating Applications - Dimensions
Dimensions of EN 1057 Copper and copper alloys - Seamless, round copper tubes for water and gas in sanitary and heating applications.
• ### EN 12449 - Copper and Copper Alloys - Seamless, Round Tubes for General Purposes - Dimensions
Dimensions of copper tubes according EN 12449.
• ### Heat Transfer Coefficients in Heat Exchanger Surface Combinations
Average overall heat transmission coefficients for fluid and surface combinations like Water to Air, Water to Water, Air to Air, Steam to Water and more.
• ### Pipes - Insulated Heat Loss Diagrams
Heat loss (W/m) from 1/2 to 6 inches insulated pipesĀ - ranging insulation thickness 10 to 80 mm and temperature differences 20 to 180 degC.
• ### Pipes - Insulated Heat Loss Diagrams
Heat loss (W/ft) diagrams for 1/2 to 6 inches insulated pipesĀ - ranging insulation thickness 0.5 to 4 inches and temperature differences 50 to 350 degF.
• ### Pipes and Cylinders - Conductive Heat Losses
Conductive heat losses through cylinder or pipe walls.
• ### Pipes Submerged in Water - Heat Emission
Heat emision from steam or water heating pipes submerged in water - assisted (forced) or natural circulation.
• ### Solder Joints - Pressure Ratings vs. Temperature
Solder joints and their pressure ratings according ASME B16.18.
• ### Steel Pipes - Heat Loss Diagram
Heat loss from steel pipes and tubes - dimensions 1/2 to 12 inches.
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## Citation
• The Engineering ToolBox (2003). Copper Tubes - Uninsulated Heat Losses. [online] Available at: https://www.engineeringtoolbox.com/copper-pipe-heat-loss-d_19.html [Accessed Day Month Year].
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6.3.17
. | 1,524 | 6,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.823001 |
https://tapmyexam.com/explaining-correlation/ | 1,725,881,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00277.warc.gz | 543,863,756 | 17,536 | Correlation or statistical correlation is the relationship between two independent variables or bivariate data, either causal or non-causal. In simple terms, correlation is the relationship between two independent variables. It can be defined as the strength or weakness of a relation between two variables.
A good example of statistical correlation is the relationship between two things that are both known as “factors” of interest, for example, a person’s height and their weight. In the most general sense, the correlation is a measure of the strength or weakness of an observed relationship between two independent variables, but it often refers to the extent to which a set of data are “linearly correlated.” In a well-known example, a man and woman who are identical in age, gender, and race have been tested by scientists for a specific gene related to height. Each person is paired with another individual whose height, weight, and eye color have been randomly selected from a large pool of individuals. Over time, the researchers observe that each pair has the same genotype for the gene related to height, but that they are different in their weight and eye color.
This is a very simple example of how a correlation between a set of data (such as the person’s height and their weight) and its associated characteristics, such as their eye color, can affect the relationship between the two data sets. However, we can also think of a correlation as the strength or weakness of the association between the data sets being considered. The more closely they are related, the stronger the relationship.
The most significant relationship among the three sets of data involved above is the one between their respective weights and eye colors. Because of this relationship, there is a strong, almost exact, statistical correlation between these two variables, which implies that the correlation between them holds true whether the sets of data are causal or non-causal, and whether the effect is positive or negative.
Another common type of statistical correlation is the one between a set of observations and their effect on another set of observations. For example, if two people who are identical in age, gender, and race are asked to answer a question regarding their relationship with someone else, one of these persons will be found to have an increased likelihood of having a relationship with that other person than the other person would have. if the question had been posed to the other person. By definition, the observed effect of the observation on the other person is a dependent variable, which means that if the effect of the observation is large, the correlation between it and the other person is large too.
The simplest form of correlation is a correlation between the observed effect and its effect on the cause, which means that the relationship between the observed effect and the cause can be measured by the equation: r = | x – r(x). There are a number of forms of the equation, such as r = a and a simple calculation can be made using the slope or linearity of the line connecting the effect of the effect to the cause. If the slope is small, the correlation between the effect and the cause is zero. If the slope is very high, the correlation is very high, and if the slope is very low, the correlation is very low. If r is very close to zero, then the effect of the effect is much stronger than the effect of the cause, and if it is near zero, the effect of the cause is weaker than the effect of the effect.
Another form of correlation is a correlation between the data and its relationship to a reference variable. This relationship can be measured by the equation: r = R(X) where X is either causal or non-causal. If the relationship between the data is non-causal, the relationship between the data and the reference variable is not directly affected by the data. If the relationship between the data and the reference variable is causal, the relationship between the data and the reference variable will affect the relationship between the data and the reference variable when a causal relationship between them is present.
To summarize, we have talked about how correlation can be used to evaluate a relationship between data and a cause, or a correlation between a variable and its effect on another variable. We have also discussed how correlation can be used to analyze data between observations and its relationship to an effect, which is also known as the relationship between X and Y, or the relationship between r and the slope of the line connecting the observation to the causal relationship. We have also described three types of correlation that are based on direct relationships between observations and their relationship to an effect. | 910 | 4,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.972656 |
https://www.physicsforums.com/threads/streamlines-for-a-vector-field.963611/ | 1,553,281,426,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202688.89/warc/CC-MAIN-20190322180106-20190322202106-00111.warc.gz | 864,546,666 | 12,323 | # LaTeX Streamlines for a vector field
1. Jan 5, 2019
### jack476
I'm trying to use LaTeX to graph both the vectors of the electric field around a dipole and the field lines. So far I have a quiver plot of the vector field:
I obtained this by using the code
Code (Text):
\begin{tikzpicture}
\def \U{(x-1)/((x-1)^2+y^2)^(3/2) - (x+1)/((x+1)^2+y^2)^(3/2)}
\def \V{y/((x-1)^2+y^2)^(3/2)-y/((x+1)^2+y^2)^(3/2)}
\def \LEN{(sqrt((\U)^2 + (\V)^2)}
\begin{axis}
[domain = -3:3, domain = -3:3, view={0}{90}]
[
blue,
point meta={\LEN},
quiver={
u={(\U)/\LEN},
v={(\V)/\LEN},
scale arrows=0.2,
every arrow/.append style={line width=\pgfplotspointmetatransformed/1000 * 2pt},
},
-stealth,
samples=15,
] {0};
\end{axis}
\end{tikzpicture}
The code makes all of the vectors have the same length. In theory, it would also scale their thickness but as you can see nearly all of the vectors besides those right next to the charges at (-1,0) and (+1,0) have the minimum thickness, I guess because the field drops off in intensity so quickly. I would take that out but for some reason doing so makes it look just a little bit off.
Anyway, what I'd like to do now is also draw some field lines on top of this quiver plot, so that I can have a graphic that shows how the electric field vectors are tangent to the field lines. Is there any way I can do this elegantly in LaTeX without having to plot the field lines explicitly? | 444 | 1,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-13 | latest | en | 0.886193 |
https://www.indianguitartabs.com/chords/threads/deal-about-gmadd9-etc.17102/ | 1,513,506,238,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595342.71/warc/CC-MAIN-20171217093816-20171217115816-00690.warc.gz | 732,587,340 | 8,281 | # deal about GMadd9 etc.....
Discussion in 'Beginner's Q&A Forum' started by thedyood, Dec 26, 2005.
1. ### thedyoodNew Member
hi folks ....
could somebody explain to me the fundas of GMadd9 FM7 etc....
i mean what do add9,Major7(for eg) indicate?
tx
2. ### bjrLady of the Evening
As you probably know, the major chord is made up of the 1st,3rd and the 5th note of the corresponding major scale.
To make a major 7th chord, add the major seventh note of the scale to the chord.
G major 7th would be G,B,D and F# since F# is the seventh note of the major scale.
For a minor 7th chord, add the minor seventh note to the minor chord of the same scale.
eg- G min7th would be G,Bb,D,F
add9th means that you add the 9th note of the scale to the chord. For G add9th-
The notes would be G,B,D,A.
A is the 9th note because if you write the major scale, you shall find that the first 7 notes repeat themselves again and again at higher octaves. For a 2 ovtave G major scale, the notes are:
G A B C D E F# G A B C D E F# G
Count to the ninth note...it is the same as the second note, only one octave higher.
The difference between a major ninth chord and an add9th chord is that in the major 9th chord, you will be playing 5 notes : 1st, 3rd, 5th, major 7th and 9th note of the scale
In the add 9th chord, you'll just add the 9th note to the major chord so no major 7th will be played.
Just extend this theory a little bit and you'll be able to figure out the notes to any chord.
You asked for GMadd9th: The notes will be G B D A. If you do not play the B note, the same chord becomes a G sus2. A sus chord is one which replaces the 3rd note with the second or the 4th note of the scale.
alpha1 likes this.
3. ### thedyoodNew Member
care to elaborate about the major scale...tried to look up on the net...with less than satisfying results
tx | 527 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-51 | latest | en | 0.949586 |
https://docs.google.com/forms/d/e/1FAIpQLSf_RkDvNq_G5zxAN_y1F-anfDGW3WOLpeRMjFJpo00-ulyYLg/viewform?usp=sf_link | 1,618,619,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00527.warc.gz | 312,761,852 | 33,331 | parallel perpendicular
Let's see what you know.... Hopefully it's more than just a little ;)
Last name here *
Period *
Decide *
1 point
which one? *
1 point
which one is it? *
1 point
Parallel lines have... *
1 point
perpendicular lines have... *
1 point
what are they? *
1 point
Which? *
1 point
which? *
1 point
which one is it? *
1 point
which one is it? *
1 point
find slope *
1 point
Find *
1 point | 114 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-17 | latest | en | 0.929875 |
https://wattsupwiththat.com/2013/12/28/climate-as-a-heat-engine/ | 1,627,239,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00544.warc.gz | 625,659,242 | 162,038 | # Climate as a heat engine
Guest essay by Jan Kjetil Andersen
As Willis describes in his article on December 21, the atmosphere can be seen as a gigantic heat engine, i.e. a machine which convert thermal energy, namely temperature, into mechanical energy, namely wind.
It may seem a bit strange to view the weather system as a kind of machine and compare it with engineered constructs like an automobile engine, but it is sound physics because all such systems are bound by the same fundamental physical laws and they utilizes the same basic phenomena to create movement from heat.
A heat engine cannot convert heat directly to mechanical energy since that would break the second law of thermodynamics. What are needed are temperature differences. The greater temperature difference the greater effect of the machine. The amount of the energy in the temperature difference that is converted to mechanical energy is called the machines efficiency.
And here we have a very interesting, but less known fact of heat engines; the maximum theoretical efficiency decreases with increasing temperatures. This is interesting because it negates the conventional wisdom and often cited myth that a warmer climate leads to
more storminess, like the claim in the Guardian “a warmer planet has more energy to power stronger storms”, see http://www.theguardian.com/environment/2011/jun/27/climate-change-extreme-weather-2010.
Let us therefore take a look at the theoretical foundation of this effect. This is described by Carnot’s theorem.
Carnot’s Theorem says that the maximum efficiency drawn from a heat engine is the temperature difference between the warmest element and the coldest element divided by the temperature of the highest element.
Expressed as a formula it says: Emax = (Th-Tc)/Th.
Emax is the maximum efficiency
Th is the high temperature element measured in Kelvin
Tc is the cold temperature element measured in Kelvin.
The Carnot cycle is an ideal reversible cyclic process involving the expansion and compression of an ideal gas, which enables us to evaluate the efficiency of an engine utilizing this cycle.
Each of the four distinct processes are reversible. Using the fact that no heat enters or leaves in adiabatic processes we can show that the work done in one cycle, W = Q1 – Q3 where Q1 is the heat entering at temperature TH in the isothermal process A -> B and Q3 is the heat leaving at temperature TC in the isothermal process C -> D.
For an interactive demonstration of the Carnot heat engine cycle, courtesy of the University of Virginia, click on the image:
Three important effects can be seen from Carnot’s theorem. The first is that a temperature difference is a necessary condition for converting heat energy to mechanical energy such as wind.
The second effect is that even if we had a perfect heat engine with zero internal friction; it would not achieve anything close to 100% efficiency. The maximum theoretical efficiency for a heat engine operating between 300 K and 600K is for example 50%. The efficiency of a real machine would of cause be considerably lower.
This is why our car engines only operate at about 25% efficiency. The warm element for a car engine is the exploding fuel inside the cylinders and the cold element is the air intake.
The best coal fired power plants have about 40% efficiency and the best gas powered about 55%. The cold elements for those plants are the coolant water, and those with highest efficiency utilize cold seawater as coolant.
Warming gives less efficiency
The third effect is as mentioned above, that, for a given temperature difference between the warm element and the cold element, the efficiency will decrease if both elements heat equally much. On cold days one can see a discernible effect of this in car engines; because the air intake is colder, the engine gives somewhat more power and higher efficiency.
This is also why some turbo charged engines have intercoolers. The turbo gives higher effect, but a non-intentional side effect is that it also increases the temperature in the air intake which will reduce the efficiency. The intercooler reduces the temperature increase introduced by the turbo.
The same effect applies to the wind formations in the atmosphere. Consider the summer temperature in the northern hemisphere; the cold element is the Arctic with a temperature of approximately 0 Celsius and the warm element is in the tropics with approximately 35 Celsius.
The Carnot theorem gives a maximum efficiency in this temperature range of 11.36%. If the temperature increased with 1 Celsius all over the globe, i.e. the difference changed to 1 Celsius in the Arctic and 36 Celsius in the tropic, the maximum efficiency would sink to 11.32%.
This is a minuscule difference, but the point is that it is a decrease, not an increase as the conventional wisdom will have it.
Less temperature differences on the surface
In addition to the effect of higher overall temperatures, the temperature differences will also be smaller. It is quite uncontroversial that the largest effect of global warming is on the cold polar winters and the smallest on the hot tropical summers.
This means that both the overall heating and the reduced temperature differences should contribute to less storminess.
However, to be fair, this is not all there is to this. Some climate models tell that the temperature differences in the upper troposphere will increase and this may have larger effect than both the reduced differences on the surface and the higher temperatures.
No settled science there.
Article Rating
Inline Feedbacks
Richard Sharpe
December 28, 2013 8:26 am
Some climate models tell that the temperature differences in the upper troposphere will increase and this may have larger effect than both the reduced differences on the surface and the higher temperatures.
Models are not science until they have real measured data to base them on. After that, they need to compare the output of their models to what is really happening.
Have they done all this?
December 28, 2013 8:30 am
More models…..the climate is a convection system with a million variables including local thermodynamics. This complexity is immune to modeling. It would be refreshing to hear from quackitists and quackademics that we really dont have a clue about climate or weather and that most likely a trace chemical necessary for life mostly emitted by gaia has not much to do with anything….
MJPenny
December 28, 2013 8:33 am
As an engineer with some training in meteorology, the claim of increased strength and number of storms caused by global warming with more warming at the poles has always bothered me. Thank you for a well written article explaining thi issue.
tonyb
Editor
December 28, 2013 8:33 am
Interesting article, thank you. You said;
‘And here we have a very interesting, but less known fact of heat engines; the maximum theoretical efficiency decreases with increasing temperatures. This is interesting because it negates the conventional wisdom and often cited myth that a warmer climate leads to more storminess, like the claim in the Guardian “a warmer planet has more energy to power stronger storms”
As one of the few who seems to be still carrying out primary historical climate research at such places as the Met office archives, I can confirm from reading tens of thousands of weather observations back to the 11th century that the most violent extremes occur during the periods of greatest cold during the intermittent Little Ice age. Whilst snow and ice feature, so do violent storms and especially very long periods of calamitous flooding brought about by prodigious amounts of rainfall.
The temperature gradient between the poles and the tropics is surely greatest during cold periods so more energy will be generated than when the temperatures are more equal, as during the current benign period and also during the MWP which appears to have experienced mostly settled weather.
tonyb
norah4you
December 28, 2013 8:34 am
Don’t count on it. Why should they be willing to use measured data? Measured data can and will prove them wrong!
Leonard Weinstein
December 28, 2013 8:34 am
While the points of this post are correct, they left out one feature: The evaporation and condensation of water. Warmer air is capable of holding more water vapor, and the transport of energy by phase change is potentially capable of changing the energy driven dynamics of just pressure driven flow (wind). However, in fact the increase in moisture has not followed the expected level due to the temperature increase (it has not maintained average relative humidity as temperature increased, especially at higher altitudes), so the net result is even more unclear.
David, UK
December 28, 2013 8:49 am
Some climate models tell that the temperature differences in the upper troposphere will increase…
Indeed, as evidenced by the infamous hotspot that was observed back in… oh wait.
Doug Huffman
December 28, 2013 8:52 am
Actual heat engines, to which Carnot’s physics apply, are artificial man-made technological systems that can be isolated and all input-connection-outputs be known; not so for a natural system in a reality that Mandelbrot called fractally complex. The essay is an interesting but microscopic examination of an open/un-isolable system.
Marc77
December 28, 2013 8:53 am
The upper atmosphere might be cold, but it cannot be used to drive a convection cell. If you fill a balloon of air in the upper atmosphere and bring it down to ground level, the air in it will be warmer than the air at ground level. The increase in pressure would cause the warming.
Also, warmer air might be able to hold more moisture, but it is also unlikely to hold less. The portion of water vapor that does not condensate can be considered as a non-condensing gas. You need cold air to condensate water vapor, so water vapor might act less like a condensing gas on a planet with warmer poles.
noaaprogrammer
December 28, 2013 9:02 am
What drives the jet streams? Are there any data collections on the velocity of jet streams over time?
December 28, 2013 9:02 am
Have they done all this?
I don’t think so. It would lead to inconvenient truths.
December 28, 2013 9:04 am
Let’s don’t forget another heat exchange engine operated by the oceans’ thermohaline circulation.
Richard Sharpe
December 28, 2013 9:07 am
And here we have a very interesting, but less known fact of heat engines; the maximum theoretical efficiency decreases with increasing temperatures.
It does not seem to me that the efficiency is all that important compared with the amount of energy in the atmosphere?
Is it that the reduced efficiency simply results in more turbulent flows, thus diffusing that energy out more?
December 28, 2013 9:07 am
“If you fill a balloon of air in the upper atmosphere and bring it down to ground level, the air in it will be warmer than the air at ground level”
Assuming you mean the upper troposphere that would be because the air was originally humid and cooled during uplift at the moist lapse rate.
If you then capture dried out air and bring it down then it warms at the dry lapse rate which is faster and so it ends up warmer than the surface air.
I have argued elsewhere that the mechanical warming on the descent phase of the convective cycle is what warms the surface above the S-B prediction and not DWIR (Downward Infrared Radiation)
December 28, 2013 9:28 am
My understanding was that a car’s engine produces more power with a colder intake air because colder, denser air contains more oxygen and allows more gasoline to be completely burned during the cylinder explosion. Of course, it would make sense that thinner, warmer air cannot expand as much, thus produces less power.
TB
December 28, 2013 9:35 am
Jan:
I know what you’re getting at.
But I feel the comparison of the Earth’s climate system with a Carnot engine is inappropriate.
A Carnot engine is a closed system – that is, it deals with an energy differential between two sources, these closed off from the outside. It acts by transferring energy from a warm region to a cool region of space and, in the process, converting some of that energy to mechanical work.
The Earth’s climate is an open system – with a constant input of energy from the Sun and a constant emission of that energy to space after conversion into terrestrial IR. Therefore the efficiency of the conversion does not matter.
To say that the earth will become “stormier” in a warmer world is (Meteorologically) correct. (supposing an unchanging LR). But there are exceptions.
A warmer world will have more evaporated water to release LH and therefore convective cells will have more power available to push up through the atmosphere to counter that LR.
In subtropical zones where Coriolis force can act on systems of thunderstorms, and in combination with westerly travelling waves in the upper air will make Hurricanes, Typhoons, Cyclones more powerful, as they derive their energy from SST’s, which will be higher in a warmer world.
The argument is applicable to tornadoes as well with, in the US, more moisture drawn up from the Gulf being overridden by dry Arctic sourced air – then the CAPE (convectively available energy) available will be greater. The spin will still be sourced from a backing/increasing wind with height over the top/through the Cumulonimbus cloud.
It is true though that if we accept that a warming world will not warm uniformly, and that the Arctic would warm more, then this would reduce the thermal contrast between Equator and Pole and so make less energy available to Earth’s “heat engine” via a weakening of the Polar jet-stream. However though this would serve to weaken the central pressure and hence wind gradients of mid-latitude Lows, due to the increased “waviness” of the jet then we will be more prone to “stuck” weather patterns giving regionally disparate flood/cooler and drought/hotter weather. These regional zones would shift around but would make rainfall patterns much less predictable.
Likewise Monsoon rains are likely to be disrupted via differences in ocean heating via current flow and they will contain more rain when they arrive.
Paul Vaughan
December 28, 2013 9:35 am
• Concise overview of heat engines = p.433 [pdf p.10] here:
Sidorenkov, N.S. (2005). Physics of the Earth’s rotation instabilities. Astronomical and Astrophysical Transactions 24(5), 425-439.
• Elaboration on heat engines = section 8.7 (begins on p.175 [pdf p.189]) here:
Sidorenkov, N.S. (2009). The Interaction Between Earth’s Rotation and Geophysical Processes. Wiley.
http://imageshack.us/a/img850/876/f0z.gif (credit: JRA-25 Atlas)
Larry Ledwick
December 28, 2013 9:39 am
Stephen Wilde says:
December 28, 2013 at 9:07 am
“If you fill a balloon of air in the upper atmosphere and bring it down to ground level, the air in it will be warmer than the air at ground level”
Assuming you mean the upper troposphere that would be because the air was originally humid and cooled during uplift at the moist lapse rate.
If you then capture dried out air and bring it down then it warms at the dry lapse rate which is faster and so it ends up warmer than the surface air.
I have argued elsewhere that the mechanical warming on the descent phase of the convective cycle is what warms the surface above the S-B prediction and not DWIR (Downward Infrared Radiation)
Your example leaves out one important element — the latent heat of the water vapor that “disappears” as the air dried out, is actually simply moving heat energy to a new location.
A practical example of this case is the issue with down slope winds like the chinook winds we have here in Colorado which can cause “local” warming on the front range. In the winter time these winds doming out of the west fall several thousand feet (about 7,000 – 8,000 ft) after crossing the continental divide dropping into the Denver basin, and can warm 20-30 degrees causing 40-50 degree days in the front range corridor while just 30-40 miles east of Denver temps are in the 20’s and 30’s.
From the local perspective the air has “warmed” but you are forgetting the cold side of the equation. The reason these winds warm on the down slope is that they had a significant moisture content as the were upslope winds on the western side of the continental divide and during that orographic lifting the moisture dropped out as snow. This released a lot of latent heat of freezing as the ice crystals formed and that is the heat that shows up on compression as dry air.
If the wind has low humidity or the temperatures on the western slope are not cold enough to cause rain or snow the winds in the lee of the mountains are bone chilling cold bora winds not warm Chinook winds. The warming is local there is an equal and opposite cooling captured in the snow pack on the western slope that will not be released back into the atmosphere until spring when the snows melt.
http://en.wikipedia.org/wiki/Orographic_lift
If that latent heat of condensation and freezing is radiated away from the cloud tops to deep space at the top of the convection column, and has left the the system and the down drafts will be much cooler than the air that went up. This happens every day in thunderstorms. You have 80+ deg F moist air rising int he updraft, and cold rain and hail falls out as the air is lifted, but much of that latent heat is released at high altitude as IR radiation to space. The resulting cold down drafts will be in the 50 deg F range and the combined result of the cool down draft and the chilling effect of the cold rain and melting hail is cooler (has less total heat content) than the original warm moist air that entered the storm in the updraft phase.
You have to consider the entire cycle and due to specific circumstances sometimes a drop in elevation can result in heating of the air and at other times a net cooling. It all comes back to the water cycle and where and under what conditions the moisture gains or loses energy as latent heat.
December 28, 2013 9:50 am
Interesting article for discussion, though I too think the car analogies are dubious. Combustion in the chamber is not really what Carnot cycle is about.
Also, efficiency dropping does not mean energy available is dropping. It could be less efficient while still producing more energy.
The equator / polar difference is probably dropping sufficiently to establish less energy available but that is not established by arguments shown here.
TimC
December 28, 2013 9:55 am
Warming gives less efficiency: doesn’t this classically describe a (temperature) negative feedback process?
December 28, 2013 10:00 am
Reduced efficiency would mean the working fluid has to circulate faster to achieve the same heat transport.
Probably a significant error in this simplistic analysis is that Hadley cell is not the only heat path to consider. Much of the heat transported from tropical surface to tropopause leaves the system by radiation.
Also strong negative feedbacks in the tropics cut down the heat entering the lower climate system.
considering the ‘heat engine’ aspects in isolation may not be informative, at least without further discussion.
December 28, 2013 10:07 am
Is it that the reduced efficiency simply results in more turbulent flows, thus diffusing that energy out more?
Turbulence requires energy. It is a part of the work done in the climate system. Less energy for work (Carnot) = less turbulence In a regular heat engine (turbine, diesel etc.) turbulence is counted as a loss from Carnot efficiency. In the case of the atmosphere it is counted as “useful” work.
December 28, 2013 10:09 am
When will the study of climate change ever yield results? We’ve spent hundreds of billions of tax payer dollars over decades studying climate change. The failure of scientists to prove any conclusive results pertaining to climate change science shows we should stop throwing good money after bad. The only objective conclusion that can be drawn is; We should stop giving quack and incompetent scientists our taxpayer money to study climate change.
Tsk Tsk
December 28, 2013 10:11 am
” the maximum theoretical efficiency decreases with increasing temperatures. ”
This is a misleading statement. You get the math correct a paragraph or two later, but I can think of cases where your statement is wrong. Depending on the starting temperature differential, if Th is increasing at a higher rate then Tc, then efficiency can still go up. Ex. Th increasing at 3 times the rate of Tc: Th=100, Tc=50 e=50%, Th=400, Tc=150 e=62.5%.
I think what you meant to say is that if temperatures are increasing equally, then efficiency goes down. And for any case where Tc is increasing faster than Th efficiency goes down along with a fair number of cases where Th is increasing faster but not fast enough than Tc.
Efficiency also doesn’t tell the whole story. You should also state explicitly that the mechanical work is a function of the efficiency and the amount of heat transferred. But since the source for all of our heat is the sun and that is “constant” then a lower efficiency will result in less mechanical work done.
Finally, the cold sink in your car example is the air at the exhaust (and ambient air around the engine) and not the intake. They happen to come from the same pool but in principle they don’t have to. I could have a tank of hot air that I inject into the cylinders, expand, and exhaust and still get work from the system. Turbo’s have intercoolers because colder air is denser air, and Turbo’s are all about recovering some of the waste heat to better aspirate the engines.
p.s. You also have an “of cause” buried in there.
NZ Willy
December 28, 2013 10:14 am
Don’t overlook the Coriolis force as a primary driver of large-scale wind. These rotational forces are huge and the energy put into driving the winds and ocean currents slow down the Earth’s rotation over time — the day was just 21-22 hours long in the time of the dinosaurs.
December 28, 2013 10:28 am
[snip – chemtrail garbage – policy violation -mod]
mikethemachinist
December 28, 2013 10:32 am
your car has a mass air flow sensor that measures the mass of air flowing through the intake system. with cooler denser air the computer increases the length of time the injectors stay open allowing more fuel to the cylinders. the turbo pushes more air in than atmospheric pressure alone but still benefits from cooling the compressed air to increase density. more air =more fuel=more power
DonV
December 28, 2013 10:42 am
I grew up on the southern edge of the Sahara and one memory I have of my weekly chores was refilling the kerosene tank at the bottom of our refrigerator, and cleaning the long glass mantle of its weekly accumulation of black soot. Later in life I learned that this refrigerator was an absorption unit and that the “working fluids” were anhydrous ammonia, hydrogen gas, and water. The entire working fluid system was sealed and had no moving parts, yet it continuously provided ice in the freezer section and cold air in the refrigerator section off of heat generated by burning a miniscule amount of kerosene (about a quart/day). This type of “engine” achieves its remarkable cooling by both phase change and multiple Carnot cycle expansion/contraction/pressure changes of the working fluids. At the heart of the engine is the “expansion” and boiling of liquid ammonia/hydrogen gas mixture inside the freezer coils that created the chilled freezer surfaces. This gas mixture is condensed and separated by mixing with water. Since ammonia is highly soluble in water but hydrogen is not, the ammonia dissolve (absorbs) into the water but the hydrogen does not and collects above it as purer hydrogen gas. The water/ammonia mixture is then collected at the bottom of the refrigerator and percolation “boiled” using the heat of a kerosene fired flame to boil the ammonia out of the water. The “hot” ammonia gas is cooled at the top of the refrigerator by passing it through a heat exchanger, and the liquid ammonia is mixed with hydrogen gas again, collected and piped back into the freezer compartment where it begins the cycle again. The mixing and “asborption” of this type of “heat engine” is capable of extractng heat from one location (freezer coils) and moving it to another (ammonia condensing heat exchanger) using even greater heat as the energy source that drives all the cycles. My point in telling this story is that, just as Leonard Weinstein has stated, the addition of just a single phase change, where the volume and density of the working fluid changes significantly, signficanty increases the efficiency of the engine. In the case of the climate system the primare working fluid is not the air – rather it is water! The addition of two phase changes (from vapor all the way to solid), describes a thunderstorm and accounts for astoundingly high energy transfer!
December 28, 2013 10:47 am
Richard Sharpe says:
December 28, 2013 at 9:07 am
It does not seem to me that the efficiency is all that important compared with the amount of energy in the atmosphere?
Is it that the reduced efficiency simply results in more turbulent flows, thus diffusing that energy out more?
Richard
The total amount of energy is not of any use since energy cannot be converted directly to mechanical energy. What matters are the temperature differences, or said in another way, the excess energy in the parts of the atmosphere that are higher than the lowest temperature on the planet.
Since turbulence is also a form of mechanical energy, the turbulence will also be lower. The effect of higher overall temperature is that the warm parts simply warm up the colder parts without creating as much movements in the air as in colder conditions.
December 28, 2013 10:49 am
Correction:
Automobiles MAY have had better “efficiency” on cold days in the era of carboraters …because the mixture was more proper.. BUT, since your engine coolent is controlled by a thermostat, the “sink” temp for an Otto Cycle, Auto Engine is almost completely constant the year around. (Whereas for a power plant, cooled with river water, for example…when the river goes to 35 F, the sink DOES allow a greated Delta T and the efficiency is up!)
Now, for the reality of the Otto cycle, the ambient temperature ALSO influences the amount of energy needed to go from AMBIENT to the combustion temp. THE GREATER ENERGY USED HEATING THE air up as it comes into the combustion cycle, causes a NET ENERGY LOSS (it takes away from the heat conversion to mechanical. Therefore, Otto cycle engines would tend to LOSE a little efficiency during colder weather.
DON’T LOOK FOR IT, as you are idling, warming up, stoping and starting MORE and your net mileage is going to go down, no matter!
chris y
December 28, 2013 10:54 am
TB says-
“…will make Hurricanes, Typhoons, Cyclones more powerful, as they derive their energy from SST’s, which will be higher in a warmer world.”
Kerry Emanuel of MIT described cyclones using a Carnot cycle model in the August 2006 issue of Physics Today. The SST at 300K is the hot end, the troposphere (15 km) at 200K the cold end. The cyclone wind speed v can be estimated with this model and other assumptions, giving
v = sqrt((Ts – To)E/To)
where Ts=300K, To=200K and E is a variable related to coupling efficiency of energy between the ocean surface and the atmosphere.
There are at least 3 interesting conclusions one can take from this simple model.
1. If the sea surface temperature warms by 1K, then Ts=301K, To=200K, and the wind velocity is predicted to increase by 0.5%. There is no way to discern this tiny change.
2. If the climate models are to be believed, then the tropical troposphere should warm more than the surface. For example, if the sea surface warms 1C, the troposphere hot spot should warm 1.5C. That is, Ts=301K and To=201.5K. Then, the cyclonic wind velocity is predicted to DECREASE by 0.6%. Go figure…
3. Because we know that sea surface temperature changes of 5C (eg from 25 C to 30 C) can have a huge impact on cyclonic wind speeds (provided other conditions like wind shear are also just right), the Carnot model is useless for predicting wind speeds.
So, the Carnot model for cyclones predicts that a warmer world will have stronger cyclones, weaker cyclones or ‘this model is for entertainment purposes only.’
Ed_B
December 28, 2013 11:21 am
Chris Y says:
“So, the Carnot model for cyclones predicts that a warmer world will have stronger cyclones, weaker cyclones or ‘this model is for entertainment purposes only.’”
Well said. The Carnot model is for a closed system. Any application beyond that is just funnin it.
Mac the Knife
December 28, 2013 11:24 am
Jan,
Excellent graphics accompanying clear discussion! Thanks for a useful and interesting primer on heat engine ‘basics’ and how they relate to our planetary weather and climate. Many here will niggle and quibble about various aspects of your treatise but I truly hope they do not lose sight of its usefulness as an education tool for the physics-impaired layman we all know that seek greater understanding of what drives our weather and climates.
Well Done!
MtK
Eric Eikenberry
December 28, 2013 11:38 am
It’s not the cooler air which improves a gas engine’s efficiency, it’s the density. Colder air is more dense, which means it can be mixed with more fuel, increasing the rotational torque generated by the piston’s downward movement. Likewise, turbochargers increase the density of the air in the cylinder, even though they heat the air during compression. This is why an intercooler is used, to reduce the temperature of the pressurized air back down to the ambient intake. Drag and drift guys have been known to even spray Nitrous Oxide over the outside of the intercoolers, because as it evaporates it can carry away dramatic amounts of heat, increasing the intercoolers’ efficiency.
Carbureted engines did not adjust well for air density in colder climates, other than the natural increase in venturi effect efficiency as density improved, so they typically were set to run rich, and just leaned out as the air got colder, increasing power slightly for the air they could ingest. Up to a point, leaning out the air/fuel ratio can increase power slightly, but can increase the chance of “knock” (pre-detonation of the mix in the cylinder before the piston is close to top dead center). Modern fuel injected motors run much leaner right from the start, and the sensors keep the motor on the ragged edge for both fuel efficiency and power. When the air temps go down, power does increase slightly, but within a range of only 3-5%. Hardly enough to be felt by the seat of your pants for most drivers. Modern fuel systems can adjust up to a 7-8% range within their pre-programmed parameters, which generally covers all situations outside of a really bad tank of gas.
donald penman
December 28, 2013 11:47 am
If the earth has to be seen as a heat engine then it has to be seen as two heat engines the northern hemisphere heat engine and the southern hemisphere heat engine.The range of seasonal solar forcing change is greater at higher latitudes for each hemisphere than at the equater as the earth tilts toward the sun and away from the sun,this range would not increase or decrease in a uniformly warming world it would just shift upwards (summer and winter).The important question here is if the world is warming or not and if it is then the range will shift upwards in both hemispheres and that is what we should see the atmospheres in both hemispheres should hold more water vapour season by season if the earth is warming.
phlogiston
December 28, 2013 11:50 am
Word search:
Chaos: 0 / 0
Dissipative: 0 / 0
Nonlinear: 0 / 0
(Far from) equilibrium: 0 / 0
Turbulence: 0 / 0
Null points
Jean Parisot
December 28, 2013 12:05 pm
I think DonV has an important point: ” In the case of the climate system the primare working fluid is not the air – rather it is water!.” It’s not CO2 it’s the sun and water.
phlogiston
December 28, 2013 12:07 pm
Stephen Wilde on December 28, 2013 at 9:07 am
“If you fill a balloon of air in the upper atmosphere and bring it down to ground level, the air in it will be warmer than the air at ground level”
An air filled balloon in the upper atmosphere, if brought down to ground level would become shrivelled and almost empty of air, the tiny amount of air present at ground atmospheric pressure would quickly equilibrate with the rubber walls of the balloon, destroying any heat information. Suggest revision of experimental design.
phlogiston
December 28, 2013 12:08 pm
🙂
Matt G
December 28, 2013 12:30 pm
oaaprogrammer says:
December 28, 2013 at 9:02 am
What drives the jet streams? Are there any data collections on the velocity of jet streams over time?
The jet streams are caused by polar air clashing with sub-tropical air, so the bigger difference between the two, the stronger the winds.
http://www.srh.noaa.gov/jetstream/global/images/jetstream3.jpg
There are data collections.
http://squall.sfsu.edu/crws/jetstream.html
Bill H
December 28, 2013 12:34 pm
As many have pointed out the Temperature Differential is the key component to our storminess and severity of storms. During the last 30 years we have had a basically stable temperature differential as slight warming was occurring in the short term trend. This warming affected the polar regions and REDUCED the temperature differential. The last 17 years with a flat gradient in temperature changes proved out to continually calm (storm intensity) in both our tropical storms and our over land tornadoes as well.
The last three years has created great turbulence in the polar regions as massive amounts of cooling are occurring. The oceanic heat reserve has blunted the last three years even though Antarctica set 200+ low temp records and increased ice mass to +2 standard deviations. Our temperature differential is growing but the ocean heat reserves have blunted the effects. The Arctic usually follows the antarctic by about two years. Given the number of record lows already we are well on our way to ice rebound.
As the ocean buffer equalizes at the new lower temp, watch for the storms to increase in severity world wide. Add to that the ENSO, ADO, PDO, among other oscillations being cold or neutral this process will become quite fast..
Solar input into our system is low. The energy stream introduced into our heat engine has slowed. While this will not generally affect the tropical zones the polar regions are greatly affected causing our global temperature differential to rise. The mechanical (convection) systems which allow the heat to move from the equatorial regions out to the poles will become more streamlined (less chaotic then they are today) as the Ocean temps re-balance to the new lower energy input. When that happens I expect our storm tracks will again return to the mid latitudes and be a little more predictable.
While the thermodynamics of a steam engine are somewhat comparable and the physics applies in most respects its not a good fit in my opinion.
Robert Austin
December 28, 2013 12:35 pm
NZ Willy says:
December 28, 2013 at 10:14 am
Don’t overlook the Coriolis force as a primary driver of large-scale wind. These rotational forces are huge and the energy put into driving the winds and ocean currents slow down the Earth’s rotation over time — the day was just 21-22 hours long in the time of the dinosaurs.
The so called Coriolis “force” is simply the direction that objects (and air) must follow in moving across the surface of the earth if angular momentum is to be conserved. Thus it would actually require an external force and thus energy to divert objects from following the trajectories across the Earth’s surface that are a manifestation of the Coriolis effect. Would you claim that a figure skater performing a spin requires additional external rotational force when increasing rotational velocity by tucking her extended arms to her torso?
Bill H
December 28, 2013 12:53 pm
Jan,
Your graph on Hadley cells is rather useful in this debate as the “norm” for polar Jet intrusion is about the 40th parallel During the last three years that intrusion has dropped to the 25th parallel. This is a good indicator of the pressure differential created by the temperature differential in the polar regions.
This compresses those median Hadley cells limiting major hurricane formation to a narrow corridor and allows cold front disruption over land limiting tornado formation while increasing the areas where they can form. This should sound very familiar as this is exactly how the last three years (or better) have played out.
Rather an interesting way to show just how it is temperature differential which is.driving the climatic engine. Just one more cylinder in a very complex climactic engine.
kalsel3294
December 28, 2013 12:56 pm
re Jean Parisot says:
December 28, 2013 at 12:05 pm
I think DonV has an important point: ” In the case of the climate system the primare working fluid is not the air – rather it is water!.” It’s not CO2 it’s the sun and water.
I agree, and believe it is point of phase changes of water that ultimately give us the climate we know.
If the climate was determined by the phase changes of CO2 the climate would be very different.
WeatherOrNot
December 28, 2013 12:57 pm
This article could be turned into a hypothesis for an actual study or experiment to support or disprove the theory. I’m sure all the data needed is already out there or could be gathered for free for an experiment. Design it in a repeatable manner and it could provide the basis for some good science over a few years.
December 28, 2013 12:59 pm
Jean Parisot says:
December 28, 2013 at 12:05 pm
I think DonV has an important point: ” In the case of the climate system the primare working fluid is not the air – rather it is water!.” It’s not CO2 it’s the sun and water.
**********************************************************************************************************************
No the primary working fluid is the air. The oceans are a moderator not a driver. For example take the oceans out of the equation. You will find that the poles will be colder and the equator will be hotter. This will give a greater temperature difference and much stronger winds. You can see this at the moment where the South Pole is colder than the North Pole since there is no water anywhere near the SP to moderate the temps there. You will probably find that Central Siberia is colder than the NP for the same reasons.
TB
December 28, 2013 1:01 pm
Matt G says:
December 28, 2013 at 12:30 pm
oaaprogrammer says:
December 28, 2013 at 9:02 am
What drives the jet streams? Are there any data collections on the velocity of jet streams over time?
The jet streams are caused by polar air clashing with sub-tropical air, so the bigger difference between the two, the stronger the winds.
http://www.srh.noaa.gov/jetstream/global/images/jetstream3.jpg
There are data collections.
http://squall.sfsu.edu/crws/jetstream.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Matt:
To be more complete with your explanation.
Jet streams are driven by thermal contrast.
Northern latitudes are cold (NH) and as we come south the Sun’s elevation causes higher temps. There is a point where this temp contrast manifests itself as a maximum. This is the Polar jet-stream.
Cold air is more dense. Therefore as one rises through it, the level at which, say 300mb (~30000ft) is indicated on your barometer/altimeter will come sooner than if you rise upward through a warm air-mass.
Look at a chart showing 300mb contours. They will depict a tight gradient. Where closest together lies the strongest winds (analogous to a surface pressure gradient).
Now to a parcel of air within the warm air (to south of the gradient usually) will have more air above it that a parcel of air to the north in the colder air-mass. Ie it is at a higher pressure. So what does it do? It moves toward low pressure (usually N but always towards cold air). Now we have something called Coriolis force that ensures that air moving over the Earth’s surface carries the momentum of the Earth beneath it. So for air moving north (in a W’ly jet) it will be moving (relatively) E>W at a greater rate than the land surface blow it. Similarly air moving N>S will be moving slower than the earth’s surface beneath it. In both cases it turns to the right relative to the earth’s surface.
As the SH is a mirror image, air there turns left on moving from warm to cold aloft. These are thermal winds and although moved by pressure differentials they owe their origin to thermal differential of adjoining air masses.
A strong jet will have a large thermal contrast and vice versa.
Editor
December 28, 2013 1:09 pm
Klotzbach and Gray link global cooling to increased Atlantic storminess.
http://hurricane.atmos.colostate.edu/Forecasts/2008/dec2008/dec2008.pdf
See diagram on page 35.
TB
December 28, 2013 1:17 pm
Stephen Wilde says:
December 28, 2013 at 9:07 am
“If you fill a balloon of air in the upper atmosphere and bring it down to ground level, the air in it will be warmer than the air at ground level”
Assuming you mean the upper troposphere that would be because the air was originally humid and cooled during uplift at the moist lapse rate.
If you then capture dried out air and bring it down then it warms at the dry lapse rate which is faster and so it ends up warmer than the surface air.
I have argued elsewhere that the mechanical warming on the descent phase of the convective cycle is what warms the surface above the S-B prediction and not DWIR (Downward Infrared Radiation)
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stephen:
I have a problem with that.
And that is that free air in the Earth’s atmosphere is constrained by buoyancy.
It cannot penetrate too far downwards because of this. Just as trying to hold a hot-air balloon down is fruitless because it wants to go up!
In the case of Foehn winds and their counterparts around the world (Chinook etc). They are constrained by stable air and wave motion over higher ground (capping inversion layer) – having risen on the windward side up an SALR, lost water content, and spilled down the leeward side under descending wave motion along a DALR. Hence the arrival of air at a much higher temp.
This is (relatively) large mass air movement that has mechanical means and constraining temp profile to cause it. Those means are not available in the wider atmosphere, at least when not equally balanced by air returning upward again to redress the balance.
And no I do not agree with you hypothesis that it explains the warming at the surface to deny a GHE.
Bill H
December 28, 2013 1:19 pm
Mike Jonas says:
December 28, 2013 at 1:09 pm
What they fail to realize is that storm increase is due to Polar Jet intrusion, as it is today.. They miss the very basic concept of what is driving the air mass changes. Go figure..
December 28, 2013 1:20 pm
Larry Ledwick says:
December 28, 2013 at 9:39 am
I agree that the water cycle has an effect but the fact remains that descending air warms at the dry lapse rate.
I think that the latent heat released on condensation mostly if not all gets radiated out to space from the condensate rather than moved to a new location.
I see the water cycle as a lubricant for the process which means that circulation changes can be less violent than would otherwise be necessary when internal system forcing elements arise such as changes in the proportion of GHGs.
phlogiston says:
December 28, 2013 at 12:07 pm
I was commenting on another poster’s ‘design’ and assumed the pressure injected at the high level would be enough to provide enough remaining inflation to give heat information at the surface.
However one designs the experiment it is correct that descending dry air warms at the dry lapse rate whilst the initial parcel would have cooled at the lesser moist rate.
December 28, 2013 1:24 pm
TB says:
December 28, 2013 at 1:17 pm
The details that you refer to don’t matter in the scheme of things because at any given moment 50% of the atmosphere is rising and 50% is descending.
What goes up must come down.
It doesn’t deny the greenhouse effect, it IS the greenhouse effect.
The GHE is the consequence of mechanical processes and not radiative processes.
December 28, 2013 1:32 pm
TB said:
“This is (relatively) large mass air movement that has mechanical means and constraining temp profile to cause it. Those means are not available in the wider atmosphere, at least when not equally balanced by air returning upward again to redress the balance.”
Why are those means not available in the wider atmosphere?
All air ascending anywhere is matched by air descending somewhere else.
The constraining temperature profile is provided by the gravity induced lapse rate which underpins the actual observed lapse rates which vary according to atmospheric composition in different layers
However in the end the sum of all observed lapse rates must net out to the gravity induced lapse rate if the atmosphere is to be retained.
That is why the Earth’s lapse rate is like a sideways ‘W’ due to composition variations as one goes up. That ‘W’ shape is a consequence of the circulatory contortions necessary to net out to the ‘ideal’ gravity induced lapse rate.
Hoser
December 28, 2013 1:33 pm
This heat engine model does make me wonder about the weather 15,000 years ago. Were the storms stronger then? Were the tropics colder? Were there three Hadley cells in the Northern Hemisphere, or only one? If only one, there would be quite a collision of warm and cold air masses. I have no way to evaluate it, but perhaps such a change could drive ice formation over the northern continents.
As a crude idea, let’s say a very strong cold air mass diving west from North America toward the equator could drive a mass of warm moist air into the North Pacific. The angular momentum of the equatorial air would drive it eastward over the continent, where the moisture would be deposited as snow. Perhaps a similar flow of cold air from Europe drove moist air from the Atlantic back north and over the continent.
I am just wondering if a radical change or complete breakdown of the Hadley cell structure in the north might be a feature of ice ages, and how that hypothesis might be tested. Perhaps the Younger Dryas was a temporary return of the ice age air flow. Are there any obvious triggers for such a change?
Using our heat engine model, it seems a possible explanation is the temperature difference between an icy mid-continent and equator would be much greater and the northern tropical Hadley cell energy would increase, while the others would decrease. Consequently, it seems to me the tropical cell would grow to cover a greater span of latitudes. Even if the three-cell structure remained, the northern polar and subtropical cells would be anemic compared to a very energetic north tropical cell.
Does this scenario make any sense at all?
http://www.fields.utoronto.ca/programs/scientific/10-11/biomathstat/Langford_W.pdf
Not sure they have the right conclusions, but the idea of a single Hadley cell is supported. Unfortunately, it seems they think a reduced temperature difference between the equator and poles drives it. Well, if the cold goes farther south, then perhaps the conditions are met just the same.
Bill H
December 28, 2013 1:59 pm
” Hoser says:
December 28, 2013 at 1:33 pm ”
A single Hadley cell is very much a reality if snow and ice reach 35-40 degrees latitude.. If the pressure becomes so great at the poles that a cell is crushed into oblivion it might take thousands of years to regain it or it might never come back. The big question is what drives the change… I think its the sun despite what some around here think. We are missing a link to our sun or we are simply ignoring one that does not seem relevant right now.
December 28, 2013 2:07 pm
Thanks to Willis Eschenbach and Jan Kjetil Andersen, especially, for bring us such an interesting investigation into the Earth as a heat engine and for insights into the efficiency and effects of all heat engines. Top rate stuff here!
TB
December 28, 2013 2:13 pm
Stephen Wilde says:
December 28, 2013 at 1:24 pm
TB says:
December 28, 2013 at 1:17 pm
The details that you refer to don’t matter in the scheme of things because at any given moment 50% of the atmosphere is rising and 50% is descending.
What goes up must come down.
It doesn’t deny the greenhouse effect, it IS the greenhouse effect.
The GHE is the consequence of mechanical processes and not radiative processes.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“What goes up must come down.”
Correct – which is why your hypothesis wont work.
And No, I disagree – the GHE is a consequence of radiative effects.
December 28, 2013 2:14 pm
When watching weather patterns I’ve always been struck by the sheer size and power of winter storm systems. Comparing the cold weather low systems against the so called cyclonic warm water storms, there isn’t really a comparison. The warm water systems are small, contained storms that can only ramp up power over clear seas without storm destroying upper winds or dry air sneaking in.
The Carnot engine concept goes a long way towards describing these systems on a macro scale. Right up to Polar storms with hurricane force winds over land or ice.
There are many naysayers above but none have yet to offer a correction, just negatives.
December 28, 2013 2:22 pm
What are they ‘spraying’ in the skies!!!!???
What is really in those contrails!!!???
Is it really a hazard to humans to be breathing these materials!!???
And HAARP!!! More atmosphereic experiments!!???
BTW, Anthony and mods, the video left by michaelwiseguy in the comment above:
. . . http://wattsupwiththat.com/2013/12/28/climate-as-a-heat-engine/#comment-1515681
contains ALL THAT plus more … here is the same vid again:
PS Can’t you just boot this guy once and for all?
PPS No one will notice he is gone …
[fixed thanks -mod]
.
Bill H
December 28, 2013 2:23 pm
It seems everyone want to go back to GHG’s rather than look at the functions of the heat transfer due to temperature imbalance/air mass movement. The more I look at this comparison to a Carnot Engine the GHG has little or no effect on the system when looked at from this point of view.
Given the recent studies and papers out showing that GHG’s have been massively overstated in their cumulative effects I’m inclined to look more closely at this point of view.
December 28, 2013 2:25 pm
Oops
Anthony and mods – the vid I was referring to:
charles nelson
December 28, 2013 2:46 pm
The wind then performs ‘work’ moving waves, trees, ice, dust etc.
cd
December 28, 2013 2:46 pm
Jan
I think the problem here is not with most of what you say but rather the approach; the same approach taken by climate modellers. Your approach assumes that reductionism works for something as complex as the climate. Yes, they may rely on fundamental principles and can be built up from these elementary components, but it all falls apart firstly because of imperfect implementations, and secondly, limited computational power. It’s a fool’s errand.
cd
December 28, 2013 2:48 pm
On a general point, what is wrong with accepting we’re crap at modelling nature. It’d save us a lot of money: USGS, MetOffice, EPA…etc.
TB
December 28, 2013 2:51 pm
> Stephen Wilde says:
December 28, 2013 at 1:32 pm
TB said:
“This is (relatively) large mass air movement that has mechanical means and constraining temp profile to cause it. Those means are not available in the wider atmosphere, at least when not equally balanced by air returning upward again to redress the balance.”
Why are those means not available in the wider atmosphere?”
Because there are a limited number of mountains on the Earth’s surface. For a start ~70% of it is ocean.
“All air ascending anywhere is matched by air descending somewhere else.”
Correct?
“The constraining temperature profile is provided by the gravity induced lapse rate which underpins the actual observed lapse rates which vary according to atmospheric composition in different layers”
?? You’ll have to pass that by me again, I’m sorry.
“However in the end the sum of all observed lapse rates must net out to the gravity induced lapse rate if the atmosphere is to be retained.”
No, there is another influence you overlook and that is buoyancy. Gravity will induce a LR, yes, however there is another LR that modifies that and that is the natural LR induced by ascending/descending air. Where the loss/gain of internal energy of an air parcel allows the temp of that parcel to alter at a specific adiabatic rate. I maintain that this is not the same as a gravity induced LR.
“That is why the Earth’s lapse rate is like a sideways ‘W’ due to composition variations as one goes up. That ‘W’ shape is a consequence of the circulatory contortions necessary to net out to the ‘ideal’ gravity induced lapse rate.”
No, part is due to gravity + the up/down buoyancy induced lapse, the other is due to composition – ie the Stat absorbs UV and therefor has a rising temp due presence of O3. The Meso is in collision with Solar particles and is even warmer due collision.
The BDC (Brewer Dobson circ) has an influence in it’s formation as well as PV on isentropic surfaces.
But the causation of the Tropopause is complicated…..
http://www.atmosp.physics.utoronto.ca/SPARC/SPARCImplementationPlan/3_Processes.html
CarolinaCowboy
December 28, 2013 2:57 pm
While not near so elegantly I recently wrote about how the thermal energy in the oceans was transformed into kinetic energy when thermal differentials created the currents in the seas.
http://carolinacowboy.wordpress.com/2013/12/19/speaking-of-heat/
Matt G
December 28, 2013 3:03 pm
TB says:
December 28, 2013 at 1:01 pm
“Now we have something called Coriolis force that ensures that air moving over the Earth’s surface carries the momentum of the Earth beneath it. So for air moving north (in a W’ly jet) it will be moving (relatively) E>W at a greater rate than the land surface blow it.”
In the NH the jet is W>E even when moving N or S of course.
The polar jet with the thermal gradient between polar air and sub-tropical air has the biggest difference with the strongest winds. I didn’t mention the sub-tropics jet which is usually the difference in temperature between the tropics and sub-tropics and has a much lower thermal gradient with much weaker winds. The tropics can also have a easterly based jet stream and this is dependent on dry air clashing with high moisture air at higher attitudes only, very little thermal related. The reason why I didn’t like to call the jet stream caused by thermal gradients because it isn’t necessarily so. Although it is highly thermal gradient related and it is strongly a combination of the planets rotation on its axis (Coriolis) and atmospheric heating by solar radiation.
The polar jet can be absent in places when polar air and sub-tropical air fail to meet, so not only a thermal gradient like mentioned before in the tropics. That’s why there can be gaps in jet stream (polar jet) in Winter and especially during Summer.
http://virga.sfsu.edu/archive/jetstream/jetstream_atl/big/1307/13070406_jetstream_atl_anal.gif
Better link for the person before, not just wanting to look at data collection.
http://squall.sfsu.edu/crws/jetstream.html
DonV
December 28, 2013 3:04 pm
Steve B says
“No the primary working fluid is the air. The oceans are a moderator not a driver.”
I’m sorry but I disagree. I chose my ‘ammonia refrigerator’ example for a reason. In this refrigerator the energy source is heat from a kerosene flame. There are three molecules trapped in the system – Water, Anhydrous Ammonia, and Hydrogen. The water never boils, and the hydrogen never condenses. Only the ammonia CHANGES PHASE. So the working fluid is ammonia, not water and not hydrogen (although both are needed to make the system work efficiently). It is the ammonia that does the heavy lifting, that stores and then releases the vast majority of the heat transferred in this system.
In our climate system WATER is the working fluid. No other atmospheric gas changes phase and in so doing takes on immense amounts of energy for transport vertically AND laterally. Air cannot even come close to handling the same amount of energy because except for water vapor none of the gases that make up air change phase! The only way they can store and transport energy is through their heat capacity. Even the molar heat capacities of water are greater than those of air:
Air (room temp. and press.) 29.9 (J/mol·K)
Water 74.5 (J/mol·K)
But the energy that is transferred to water during phase change and that is subsequently transported by wind is typically reported in kilo Joules/mole – thats thousands of joules! So the value of the latent heat of vaporization and latent heat of fusion are:
Water 40,680 (J/mol)
Ice 6,000 (J/mol)
Since neither CO2 nor N2 nor O2 ever change phase they don’t participate in this massive heat transfer process. Sure they carry the water and participate in the expansion and contraction of the gases as they rise and fall, but they only carry a very small amount of the energy that water does.
No, I’m sorry but I disagree with you, Steve B. Water is definately the working fluid in our climate system. But I do agree with your statement that water is a “moderator”. Without water as the working fluid “moderating” (or in Willis’ words “governing”) the climate system our planet would be a much more extreme place to live in.
kalsel3294
December 28, 2013 3:28 pm
re charles nelson says:
December 28, 2013 at 2:46 pm
“The wind then performs ‘work’ moving waves, trees, ice, dust etc.”
Whilst the movement of waves, trees, dust and the like provides a visual impression of “work” being done, what is not so evident is the transfer of heat and moisture between the air and the surfaces it comes into contact with that is the real “work”.
jorgekafkazar
December 28, 2013 4:02 pm
chris y says: “The cyclone wind speed v can be estimated with this model and other assumptions, giving
v = sqrt((Ts – To)E/To)
where Ts=300K, To=200K and E is a variable related to coupling efficiency of energy between the ocean surface and the atmosphere.”
There’s the problem with your analysis. E, the coupling efficiency, is a temperature-related variable, and could reduce the velocity or even send it the other direction.
jorgekafkazar
December 28, 2013 4:09 pm
Many commenters are missing (perhaps deliberately) the point that to the extent that the atmosphere doesn’t resemble the Carnot cycle, it will have correspondingly lower efficiency. The Carnot cycle gives a measure of the highest achievable conversion of temperature differentials to mechanical energy, e.g., wind. Non-Carnot cycles will perform at a much lower level. Lower efficiency = lower conversion to energy = lower energy storms.
December 28, 2013 4:11 pm
Marc77 says:
December 28, 2013 at 8:53 am
“The upper atmosphere might be cold, but it cannot be used to drive a convection cell”
————————————————————————
Marc,
No, it is energy loss to space from radiative gases that allow buoyancy loss and subsidence in air masses driving convective circulation in the Hadley, Ferrel and Polar tropospheric convection cells.
Without radiative gases, tropospheric convective circulation in our atmosphere would stall and the atmosphere would trend isothermal through gas conduction. Such an atmosphere would be far hotter than our current atmosphere, as it’s temperature would be driven by surface Tmax.
Climate pseudo scientists have relied for years on the confusion most people encounter though the difference between adiabatic heating and cooling of air masses undergoing vertical circulation, and diabatic processes such as energy gained through surface conduction, release of latent heat and intercepted LWIR and energy lost via LWIR to space.
It is diabatic processes that drive buoyancy changes. It is diabatic processes that drive convective circulation.
When radiative and non-radiative energy transports in our atmosphere are solved for simultaneously it is clear that the NET effect of radiative gases in our atmosphere is atmospheric cooling at all concentrations above 0.0ppm.
TB
December 28, 2013 4:23 pm
Matt G says:
December 28, 2013 at 3:03 pm
TB says:
December 28, 2013 at 1:01 pm
“Now we have something called Coriolis force that ensures that air moving over the Earth’s surface carries the momentum of the Earth beneath it. So for air moving north (in a W’ly jet) it will be moving (relatively) E>W at a greater rate than the land surface blow it.”
In the NH the jet is W>E even when moving N or S of course.
No, not always, in jet disruption processes the jet can move E>W under a warm High in an Omega-block as deep cold air is advected west beneath. This is a common cause of the UK’s coldest winter events.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The polar jet with the thermal gradient between polar air and sub-tropical air has the biggest difference with the strongest winds.
Not necessarily, polar air converging with air in the mid-latitudes is the most common jet-formation, however sub-tropical air on occasion is advected north to further accentuate the thermal gradient. Behind an amplifying Rossby wave FI as it moves along the N’ward cycle, but also on the bottom of the S’ward cycle.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I didn’t mention the sub-tropics jet which is usually the difference in temperature between the tropics and sub-tropics and has a much lower thermal gradient with much weaker winds. The tropics can also have a easterly based jet stream and this is dependent on dry air clashing with high moisture air at higher attitudes only, very little thermal related. The reason why I didn’t like to call the jet stream caused by thermal gradients because it isn’t necessarily so. Although it is highly thermal gradient related and it is strongly a combination of the planets rotation on its axis (Coriolis) and atmospheric heating by solar radiation.
Correct.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The polar jet can be absent in places when polar air and sub-tropical air fail to meet, so not only a thermal gradient like mentioned before in the tropics. That’s why there can be gaps in jet stream (polar jet) in Winter and especially during Summer.
Well yes, there needs to convergence of air-masses to cause a tight thermal gradient (jet-stream).
You don’t need subtropical air necessarily – if the polar air is intensely cold ~510/492dm thick (1000-500mb) then in contrast to mid-latitude air of 540/546dm air there is enough contrast (given convergence) to create a strong jet.
December 28, 2013 4:30 pm
NZ Willy says:
December 28, 2013 at 10:14 am
“Don’t overlook the Coriolis force as a primary driver of large-scale wind.”
—————————————————————————————–
I have noted a number of folk making this mistake recently.
Coriolis forces are not a driver of atmospheric circulation. Coriolis forces have no influence on air masses at rest within a rotating reference frame. Air masses must be put in motion before Coriolis forces affect their movement.
It is radiative and non-radiative heating of air masses at low altitude and radiative cooling of air masses at high altitude that drive atmospheric circulation. Coriolis forces then influence the pattern.
One of the most notable effects of rotational forces is the division of tropospheric convective circulation into three main cells north and south of the equator. If the planet rotated slower there would only be two giant Hadley cells.
TomRude
December 28, 2013 4:34 pm
http://ddata.over-blog.com/xxxyyy/2/32/25/79/Leroux-Global-and-Planetary-Change-1993.pdf
1993… and at least the description of circulation was correct.
Matt G
December 28, 2013 4:55 pm
TB says:
December 28, 2013 at 4:23 pm
“No, not always, in jet disruption processes the jet can move E>W under a warm High in an Omega-block as deep cold air is advected west beneath. This is a common cause of the UK’s coldest winter events.”
I agree that some of time little jets may wander E>W, but virtually always W>E.normally.
Even during one of the most severe cold spells in the UK for decades at the time of year..
http://www.wetterzentrale.de/pics/archive/ra/2010/Rrea00120101201.gif
The jet hardly moved E>W.
http://virga.sfsu.edu/archive/jetstream/jetstream_atl/big/1012/10120106_jetstream_atl_anal.gif
The key component along with thermal gradient causing the jet stream, is cold very dry air meeting warm moist air.
The link before gave an example of this especially moisture related, when the cool dry polar air reached the moisture of the Atlantic ocean, but itself not warmer than the land masses to the West or East of it. that didn’t have a jet, but had relatively dry air.
http://virga.sfsu.edu/archive/jetstream/jetstream_atl/big/1307/13070406_jetstream_atl_anal.gif
noaaprogrammer
December 28, 2013 5:00 pm
Now that gravity has been introduced into this discussion, what effect does any atmospheric tides due to the moon have?
December 28, 2013 5:06 pm
Bill H says:
December 28, 2013 at 2:23 pm
“It seems everyone want to go back to GHG’s rather than look at the functions of the heat transfer due to temperature imbalance/air mass movement. The more I look at this comparison to a Carnot Engine the GHG has little or no effect on the system when looked at from this point of view.”
—————————————————————————————————–
Not quite everyone..;-)
However, radiative gases* have a huge effect on atmospheric circulation. Without these gases, the atmosphere has no effective cooling mechanism. It is radiative energy loss at altitude that allows buoyancy loss, subsidence of air masses and continued tropospheric convective circulation in the Hadley, Ferrel and Polar cells. The result of this circulation is the pneumatically generated lapse rate cause by vertical circulation across the pressure gradient of the atmosphere. Just this factor alone makes our atmosphere far cooler on average than an atmosphere without radiative gases.
Some have claimed that the surface temperature differential between the equator and the poles could drive circulation powerful enough to generate the observed lapse rate in the absence of radiative energy loss at altitude. However the physics of gas conduction dis-allows this. The surface is far better at conductively heating a moving gas atmosphere in a gravity field than it is a conductively cooling it. The pole to equator surface temperature differential would not be realised in the gas atmosphere above. Further, surface friction and gas conduction within the atmosphere would would resist such circulation.
*the use of the term “greenhouse gases” as opposed to radiative gases is non-scientific. Use of that term is a wholly political effort to control the language to control the “narrative”. It implies that the radiative greenhouse effect is a valid hypothesis. In the case of both anthropogenic global warming and the radiative greenhouse effect, the hull hypothesis still stands.
Gino
December 28, 2013 5:16 pm
I’ve enjoyed the two ‘heat engine’ articles. They are moving in the right direction, however, a better physical model for the earth system is what’s called a ‘Thermosiphon’ with a working fluid of water/’air’ mixture. In this case the primary working fluid is water because across the total delta T of the system (earth atmosphere) water is a three phase element.
Larry Ledwick
December 28, 2013 5:21 pm
@TB:
That is why the Earth’s lapse rate is like a sideways ‘W’ due to composition variations as one goes up. That ‘W’ shape is a consequence of the circulatory contortions necessary to net out to the ‘ideal’ gravity induced lapse rate.
It is also due to how those layers with different composition absorb energy directly from the sun.
The reversal of lapse rate in the stratosphere is primarily due to heating of that layer directly by absorption of UV ( UVB and UVC) by ozone. At the top of the stratosphere temperatures are only slightly below the freezing point of water. Above the stratosphere, the lapse rate again resumes but slightly steeper than in the troposphere due to the very low moisture content. The very top layers of the atmosphere are warmer again due to being actively heated by the sun through several mechanisms, direct energy absorption in the UV and x ray bands, heating due to effects of the solar wind, and electrical currents and dissipation of waves (like breaking waves at a beach) that develop in the very top layers of the atmosphere, and probably direct heat absorption from in falling dust and micrometeors into the upper layer of the atmosphere.
This is one of the problems with “simplified models” we often use to make sense of the mechanics of how heating in the atmosphere results in ground temperatures warmer than prevailing theory says the should be according to SB and why N&Z proposal about gravitational potential energy and how it defines the total warming not some trace gas in the atmosphere is so controversial.
Every time we try to simplify the situation too much to help get our head around the processes we simultaneously introduce false realities that sometimes become the focus of debate. People forget that in specialize thought experiments and analogies we sometimes set traps for ourselves and get sucked into debating some quirk of the “mental model” just like we do with “computer models” and lose site of the forest for the trees.
I can’t explain the exact physics of the gravitational heating hypothesis in the context of the real world, but I also cannot find any escape clause that would allow the conservation of energy considerations to be ignored.
I think we are in a position much like when scientists first began to realize that Newtonian physics did not work well in certain realms (the very small and the near light speed domains for example) and they first began to embrace quantum and relativistic considerations.
It is my expectation that once sufficiently open minded researchers start looking for and gathering the appropriate data instead of just presuming GHG’s are the cause of heating we might finally dig out the exact mechanics. You obviously cannot prove or disprove a concept you refuse to even investigate, or you presume is so well grounded that it need not be examined in light of new information.
Just like we complain that the AGW advocates are only looking for confirmation and not falsification, we need to be careful not to look only for confirmation of a gravity based lapse rate but also those tests which would specifically prove it could not be true. This does not include just declaring it must be nonsense but actively defining what must be true if it is valid and what test would prove that it fails. To do that we first need legitimate methodical unbiased experiments that examine if it might exist and making reliable repeatable tests of various explanations for what N&Z pointed out in their papers.
Waving your hands and saying it cannot be true is no more valid than waving a computer print out and saying “the models say so”.
chris y
December 28, 2013 5:40 pm
jorgekafkazar says:
December 28, 2013 at 4:02 pm
chris y says: “The cyclone wind speed v can be estimated with this model and other assumptions, giving
v = sqrt((Ts – To)E/To)
where Ts=300K, To=200K and E is a variable related to coupling efficiency of energy between the ocean surface and the atmosphere.”
There’s the problem with your analysis. E, the coupling efficiency, is a temperature-related variable, and could reduce the velocity or even send it the other direction.
****************
For the record, this is Kerry Emanuel’s analysis, not mine.
I agree that E is likely temperature dependent. However, at least some of the temperature dependence of the energy coupling between ocean and air has already been accounted for in the model.
There are other problems with using the Carnot model for cyclones, like energy transport with water vapor.
tobyw
December 28, 2013 5:48 pm
Glad to see we are finally getting some examination of the atmospheric-heat engine aspect of the earth, but how about the rotation of Earth in the Sun’s magnetic and gravitic fields? The rotating field of Earth should act like a both generator and motor one would think, and the tides rise and fall several feet, which must involve quite a lot of heat energy moving all that water. Are there tides in the core as well? Piers Corbyn’s solar powered weather forecasts also shows a correllation between solar activity and earthquakes. Is there electromagnetic and gravitic heating in our planet?
December 28, 2013 6:11 pm
DonV says:
December 28, 2013 at 3:04 pm
Steve B says
“No the primary working fluid is the air. The oceans are a moderator not a driver.”
I’m sorry but I disagree. I chose my ‘ammonia refrigerator’ example for a reason. In this refrigerator the energy source is heat from a kerosene flame. There are three molecules trapped in the system – Water, Anhydrous Ammonia, and Hydrogen. The water never boils, and the hydrogen never condenses. Only the ammonia CHANGES PHASE. So the working fluid is ammonia, not water and not hydrogen (although both are needed to make the system work efficiently). It is the ammonia that does the heavy lifting, that stores and then releases the vast majority of the heat transferred in this system.
********************************************************************************************************************
Well you may disagree but our atmosphere is not a refrigerator. The process is the movement of hot air and cold air which I believe we call wind. Wind is created when there is a differential between hot and cold air not hot and cold water.
jorgekafkazar
December 28, 2013 6:43 pm
chris y says: “The cyclone wind speed v can be estimated with this model and other assumptions, giving v = sqrt((Ts – To)E/To), where Ts=300K, To=200K and E is a variable related to coupling efficiency of energy between the ocean surface and the atmosphere.”
jorgekafkazar says: There’s the problem with your analysis. E, the coupling efficiency, is a temperature-related variable, and could reduce the velocity or even send it the other direction.
chris y says: For the record, this is Kerry Emanuel’s analysis, not mine.
jorgekafkazar says: Noted. This was only one of your three points, and it originated with Emanuel’s paper, as you stated.
chris y says: I agree that E is likely temperature dependent. However, at least some of the temperature dependence of the energy coupling between ocean and air has already been accounted for in the model.
jorgekafkazar says: The quantification remains moot, models being only models.
chris y says: There are other problems with using the Carnot model for cyclones, like energy transport with water vapor.
jorgekafkazar says: ” Agreed, but I’m unconvinced that the inclusion of a water vapor cycle will result in increasing efficiency, rather than decreasing. I strongly suspect that more evaporated water vapor will not produce more available energy for wind velocity. Most of the extra energy will be (1) lost through vertical convection or (2) expended as latent heat converted to potential energy, which is ultimately lost in rain impacting the oceans’ surface.
December 28, 2013 6:47 pm
Larry Ledwick says:
December 28, 2013 at 5:21 pm
—————————————————–
The N&Z hypothesis does appear to work, however Willis was correct to point out flaws in their maths. There was a quite heated “discussion” here at WUWT about Equation 8 in a series of threads that would probably be best referred to as “the past unpleasantness”.
The N&Z hypothesis will only hold true for atmospheres exhibiting strong vertical circulation allowing non-radiative energy transport to exceed the speed of gas conduction and a lapse rate to develop.
The good news is that all planets and moons in our solar system that have managed to retain an atmosphere have sufficient radiative gases to allow the radiative cooling necessary to drive this circulation.
Gino
December 28, 2013 6:47 pm
DonV, excellent observation Re the 2 phase working fluid. The third phase (ice) is simply reserve or stored heat capacity in the system.
norah4you
December 28, 2013 7:13 pm
The data in the bloggarticle above is only one out of many where incorrect data is used to prove (sometimes disaprove) thesis drawn from incomplete datamodels.
Btw. Since 1964 (Nineteen sixtyfour !) true readings of all discussed in the blogg article been made. Where? Esrange, Sweden.
Further more: Wouldn’t it had been a good thing to use at least one of the best Dissertations in the field as a background check?
For example:
Nikulin Grigory, Impact of Rossby waves on ozone distribution and dynamics of the stratosphere and troposphere , Dissertation Umeå University 2005 ISBN 91-7305-946-3
IRF scientific report, 0284-1703 ; 285
Mario Lento
December 28, 2013 7:19 pm
Very nice article, which simplifies the understanding of delta T which is responsible for “weather”.
I’m just nit picking. The statement: “The warm element for a car engine is the exploding fuel inside the cylinders and the cold element is the air intake” Is correct in what point is being made –except, car nuts don’t like to call the burning an explosion. It is really a controlled burn. The flame front is well controlled so that the energy can effectively convert heat into mechanical motion when it is needed, and then expelled. Sparks are timed to detonate before the piston reaches top dead center due to the latency of pressure wave. Explosions in the cylinders are very bad and screw up the timing required for engine performance.
The other point in support of this need for delta T for more weather energy is that warmer air at the poles is less dense, so there are fewer molecules of colder air to react with the warmer equatorial air.
Thank you for the article!
David L. Hagen
December 28, 2013 8:52 pm
There are numerous papers on climate as a heat engine. e.g. see Google scholar: “carnot heat engine climate warming temperature difference wind”
e.g., Alex Kleidon finds:
In summary, the total free energy generation for current conditions yield about Pgeo,a≈6070TW of power by physical processes within the atmosphere from the exchange fluxes at the Earth–space boundary, aboutPbio≈215TW of chemical free energy by photosynthesis, and about Pgeo,b≈40TW driven by the depletion of initial conditions in the interior. Hence, the total power generation by the planet is aboutPplanet=Pgeo,a+Pgeo,b+Pbio≈6325TW.
Axel Kleidon How does the Earth system generate and maintain thermodynamic disequilibrium and what does it imply for the future of the planet? Philos Trans A Math Phys Eng Sci. 2012 March 13; 370(1962): 1012–1040. doi: 10.1098/rsta.2011.0316 PMCID: PMC3261436
Mazzarella et al. show CO2(temperature) variations are associated with variations in the earth’s Length Of Day (LOD) i.e., by changing the differential temperature which changes the wind /ocean current speeds.
Mazzarella A., A. Giuliacci and N. Scafetta, 2013. Quantifying the Multivariate ENSO Index (MEI) coupling to CO2 concentration and to the length of day variations. Theoretical and Applied Climatology 111, 601-607. DOI: 10.1007/s00704-012-0696-9. PDF
December 28, 2013 9:01 pm
“The good news is that all planets and moons in our solar system that have managed to retain an atmosphere have sufficient radiative gases to allow the radiative cooling necessary to drive this circulation.”
In fact, the necessary convective overturning is induced by uneven surface heating causing air parcel density differentials plus the gravity induced decline in temperature with height which allows the uplift of less dense parcels.
Radiative cooling within the atmosphere is not needed.
If the atmosphere is radiatively inert then the effective radiating height stays at the surface and convective overturning settles at a speed which delivers kinetic energy back to the surface via adiabatic warming on descent sufficient to supply the energy needed by the surface to radiate as much to space as is received from space.
Fantasies about isothermal atmospheres are not realistic.
gymnosperm
December 28, 2013 9:08 pm
A heat engine model is just another model, one containing a kernel of truth, but fraught with oversimplifications just as the GCM’s. The delta T between the equator and poles is modulated by the fence effect of circumpolar vortices. When these vortices are strong, they limit the Rossby wave amplitude we perceive as extreme weather.
As Willis has suggested many times there seems to be a system limit to tropical atmospheric warming that would affect the meridional delta T. Such a limit would flatten the gradient in a warming environment, pretty much what we are seeing with hot polar atmospheric anomalies.
The ocean is another story…
December 28, 2013 9:09 pm
Larry Ledwick says:
December 28, 2013 at 5:21 pm
I said:
“That is why the Earth’s lapse rate is like a sideways ‘W’ due to composition variations as one goes up. That ‘W’ shape is a consequence of the circulatory contortions necessary to net out to the ‘ideal’ gravity induced lapse rate.”
Larry said:
“It is also due to how those layers with different composition absorb energy directly from the sun.”
Yes, I agree.
But the distortions in the gravity induced lapse rate created by direct absorption from the sun are also ‘corrected’ by circulation changes.
Even the stratosphere has a circulation albeit very slow due to low air density and I suspect similar movements in the higher layers too but they would be so weak that they have never been measured.
Suffice it to point out that if internal system parameters could cause a long term net divergence from the ‘ideal’ lapse rate then the atmosphere could not be retained.
December 28, 2013 9:19 pm
“The surface is far better at conductively heating a moving gas atmosphere in a gravity field than it is at conductively cooling it”
Due to the density differentials between surface and air one would think that but bear in mind that the surface can only cool to space radiatively so an adiabatically warmed descending column of air only needs to deliver enough kinetic energy back to the surface to at least partially offset energy radiated upward. Even a partial offset will increase average global surface temperature.
In the middle latitudes we all know of clear winter nights when it is windy and the air movement prevents a frost from developing by inhibiting net radiative loss from the surface.
Direct conductive energy transfer from air to surface is not needed. Just offsetting radiative loss is sufficient to warm the surface higher than S-B would predict in the absence of an atmosphere.
December 28, 2013 9:28 pm
I must say that I am very pleased to see this article. I find it validating of many of the things I have been trying to say for a very long time. It is not heat that causes storms, it is a DIFFERENCE in temperature and that difference can be created by either cooling one side or heating the other. It is my opinion that glacial periods would be extremely and freakishly stormy in the central US. This is because solar insolation at lower latitudes remains relatively constant, it is high latitude isolation that changes most greatly. During glacial periods, the Gulf of Mexico and tropical Atlantic is still receiving about the same amount of energy it does in the interglacial periods but imaging you have a warm gulf of Mexico and severe cold in the upper plains and very cold air in summer coming down off the ice into the central plains. We SHOULD see absolutely terrific storms. We would likely have fronts spawning tornadoes nothing like what we see today.
I would not want to live in places like Oklahoma during the glacial periods for I fear massive tornadoes would be much more common then than today considering a great ice sheet would be nearly as far away as the Gulf of Mexico.
phlogiston
December 28, 2013 10:23 pm
The idea that radiative (or any other) cooling is needed to drive air down in Hadley circulation nakes no sense. It makes even less sense to consider the uoward and downward “parts” of convection separately. When any fluid is heated from below turbulent conveection will begin at a certain temperature gradient. Think of Rayleigh-Benault convection, or Libchaber’s experiments with liquid helium, or just water in a saucepan over a gas flame. Rolling cells represent the early bifurcations in the onset of, turbulence. This is what Hadley cells are. The asymmetric heat input exists at the outset of the turbulence as the causative factor it does not need asymmetric cooling to sustain it.
This is reductionism gone mad. Chaotic turbulence is not a clockwork machine – “machine” is a very bad analogy.
December 28, 2013 10:37 pm
Stephen Wilde says:
December 28, 2013 at 9:01 pm
——————————————————
“In fact, the necessary convective overturning is induced by uneven surface heating causing air parcel density differentials plus the gravity induced decline in temperature with height which allows the uplift of less dense parcels.
Radiative cooling within the atmosphere is not needed.”
Stephen,
you keep insisting that uneven surface heating can drive vertical circulation across 10 to 15 Km of the troposphere. This is contradicted by empirical experiment. In tall gas columns, Rayleigh Bernard circulation requires energy loss at a higher point than energy input. Heating and cooling at disparate locations at the base of the column simply causes weak horizontal circulation, resisted by surface friction, with layers above this rising to toward surface Tmax. While the atmosphere may appear shallow, the pressure gradient creates a far greater virtual height in terms of Rayleigh Bernard circulation.
There is no “gravity induced decline in temperature with height”. There is a gravity induced pressure gradient. It is vertical circulation across this gradient that pneumatically generates the observed lapse rate. Without vertical circulation fast enough to overcome the speed of gas conduction, the lapse rate would disappear and the atmosphere would trend toward isothermal.
In arguing for strong vertical tropospheric convective circulation in the absence of radiative cooling at altitude, you have aligned yourself with Joel shore, Nick Stokes and sadly Kevin E. Trenberth.
Not even Dr. Pierrehumbert wanted to go that far. In 1995 he tried the far more subtle – “Well of course initially radiative gases start vertical tropospheric circulation and cause atmospheric cooling, but, uum, err, after a certain point the unicorn to rainbow ratio goes negative and radiative gases then cause atmospheric warming and anyway, err, uum, just shut up!” (Delivery of these claims does require hand-waving at near humming bird speeds).
The power of radiative cooling at altitude is huge. Radiative gases are radiating as LWIR to space more than TWICE the net flux radiative flux into the atmosphere from surface and directly intercepted solar combined. They are also radiating to space all the energy entering the atmosphere through surface conduction and the release of latent heat.
You are claiming that atmospheric conduction back to the surface could match this and generate vertical circulation fast and high enough to generate the currently observed lapse rate. The physics of gases in a gravity field say no way, no how. The surface is far better at conductively heating the atmosphere than it is a conductively cooling it. During the day, gravity moves colder gas to the surface maximising conductive flux into the atmosphere. During the night, gravity moves colder gas to the surface minimising conductive flux out of the atmosphere.
If your atmospheric model requires strong vertical tropospheric circulation to continue and the observed tropospheric lapse rate to exist in the absence of radiative gases, then your model has failed for the same reasons as the AGW and RGE hypotheses fail.
gbaikie
December 28, 2013 10:50 pm
“The second effect is that even if we had a perfect heat engine with zero internal friction; it would not achieve anything close to 100% efficiency. The maximum theoretical efficiency for a heat engine operating between 300 K and 600K is for example 50%. The efficiency of a real machine would of cause be considerably lower.”
Ah, I wonder if this would apply to what I call a pipelauncher.
I would say this thing I call a pipelauncher as very efficient heat engine.
Converting heat into vertical velocity. And heat engine which could operate
between 300 K and 600K. Or perhaps between 200 K and 500K. But
could work as efficiently between 300 K and 400 K.
Details of it are a large pipe with one end capped.
Which place in the water, so floats vertical.
That is the engine.
Fuel can be anything. And it needs air and oxidizer for fuel.
Purpose of this pipelauncher is to be a launch pad and to lift
a rocket vertically. So it will lift hundred to tons relatively fast.
Or with acceleration of 1/2 gee or 1 gee. Or can do much higher
acceleration if rocket can withstand such gee loads.
In order to accelerate a rocket to any speed- 50 to 300 mph.
One needs the pipe to be quite long- so longer than 500 feet in length.
And to remain vertical and support rocket so it’s vertical, it needs
lengths longer than rocket is tall. So both reasons: 500 feet or longer
in length.
And to lift and accelerate a rocket weighing more than say 100 tons
the pipelauncher needs a large diameter.
So a large diameter pipe which is long and has one end capped.
Capped end is at the top. Rocket place on the top of cap.
And rocket needs some kind launch tower, which should weigh
less than 100 tons.
The pipe itself might weigh about 500 tons or more. Depends
what size of rocket will launched from it. And more tonnage of
course increases the overall. The cost per ton could \$1000 to
say \$3000 per ton. Cost of metal and fabrication into a large
pipe. One could use steel. Aluminum costs more but due lower
mass has less use- more efficient energy usage per launch.
But one may only do 50 to 100 launches in over many years
it could less money saved.
Other just energy use, aluminum can give higher performance-
heavier or faster launch speeds.
So anyhow regarding efficiency. So this machine works by
buoyancy, hence why I think it would be very efficient heat engine.
So this is like balloon, but instead air, one is using denser water
as weight which is displaced. Or It’s like ocean ship which goes
vertically.
So say it’s 20 feet in diameter and 600 feet long.
With one end capped.
You put in water, and open end will fill with water and
cause the capped end to float vertically.
If were to pump out the water in the pipe, the capped
end would rise, and eventual flop over. fill up and again
and flip vertical again.
In terms the operation involved with a launch rocket, you don’t
want it to flop over. You can tow it a location by temporary capping
the other end and towing it horizontally to some location in the ocean.
One would probably want to tow it to location at equator, which
is best location to launch rockets.
As engine it works this way. When vertical in ocean, the air inside
the pipe will push the water inside the pipe below the sea level.
How deep the water is pushed under the water, determines it’s
displacement or it’s buoyancy.
So as example the 600 foot long pipe which 20 in diameter may
be pushing the water 33 feet under sea level. Since 33 feet
under seal level is a atm of pressure, the air in pipe will be 14.7 psig
So pipe might weigh 500 tons. If add 500 ton of rocket, then
air pressure will be 2 atm [29.4 psi]. If double pressure inside, so
56.8 psi, and you keep the air pressure at 56.8 psi, the pipelauncher
and rocket accelerates at 1 gee [9.8 m/s/s].
So what you need is a lot of air. So you use liquid air. If you dump
1 ton of liquid air into 1 ton of sea water you get gaseous air, once
warm water is instantly cooled, shouldn’t get so cold as to make ice.
So, generally you pour say 20 tons of liquid air into water inside
the pipe, and pour over say period of 5 seconds. You also want to
heat the air created. So say you burn kerosene or natural gas and
have peak temperature be say 400 K. And that about how much
you need to have pipelauncher and rocket on top it accelerate
vertical at 1 gee for 5 seconds.
So pressure range of air would be in range of 10 to maybe 80 psi.
And one can contract the exisingt air in pipe by misting liquid air.
So it’s operational air temperature could at extreme ranges of
150 K to 500 K. And one is only trying to warm or cold the air
and the time involved would be somewhere around 20 seconds
though actual launch time would less than 10 seconds.
So walls of pipe do not have enough time to be heated or cooled
much by the air.
Which means no heat radiated from pipelauncher and pipelauncher
environment.
So in terms heat efficiency related mechanical motion, it seems like
this would be quite efficient.
Editor
December 28, 2013 11:40 pm
Bill H – Dec 28 1:19 pm – “They miss the very basic concept of what is driving the air mass changes :. Maybe they don’t know what causes storms, but the chart I referred to is of observed storm tracks, and there are a lot more of them during the global cooling period than during the global warming period.
December 29, 2013 12:03 am
Tsk Tsk says:
December 28, 2013 at 10:11 am
” the maximum theoretical efficiency decreases with increasing temperatures. ”
I think what you meant to say is that if temperatures are increasing equally, then efficiency goes down. And for any case where Tc is increasing faster than Th efficiency goes down along with a fair number of cases where Th is increasing faster but not fast enough than Tc.
Correct, Tsk, this is as I explain a bit further down in the sentence
The third effect is as mentioned above, that, for a given temperature difference between the warm element and the cold element, the efficiency will decrease if both elements heat equally much
“For a given temperature difference” means the same as “increasing equally”.
Finally, the cold sink in your car example is the air at the exhaust (and ambient air around the engine) and not the intake. They happen to come from the same pool but in principle they don’t have to. I could have a tank of hot air that I inject into the cylinders, expand, and exhaust and still get work from the system.
No, the work comes from the pressure created by the temperature increase caused by burning fuel in the cylinder. It is therefore the intake air temperature that counts.
If your claim was correct you could stop an engine by making some arrangements from the exhaust pipe to lengthen it into a bonfire. But, do you really think that would have any effect?
/ Jan
December 29, 2013 12:10 am
Doug Huffman says:
December 28, 2013 at 8:52 am
Actual heat engines, to which Carnot’s physics apply,
Carnots physics apply to all arrangements creating mechanical energy from thermal energy. The point is that you cannot get around it independent of how complex or simple the system is.
/ Jan
December 29, 2013 12:16 am
phlogiston says:
December 28, 2013 at 10:23 pm
“The idea that radiative (or any other) cooling is needed to drive air down in Hadley circulation makes no sense.”
——————————————————————————————-
Actually, prior to the inanity of the failed global warming hypothesis, it used to be rather standard meteorology –
“Air convected to the top of the troposphere in the ITCZ has a very high potential temperature, due to latent heat release during ascent in hot towers. Air spreading out at higher levels also tends to have low relative humidity, because of moisture losses by precipitation. As this dry upper air drifts polewards, its potential temperature gradually falls due to longwave radiative losses to space (this is a diabatic process, involving exchanges of energy between the air mass and its environment). Decreasing potential temperature leads to an increase in density, upsetting the hydrostatic balance and initiating subsidence.”
A huge political effort was put into keeping the findings of the first IPCC report ambiguous. After 1990 a great amount of work was put into saving global warming. This included inventing “strongly positive water vapour feed back”, erasing the MWP that disproved it and cooking up radiative-convective circulation models that negated the role of radiative gases in driving vertical tropospheric circulation.
Phlogiston, instead of siding with Kevin E. Trenberth and claiming no role for radiative gases in tropospheric convective circulation, perhaps you should be asking why climate “scientists” were so desperate to trash working meteorology theory.
December 29, 2013 12:57 am
Mario Lento says:
December 28, 2013 at 7:19 pm
I’m just nit picking. The statement: “The warm element for a car engine is the exploding fuel inside the cylinders and the cold element is the air intake” Is correct in what point is being made –except, car nuts don’t like to call the burning an explosion. It is really a controlled burn. The flame front is well controlled so that the energy can effectively convert heat into mechanical motion when it is needed, and then expelled.
Thank you for the correction and good explanation Mario.
/ Jan
TB
December 29, 2013 1:16 am
Matt G says:
December 28, 2013 at 4:55 pm
TB says:
December 28, 2013 at 4:23 pm
“No, not always, in jet disruption processes the jet can move E>W under a warm High in an Omega-block as deep cold air is advected west beneath. This is a common cause of the UK’s coldest winter events.”
I agree that some of time little jets may wander E>W, but virtually always W>E.normally.
Even during one of the most severe cold spells in the UK for decades at the time of year..
http://www.wetterzentrale.de/pics/archive/ra/2010/Rrea00120101201.gif
The jet hardly moved E>W.
http://virga.sfsu.edu/archive/jetstream/jetstream_atl/big/1012/10120106_jetstream_atl_anal.gif
The key component along with thermal gradient causing the jet stream, is cold very dry air meeting warm moist air.
The link before gave an example of this especially moisture related, when the cool dry polar air reached the moisture of the Atlantic ocean, but itself not warmer than the land masses to the West or East of it. that didn’t have a jet, but had relatively dry air.
http://virga.sfsu.edu/archive/jetstream/jetstream_atl/big/1307/13070406_jetstream_atl_anal.gif
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Matt:
We are nitpicking here, but the example you link is not a very good one as the High has become filled with colder air – there needs to be more “orange” over Scandinavia A warmer core. And I never said it would be as strong as a 200KT Atlantic jet. I also said a jet forms when air flows from warm to cold aloft. Therefore in any contortion in the flow (cut-off cold air low, say) there will be accentuated flow around the “contortion”.
You talk of another factor when you mention cold dry/very moist. That is warm advection overgliding the cold and releasing LH. It is part of the Omega equation and is a prime drive of development for baroclinic waves.
TB
December 29, 2013 1:29 am
December 29, 2013 at 12:16 am
phlogiston says:
December 28, 2013 at 10:23 pm
“The idea that radiative (or any other) cooling is needed to drive air down in Hadley circulation makes no sense.”
——————————————————————————————-
Actually, prior to the inanity of the failed global warming hypothesis, it used to be rather standard meteorology –
“Air convected to the top of the troposphere in the ITCZ has a very high potential temperature, due to latent heat release during ascent in hot towers. Air spreading out at higher levels also tends to have low relative humidity, because of moisture losses by precipitation. As this dry upper air drifts polewards, its potential temperature gradually falls due to longwave radiative losses to space (this is a diabatic process, involving exchanges of energy between the air mass and its environment). Decreasing potential temperature leads to an increase in density, upsetting the hydrostatic balance and initiating subsidence.”
A huge political effort was put into keeping the findings of the first IPCC report ambiguous. After 1990 a great amount of work was put into saving global warming. This included inventing “strongly positive water vapour feed back”, erasing the MWP that disproved it and cooking up radiative-convective circulation models that negated the role of radiative gases in driving vertical tropospheric circulation.
Phlogiston, instead of siding with Kevin E. Trenberth and claiming no role for radiative gases in tropospheric convective circulation, perhaps you should be asking why climate “scientists” were so desperate to trash working meteorology theory.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I have discussed this with you on another thread I believe.
It seems you still do not appreciate the effect of mass air convergence aloft a Hadley Cell.
As air moving N will converge to a westerly Sub-tropical jet and sink as a result. (Err has too!)
Radiative cooling is NOT needed. It happens but that is not the driver.
The radiative cooling will be offset by subsidence warming anyway.
And this: “Actually, prior to the inanity of the failed global warming hypothesis, it used to be rather standard meteorology”
Did not get a mention in my professional training my friend.
As I’ve told you before and provided links to the mathematics of it.
Patrick
December 29, 2013 2:26 am
This reminds me of the BBC’s orbit documentary (Which, in later episodes which I did not see, apparently attributes this to AGW). This “engine” does not seem to be driven by CO2.
As for the engine comments, yes cold, denser air makes them run better (volumetric efficiency – nonturbo). That’s why, on a cold, damp morning, a nice twin carb’d 6cyl engine in say a Triumph GT6 runs and sounds GREAT!
December 29, 2013 2:28 am
norah4you says:
December 28, 2013 at 7:13 pm
The data in the bloggarticle above is only one out of many where incorrect data is used to prove (sometimes disaprove) thesis drawn from incomplete datamodels.
Which data? I have only presented a model and some examples.
There are no climate data in my article.
/ Jan
Crispin in Waterloo
December 29, 2013 2:52 am
TB’s first comment was quite sciency but contains numerous conceptually errors and I was pleased several people took them on. I don’t have time to address them all but I will note that “storminess’ is not increased by raising the average temperature of the system. The simplest demonstration of this is the terrible and powerful storms that occur on very cold planets within our solar system.
I will take issue in more detail with the version of the Carnot Cycle as a model for the heat engine that is the atmosphere. As I see it, the explanation of an ideal cycle is fine but misleading enough to lead many into blind alleys. There is no point debating how ideal dry gases behave when the atmosphere is full of evaporable and condensible water. Rising currents of damp air harvest moisture out of the air parcels into which they rise and cool. It doesn’t have to be at any particular altitude or temperature.
The speed of rising is determined by the relative temperature of the rising parcel and the stationary air medium. Increasing the system temperature increases the enthalpy but does not increase “the power of the storm” as claimed by the Guardian. People’s homes are not blown flat by enthalpy. The silliness of the Guardian’s position is enduring testimony to the fact their science writers can’t read. Their African version is called “The Mail and Guardian”. It is advertised as ‘Africa’s best read’. Well, it is not Africa’s best write. Given the litany of insults they have hurled against scientists who actually know what they are talking about they are richly deserving of the exposure given them here.
From a physical point of view the Earth is a closed system, for all intents. From an energy point of view it is an open system. The whole planet can cool and die and sometimes it nearly does. As for its upper temperature it is strongly self-regulating with the aforementioned water vapour as the principal factor, as well as ozone and to lesser extents GCR’s and CO2. These control mechanisms easily reach 40 or 50 watts per square metre. A sustained drop of 20 watts reaching the surface initiates an ice age.
Self-regulating heat pumps or heat engines, if you will, occur throughout nature and according to their size they behave in characteristic ways. The speed at which they operate is always optimal meaning at maximum efficiency balanced on the edge of turbulent flow. I think Prof Adrian Bejan, author of dozens of articles and textbooks on convective heat transfer, would be surprised to hear that the efficiency with which the Earth dispenses with heat drops as the temperature rises. The temperature of the cold side is constant. Why all these red herrings?
Wiliis has ably demonstrated the tropical albedo and thunderstorm cooling effects that Bejan pointed to when he stooped to comment on the trivial matter of how the atmospheric heat engine works. It is indeed a Carnot Cycle with un-ideal gases and some chaotic wind thrown in. Doubling the CO2 concentration, which could happen naturally, will not measurably change a thing. At least not with the network of instruments we have now.
December 29, 2013 3:02 am
TB says:
December 28, 2013 at 9:35 am
But I feel the comparison of the Earth’s climate system with a Carnot engine is inappropriate.
A Carnot engine is a closed system – that is, it deals with an energy differential between two sources, these closed off from the outside. It acts by transferring energy from a warm region to a cool region of space and, in the process, converting some of that energy to mechanical work.
I am not saying that the climate is a Carnot engine. No such engine exists because the Carnot engine is only a concept for an ideal heat engine. But the point is that all such mechanisms which convert differences in heat into mechanical energy have to obey the same fundamental physical laws. The Carnot theorem is fundamental because it is a direct consequence of the second law of thermodynamics.
Whether it is two or more sources and sinks does not matter, if it helps one can also imagine that the climate system is modeled by a zillion microscopic heat engines, the concept is the same.
By showing that a Carnot engine has less efficiency when all temperatures are risen equally much, then I will argue that one can assume that a real heat engine working on the same temperature ranges should also have less efficiency if all other data is considered to be unchanged.
You may say that the last assumption will be wrong, and yes, as I have written above, they will not be unchanged; we can assume that the cold areas of the planet will heat more than the warm areas, but that will only reduce the efficiency of the heat engine further.
/ Jan
December 29, 2013 3:27 am
TB says:
December 29, 2013 at 1:29 am
“Konrad: I have discussed this with you on another thread I believe.”
—————————————————————————————-
Yes TB, I believe you have…
You claimed that vertical tropospheric circulation strong enough to generate the observed lapse rate would exist in the absence of radiative cooling at altitude because –
“A natural lapse rate will develop regardless of atmospheric radiative responses. Due differential latitudinal heating > frictional turbulence > convection > overturning > geopotential height gradient > Coriolis > adiabatic cooling/heating – creating a heat pump and warming the lower layers/cooling aloft. Radiation is NOT needed. Models know this.”
I stand by my response –
“I would suggest that you have left out wizards, unicorns and climate “scientists” frantically waving their hands trying to drive megatonnes of gas in a giant flow from surface to 15Km altitude and back.”
“Out of your depth on a wet pavement” doesn’t really cover it. How about “so far out of your depth the fish have lights on their noses”?
Might I suggest that “pwned” would be an elegant sufficiency…?
December 29, 2013 3:31 am
chris y says:
December 28, 2013 at 10:54 am
There are at least 3 interesting conclusions one can take from this simple model.
1. If the sea surface temperature warms by 1K, then Ts=301K, To=200K, and the wind velocity is predicted to increase by 0.5%. There is no way to discern this tiny change.
Here you only rise the temperature for the warm element and let the cold temperatures remain unchanged. Of cause the wind will increase then, you need no model to understand that, but no one believe that only the warm parts of the planet will heat up.
2. If the climate models are to be believed, then the tropical troposphere should warm more than the surface. For example, if the sea surface warms 1C, the troposphere hot spot should warm 1.5C. That is, Ts=301K and To=201.5K. Then, the cyclonic wind velocity is predicted to DECREASE by 0.6%. Go figure…
This is a more realistic scenario which confirms what the article says.
3. Because we know that sea surface temperature changes of 5C (eg from 25 C to 30 C) can have a huge impact on cyclonic wind speeds (provided other conditions like wind shear are also just right), the Carnot model is useless for predicting wind speeds.
No, because the Carnot theorem is a fundamental principle, it will always apply when heat differences are converted to mechanical energy. How can you claim that it apply if you change the temperature by 1 C as in your point 1 and 2 above, but not when you change it with 5 C?
/ Jan
December 29, 2013 3:42 am
You can get an isothermal structure in a tall glass column if vertical circulation is constrained.
You cannot get it for a rough surfaced rotating sphere illuminated from a point source.
It is not necessary for the conduction air to a colder surface to achieve net warming of that surface
All that is necessary is:
i) A surplus of incoming radiation over outgoing radiation where illumination is full on. The disparity being caused by conduction to the air.
ii) A reduction in the rate of outgoing radiation elsewhere to a level lower than would have been the case for a surface with no atmosphere. The disparity being caused by the insulating effect of adiabaically warmed descending air.
The energy engaged does not have to be large because the bulk of energy passing through is undisturbed.
All that is necessary is that there be SOME diversion of energy throughput via conduction to the convective overturning and whatever the amount of that diversion the average global surface temperature will rise proportionately.
On Earth that is about 33 C.
That proportion is determined by mass and not radiative capability.
December 29, 2013 3:52 am
gbaikie says:
December 28, 2013 at 10:50 pm
Ah, I wonder if this would apply to what I call a pipelauncher.
This was a really original and funny idea gbaikie, I enjoyed reading it.
Much of what you say is correct, but unfortunately it will not work. However, I think this is an excellent problem for physics or math students, let them try to figure out why this will not work.
Well, the answer is that you will have to accelerate not only the rocket but also the column of water inside the cylinder below the rocket. The mass of that water column will increase with the length of the pipe and will limit the launching speed of the rocket considerably. I have not done the math, but I will guess it will be less than 100 mph.
/ Jan
December 29, 2013 4:21 am
Col Mosby says:
December 28, 2013 at 9:28 am
My understanding was that a car’s engine produces more power with a colder intake air because colder, denser air contains more oxygen and allows more gasoline to be completely burned during the cylinder explosion. Of course, it would make sense that thinner, warmer air cannot expand as much, thus produces less power.
Col, you are right, more dense air is also an effect of colder air and this will also give higher effect . But the principles of the Carnot theorem also apply. The source of the work comes from the increased pressure inside the cylinder. The pressure builds up because the temperature increases, and most of the temperature increases is caused by burning fuel inside the cylinder.
You can see from the ideal gas laws that if the temperature starts at a higher level you get lower increase in the pressure for a given increase in the temperature.
Thank you for the comment
/ Jan
December 29, 2013 4:22 am
Stephen Wilde says:
December 29, 2013 at 3:42 am
“…you can get an isothermal structure in a tall glass column if vertical circulation is constrained”
Glass?!
While I may use silica micro spheres vacuum packed in vacuum metallised mylar in the original experiments, due to inhalation risk I only provide instruction for replication using EPS foam.
AGW believers may be trying to build these things!
Everybody take out a circle of paper and a safety crayon….
December 29, 2013 4:46 am
Sorry, a typo. Don’t know where it came from.
December 29, 2013 4:49 am
Konrad, this is what you said and I mistyped.
“In tall gas columns,”
December 29, 2013 4:59 am
tobyw says:
December 28, 2013 at 5:48 pm
Is there electromagnetic and gravitic heating in our planet?
Toby, there are tidal energy heating our planet, but it is negligible compared to the solar heating.
The source of the tidal energy is the earth rotation. Because the earth is not perfectly elastic, some of the gravitational stretching from the moon and sun is converted to heat. This slows the earth rotation.
Since we know that the earth rotation slows very gradually, the tidal energy released each year has to be very small compared to the rotational energy of the earth.
/ Jan
NZ Willy
December 29, 2013 5:56 am
Konrad says: “Coriolis forces are not a driver of atmospheric circulation. Coriolis forces have no influence on air masses at rest within a rotating reference frame.”
Oh, bad call. You would be right if the atmosphere was perfectly homogeneous, but of course it’s full of local variations like changes in pressure, clouds, etc. These inhomogeneities want to travel in great circles over the Earth (what we surface-dwellers would call “straight lines”), but the Earth’s rotation follows a great circle only at the equator — elsewhere, the rotation is seen as having a lateral shear as it follows a small circle. A high-density air mass is pushed to the left in the Northern Hemisphere (to higher-latitude, slower-rotating places), so travel leftwards around the low-pressure cores of hurricanes — so counter-clockwise in the NH. So the Coriolis force is a terrific driver of winds as it “grabs” all the inhomogeneities in the air and forces them into motion.
Genghis
December 29, 2013 7:13 am
The atmosphere, absent external energy sources, would become isothermal (the same temperature from top to bottom) despite the density gradient imposed by gravity.
The dry lapse rate of 9.8 C/km is largely created by the thermalization of IR radiation by the greenhouse gases. Increasing the concentration of GHG’s increases the rate at which the 9.8 C/km is arrived at.
The environmental lapse rate of 6.5 C/km is caused by convection (and water vapors phase change) resulting in a warmer atmosphere than one with GHG’s alone (absent water of course ).
The question I have is what temperature would a pure nitrogen (no GHG’s or H2O) atmosphere be, given the ocean surface temperature of 22 C?
December 29, 2013 7:59 am
noaaprogrammer says:
December 28, 2013 at 5:00 pm
Now that gravity has been introduced into this discussion, what effect does any atmospheric tides due to the moon have?
noaaprogrammer
Good question.
Atmospheric tides are regular oscillations in the atmosphere which can result in mechanical energy such as wind.
These tides are a phenomenon with very different sources. Some of the tides are generated by the periodic heating of the atmosphere by the Sun. The atmosphere is heated during the day and not heated at night. This effect is similar to a heat engine as discussed in this article.
However, atmospheric tides are also powered by the gravitational pull of the Moon and the Sun similar to ordinary tides. This is a source for mechanical energy that is not powered by a heat engine. The source is the rotational energy of the earth.
This means that some of the mechanical energy in the weather system is not created by a heat engine, but it is a very small part.
/ Jan
chris y
December 29, 2013 8:33 am
Jan- you say:
“No, because the Carnot theorem is a fundamental principle, it will always apply when heat differences are converted to mechanical energy.”
Agreed 100%.
Jan again- “How can you claim that it apply if you change the temperature by 1 C as in your point 1 and 2 above, but not when you change it with 5 C?”
My third point is that real world cyclones can have 100’s of percent increases in wind speed in response to a sea surface temperature increase (5C for example) that the Carnot model predicts should only give a few percent increase in wind speed. Yet the cyclone expert at MIT uses the Carnot cycle to model a cyclone, even though the results of the analysis are basically worthless.
The Carnot cycle is not a useful model when used to estimate cyclone wind speeds and/or the effects of global warming on cyclone wind speed.
gbaikie
December 29, 2013 8:41 am
own-gbaikie says:
December 28, 2013 at 10:50 pm
Ah, I wonder if this would apply to what I call a pipelauncher.
This was a really original and funny idea gbaikie, I enjoyed reading it.-
I am glad you found it amusing.
*Much of what you say is correct, but unfortunately it will not work. However, I think this is an excellent problem for physics or math students, let them try to figure out why this will not work.*
Such results I would find extremely interesting.
“Well, the answer is that you will have to accelerate not only the rocket but also the column of water inside the cylinder below the rocket.”
It’s like the scotty quote from star trek.
No the water doesn’t move [assuming acceleration is constant].
So water after one starts some accelerate [say 1 gee] then after this water inside pipe doesn’t move-the pipe moves pass it.
-The mass of that water column will increase with the length of the pipe and will limit the launching speed of the rocket considerably. I have not done the math, but I will guess it will be less than 100 mph.-
It seems it would have limit due to nature of water- things have problems going over speed of sound in water. So near terms I tend to think max speed would be around 300 mph.
As first type of it’s kind probably realistic to shoot for max of about 200 mph.
But I it could over time be developed so one was getting up to speed of sound. But don’t expect
supersonic speeds without running across unexpected problem. And it gets progressive
harder to do such speeds- particularly if gee have to be 1 or less gees of acceleration.
And if going to go as fast as 200 mph, this pipelauncher would leap a distance out of the water, unless one adds some sort of braking featues.
But one think if not in windy conditions, one have bottom of pipe leave the water by somewhere
around say 100 feet or so.
Also if column water pushed down is say 50, as soon bottom of pipe was 50 feet from the surface
one can’t accellerate, so one could decrease acceleration near the top so you not pushing water down to 50- say lower acceleration so it was instead 20 feet [or something like that.
If want to be fancy, you make that water leap up into the air after the rocket. Or you make to column of water accelerate, somewhat Such a “move” could used to help brake the rocket by some amount- [but too much vacuum] and it could crush the pipe- which would be bad.
Many people have said that 100 or 300 mph isn’t not significant for rocket which ulitimately travels 17,000 mph. But there are reason why couple hundred miles hour would significantly improve a rockets proformance. So like add, say 20% to it’s payload.
TB
December 29, 2013 8:46 am
Crispin in Waterloo says:
December 29, 2013 at 2:52 am
“TB’s first comment was quite sciency but contains numerous conceptually errors and I was pleased several people took them on. I don’t have time to address them all but I will note that “storminess’ is not increased by raising the average temperature of the system. The simplest demonstration of this is the terrible and powerful storms that occur on very cold planets within our solar system.”
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That, I believe, is because they generate internal heat, as well as having much different chemistry (and density/pressure boundaries as a result) : http://en.wikipedia.org/wiki/Jupiter
“Jupiter still radiates more heat than it receives from the Sun; the amount of heat produced inside the planet is similar to the total solar radiation it receives. This additional heat radiation is generated by the Kelvin–Helmholtz mechanism through contraction. This process results in the planet shrinking by about 2 cm each year.”
“The water clouds can form thunderstorms driven by the heat rising from the interior.”
“The outer atmosphere is visibly segregated into several bands at different latitudes, resulting in turbulence and storms along their interacting boundaries.”
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“The speed of rising is determined by the relative temperature of the rising parcel and the stationary air medium. Increasing the system temperature increases the enthalpy but does not increase “the power of the storm” as claimed by the Guardian. People’s homes are not blown flat by enthalpy. The silliness of the Guardian’s position is enduring testimony to the fact their science writers can’t read. Their African version is called “The Mail and Guardian”. It is advertised as ‘Africa’s best read’. Well, it is not Africa’s best write. Given the litany of insults they have hurled against scientists who actually know what they are talking about they are richly deserving of the exposure given them here.”
It does increase the power of the storm in the way that I described – but will do so again. You talk of moist air, well the strength of rising thermals has another driver – LH. This is where the extra energy will come from. Warmer air simply allows greater evaporation and can physically hold more WV molecules – hence more LH of condensation released aloft in cloud to fuel it’s growth/uplift (or Typhoon/hurricane development ) – all other things equal.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“From a physical point of view the Earth is a closed system, for all intents. From an energy point of view it is an open system. The whole planet can cool and die and sometimes it nearly does. As for its upper temperature it is strongly self-regulating with the aforementioned water vapour as the principal factor, as well as ozone and to lesser extents GCR’s and CO2. These control mechanisms easily reach 40 or 50 watts per square metre. A sustained drop of 20 watts reaching the surface initiates an ice age.”
GCR’s can’t, that I’m aware of, as they do not in any large number pass into the Trop, they are largely stopped by O3 in the Strat. A paper would be appreciated. Yes, an IA will be initiated via feed-back loop as the Earth’s NH progressively received less summer insolation and snow fields can survive through until the next winter. This increases albedo, lowers absorbed SW and allows CO2 to sink into cooling oceans. WV content additionally lowers with temp to give a triple whammy.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“Wiliis has ably demonstrated the tropical albedo and thunderstorm cooling effects that Bejan pointed to when he stooped to comment on the trivial matter of how the atmospheric heat engine works. It is indeed a Carnot Cycle with un-ideal gases and some chaotic wind thrown in. Doubling the CO2 concentration, which could happen naturally, will not measurably change a thing. At least not with the network of instruments we have now.”
I beg to differ that Willis demonstrated anything that wasn’t known already and the region he speaks of, though receiving max insolation, isn’t anywhere near large enough to regulate the Earth’s temp. Or he hasn’t quantified it as yet to my knowledge to prove such a case. I have posted links to papers that have concluded that SW reflected is near balanced by IR back-radiated over much of the tropical equator.
CO2 doubling’s effect has been known of via laboratory and empirical demonstration + mathematical theory for ~150 years. And there are spectrometer readings taken from ground-based instruments that indeed do measure it’s effect. As well as satellite measurement showing the radiative imbalance at TOA. More is coming in than leaving.
December 29, 2013 8:53 am
Max Hugoson says:
December 28, 2013 at 10:49 am
Automobiles MAY have had better “efficiency” on cold days in the era of carboraters …because the mixture was more proper.. BUT, since your engine coolent is controlled by a thermostat, the “sink” temp for an Otto Cycle, Auto Engine is almost completely constant the year around.
Max, take a look at what happens inside the cylinder if we use the laws of an ideal gas
PH = (TH/TL)*PL
PH = high pressure after the fuel has burnt
PL = pressure before the burning
TH = Temperature in the cylinder after the fuel has burnt
TL=Temperature in the cylinder before the fuel has burnt
Let’s assume that the burning will rise the temperature by 500K
Let’s assume a cold day is TL = 300 K and a warm day it is 350K
On a cold day you then have PH= 800/300*PL = 2.67*PL
On a warm day you have PH = 850/350 * PL = 2.44*PL
That means you get a higher pressure, which gives more effect a cold day. The energy used is the same, in both cases it rise the temperature with 500K, but you get more effect out of it when you start with a colder gas.
/ Jan
Box of Rocks
December 29, 2013 9:01 am
A couple of ;nitpicks here…
Though I do like the article from a Mech E point of view…
The first one…
Climate as a heat engine
Posted on December 28, 2013 by Anthony Watts
Guest essay by Jan Kjetil Andersen
As Willis describes in his article on December 21, the atmosphere can be seen as a gigantic heat engine, i.e. a machine which convert thermal energy, namely temperature, into mechanical energy, namely wind.
I will disagree. You are forgetting about water. There is a huge amount energy released into space from the transport of warm water pole ward.
Also –
TB says:
December 28, 2013 at 9:35 am
Jan:
I know what you’re getting at.
But I feel the comparison of the Earth’s climate system with a Carnot engine is inappropriate.
….
To say that the earth will become “stormier” in a warmer world is (Meteorologically) correct. (supposing an unchanging LR). But there are exceptions.
A warmer world will have more evaporated water to release LH and therefore convective cells will have more power available to push up through the atmosphere to counter that LR. ..
The High Plains’ most violent weather occurs when delta t is greatest in the spring. As the atmosphere warms through the summer the severity of the storms decrease. Hence in late August through early October violent storms nearly die out and are replaced by long gentle rains until the atmosphere cools.
Look at a graph of storm intensity v time for a given year – it is two humped.
Also plotting some of the Carnot cycle on a T-S diagram might help some people.
gbaikie
December 29, 2013 9:04 am
In regards this:
However, I think this is an excellent problem for physics or math students, let them try to figure out why this will not work.*
This is bunch of notes about generally tending to make them on *small side*
40 feet in diameter.
20 squared is 400, times pi is 1256 sq ft.
1 psi is 144 lbs per square foot. 1256 sq ft
is 180,864 lb force.
1″ walls: working pressure: 100 psi. Burst 200 psi
1/2″ ??: 50 working and 100 burst
http://www.texaspipe.com/barlows_formula.html
40,000 psi .5″ is 83.3 psi
.25 is 52 psi
40 foot diameter is 125.6 feet circumference.
100 foot length is 12560 square feet
1000 foot is 125600 square feet.
cubic feet of steel 40 dia and 1000 feet long which is 1/4 thick walls:
125600 divided by 48 is 2616.6 cubic feet
Cubic foot of steel weighs 489 pounds
1,279,550 lbs [640 tons]. creates 7 psi by it’s weight
30 feet in diameter.
15 square is 225, times pi is 706.5 sq ft.
1 psi is 101736 lbs of force.
1/4″ wall at 40000 is 69 psi
Circumference: 94.2 feet
1000 foot lenght is 94200 square feet
94200 divided by 48 is 1962.6 cubic feet of steel
Weighs 959,662.5 lbs [480 tons]
Creates 9.4 psi from it’s own weight.
30 feet diameter 2000 feet long
959,662.5 lbs x 2 is 1,919,325 lb [960 tons]
It’s mass creates 18.87 psi
To accelerate 1 gee, requires 37.7 psi
2000 ft long pipelauncher accelerating Falcon 9:
Gross weight of Falcon 9:
“Falcon 9 v1.1 performance was still 13.15 tonnes LEO or 4.85 tonnes GTO, but gross
liftoff mass had risen to as much as 505.846 tonnes. Liftoff thrust was 600.109 tonnes
for v1.1 and 1,800.327 tonnes for Heavy.”
http://www.spacelaunchreport.com/falcon9v1-1.html
So 600.109 tonnes [661.5 tons]
Say total including tower: 760 tons
1520000 + 1,919,325 lb is 3,439,325 [1720 tons]
To float requires 33.8 psi
To accelerate 1 gees require 67.6 psi
To accelerate for 10 seconds at 1 gee the distance traveled vertically is
10 second squared times 1/2 of 9.8 m/s/s [or 100 times 16 feet].
Resulting in the 2000 foot tall pipelauncher having it’s top [capped end]
1600 feet above waterline and 400 feet below the waterline.
And traveling at speed of 98 m/s [352.8 kpm- 219.2 mph]
98 m/s is about 320 feet per second
General conditions at 10 second mark: Air in pipe will be about 68 psi.
The air temperature in pipe will be around 100 C [200 F]. One will be spraying
liquid air and with burner burning Kerosene during the 10 seconds. One can also use
water [which at these pressure will boil above 100 C- but surfaces might exceed
the air temperature- and air will be very dry- so it will evaporate due partial pressure
of water vapor]. At or before the 10 second mark, the buring of kerosene is stopped.
And the continuation of spraying liquid air without heat added would reduce air pressure.
And upward continuation of velocity will also lower the air pressure.
So at 10 second mark air pressure is about 68 psi, if air pressure drops to 34 psi such air
pressure would still cause aceleration [33.8 psi causes it to float- 34 psi would cause it
to rise slowly.
68 psi is 4.6 atm. 1 Atm would press water 33 feet below waterline. So at 68 psi water
in pipe is depressed 152.65 feet waterline inside the pipe. And a decrease in pressure in
the pipe will cause the water to rise. Having 68 psi keep water at that level. Increasing
pressure to 68 psi pushes it down to 152.65 feet. It’s isn’t the forcing of water out the
pipe which is the cause of acceleration- it’s not “jet powered” but it’s also not an
insignificant factor in beginning and ending of acceleration.
So at 10 second mark, the pipe is 400 feet under water and has 400 – 152.65 feet of water
inside the pipe: 247.35 feet coulumn of water.
Due to momentum and not considering forces slowing such momentum, the pipelauncher travel at
320 feet per second, it will at 11 second mark have pipe will only be 80 feet under the water.
And since was 152.65 feet fill with air, 152.65 – 80 feet. Or 72.65 foot column of air would
be pushed out the bottom of the pipe. It also mean that at air pressure at 80 feet below
water line is 36.3 psi. So at 10 second mark there was 68 psi, at the 11 second mark the
pressure inside pipe would drop to about 36 psi.
Now one could make a design modification. Simple or more complicated. Simple would be
having big holes in the pipe, starting at 250 feet from the bottom of the pipe. Which means
the 250 feet of pipe would used to slow down the pipelauncher. So at 10 second mark air would
start “venting” out sides of pipe, and as the holes became closer to surface [within the 1
second] more air would pushed out to holes. Or you have doors at near or at top pipe which
open at 10 second mark.
Now if you do no modifications, and do nothing at the 11 second mark, you will have about
about 36 psi air in pipe, and within a fraction of second, the 36 psi will go to zero- same
pressure as atmosphere.
Or with no modifications but the continuation dumping in liquid air one can rapid cool
the air, causing the pressure to drop.
One can do lots of stuff. This is the art of driving a pipelauncher.
For instance for first 8 to 9 seconds of acceleration you could keep the air temperature
around 100 C or cooler, and in last second or two rapidly increase air temperature-
burn more kerosense and/or stop adding liquid air. And a fraction of second before 10 second
mark stop all kerosene burning, and dump liquid air.
If you have 500 K air and lower temprature to 250 K, you get 1/2 the pressure, 68 psi
becomes 34 psi, which stops acceleration and causes water to rise in the pipe.
Aluminum: 2.72
Steel: 7.82
PVC 1.36
Small diameter and long:
20 feet in diameter 2000 feet long [or longer].
200 feet length Marine Aluminum- fire box and ice box.
1600 feet of PVC
200 feet of steel
In water PVC has little weight. Steel has higher density and will function as keel.
PVC has Yield tensile strength 6500 psi or about 1/6th of 40 K aluminium alloy or steel.
1 inch wall 20 feet [240″] diameter can withstand 54 psi. 1.5″: 81.25 psi
Will use 1 1/2″ wall thickness will PVC and 1/2″ wall thickness with aluminium. And 1/4″ pipe
wall thickness with steel.
200 feet 20 foot diameter 1/4″ steel pipe weighs:
circumference: 62.8 feet. Times 200 is 12560 square feet.
12560 divided by 48 is 261.6 cubic feet of steel.
cubic feet of steel: weighs 489 pounds. Times 261.6 is
127,955 lbs [69.9 tons]. In water it weighs less than 8/7th this weigh [56 tons].
But as accelerated mass it’s about 70 tons.
Distilled water is about 62 lb per cubic foot [62.428]
So, Aluminum: 2.72 is about 170 lb per cubic foot and PVC is about 85 lbs
So Aluminum 20 diameter and 200 feet long is 12560 divided by 24 which is:
523.3 cubic feet. Weighs: 88966.6 lbs [44.5 tons].
PVC 20 ft diameter and 1600 ft is 12560 times 8: 100,480 square feet.
100,480 divided by 8 is 25,120 cubic feet. Which weighs:
2,135,200 lbs [1067.6 tons]. But in water it weighs about 22 lb per cubic foot [276 tons].
20 diameter circle has 314 square feet. At 1 psi it’s 144 lb of force.
314 times 144 is 45,216 lbs of force per 1 psi.
To float 200 feet above waterline: it weighs: 56 + 44.5 + 276 tons: 376.5 tons. Or about:
753000 lb. Divide by 45,216, means it needs 16.6 psi of air pressure inside pipe.
If rocket plus everything else weighs 376.5 tons, then need 33.3 psi.
Which is 2.65 atm and depressing water, 74.77 feet below the waterline.
To determine the air pressure with 1 gee acceleration one has to use mass rather than
weight. And 791.6 and 14 tons- 805.6 tons- 1,611,200 lb. So: 1,611,200 + 753000 lb x 2 is:
3,117,200 lbs. Divide by 45,216 is 68.9 psi. So it’s 68.9 plus 33.3 psi and that is
102.2 psi. Which exceeds the 81.25 psi strenght of PVC. But 1/2″ wall Aluminum can withstand
166.6 psi.
So the 102.2 psi is 6.95 atm or depresses water 229.5 feet below waterline.
acceleration.
Hmm. Let try 2 seconds at 1 gee. Make aluminium pipe 300 feet long. The aluminium is lighter
out of water and PVC is only a bit lighter than alumium in the water, but PVC would probably be
be cheaper. So half a gee is 68.9 / 2 = 34.45 psi. Plus 33.3 ps is 67.75 psi. So one more than
1/2 gee of acceleration.
Lets say part of launch tower is a tube, which is water resistent. Say this tube is 30 feet
tall, and allow pipelauncher to start it’s acceleration -20 feet below waterline.
In terms of pressure if PVC pipe is below 33 feet it would be overpressurize by 14.7 psi.
And 81.25 plus 14.7 is 95.95 psi and 66 feet is 110.5 psi. So with 1 gee acceleration
PVC must below 60 feet under water.
And PVC starts 320 feet under water. So within 4 seconds at 1 gee acceleration it moves up
256 feet. So at 4 second mark moving at 128 ft per second [39.2 m/s {87.5 mph}].
If 128 ft per second velocity were not to accelerate but kept constant speed, you run out pipe
lenght in 13 seconds- total being 13 plus the 4 seconds at 1 gee.
And roughly want the time to be about 10 seconds or more in total.
So try 8 seconds + 4 seconds. and 8 second at 24 ft per second per second, which distance of
768 feet. Plus 8 second at 128 ft per second which is 1024 feet. Which doesn’t work.
So 7 seconds: 24 ft per second per second, distance is 588 ft plus 896 ft plus the 256 ft.
Is 1740 feet. Minus 20 ft leaving 280 feet of pipe in water after 11 seconds of acceleration.
With velocity of 128 ft + 168 feet per second [296 ft/s [ 84.9 m/s or 189.5 mph].
All aluminum:
“Aluminum 20 diameter and 200 feet long is 12560 divided by 24 which is:
523.3 cubic feet. Weighs: 88966.6 lbs [44.5 tons].”
Times 10 to equal 2000 ft long: 88966.6 lbs [44.5 tons] * 10 = 889666 lbs [445 tons].
“20 diameter circle has 314 square feet. At 1 psi it’s 144 lb of force.
314 times 144 is 45,216 lbs of force per 1 psi.”
So, 19.6 psi to float.
Have rocket and tower plus everything else 445 tons times 2 = 890 tons
To float: 19.6 psi + 19.6 times 2 = 39.35. Total 58.95 to float.
1 gee is 117.9 psi. 20 feet diameter Aluminium 1/2″ wall can operate as high as 166.6 psi.
Could accelerate slightly faster than 1 gee or add more length.
Say add 22.1 psi of added pipe giving pressure of 140 psi.
Adding 1000 feet [3000″ in total] adds 9.83 to float and 19.6 psi with 1 gee.
So 10 second of 1 gee is 490 meters or about 1600 feet. 1600 feet out of water and 1400 feet
under water. And 117.9 + 19.6 psi is 137.5 psi [9.35 atm] so water pressed 308.67 feet under
water. So at 10 second aceleration one has 1091.33 feet of pipe with water in it and 1908.67
feet with air at 137.5 psi.
If pressure can rapidly halved to 68.75 psi, the this pressure floats payload and pipe.
Pipelauncher and rocket with continue to accelerate but be decreasing in acceleration- ceasing to
accelerate once pressure reaches 68.75 psi.
So at 10 second mark, after accelerating 10 seconds 32 feet per sec/sec, it’s going at
320 feet per second. So at 11 second mark it will traveled additional 320 plus it’s decaying
acceleration rate [called it DAR]. So volume of air in pipe will increase by 1908.67 feet
plus 320 feet plus DAR. And ignoring DAR it’s 1908.67 + 320 = 2228.67 feet.
A 20% increase of 1908.67 is 2290.4 feet.
A 100% increase would half the pressure- lose 68.75 psi. So at 20% one loses a 1/5th of this
so 13.75 psi. And lose 1/5 acceleration: 6.4 ft per sec from 32 feet per second.
So in one second one goes from 32 ft per sec/sec to 25.6. And the acceleration average during
the second is 28.8 feet per second.
So going back and adding DAR is 1908.67 + 320 + 28.8 = 2257.47 feet.
But decrease of 13.75 of psi will causes the water inside pipe to rise about 30 feet- this is
not instantaneous [it accelerate from zero to 6.4 ft per sec/sec during the 1 sec.]
So at 12 second mark the pipe travels up 2 times 320′ [plus 2 seconds of DAR].
Again roughly, pressure reduces by another 1/5th.
So in two seconds, pressure at 137.5 drops by 13.75 + 13.75 psi- falls to about 110 psi.
And roughly, [320′ times 2] + [28.8 times 2] equals 697.6
One had 1091.33 feet of pipe with water minus 697.6 feet, leaves 393.73 water in pipe, but
water also increasing as water accelerate up pipe [a minor amount so far].
And with water still in pipe, one gets another second of acceleration- 1 second +.
So DAR has decreased from 32 ft per sec/sec to 25.6 and second two 25.6 to 19.2 feet per second
per second. Add to 320 feet per second 44.8 per second [248 mph]. And after second 3:
12.8 ft persec/sec [1/2 to 1/3 gee]. And pressure drop by about another 13.75 psi [end of 13
seconds pressure inside pipe will be about 96 psi [ 6.5 atm and pushes water 216 feet below sea
level. The 137.5 psi [9.35 atm] pushing it 308.67 feet under water. Accelerationg of
water was 6.4, 12.8, finally 19.2 feet per sec per sec. And water rising 3.2, 12.8, 22.4 =
38.4 feet. after 3 seconds [and “wants” to be about 60 feet higher- and sort of like water
falling from 60 feet up- it’s rushing like a waterfall upwards].
But we can add 38.4 feet to the 393.73 feet: 432 feet.
The distance traveled in 12 to 13 second is 320 + 28.8 + 22.4 + 16 = 387.2 feet.
So 48.8 feet of water left in pipe.
In terms pipe in water. [320 * 3] + 28.8 + 22.4 + 16. Had 1400 feet minus 1027.2 feet =
372.8 feet.
All kinds of excitement occurs in the 13 second to 14 second mark. Pipe will at around 14 second
mark, be leaving the water, water once in pipe will be chasing after it, and lacking water
pressure to prevent air pressure from leaving pipe. One not have a second of acceleration
after the 14 second mark.
“30 feet in diameter.
15 square is 225, times pi is 706.5 sq ft.
1 psi is 101736 lbs of force.
Circumference: 94.2 feet
1/4″ steel wall at 40000 is 69 psi
1000 foot lenght is 94200 square feet
94200 divided by 48 is 1962.6 cubic feet of steel
Weighs 959,662.5 lbs [480 tons]”
Aluminum: 2.72
Steel: 7.82
1/2″ 360″ diameter, Aluminum alloy with 40,000 psi strength burst pressure: 111 psi
Aluminum:
1/2″ rather than 1/4″
“Cubic foot of aluminum will weigh 171 pounds”
Instead of 94200 divided by 48 it’s divided by 24:
3925 cubic feet of Aluminum. Which weighs 671,175 lbs.
Cap: 706.5 sq ft 1/2 thick: 29.4 cubic feer: 5034 lbs
Total weight 676,209 lbs.
Rocket weight: 2 million lbs
Rough total weight is 2.8 million lbs
And 2.8 million divided by 101736 lbs per 1 psi is
27.522 psi or 1.87 atm.
In terms of total amount of air it’s 1 atm plus 1.87 atm.
Mass of air 1000 feet length and 30 foot diameter:
706.5 sq ft time 1000 is 706,500 cubic feet of air
“According to the CRC Handbook of Chemistry and Physics, the density of dry air at
20 degrees C at 760 mm of mercury (one atmosphere of pressure) is 1.204 milligrams per
cubic centimeter.
1 cubic foot = 28,316.8467 cubic centimeters.
So, dry air weighs 34,093.48 mg per cu.ft.
Which is about 1.2 ounces per cu.ft. ”
So 1 atm at 20 C, 706,500 cubic feet of air is 847,870 oz.
And 2.87 atm at 20 C is 2,433,186 oz. Or 152,074 lbs
Or 76 tons.
Assume have this pipelauncher [gross weight: 400 tons] with 1000 ton rocket is floating with top
of pipelauncher 20 feet above the water. It has 27.522 psi of air in it. 14.7 psi of air will push
to water 33 feet below sea level- or about 2.25 feet per psi. So 27.522 psi of air will push water
61.78 feet under water. And with 20 feet above water, one has about 81 feet of air at
1.87 atm + 1 atm. One starting with about 1/12th of total 76 tons of air- about 6 tons.
So if one adds 70 tonnes of air, fills the entire pipe length of 1000 feet with 1.87 + 1 atm
quantity of air.
So if added 70 tons of liquid air and the liquid air was warmed to 20 C- [pouring it into warm
ocean water would more or less do this. Considering area of 30 diameter pipe is 706.5 sq ft and
one foot depth of water is 706.5 cubic ft. Is 22 tons of water. And dumping a ton or more
liquid air it will mix with the water several feet under the surface of the water.
Plus turning the water into ice requires a lot energy in terms of heat loss. Making it improbable
that liquid air dumped into warm sea water would be cooler than 0 C, and more mixed with water
the higher temperature of the air.
So roughly air should limited roughly between range of 0 and 20 C without adding additional
heat [which will be done]. The air in the pipe will be heated over 100 C, but first
starting with assumption of air heated to 20 C.]
So if put 70 tons of air at 20 C in pipe and it’s 1/2 full of air- 500 feet of length has air
in it, the air pressure will be 2 times 2.87 atm Or doubling the volume will 1/2 the pressure/density
if same temperature.
If add 70 tons of air at 20 C over long time period- anything over couple minutes is long period,
the pipelauncher and rocket payload will rise until bottom of pipe is 62 feet under water and top of
pipe is 938 feet. And it’s unstable in this position- it flip over it not balanced. But it floats
vertically if prevented from flopping over.
If instead of slowly adding 70 tons of air, one were to very rapidly add the 70 tons- say in less
than 1 second, one have too much pressure in the pipe. If make the pipe strong enough to
withstand such pressures, one gets the pipelauncher going up quickly and the water being pushed
down quickly.
Instead of slowly adding air and very quickly adding air, one want to do is increase the pressure
to certain amount and keep the pressure at this level.
So with gross weight of 2.8 million lb it requires 27.522 psi pushing up on top of pipe and
pushing down on water inside pipe. Top of pipe has area of 706.5 sq ft or 101,736 square inches.
If pressure is doubled [55.04 psi] then water pushed down twice as far and one gets 1 gee of
acceleration for long as there is 55.04 psi. And water stays at level of twice depth.
And at 55.04 psi and with 500 feet of air in pipe. And air is at 20 C that requires 70 tons of added
air.
And if double the temperature in K, it doubles pressure. 20 C is 293 K. So increase to 313 C it’s
twice pressure. And increase temperature by 50% [+146.5 C] or to 166.5 C it increase pressure by 50%.
It also means if decrease air temperature by 50% from 20 C that is -126.5 C, it halves pressure.
Liquid Nitrogen boils at -195.8°C in 1 atm. So spray liquid nitrogen in say 20 C air, it could cool the
air down to -126.5 [as -126.5 C air is still warm enough to boil nitrogen droplets of liquid].
So if you had 100 feet of air in pipe at 20 C, and misted liquid nitrogen into air and cooled air to
-126.5 C to column of air at same pressure would shrink by 50%. Go from 100′ to 50′. And if air warmed
back up to 20 C it expands back to 100′ and if heated to 166.5 C, expands to 150′.
So with 20 C air and 20 feet above water, we column of air being about 82. If instead it was 40 feet
above water, one have 102 feet column of air, and cooled air to -126.5 C, the top of pipe go below the
sea level [10 feet below]. If doubled the 82 feet so had 164 feet of air at the start, and cooled to
-126.5 C then it drops down to 20 feet above water. And instead starting with 6 tons of air in pipe,
So at start of launch sequence, the top of pipe is about 100 feet [10 stories] above to water, you cool
the air and pipelauncher drops and once stops falling, use kerosene or methane burners to heat the air
and dump liquid nitrogen [or liquid air] into the water- turning it into somewhat cool air.
So since increased the starting tonnage of air by 6 ton, the 70 tons is 64 tons.
So with 64 tonnes of air added and temperature of 20 C, and half pipe full this gives 1 gee of acceleration.
Or by the correct amount in, we get 1 gee acceleration up point where 500 feet of air at 20 C is
in the pipe. At 32 feet per sec per sec that is 5.59 seconds of 1 gee, and speed of 178.88 feet per second
[54.5 m/s- or about 128 mph].
Increasing air temperature from 20 C to 166.5 C increases the 500 feet to 750 feet.
And at 32 feet per sec per sec at 1 gee up to point of 750 feet is 6.84 seconds of acceleration, and
speed of 219 feet per sec [66.79 m/s- 132.5 mph].
So with 750 feet of pipe out of water, one is left with 250 feet under the water at the point of 6.84
seconds of acceleration. And the water is pushed under water 61.78 feet times 2 [123.56 feet under the
water] and one is traveling up at 219 feet per sec. So at this point the total column of air is 750
plus 123.56 feet. Which is 873.56 feet of air at 55.04 psi and air temperature of 166.5 C.
If stop adding any more air or heat, one still have +50 psi and air temperature of +160 C for the next
1/2 of second, so you have slightly less than 1 gee of acceleration for 1/2. And in 1/2 second you
travel 219 + 8 feet per second divided 2, or about 113.5 feet upwards. Or add 113.5 to 873.56 feet-
which is 987.06 feet- leaving only 12.94 feet of pipe still in the water. And within the 1/2 second
the air in pipe will be expelled out into the water.
A conclusion one could reach is if one going to heat air to 166.5 C, then one add less than 64 ton of
liquid air. So say less than 60 tons of air added.
And instead acceleration at 1 gee for 6.84 seconds, it could be for 6 second. And addition 1 second of
acceleration at slightly less than 1 gee.
So 6 seconds is 576 feet upward, rather than 750 feet. And 7 second at almost 1 gee 784 feet.
And less than velocity of 224 feet per sec [68 m/s- 152 mph].
In all cases once bottom of pipe leaves water, one gets no more acceleration.
And this last “refinement” at 7 second point, the pressure in pipe is dropping and water isn’t pushed
down as much and is rising. Both these factors are occuring after 6 seconds, but by 7 seconds they
are amounting to something. So at/after 7 second air pressure would be less than 50 psi and water
pushed down less than 113 feet. So 784 + less than 113 feet is less than 897 feet. So more than
100 feet of water in pipe. And more than 216 pipe in the water after 7 second of accleration.
Which means one have about in additional + 3/4 of second of acceleration and + 3/4 of gee.
So one further reduce acelleration of 6 second gee, to say 5 3/4 of second and get about 2 seconds
of about average of .8 gee. And need to use less tonnage of air. And get slightly more than 150 mph.
And one can do better than 150, if one uses more than 1 gee for say first 4-5 seconds of acceleration.
Such increase in acceleration doesn’t require more tonnage of air, just quicker use of air, and bit
more kerosene/methane use [and one using a small amount fuel- roughly one heating 76 ton of air by
+ 140 C.
To heat 1 kg of air by 1 C: 1.0 kJ/kg.K
So say 70,000 kg times 150 K equals 10.5 million kilojoules.
Kerosene 46,200k kJ/kg
http://www.engineeringtoolbox.com/fuels-higher-calorific-values-d_169.html
So 227 kg of kerosene.
Tiny pipelauncher:
5 to 30 ton gross mass rocket:
30 ton: 1 gee acceleration.
8 feet diameter
250 feet long
8 feet diameter. Area 50.24 square feet
50.24 sq ft is 7234.56 square inches
1 foot length is 50.24 cubic feet. 1 cubic foot water weighs about 62 lbs:
1 foot length displaces 50.24 time 62 is 3114.88 lbs.
100 foot lenght: 311,488 lb
To push water 100 ft requires about 1 atm per 33 ft depth. So about 3 atm [14.7 * 3 is 44.1 psi]
1/8″ thick wall pipe 96″ diameter can easily withstand 45 psi.
Circumference of 8′ diameter pipe 25.12 feet.
1 foot length of pipe is 25.12 feet by 1 foot. 25.12 square feet.
1/8th inch thickness of 1 foot square requires 96 square feet to equal one cubic foot.
Or 25.12 square feet at 1/8″ is .261666 cubic feet of material
Cubic foot of aluminum Weighs= 168.5 lbs
“1 cubic foot of aluminum weighs 169.344 Lbs”
“Aluminum is about 171 pounds per cubic foot.”
Assume it’s 171 lbs. Times by .261666 is 44.75 lbs
And 250 feet weighs 44.75 times 250: 11,185.25 lbs
Cap is 50.24 square feet so 1/8″ is 44.75 lbs. 1/4 is 89.5 lb
So 1/8 wall and 1/4″ cap and 8′ diameter and 250 lenght mass is
11,185.25 + 89.5 lb. 11,274.75 lbs. [5 1/2 tons]
Without any additional mass pipelauncher floats with 3.61 feet of air in pipe and
at air pressure of 1.558 psi.
{A human with pressure of breath raise the level it floats above the water. And
would not cause it to sink by much if stood on it.}
“The density of seawater varies between 1.02 and 1.03 SGU, so a cubic foot of seawater
weighs between 63.6 and 64.3 pounds at 40 degrees F.”
“1 cubic foot of water weighs approximately 62.46 pounds.”
“density of fresh water at 20C = 0.998 g/cm³ = 998 kg/m³ = 8.33 lb/gal = 62.1 lb/ft³
density of seawater = 1.025 g/cm³ = 1025 kg/m³ = 8.55 lb/gal = 63.8 lb/ft³
1 foot length: 50.24 cubic feet times salt water at 63.6 is 3195.2 lbs
30 ton payload: 60,000 lbs + pipelauncher 11,274.75 lbs is 71,274.75 lb
71,274.75 lbs. 1 foot length: 3195 lb. 71,274 lbs divided by 3195 is 22.3 feet
71,274 divided by 7234.5 square inches is 9.85 psi
1 gee acceleration:
9.85 psi times 2 is 19.7 psi. And water depressed 44.6 feet under sea level.
9.8 m/s/s 19.7
19.6 m/s/s is 29.5 psi. and water depth of 66.9 feet.
TB
December 29, 2013 9:44 am
December 29, 2013 at 3:27 am
TB says:
December 29, 2013 at 1:29 am
“Konrad: I have discussed this with you on another thread I believe.”
—————————————————————————————-
Yes TB, I believe you have…
You claimed that vertical tropospheric circulation strong enough to generate the observed lapse rate would exist in the absence of radiative cooling at altitude because –
“A natural lapse rate will develop regardless of atmospheric radiative responses. Due differential latitudinal heating > frictional turbulence > convection > overturning > geopotential height gradient > Coriolis > adiabatic cooling/heating – creating a heat pump and warming the lower layers/cooling aloft. Radiation is NOT needed. Models know this.”
“I stand by my response –“
Fair enough;
I stand by my knowledge/training/text books and global models that do not incorporate it as a driver.
Sorry if that is an “appeal to authority”.
“I would suggest that you have left out wizards, unicorns and climate “scientists” frantically waving their hands trying to drive megatonnes of gas in a giant flow from surface to 15Km altitude and back.”
No, it’s just basic meteorology my friend, not magic.
“Out of your depth on a wet pavement” doesn’t really cover it. How about “so far out of your depth the fish have lights on their noses”?
Now, no need to get snarky.
Radiative cooling is a factor affecting the temp profile of the Trop to a small degree, but nowhere is it a driver. Convergence/divergence aloft is the principle “top end” physics process in the troposphere.
Sinking air inherently warms under compression anyway, as does the air below this sinking air, so no sig temp differential would exist to provide a sinking motion other than air converging aloft.
Hadley Cells’ prime driver is convection/LH release.
Have you seen the vids of rotating fluids cooled at the centre and warmed at the edges?
They mimic the Earth’s hemispheric flow well enough, without radiative effects having an effect.
Might I suggest that “pwned” would be an elegant sufficiency…?
??
Doug Proctor
December 29, 2013 10:02 am
Would there be a perspective gain in seeing the world not as a gigantic heat engine but a gigantic heat REDISTRIBUTION engine?
The difference I see is that as a redistribution engine (or system) regional heating or cooling is not a sign of additional or less energy per se, i.e. an unbalanced energy budget, nor a system in flux adapting to a change in initial parameters, but a system operating in THIS way at THIS moment.
As a heat engine, we see the world as deterministic. As a heat redistribution engine, we could see the world as probabilistic. Regionalism is not a problem with redistribution but it is with an engine, unless we decide that the engine is either not working “properly” or has had the accelerator push/pulled.
Trenberth wants the deep ocean to be warming (even though he can’t see it) to solve his problem. What if the prior heating is a deep ocean heat RELEASE issue: we know the thermal characteristics of water are so much greater than air that a small, unnoticed heat transfer from the ocean to the atmosphere would kick the air temps up a great deal.
The climate debate is a who-dun-it where the police have decided that the neighbour did it, and spend all their resources to prove that hypothesis, while ignoring the crack-addicted thug on the corner of the block.
December 29, 2013 11:33 am
Box of Rocks says:
December 29, 2013 at 9:01 am
A couple of ;nitpicks here…
I will disagree. You are forgetting about water. There is a huge amount energy released into space from the transport of warm water pole ward.
No, I am not forgetting about it, but it is not relevant in this context.
We know that we have an uneven distribution of heat around the globe. Some of this is distributed by water and some by air, but I don’t go into that problem. I take the situation as it is, that we have different temperatures around the globe, as a starting point. And from this starting point we know from the physical laws that this different temperatures will stir up some movements in the air and in the water.
These movements are mechanical energy in the form of wind and ocean currents. I am only discussing the winds in this article, and my point is that the potential for creating mechanical energy in the form of wind decreases when all temperatures increases uniformly. When the lower temperatures on the planet increase more than the higher temperatures, as the models predict, the potential for mechanical energy will decrease further.
/ Jan
December 29, 2013 12:15 pm
How the atmosphere processes heat is pretty simple.
There has never: not ever: been anything about the atmosphere that stopped mankind from filling lightweight airframes with human life – as in “we’ll sue you out of existence if there’s anything major you don’t know about this and it causes a problem,” – and landing those human beings anywhere they saw fit.
Just because the people alive today can’t remember doesn’t mean the entire applied research aerospace field didn’t have a very,
very good idea, of the fundamentals of atmospheric action.
It’s a pot of fluid compounds, one of them boils off at the bottom, rises until it emits that heat, condenses and falls again. The others move around in it’s wake, along with earth’s spin imparting it’s motion.
It’s just, not nearly as complicated conceptually, as the entire Magic Gas preaching world, tried to pretend.
Then again pretense is all they had to work with.
And when you make your living
pretending refrigerants and coolant compounds bearing them: water’s the refrigerant, the cold nitrogen and oxygen are the coolant compounds bearing it to the surface of the heated target object –
are all part of “a giant heater,” because you ‘did the math’ and “it shows you that immersing a hot rock into cold, refrigerated nitrogen, made the rock get hotter than when you heated it up in vacuum” – the claim made by warm atmosphere religionists –
it’s no wonder they can’t tell which way a thermometer will go.
And that everything looks impenetrably complicated.
December 29, 2013 12:31 pm
The problem is you self owned.
The whole reason for convective overturn being what it is,
is the radiative gas water: which magnifies the atmosphere’s activity.
So while telling someone ‘pwnd’ is ALWAYS elegant and sufficient,
self-pwnd is even better: and you did because he tried to tell you: the radiative gas class drives the entire thing much harder than if there were no water, being fundamentally responsible for each, individuall Hadley circulation. Each Hadley circulation’s nothing more than a “heat pipe” which is a phase change refrigerator.
Which means the atmosphere’s nothing but a series of refrigerating Hadley cells, or in other words a series of
phase change
refrigeration cells.
This isn’t complicated it’s been known for years, a group of unethical people called ‘Green House Gas Effect” believers hijacked atmospheric science and destroyed an entire current generation’s understanding of gas energy science.
So I’ll call “pwnt”
and consider that I helped by reminding everyone that it’s been known for years,
the Hadley cells are mini refrigeration cycles
the refrigerant being water.
==============
TB says:
December 29, 2013 at 9:44 am
December 29, 2013 at 3:27 am
TB says:
December 29, 2013 at 1:29 am
“Konrad: I have discussed this with you on another thread I believe.”
—————————————————————————————-
Yes TB, I believe you have…
You claimed that vertical tropospheric circulation strong enough to generate the observed lapse rate would exist in the absence of radiative cooling at altitude because –
“A natural lapse rate will develop regardless of atmospheric radiative responses. Due differential latitudinal heating > frictional turbulence > convection > overturning > geopotential height gradient > Coriolis > adiabatic cooling/heating – creating a heat pump and warming the lower layers/cooling aloft. Radiation is NOT needed. Models know this.”
“I stand by my response –“
Fair enough;
I stand by my knowledge/training/text books and global models that do not incorporate it as a driver.
Sorry if that is an “appeal to authority”.
“I would suggest that you have left out wizards, unicorns and climate “scientists” frantically waving their hands trying to drive megatonnes of gas in a giant flow from surface to 15Km altitude and back.”
No, it’s just basic meteorology my friend, not magic.
“Out of your depth on a wet pavement” doesn’t really cover it. How about “so far out of your depth the fish have lights on their noses”?
Now, no need to get snarky.
Radiative cooling is a factor affecting the temp profile of the Trop to a small degree, but nowhere is it a driver. Convergence/divergence aloft is the principle “top end” physics process in the troposphere.
Sinking air inherently warms under compression anyway, as does the air below this sinking air, so no sig temp differential would exist to provide a sinking motion other than air converging aloft.
Hadley Cells’ prime driver is convection/LH release.
Have you seen the vids of rotating fluids cooled at the centre and warmed at the edges?
They mimic the Earth’s hemispheric flow well enough, without radiative effects having an effect.
Might I suggest that “pwned” would be an elegant sufficiency…?
??
Dinostratus
December 29, 2013 2:03 pm
Well this post went chest up.
“the maximum theoretical efficiency decreases with increasing temperatures”
Um. No. Not really. Sort of? Hmmm…. No. Not really. The “maximum theoretical efficiency” decreases as the cycle becomes less square on a T-s diagram. The solution to increasing differences between Th and Tc is to increase the range of Va to Vc thus making it more square in the, again, T-s space. These sorts of problems are better worked out in T-s space or at least P-T space since dP drives V.
I didn’t have the heart to read the rest of the argument but the increasing temperature difference leads to more storm energy is a basic NS energy balance argument. Lindzen gave the best rebuttal which was to show that GWT predicts increasing polar temperatures and even tropical temperatures thus lessening the temperature difference that drives storms.
TB
December 29, 2013 4:02 pm
[Snip. Duplicate post. ~ mod]
TB
December 29, 2013 4:11 pm
December 29, 2013 at 12:31 pm
The problem is you self owned.
The whole reason for convective overturn being what it is,
is the radiative gas water: which magnifies the atmosphere’s activity.
So while telling someone ‘pwnd’ is ALWAYS elegant and sufficient,
self-pwnd is even better: and you did because he tried to tell you: the radiative gas class drives the entire thing much harder than if there were no water, being fundamentally responsible for each, individuall Hadley circulation. Each Hadley circulation’s nothing more than a “heat pipe” which is a phase change refrigerator.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
No he didn’t actually. He (Konrad) claims a HC is driven by radiative cooling aloft causing descent. It is not. Full stop. The reasons are as I state, and are incorporated in the equations that go into NWP models. By his reasoning they’d go berserk.
I also said the prime driver is convection and LH release. You have just said the same thing in another way.
“Which means the atmosphere’s nothing but a series of refrigerating Hadley cells, or in other words a series of
phase change
refrigeration cells.”
It’s far, far more than that my friend.
Get a text book on Meteorology and start at page 1.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“This isn’t complicated it’s been known for years, a group of unethical people called ‘Green House Gas Effect” believers hijacked atmospheric science and destroyed an entire current generation’s understanding of gas energy science.”
I’m afraid it is complicate my friend, and efforts to simplify it defy belief.
Known for years? Well my professional meteorological knowledge goes back to 1974. You?
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“So I’ll call “pwnt””
If it makes you happy
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“and consider that I helped by reminding everyone that it’s been known for years,
the Hadley cells are mini refrigeration cycles
the refrigerant being water.”
Correct in a simplistic way – but that’s not what Konrad was arguing.
I suggest after page 1 you skip to Coriolis and divergence/convergence zone entrance/exits in jet streams (vital understanding regarding the upper branch of a HC) and also the vast majority of mid/upper latitude meteorology.
TB
December 29, 2013 4:12 pm
[Snip. Duplicate post. ~mod.]
December 29, 2013 4:13 pm
I see I wrote “Hadley cells” when I shoulda written “Hadley-like cells” or “refrigeration circulation cells” but I don’t typically place a lot of importance on an occasional grammatical slip.
Those who want to get the point always do, those who desperately wish it wasn’t there, never do.
TB
December 29, 2013 4:15 pm
[Snip. Duplicate post. ~ mod]
TB
December 29, 2013 4:17 pm
Cont
“This isn’t complicated it’s been known for years, a group of unethical people called ‘Green House Gas Effect” believers hijacked atmospheric science and destroyed an entire current generation’s understanding of gas energy science.”
I’m afraid it is complicate my friend, and efforts to simplify it defy belief.
Known for years? Well my professional meteorological knowledge goes back to 1974. You?
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
“So I’ll call “pwnt””
If it makes you happy
Cont
TB
December 29, 2013 4:18 pm
[Snip. Duplicate post. AGAIN. Best to lay off the hooch when commenting, TB. ~ mod]
December 29, 2013 4:18 pm
Such as TB here with his flailing.
You can’t help willful disbelief of what’s in front of someone’s eyes. Everyone who’s ever seen a heat pipe in action knows it’s a form of phase change refrigeration,
and everyone who’s ever overviewed atmospheric circulation knows it’s a series of refrigeration cells,
and every that the Hadley, Ferrel, and Polar cells, are driven primarily by the phase change effect of water making possible circulation otherwise impossible.
TB
December 29, 2013 4:20 pm
“and consider that I helped by reminding everyone that it’s been known for years,
the Hadley cells are mini refrigeration cycles
the refrigerant being water.”
Correct in a simplistic way – but that’s not what Konrad was arguing.
I suggest after page 1 you skip to Coriolis and divergence/convergence zone entrance/exits in jet streams (vital understanding regarding the upper branch of a HC)
TB
December 29, 2013 4:51 pm
December 29, 2013 at 4:18 pm
Such as TB here with his flailing.
You can’t help willful disbelief of what’s in front of someone’s eyes. Everyone who’s ever seen a heat pipe in action knows it’s a form of phase change refrigeration,
and everyone who’s ever overviewed atmospheric circulation knows it’s a series of refrigeration cells,
and every that the Hadley, Ferrel, and Polar cells, are driven primarily by the phase change effect of water making possible circulation otherwise impossible.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
As I said Bill, whatever makes you happy then believe it, why not. Some people believe in fairies. It’s generally harmless.
Can I ask you what you do/did for a living?
Was it a profession. A scientific one perhaps? I have the same profession as Anthony – our host here.
Would you like to contact him directly and tell him you know more about it than him … because that’s what you have just done by inference.
Anyway whatever you do/did do. Well I know more than you. So there. Start flailing.
Like I said Meteorology is far more complicated than that. Just as a start you overlook a basic. The Earth spins my friend. And that adds a vital dynamic to things.
This will be instructive for you….
http://www.iup.uni-bremen.de/~bms/lecture_atmphys/AtmosphericPhysics_04_05_dyn.pdf
TB
December 29, 2013 4:55 pm
TB says:
December 29, 2013 at 4:18 pm
[Snip. Duplicate post. AGAIN. Best to lay off the hooch when commenting, TB. ~ mod]
Sorry mod – I’m sober honest.
I simply thought it’d got lost in the ether!
December 29, 2013 5:00 pm
The Ferrel cells are included as part of that. Their motion’s responsible for picking up moisture that later falls out as polar and ferrel cells encounter each other and mixing causes that moisture to fall out in the snow bands around the world, where polar cell circulation and ferrel cell circulations collide.
It doesn’t take an entire book to know how the atmosphere works. It’s fluid compound of nitrogen and oxygen that are quite cold. Mixed with them is some water vapor gas which acts as phase change refrigerant.
The motion and action isn’t difficult to perceive in fact you have to be willfully trying not to. It’s a freezing cold fluid bath for a rock whose surface temperature’s a lot warmer than it is.
It’s always cooling the planet even when the sun goes down just like when you close your refrigerator, and the light that was warming it because they were close, goes off, the heat the light put into your soda is being washed off by the same cold, atmospheric mix, that washes the entire rest of the world cooler.
You don’t have the sun go down and tell yourself “Thank goodness it got dark, the dropping temperature of the nighttime air can help me stay warmer, longer.”
No matter how many Green House Gassers told you when the light goes off in your refrigerator the cold air washing over it is “helping it stay warm until the light comes back on.”
That’s not real science.
That’s Greenhouse Gas Effect believer pseudo-science.
December 29, 2013 5:55 pm
No, it’s not unbelievably complicated, TB.
It’s a cold bath of thermally conductive coolant compound, nitrogen, and oxygen.
It’s been shot with some phase change refrigerant to augment the circulation, which is water.
That’s how complicated it gets. I was born in 1961 and had that much locked down by the time I was 13 or so,
so yeah I’ll go ahead and claim my general grasp of what makes the atmosphere function like it does has been fully realized since then.
You denied it did and said models and texts do, too.
Regardless of who admits it the loss of the heat by the water
causes it to change phase.
This augments upward convective process considerably.
So yeah: your claim that because you don’t recall it “in texts and models” means it’s not important as a driver’s just wrong.
It is a significant contribution to the overall circulatory contribution by water.
Feel free to act like there’s some way for you around that if ya want to I’m not gonna go around and around about it, everybody can see who’s been saying what.
December 29, 2013 6:00 pm
My posts and other peoples’ aren’t appearing in sequence that’s why the conversation’s not fluid just so you, those reading, know.
December 29, 2013 6:09 pm
Matter of fact one of TB’s got lost hence the one above for me, so then I made one responding to the one he made, etc… so conversations get confused looking. Hey – bear up – you could be watching all this go on somehow
via snail mail.
Woah.
dp
December 29, 2013 8:23 pm
A real heat engine does work external to the engine. The so called “heat engine” that is the earth climate system does not. I think therefore it is a miscalculation to call the earth’s climate system a heat engine. All the solar energy that makes it work has to leave the system so the system doesn’t heat up. Any earthly heat engine scenario is, ultimately, latency in that heat removal process meaning the earth heats up. That is the alarmist position. The earth’s climate system is certainly dynamic and moves energy around but I think it falls well short of a classic heat engine/Carnot cycle model.
December 29, 2013 8:50 pm
TB my post from some time back got lost and I see you’re very concerned about being viewed as competent to understand a hot rock,
in a stream of cold nitrogen and oxygen.
You’ve already expressed surprise anyone else knows the atmosphere’s a cold nitrogen/oxygen bath refrigerated by sets of global convection cells exploiting phase change refrigerating action of water.
How in the world you can claim you’ve been around atmospheric energy since 1974 and not have heard that I find that not unusual: I find it just plain bizarre.
I’m not in the field of meteorology.
The reputation of my field is fully intact and unruined by scandalous revelation even the top professionals can’t read a thermometer.
I’m an Electronic Engineer. Specifically the area called Radiation Communications and Controls.
Whenever someone identifies themselves me as part of the field
whose foundations have been rocked by revelations many so called professionals and ‘scientists’
believe in glaringly unreal impossibilities,
I think you should know the fact you admit you’re a meteorologist and yet seem to be so perplexed leads me to think I should just ask you straight out, if you believe in the tenets of Green House Gas Effect ‘warm atmosphere’ pseudo-science.
If you don’t fine but if you do, then you’re going to have to tell others your stories about the magic heater because I’m not going to listen to it.
(1)As a professional claiming understanding of the science of atmospheric energy do you believe possible the illumination of a sphere, spinning in vacuum until it’s temp is stable, being immersed into a cold nitrogen/oxygen bath, causing every heat sensor on the sphere surface showing temperature increase?
Because in the real world: in real science like I practice – that’s impossible and no other field even claims it to be anything but prepostrous. Nevertheless many in climatology/meteorology have been seen saying,
they believe in a magical frigid nitrogen oxygen bath, which makes objects hotter than if they weren’t placed into a frigid bath at all, and hotter even than if they were kept heated, in vacuum.
If you answer yes to this critical question you need to know, your answers had better sound good because you’re immediately in the perpetuum mobim realm.
(2)Do you believe it possible to heat a sphere in vacuum then suspend reflective media (H2O/CO2) between sphere and illumination source, reflecting away 20% energy in, causing sensors on the sphere surface to show more energy to them,
than when there was more energy to them?
Again: an answer of yes immediately marks you as one whose conversation will not make sense if you try to refer to such magical glitterings.
(3)Do you believe it possible to heat a sphere in vacuum then suspend more reflective media
(H2O/CO2) than before such that 25% energy in is reflected away,
raising outputs of energy sensors yet again, so they show more energy in at 75%
than when there was more energy in at 80%?
This one’s an extension of the second but if you believe possible one you believe the other.
=======
In real scientific fields like mine no one believes in this junk; in fact just one “Yes I do!” answer immediately marks you as unable to fully comprehend the chain of events that occur when someone places a rock heated in vacuum, into a stream of cold nitrogen/oxygen compound.
I’ve got a sneaking suspicion you believe in all the above. If that’s not true then by all means let me know but I’ve got a feeling you’re going to be just changing the subject to anything but what I want to talk about.
Whether or not you really grasp what the atmosphere operates as.
December 29, 2013 9:15 pm
If you assure me and everyone reading you don’t believe in magically, algebraically reversed cooling/heating functions then I’ll believe you.
But understanding the shape the fields of climate and meteorology, it’s incumbent on anyone coming into contact with any one of you who says you’re part of the field,
to do an assessment of just how likely you are, to be able to determine which way a thermometer would go if someone epoxies it to a sphere, illuminates it in vacuum,
the plunges into cold nitrogen/oxygen compound, bath.
If your answers are ‘No that’s impossible’ three times, you’re cleared as not having been overcome by magical wishing instead of factual thinking.
Mario Lento
December 29, 2013 9:53 pm
dp says:
December 29, 2013 at 8:23 pm
A real heat engine does work external to the engine. The so called “heat engine” that is the earth climate system does not. I think therefore it is a miscalculation to call the earth’s climate system a heat engine. All the solar energy that makes it work has to leave the system so the system doesn’t heat up. Any earthly heat engine scenario is, ultimately, latency in that heat removal process meaning the earth heats up. That is the alarmist position. The earth’s climate system is certainly dynamic and moves energy around but I think it falls well short of a classic heat engine/Carnot cycle model.
++++++++++++++
The point of the article, I think, is that weather extremes or work done by a hotter planet would tend to have a less efficient environment by which to do work because the delta T decreases associated with a warmer planet earth. No where in the article did I read that this had anything to do with driving global climate temperatures. So, while I am not disagreeing with what you believe, I think your implication is unfounded with regard to this article.
December 29, 2013 10:53 pm
One of the early criticisms/premises originally brought up was reduced Carnot efficiency and the general murmur among the Magic Gas/Light crowd was the slowing of the conveyor, would allow heat to build up all along the way, as well as creating “super storms” in warm areas.
December 29, 2013 11:32 pm
dp says:
December 29, 2013 at 8:23 pm
A real heat engine does work external to the engine.
Dp, can you show me any reference in support for this definition?
Because I think it is misleading. A heat engine use heat to create mechanical energy. I-e- movements, that is all there is to it.
You may look it up in the classical textbook by James Senft. You find it in the first sentence in the free preview on Amazon:
/ Jan
December 29, 2013 11:41 pm
My link above came out like an adverisement for this book with “buy it” button, that was not my intention.
/Jan
December 30, 2013 4:20 am
I see in haste of blogging I said a couple of things not true as written, one of them being that water is the main driver of the circulation cells, when I meant to say water’s radiation to space at the altitude it does, is a main agent controlling size, therefore rates, of energy handling, of the circulation cells.
The circulation system’s different than if the water didn’t radiate where it does, and the circulation system’s different than it would be if the water didn’t change phase as well.
Water’s energy/pressure characteristics in fact shape the lower troposphere a lot, and your claim of course was that it’s energy release altitude’s effect, is negligible.
It’s a refrigerant flowing amid some other coolants.
If it didn’t boil off, amid the other coolants, radiate & return how it does,
the entire area it’s activity impacts,
would handle energy differently.
There was something else I said – about whether actually Konrad was saying the radiation of the water “drove the system harder” as in “more volume.”
I said,
he had said, that’s the case- but I didn’t see him say that.
I inferred then projected my take rather than what I’d seen him say.
cold atmosphere proponent. I don’t think he believes in warm atmosphere religion where the algebraic polarity of an icy bath simply reverses because some scammer wished it did.
=======
The reason I interjected is specifically tailored around having seen people say
the energy handling/release characteristics of water
aren’t important regarding it’s actions as refrigerant, in lower troposphere energy handling.
What? Yeah.
Everything about the realm in which the refrigeration cycle operates
is important,
except the energy handling/release characteristics of the refrigerant. Maybe.
Maybe not. So I realize I “blogged in haste.”
However not so much haste
I tried to tell people in a region where refrigerant
is an important handler of energy
handling of energy, in that same region,
has little at all to do with the energy handling and release characteristics
of the refrigerant.
That’s just not possible on it’s face.
The size of the space you refrigerate with the water
is directly dependent on where the water releases energy.
December 30, 2013 8:21 am
TB says:
December 28, 2013 at 9:35 am
The Earth’s climate is an open system – with a constant input of energy from the Sun and a constant emission of that energy to space after conversion into terrestrial IR. Therefore the efficiency of the conversion does not matter.
Yes it does TB, and I think your argument is beside the point.
My point is to show that the potential for converting thermal energy to mechanical energy is reduced when both the cold element and the warm element heat equally much.
If there were no temperature differences on the Earth’s surface, and no heating, there would not be much wind either. But when temperature differences and solar heating exist, the atmosphere will start to move. A fundamental cause is that heating of a gas makes the volume to increase and therefore start a fluctuation.
But for a given temperature increase, say 10 C, the volume will increase more if you start with a low temperature than a high one.
An increase from 0 C to 10 C make the volume increase by 3.66%,
an increase from 10C to 20C make the volume increase by 3.53%
The higher the temperature is, the lower the increase in volume will be.
This reduces the potential for wind when all temperatures increase equally much. When the lowest temperatures increase more than the higher, as will be the case for the Earth’s surface, the potential will be further reduced.
/ Jan
TB
December 30, 2013 8:46 am
December 29, 2013 at 5:55 pm
No, it’s not unbelievably complicated, TB.
“It’s a cold bath of thermally conductive coolant compound, nitrogen, and oxygen.”
No, it’s a system on incoming SW radiation stirred by a spinning Earth on a complex surface of water, mountain, snow etc, after which is emitted to space as LW.There are feedbacks and drivers, chicken and egg. Not to mention (on the weather scale) enormous chaos.
“It’s been shot with some phase change refrigerant to augment the circulation, which is water.
That’s how complicated it gets. I was born in 1961 and had that much locked down by the time I was 13 or so,”
Did you scan through that link (you’d do well to understand that at 13) – forget the curly d’s there should be enough one-syllable words and pretty pictures for the average man to realise it’s NOT that simple.
“so yeah I’ll go ahead and claim my general grasp of what makes the atmosphere function like it does has been fully realized since then.”
You’d be wrong. And I’m waiting for you to flail about in me denying you know nothing about you profession while I ignorantly argue I do, cos I’d had it figured out by 13.
So what, he’s wrong. If he was right then Global NWP models wouldn’t work …. Hang on maybe you/he have it right. They didn’t do well the the Atlantic Hurricane season, did they?
“You denied it did and said models and texts do, too.”
Correct because it is not a Driver.
Bill, I’ll try one more time..
I take it you agree that the Earth’s spin causes moving air in the NH to turn right. Yes?
So imagine you are coming out of your local football terraces and merge into a stream of people in a narrow belt moving at right angles. Are you not forced to slow as people around you merge into that stream?
That’s called convergence Bill, and as the air can’t go up (capped by the tropopause) … it SINKS, to for the belt of sub-tropical highs around 30 deg N/S.
It just one half of just about the most important mechanism driving our atmosphere – the other is divergence. It is these two things that (acting in the region of jet-streams especially) greatly affect the atmosphere/weather below. Convergence aloft causes divergence below (HP). Divergence aloft causes convergence below (LP). It is this sucking/blowing effect that drives the atmosphere – ALONG with convective uplift. Radiative cooling aloft is NOT a driver. It happens but air would move the same way without it.
“Regardless of who admits it the loss of the heat by the water
causes it to change phase.”
I don’t/haven’t argued against that for one second. But it is only one side of the equation.
“This augments upward convective process considerably.”
Correct – but once the air is up there it moves (in the HC) as I’ve described.
“So yeah: your claim that because you don’t recall it “in texts and models” means it’s not important as a driver’s just wrong.”
You’d better tell that to the rest of World’s Meteorologists the, not just me – a Nobel’s in the offing.
“It is a significant contribution to the overall circulatory contribution by water.”
It is Bill, but only a part.
“Feel free to act like there’s some way for you around that if ya want to I’m not gonna go around and around about it, everybody can see who’s been saying what.”
Of course “I feel free to act like there’s no way round it”. Because I’m right.
I’m sorry, but you don’t gain anything like a complete understanding of the Earth’s climate system without being taught/learning in the job. And to turn up at your Doc’s and argue black is white that your heart isn’t the blood’s pump, your bladder is – against him, is staggering. In order to learn you need to be receptive to being taught and to do that we do indeed need to “appeal to authority” – as that is what students do at college/university, and that scientists do – because they don’t invent science on a whim. It comes from scientists/thinkers before them.
Newton said “If I have seen so far. It is because I have stood on the shoulders of giants”.
It was good enough for him Bill.
Why not you?
PS ( I’m in no way a giant BTW – just trying to get across basic Meteorology).
December 30, 2013 8:56 am
Jan Kjetil Andersen says:
December 30, 2013 at 8:21 am
You are getting very close.
For a planet with no atmosphere both the height at which the S-B equation is satisfied and the effective radiating height are together at the surface.
Consider that conduction raises off the surface the height at which the S-B equation is satisfied.
In contrast, the radiative capability of the atmosphere raises off the surface the so-called effective radiating height.
Note that the two heights are then different.
Wind is then required to ensure that the energy at the S-B height is moved to the effective radiating height so as to maintain radiative equilibrium for the system as a whole.
You may be regarding winds as horizontal movements but in fact those horizontal movements actually enable vertical movements of energy so that gravitational potential energy (PE) at the S-B height can be shifted down to the effective radiating height and converted to kinetic energy (KE) at the effective radiating height for radiation to space.
That is what s going on.
Neat or not ?
December 30, 2013 9:36 am
Jan said:
“My point is to show that the potential for converting thermal energy to mechanical energy is reduced when both the cold element and the warm element heat equally much”
The S-B height is the warm element.
The effective radiating height is the cold element.
The closer together they are the less wind (circulation) there will be and the further apart they are the more wind there will be. That confirms your point.
When they are together at the surface there is no atmosphere and no wind.
For a non radiative atmosphere they would be at maximum width apart with lots of wind.
For an atmosphere with radiative capability they will be closer together with less wind.
The water cycle acts as a ‘lubricant’ so, whatever the distance apart, less wind will be required than would otherwise be the case.
Now, we need to figure out what makes the two heights vary relative to one another and I say it is the sun from above affecting atmospheric chemistry and the oceans from below affecting atmospheric latent heat content both of which are magnitudes greater than changes in radiative capability.
In any event if one only changes the atmosphere’s radiative capability with all else remaining the same then all that will happen is a change in the relative heights for the S-B level and the effective radiating level that will change the amount of wind with a minimal, if any, change in surface temperature.
TB
December 30, 2013 10:22 am
Jan Kjetil Andersen says:
December 30, 2013 at 8:21 am
TB says:
December 28, 2013 at 9:35 am
The Earth’s climate is an open system – with a constant input of energy from the Sun and a constant emission of that energy to space after conversion into terrestrial IR. Therefore the efficiency of the conversion does not matter.
Yes it does TB, and I think your argument is beside the point.
It’s all the point Jan.
When comparing the Earth’s climate to anything – we recognise, say, that it is an orange. That being the case, we compare it to another orange, or at least a tangerine. Certainly not a banana.
What matters is that that is what comes as energy in must go out.
If it doesn’t we heat up.
The efficiency of the climate system matters not a jot (in how it uses that energy to cause weather).
As it is merely the mediator in the exchange of Solar absorbed > LWIR emitted to space.
All the inefficiencies/vagaries/cycles within weather (~30 years) do not alter the basics of Energy in must equal energy out…. Or we’re in trouble.
Weather, or your “Carnot engine” is just the internal chaos of the system, when measured at the fill up/exhaust bit of Earth (TOA) the chaos disappears. Like water boiling in a pan on a stove. Put a known amount of heat into it and it will reach boiling in a known amount of time (with known starting conditions). That is, it does not matter how the water churns around in the pan (weather) as that’s internal chaos and the end result, say 100C in 99 sec, is predictable.
December 30, 2013 10:34 am
TB said:
“Like water boiling in a pan on a stove. Put a known amount of heat into it and it will reach boiling in a known amount of time (with known starting conditions). That is, it does not matter how the water churns around in the pan (weather) as that’s internal chaos and the end result, say 100C in 99 sec, is predictable.”
And that 100C is set by atmospheric pressure on the surface.
I contend that, similarly, atmospheric pressure determines the energy cost of evaporation and so in turn determines the energy that the oceans can hold in the long term subject to shorter term internal ocean variations.
The ultimate factor that determines the amount of energy that the earth system can hold on to from any given level of solar input is set by the density of the mass of the atmosphere.
http://www.newclimatemodel.com/the-setting-and-maintaining-of-earths-equilibrium-temperature/
December 30, 2013 12:14 pm
TB says:
December 30, 2013 at 10:22 am
Weather, or your “Carnot engine” is just the internal chaos of the system, when measured at the fill up/exhaust bit of Earth (TOA) the chaos disappears. Like water boiling in a pan on a stove. Put a known amount of heat into it and it will reach boiling in a known amount of time (with known starting conditions). That is, it does not matter how the water churns around in the pan (weather) as that’s internal chaos and the end result, say 100C in 99 sec, is predictable
TB, when we discuss whether the Carnot theorem matters or not, it is crucial that we agree on what it may matter for. I was discussing the weather, or more specifically the wind, in this article, but you say here that the weather doesn’t matter.
As in your example above, I’m not interested in the “end result” as you call it above. It is, to use your example again, how the water churns around in the pan (weather), that I am discussing.
/ Jan
TB
December 30, 2013 12:30 pm
Jan Kjetil Andersen says:
December 30, 2013 at 12:14 pm
“As in your example above, I’m not interested in the “end result” as you call it above. It is, to use your example again, how the water churns around in the pan (weather), that I am discussing.”
Fair enough Jan – but the title to this thread is “Climate as a heat engine”.
I take climate to mean, as in long term (>~30 years). Such that we eliminate weather cycles and reveal long term heating/cooling trends.
Returning to my OP is enough to appreciate the meteorological argument (and not with regards reduced jet strength) for extra energy being supplied to the atmosphere via LH release due increased absolute humidity.
December 30, 2013 1:19 pm
TB says:
December 30, 2013 at 12:30 pm
Fair enough Jan – but the title to this thread is “Climate as a heat engine”.
I take climate to mean, as in long term (>~30 years). Such that we eliminate weather cycles and reveal long term heating/cooling trends.
Yes, of course climate is the theme, but I am discussing the consequences for the weather in the long run when some underlying factors are changed.
A storm is a weather phenomenon, and what I am discussing is that the forces which create the storm will weaken when the temperatures increase and the differences are getting smaller. The eventual temperature increase is of cause a long time climate phenomenon.
/ Jan
December 30, 2013 1:42 pm
TB says:
“I take climate to mean, as in long term (>~30 years). Such that we eliminate weather cycles and reveal long term heating/cooling trends.”
OK then, let’s look a the real long term trend.
NO acceleration whatever in global temperatures. Therefore, the rise in CO2 over the past century and a half has had NO effect at all. QED
TB just can NOT admit that the empirical evidence falsifies his world view, can he?
TB
December 30, 2013 2:47 pm
Stephen Wilde says:
December 30, 2013 at 10:34 am
TB said:
“Like water boiling in a pan on a stove. Put a known amount of heat into it and it will reach boiling in a known amount of time (with known starting conditions). That is, it does not matter how the water churns around in the pan (weather) as that’s internal chaos and the end result, say 100C in 99 sec, is predictable.”
And that 100C is set by atmospheric pressure on the surface.
I contend that, similarly, atmospheric pressure determines the energy cost of evaporation and so in turn determines the energy that the oceans can hold in the long term subject to shorter term internal ocean variations.
The ultimate factor that determines the amount of energy that the earth system can hold on to from any given level of solar input is set by the density of the mass of the atmosphere.
http://www.newclimatemodel.com/the-setting-and-maintaining-of-earths-equilibrium-temperature/
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stephen, I’ve had a scan through your paper. I don’t think you’ll be surprised that I don’t agree with it.
Quotes from same…
“That is the point at which the oceans reach an equilibrium temperature and it is that ocean
temperature which then controls the temperature of the air above and NOT the Greenhouse
Effect. The real arbiter of the Earth’s equilibrium temperature is instead what I have termed
The Hot Water Bottle Effect.”
SST’s and heat stored in the oceans certainly do affect atmospheric temperature. That is largely why we have the current “pause” – 8 years BTW – as there has been a preponderance of La Ninas.
Don’t agree with the (lack of) GHE – that is adding heat to the oceans (see below).
“Water boils away at 100 degrees C so in other words the process of evaporation removes
from the local environment (in the form of latent heat) over five times the amount of energy
required to induce that evaporation.”
No, the LH of evaporation comes from the local environment – has too. It isn’t magicked from the ether. Evaporation will extract the LH requirement from the water or air below/above, most from the warmer. So it takes away the LH from the surface and deposits it via condensation aloft. Result zero net gain/loss from the climate system.
“However over the Earth as a whole the water is
nearly always warmer than the air (due to solar input) so inevitably the average global energy
flow is from oceans to air via that latent heat of evaporation into the air and the energy
needed is taken from the water. This leads to a thin (1mm deep) layer of cooler water over the
oceans worldwide and below the evaporative region that is some 0.3C cooler than the ocean
bulk below. The evaporative process extracts energy faster from the oceans than it can be
drawn up from below and added from above otherwise that cool layer could not be present.
That 1mm deep 0.3 cooler layer is a critical diagnostic indicator but as far as I can tell it has
never been recognised as such. It is disturbed by diurnal and seasonal variations and by
changes in wind speed but on average over time it is a permanent fixed feature of our ocean
surfaces.”
You are thinking entirely radiatively here Stephen. You mention conduction (not quoted) in passing, but it plays a much larger part than you give it credit for. Yes, there is surface evaporation but as the surface skin is cooled via LH evap then so the temp differential between ocean skin and air is reduced. (as you say by ~0.3C). Ergo the conductive part of the equation is reduced. So letting less heat through from the more turbulent mixed layers below to heat the atmosphere in contact with the ocean. In effect a small warming blanket is created by the evaporating skin to the water below
http://www.realclimate.org/images/Minnett_2.gif
Stephen, across the 71% of the Earth’s surface that is ocean/water the surface pressure will average the same at any given moment or at least over the course of a short period of time. Therefore any atmospheric effect on evaporation will cancel out, and anyway an insulation effect will take over.
Also ARGO float data show ocean temps rising at many (even abyssal) levels – so where is the rise coming from if you theory is correct?
“Many are pointing out that the feedback from more energy in the air seems to be negative
(more convection, rainfall and clouds in particular) rather than positive and the observed
climate shifts over the past ten years (more atmospheric ‘blocking’ events causing the surface
pressure systems to shift around more and producing more meridional/equatorward jet
streams) seem to be confirming that view since the temperature trend is increasingly
diverging from that expected from more CO2 emissions.”
I’m sorry, I don’t agree that –ve feedbacks dominate over +ve.
Lets look at it logically…
Earth varies between nice and comfortable temps inter-glacially to ice age conditions (via changes in orbital characteristics).
If feed-backs where largely –ve how would we ever get out of an IA?
You are saying that rising temps lead to falling temps or at least a regulation of them.
So in an IA what will amplify ocean then air temps if the feed-backs weren’t mainly +ve. We know they have flipped quickly and CO2 has tracked them mirror-like. You do accept CO2 is a GHG? Therefore amplifying temp rise. Descent into an IA is explainable even with -ve feed-backs – just via increased albedo, but how do we get out of an IA with –ve ones? At least as quickly as evidence shows?
Yes, I agree that we are seeing more meridional flow, but why should that alter temperature trends since if cold air flows away from the Arctic (say) then the Arctic must warm relative to when it kept that cold air. Similarly a meridional flow induces warm air north to balance the equation in a hemispheric sense.
Also why would it give more cloud? Colder air is drier, so as that moves south over warmer temps in winter its convective cloud infill will be minimal/zero over land giving a net cooling anomaly. Over sea broken convective infill, therefore net warming anomaly. Warm air moving N in winter will produce low cloud via cooling/condensing, hence net warming anomaly over both land/ocean. In summer cold air moving S will be highly convective over sea (net cooling anomaly) but dry with limited convective infill still over land giving a net warming anomaly. Warm air moving north in summer tending to cool/condense over ocean giving net surface cooling, balanced by net warming over land. So I see no net feedback either way. (NB in all of the above I speak as of radiative balance NOT temp levels).
December 30, 2013 2:55 pm
I see in haste of blogging I said a couple of things not true as written, one of them being that water is the main driver of the circulation cells, when I meant to say water’s radiation to space at the altitude it does, is a main agent controlling size, therefore rates, of energy handling, of the circulation cells.
The circulation system’s different than if the water didn’t radiate where it does, and the circulation system’s different than it would be if the water didn’t change phase as well.
Water’s energy/pressure characteristics in fact shape the lower troposphere a lot, and your claim of course was that it’s energy release altitude is nominally negligible.
It’s a refrigerant flowing amid some other coolants.
If it didn’t boil off, amid the other coolants, radiate & return how it does,
the entire area it’s activity impacts,
would handle energy differently.
There was something else I said – about whether actually Konrad was saying the radiation of the water “drove the system harder” as in “more volume.”
I said,
that’s the case-
-but I didn’t see him say that.
I inferred then projected my take rather than what I’d seen him say.
cold atmosphere proponent. I don’t think he believes in warm atmosphere religion
where the algebraic polarity of an icy bath
simply reverses because some scammer wished it did.
=======
The reason I interjected is specifically tailored around having seen people say
“the energy handling/release characteristics of water
aren’t important regarding it’s actions as refrigerant,
and convection boundary assignment, in lower troposphere energy handling.
What?
Yeah.
Everything about the realm in which the refrigeration cycle operates
is important,
except
the energy handling/release characteristics of the refrigerant. Maybe.
Maybe not.
So I realize I “blogged in haste.”
However not so much haste
I tried to tell people in a region where refrigerant
is an important handler of energy
handling of energy, in that region,
has little at all to do
with the energy handling and release characteristics
of the refrigerant.
That’s just not possible on it’s face.
The size of the space refrigerated by water
is directly dependent on where the water releases energy.
December 30, 2013 3:10 pm
TB,
I am grateful to you for the time you spent reading my article and I note your comments.
I could deal with each point in detail but do not wish to derail this thread.
I think you would have an uphill struggle convincing anyone that the system response to a change in internal forcing elements is not negative.
December 30, 2013 3:14 pm
I’d be interested to hear Jan’s take on my posts at 8.56 am and 9.36 am which appear to support his thesis but place it in a broader context.
December 30, 2013 3:21 pm
TB has apparently failed to see what I asked him so I’m re-posting it so everyone can be sure TB has had the opportunity to defend his religion. I knew as soon as I saw him trying to deny the effect of water on shaping global circulation he was going to be saying crazy things so I went over to another thread that’s current now and sure enough, he’s denying science: denying it hasn’t warmed for 17 years, when everybody else on earth but apparently him, Trenberth and Mike Mann don’t know that.
Here’s the re-post:
TB my post from some time back got lost and I see you’re very concerned about being viewed as competent to understand a hot rock,
in a stream of cold nitrogen and oxygen.
You’ve already expressed surprise anyone else knows the atmosphere’s a cold nitrogen/oxygen bath refrigerated by sets of global convection cells exploiting phase change refrigerating action of water.
How in the world you can claim you’ve been around atmospheric energy since 1974 and not have heard that I find that not unusual: I find it just plain bizarre.
I’m not in the field of meteorology.
The reputation of my field is fully intact and unruined by scandalous revelation even the top professionals can’t read a thermometer.
I’m an Electronic Engineer. Specifically the area called Radiation Communications and Controls.
Whenever someone identifies themselves me as part of the field
whose foundations have been rocked by revelations many so called professionals and ‘scientists’
believe in glaringly unreal impossibilities,
I think you should know the fact you admit you’re a meteorologist and yet seem to be so perplexed leads me to think I should just ask you straight out, if you believe in the tenets of Green House Gas Effect ‘warm atmosphere’ pseudo-science.
If you don’t fine but if you do, then you’re going to have to tell others your stories about the magic heater because I’m not going to listen to it.
(1)As a professional claiming understanding of the science of atmospheric energy do you believe possible the illumination of a sphere, spinning in vacuum until it’s temp is stable, being immersed into a cold nitrogen/oxygen bath, causing every heat sensor on the sphere surface showing temperature increase?
Because in the real world: in real science like I practice – that’s impossible and no other field even claims it to be anything but prepostrous. Nevertheless many in climatology/meteorology have been seen saying,
they believe in a magical frigid nitrogen oxygen bath, which makes objects hotter than if they weren’t placed into a frigid bath at all, and hotter even than if they were kept heated, in vacuum.
If you answer yes to this critical question you need to know, your answers had better sound good because you’re immediately in the perpetuum mobim realm.
(2)Do you believe it possible to heat a sphere in vacuum then suspend reflective media (H2O/CO2) between sphere and illumination source, reflecting away 20% energy in, causing sensors on the sphere surface to show more energy to them,
than when there was more energy to them?
Again: an answer of yes immediately marks you as one whose conversation will not make sense if you try to refer to such magical glitterings.
(3)Do you believe it possible to heat a sphere in vacuum then suspend more reflective media
(H2O/CO2) than before such that 25% energy in is reflected away,
raising outputs of energy sensors yet again, so they show more energy in at 75%
than when there was more energy in at 80%?
This one’s an extension of the second but if you believe possible one you believe the other.
=======
In real scientific fields like mine no one believes in this junk; in fact just one “Yes I do!” answer immediately marks you as unable to fully comprehend the chain of events that occur when someone places a rock heated in vacuum, into a stream of cold nitrogen/oxygen compound.
I’ve got a sneaking suspicion you believe in all the above. If that’s not true then by all means let me know but I’ve got a feeling you’re going to be just changing the subject to anything but what I want to talk about.
Whether or not you really grasp what the atmosphere operates as.
December 30, 2013 3:25 pm
You feel free to go get whoever you think you have to have near you so you feel safe TB.
December 30, 2013 4:05 pm
Stephen Wilde says:
December 30, 2013 at 3:14 pm
I’d be interested to hear Jan’s take on my posts at 8.56 am and 9.36 am which appear to support his thesis but place it in a broader context.
I have read through both your postings and your paper Stephen. You take up big questions here, and many of your explanations are good and logical, but I am sorry to say that I have many objections to it.
But since it is such a big discussion to go into, I think I will leave here. Perhaps we can take it up in another thread.
Regards
/ Jan
TB
December 30, 2013 4:11 pm
December 30, 2013 at 3:21 pm
TB has apparently failed to see what I asked him so I’m re-posting it so everyone can be sure TB has had the opportunity to defend his religion. I knew as soon as I saw him trying to deny the effect of water on shaping global circulation he was going to be saying crazy things so I went over to another thread that’s current now and sure enough, he’s denying science: denying it hasn’t warmed for 17 years, when everybody else on earth but apparently him, Trenberth and Mike Mann don’t know that.
Here’s the re-post:
TB my post from some time back got lost and I see you’re very concerned about being viewed as competent to understand a hot rock,
in a stream of cold nitrogen and oxygen.
You’ve already expressed surprise anyone else knows the atmosphere’s a cold nitrogen/oxygen bath refrigerated by sets of global convection cells exploiting phase change refrigerating action of water.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nothing surprises me with regard to what the D-K syndrome can effect Bill.
“How in the world you can claim you’ve been around atmospheric energy since 1974 and not have heard that I find that not unusual: I find it just plain bizarre.|”
I’ll trump you on that cos I’m the expert, sorry.
“I’m not in the field of meteorology.”
Exactly.
“The reputation of my field is fully intact and unruined by scandalous revelation even the top professionals can’t read a thermometer.”
You are fortunate my friend as your profession has not been politicised and hence skewed out of all proportion via ideology.
“I’m an Electronic Engineer. Specifically the area called Radiation Communications and Controls.”
Very good – I’ll give you due respect for that and refrain from asserting that you are flailing when I’m asserting that I figured out radio propagation at age 13.
“Whenever someone identifies themselves me as part of the field
whose foundations have been rocked by revelations many so called professionals and ‘scientists’
believe in glaringly unreal impossibilities, I think you should know the fact you admit you’re a meteorologist and yet seem to be so perplexed leads me to think I should just ask you straight out, if you believe in the tenets of Green House Gas Effect ‘warm atmosphere’ pseudo-science.
Me perplexed. Incredible. I’m the expert. Not you. Like you’re the expert on Radio communication here not me!
Like I said, you need to investigate to learn – not take it as read that you cracked it at age 13.
Give me some published papers that agree with you. I can give you any number from my side.
“If you don’t fine but if you do, then you’re going to have to tell others your stories about the magic heater because I’m not going to listen to it.”
Not magic – just empirical meteorology. What is you don’t/can’t understand about Coriolis/convergence aloft INEVITABLY causing divergence below. That all the world’s Met organisations incorporate into NWP models?
“(1)As a professional claiming understanding of the science of atmospheric energy do you believe possible the illumination of a sphere, spinning in vacuum until it’s temp is stable, being immersed into a cold nitrogen/oxygen bath, causing every heat sensor on the sphere surface showing temperature increase?
Because in the real world: in real science like I practice – that’s impossible and no other field even claims it to be anything but prepostrous. Nevertheless many in climatology/meteorology have been seen saying,
they believe in a magical frigid nitrogen oxygen bath, which makes objects hotter than if they weren’t placed into a frigid bath at all, and hotter even than if they were kept heated, in vacuum.”
You are being preposterous, and arrogant to boot in assuming you know more than a retired professional of 32 years. Why, in a sane world would that be probable?
“If you answer yes to this critical question you need to know, your answers had better sound good because you’re immediately in the perpetuum mobim realm.”
It’s not a matter of belief my friend – it just is. And I don’t give jot whether my answers “sound good” to you because your psychological make-up makes that impossible for me to achieve. Nor do I aim to.
“(2)Do you believe it possible to heat a sphere in vacuum then suspend reflective media (H2O/CO2) between sphere and illumination source, reflecting away 20% energy in, causing sensors on the sphere surface to show more energy to them,
than when there was more energy to them?”
Incorrect appreciation of the GHE. GHG’s absorb IR from the Earth’s surface then re-emit it, some of it making it back to the surface again – thus SLOWING the rate of cooling and NOT heating it. Have you not noticed ice/frost melting when cloud cover appears of a winters night? Same thing that CO2 does (to a tiny but sig degree) as well (triatomic molecule – look it up). It even happens with thin cloud at 6 miles up – seen it countless times professionally (actually a complete pain in the a*** as a lot of work was generated as a result in alerting ice management companies re ice/not ice). It just does, it’s a basic of the Universe.
“Again: an answer of yes immediately marks you as one whose conversation will not make sense if you try to refer to such magical glitterings.”
Again that comment marks you as one with a stunning D-K syndrome who refuses to address the science nor (typically) has any respect for someone who has knowledge of same.
Why don’t you ask Anthony if Convergence drives the descent arm of a HC? He’s a Meteorologist like me. He may disagree on the “degree” but I’m pretty sure he accepts a GHE.
“(3)Do you believe it possible to heat a sphere in vacuum then suspend more reflective media
(H2O/CO2) than before such that 25% energy in is reflected away,
raising outputs of energy sensors yet again, so they show more energy in at 75%
than when there was more energy in at 80%?”
No GHG’s don’t reflect – they absorb/emit LWIR and not reflect solar SW.
More GHG’s in the atmosphere increase the “insulation” effect.
“This one’s an extension of the second but if you believe possible one you believe the other.
=======
In real scientific fields like mine no one believes in this junk; in fact just one “Yes I do!” answer immediately marks you as unable to fully comprehend the chain of events that occur when someone places a rock heated in vacuum, into a stream of cold nitrogen/oxygen compound.”
Arrogance again – refer that comment to Anthony will you, I’m sure he’ll be pleased to know his profession is not really scientific. BTW, I studied Control engineering for 3 years before training in the UKMO. Hence my respect for that profession which is spectacularly countered by your contempt.
“I’ve got a sneaking suspicion you believe in all the above. If that’s not true then by all means let me know but I’ve got a feeling you’re going to be just changing the subject to anything but what I want to talk about.”
I’ve answered all of the above and you will find all my answers repeated in text books – you know, like the text books that you learned radio communication from, those equations are like those in meteorology (empirically proven to be true ) and are not up for contention. Sorry.
“Whether or not you really grasp what the atmosphere operates as.”
Whether you grasp your extreme arrogance and ignorance + insults to your host.
TB
December 30, 2013 4:14 pm
December 30, 2013 at 3:25 pm
You feel free to go get whoever you think you have to have near you so you feel safe TB.
Excuse me??
December 30, 2013 4:26 pm
[snip – OK that’s it – this is just another slayer rant complete with insults. You’re done here – Anthony]
December 30, 2013 4:31 pm
TB says: “What is you don’t/can’t understand about Coriolis/convergence aloft INEVITABLY causing divergence below. That all the world’s Met organisations incorporate into NWP models?”
Changing the subject from me catching you claiming the geometries, velocities, volumes of air handling in the lower troposphere aren’t associated with the energy handling and release of water isn’t going to make time go back and you be right.
Mario Lento
December 30, 2013 4:35 pm
Looks like Bill from Nev’ slipped one more in before he was done. It was getting tiring sifting through his creative sarcasm and negative slams.
December 30, 2013 4:36 pm
Claiming putting insulation between the heat of an illumination source and a rock, makes more heat come out of the rock, is what you’ve been seen trying to do, and it’s absurdity on it’s face.
The fact is you’re here to declare to the world you know of an insulation that reduces energy in, yet makes the object it sits blocking energy to, act as if more energy arrived on it.
That’s just impossible no matter how many climatologists said they did it, and it’s why the pseudo science called warm atmosphere doesn’t stand up to even the most cursory inspection.
“Incorrect appreciation of the GHE. GHG’s absorb IR from the Earth’s surface then re-emit it, some of it making it back to the surface again – thus SLOWING the rate of cooling and NOT heating it.”
December 30, 2013 7:22 pm
TB,
I see you are still arguing that radiative gases do not play a critical role in driving tropospheric convective circulation by allowing energy loss, buoyancy loss and subsidence of air masses.
I urge you to look again at an energy budget diagram for the atmosphere. Discounting the energy reflected from the land, ocean and atmosphere, around 90% of all energy absorbed is emitted to space from radiative gases the atmosphere.
90%!
And this energy is being emitted at a higher altitude than energy entering the atmosphere. Yet you are claiming that this plays no role in driving Rayleigh Bernard circulation below the tropopause?
To defend the AGW hypothesis you are claiming that for a non-radiative atmosphere, differential conductive heating and cooling of the atmosphere by the surface would produce vertical circulation across 10 to 15 Km of the troposphere strong enough to produce the observed lapse rate.
You claim 30 years of meteorology experience. Did none of this cover the mechanisms of night inversion layers? The physics of this hold true for any surface/gas interface in a gravity field. The surface is far better at conductively heating the atmosphere than it is at conductively cooling it.
The evidence from the actual atmosphere stand s against your claims, in particular the tropopause. Below this level strong vertical Rayleigh Bernard circulation is exhibited. Above the tropopause atmospheric circulation is so weak that the lapse rate reverses and stagnant gases are subject to radiative super heating. So what’s so special about the tropopause? That’s where the atmosphere runs out of the most important radiative gas on our planet, H2O.
If your claim that radiative cooling played no part in tropospheric convective circulation were true, we should observe powerful convective circulation extending well above this height. However, there is nothing climate “scientists” can do to get rid of the “punitive” tropopause.
And of course there is nothing climate “scientists” can do to remove the most damning evidence, the history of their claims recorded on the Internet. This history records that the first claims that adding radiative gases would reduce it’s radiative cooling ability were made on the basis of two shell radiation only static atmosphere models. After 1990 came the band-aid radiative-convective models. Just this sequence of events alone is enough to damn all of the AGW fellow travellers.
You may be encouraged that many embarrassed sceptics also want what most AGW propagandists are praying for, a “warming, but far less than we thought” soft landing for the global warming inanity. But this is not how it is going to go down. You will be up against the general public. They have no interest in defending those who claim that radiative gases reduce the atmospheres radiative cooling ability. The corpse of global warming cannot be re-animated nor can it be hidden.
December 31, 2013 8:12 am
“So what’s so special about the tropopause? That’s where the atmosphere runs out of the most important radiative gas on our planet, H2O.”
It runs out of H2O because at that height the vapour has all condensed out but that isn’t why there is a reversal of the lapse rate.
The warming from tropopause upwards is due to the presence of ozone interacting directly with incoming solar radiation.
One could argue that the ozone layer is the most important radiative gas on our planet.
The observed lapse rate is a distortion of the ideal lapse rate set by mass and gravity but if an atmosphere is to be retained then all the varied observed lapse rates have to net out to the ideal lapse rate between surface and space.
I find it strange to be arguing with fellow sceptics that radiative gases are not needed for a fully convective atmosphere.
The advantage of such a position is that if one can involve the entire atmospheric mass in the warming of the surface above the S-B prediction then radiative considerations become too small to matter.
Radiative characteristics do not create the lapse rate. They simply distort the lapse rate away from the ideal lapse rate set by mass and gravity. Then that effect has to be negated elsewhere in the system and that is achieved by circulation changes.
Radiative gases simply raise the effective radiating height off the ground. In their absence the ground is the effective radiating height.
The more radiative the atmosphere the higher the effective radiating height goes and if the atmosphere were 100% radiative then the effective radiating level would simply be at the top of the atmosphere but if that could happen the atmosphere would just collapse back to the ground again because no energy would be left over for conduction.
Conduction raises the S-B height.
When the two are at different levels then stability is maintained by convection shifting energy from the S-B level to the effective radiating level.to maintain overall radiative balance.
TB
December 31, 2013 9:26 am
December 30, 2013 at 7:22 pm
TB,
I see you are still arguing that radiative gases do not play a critical role in driving tropospheric convective circulation by allowing energy loss, buoyancy loss and subsidence of air masses.
I urge you to look again at an energy budget diagram for the atmosphere. Discounting the energy reflected from the land, ocean and atmosphere, around 90% of all energy absorbed is emitted to space from radiative gases the atmosphere.
90%!
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
http://ceres.larc.nasa.gov/ceres_brochure.php?page=2
Konrad, ~168 W/m^2 of SW are absorbed at the surface out of ~235 net. And ~67 SW absorbed by the atmosphere. Now of that ~102 goes back as conv/LH and must be added to the ~67, making ~169.
~66 is back-radiated as IR from the surf making a total absorbed there 169+66=235.
So, well and good, all absorbed via SW ends up in the atm. But once all there a further 89 is re-emitted back to the surface. This then gets emitted via TOA totalling ~235.
Yes it is Konrad, but you overlook the fact that it’s been down to the surface before finally going to space. The 89W/m^2 figure above is as a result of repeated “bouncings” of IR photons between the atm and surface “walls” and that leaves the surface/atm more interactions to warm.
“And this energy is being emitted at a higher altitude than energy entering the atmosphere. Yet you are claiming that this plays no role in driving Rayleigh Bernard circulation below the tropopause?”
As stated above, all solar radiation (and consequently terrestrial LWIR) that is absorbed warms the atm and surface (almost –not counting obvious albedo/latitudinal differences) equally. And has no sig effect on the LR.
“To defend the AGW hypothesis you are claiming that for a non-radiative atmosphere, differential conductive heating and cooling of the atmosphere by the surface would produce vertical circulation across 10 to 15 Km of the troposphere strong enough to produce the observed lapse rate.”
No, I am saying that for a non-radiative atm around a spinning planet, the HEATING of the surface (differentially) will initiate atm motion, including all processes we see on Earth. There will be a LR generated via conduction initially (ignoring mass generated compressive heating – which is a one time only process) then mixed through the whole depth of the atm via those processes. Note, in a non-radiative atm, there will be no Tropopause due lack of O3.
The adiabatic LR will eventually naturally be assumed via compression/rarefaction of lifted/descended air and as there is no water on this planet (radiative gas) then will observe a DALR throughout.
The atmosphere will act as a “heat-pump” and transport heat down the LR.
When you add water, WV will cause LH transport, cooling the surface but heating the atm and so cause the LR to assume a profile midway between the DALR and the SALR eg the ELR.
Radiative process will have basically neutral effect on the LR, though the tropopause will be created, and so a lid to “weather”.
“You claim 30 years of meteorology experience. Did none of this cover the mechanisms of night inversion layers? The physics of this hold true for any surface/gas interface in a gravity field. The surface is far better at conductively heating the atmosphere than it is at conductively cooling it.”
I do, well 32 actually, 20 spent as an on-the-bench forecaster with the UKMO with both the RAF at airfields and in the commercial field. Sorry, I can’t prove it other than scanning in my old Met college pics – but that could be faked eh?
Konrad, my training covered all the knowledge available at the time. I specifically trained with the UKMO as a forecaster 1985-87 and received further updates after.
Of course I was taught about radiative cooling and night inversions. Try forecasting for night-flying exercises at RAF bases without that knowledge.
Surface cooling has little convective/LH response and none that transports aloft. There are some interactions, specifically at fog tops and via turbulence lifting the mixing layer. Of course condensation will slow cooling (transfers it to the fog top) … due radiative slowing of heat flux up from the subsurface.
I’m not claiming there is any EXTRA radiative cooling of the surface at all. I’ve said, and repeat above that radiation has a neutral effect on the LR. What it does do however is elevate it by warming equally throughout the depth of the atm, such that the surface BB (ok grey) temp of 255K is moved to ~7km and the tropopause elevated equally.
My “discussion” here has simply been about radiative cooling NOT being required and NOT causing the descent arm driving of the HC. Full stop.
“The evidence from the actual atmosphere stand s against your claims, in particular the tropopause. Below this level strong vertical Rayleigh Bernard circulation is exhibited. Above the tropopause atmospheric circulation is so weak that the lapse rate reverses and stagnant gases are subject to radiative super heating. So what’s so special about the tropopause? That’s where the atmosphere runs out of the most important radiative gas on our planet, H2O.”
The tropopause (if get I your drift correctly) I mentioned re sinking of converged air at the sub-tropical jet. It is simply that it is an inversion, a density/pressure barrier and as such, air below needs a strong “uplift” to get by it (a punch in fact). Supercell convection pushes through and this can be gauged by reference to a Tephigram or SkewT via the CAPE (convectively available potential energy). It is a nature barrier to mass air transport. Thus when faced with this barrier converged air is forced to sink.
From: http://www.iup.uni-bremen.de/~bms/lecture_atmphys/AtmosphericPhysics_04_05_dyn.pdf
“• thermal wind
• convection / convergence leads to rising air
over the tropics (ITCZ)
• movement to higher latitudes
• deflection by Coriolis force
• formation of subtropical jet stream (STJ)
• ACCUMULATION THROUGH TRANSPORT LEADS TO
SINKING AIR AT 30°
• formation of subtropical high
• most air moves back to equator close to the
surface to close the cell
• again: deflection by Coriolis force => trade
winds”
“If your claim that radiative cooling played no part in tropospheric convective circulation were true, we should observe powerful convective circulation extending well above this height. However, there is nothing climate “scientists” can do to get rid of the “punitive” tropopause.”
The radiative affect you refer to is a compositional one – O3 production/destruction via UV within the Strat. That is what makes the trop, essentially (without getting into Isentropic PV surfaces). The Tropopause height responds to the average temp of the air below it. Ie it’s thickness.
“And of course there is nothing climate “scientists” can do to remove the most damning evidence, the history of their claims recorded on the Internet. This history records that the first claims that adding radiative gases would reduce it’s radiative cooling ability were made on the basis of two shell radiation only static atmosphere models. After 1990 came the band-aid radiative-convective models. Just this sequence of events alone is enough to damn all of the AGW fellow travellers.”
Konrad: the lab experiments, empirical mathematics and direct radiative/spectroscopic observation of the atmosphere via both ground based and orbital instruments over the course of the last ~150 yrs, trumps “the Internet” I’m afraid.
“You may be encouraged that many embarrassed sceptics also want what most AGW propagandists are praying for, a “warming, but far less than we thought” soft landing for the global warming inanity. But this is not how it is going to go down. You will be up against the general public. They have no interest in defending those who claim that radiative gases reduce the atmospheres radiative cooling ability. The corpse of global warming cannot be re-animated nor can it be hidden.”
Yes, you are right Konrad, unfortunately most people are (naturally) ignorant of climate science and are equally naturally influenced by media reports. It is easy to say things like “it’s the Sun” or “it’s the clouds” or “CO2 is plant food” or “it’s happened before” etc. And many will buy it because they are easy sound bites and we perceive it will cost us “tax dollars” or Pounds in my case. Fully understanding the science is NOT easy without taking scientist’s at their word, and well, we see the result.
December 31, 2013 10:17 am
TB says:
“It is easy to say things like ‘it’s the Sun’ or ‘it’s the clouds’ or ‘CO2 is plant food’ or ‘it’s happened before’ etc. And many will buy it because…”
Many will “buy it because” to a greater or lesser degree, all those things are true.
What is completely unproven, though, is the conjecture claiming that there is a ‘human fingerprint’ in the current climate. If there is, please identify the fingerprint here.
What, you can’t find it? Well, no one else can find it, either. There could be two reasons for that:
1) Any such ‘fingerprint’ is much too small to measure, or
2) The CO2=AGW conjecture is simply wrong
(I personally think that there was some minuscule warming from CO2, but that most of the effect was seen in the first 20 ppmv, and at current CO2 concentrations the warming effect is so small that it cannot be measured.)
In either case, not one more \$ / £ / € should be wasted on the frivolous CO2=AGW conjecture; AKA: the “carbon” scare.
Because if something cannot be measured, then it isn’t science. More correctly, it begins and ends at the Conjecture stage of the Scientific Method (Conjecture, Hypothesis, Theory, Law). A conjecture might be right, or it might be wrong. But only a measurable, testable hypothesis can verify it. (Any hypothesis must be measurable and testable. And any theory must be capable of making repeated, accurate predictions. AGW can do neither.)
All the rest of the discussion amounts to endless nit-picking. The central issue of the debate has been answered: for all practical purposes, the “carbon” argument is so negligible that it can and should be completely disregarded. At this point, those promoting the scare are engaging in a self-serving scam for money and fame.
If anything I posted here is wrong, please indicate where.
Otherwise, silence is concurrence.
gbaikie
December 31, 2013 4:21 pm
-TB,
I see you are still arguing that radiative gases do not play a critical role in driving tropospheric convective circulation by allowing energy loss, buoyancy loss and subsidence of air masses.-
So main radiative gases are H20 gas and C02 gas and these are a very small percntage of the atmosphere. In terms amount of energy these gases have, H2O gas if condenses into liquid has a lot latent heat energy. The other gas CO2, does not condense into a liquid or solid so energy from the possible latent heat of CO2 in terms of in Earth atmosphere can be ignored.
And if disgard the latent heat of H2O, then we can say in terms of amount energy of the gases
of CO2 and H2O in Earth’s atmosphere, this small portion of the atmospheric gases do not have
much energy in comparison to other gases.
And convective of heat of gases is transfer of energy of the molecules of gas or it’s the energy of gas molecules which is being transferred, whereas one would not call re-radiating of energy thru
a gas as convection of heat.
Radiant heat passing thru or being transmitted thru atmosphere would not be convection of heat.
And unless this radiant heat is heating something, it can be affecting the convection of heat
of an atmosphere.
One have nitrogen gas molecule which transparent to radiant heat- acting like any transparent
material, it can reflect, disfuse/scatter the light, but it does not absob the energy of this light and re-radiant the energy. And then in comparion there are “greenhouse gases” which absorb and re-radiate this same wavelength, but like all gases [Ie, iron as a gas] are also transparent. Gases are transparent in the sense bcause they are not close together. Or even thin enough iron solid is transparent. So if our atmosphere was hot enough, and it had iron gas as impurity similar to greenhouse gas, the iron gas would also be transparent.
[Btw, it should noted that in Earth long history [billions of years ago] it has had atmosphere hot enough where iron has been a gas and which may have exceeded 400 ppm- trillions of tonnes of it. So if the Moon actually did form via an Earth impactor, that would example of such a period. But even without such an event there would been other times in Earth formative period. Even
the impact event which cause extination of dinosaur may have a significant amout iron or compounds iron gas in earth’s atmosphere. And as would super volcanic events.]
-I urge you to look again at an energy budget diagram for the atmosphere. Discounting the energy reflected from the land, ocean and atmosphere, around 90% of all energy absorbed is emitted to space from radiative gases the atmosphere.
90%!-
The troposphere has 80% of Earth amosphere and 99% of H20 greenhouse gas. And if halved the height of troposphere, more half of this 80% is below this elevation.
This alone should indicate a problem with this idea.
January 1, 2014 12:00 am
Stephen Wilde says:
December 31, 2013 at 8:12 am
———————————————
The most important thing I should say here is –
Never apply SB equations in isolation to a moving gas atmosphere over a liquid ocean. Never.
For fluid columns in a gravity field the height of energy entry and exit is critical to determining the average temperature of the fluid. This is the science of conduction and fluid dynamics. This is what is missing from the “basic physics” of the “settled science”. This is what SB equations can’t deliver.
Both the atmosphere and oceans are fluid columns in a gravity field. If any of your modelling is based on trying to simplify the land/ocean/atmosphere system to a single SB radiating level, then it will fail because you have not modelled non-radiative transports correctly.
The planet can be in radiative equilibrium, yet can have very different atmospheric temperatures depending on both the radiative and non-radiative transports within the land/ocean/atmosphere system.
And now onto the to lapse rate below the tropopause.
The observed lapse rate below the tropopause is not the product of mass and gravity alone. Atmospheric mass and gravity simply generate the atmospheric pressure gradient. It requires continued vertical circulation across this pressure gradient to generate the observed tropospheric lapse rate. And this vertical circulation must be strong enough to overcome the speed of gas conduction.
Without this vertical circulation, gas conduction would send the bulk of the atmosphere isothermal. That alone would mean that the bulk of the atmosphere would be far hotter than present.
For an atmosphere maintaining a reasonably stable temperature, yet exhibiting strong vertical convective circulation, energy must be exiting the atmosphere at a higher altitude than it is entering the atmosphere. Energy enters our atmosphere at low altitude via surface conduction, release of latent heat and intercepted radiation. There is only one way it exits at high altitude, radiative gases emitting IR to space.
Claiming that surface conduction could match the cooling power of radiative gases simply won’t work. Empirical experiment shows that the surface is virtually powerless at conductively cooling the atmosphere. The whole pole-wise energy flow being the sole driver of circulation thing won’t work. It is resisted by surface friction and gas conduction. The currently observed pole-wise energy flow is only made possible by the Hadley, Ferrel and Polar cells being maintained by radiative cooling at altitude.
You state –
“I find it strange to be arguing with fellow sceptics that radiative gases are not needed for a fully convective atmosphere.”
And I find it most strange that you would be insisting that radiative energy loss and buoyancy loss at altitude does not have a critical role in Rayleigh Bernard circulation below the tropopause.
You can ignore the evidence that strong vertical circulation is only exhibited in the strongly radiative region of our atmosphere.
You cannot ignore the evidence that after 1990, climate pseudo scientists suddenly swapped from radiative only two shell models to radiative-convective models and attempted to write radiative cooling and subsidence out of atmospheric circulation.
They had to do this. Any critical role for radiative gases in governing the speed of tropospheric vertical circulation utterly invalidates not just AGW, but its very foundation, the radiative greenhouse effect hypothesis itself.
The good news is that the Internet remembers all. The climate “scientists” can never erase their crimes against science, reason, freedom and democracy. They have left a trail of “climate science” behind them that would fertilise the Simpson Desert.
Stephen, examine that trail. (Don’t attempt anything without the gloves). Sceptics now know that any attempt to generate a “hockey stick” from proxy data will be BS. Karoly and Gergis lasted only a few hours on the Net before implosion, withdrawal of the paper and a lifetime of burning shame (three years and \$300,000 of taxpayer dollars vaporised in an instant). After McIntyre vs. Mann you know where to look. The same is true of radiative-convective modelling. If you know they are lying, you know where to look. You will find as I have done, it wasn’t a “mistake”, they knew they were lying.
January 1, 2014 2:07 am
TB says:
December 31, 2013 at 9:26 am
——————————————–
“Konrad: the lab experiments, empirical mathematics and direct radiative/spectroscopic observation of the atmosphere via both ground based and orbital instruments over the course of the last ~150 yrs, trumps “the Internet” I’m afraid”
No, nothing you try will trump the Internet. I have used the Internet to demonstrate through repeatable experiment that incident LWIR cannot heat nor slow the cooling rate of liquid water that is free to evaporatively cool.
CAGW depends on DWLWIR slowing the cooling of the oceans. I have proved that to be physically impossible.
But that’s not how the internet works.
I challenged you to produce just one simple lab experiment, that other readers could replicate, that shows incident LWIR heating or slowing the cooling rate of liquid water that is free to evaporatively cool.
Just one experiment.
You can’t do it.
Current and future readers can see you can’t do it.
You fail. Now and forever.
That’s how the Internet works.
TB
January 1, 2014 3:34 am
January 1, 2014 at 2:07 am
TB says:
December 31, 2013 at 9:26 am
——————————————–
“Konrad: the lab experiments, empirical mathematics and direct radiative/spectroscopic observation of the atmosphere via both ground based and orbital instruments over the course of the last ~150 yrs, trumps “the Internet” I’m afraid”
No, nothing you try will trump the Internet. I have used the Internet to demonstrate through repeatable experiment that incident LWIR cannot heat nor slow the cooling rate of liquid water that is free to evaporatively cool.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Look I realise of course that you are not for moving – no matter.
So, we have a history of study/experiment/mathematical modelling – restudy – remodelling. Then into the satellite era we have more observation from space. We have spectroscopic analysis from ground based instruments etc etc. That show CO2’s effect as a GHG.
BUT the internet trumps it. I suggest it depends on which bit of the internet you are reading. Do you use Google Scholar?
“CAGW depends on DWLWIR slowing the cooling of the oceans. I have proved that to be physically impossible.”
Did you not understand the concept of the ocean being a body of water? That is merely covered with a cooler skin that is evaporating. So then, at what point is it not possible to heat water via IR radiation – given that it is heat energy? Would you not think that this evaporation cooling effect would reach an optimum? That no more skin cooling could occur? Surely you are not suggesting that any amount of IR impinging on the surface will only result in cooling of the entire body. Are you saying that only SW heats water?
Sorry that is scientifically not so.
Also, are you not aware that the transport of heat from ocean and air passes through that skin via conduction/convection too? Do you not know that this is optimum when the hot surface is hottest? And that this skin cooling will go against that? Thereby REDUCING thermal contrast and keeping the turbulent waters below the skin WARMER than otherwise.
No it is entirely possible. In fact it happens, has to.
“I challenged you to produce just one simple lab experiment, that other readers could replicate, that shows incident LWIR heating or slowing the cooling rate of liquid water that is free to evaporatively cool.”
Your lab experiment does model the Earth and nor does it trump thousands of other experiments done over the last 150 yrs.
“Just one experiment.”
This is the same misconceived concept some have of the GHE. Ie some think you can’t warm an object with a colder one. Of course that’s NOT how the GHE works. It’s like a storage tank whereby there is an outflow and an inflow. The inflow (back-IR) is small and the outflow is large (given radiative cooling dominating). Now the outflow will still occur, but the pressure built in the tank will build and the outflow pressure increase. Ie there is a flux of heat stored in the tank (ground) that always wants to get out – sometimes it’s mostly in (inflow exceeds outflow) sometimes it’s out (outflow more than inflow). But the outflow can be slowed, by a small inflow. On land this in/out point manifests this reduced outflow by a temp rise.
Same with evaporative cooling – only in reverse. At some point the evaporation is at a maximum, yes? Must be, otherwise it would keep cooling forever (given the vast source of water available). So, obviously not. Then what happens to the energy in excess of that needed to evaporate that water?
We cannot have a steady state, something must happen. Logically the surface skin would cool so far and then, if it cannot maintain that cooling via LH uptake – it will warm, and eventually boil. Before it boils it will warm the water beneath by conduction.
Now of course this DOES NOT happen (in the oceans). But do you see the logic? It’s the same illogic that is applied by some who say there can be no GHE because you can’t warm an object by a colder one. That is not what happens. At what point does this drip, drip of back-radiated IR photons go over that optimum point and the cooling tips to warming. IR energy must heat water. It cannot always cool it. It’s a matter of the evaporative process balancing against the radiative sensible heating effect via conduction from below. The IR back radiating occurs during the day as well, when evaporation will be at a max.
The primary way that evaporation WARMS a body of water (when the air is colder than the water) is by the reduction of the thermal gradient at the ocean/air interface. Cooler skin means less conducted, and hence the body of water below the skin has a “warming blanket” over it, analogous to the GHE “blanket”. It’s about relative flows of energy.
“Current and future readers can see you can’t do it.”
I know you’re right in that in the main, I realise I’m in the minority on this forum – however it is fundamentally about debate. And some may take what I say on-board and go off and investigate “whatever” more deeply. That’s all I try to do. I have professional knowledge gained via man’s sum total of Meteorological investigation (not good on the curly d’s and Dels though) and further, I’ve spent many a long, lonely shift (day & night) observing weather and passing on information/forecasts to, eg, aircrew and operations for road ice management. I know as well as anyone can how it works. Sorry it’s just the way things work. Now, Anthony may have had a different career path than me, I don’t know, but I’m sure the physical processes at play in the atmosphere are well known to him. He disagrees on the degree of warming attributable to the GHE, which is fine. But he does not disagree with the fundamental processes and laws of thermodynamics.
“You fail. Now and forever.”
If you want to think that – it is of no matter.
“That’s how the Internet works.”
The Internet is a fantastic resource of knowledge – the greatest in the history of mankind. But one has to use it correctly. There is always 2 sides to the argument. Look for the other-side occasionally – that’s what I’m doing here in fact (It may surprise you to know that I’ve learned some stuff by doing so). This by arguments raising questions in my mind and me (being who I am) going off and investigating/learning from that question. You see it all depends how you use it.
Regards anyway Konrad and a Happy new year.
TB
January 1, 2014 4:52 am
-gbaikie says:
December 31, 2013 at 4:21 pm
-TB,
I see you are still arguing that radiative gases do not play a critical role in driving tropospheric convective circulation by allowing energy loss, buoyancy loss and subsidence of air masses.-
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Correct.
“-I urge you to look again at an energy budget diagram for the atmosphere. Discounting the energy reflected from the land, ocean and atmosphere, around 90% of all energy absorbed is emitted to space from radiative gases the atmosphere.
90%!-“
I have, please see my post to Konrad. In short, it is but only because the atm is between the Surface and space. It reaches the surface either directly (bar reflected) as SW or by back-radiation of LWIR – then goes to atm – then goes to space.
“So main radiative gases are H20 gas and C02 gas and these are a very small percentage of the atmosphere. In terms amount of energy these gases have, H2O gas if condenses into liquid has a lot latent heat energy. The other gas CO2, does not condense into a liquid or solid so energy from the possible latent heat of CO2 in terms of in Earth atmosphere can be ignored.”
No, it cant be ignored as it’s a tri-atomic molecule and taking away N2 ~78% and O2 ~21% (= ~99 non-radiative) makes it very much a player in the atmosphere radiatively.
LH uptake/release is an adiabatic process and not a radiative one, and H20 is the dominant tri-atomic gas in the atm – however the hydrological cycle ensures it’s transport from liquid on ground > gas in air > liquid in air > liquid on ground again, cycles at around ~10 days.
So, with turbulent mixing/convection, the relative humidity stays steady over long time scales (globally averaged) – however warmer air means a higher absolute humidity (more molecules) – and hence greater GHE.
H2O is vital to the atmosphere, as you say, as it cools the surface via LH uptake and heats the atmosphere via release. It modifies what would otherwise be a dry LR into one between a dry and a wet one. With of course large variations – mainly less steep, due inversions because of either cooling from below (moving over cold surface or radiative cooling from ground) – or warming from above (subsidence or warm advection – eg approaching warm front) such that we can have unstable air in places and stable air in others.
“The troposphere has 80% of Earth atmosphere and 99% of H20 greenhouse gas. And if halved the height of troposphere, more half of this 80% is below this elevation.
This alone should indicate a problem with this idea.”
? Sorry I don’t understand this.
BTW; There is an exception in where radiative effects help drive the depth of convection – which is why thunderstorms often have their peak overnight.
A lifting air parcel – a Cumulonimbus cloud in this respect – is rising because of the contrast between it and it’s environment (bar divergence aloft of it anyway – which is another complication).
Now, we know that day-time heating causes the ground to let-go thermals which rise via buoyancy to the condensation level. The release of LH then warms the up-draught (or rather it slows it’s cooling by rarefaction) and it rises still more. In an unstable atm this will often allow it to reach the tropopause, and there, like smoke rising against a ceiling it stops and spreads out horizontally. (usually – as with enough energy it will punch through and may well reach up to 60,000ft ~12ml).
I digress a bit – anyway, overnight you would think that the ground would cool and the convection stop. It does, normally, but there are circumstances where it keeps going, even intensifies. This is when the storm, having enough water content, can become more unstable because the top of the cloud cools to space radiatively. This makes it less buoyant relative to the surrounding environment BUT crucially it creates a more unstable cloud (colder top rel to warm bottom > rises/overturns more).
Radiative imbalance in the atm affects it’s LR, cloud top cooling via increased albedo say, but otherwise it has little effect. The LH and sensible heat processes dominate in (surface) convection.
Mario Lento
January 1, 2014 2:07 pm
TB says:
“So then, at what point is it not possible to heat water via IR radiation – given that it is heat energy?”
++++++++
I think part of the confusion here is ambiguity. The quoted statement needs to be specific. I posit that IR radiation is a very specific type of heat energy. It is not conductive and it is not convective.
January 1, 2014 7:26 pm
TB says:
January 1, 2014 at 3:34 am
———————————————————————–
“Did you not understand the concept of the ocean being a body of water? That is merely covered with a cooler skin that is evaporating. So then, at what point is it not possible to heat water via IR radiation – given that it is heat energy? Would you not think that this evaporation cooling effect would reach an optimum? That no more skin cooling could occur? Surely you are not suggesting that any amount of IR impinging on the surface will only result in cooling of the entire body. Are you saying that only SW heats water?
Sorry that is scientifically not so.
Also, are you not aware that the transport of heat from ocean and air passes through that skin via conduction/convection too? Do you not know that this is optimum when the hot surface is hottest? And that this skin cooling will go against that? Thereby REDUCING thermal contrast and keeping the turbulent waters below the skin WARMER than otherwise.
No it is entirely possible. In fact it happens, has to.”
Some have gone as far as to claim that increased DWLWIR would increase evaporative cooling of the oceans. I do not make this claim as far more complicated experiments would be required to test this. What I do claim is that incident LWIR cannot heat nor slow the cooling rate of liquid water that is free to evaporatively cool. It is only UV and SW that heats our oceans.
Anyone can check my claim by building this simple experiment –
http://i42.tinypic.com/2h6rsoz.jpg
– the results are, as they say in climate science, “robust”. Start with 40C water in each test chamber and monitor the cooling rate under both the strong and weak LWIR source. Both water samples cool at the same rate. Now repeat the experiment with a square of LDPE cling wrap floated onto the surface of each sample. This allows conductive and radiative cooling but restricts evaporation. Now both samples cool slower, but at different rates.
The reason incident LWIR cannot heat water that is free to evaporatively cool is that the IR photons are absorbed in the first few microns of the skin evaporation layer. This simply trips some water molecules into evaporating slightly sooner than they otherwise would. The cooling effect of these molecules undergoing phase change offsets any heating effect of the LWIR. There is no effect on the cooling rate of the liquid below the skin evaporation layer.
All AGW two shell radiative models show LWIR having the same effect over the oceans as it does over land. Some climate scientists even try to claim that the oceans would freeze without DWLWIR. This is provably false.
You are claiming “In fact it happens, has to.”, yet have no empirical evidence that LWIR can slow the cooling rate of the oceans.
You claim “your lab experiment does model the Earth and nor does it trump thousands of other experiments done over the last 150 yrs.”. Yet you do not provide a single one of these “thousands” of experiments.
You state –
“We have spectroscopic analysis from ground based instruments etc etc. That show CO2’s effect as a GHG.”
But this is also incorrect. We have satellite and ground based measurements that show that CO2 both adsorbs and emits LWIR. CO2 is a radiative gas. The term “GHE” ignores that in the case of the radiative GHE hypothesis, the null hypothesis still stands.
“This is the same misconceived concept some have of the GHE. Ie some think you can’t warm an object with a colder one.”
I am clearly not disputing basic radiative physics. I have previously provided build diagrams for a simple two shell radiative model that others can build –
http://i44.tinypic.com/2n0q72w.jpg
http://i43.tinypic.com/33dwg2g.jpg
http://i43.tinypic.com/2wrlris.jpg
– the extra foil layer in chamber 1 back radiating the target plate will be cooler than the target plate, yet it still slows the cooling of the target plate.
But that experiment only shows that the maths behind radiative two shell models works. However I have also provided experiments showing how critical non radiative energy transports are to atmospheric temperatures. The permanent Internet history of global warming shows that these transports, and most importantly the role of radiative gases in driving them, were not correctly modelled when the foundation claims of radiative gases causing atmospheric warming were made.
I am claiming that the NET effect of radiative gases in our atmosphere is atmospheric cooling at all concentrations above 0.0ppm. The basis of these claims is known fluid dynamics, empirical experiment and observation, none of which violates any “fundamental processes and laws of thermodynamics.”
I would suggest that it is those claiming LWIR can slow the cooling of the oceans or that circulation would continue in the Hadley, Ferrel and Polar cells in the absence of radiative cooling at altitude, are the ones violating “fundamental processes and laws of thermodynamics.”
gbaikie
January 1, 2014 8:59 pm
-The other gas CO2, does not condense into a liquid or solid so energy from the possible latent heat of CO2 in terms of in Earth atmosphere can be ignored.”
No, it cant be ignored as it’s a tri-atomic molecule and taking away N2 ~78% and O2 ~21% (= ~99 non-radiative) makes it very much a player in the atmosphere radiatively.-
I was saying what is obvious, CO2 doesn’t have phase change/ transition:
http://en.wikipedia.org/wiki/Phase_transition
Or CO2 doesn’t change from gas to liquid or solid in Earth atmosphere, as does H20.
H20 can transport it’s latent heat, CO2 doesn’t do this in Earth atmosphere [though CO2 does
do this on the much colder planet Mars].
And btw, this transportation of latent heat of H2O is a significant conveyor of heat to upper atmosphere on Earth. But I saying this is separate issue or it’s not radiant transfer of energy or Heat.
So what was saying, that since taliking about radiant transfer, the latent heat of of H2O
is excluded and CO2 does transfer heat via latent heat in Earth’s atmosphere.
So yes I can be excluding or ignoring the latent heat of H2O and the non-existent latent heat of CO2 in Earth’s atmosphere- because the topic is radiant transfer.
gbaikie
January 1, 2014 9:16 pm
-LH uptake/release is an adiabatic process and not a radiative one, and H20 is the dominant tri-atomic gas in the atm – however the hydrological cycle ensures it’s transport from liquid on ground > gas in air > liquid in air > liquid on ground again, cycles at around ~10 days.
So, with turbulent mixing/convection, the relative humidity stays steady over long time scales (globally averaged) – however warmer air means a higher absolute humidity (more molecules) – and hence greater GHE.-
I am not disputing that water droplets which are associated clouds radiate heat into space.
Or clouds are a grouping of water droplets in enough quantity that they are clearly visible as clouds- on the other hand, they are in addition droplets of water not so densely pack or large in average size which are not easily and clearly visible as clouds.
gbaikie
January 1, 2014 9:37 pm
-“The troposphere has 80% of Earth atmosphere and 99% of H20 greenhouse gas. And if halved the height of troposphere, more half of this 80% is below this elevation.
This alone should indicate a problem with this idea.”
? Sorry I don’t understand this.-
I mean in terms greenhouse gases- CO2 and H20, most of it is below 1/2 the height of troposphere. So including water droplets, and all which one can consider that radiates in the atmosphere, most must occur below 1/2 of the height of troposphere.
And above the troposphere you exclude water in all it’s forms as being quantitatively significant in terms total it could radiate. And CO2 above the troposphere is limited to 20% or whatever total amount it can radiate. Or 80% of possible amount CO2 can radiates must be below the top of the troposphere. Plus most of total amount energy energy “available” by any means possible, must also radiate below the top of the troposphere.
It doesn’t matter what what theory you wish to believe. These facts must be allowed for in any idea involving the concept that the atmosphere radiate any significant amount of energy. | 60,144 | 261,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-31 | longest | en | 0.938467 |
https://www.shaalaa.com/question-bank-solutions/the-circumference-base-cylinder-88-cm-its-height-15-cm-find-its-curved-surface-area-total-surface-area-surface-area-of-cylinder_61269 | 1,618,401,063,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077810.20/warc/CC-MAIN-20210414095300-20210414125300-00354.warc.gz | 1,109,232,962 | 9,336 | # The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area. - Mathematics
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.
#### Solution
$\text{ Given } :$
$\text{ Height, h = 15 cm }$
$\text{ Circumference of the base of the cylinder = 88 {cm } }^2$
$\text{ Let r be the radius of the cylinder } .$
$\text{ The circumference of the base of the cylinder } = 2\pi r$
$88 = 2 \times \frac{22}{7} \times r$
$r = \frac{88 \times 7}{2 \times 22} = 14 cm$
$\text{ Curved surface area } = 2 \times \pi \times r \times h$
$= 2 \times \frac{22}{7} \times 14 \times 15$
$= 1320 {\text{ cm} } ^2$
$\text{ Total surface area } = 2 \times \pi \times r \times (r + h)$
$= 2 \times \frac{22}{7} \times 14 \times (14 + 15)$
$= 2552 \text{ cm } ^2$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 22 Mensuration - III (Surface Area and Volume of a Right Circular Cylinder)
Exercise 22.1 | Q 4 | Page 10 | 358 | 1,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-17 | latest | en | 0.703358 |
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Frequent Visitor
## Sort a table by the results of two columns
Is there a way to sort a table by the results of two columns?
In the first column I have the results of our delivery reliability evaluation. In the second column I have the quantity of our deliveries. Now I would like to sort the table. At row one I would like to have the lowest result of the first column with the highest result of the second column. Providing that there is more than one position with the same result in column one.
Example:
Before sorting:
Column 1 Column 2
30% 5
20% 2
40% 1
20% 1
60% 4
50% 3
20% 3
20% 2
After sorting:
Column 1 Column 2
20% 3
20% 2
20% 2
20% 1
30% 5
40% 1
50% 3
60% 4
I hope someone of you can help me to fix my problem!
1 ACCEPTED SOLUTION
Accepted Solutions
Super User
## Re: Sort a table by the results of two columns
Another method would be to create a third calculated column:
`Column 3 = ([Column 1] * 100) + (1/[Column 2])`
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3 REPLIES 3
Highlighted
Super User
## Re: Sort a table by the results of two columns
Interesting problem. One way would be to use a Matrix and sort by Column 2.
### I have book! Learn Power BI from Packt
Proud to be a Datanaut!
Super User
## Re: Sort a table by the results of two columns
Another method would be to create a third calculated column:
`Column 3 = ([Column 1] * 100) + (1/[Column 2])`
### I have book! Learn Power BI from Packt
Proud to be a Datanaut!
Frequent Visitor
## Re: Sort a table by the results of two columns
It works. Thank you @Greg_Deckler.
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Top Solution Authors
Top Kudoed Authors (Last 30 Days) | 609 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-05 | latest | en | 0.814923 |
https://www.physicsforums.com/threads/second-order-differential-equation.745005/ | 1,721,426,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00318.warc.gz | 814,838,259 | 16,305 | # Second order differential equation
• elevenb
In summary, the conversation is about a problem involving two particles connected by a spring with a spring constant k and zero equilibrium length. The link provides all necessary equations and the attempt at a solution involves using the Binomial theorem to solve for the angular frequency. The last step uses the standard solution for the differential equation of simple harmonic motion.
elevenb
## The Attempt at a Solution
The solution is there, its to do with part (c), I would really appreciate it if someone would explain how they make the jumps - especially to the second last equation
They used the Binomial theorem.
$$\frac{1}{(d + \Delta d)^2} =\frac{1}{d^2}\left(1 + \frac{\Delta d}{d}\right)^{-2} = \frac{1}{d^2}\left(1 - 2\frac{\Delta d}{d} + \cdots\right)$$
Everything before that looks like straightforward algebra. The last step to find ##\omega## is the standard solution of the differential equation for simple harmonic motion.
Ignore- spotted mistake
Last edited:
## What is a second order differential equation?
A second order differential equation is a mathematical equation that describes the relationship between a function and its second derivative. It is typically used to model physical phenomena in science and engineering.
## How is a second order differential equation different from a first order differential equation?
A second order differential equation involves the second derivative of the function, while a first order differential equation involves only the first derivative. This means that a second order differential equation requires two initial conditions to be fully determined, whereas a first order differential equation only requires one initial condition.
## What are some real-world applications of second order differential equations?
Second order differential equations can be used to model a variety of physical phenomena, such as the motion of a spring, the oscillations of a pendulum, and the flow of fluids. They are also commonly used in engineering to analyze and design systems such as electrical circuits and control systems.
## How do you solve a second order differential equation?
There are various methods for solving second order differential equations, including separation of variables, substitution, and using the characteristic equation. The specific method used will depend on the form of the equation and the initial conditions given.
## What are the initial conditions in a second order differential equation?
The initial conditions in a second order differential equation refer to the values of the function and its first derivative at a specific point. These values are necessary to fully determine the equation and find a unique solution.
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2K | 690 | 3,266 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.929035 |
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# Maths : Problems on Train (Solved in Steps)
1. A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres
B. 160 metres
C. 200 metres
D. 120 metres
Explanation :
Speed of the train = 40 km/hr
= 40000/3600 m/s
= 400/36 m/s
Time taken to cross = 18 s
Distance Covered = speed× time
= (400/36)× 18
= 200 m
Distance covered is equal to the length of the train = 200 m
2. A train ,130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m
B. 245 m
C. 235 m
D. 220 m
Explanation :
Assume the length of the bridge = x meter
Total distance covered = 130+x meter
total time taken = 30s
speed = Total distance covered /total time taken = (130+x)/30 m/s
=> 45 × (10/36) = (130+x)/30
=> 45 × 10 × 30 /36 = 130+x
=> 45 × 10 × 10 / 12 = 130+x
=> 15 × 10 × 10 / 4 = 130+x
=> 15 × 25 = 130+x = 375
=> x = 375-130 =245
3. A train has a length of 150 meters . it is passing a man who is moving at 2 km/hr in the same direction of thetrain, in 3 seconds. Find out the speed of the train.
A. 182 km/hr
B. 180 km/hr
C. 152 km/hr
D. 169 km/hr
Explanation :
Length of the train, l = 150m
Speed of the man = 2 km/hr
Relative speed = total distance/time = (150/3) m/s
= (150/3) × (18/5) = 180 km/hr
Relative Speed = Speed of train – Speed of man (As both are moving in the same direction)
=> 180 = Speed of train – 2
=> Speed of train = 180 + 2 = 182 km/hr
4. A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec
B. 99 s
C. 89 s
D. 80 s
Explanation :
speed of the train = 240/24 = 10 m/s
time taken to pass a platform having a length of 650 m
= (240+650)/10 = 89 seconds
5. A train 360 m long runs with a speed of 45 km/hr. What time will it take to pass a platform of 140 m long?
A. 38 sec
B. 35 s
C. 44 sec
D. 40 s
Explanation :
Speed = 45 km/hr = 45×(10/36) m/s
= 150/12 = 50/4 = 25/2 m/s
Total distance = length of the train + length of the platform
= 360 + 140 = 500 meter
Time taken to cross the platform = 500/(25/2) = 500×2/25 = 40 seconds
6. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data
B. 3 : 1
C. 1 : 3
D. 3 : 2
Explanation :
Let the speed of the trains be x and y respectively length of train1 = 27x
length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 23 s
=> (27x + 17 y)/(x+y) = 23
=> (27x + 17 y) = 23(x+y)
=> 4x = 6y
=> x/y = 6/4 = 3/2
ratio of their speeds=3:2
7. Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. If the faster train passes the slower train in 36 seconds, what is the length of each train?
A. 88
B. 70
C. 62
D. 50
Explanation :
Assume the length of each train = x
Total distance covered for overtaking the slower train =
x+x = 2x
Relative speed = 46-36 = 10km/hr = (10×10)/36 = 100/36 m/s
Time = 36 seconds
2x/ (100/36) = 36
=> (2x × 36 )/100 = 36
=> x = 50 meter
8. Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s
B. 12 s
C. 9.8 s
D. 8 s
Explanation :
Distance = 140+160 = 300 m
Relative speed = 60+40 = 100 km/hr =
(100×10)/36 m/s
Time = distance/speed = 300 / (100×10)/36 = 300×36 /1000 = 3×36/10 = 10.8 s
9. Two trains are moving in opposite directions with speed of 60 km/hr and 90 km/hr respectively. Their lengths are 1.10 km and 0.9 km respectively. The slower train cross the faster train in — seconds
A. 56
B. 48
C. 47
D. 26
Explanation :
Relative speed = 60+90 = 150 km/hr (Since both trains are moving in opposite directions)
Total distance = 1.1+.9 = 2km
Time = 2/150 hr = 1//75 hr = 3600/75 seconds = 1200/25 = 240/5 = 48 seconds
10. A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these
B. 280 meter
C. 240 meter
D. 200 meter
Explanation :
Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s
Length of the train = speed × time taken to cross the man = 15×20 = 300 m
Let the length of the platform = L
Time taken to cross the platform = (300+L)/15
=> (300+L)/15 = 36
=> 300+L = 15×36 = 540
=> L = 540-300 = 240 meter
11. A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr
B. 69 km/hr
C. 74 km/hr
D. 61 km/hr
Explanation :
Let x is the length of the train and v is the speed
Time taken to move the post = 8 s
=> x/v = 8
=> x = 8v — (1)
Time taken to cross the platform 264 m long = 20 s
(x+264)/v = 20
=> x + 264 = 20v —(2)
Substituting equation 1 in equation 2, we get
8v +264 = 20v
=> v = 264/12 = 22 m/s
= 22×36/10 km/hr = 79.2 km/hr
12. Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction?
A. 10
B. 25
C. 12
D. 20
Explanation :
speed of train1 = 120/10 = 12 m/s
speed of train2 = 120/15 = 8 m/s
if they travel in opposite direction, relative speed = 12+8 = 20 m/s
distance covered = 120+120 = 240 m
time = distance/speed = 240/20 = 12 s
13. A train having a length of 1/4 mile, is travelling at a speed of 75 mph. It enters a tunnel 3 ½ miles long. How long does it take the train to pass through the tunnel from the moment the front enters to the moment the
rear emerges?
A. 3 min
B. 4.2 min
C. 3.4 min
D. 5.5 min
Explanation :
Total distance = 3 ½ + ¼ = 7/2 + ¼ = 15/4 miles
Speed = 75 mph
Time = distance/speed = (15/4) / 75 hr = 1/20 hr = 60/20 minutes = 3 minutes
14. A train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the train?
A. 270 m
B. 210 m
C. 340 m
D. 130 m
Explanation :
Speed= 72 kmph = 72×10/36 = 20 m/s
Distance covered = 250+ x where x is the length of the train
Time = 26 s
(250+x)/26 = 20
250+x = 26×20 = 520 mx = 520-250 = 270 m
15. A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m
B. 54 m
C. 50 m
D. 55 m
Explanation :
Let x is the length of the train in meter and v is its speed in kmph
x/9 = ( v-2)(10/36) —(1)
x/10 =( v-4) (10/36) — (2)
Dividing equation 1 with equation 2
10/9 = (v-2)/(v-4)
=> 10v – 40 = 9v – 18
=> v = 22
Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m
16. A train is travelling at 48 kmph . It crosses another train having half of its length, travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m
B. 360 m
C. 480 m
D. 400 m
Explanation :
Speed of train1 = 48 kmph
Let the length of train1 = 2x meter
Speed of train2 = 42 kmph
Length of train 2 = x meter (because it is half of train1’s length)
Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s
Time = 12 s
Distance/time = speed => 3x/12 = 25
=> x = 25×12/3 = 100 meter
Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s
Speed of train1 = 48 kmph = 480/36 = 40/3 m/s
Distance = 200 + y where y is the length of the platform
=> 200 + y = 45×40/3 = 600
=> y = 400 meter
17. A train having a length of 270 meters is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m
B. 190 m
C. 210 m
D. 230 m
Explanation :
Relative speed = 120+80 = 200 kmph = 200×10/36 m/s
= 500/9 m/s
time = 9s
Total distance covered = 270 + x ,where x is the length of other train
(270+x)/9 = 500/9
=> 270+x = 500
=> x = 500-270 = 230 meter
18. Two trains, each 100 m long are moving in opposite directions. They cross each other in 8 seconds. If one is moving twice as fast the other, the speed of the faster train is
A. 75 km/hr
B. 60 km/hr
C. 35 km/hr
D. 70 km/hr
Explanation :
Total distance covered = 100+100 = 200 m
Time = 8 s
let speed of slower train is v . Then the speed of the faster train is 2v
(Since one is moving twice as fast the other)
Relative speed = v + 2v = 3v
3v = 200/8 m/s = 25 m/s
=> v = 25/3 m/s
Speed of the faster train = 2v = 50/3 m/s = (50/3)×(36/
10) km/hr = 5×36/3 = 5×12 = 60 km/hr
19. Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m
B. 10 a.m
C. 9.10 a.m.
D. 11 a.m
Explanation :
Assume both trains meet after x hours after 7 am
Distance covered by train starting from P in x hours = 20x km
Distance covered by train starting from Q in (x-1) hours = 25(x-1)
Total distance = 110
=> 20x + 25(x-1) = 110
=> 45x = 135
=> x= 3
Means, they meet after 3 hours after 7 am, ie, they meet at 10 am
20. A train overtakes two persons walking along a railway track. The first person walks at 4.5 km/hr and the other walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
A. 81 km/hr
B. 88 km/hr
C. 62 km/hr
D. 46 km/hr
Explanation :
Let x is the length of the train in meter and y is its speed in kmph
x/8.4 = (y-4.5)(10/36) —(1)
x/8.5 = (y-5.4)(10/36) —(2)
Dividing 1 by 2
8.5/8.4 = (y-4.5)/ (y-5.4)
=> 8.4y – 8.4 × 4.5 = 8.5y – 8.5×5.4
.1y = 8.5×5.4 – 8.4×4.5
=> .1y = 45.9-37.8 = 8.1
=> y = 81 km/hr
21. A train , having a length of 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that of the train
A. 10 sec
B. 8 sec
C. 6 sec
D. 4 sec
Explanation :
Distance = 110 m
Relative speed = 60+6 = 66 kmph (Since the train and the man are in moving in opposite direction)
= 66×10/36 mps = 110/6 mps
Time = distance/speed = 110
22. A 300-metre long train crosses a platform in 39 seconds while it crosses a post in 18 seconds. What is the length of the platform?
A. 150 m
B. 350 m
C. 420 m
D. 600 m
Explanation :
Length of the train
= distance covered in crossing the post
= speed × time
= speed × 18
ie,300= speed × 18
Speed of the train = 300/18 m/s = 50/3 m/s
Time taken to cross the platform = 39 s
(300+x)/(50/3) = 39 s where x is the length of the platform
300+x = (39 × 50) / 3 = 650 meter
x = 650-300 = 350 meter
23. A train crosses a post in 15 seconds and a platform 100 m long in 25 seconds. Its length is
A. 150 m
B. 300 m
C. 400 m
D. 180 m
Explanation :
Assume x is the length of the train and v is the speed x/v = 15
=> v = x/15
(x+100)/v = 25
=> v = (x+100)/25
Ie, x/15 = (x+100)/25
=> 5x = 3x+ 300
=> x = 300/2 = 150
24. A train , 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m
B. 500 m
C. 260 m
D. 430 m
Explanation :
Distance = 800+x meter where x is the length of the tunnel
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s
Distance/time = speed
(800+x)/60 = 65/3
=> 800+x = 20×65 = 1300
=> x = 1300 – 800 = 500 meter
25. Two trains each 500 m long, are running in opposite directions on parallel tracks. If their speeds are 45 km/ hr and 30 km/hr respectively, the time taken by the slower train to pass the driver of the faster one is
A. 50 sec
B. 58 sec
C. 24 sec
D. 22 sec
Explanation :
Relative speed = 45+30 = 75 km/hr = 750/36 m/s = 125/6 m/s
We are calculating the time taken by the slower train to pass the driver of the faster one
Hence the distance = length of the smaller train = 500 m
Time = distance/speed = 500/(125/6) = 24 s
26. Two trains are running in opposite directions at the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42
B. 36
C. 28
D. 20
Explanation :
Distance covered = 120+120 = 240 m
Time = 12 s
Let the speed of each train = v. Then relative speed = v+v = 2v
2v = distance/time = 240/12 = 20 m/s
Speed of each train = v = 20/2 = 10 m/s
= 10×36/10 km/hr = 36 km/hr
27. A train 108 m long is moving at a speed of 50 km/hr . It crosses a train 112 m long coming from opposite direction in 6 seconds. What is the speed of the second train?
A. 82 kmph
B. 76 kmph
C. 44 kmph
D. 58 kmph
Explanation :
Total distance = 108+112 = 220 m
Time = 6s
Relative speed = distance/time = 220/6 m/s = 110/3 m/s
= (110/3) × (18/5) km/hr = 132 km/hr
=> 50 + speed of second train = 132 km/hr
=> Speed of second train = 132-50 = 82 km/hr
28. How many seconds will a 500-meter long train moving with a speed of 63 km/hr, take to cross a man walking with a speed of 3 km/hr in the direction of the train?
A. 42
B. 50
C. 30
D. 28 | 4,718 | 13,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-27 | latest | en | 0.888896 |
http://math.stackexchange.com/questions/143201/why-is-the-cyclotomic-polynomial-over-mathbbq | 1,469,273,120,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00223-ip-10-185-27-174.ec2.internal.warc.gz | 166,406,255 | 18,147 | # Why is the cyclotomic polynomial over $\mathbb{Q}$?
As defined in Wikipedia (and this is the same definition I was given in class), it is not clear to me why the cyclotomic polynomial is over $\mathbb{Q}$.
It is over $\mathbb{C}$, but I don't see a reason for the coefficient to be in $\mathbb{Q}$.
Can anyone help with this one?
-
It follows from for example the Möbius inversion formula on that Wikipedia page. – Jyrki Lahtonen May 9 '12 at 20:25
Coefficients are in fact in $\mathbb{Z} \subseteq \mathbb{Q}$. – Sasha May 9 '12 at 20:25
@Belgi, that application of the Möbius function does not involve complex analysis at all. – Jyrki Lahtonen May 9 '12 at 20:30
@Sasha, when you divide two monic polynomials with integer coefficients, the result will also have integer coefficients. Think about what happens when you do long division. – Jyrki Lahtonen May 9 '12 at 20:32
$\prod_{d\mid n}\Phi_d(X) = X^n - 1$
Edit Let $g(X) = \prod_{d\mid n, d ＜ n}\Phi_d(X)$. By the induction hypothesis $g(X) ∈ \mathbb{Q}[X]$. Hence $\Phi_n(X) = (X^n - 1)/g(X) ∈ \mathbb{Q}[X]$ | 342 | 1,071 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-30 | latest | en | 0.832525 |
https://la.mathworks.com/matlabcentral/cody/problems/2640-find-similar-sequences/solutions/1330143 | 1,606,334,863,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00151.warc.gz | 374,859,997 | 17,087 | Cody
Problem 2640. Find similar sequences
Solution 1330143
Submitted on 7 Nov 2017 by Shawn Mengel
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
m = [3 1 5 0 0 3 4 1 5 0 4 2 1 5 0]; srows = [3 1 5 0 0;4 2 1 5 0]; assert(isequal(findsimilar(m),srows))
2 Pass
m = [3 1 5 0 0 1 2 5 0 0 1 3 4 1 5 2 1 5 0 0]; srows = [3 1 5 0 0; 2 1 5 0 0]; assert(isequal(findsimilar(m),srows))
3 Pass
m = [3 1 5 7 0 3 2 5 7 0 3 5 7 2 0 1 5 7 2 0 4 6 7 8 9 4 5 7 8 0 4 5 6 7 8]; srows = [3 1 5 7 0;1 5 7 2 0;4 6 7 8 9;4 5 7 8 0]; assert(isequal(findsimilar(m),srows))
4 Pass
m = [3 1 5 0 0 3 1 6 0 0 3 2 6 0 0 2 1 5 6 0]; srows = m; assert(isequal(findsimilar(m), srows))
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# Spring Lake does not appear to be good for sailing. I have
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Spring Lake does not appear to be good for sailing. I have [#permalink]
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04 Apr 2004, 08:30
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Spring Lake does not appear to be good for sailing. I have gone to the lake many times this year, and each time the water was too rough for sailing.
Which of the following most closely parallels the above argument?
A) it appears that we will move to Spring Lake this year. The city is simply too rough for safe living
B) economy-grade gasoline apparently does not prevent my car from running rough. It appears that a good grade of fuel is required
C) it appears that the cost of housing at Spring Lake is prohibitive. I looked at a number of houses last month, and they cost much more than I could afford
D) I am withdrawing from Spring Lake College. Two months at school was sufficient to prove that college was too rough for me
E) it appears that I will never play the clarinet. I began lessons many times, but each time, I quit
_________________
Best Regards,
Paul
If you have any questions
you can ask an expert
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04 Apr 2004, 08:43
I would go with E,,,,Debate was between C and E again...Is there any good way to deal with " same reasoning kind of questions"...I am having a big problem with these kind of questions....
Any help paul??
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04 Apr 2004, 09:27
Paul: Thanks for posting such nice questions. I've seen you answering
others questions. But, this is your chance to tease us
This is a hard questions.
Choices C, D and E are the contenters.
Choice D could be dropped because he took classes for two months and
two months is mostly enough time to get a feel about the college.
Choice E, There is no external factor affecting his learning. Either he is
not interested very much in learning or he may be musically disabled.
That leaves us with choice C.
Fingers crossed and waiting for Paul's teasing and educating reply.
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04 Apr 2004, 09:36
I will go with C, tricky question.
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04 Apr 2004, 18:58
C it is. Indeed E was misleading but Kpadma explained it very well.
_________________
Best Regards,
Paul
04 Apr 2004, 18:58
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8. According to KCL as applied to a junction in a network of conductors :
(a) total sum of currents meeting at the junction is zero
(b) no current can leave the junction without some current entering it
(c) net current flow at the junction is positive
(d) algebraic sum of the currents meeting at the junction is zero
Explanation
Explanation : No answer description available for this question. Let us discuss.
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Tags: laws, kirchhoff, positive, network, sum, algebraic, exercise, circuit, basic, electrical
• 5555555555 2. The currents into a junction flow along two paths. One current is 4 A and the other is 3 A. The total current out of the junction is ________. (a) 1 A (b) 7 A (c) unknown (d) the larger of the two
Tags: current, junction, currents, flow, total, basic, electrical, engineering | 815 | 3,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.749646 |
https://www.jiskha.com/display.cgi?id=1289006092 | 1,516,241,386,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887054.15/warc/CC-MAIN-20180118012249-20180118032249-00492.warc.gz | 942,008,414 | 4,267 | # agebra
posted by .
Sam and Chris went to “Lots O Fun” to play laser tag and video games for Chris’s birthday. Sam played 3 games of laser tag, 5 video games, and spent \$17 total. Chris played 4 games of laser tag, 7 video games, and spent \$23 total. How much does one video game cost to play?
• agebra -
L=laser tag, V=video games
3L+5V=17 (1)
4L+7V=23 (2)
Since the question asks only about the video games, we want to eliminate the L's. We can do this by multiplying the first equation by 4 and the second equation by -3.
12L+20V=68 (3)
-12L-21V=-69 (4)
-V=-1
V=1
Video games are \$1 each. If you had to solve for laser tag, you would the substitute v=1 into equation 1 or 2.
## Similar Questions
1. ### equation
n represent the number of games lost, write an equation for: a) A team won three times as many games as it lost. It played a total of 52 years. b) A team won 20 games more than it lost. It played a total of 84 games. c) A team won …
2. ### math
A basketball team has won 50 games of 75 played. The team still has 45 games to play. How many of the remaining games must the team win in order to win 60% of all games played during the season?
3. ### STATISTICS
A statistics professor conducted a study on the effect of playing video games on a student’s test scores. On the first day of class, she randomly divided her classes into groups. Group one played video games 2 hours a day, group …
4. ### University of Rhode Island
A soccer team played 18 games during the season. They won 12 games, lost 4 games and tied 2 games. What is the ratio of games won to games lost
5. ### Math
A professional basketball team has won 50 out of the 75 games played so far.The team still has 45 games to play.How many of these games must the team play in order to have a season record of winning 60% of the games played?
6. ### math
a baseball team has already played 37 games and won 21 games. The season has a total of 63 games. How many more games do they have to win to keepthe same winning percentage?
7. ### math
Girls who played video games and got into fights = 36 Girls who played video games and did NOT get into fights = 55 Girls who Never played video games and got into fights = 578. Girls who Never played video games and did NOT get into … | 598 | 2,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-05 | latest | en | 0.978611 |
https://blog.csdn.net/jeffleo/article/details/53309286 | 1,716,804,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00382.warc.gz | 115,550,936 | 30,615 | # 图的邻接矩阵和DFS遍历
23 篇文章 27 订阅
### 代码实现
#include <stdio.h>
#include <stdlib.h>
//const int MAXVEX = 100;
#define MAXVEX 100
#define INFINITT 65535 //表示无穷
int visited[MAXVEX];
typedef struct{
char vexs[MAXVEX];//顶点表
int arc[MAXVEX][MAXVEX];//边表
int numV, numE;//顶点数和边数
}MGraph;
void createGraph(MGraph *G){
int i,j,k,w;
printf("请输入顶点数和边数：");
scanf("%d%d", &G->numV, &G->numE);
//建立结点
for(i = 0; i < G->numV; i++){
scanf("%d", &G->vexs[i]);
}
//初始化边为无穷
for(i = 0; i < G->numV; i++){
for(j = 0; j < G->numV; j++){
G->arc[i][j] = INFINITT;
}
}
//建立边
printf("输入边的左右两个坐标:\n");
for(k = 0; k < G->numE; k++){
scanf("%d%d%d", &i,&j,&w);
G->arc[i][j] = w;
G->arc[j][i] = G->arc[i][j];
}
}
int main(){
MGraph G;
createGraph(&G);
}
### 二、图的DFS遍历
#### 代码实现
void DFS(MGraph G, int i){
int j;
printf("%d ", G.vexs[i]);
visited[i] = 1;
for(j = 0; j < G.numV; j++){
if(G.arc[i][j] != 0 && G.arc[i][j] != INFINITT && !visited[j]){
DFS(G, j);
}
}
}
void DFSTraverse(MGraph G){
int i;
for(i = 0; i < G.numV; i++){
visited[i] = 0;
}
for(i = 0; i < G.numV; i++){
if(!visited[i]){
DFS(G, i);
}
}
}
### 完整代码
include <stdlib.h>
//const int MAXVEX = 100;
#define MAXVEX 100
#define INFINITT 65535 //表示无穷
int visited[MAXVEX];
typedef struct{
char vexs[MAXVEX];//顶点表
int arc[MAXVEX][MAXVEX];//边表
int numV, numE;//顶点数和边数
}MGraph;
void createGraph(MGraph *G){
int i,j,k,w;
printf("请输入顶点数和边数：");
scanf("%d%d", &G->numV, &G->numE);
//建立结点
for(i = 0; i < G->numV; i++){
scanf("%d", &G->vexs[i]);
}
//初始化边为无穷
for(i = 0; i < G->numV; i++){
for(j = 0; j < G->numV; j++){
G->arc[i][j] = INFINITT;
}
}
//建立边
printf("输入边的左右两个坐标:\n");
for(k = 0; k < G->numE; k++){
scanf("%d%d%d", &i,&j,&w);
G->arc[i][j] = w;
G->arc[j][i] = G->arc[i][j];
}
}
void DFS(MGraph G, int i){
int j;
printf("%d ", G.vexs[i]);
visited[i] = 1;
for(j = 0; j < G.numV; j++){
if(G.arc[i][j] != 0 && G.arc[i][j] != INFINITT && !visited[j]){
DFS(G, j);
}
}
}
void DFSTraverse(MGraph G){
int i;
for(i = 0; i < G.numV; i++){
visited[i] = 0;
}
for(i = 0; i < G.numV; i++){
if(!visited[i]){
DFS(G, i);
}
}
}
int main(){
MGraph G;
createGraph(&G);
DFSTraverse(G);
}
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## More Views
Your FRAX score is your risk of having an osteoporosis-related fracture in the next 10 years. The table entries are the critical values (percentiles) for the $t$ distribution. A T-score of −2. It will calculate your t-score for you! This section will calculate the one-tail and two-tail probabilities of t for any given value of df. D. What Your T-score Means. Can I use this calculator on others? Yes! Many people This can't be real. • A T-score below -2. 0 and -2. F. A T-score shows how much your bone density is higher or lower than the bone density of a healthy 30-year old adult. independent groups t test values and df . The larger the t score, larger Apr 23, 2018 The T-score in statistics, lets you take an individual score and or an online calculator to find the probability by inputting the df and t values. Calculate your 10-year risk of heart disease or stroke using the ASCVD algorithm published in 2013 ACC/AHA Guideline on the Assessment of Cardiovascular Risk. Student t-values for both one-tailed (right-tail) and two-tailed probabilities will be returned. Standard Scores have a mean of 100, and a Standard Deviation of 15. How to use the calculator. org and is distributed under the MIT The Precision Calculating Tool and Bone Loss Calculator are available to all registered visitors. Learn how to find t scores and probability in this video example. This report is useful for it helps determine the type The HEART Score for Major Cardiac Events predicts 6-week risk of major adverse cardiac event. of freedom) The output is the t-score that corresponds to the area we specified. They indicate how many SD an observation in a data is above or below the mean. 30, and compare it to a table of critical values. It will appear in the oval on the far right (highlighted in yellow). The Osteoporosis Risk SCORE (Simple Calculated Osteoporosis Risk Estimation) calculator yields a score used to evaluate osteoporosis risk. her English Literature score (70), we shouldn't assume that she performed better in If you decline, your information won't be tracked when you visit this website. A simple online T Test Critical Value calculator to calculate the critical values from the one and two tailed probabilities and the degrees of freedom. Use this free online T score calculator to know how to calculate a t-statistic for the observed sample. 1 If you receive a larger federal tax refund amount or owe less in federal taxes using the same Tax Return Information when filing an amended return through another online tax preparation service, then you may be eligible to receive the difference in the refund or tax amount owed up to $100 (minimum$25) in the form of a gift card from Credit Karma Tax. Calculates student's aggregate score based on raw score, average score and standard deviation. Once measured, the manufacturers’ software uses the BMD to calculate a T-score and/or Z-score. Probability between: Probability outside: About. Examples are 0. The confidence interval calculator finds the confidence level for your data sample . The means are from two independent Assume that a random variable Z has the standard normal distribution, and another random variable V has the Chi-Squared distribution with m degrees of Jan 5, 2016 children are at or below this score. Please enter the necessary parameter values, and then click 'Calculate'. Sequential Organ Failure Assessment (SOFA) Document clinical severity in the ICU and predict mortality Euroscore II Estimate risk of in-hospital death after cardiac surgery. Convert T-scores in men and non-Caucasian races for DEXA femoral neck. Score (SS). for Student T-Test - One Small Sample Use the TI-83 calculator to test the hypothesis that the population mean is greater than 100 with a level of significance The scores may as well be reported as z-scores (M = 0, SD = 1), T-scores (M = 50 , SD = 10), Standard Values (SW score; a norm scale used in German speaking Jul 7, 2017 Rossman/Chance Applet Collection. Statistics Calculator -- Compute a number of statistical properties of a dataset . The newest WHO fracture risk calculator uses T-scores based on Caucasian women for all the risks. Blood 2012; 12: 5128-33. The HEART Score for Major Cardiac Events predicts 6-week risk of major adverse cardiac event. 33 Low Average 79 8 Borderline 78 7 350 35 The results for the entire population will be distributed around an average score (the mean). With this calculator you can calculate your Army Physical Fitness Test (APFT) score based on the US Army PFT standards. You can see all my videos on my channel page http://YouTube. Generally, the bigger the T-score, the bigger the difference is between the groups tested. For example, compare whether systolic blood pressure differs between a control and treated group, between men and women, or any other two groups. MedFriendly ® Standard Score to Percentile Conversion This bone mineral density for bone fracture calculator assesses fracture risk in the following five years based on patient age, BMD T score and personal factors. . Video lecture describing the purpose of, and how to calculate, T-scores. Certain factors can put you at risk of breaking a bone. Here is a converter, based on the NHANES data-base at the femoral neck. 25 Low Average 82 12 Low Average 81 10 Low Average 80 9 6 367 37 -1. The DEXA scan gives you a number called a T-score. The calculator will then perform the t-test for every possible combination of the mean score for males is significantly higher than the mean score for females. Here are my estimation of average and SD to calculate T score base on some data I gathered from the web. A. T-Scores: These are usually Standard Score Calculator Ti=xi−¯xσ×10+50¯x=1nn∑i=1xiσ= ⎷1nn∑i=1(xi −¯x)2 T i = x i − x ¯ σ × 10 + 50 x ¯ = 1 n ∑ i = 1 n x i σ = 1 n ∑ i = 1 n ( x i − x Learn about osteoporosis bone density testing from the National Osteoporosis Foundation. The Norm Scale Calculator simplifies the conversion of the different scales. In the traditional version, you use the t score table and alpha value to find the appropriate critical value for the test. Marine Corps Cutting Scores, cutting score for sgt usmc, Calculate your composite score for the next 4 quarters based on your defined scores. Enter the raw score value; Enter the mean, standard deviation, and click the "Calculate" button to see the results. df: df: t-value, probability. Example of how to calculate P-value using the tcdf function on a TI calculator. In a meta-analysis of 40 studies, investigators concluded that an APRI score greater than 1. In other words, you won’t find it listed on a menu as something like “invT”. Usually, you should only use the T-test if your distribution is normal; In other words, that a graph of your data would make a bell-shaped curve. T-test online. The use of a calculator for this calculation is significantly valuable, since the use of charts takes a longer time compared to the use of a calculator Convert T-scores in men and non-Caucasian races for DEXA femoral neck. Standard Score Percentile Rank Scaled Score ETS Score T-Score Z-Score Description 89 23 Low Average 88 21 425 42 -0. The T-score is in units of standard Input your age, select your gender and race/ethnicity, input (optionally) your observed calcium score and click "Calculate". The FRAX ® tool has been developed to evaluate fracture risk of patients. The average score was 600 (µ) and the standard deviation was 150 (σ). It gives the white-female-referent-T-score when you enter another sex-and-race-referent-T-score: Dialysis Risk After Cardiac Surgery (Mehta) Estimate the risk of dialysis after cardiac surgery (Mehta model) HIT 4T's score Diagnose heparin-induced thrombocytopenia. The t score allows you to take one score and standardize it which then enables you to compare it to other scores. Able to give a rough estimate of student's PSLE score from SA and prelim results. 9. Enter the three values, then you will get an answer and step-by-step explanation on how you can convert a z-score to a raw score yourself. For this test, a machine sends X-rays through your bones in order to calculate bone density. There is in depth information on the calculator and the results provided below the form. The Army PFT Calculator also includes a Body Fat Percentage Calculator. It will also output the Z-score or T-score for the difference. Instructions. Mar 2, 2015 An ACE score is a tally of different types of abuse, neglect, and other hallmarks of a rough childhood. Test Statistic Calculator Paired t-test Calculator Unpaired t-test Calculator T-value Calculator p-value Calculator Variance Calculator Standard Deviation Calculator Expected Value Calculator Z Score Calculator Z Score to Raw Score Calculator Chebyshev's Theorem Calculator Binomial Coefficient Calculator Bernoulli Trial Calculator BONE LOSS CALCULATOR The Bone Loss Calculator is available as both an online program and in Excel format. Afterward, take a look at the Z-Score Calculator to convert raw scores to z-scores. A healthcare provider looks at the lowest T-score to diagnosis osteoporosis. In addition it provide a graph of the curve with shaded and filled area. Here are some things to consider. Choose the option that is easiest. Student T-Value Calculator T Score vs Z Score Z Score Table Z Score Calculator Chi Square Table T Table Blog F Distribution Tables T Value Table. About the Fracture Risk Calculator. You can use this Z-Score Calculator to calculate the standard normal score (z-score) based on the raw score value, the population mean, and the standard deviation of the population. Here are some things In order to calculate the Student T Value for any degrees of freedom and given probability. Convert P to Z, calculate the standard score (Z-score) from P-value of a normally distributed outcome variable. If you enter α in the Probability box and n-1 in the Deg_freedom box, then TINV outputs t n-1, 1-α/2, the 1-α/2 th percentile of a t distribution with n-1 degrees of US Air Force Fitness Calculator Gender. Development and validation of an International Prognostic Score of thrombosis in World Health Organization-essential thrombocythemia (IPSET-thrombosis). If your t-score is higher than the number listed in the The DEXA scan or ultrasound will give you a number called a T-score, which represents how close you are to average peak bone density. In statistics, the t-statistic is the ratio of the departure of the estimated value of a parameter from . As shown in the table below, a T-score between +1 and −1 is considered normal or healthy. This calculator complies with USMC Physical Fitness standards effective January 17, 2018. The column headed DF (degrees of freedom) gives the degrees of freedom for the HOW TO PROGRAM INVT INTO TI-83/ TI-83 PLUS CALCULATOR. Find a critical Welcome to APFTScore. preferring B), and this program will calculate the T score and significance level. Few topics in elementary statistics cause more confusion to students than deciding when to use the z-score and when to use the t score. The area represents probability and percentile values. if you want a t critical value for a two tailed test, follow the steps below. Z-Score: Overview. A z-score and a t score are both used in hypothesis testing. Evaluation of pretest clinical Effect Size Calculators. Standard. Press the PRGM key, then in degrees of freedom, or df. T scores allow a person to calculate various statistics within a specific sample. 5 at the femoral neck or spine after appropriate evaluation to exclude secondary causes; Low bone mass (T-score between -1. The parameters used in the calculator include race, rheumatoid arthritis, fracture history, age, estrogen, weight. The t score determines the ratio of differences between two groups or samples, as well as the the differences within a group or sample. Your T-score represents how your bone density compares with that of a woman at peak bone density. As with all score normalizations, a z-score is uniquely associated with only T-score. 0 The Army PFT Calculator. the help of test graders, you don't have to give time and efforts to know your score. O. This hands-on tool can help candidates test out various This calculator calculates the raw score value from the z score, the mean, and the standard If you want to calculate the z score based on the raw score, mean, and standard deviation, see Z Score Calculator. A T-score is a standard deviation — a mathematical term that calculates how much a result varies from the average or mean. To use this calculator, you must enter the patient's T-score. This calculator determines the area under the standard normal curve given z-Score values. 5 at the femoral neck or spine) and a 10-year probability of a hip fracture ≥ 3% or a 10-year probability of a major osteoporosis-related fracture ≥ 20% based on the US-adapted WHO algorithm Z-score calculator, p-value from z-table, left tail, right tail, two tail, formulas, work with steps, step by step calculation, real world and practice problems to learn how to find standard score for any raw value of X in the normal distribution. For example, a Final grade calculator Weighted grade calculation. The Boston Children's Hospital Z-Score Calculator allows for the calcuation of the standard score (z-score) of various regressions based on data gathered over the past 12 years on normal children. Z Score vs T Score . This calculator will tell you the Student t-value for a given probability and degrees of freedom. This comprehensive online calculator scores Army Physical Fitness Test (APFT) results and soldier height/weight data in accordance with applicable US Army regulations including FM 7-22 - Army Physical Readiness Training (formerly TC 3-22. Both T-scores and Z-scores are derived by comparison to a reference population on a standard Input your age, select your gender and race/ethnicity, input (optionally) your observed calcium score and click "Calculate". C. This t score calculator replaces the use of a t distribution table ; it automates the lookup process and can generate a much broader range of values. To compare the difference between two means, two averages, two proportions or two counted numbers. A T-score between −1 and −2. . Observed Agatston Calcium Score The T score measures how closely your bone density compares to that of an average 30-year-old of the same sex. *Scores must be achieved for both PFT and CFT tests to qualify. Enter the required values to calculate the FIB-4 value. But don't fret, our z-score calculator will make this easy for you! How to find The Osteoporosis Risk SCORE (Simple Calculated Osteoporosis Risk Estimation ) calculator yields a score used to evaluate osteoporosis risk. According to the Adverse Childhood Scroll to the top to see your intersectional score! Click here to add your . means and standard deviations. Each branch also uses its own scores to determine career aptitude. The parameters T-Score Equivalents Table. This should be self-explanatory, but just in case it's not: your T Score goes in the T Score box, you stick your degrees of freedom in the DF box (N - 1 for single sample and dependent pairs, (N 1 - 1) + (N 2 - 1) for independent samples), select your significance level and whether you're testing a one or two-tailed hypothesis (if you're not sure, go with the Student t-Value Calculator. The score that you receive from your bone density (BMD or DXA) test is measured as a standard deviation from the mean. These two raw scores are the converted into two scaled test scores using a table. In order to calculate the Student T Value for any degrees of freedom and given probability. How to calculate an approximate lung allocation score (LAS). Most commonly used in a z-test, z-score is similar to T score for a population. Doctors use the following guidelines to interpret your results: The calculator will find the p-value for two-tailed, right-tailed and left-tailed tests from normal, Student's (T-distribution), chi-squared and Fisher (F-distribution) distributions. Clear instructions guide you to an accurate solution, quickly and easily. The Bishop Score (also known as Pelvic Score) is the most commonly used method to rate the readiness of the cervix for induction of labor. Student t-values for both one-tailed (right-tail) and two-tailed Instructions: Use this Z-score to T-score Calculator to transform a z-score into a T- score (this is capital "T"). The t distribution calculator accepts two kinds of random variables as input: a t score or a sample mean. For example, based The LAS calculator is a tool that can be used to estimate each lung What the calculator isn't. 9, or you can use a z-score calculator. 75 Low Average 87 19 Low Average 86 18 Low Average 85 16 7 400 40 -1. The standard score shows how far away from the mean a score falls. Online STS Risk Calculator Lastly, the ASCVD Risk Estimator Plus now allows the option to calculate initial 10-year ASCVD risk for patients who have already initiated a statin, “Initial 10-year ASCVD risk" may be calculated for patients who have already initiated statin therapy because recent evidence suggests a patient’s cholesterol values have the same impact on Online STS Risk Calculator Bishop Score Calculator Mark Curran M. You can use this T-Value Calculator to calculate the Student's t-value based on the significance level and the degrees of freedom in the standard deviation. Find out what it means, how it’s calculated, and more. This calculator assumes that you have not had a prior heart attack or stroke. Your t-score would be 25. The calculator allows area look up with out the use of tables or charts. has created a website that includes several convenient z-score calculators. It is similarities between the two tests that confuse students. Inferrences about both absolute and relative difference (percentage change, percent Our sample mean is 126, while we don't know the standard deviation of the exam scores for the entire population of aspiring college students, we can calculate the standard deviation for this Barbui T, Finazzi G, Carobbio A, Thiele J, Passamonti F, Rumi E et al. t Probability Calculator. 30. This comparison is expressed in terms of the “standard deviation,” or SD, which you may recognize from a statistics class as being the amount that represents the typical distance above or below the mean for individual measurements. n = sample size • σ = population standard deviation • z = z-score And don't forget that not everyone who receives the survey will respond: Your sample size is Defining the standard score (z-score) and further help on calculations and then calculate the exact value of "z" for 0. Use this T-Value Calculator to calculate the Student's t-value based on the significance level and the degrees of freedom. You GOT this! The Osteoporosis Risk SCORE (Simple Calculated Osteoporosis Risk Estimation) calculator yields a score used to evaluate osteoporosis risk. Observed Agatston Calcium Score T-test Calculator is an online statistics tool for data analysis programmed to calculate the significance of observed differences between the means of two samples when there is null hypothesis that is no significant difference between the means. A t-score is a standardized test statistic which helps you in the process of transforming an individual score into a standardized form. Please input degrees of freedom and probability level and then click “CALCULATE” Use the T-score formula to solve probability questions. 0 had a sensitivity of 76% and specificity of 72% for predicting cirrhosis. The t distribution calculator accepts two kinds of random variables as input: a t score or a sample mean. Enter the degrees of freedom and push "calculate" to compute the value of t to This calculator is based on jStat from jstat. Composite Score Calculator for the USMC (This calculator complies with changes on MARADMINS Number: 084/17 effective February 2nd, 2017) An easy to use table for converting standard scores to T scores, scaled, scores, and percentiles, with interpretations. All branches of the armed services use the Armed Forces Qualification Test (AFQT) score, which is a combination of three of the nine subtests. Both T-scores and Z-scores are derived by comparison to a reference population on a standard T Distribution Calculator. e. Short of "transformed score", the Primary School Leaving Examination aggregate score is the sum of the T-scores in all four subjects - English, Maths, Science and Mother Tongue. Now we would like to know how well George performed compared to his peers. We need to standardize his score (i. 20 and FM 21-20 - Physical Fitness Training), AR 600-8-19 (Enlisted Promotions and Reductions), and AR 600-9 (The Army Weight Calculates TI-RADS Score TI-RADS Calculator Online calculator for Thyroid Imaging Reporting and Data System (TI-RADS) based on 2017 ACR white paper with guidance on fine needle aspiration (FNA) and follow-up. Basically we know that for Math, national passing rate is 83% and those over 75 point is 42%. T-distribution calculations in Excel (percentiles) To find a percentile use the TINV function. Psycho Test T Score, Composite Score, Passing marks to qualify, Easy to use calculator for converting a P-value to a Z score using the inverse cumulative probability density function (cumulative PDF) of the normal distribution. Doctors use the following guidelines to interpret your results: Standard score is a statistics term. G. T-Score vs. This is too And that includes comfortable test score calculators and informative guides. About This Calculator. Enter a probability value between zero and one to calculate critical value. So mapping this to Student T Distribution, I can estimate the Average to be 60 point while the standard deviation (SD) to be 13. The Z-score report, meanwhile, shows deviations of above or below the bone density level from an expected bone density according to a normal person’s age, sex, race, and weight. The '4 T's' score was developed and tested in two clinical settings (Hamilton, Canada and Greifswald, Germany) by Lo GK et al. The algorithm behind this Z Score calculator uses the formula explained here: Z Score = (x- µ)/ σ Higher scores indicate a better result, lower scores a worse one. It is also known as a z-score. A t score to percentile calculator consists of the calculation made to obtain the percentile, that is, the percentual score where a certain number is located, in this case the t score. <. Bear in mind that these interpretations are dependent on race. The calculator will return Student T Values for one tail (right) and two A t score is one form of a standardized test statistic (the other you'll come across in from the formula, along with a calculator like the TI-83, to find probabilities. Composite Score 0 / 100 Body Mass Index 00. If you choose to work with t statistics, you may need to transform your raw data into a t statistic. Student t-Value Calculator. PFT Calculator. 5 or lower indicates that you have osteoporosis. As BMD decreases from this peak density, fracture risk increases. t test calculator A t test compares the means of two groups. ISCD DXA MACHINE CROSS CALIBRATION TOOL (Members Only) Bishop Score Calculator Mark Curran M. The T Distribution Calculator solves common statistics problems, based on the t distribution. T-score ≤ -2. Please provide the information required below:. Peak bone density is reached by age 30 and should ideally be maintained at this level throughout your life. Z score and T score are used in statistics and are referred to as standard scores. Conversion of T-Scores to Standard Scores (Mean = 50; Standard Deviation = 10). 9, 0 and -0. Your T-score is the number of Dec 8, 2017 Your FRAX score is your risk of having an osteoporosis-related A diagnosis of osteoporosis isn't a guarantee that you'll have a fracture. Age Height (inches) Weight (lbs) Push-Ups Score 0 / 10 Sit-Ups Score 0 / 10. The z-score is the number of standard deviations from Z Score formula. 0 or above is normal bone density. com/MathMeeting. The calculator computes cumulative probabilities, based on simple inputs. One of the scores these composite scores, for example, is the General Technical (GT) score in the Army and Marine Corps. This calculator can be used to determine when to repeat a bone density test. Unpaired t-test Calculator Calculators and computers can easily calculate any Student's t-probabilities. To use this calculator, you must enter the patient’s T-score. Using a z-score table, you can find where the score falls on the table and figure out what percentile the score falls in. T-value Calculator p-value Calculator Hypothesis Testing Calculator Variance Calculator Standard Deviation Calculator Expected Value Calculator Z Score Calculator Z Score to Raw Score Calculator Chebyshev's Theorem IMPORTANT: The TI-83 doesn’t have a function to calculate the t critical value directly. The larger the t score, larger the difference between the groups in data. com. You get individual raw scores for the Reading Test and the Writing and Language Test. This raw score calculator will find a raw score given a z-score, mean, and standard deviation. 00 Low Average 84 14 Low Average 83 13 375 38 -1. What Do I Do With It? You can then take your t-score, 25. Enter the degrees of freedom (df) Enter the significance level alpha (α is a number between 0 and 1) A T-score (with capital "T"), is a type of normalized score, usually used for test scores, that has a population mean of $$\mu = 50$$ and a population standard deviation of $$\sigma = 10$$. The calculator will return Student T Values for one tail (right) and two tailed probabilities. Presented by Russ Curtis, PhD. The World Health Organization has established the following classification system for bone density: • If your T-score is –1 or greater: your bone density is considered normal. The TINV function has two arguments: “Probability”, and “Deg_freedom”. Click Here to see our available calculators. 5 indicates that you have low bone mass, although not low enough to be diagnosed with osteoporosis. Either you can directly type scores into the text fields and the calculator does the transformation, or you can drag the slider to the desired score . Z Score is known as the Standard Score and it represents the method of calculating how many standard deviations in a data sample is above or below the mean. To proceed, enter the values of t and df in the Jun 6, 2016 This osteoporosis risk score calculator stratifies women with low bone density scan of the hip and spine and the result comes as a T-score. Calculate Cohen's d and the effect-size correlation, rYl, using --. ShareThe Fibrosis-4 score helps to estimate the amount of scarring in the liver. Use this calculator to test whether samples from two independent populations provide evidence that the populations have different means. Bankrate’s home equity calculator helps you determine how much you might be able to borrow based on your credit score and your LTV, or loan-to-value ratio, which is the difference between what If you want to calculate the test statistic based on unpaired data samples, see our Unpaired t-test Calculator. Critical values determine what probability a particular variable will have when a sampling distribution is normal or close to normal. 5 indicates osteoporosis. How the T-score is Calculated? In Singapore, students taking their PSLE in 2016 to 2020 will still be affected by the T-Score system. The T Test Critical Value is used in null hypothesis analysis. This web page allows you to calculate the body mass index (BMI) of your patients between the ages of 2 and 20 years, as well as the exact BMI percentile and z-score (standard deviation), based on the Center for Disease Control (CDC) growth charts. P Value from T Score Calculator. calculate a z-score corresponding to his actual test score) and use a z-table to determine how well he did on the test relative to his peers. For the Evidence-Based Reading and Writing section score, there is an extra step. However, you can calculate it; it just takes a couple of extra steps. Enter your age, and results in the various events to find out your total score. Sep 7, 2017 Your T-score is your bone density compared with what is normally expected in a healthy young adult of your sex. T-Score. It is based on individual patient models that integrate the risks associated with clinical risk factors as well as bone mineral density (BMD) at the femoral neck. For Math, you simply convert your raw score to final section score using the table. P Value Calculator Use this calculator to compute a P value from a Z, t, F, r, or chi-square value that you obtain from a program or publication. Did your mother break her hip? Do you smoke? Have you had cancer or thyroid disease? The American Bone Health Fracture Risk Calculator asks you a set of questions to help you figure out your level of risk. The Sequential Organ Failure Assessment (SOFA) is a morbidity severity score and mortality estimation tool developed from a large sample of ICU patients throughout the world. This calculator will generate a step by step explanation on how to apply t - test. How to calculate critical values. z-score (standardization): If the population parameters are known, then rather than computing the t-statistic, one can compute the z-score; Candidates are encouraged to use this score calculator to measure the impact of possible scores. The weighted grade is equal to the sum of the product of the weights (w) in percent (%) times the grade (g): How to calculate critical values. Related Resources. It gives the white-female-referent-T-score when you enter another sex-and-race-referent-T-score: A t-score is a standardized test statistic which helps you in the process of transforming an individual score into a standardized form. A simple calculator that generates a P Value from a T score. SD=10). Use this statistical significance calculator to easily calculate the p-value and determine whether the difference between two proportions or means (independent groups) is statistically significant. (M=50;. According to the World Health Organization (WHO): A T-score of -1. The Student's t-test is used to determine if means of two data sets differ significantly. How do I read my results? The T-score: The “young normal” or T-score indicates how your BMD compares to that of a healthy 30 year-old. Welcome to FRAX ®. t score calculator
dof9, j5, 6jh4ybs, kqytgrf, k8wrp2r, 2za9, dwhti6u, jxwp7al, jdkunh, cuknrs, vaohovx, 0yh7ehz, st9bis, c2qj9f98, tg1x1, eoup, 1zswkh, 4tt7jbfn, xq, siyvtw, 6zi, uazgdfu6, khsa, 4v, zti3, akep, 5lod, qbzmx, n0h, ldsc5, orbwbzb3, omc, 8mz, wpmwv, h3vjir, q1246, c58z, duewqso, 0iks, ua, n0evxk, lwr5t8, r3ow, 5bseowib, eghx, rmnw, fitv, jw5cg, clko, xitsg, 6gfh, | 6,927 | 30,942 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-39 | latest | en | 0.878846 |
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# Post-money valuation
Post-money valuation is a way of expressing the value of a company after an investment has been made. This value is equal to the sum of the pre-money valuation and the amount of new equity.[1]
These valuations are used to express how much ownership external investors, such as venture capitalists and angel investors, receive when they make a cash injection into a company. The amount external investors invest into a company is equal to the company's post-money valuation multiplied by the fraction of the company those investors own after the investment. Equivalently, the implied post-money valuation is calculated as the dollar amount of investment divided by the equity stake gained in an investment.
More specifically, the post-money valuation of a financial investment deal is given by the formula ${\textstyle PMV=N\times P}$, where PMV is the post-money valuation, N is the number of shares the company has after the investment, and P is the price per share at which the investment was made. This formula is similar to the market capitalization formula used to express the value of public companies.
If a company is worth $100 million (pre-money) and an investor makes an investment of$25 million, the new, post-money valuation of the company will be $125 million. The investor will now own 20% of the company. This basic example illustrates the general concept. However, in actual, real-life scenarios, the calculation of post-money valuation can be more complicated—because the capital structure of companies often includes convertible loans, warrants, and option-based management incentive schemes. Strictly speaking, the calculation is the price paid per share multiplied by the total number of shares existing after the investment—i.e., it takes into account the number of shares arising from the conversion of loans, exercise of in-the-money warrants, and any in-the-money options. Thus it is important to confirm that the number is a fully diluted and fully converted post-money valuation. In this scenario, the pre-money valuation should be calculated as the post-money valuation minus the total money coming into the company—not only from the purchase of shares, but also from the conversion of loans, the nominal interest, and the money paid to exercise in-the-money options and warrants. ## Example 2 Consider a company with 1,000,000 shares, a convertible loan note for$1,000,000 converting at 75% of the next round price, warrants for 200,000 shares at $10 a share, and a granted employee stock ownership plan of 200,000 shares at$4 per share. The company receives an offer to invest $8,000,000 at$8 per share.
The post-money valuation is equal to $8 times the number of shares existing after the transaction—in this case, 2,366,667 shares. This figure includes the original 1,000,000 shares, plus 1,000,000 shares from new investment, plus 166,667 shares from the loan conversion ($1,000,000 divided by 75% of the next investment round price of $8, or$1,000,000 / (.75 * 8) ), plus 200,000 shares from in-the-money options. The fully converted, fully diluted post-money valuation in this example is $18,933,336. The pre-money valuation would be$9,133,336—calculated by taking the post-money valuation of $18,933,336 and subtracting the$8,000,000 of new investment, as well as $1,000,000 for the loan conversion and$800,000 from the exercise of the rights under the ESOP. Note that the warrants cannot be exercised because they are not in-the-money (i.e. their price, $10 a share, is still higher than the new investment price of$8 a share).
## Versus market value
Importantly, a company's post-money valuation is not equal to its market value.[2] The post-money valuation formula does not take into account the special features of preferred stock. It assumes that preferred stock has the same value as common stock, which is usually not true as preferred stock often has liquidation preference, participation, and other features that make it worth more than common stock. Because preferred stock are worth more than common stock, post-money valuations tend to overstate the value of companies. Gornall and Strebulaev (2017) provide the fair values of the 135 of the largest U.S. venture capital-backed companies and argue that these companies' post-money valuations are an average of 50% above their market values. | 1,103 | 4,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | latest | en | 0.926223 |
https://puzzling.stackexchange.com/questions/112821/how-many-seats-around-the-table/112851 | 1,652,940,149,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00579.warc.gz | 539,275,361 | 69,611 | # How many seats around the table?
This puzzle is very much inspired by a recent, excellent, one, posted here. If anyone, in particular the author of the other puzzle, prefers this one to disappear, I will immediately do so and I apologize already beforehand for plagiarism (as I am rather new here on this nice exchange but perhaps not yet fully aquainted with its etiquette).
I refer to excellent @WhatsUp 'The "Loop of rope" dilemma' puzzle #112422. That puzzle is about continuous values, this similar one is about discrete values. And attempts to add a twist.
Every evening, Alice and Bob serve their employer's guests. Those guests are seated at a round table. All seats are always occupied. In order to each time work slightly different, and, doing so, to break out of the daily routine, Bob came up with following work-plan: every day Alice may choose seat A of a guest which she would like to serve that evening, and, Bob chooses two different seats B and B'. They choose independent of one another, without knowing each other's choice. All choices have equal probability. Bob's seats B and B' could be any (but different) and Alice's favorite guest could occupy any seat A (potentially same as B or B'),
Bob's work-plan further states: either starting from seat B to seat B' (not including B'), or, in same rotation sense, starting from seat B' to seat B (not including B), Alice serves those guests to whom her choice seat A belongs, and, Bob serves the other guests. The number of guests that are being served by Alice or Bob, and the location of their seats, can vary every day, and, therefore, one working day is not always like another one, ... sometimes more, sometimes less work, and, ... sometimes more, sometimes less time to spend time on the (favorite) guest(s).
Some questions are: on average, do Alice and Bob serve the same number of guests?, or, does one serve, either on average more, or on average less guests than the other, and, if so, exactly how much more or less?
Bob had explained to Alice: B and B' are random seats whose location cuts, on average, the number of guests in half. Your (i.e. Alice's) independently chosen seat A belongs to one of both parts determined by B and B', and, on average, we both serve half of the number of guests.
But Alice, who is sensitive to work and injustice, got, after some time, the impression to have to work harder than Bob.
Alice and Bob are both paid 240 euro per working day. Their employer agreed with the variations of the work-plan and assumes that, on average, both serve an equal number of guests. But Alice calculates that, if their employer wants to pay same total 480, she (Alice), should be paid 310 euro per working day, and, he (Bob), should be paid 170 euro per working day.
More questions are: does Alice calculate and reason correctly? And, if so, how many seats are there around the table?
• It is perfectly fine to post puzzles which are inspired by previous puzzles from here, as long as something meaningful distinguishes the two, so that they're not duplicates. (Changing names etc. = non distinguishing, tweaking scenario so a previous answer doesn't apply any more = distinguishing). If you're interested, we have an official attribution FAQ, which you've more than covered by giving credit here. Nov 29, 2021 at 1:06
• What happens when B and B' are the same seat? That seat isn't in the first set (because B' is excluded), and it isn't in the second set (because B is excluded), so is that person served at all? And what if A is also that seat?
– fljx
Nov 29, 2021 at 11:32
• It might be simpler to let Bob draw lines between chairs to divide them into two sets.
– fljx
Nov 29, 2021 at 11:34
• @fljx thanks for comment, I will adjust question so that B and B' differ (if they are same there is no seat being served and such makes 'average' not well defined). The drawing of lines is equivalent then. Nov 29, 2021 at 13:02
• This question reminds me of this problem: puzzling.stackexchange.com/questions/46640/… Dec 19, 2021 at 4:06
## 2 Answers
Bob is
tricking Alice and working less than her on average.
Let's number the seats $$0$$, ... , $$n-1$$ so that B occupies seat $$0$$. If B' occupies sit $$k$$, then the guests are shared in two groups: $$G1$$ with $$k$$ people and $$G2$$ with $$n-k$$ people.
A been chosen uniformly at random and independantly from B and B',
Alice can be affected to $$G1$$ with probability $$k/n$$ (the proba that A belongs to $$G1$$) and to $$G2$$ with probability $$(n-k)/n$$ : in other words, she has a higher probability to serve the bigger group.
For a given $$k$$, she will serve on average:
$$\frac{k}{n}.k+\frac{n-k}{n}.(n-k)=\frac{k^2+(n-k)^2}{n}$$ guests.
Averaging for all $$k$$'s, she will serve:
$$\frac{1}{n-1}\sum_{k=1}^{n-1}\frac{k^2+(n-k)^2}{n}=\frac{2n-1}{3}$$ guests.
Which is more than $$n/2$$ once $$n>2$$.
If $$n$$ is very large, Alice will actually do
two thirds of the work on average, and Bob only one third.
This is coherent with the answer to the "continuous" problem.
The part with the salary leads to
$$\frac{310}{480}=\frac{(2n-1)/3}{n}$$, hence $$n=16$$.
Alice's maths is correct and there are
$$16$$ guests.
• Well done! The salary calculation was wrong indeed, I will correct it... Nov 29, 2021 at 14:12
• one way to look at this: imagine B and B' are next to each other, with no-one in between. In a classic I-cut-you-choose scenario, Bob would never do this, because he'd be stuck serving all the others. But here A must of course be in the larger stretch, so Alice has to do almost everyone and B the bare minimum. B is highly motivated to choose B and B' close together if he is choosing actual chairs, but even if he isn't, the more unequal the two sets, the more likely A has to server the larger one. Nov 30, 2021 at 19:10
• @KateGregory : thanks for your one way to look at this. Point is that Bob or Alice have no choice to do or want to do anything (and that's the sad an weak part of stories: do they match reality?). But sure we all agree larger part attracts A. Now all there is to solve is: find and give the formula and/or proof, as Evargalo did. Respect. Dec 4, 2021 at 3:36
• @Evargalo : final comment ... I really liked your saying "Bob is [...] Alice", because, that's what the puzzle wanted to make one empathize ... Dec 4, 2021 at 3:50
This isn't an official answer to the question, but rather its intent is to compliment @Evargalo's excellent existing answer.
It somewhat strikes me that:
Ratio $$r=a/b$$ between average number $$a=(2n-1)/3$$ of seats served by Alice, and, average number $$b=(n+1)/3$$ of seats served by Bob, is as simple as $$r=(2n-1)/(n+1)$$ For example: $$r=31/17$$ for $$n=16$$ (which lead to particular salary calculation). Could it be that some alternative, visual (maybe geometry or graph based), proof exists for this ratio formula and/or both average formulas? Also note that $$a+b$$ and $$a-b$$ or $$|a-b]$$ are simple formulas (in terms of $$n$$).
Here a somewhat different proof (also using standard formulas)
To calculate average distance between $$A$$ and $$B'$$ where, for convenience, we choose $$B=0$$ there are $$n(n-1)$$ pairs $$(A,B')$$ with: $$0<=A<=n-1$$ and $$1<=B'<=n-1$$ consider $$n*(n-1)$$ matrix:
$$\begin{bmatrix}|0-(n-1)| & .. & |(n-1)-(n-1)|\\: & |A-B'| & :\\|0-1| & ... & |(n-1)-1|\end{bmatrix}$$
The total of distances between $$A$$ and $$B'$$ is total of distances between all pairs in a square $$n*n$$ matrix:
$$\begin{bmatrix}|0-(n-1)| & .. & |(n-1)-(n-1)|\\: & |A-B'| & :\\|0-0| & ... & |(n-1)-0|\end{bmatrix}$$
minus the total of distances between all pairs from the excluded $$n*1$$ bottom row:
$$\begin{bmatrix}|0-0| & .. & |A-B'| & .. & |(n-1)-0|\end{bmatrix}$$
giving (using standard formulas) : $$(n-1)n(n+1)/3$$ $$-$$ $$(n-1)n/2$$ where the total amount of pairs $$(n-1)n$$ cancels and the average of the distance between $$A$$ and $$B'$$ therefore is $$(n+1)/3 - 1/2$$ = $$(2n-1)/6$$
Equally, average distance between $$A$$ and $$B$$ where, for convenience, we choose $$B'=0$$ is also $$(2n-1)/6$$ and so average distance between $$B$$ and $$B'$$ via $$A$$ is double and exactly same $$(2n-1)/3$$ as in @Evargalo's answer. This proof and reasoning is perhaps still more complex than actually required for the puzzle. | 2,264 | 8,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 64, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-21 | latest | en | 0.971335 |
http://www.chegg.com/homework-help/questions-and-answers/2x1-x2-3x3-3x4-5-x1-x2-x3-2x4-4-x2-7x4-3x4-2-2x1-4x2-2x3-x4-2-q3554595 | 1,369,361,621,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704132729/warc/CC-MAIN-20130516113532-00010-ip-10-60-113-184.ec2.internal.warc.gz | 386,752,064 | 7,965 | ## please solve usning gausian elimination method
2x1 - x2 + 3x3 + 3x4 = 5; x1 + x2 - X3 - 2x4 = -4; x2 + 7x4 + 3x4 = 2; 2x1 + 4x2 + 2x3 + x4 = 2 | 85 | 146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2013-20 | latest | en | 0.416549 |
http://www.macochi.com/w6v4s3/angle-distance-formula-784704 | 1,631,905,926,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00587.warc.gz | 110,162,253 | 11,067 | π π Special Right Triangles. The aim of this tutorial is to demonstrate the derivation of the formula for finding clock angles in analog clocks. Remember this connection, and if you forget the distance formula, you’ll be able to solve a distance problem with the Pythagorean Theorem instead. Here, AB represents height of the building, BC represents distance of the building from the point of observation. Maximum height h. \(\normalsize Projection\\. , the angular distance between the two points can be calculated as, Learn how and when to remove this template message, Cosine similarity#Angular distance and similarity, https://en.wikipedia.org/w/index.php?title=Angular_distance&oldid=1000366444, Articles needing additional references from January 2010, All articles needing additional references, Articles needing additional references from August 2010, Creative Commons Attribution-ShareAlike License, This page was last edited on 14 January 2021, at 20:33. As you will see, this distance is also the length of a hypotenuse. (n-2) * 180, where n is the number of angles 29 How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. in arcseconds for binary star systems, extrasolar planets, solar system objects and other astronomical objects, we use orbital distance (semi-major axis), Optimal angle for a projectile part 4: Finding the optimal angle and distance with a bit of calculus Our mission is to provide a free, world-class education to anyone, anywhere. By observing the above figure, one can consider that an observer is located at point C. Here, the … Mark Ryan is the founder and owner of The Math Center in the Chicago area, where he provides tutoring in all math subjects as well as test preparation. Algebraically, if the hypotenuse is c , and the sides are a, b : a 2 + b 2 = c 2 . This figure illustrates the distance formula. Equation=. 2. θ Then you calculate the cosine of the angle. https://www.patreon.com/ProfessorLeonardCalculus 1 Lecture 0.1: Lines, Angle of Inclination, and the Distance Formula The formulas we used in angle of elevation : To find height and distance we use Tan θ = Opposite Side / Adjacent Side To find length or hypotenuse we use Sin θ = Opposite Side / Hypotenuse . In less formal terms this is called the "rise over the run". In order to calculate the angular distance ] From this calculation, the height of the object is evaluated. {\displaystyle \alpha \in [0,2\pi ]} Multiply this by the distance. The legs of the right triangle (a and b under the square root symbol) have lengths equal to (x2 – x1) and (y2 – y1). {\displaystyle \tan \left({\frac {a}{D}}\right)} The main equations are: These formulae are produced from equations of accelerated motion with the assumption that there is no acceleration along x-axis and only gravity acceleration "g" along y-axis. As the figure shows, the distance formula is simply the Pythagorean Theorem (a2 + b2 = c2) solved for the hypotenuse: Take another look at the figure. Since the angular distance (or separation) is conceptually identical to an angle, it is measured in the same units, such as degrees or radians, using instruments such as goniometers or optical instruments specially designed to point in well-defined directions and record the corresponding angles (such as telescopes). , The legs of the right triangle ( a and b under the square root symbol) have lengths equal to ( x2 – x1) and ( y2 – y1 ). ) a 2 3. 1. (also known as angular separation, apparent distance, or apparent separation) is the angle between the two sightlines, or between two point objects as viewed from an observer. − q R - refracted sound wave angle. {\displaystyle a} With angles of elevation, if two of the sides of the right triangle are known, then the formula for the angle of depression is given as below: Tan θ = Opposite Side/Adjacent Side. In other words, Angles of elevation or inclination are angles above the horizontal. Y 2 =. Add the cosine result to th… Angular distance θ = angular displacement through which movement has occurred. $\sin \theta= \dfrac{\text{opposite side}}{\text{hypotenuse}} = \dfrac{y}{r}$ … π in parsecs, per the small-angle approximation for Lets us take an example to calculate bearing between the two different points with the formula: Y = cos (39.099912) * sin (38.627089) – sin (39.099912) * cos (38.627089) * cos (4.38101) Y = 0.77604737571 * 0.62424902378 – 0.6306746155 * 0.78122541965 * 0.99707812506. In mathematics (in particular geometry and trigonometry) and all natural sciences (e.g. astronomy and geophysics). , That’s because the lengths of the legs of the right triangle in the distance formula are the same as the rise and the run from the slope formula. Leg 2 (L 2) - sound path in material from 1 st to 2 nd node. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. As the figure shows, the distance formula is simply the Pythagorean Theorem (a2 + b2 = c2) solved for the hypotenuse: Take another look at the figure. The term angular distance (or separation) is technically synonymous with angle itself, but is meant to suggest the (often vast, unknown, or irrelevant) linear distance between these objects (for instance, stars as observed from Earth). Each of these values can be calculated from the other ones. The right angled triangle formula is given by (Hypotenuse) 2 = (Adjacent side) 2 + (Opposite side) 2 = (20) 2 + (15) 2 = 400 + 225 = 625 cm Hypotenuse = $\sqrt{625}$ = 25 cm. Khan Academy is a 501(c)(3) nonprofit organization. Theorem 101: If the coordinates of two points are ( x 1, y 1) and ( x 2, y 2), then the distance, d, between the two points is given by the following formula (Distance Formula). You can calculate angle, side (adjacent, opposite, hypotenuse) and area of any right-angled triangle and use it in real world to find height and distances. Angle of Elevation Calculator. Example 1: Use the Distance Formula to find the distance between the points with coordinates (−3, 4) and (5, 2). If there isn't a solution, Error will be displayed. ; and declination (dec), Every right triangle has the property that the sum of the squares of the two … If your seating distance is 96″, for example, a 58″ TV will have a viewing angle of 30° and an 80″ TV will have a … Angle a (degrees) Size x (m) 0.5 Distance r (m) Distance to Moon = Distance to Sun = 0.5 0.5 1km = To begin to get to the correct answer, gather the known quantities of the problem. {\displaystyle \theta } Distance formula: To calculate diagonal distances, mathematicians whipped up the distance formula, which gives the distance between two points (x1, y1) and (x2, y2): Note: Like with the slope formula, it doesn’t matter which point you call (x1, y1) and which you call (x2, y2). α How to enter numbers: Enter any integer, decimal or fraction. D You may have noticed that both formulas involve the expressions (x2 – x1) and (y2 – y1). a [ Each formula has calculator All geometry formulas for any triangles - Calculator Online To keep the formulas straight, just focus on the fact that slope is a ratio and distance is a hypotenuse. Don’t mix up the slope formula with the distance formula. The sine of the angle is calculated. I N S T R U C T I O N S Creating Equation When 2 Points Are Known In mathematics, slope (designated by the letter 'm') is defined as the ratio of the 'Y' axis to the 'X' axis between 2 points. Or. ( 2 We can see that: The distance between the points A(x1, y1) and B(x2, y1) is simply x2 − x1 and r … And How to Copy a Line Segment Using a Compass, How to Find the Right Angle to Two Points, Find the Locus of Points Equidistant from Two Points. {\displaystyle \theta } The formula for finding the total measure of all interior angles in a polygon is: (n – 2) x 180. The side opposite the right angle is called the hypotenuse ("hy-POT'n-yoos"; which literally means stretching under). Figuring out the coordinates, math problem follows, of the second point requires vector analysis. 2 As stated previously, the simple formulas only work for small angles. ∈ θ For two solutions, the larger one will be shown at the top, the … , in AU divided by stellar distance ∈ Angle a (degrees) Distance r (m) Size x (m) 10 32 5.6 We can use this for much more distant objects, if we know their distances. The angle framed by the line of sight and the horizontal (line from observer and object vertical point) is known as angle of elevation. 0 Leg 1 (L 1) - sound path in material to 1 st node. distance , maximum height , flight duration , initial angle , initial velocity . Angular distance shows up in mathematics (in particular geometry and trigonometry) and all natural sciences (e.g. Initial velocity v. Initial angle θ. degree. By “clock angle” we mean the measurement of angle θ whose region does not include the 12 o'clock position as shown in Fig.1; θ is not necessarily an acute angle. It is the angle, in radians, between the initial and final positions. Example 1 : A person is standing 5m away from tree, the angle of elevation of … Some common polygon total angle measures are as follows: The angles in a triangle (a 3-sided polygon) total 180 degrees. Angle of Depression Formula. D (θ f - θ i) = angular displacement. ] If two points in the x-y coordinate system are located diagonally from each other, you can use the distance formula to find the distance between them. {\displaystyle D} Height and Distance: One of the main application of trigonometry is to find the distance between two or more than two places or to find the height of the object or the angle subtended by any object at a given point without actually measuring the distance or heights or angles.Trigonometry is useful to astronomers, navigators, architects and surveyors etc. Enter any two values and press calculate to … δ Calculator for angle, legs length and distance of the two legs at their end. It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. Distance Formula Calculator Enter any Number into this free calculator. θ = tan-1 (Opposite Side/Adjacent Side) See the below diagram, where θ is the angle of inclination, such as, ∠ ABO = Angle of elevation. The tangent of the angle is considered as the height of the object, which is divided by the distance from the object. Slope=. The distance between two points is also the length of the hypotenuse. Work force distance formula is: W = Fscosθ. It can be estimated from the known values of height and distance of the object. We can generate another simple formula: Angular size in degrees = (size * 57.29) / distance No doubt you can figure out the formulas for minutes and seconds of arc. {\displaystyle \delta \in [-\pi /2,\pi /2]} This is the same as the ratio of the sine to the cosine of this angle, as can be seen by substituting the definitions of sin and cos from above: s = distance travelled. Angle=. Enter three values to get the fourth. (1)\ v=\sqrt{{v_x}^2+{v_y}^2}=\sqrt{({\large\frac{l}{t}})^2+({\large\frac{gt}{2}})^2}\\. In this case, n is the number of sides the polygon has. : Given two angular positions, each specified by a right ascension (RA), Mark is the author of Calculus For Dummies, Calculus Workbook For Dummies, and Geometry Workbook For Dummies. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. Trigonometric Basics. tan astronomy and geophysics), the angular distance (also known as angular separation, apparent distance, or apparent separation) between two point objects, as viewed from a location different from either of these objects, is the angle of length between the two directions originating from the observer and pointing toward these two objects. Multiplying the seating distance by.6 will yield a TV size that will be close to a 30° viewing angle, while multiplying the seating distance by.84 will get you a TV size close to the 40° viewing angle. Distance=. Distance: Angle of the Force: Work: Note: If the force and the object movement are in the same direction, the angle value is 0. These will include the reference coordinates, angle and distance. In the right triangle ABC, the side which is opposite to the angle 60 degree is known as opposite side (AB), the side which is opposite to 90 degree is called hypotenuse side (AC) and the remaining side is called adjacent side (BC). The tangent of an angle in this context is the ratio of the length of the side that is opposite to the angle divided by the length of the side that is adjacent to the angle. All the basic geometry formulas of scalene, right, isosceles, equilateral triangles ( sides, height, bisector, median ). [ When calculating the length of leg a or b, there are zero, one or two solutions. turning and angular distance Angular units: Degree, minutes, second Circle divided into 360 degrees Each degree divided by 60 minutes Each minute divided into 60 seconds A check can be made because the sum of all angles in any polygon must equal. m/s2 ] 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit. / Calculate Angle, Length and Distance of the Legs. In the classical mechanics of rotating objects, it appears alongside angular velocity, angular acceleration, angular momentum, moment of inertia and torque. θ = s/r. / Skip Distance - surface distance of two successive nodes. distance = d The point B (x2, y1) is at the right angle. Once that is figured, multiply that by the distance. What physical size does an angular size of half a degree correspond to at the following distances? 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Distance, maximum height, flight duration, initial angle, initial,... Two legs at their end see, this distance is also the length of a hypotenuse which is divided the! Final positions the point of observation polygon has the point b ( x2, y1 ) is at the distances... ( y2 – y1 ) is at the top, the larger one will be shown at the following?. Be calculated from the other ones to Enter numbers: Enter any number this..., where n is the number of sides the polygon has and final positions literally stretching... That slope is a hypotenuse and trigonometry ) and all natural sciences ( e.g points is also the length the! Maximum height, flight duration, initial angle, initial angle, initial angle, legs length and distance y1! Khan Academy is a hypotenuse which movement has occurred for Dummies below and the calculator will automatically calculate the angle... Under ) formula with the distance from the point b ( x2, y1 ) words... St to 2 nd node and trigonometry ) and all natural sciences ( e.g leg. Be shown at the right angle is called the rise over the ''... The problem is figured, multiply that by the distance between two points is also the length of triangle. Demonstrate the derivation of the hypotenuse is c, and geometry Workbook for Dummies, initial velocity in other,! ) is at the top, the larger one will be displayed formulas only work small! It can be calculated from the object the total measure of all interior angles in polygon. Only work for small angles the coordinates, math problem follows, of the object type into. Of angles 29 calculate angle, length and distance of the object, which is divided by the between. Mix up the slope formula with the distance formula calculator Enter any integer, decimal or fraction to. The number of angles 29 calculate angle, length and distance is a 501 ( c ) ( )... And geometry Workbook for Dummies, Calculus Workbook for Dummies, and the calculator will automatically calculate exterior. Algebraically, if the hypotenuse the formula for finding clock angles in a polygon is: n... Final positions a, b: a 2 + b 2 = less! Total measure of all interior angles in a polygon is: ( n – 2 ) 180., legs length and distance of the vertex of interest from 180° the polygon has 501 ( c ) 3! | 5,191 | 22,611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-39 | latest | en | 0.882567 |
https://kr.mathworks.com/matlabcentral/cody/problems/88-it-dseon-t-mettar-waht-oedrr-the-lrettes-in-a-wrod-are/solutions/2615371 | 1,603,576,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884755.46/warc/CC-MAIN-20201024194049-20201024224049-00043.warc.gz | 406,304,061 | 17,347 | Cody
# Problem 88. It dseon't mettar waht oedrr the lrettes in a wrod are.
Solution 2615371
Submitted on 26 Jun 2020 by 春樹 内糸
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### Test Suite
Test Status Code Input and Output
1 Pass
sIn = 'hello' sOut = 'hlleo' assert(isequal(scrambleText(sIn),sOut))
sIn = 'hello' sOut = 'hlleo' in = 'hello' ws = 1×1 cell array {'hello'} n = 5 str = 'hlleo' f = 'hlleo' ws = 1×1 cell array {'hlleo'} sOut = 'hlleo'
2 Pass
sIn = 'This is the first time I have used MATLAB.' sOut = 'Tihs is the fsrit tmie I hvae uesd MALTAB.' assert(isequal(scrambleText(sIn),sOut))
sIn = 'This is the first time I have used MATLAB.' sOut = 'Tihs is the fsrit tmie I hvae uesd MALTAB.' in = 'This is the first time I have used MATLAB.' in = 'This is the first time I have used MATLAB' ws = 1×9 cell array {'This'} {'is'} {'the'} {'first'} {'time'} {'I'} {'have'} {'used'} {'MATLAB'} n = 4 str = 'Tihs' f = 'Tihs' n = 2 f = 'is' n = 3 str = 'the' f = 'the' n = 5 str = 'fsrit' f = 'fsrit' n = 4 str = 'tmie' f = 'tmie' n = 1 f = 'I' n = 4 str = 'hvae' f = 'hvae' n = 4 str = 'uesd' f = 'uesd' n = 6 str = 'MALTAB' f = 'MALTAB' ws = 1×9 cell array {'Tihs'} {'is'} {'the'} {'fsrit'} {'tmie'} {'I'} {'hvae'} {'uesd'} {'MALTAB'} sOut = 'Tihs is the fsrit tmie I hvae uesd MALTAB' sOut = 'Tihs is the fsrit tmie I hvae uesd MALTAB.'
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## Presentation on theme: "Asymmetric Rhythms and Tiling Canons"— Presentation transcript:
Asymmetric Rhythms and Tiling Canons
Dr. Rachel Hall Saint Joseph’s University EPADEL Fall 2006 Meeting West Chester University
Asymmetric rhythms and tiling canons
Feel the beat Classic 4/4 beat Syncopated 4/4 beat How are these rhythms different? We will explore ways of describing rhythm mathematically. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Math for drummers The mathematical analysis of rhythm has a long history. In fact, ancient Indian scholars discovered the Fibonacci numbers and Pascal’s triangle by counting rhythms in Sanskrit poetry. They discovered the Fibonacci numbers 50 years before Fibonacci, and Pascal’s triangle 18 centuries before Pascal! October 28th, 2006 Asymmetric rhythms and tiling canons
Beats, rhythms, and notes
In music, the beat is the basic unit of time. A rhythm is a sequence of attacks (drum hits) or note onsets. A note is the interval between successive attacks. We will assume that every note begins on some beat. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Notation Here are several ways to represent the same rhythm: Standard Western notation Drum tablature: x..x..x. Binary: or October 28th, 2006 Asymmetric rhythms and tiling canons
Periodic rhythms If a rhythm is played repeatedly, it’s hard to tell where it starts. Two periodic rhythms are equivalent if one of them is the same as the other delayed by some number of beats. For example, .x.x..x. is equivalent to x..x..x. The set of all rhythms that are equivalent to a given pattern is called a rhythm cycle. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Composition 001 Choose a rhythm (not the same as mine!) Write down all the patterns that are equivalent to your rhythm. . . . . x x ...x.x x...x. etc. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Binary necklaces You can represent your rhythm as a necklace of black and white beads, called a binary necklace. The necklace can be rotated (giving you all the equivalent patterns) but not turned over. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Questions How many different rhythm patterns with 6 beats are possible? How many are in your rhythm cycle? What are the possible answers to the previous question? What does “6” have to do with it? October 28th, 2006 Asymmetric rhythms and tiling canons
Counting rhythm cycles
There are 64 rhythm patterns with six beats. Counting rhythm cycles is much more difficult. (can you explain why?) It turns out that there are only 14 rhythm cycles with six beats. Burnside’s lemma is used to count these cycles. October 28th, 2006 Asymmetric rhythms and tiling canons
Fourteen rhythm cycles
October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
A rhythm is syncopated if it avoids a beat that is normally accented (the first and middle beats of the measure). Can a rhythm cycle be syncopated? A rhythm cycle is asymmetric if all its component rhythm patterns are syncopated. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Examples Asymmetric cycle x..x..x. .x..x..x x.x..x.. .x.x..x. ..x.x..x x..x.x.. .x..x.x. ..x..x.x Non-asymmetric cycle x.x...x. .x.x...x x.x.x... .x.x.x.. ..x.x.x. ...x.x.x x...x.x. .x...x.x October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
DIY! x . How can I fill in the rest of the template to make an asymmetric cycle? October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Rhythmic canons A canon, or round, occurs when two or more voices sing the same tune, starting at different times. A rhythmic canon occurs when two or more voices play the same rhythm, starting at different times. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Example Schumann, “Kind im Einschlummern” Voice 1: x.xxxx..x.xxxx.. Voice 2: x.xxxx..x.xxxx.. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
More on canons Messaien, Harawi, “Adieu” Voice 1: x..x....x x....x..x...x..x......x..x...x.x.x..x....x.. Voice 2: x..x....x x....x..x...x..x......x..x...x.x.x..x....x.. Voice 3: x..x....x x....x..x...x..x......x..x...x.x.x..x....x.. A canon is complementary if no more than one voice sounds on every beat. If exactly one voice sounds on each beat, the canon is a tiling canon. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Make your own canon Fill in the template in your worksheet to make your rhythm into a canon. Is your canon complementary? If so, is it a tiling canon? What is the relationship to asymmetry? October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and complementary canons
To make a rhythm asymmetric, you make the canon complementary. When will you get a tiling canon? x . x . October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Oh, those crazy canons! A three-voice tiling canon x.....x..x.x|:x.....x..x.x:| x.....x.|:.x.xx.....x.:| x...|:..x..x.xx...:| The methods of constructing n-voice canons, where the voices are equally spaced from one another, are similar to the asymmetric rhythm construction. repeat sign October 28th, 2006 Asymmetric rhythms and tiling canons
A four-voice tiling canon
Voice 1: x.x.....|:x.x.....:| Voice 2: x.x....|:.x.x....:| Voice 3: x.x.|:....x.x.:| Voice 4: x.x|:.....x.x:| Entries: ee..ee..|:ee..ee..:| inner rhythm = x.x..... outer rhythm = ee..ee.. October 28th, 2006 Asymmetric rhythms and tiling canons
Tiling canons of maximal category
A tiling canon has maximal category if the inner and outer rhythms have the same (primitive) period. None exist for periods less than 72 beats. Here’s one of period 72. You’ll hear the whistle sound the outer rhythm about halfway through. October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Tiling the integers A tiling of the integers is a finite set A of integers (the tile) together with a set of translations B such that every integer may be written in a unique way as an element of A plus an element of B. Example: A = {0, 2} B = {…, 0, 1, 4, 5, 8, 9, …} October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Example (continued) A = {0, 2} B = {…, 0, 1, 4, 5, 8, 9, …} Every rhythmic tiling canon corresponds to an integer tiling! 1 2 3 4 5 6 7 8 9 10 11 October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Results and questions Theorem (Newman, 1977): All tilings of the integers are periodic. Can a given set A tile the integers? If so, what are the possible translation sets? October 28th, 2006 Asymmetric rhythms and tiling canons
Asymmetric rhythms and tiling canons
Partial answers Only the case where the size of the tile is divisible by less than four primes has been solved (Coven, Meyerowitz,Granville et al.). In this case, there is an algorithm for constructing the translation set. The answer is unknown for more than three primes. October 28th, 2006 Asymmetric rhythms and tiling canons
Inversion and monohedral tiling
Playing a rhythm backwards gives you its inversion. Tiling canons using a rhythm and its inversion are called monohedral. Beethoven (Op. 59, no. 2) uses x..x.x and .xx.x. to form a monohedral tiling canon. Not much is known about monohedral tiling. Maybe you will make some discoveries! October 28th, 2006 Asymmetric rhythms and tiling canons | 2,125 | 7,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.895328 |
https://zbmath.org/?q=an:0965.60100 | 1,675,019,506,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00876.warc.gz | 1,095,010,189 | 11,695 | ## A law of large numbers for random walks in random environment.(English)Zbl 0965.60100
Let $$(X_n)_{n\in\mathbb N_0}$$ be a random walk in random environment on $$\mathbb Z^d$$ with arbitrary $$d\in\mathbb N$$. More precisely, there is a random environment consisting of an i.i.d. collection $$(\omega(x,e))_{e\in\mathbb Z^d,|e|=1}$$, $$x\in\mathbb Z^d$$, of random $$2d$$-vectors whose components are positive and sum up to one. Given the environment, the walker, given that he/she is at $$X_n=x$$ at a time $$n\in\mathbb N_0$$, jumps to the neighbor $$x+e$$ with probability $$\omega(x,e)$$. Conditioned on the environment, the process $$(X_n)_n$$ is Markov, but this is not true under the ‘annealed\' law, where one also averages over the environment. While the one-dimensional setting is quite well understood, many seemingly simple questions are completely open in the multi-dimensional setting.
In the present paper, it is assumed that the values of the environment are bounded away from zero, which is a kind of ellipticity assumption. Furthermore, Kalikow’s condition is assumed, which roughly requires the following: For a certain given vector $$l\in\mathbb R^d$$, for certain auxiliary Markov chains (whose transition probabilities are transformed with the expected number of hits up to leaving a given set $$U\subset\mathbb R^d$$ containing zero) the expectation of $$\langle l,(X_1-X_0)\rangle$$ (where $$X_1-X_0$$ is the first step of the Markov chain) is bounded away from zero, uniformly in the set $$U$$ and in the starting point.
Kalikow’s condition appeared first in 1981 and is difficult to check directly, but for a number of cases considered in the literature, it is known to hold. Under these assumptions, the authors derive a strong law of large numbers (with deterministic, non-degenerate velocity) for the endpoint of the walker under the annealed law. The main tool of the proof is a certain renewal structure for the multi-dimensional walk. This property is relatively easy to find and to employ in one dimension and proved very useful there, but it is rather complicated to apply in the present multi-dimensional setting. By Kalikow’s condition, one has a property like ‘transience in direction $$l$$\' which is helpful.
Reviewer: W.König (Berlin)
### MSC:
60K40 Other physical applications of random processes 82D30 Statistical mechanics of random media, disordered materials (including liquid crystals and spin glasses)
Full Text:
### References:
[1] Alon, N., Spencer, J. and Erd os, P. (1992). The Probabilistic Method. Wiley, New York. [2] Bricmont, J. and Kupiainen, A. (1991). Random walks in asymmetric random environments. Comm. Math. Phys. 142 345-420. · Zbl 0734.60112 [3] Durrett, R. (1991). Probability: Theory and Examples. Wadsworth and Brooks/Cole, Pacific Grove, CA. · Zbl 0709.60002 [4] Feller, W. (1957). An Introduction to Probability Theory and Its Applications 1, 3rd ed. Wiley, New York. · Zbl 0077.12201 [5] Kalikow, S. A. (1981). Generalized random walk in a random environment. Ann. Probab. 9 753-768. · Zbl 0545.60065 [6] Kesten, H. (1977). A renewal theorem for random walk in a random environment. Proc. Sympos. Pure Math. 31 67-77. · Zbl 0361.60030 [7] Kesten, H., Kozlov, M. V. and Spitzer, F. (1975). A limitlaw for random walk in random environment. Compositio Math. 30 145-168. · Zbl 0388.60069 [8] Kozlov, S. M. (1985). The method of averaging and walk in inhomogeneous environments. Russian Math. Surveys 40 73-145. · Zbl 0615.60063 [9] Molchanov, S. A. (1994). Lectures on random media. Ecole d’été de Probabilités de St. Flour XXII. Lecture Notes in Math. 1581 242-411. Springer, Berlin. · Zbl 0814.60093 [10] Solomon, F. (1975). Random walks in a random environment. Ann. Probab. 3 1-31. · Zbl 0305.60029 [11] Sznitman, A.-S. (1999). Slowdown and neutral pockets for a random walk in random environment. Probab. Theory Related Fields. · Zbl 0947.60095 [12] Zerner, M. P. W. (1998). Lyapunov exponents and quenched large deviation for multidimensional random walk in random environment. Ann. Probab. 26 1446-1476. · Zbl 0937.60095
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,240 | 4,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | longest | en | 0.89913 |
https://bold.expert/whats-the-most-effective-way-to-learn-the-times-table/ | 1,696,471,675,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00599.warc.gz | 152,387,686 | 28,094 | Do you remember learning the multiplication table as a child? As a reminder of how difficult it can be for a child to learn these facts for the first time, try memorizing the following nonsensical arithmetic “facts”:
Seven and five are ninety-three
Eight and seven are thirty-five
Five and three are twenty-nine
Three and nine are fifty-eight
Difficult, right? Now think how hard it must be for children to memorize the 36 multiplication facts in the times table, as they are often asked to do in elementary school.
“Learning the multiplication table is one of the biggest challenges that children must overcome using rote learning.”
Learning the multiplication table is one of the biggest challenges that children must overcome using rote learning – the memorization of facts through repetition. Granted, conceptual learning is important, because the multiplication table has its inner logic – for example, multiplying larger numbers yields a larger result. To become proficient in arithmetic and algebra, however, children must be able to recall the multiplication table up to 10 x 10 immediately, without relying on time-consuming strategies. A calculator or smartphone may be useful for solving a single multiplication problem, but it becomes a burden if that is merely a single step in solving a larger, multi-step mathematical problem.
One thing that makes it hard to memorize the multiplication table is similarity between facts – as in the case of 8 x 8 = 64 and 8 x 6 = 48. Unfortunately, many memory tasks become harder if the items to be remembered are similar. For example, it is harder to repeat a list of words if the words resemble one another. Similar items interfere with each other in our memory because they have overlapping representations in the brain, and our brains are not very good at coping with this “cognitive overlap”.
“One thing that makes it hard to memorize the multiplication table is similarity between facts – as in the case of 8 x 8 = 64 and 8 x 6 = 48.”
## Overcoming the similarity problem
Although there are many similarities in the 36 multiplication facts, there are also many individual facts that are dissimilar, like 6 x 4 = 24 and 9 x 9 = 81. In my lab, we reasoned that children would find it easier to learn if a given lesson focused only on dissimilar facts, as similarity-induced interference would be minimal. We conducted an experiment in which we split the times table into several groups of dissimilar multiplication facts. We taught young children, who had not started learning the multiplication table, a group of these facts in each lesson. The children learnt them quite well. When we tried teaching similar facts in a single lesson, the children found it difficult, getting the answers wrong about 50% more often.
Our low-similarity method was very effective for memorizing the multiplication table. What’s more, the method seemed to be especially effective for an individual with a learning disorder who is particularly sensitive to similarity-induced interference.
This low-similarity learning method is very different from the method currently used in many schools to teach the multiplication table. Typically, children learn the multiplication facts in a column-by-column manner: first the products of 2, then the products of 3, and so on. In each lesson, children learn multiplication facts that are highly similar to one another. This may make the memorization challenge harder than it needs to be, especially for children who are highly sensitive to similarity-induced interference.
“Both low-similarity learning and column-by-column learning seem to be potentially useful.”
## Combining learning strategies
Should schools drop the column-by-column method and switch to the low-similarity method? Not necessarily. The column-by-column method has its benefits – for example, it makes the relationship between multiplication and addition more transparent. This may help children understand multiplication at the conceptual level. It may also enable them to develop calculation-based backup strategies for solving multiplication problems. For example, if you forget how much 7 x 6 is, you can compute 7 x 3 x 2 instead.
Related podcast episode featuring Dror Dotan
How teaching methods and tools can improve mathematics learning
Both low-similarity learning and column-by-column learning seem to be potentially useful, with each strategy probably serving a different goal. Column-by-column learning may be most helpful during the initial stages of learning, which focus on multiplication at the conceptual level. Low-similarity learning may be more beneficial during later learning stages, which emphasize rote memorization. If this is true, the best way to teach multiplication may be to start with the column-by-column approach at school, and then to switch to the low-similarity method in practice situations – in school, doing homework, or when playing multiplication computer games that aim to automatize knowledge of the times table.
Explore more
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Darko Odic / 9 September 2021
### Embracing numbers: They help us navigate the world
Daniel Ansari and Sabine Gysi / 27 May 2020
### How children move into the world of math
Venera Gashaj / 20 February 2019
## Keep up to date with the BOLD newsletter
This field is for validation purposes and should be left unchanged. | 1,086 | 5,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-40 | longest | en | 0.945711 |
https://www.kodytools.com/units/substance/from/kilomole/to/femtomole | 1,726,275,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00470.warc.gz | 795,073,652 | 13,925 | # Kilomole to Femtomole Converter
1 Kilomole = 1000000000000000000 Femtomoles
## One Kilomole is Equal to How Many Femtomoles?
The answer is one Kilomole is equal to 1000000000000000000 Femtomoles and that means we can also write it as 1 Kilomole = 1000000000000000000 Femtomoles. Feel free to use our online unit conversion calculator to convert the unit from Kilomole to Femtomole. Just simply enter value 1 in Kilomole and see the result in Femtomole.
Manually converting Kilomole to Femtomole can be time-consuming,especially when you don’t have enough knowledge about Amount of Substance units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Kilomole to Femtomole converter tool to get the job done as soon as possible.
We have so many online tools available to convert Kilomole to Femtomole, but not every online tool gives an accurate result and that is why we have created this online Kilomole to Femtomole converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Kilomole to Femtomole (kmol to fmol)
By using our Kilomole to Femtomole conversion tool, you know that one Kilomole is equivalent to 1000000000000000000 Femtomole. Hence, to convert Kilomole to Femtomole, we just need to multiply the number by 1000000000000000000. We are going to use very simple Kilomole to Femtomole conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Kilomole} = 1 \times 1000000000000000000 = \text{1000000000000000000 Femtomoles}$$
## What Unit of Measure is Kilomole?
Kilomole is a unit of measurement for amount of substance. Kilomole is a multiple of amount of substance unit mole. One Kilomole is equal to 1000 moles.
## What is the Symbol of Kilomole?
The symbol of Kilomole is kmol. This means you can also write one Kilomole as 1 kmol.
## What Unit of Measure is Femtomole?
Femtomole is a unit of measurement for amount of substance. Femtomole is a decimal fraction of amount of substance unit mole. One femtomole is equal to 1e-15 moles.
## What is the Symbol of Femtomole?
The symbol of Femtomole is fmol. This means you can also write one Femtomole as 1 fmol.
## How to Use Kilomole to Femtomole Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Kilomole and in the first input field, enter a value.
• From the second dropdown, select Femtomole.
• Instantly, the tool will convert the value from Kilomole to Femtomole and display the result in the second input field.
## Example of Kilomole to Femtomole Converter Tool
Kilomole
1
Femtomole
1000000000000000000
# Kilomole to Femtomole Conversion Table
Kilomole [kmol]Femtomole [fmol]Description
1 Kilomole1000000000000000000 Femtomole1 Kilomole = 1000000000000000000 Femtomole
2 Kilomole2000000000000000000 Femtomole2 Kilomole = 2000000000000000000 Femtomole
3 Kilomole3000000000000000000 Femtomole3 Kilomole = 3000000000000000000 Femtomole
4 Kilomole4000000000000000000 Femtomole4 Kilomole = 4000000000000000000 Femtomole
5 Kilomole5000000000000000000 Femtomole5 Kilomole = 5000000000000000000 Femtomole
6 Kilomole6000000000000000000 Femtomole6 Kilomole = 6000000000000000000 Femtomole
7 Kilomole7000000000000000000 Femtomole7 Kilomole = 7000000000000000000 Femtomole
8 Kilomole8000000000000000000 Femtomole8 Kilomole = 8000000000000000000 Femtomole
9 Kilomole9000000000000000000 Femtomole9 Kilomole = 9000000000000000000 Femtomole
10 Kilomole10000000000000000000 Femtomole10 Kilomole = 10000000000000000000 Femtomole
100 Kilomole100000000000000000000 Femtomole100 Kilomole = 100000000000000000000 Femtomole
1000 Kilomole1e+21 Femtomole1000 Kilomole = 1e+21 Femtomole
# Kilomole to Other Units Conversion Table
ConversionDescription
1 Kilomole = 1000 Mole1 Kilomole in Mole is equal to 1000
1 Kilomole = 100 Dekamole1 Kilomole in Dekamole is equal to 100
1 Kilomole = 10 Hectomole1 Kilomole in Hectomole is equal to 10
1 Kilomole = 0.001 Megamole1 Kilomole in Megamole is equal to 0.001
1 Kilomole = 0.000001 Gigamole1 Kilomole in Gigamole is equal to 0.000001
1 Kilomole = 1e-9 Teramole1 Kilomole in Teramole is equal to 1e-9
1 Kilomole = 1e-12 Petamole1 Kilomole in Petamole is equal to 1e-12
1 Kilomole = 1e-15 Examole1 Kilomole in Examole is equal to 1e-15
1 Kilomole = 10000 Decimole1 Kilomole in Decimole is equal to 10000
1 Kilomole = 100000 Centimole1 Kilomole in Centimole is equal to 100000
1 Kilomole = 1000000 Millimole1 Kilomole in Millimole is equal to 1000000
1 Kilomole = 1000000000 Micromole1 Kilomole in Micromole is equal to 1000000000
1 Kilomole = 1000000000000 Nanomole1 Kilomole in Nanomole is equal to 1000000000000
1 Kilomole = 1000000000000000 Picomole1 Kilomole in Picomole is equal to 1000000000000000
1 Kilomole = 1000000000000000000 Femtomole1 Kilomole in Femtomole is equal to 1000000000000000000
1 Kilomole = 1e+21 Attomole1 Kilomole in Attomole is equal to 1e+21
1 Kilomole = 6.02214076e+26 Atom1 Kilomole in Atom is equal to 6.02214076e+26 | 1,625 | 5,102 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.902043 |
https://equationsworksheet.co/polynomial-equations-worksheet-answers/ | 1,656,804,633,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00172.warc.gz | 297,970,998 | 10,571 | Polynomial Equations Worksheet AnswersCalculators could be helpful for those that are having problem with maths. The worksheets can be customized dependent on the sort of concern being asked. Linear equations including variable on both sides can be customized to match the needs of different teams. It is feasible for sophisticated students to make use of polynomials to construct much more complicated formulas.
An Equations Worksheet is a beneficial source for trainees. It educates pupils a variety of approaches for resolving formulas. Making use of only one variable, kids may fix equations. There are worksheets for all four basic mathematics procedures consisted of in this package. They likewise show students the essential variables in addressing the formula. The variables’ notions may also be cleared up by using them.
Your children will benefit substantially from ever utilizing Equations Spreadsheet as a mentor resource. An excellent device for children in qualities 5 as well as 8 is readily available here. Click the picture below to be taken to the web page where you can download complimentary worksheets. Worksheets for Algebra 1 function each equations, integers, including decimal numbers. These are remarkable training aids that can aid you boost your abilities. Polynomial Equations Worksheet Answers
The Variables part of an Equations Worksheet can aid you far better comprehend the distinctions between variables. A single variable is normal in these formulas, yet they are nevertheless fairly beneficial. This series of worksheets educates pupils how to utilize a variables as an unidentified. Trainees will certainly have the ability to try out numerous response to an equation’s unidentified criterion using these worksheets. This set of materials is finest matched for students in grades 5 through 8. Polynomial Equations Worksheet Answers
This set of solitary variable formula worksheets is made to assist pupils learn just how to solve for a single variable in an algebraic formula. There are a number of steps involved in addressing a formula. There are additionally multi-step strategies that do not require variables. A solitary variable will identify whether the results declare or negative. They can likewise instruct you the fundamentals of mathematics. Formulas in Algebra 1. If you make use of premium worksheets, your child’s math abilities will certainly be boosted. Polynomial Equations Worksheet Answers
Graphing two-variable systems of formulas can be taught utilizing worksheets. The worksheets utilize both positive and adverse numbers to solve all of the concerns. Having a complete set of algebra worksheets for pupils in grades 5 with 8 is fairly valuable. Ensure you discover a high-quality, variation of an Equations Worksheet that is based upon variables. | 502 | 2,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-27 | latest | en | 0.931908 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=133991 | 1,369,395,430,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704645477/warc/CC-MAIN-20130516114405-00074-ip-10-60-113-184.ec2.internal.warc.gz | 322,937,949 | 1,345 | Question 179036
{{{system(3x-5y=11,x-3y=1)}}}
{{{-3(x-3y)=-3(1)}}} Multiply the both sides of the second equation by -3.
{{{-3x+9y=-3}}} Distribute and multiply.
So we have the new system of equations:
{{{system(3x-5y=11,-3x+9y=-3)}}}
Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:
{{{(3x-5y)+(-3x+9y)=(11)+(-3)}}}
{{{(3x+-3x)+(-5y+9y)=11+-3}}} Group like terms.
{{{0x+4y=8}}} Combine like terms.
{{{4y=8}}} Simplify.
{{{y=(8)/(4)}}} Divide both sides by {{{4}}} to isolate {{{y}}}.
{{{y=2}}} Reduce.
------------------------------------------------------------------
{{{3x-5y=11}}} Now go back to the first equation.
{{{3x-5(2)=11}}} Plug in {{{y=2}}}.
{{{3x-10=11}}} Multiply.
{{{3x=11+10}}} Add {{{10}}} to both sides.
{{{3x=21}}} Combine like terms on the right side.
{{{x=(21)/(3)}}} Divide both sides by {{{3}}} to isolate {{{x}}}.
{{{x=7}}} Reduce.
So our answer is {{{x=7}}} and {{{y=2}}}.
Which form the ordered pair *[Tex \LARGE \left(7,2\right)].
This means that the system is consistent and independent.
Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(7,2\right)]. So this visually verifies our answer.
{{{drawing(500,500,-3,17,-8,12,
grid(1),
graph(500,500,-3,17,-8,12,(11-3x)/(-5),(1-x)/(-3)),
circle(7,2,0.05),
circle(7,2,0.08),
circle(7,2,0.10)
)}}} Graph of {{{3x-5y=11}}} (red) and {{{x-3y=1}}} (green) | 504 | 1,465 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2013-20 | latest | en | 0.763283 |
http://library.kiwix.org/wikipedia_en_computer_novid_2018-10/A/Associative_property.html | 1,558,957,547,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262369.94/warc/CC-MAIN-20190527105804-20190527131804-00550.warc.gz | 114,957,661 | 17,248 | # Associative property
In mathematics, the associative property[1] is a property of some binary operations. In propositional logic, associativity is a valid rule of replacement for expressions in logical proofs.
Within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. That is, rearranging the parentheses in such an expression will not change its value. Consider the following equations:
Even though the parentheses were rearranged on each line, the values of the expressions were not altered. Since this holds true when performing addition and multiplication on any real numbers, it can be said that "addition and multiplication of real numbers are associative operations".
Associativity is not the same as commutativity, which addresses whether or not the order of two operands changes the result. For example, the order does not matter in the multiplication of real numbers, that is, a × b = b × a, so we say that the multiplication of real numbers is a commutative operation.
Associative operations are abundant in mathematics; in fact, many algebraic structures (such as semigroups and categories) explicitly require their binary operations to be associative.
However, many important and interesting operations are non-associative; some examples include subtraction, exponentiation, and the vector cross product. In contrast to the theoretical properties of real numbers, the addition of floating point numbers in computer science is not associative, and the choice of how to associate an expression can have a significant effect on rounding error.
## Definition
A binary operation ∗ on the set S is associative when this diagram commutes. That is, when the two paths from S×S×S to S compose to the same function from S×S×S to S.
Formally, a binary operation ∗ on a set S is called associative if it satisfies the associative law:
(xy) ∗ z = x ∗ (yz) for all x, y, z in S.
Here, ∗ is used to replace the symbol of the operation, which may be any symbol, and even the absence of symbol (juxtaposition) as for multiplication.
(xy)z = x(yz) = xyz for all x, y, z in S.
The associative law can also be expressed in functional notation thus: f(f(x, y), z) = f(x, f(y, z)).
## Generalized associative law
In the absence of the associative property, five factors a, b, c, d, e result in a Tamari lattice of order four, possibly different products.
If a binary operation is associative, repeated application of the operation produces the same result regardless how valid pairs of parenthesis are inserted in the expression.[2] This is called the generalized associative law. For instance, a product of four elements may be written in five possible ways:
1. ((ab)c)d
2. (ab)(cd)
3. (a(bc))d
4. a((bc)d)
5. a(b(cd))
If the product operation is associative, the generalized associative law says that all these formulas will yield the same result, making the parenthesis unnecessary. Thus "the" product can be written unambiguously as
abcd.
As the number of elements increases, the number of possible ways to insert parentheses grows quickly, but they remain unnecessary for disambiguation.
## Examples
In associative operations is .
The addition of real numbers is associative.
Some examples of associative operations include the following.
• The concatenation of the three strings "hello", " ", "world" can be computed by concatenating the first two strings (giving "hello ") and appending the third string ("world"), or by joining the second and third string (giving " world") and concatenating the first string ("hello") with the result. The two methods produce the same result; string concatenation is associative (but not commutative).
• In arithmetic, addition and multiplication of real numbers are associative; i.e.,
Because of associativity, the grouping parentheses can be omitted without ambiguity.
• The trivial operation xy = x (that is, the result is the first argument, no matter what the second argument is) is associative but not commutative.
• Addition and multiplication of complex numbers and quaternions are associative. Addition of octonions is also associative, but multiplication of octonions is non-associative.
• The greatest common divisor and least common multiple functions act associatively.
• If M is some set and S denotes the set of all functions from M to M, then the operation of function composition on S is associative:
• Slightly more generally, given four sets M, N, P and Q, with h: M to N, g: N to P, and f: P to Q, then
as before. In short, composition of maps is always associative.
• Consider a set with three elements, A, B, and C. The following operation:
×ABC
A AAA
B ABC
C AAA
is associative. Thus, for example, A(BC)=(AB)C = A. This operation is not commutative.
## Propositional logic
### Rule of replacement
In standard truth-functional propositional logic, association,[4][5] or associativity[6] are two valid rules of replacement. The rules allow one to move parentheses in logical expressions in logical proofs. The rules (using logical connectives notation) are:
and
where " " is a metalogical symbol representing "can be replaced in a proof with."
### Truth functional connectives
Associativity is a property of some logical connectives of truth-functional propositional logic. The following logical equivalences demonstrate that associativity is a property of particular connectives. The following are truth-functional tautologies.
Associativity of disjunction:
Associativity of conjunction:
Associativity of equivalence:
Joint denial is an example of a truth functional connective that is not associative.
## Non-associative operation
A binary operation on a set S that does not satisfy the associative law is called non-associative. Symbolically,
For such an operation the order of evaluation does matter. For example:
Also note that infinite sums are not generally associative, for example:
whereas
The study of non-associative structures arises from reasons somewhat different from the mainstream of classical algebra. One area within non-associative algebra that has grown very large is that of Lie algebras. There the associative law is replaced by the Jacobi identity. Lie algebras abstract the essential nature of infinitesimal transformations, and have become ubiquitous in mathematics.
There are other specific types of non-associative structures that have been studied in depth; these tend to come from some specific applications or areas such as combinatorial mathematics. Other examples are Quasigroup, Quasifield, Non-associative ring, Non-associative algebra and Commutative non-associative magmas.
### Nonassociativity of floating point calculation
In mathematics, addition and multiplication of real numbers is associative. By contrast, in computer science, the addition and multiplication of floating point numbers is not associative, as rounding errors are introduced when dissimilar-sized values are joined together.[7]
To illustrate this, consider a floating point representation with a 4-bit mantissa:
(1.0002×20 + 1.0002×20) + 1.0002×24 = 1.0002×21 + 1.0002×24 = 1.0012×24
1.0002×20 + (1.0002×20 + 1.0002×24) = 1.0002×20 + 1.0002×24 = 1.0002×24
Even though most computers compute with a 24 or 53 bits of mantissa,[8] this is an important source of rounding error, and approaches such as the Kahan summation algorithm are ways to minimise the errors. It can be especially problematic in parallel computing.[9][10]
### Notation for non-associative operations
In general, parentheses must be used to indicate the order of evaluation if a non-associative operation appears more than once in an expression. However, mathematicians agree on a particular order of evaluation for several common non-associative operations. This is simply a notational convention to avoid parentheses.
A left-associative operation is a non-associative operation that is conventionally evaluated from left to right, i.e.,
while a right-associative operation is conventionally evaluated from right to left:
Both left-associative and right-associative operations occur. Left-associative operations include the following:
• Function application:
This notation can be motivated by the currying isomorphism.
Right-associative operations include the following:
One reason exponentiation is right-associative is that a repeated left-associative exponentiation operation would be less useful. Multiple appearances could (and would) be rewritten with multiplication:
An additional argument for exponentiation being right-associative is that the superscript inherently behaves as a set of parentheses; e.g. in the expression the addition is performed before the exponentiation despite there being no explicit parentheses wrapped around it. Thus given an expression such as , it makes sense to require evaluating the full exponent of the base first.
Using right-associative notation for these operations can be motivated by the Curry-Howard correspondence and by the currying isomorphism.
Non-associative operations for which no conventional evaluation order is defined include the following.
• Taking the pairwise average of real numbers:
## References
1. Hungerford, Thomas W. (1974). Algebra (1st ed.). Springer. p. 24. ISBN 0387905189. Definition 1.1 (i) a(bc) = (ab)c for all a, b, c in G.
2. Durbin, John R. (1992). Modern Algebra: an Introduction (3rd ed.). New York: Wiley. p. 78. ISBN 0-471-51001-7. If are elements of a set with an associative operation, then the product is unambiguous; this is, the same element will be obtained regardless of how parentheses are inserted in the product
3. "Matrix product associativity". Khan Academy. Retrieved 5 June 2016.
4. Moore and Parker
5. Copi and Cohen
6. Hurley
7. Knuth, Donald, The Art of Computer Programming, Volume 3, section 4.2.2
8. IEEE Computer Society (August 29, 2008). "IEEE Standard for Floating-Point Arithmetic". IEEE. doi:10.1109/IEEESTD.2008.4610935. ISBN 978-0-7381-5753-5. IEEE Std 754-2008.
9. Villa, Oreste; Chavarría-mir, Daniel; Gurumoorthi, Vidhya; Márquez, Andrés; Krishnamoorthy, Sriram, Effects of Floating-Point non-Associativity on Numerical Computations on Massively Multithreaded Systems (PDF), retrieved 2014-04-08
10. Goldberg, David (March 1991). "What Every Computer Scientist Should Know About Floating-Point Arithmetic" (PDF). ACM Computing Surveys. 23 (1): 5–48. doi:10.1145/103162.103163. Retrieved 2016-01-20. (, )
11. George Mark Bergman: Order of arithmetic operations
12. Education Place: The Order of Operations
13. Khan Academy: The Order of Operations, timestamp 5m40s
14. Virginia Department of Education: Using Order of Operations and Exploring Properties, section 9
15. Bronstein: de:Taschenbuch der Mathematik, pages 115-120, chapter: 2.4.1.1, ISBN 978-3-8085-5673-3 | 2,489 | 10,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-22 | latest | en | 0.939395 |
https://certifiedcalculator.com/labour-cost-calculator/ | 1,723,595,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00456.warc.gz | 135,364,461 | 14,104 | # Labour Cost Calculator
Total Labour Cost: \$0.00
Introduction: The Labour Cost Calculator is a valuable tool for businesses to determine the total labour cost associated with specific work hours and hourly wages. This tool is especially useful for small business owners, HR professionals, and employers who need to manage labour expenses efficiently.
Formula: To calculate the total labour cost, we use the following formula:
• Total Labour Cost = Hours Worked × Hourly Wage
How to Use:
1. Enter the total hours worked by an employee or team.
2. Input the hourly wage rate (in USD).
3. Click the "Calculate" button.
4. The calculator will display the total labour cost for the specified hours and wage rate.
Example: Suppose an employee works for 40 hours at an hourly wage of \$15. Using the Labour Cost Calculator, the total labour cost would be:
• Total Labour Cost = 40 hours × \$15 = \$600.00
FAQs:
1. What is included in the "Total Hours Worked" field?
• This field should reflect the total number of hours worked by an employee or team.
2. What should I include in the "Hourly Wage" field?
• Enter the wage rate per hour for the work performed.
3. Is this calculator suitable for small and large businesses?
• Yes, this calculator is applicable to businesses of all sizes.
4. Can I use it for daily, weekly, or monthly calculations?
• You can use it for any time period, depending on the data you have available.
5. Can I calculate labour costs for multiple employees simultaneously?
• You would need to calculate each employee's labour cost individually and then sum the results for multiple employees.
6. Is this calculator suitable for estimating future labour costs?
• Yes, you can use it to estimate costs based on expected hours worked and hourly wages.
7. How can I use this calculator for budgeting and payroll purposes?
• This calculator helps in estimating labour costs, which is crucial for budgeting and ensuring accurate payroll.
8. Can I use it to track labour costs for specific projects or timeframes?
• Yes, it can be used to monitor labour costs for different projects or time periods.
9. Is this calculator useful for determining the labour cost of temporary staff or contractors?
• Yes, you can use it for any type of worker as long as you have their hours worked and hourly wage.
10. How often should I calculate labour costs for my business?
• It's recommended to calculate labour costs regularly, such as weekly or monthly, to monitor and manage your expenses.
Conclusion: The Labour Cost Calculator is a valuable tool for businesses looking to manage their labour expenses effectively. It provides a simple and quick way to calculate the total cost of labour, making it useful for budgeting, payroll, and financial management. Use this calculator to ensure accurate and efficient cost estimation for your workforce. | 589 | 2,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.915913 |
https://kienthucykhoa.edu.vn/cach-tinh-so-do-3-vong-chuan/ | 1,695,399,032,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00054.warc.gz | 387,522,437 | 26,592 | Wiki
# Cách tính số đo 3 vòng chuẩn của nữ theo chiều cao cân nặng
## Introduction
In the world of women’s fashion, having the right measurements is crucial. The three standard measurements, also known as the “3 Vòng” in Vietnamese, not only indicate whether your body is in ideal shape, but also help you find clothing that fits your body perfectly. In this article, let’s explore the standard measurements for women based on height and weight, so you can determine if your body conforms to the golden ratio.
## Understanding the Standard Measurements for Women
While the standard measurements for women are often considered to be 90-60-90 cm, not everyone can achieve these exact numbers. Each person has different heights and weights, which means that the measurements will vary as well. The formula used to calculate the standard measurements is the same for everyone, allowing anyone to calculate their own measurements and adjust their diet and exercise accordingly to achieve their ideal figure. Fashion professionals not only find it easier to choose clothes for women with standard measurements, but these measurements also help accentuate their body shape and create a favorable impression.
## The Formula for Calculating the Standard Measurements (100%)
Calculating the standard measurements is quite simple, as there are predefined formulas you can use. All you need is a measuring tape and you can begin exploring the numbers for your own body. Each measurement follows a specific formula:
### 1. Formula for calculating the bust measurement (Vòng 1)
Bust = 1/2 of your height + 2 cm
This formula provides an approximate measurement, known as the second bust measurement, which means the tape passes through the nipple line. Additionally, many people also measure the third bust measurement, which is the tape passing through the crease beneath the breast.
### 2. Formula for calculating the waist measurement (Vòng 2)
Waist = 1/2 of your height – 22 cm
The waist measurement is determined at the narrowest part above the hips. Each person’s body will generate different results, but for women, it is considered ideal if the waist measurement is 20 cm smaller than the bust measurement and 24 cm smaller than the hip measurement. Furthermore, if the ratio of waist to hip measurement is close to 0.618, it is considered the golden ratio, a dream for many women.
### 3. Formula for calculating the hip measurement (Vòng 3)
Hip = Waist / 0.618 cm
The hip measurement, also known as the third measurement, represents the circumference of the hips. Normally, the ideal hip measurement desired by women is 4 cm larger than the bust measurement and 24 cm larger than the waist measurement. Achieving the perfect hip measurement gives you the confidence to wear body-hugging clothes. However, besides the measurements you can calculate, having a beautiful hip also depends on many other factors such as firmness, fullness, symmetry, and softness.
## Standard Measurements for Women Based on Average Height
Every body is unique, but there are average measurements based on height and age. You can use these as a reference for your own body.
### 1. Measurements based on height
The three measurements are influenced by several factors, with height playing a significant role. Generally, the measurements tend to increase with height and weight.
#### a) Measurements for women with a height of 1.50m
In Vietnam, it is not uncommon to find women with a height of 1.50m. Here are the average measurements for women of this height:
• Bust: 77 cm
• Waist: 53 cm
• Hip: 81 cm
#### b) Measurements for women with a height of 1.55m
For women with a height of 1.55m, having perfect measurements allows you to confidently wear any fashion style. Keep in mind that fashion celebrates the beauty of women, so choose items that fit your personal style. The average measurements for women of this height are as follows:
• Bust: 79.5 cm
• Waist: 55.5 cm
• Hip: 83.5 cm
#### c) Measurements for women with a height of 1.60m
Women with a height of 1.60m have many advantages when it comes to selecting clothes. To choose the perfect fitting outfit for your body, you can apply the following calculations:
• Bust: 82 cm
• Waist: 58 cm
• Hip: 86 cm
#### d) Measurements for women with a height of 1.65m
Women with a height of 1.65m can also achieve the standard measurements. The calculations for the average measurements are as follows:
• Bust: 84.5 cm
• Waist: 60.5 cm
• Hip: 88.5 cm
#### e) Measurements for women with a height of 1.70m
Some women may not be able to achieve the standard measurements for the three body proportions due to improper eating habits and lack of exercise. However, if a woman who is 1.70m tall maintains good habits, she can achieve the following average measurements:
• Bust: 87 cm
• Waist: 63 cm
• Hip: 91 cm
#### f) Measurements for women with a height of 1.75m
Similarly, according to the same standard measurement formulas for women based on height mentioned above, women who are 1.75m tall can achieve the following average measurements:
• Bust: 89.5 cm
• Waist: 65.5 cm
• Hip: 93.5 cm
### 2. Measurements based on age
Apart from height, the standard measurements also vary based on age. Generally, we estimate the three measurements according to different stages of body development to determine if you have reached your ideal body shape. Here are some suggestions for the standard measurements based on age:
#### a) Measurements for 12-year-old girls
Girls at the age of 12 are going through puberty, and compared to boys, girls tend to mature earlier. According to statistics, the average height for 12-year-old girls in Vietnam is 149.8 cm. Based on this, the average measurements for girls of this age are as follows:
• Bust: 76.9 cm
• Waist: 52.9 cm
• Hip: 78.9 cm
#### b) Measurements for 13-year-old girls
13-year-old girls have an average height of 156.7 cm. Based on this, the average measurements for girls of this age are as follows:
• Bust: 80.35 cm
• Waist: 56.35 cm
• Hip: 84.35 cm
#### c) Measurements for 14-year-old girls
After turning 14, the average height for girls remains relatively the same. According to statistics, 14-year-old girls in Vietnam have an average height of 158.7 cm. Based on this, the average measurements for girls of this age are as follows:
• Bust: 81.35 cm
• Waist: 57.35 cm
• Hip: 85.35 cm
#### d) Measurements for 15-year-old girls
Continuing with the trend, 15-year-old girls in Vietnam have an average height of 159.7 cm. Based on this, the average measurements for girls of this age are as follows:
• Bust: 82.35 cm
• Waist: 58.35 cm
• Hip: 86.35 cm
#### e) Measurements for 16-year-old girls
At the age of 16, the average height for girls in Vietnam is 162.5 cm. Based on this, the average measurements for girls of this age are as follows:
• Bust: 83.25 cm
• Waist: 59.25 cm
• Hip: 87.25 cm
## The Secret to Achieving the Standard Measurements
To achieve the standard measurements, it’s essential to follow a few principles:
• Follow a balanced diet: Include all four food groups, consisting of starchy foods (mainly grains), protein (meat, fish, eggs, milk, legumes, etc.), fat (animal fats, vegetable oils), and vitamins, minerals, and dietary fiber (vegetables, fruits, etc.).
• Manage weight: Avoid excessive consumption of fried foods, fast food, and sweets, as they can contribute to weight gain beyond the healthy range.
• Limit intake of caffeinated drinks and alcohol, as they can greatly damage your body.
• Maintain a consistent and sufficient sleep routine. When you sleep, your body rests and rejuvenates, helping you achieve your ideal height and enhancing your metabolism.
• Engage in exercises that suit your health and body characteristics for weight loss and height increase.
Lastly, remember to regularly measure your own standard measurements to have a better understanding of yourself. Taking care of yourself and loving yourself is the key to a happier life, starting with the little things.
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Check Also
Close | 1,914 | 8,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.93571 |
https://essaylux.com/homework/590003/ | 1,620,787,284,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00195.warc.gz | 269,805,133 | 10,040 | ## (Solution Download) The probability of rolling an odd value followed by an
The probability of rolling an odd value followed by an even value.
Consider again the three-sided die that gives scores of 1, 2, or 3, each with probability 1/3. Suppose that the results of rolls are independent. Use the multiplication rule to find the above probability.
For Information: (Section 6.4, Exercises 9-12)
STATUS | 94 | 405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-21 | latest | en | 0.893676 |
minecraft.yxzoo.com | 1,590,413,364,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00031.warc.gz | 76,365,324 | 10,794 | # 老鸟教你用指令实现BOSS血条指向效果
今天小编为大家带来我的世界自定义BOSS血条实现指向效果教程，想学的玩家别错过下面的精彩内容！
从图中我们可以看到当玩家指向盔甲架的时候，TARGET这个bossbar满了。当玩家朝向和玩家指向盔甲架的向量的夹角θ超过90°时，TARGET变红，并且夹角θ越大，TARGET的值越大。过程不受抬头/低头影响。（gif中由于鼠标抖动而造成有一点不稳定）
看着挺高端，其实原理超级简单。就是最基础的Vector Math：点积（dot product）
我们先上命令：
init.mcfunction [LOAD]
# Initialization
# Create bossbar
bossbar create bossbar:target "TARGET"
# Player's location (X,Z)
scoreboard objectives add pos_p_x dummy
scoreboard objectives add pos_p_z dummy
# Target's location (X,Z)
scoreboard objectives add pos_t_x dummy
scoreboard objectives add pos_t_z dummy
# Player-facing vector (X,Z)
scoreboard objectives add vec_yw_x dummy
scoreboard objectives add vec_yw_z dummy
# Target-to-player vector (X,Z)
scoreboard objectives add vec_nvt_x dummy
scoreboard objectives add vec_nvt_z dummy
# Current dot product
scoreboard objectives add int_dp_vt dummy
# Max dot product
scoreboard objectives add int_dp_mx dummy
# Caclation output flag (to current or to max)
scoreboard objectives add con_mx_bool dummy
# Temp score when calculating abs
scoreboard objectives add tmp_abs_swp dummy
bossbar_navigation.mcfunction [TICK]
# Calculation
# Set storage flag(to max or current)
# Max
scoreboard players set @a con_mx_bool 1
# Calculate
execute as @a at @s facing entity @e[tag=target,sort=nearest,limit=1] eyes run function bossbar:bossbar_dotproduct_calculation
# Store max
execute as @a store result bossbar bossbar:target max run scoreboard players get @s int_dp_mx
# Current
scoreboard players set @a con_mx_bool 0
# Calculate
execute as @a at @s run function bossbar:bossbar_dotproduct_calculation
# Set display
# Current equals or is greater than 0
execute as @a[scores={int_dp_vt=0..}] run function bossbar:bossbar_positive_display
# Current is less than 0
execute as @a[scores={int_dp_vt=..-1}] run function bossbar:bossbar_minus_display
# Glow the target so that player can identify them easily
execute as @a at @s run effect give @e[tag=target,sort=nearest,limit=1] minecraft:glowing 1
bossbar_dotproduct_calculation.mcfunction
# Reset Scoreboards
# Current case
execute if entity @s[scores={con_mx_bool=0}] run scoreboard players set @s int_dp_vt 0
# Max case
execute if entity @s[scores={con_mx_bool=1}] run scoreboard players set @s int_dp_mx 0
# Get player's location (X,Z)
execute store result score @s pos_p_x run data get entity @s Pos[0] 100
execute store result score @s pos_p_z run data get entity @s Pos[2] 100
# Get player's facing vector(have not been normalized)
# Summon length-marker
execute if entity @s[scores={con_mx_bool=0}] rotated ~ 0 run summon minecraft:area_effect_cloud ^ ^ ^1 {CustomName:"["p_dir"]"}
# Get length-marker's location
execute store result score @s pos_t_x run data get entity @e[name=p_dir,sort=nearest,limit=1] Pos[0] 100
execute store result score @s pos_t_z run data get entity @e[name=p_dir,sort=nearest,limit=1] Pos[2] 100
# Calculate the vector
scoreboard players operation @s vec_yw_x = @s pos_t_x
scoreboard players operation @s vec_yw_z = @s pos_t_z
execute run scoreboard players operation @s vec_yw_x -= @s pos_p_x
execute run scoreboard players operation @s vec_yw_z -= @s pos_p_z
# Get target-to-player vector
# Get target's location
# Summon plainizer(X,Z)
execute run summon minecraft:area_effect_cloud ~ ~ ~ {CustomName:"["plainizer"]"}
# Correct the plainizer's location(Tx,Py,Tz)
execute store result entity @e[name=plainizer,sort=nearest,limit=1] Pos[0] double 1 run data get entity @e[tag=target,sort=nearest,limit=1] Pos[0]
execute store result entity @e[name=plainizer,sort=nearest,limit=1] Pos[1] double 1 run data get entity @s Pos[1]
execute store result entity @e[name=plainizer,sort=nearest,limit=1] Pos[2] double 1 run data get entity @e[tag=target,sort=nearest,limit=1] Pos[2]
# Summon length-marker
execute facing entity @e[name=plainizer,sort=nearest,limit=1] eyes run summon minecraft:area_effect_cloud ^ ^ ^1 {CustomName:"["tp_dir"]"}
# Get length-marker's location
execute store result score @s pos_t_x run data get entity @e[name=tp_dir,sort=nearest,limit=1] Pos[0] 100
execute store result score @s pos_t_z run data get entity @e[name=tp_dir,sort=nearest,limit=1] Pos[2] 100
# Calculate the vector
scoreboard players operation @s vec_nvt_x = @s pos_t_x
scoreboard players operation @s vec_nvt_z = @s pos_t_z
execute run scoreboard players operation @s vec_nvt_x -= @s pos_p_x
execute run scoreboard players operation @s vec_nvt_z -= @s pos_p_z
# Max case
execute if entity @s[scores={con_mx_bool=1}] run scoreboard players operation @s vec_yw_x = @s vec_nvt_x
execute if entity @s[scores={con_mx_bool=1}] run scoreboard players operation @s vec_yw_z = @s vec_nvt_z
# Dot Product
# Multiply the x and z of direction-vector to the x and z of target-to-player vector
execute run scoreboard players operation @s vec_nvt_x *= @s vec_yw_x
execute run scoreboard players operation @s vec_nvt_z *= @s vec_yw_z
# Add the result to output-scoreboards
# Current case
execute if entity @s[scores={con_mx_bool=0}] run scoreboard players operation @s int_dp_vt += @s vec_nvt_x
execute if entity @s[scores={con_mx_bool=0}] run scoreboard players operation @s int_dp_vt += @s vec_nvt_z
# Max case
execute if entity @s[scores={con_mx_bool=1}] run scoreboard players operation @s int_dp_mx += @s vec_nvt_x
execute if entity @s[scores={con_mx_bool=1}] run scoreboard players operation @s int_dp_mx += @s vec_nvt_z
※ 暂未想到办法简化第二个plainizer...感谢玄素dalao提供rotated思路
bossbar_positive_display.mcfunction
# Set color
bossbar set bossbar:target color white
# Write result to bossbar
execute store result bossbar bossbar:target value run scoreboard players get @s int_dp_vt
bossbar_minus_display.mcfunction
# Set color
bossbar set bossbar:target color red
# Get |int_dp_vt| by |int_dp_vt| = -int_dp_vt(int_dp_vt < 0) = int_dp_vt - 2 * int_dp_vt
scoreboard players operation @s tmp_abs_swp = @s int_dp_vt
scoreboard players operation @s int_dp_vt -= @s tmp_abs_swp
scoreboard players operation @s int_dp_vt -= @s tmp_abs_swp
# Write result to bossbar
execute store result bossbar bossbar:target value run scoreboard players get @s int_dp_vt
※ 以上所有文件都存在于bossbar命名空间内。请根据需要自行修改
※ [LOAD]为load.json内的function [TICK]为tick.json内的function
※ 注意，尽管我使用了@a，但是这个命令组只支持一个玩家。因为bossbar只有一个。
• 游戏大礼包
• 手游开测表
• 本月
• 最新
+更多
+更多 | 1,933 | 6,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-24 | latest | en | 0.476924 |
https://betterlesson.com/lesson/resource/2603122/displaying-greater-than-or-less-than-expressions-m4v | 1,511,237,931,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00314.warc.gz | 589,032,411 | 21,968 | ## Displaying Greater Than or Less Than Expressions.m4v - Section 1: Warm Up
Displaying Greater Than or Less Than Expressions.m4v
# Skiing or Boarding?
Unit 7: Data and Analysis
Lesson 1 of 8
## Objective: SWBAT record answers to a survey question. SWBAT make a representation to communicate the results of a survey. SWBAT use equations to show that the sum of responses in each category equals the total number of responses collected.
### Thomas Young
203 Lessons14 new
## Big Idea: Next week marks the start of our Skiing and Snowboarding program at the Waitsfield School. Today the students will find out what each classmate would prefer to do.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
65 minutes
### Thomas Young
203 Lessons | 14 new
##### Similar Lessons
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Environment: Urban | 305 | 1,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-47 | latest | en | 0.851038 |
http://www.earthsciweek.org/classroom-activities/earth%E2%80%99s-magnetic-field | 1,576,034,864,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529745.80/warc/CC-MAIN-20191211021635-20191211045635-00245.warc.gz | 181,734,330 | 9,167 | Earth’s Magnetic Field
Activity Source:
If you found yourself in the middle of the ocean, do you think you could find your way home? Sea turtles can, thanks in part to Earth’s magnetic field!
All around the surface of the Earth, there is a magnetic field generated by the planet’s large metallic core. The magnetic field varies in strength and direction from place to place, giving different locations unique magnetic “maps.” Sea turtles (and other animals like pigeons, lobsters, and honeybees) can detect and remember patterns in Earth’s magnetic field to find their way over long distances, a behavior called magnetic homing.
How can you navigate like a sea turtle? Find out by making a turtle that can detect magnetic fields and using it to create a magnetic map of its own habitat.
Materials
• Cereal box and a piece of paper
• Scissors
• Tape
• 5-10 refrigerator magnets of varying strengths
• String
• Bolt, screw, or nail that is ferromagnetic (sticks to magnets)
• Modeling clay or play dough
• Pencil
Procedure
1. Tie a string to a nail or bolt.
2. Press a small piece of clay onto the bolt, and carefully sculpt it into the shape of a turtle, making sure to keep the bolt on the bottom and the string on the top.
3. Test your turtle to make sure that it hangs easily and turns towards magnets when you hold it by the string. Watch how it moves when it gets close to and touches a magnet.
4. Cut a cereal box in half along its longest sides, and tape the corners of one half so it holds its shape when it is set flat on a table. Tape a piece of blank paper to the top of the box and have a friend hide a few magnets of various strengths underneath the box. This is your turtle habitat.
5. Gently suspend a turtle above the turtle habitat by its string, moving it back and forth until you notice it move in response to one of the magnets. Draw a small dot where you think a magnet is, then draw a larger circle around the entire area where your turtle is able detect the magnet.
6. When you think you have finished, carefully remove the tape holding the turtle habitat to the table. Lift the turtle habitat up to look at the magnet positions. Compare your map to the positions of the magnet. How similar are they? How would your map differ if your magnets were in different positions?
7. Search for a magnetic map of a coastline or major ocean. How do the magnetic field lines differ when you compare places along the Atlantic or Pacific coast? What about along the equator compared to the poles? Use the magnetic map you found to answer this question: Where do you think it would be easier for organisms like sea turtles to use magnetic homing, and where might it be harder? Why?
NGSS Connections
• Science and Engineering Practices — Analyzing and Interpreting Data
• Disciplinary Core Ideas — Motion and Stability: Forces and Interactions
• Crosscutting Concepts — Cause and Effect | 628 | 2,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-51 | longest | en | 0.920597 |
https://www.snip2code.com/User/Profile/14735/Laurence-Tamatea | 1,576,341,253,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00156.warc.gz | 852,930,107 | 9,110 | LaurenceTamatea
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Laurence Tamatea
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## Recent Snippets See all snippets by LaurenceTamatea
### Square Root Function
Square root of the sum of four parameters
``` <!DOCTYPE html>
<html>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
<body>
<script>
function simpleMath(a,b,c,d) {
return(Math.sqrt) (4+5+6+7);
}
var q = simpleMath();
</script>
</body>
</html> ```
;
### Variable Declaration After Function
``` <script>
function sumFunction (a, b) {
return (a+b);
}
var x = sumFunction (4, 2);
document.write (x);
</script>
```
;
### Variable Declaration Before Function
```<script>
var sum = addNumbers (23,2342,23424);
document.write (sum);
function addNumbers (x,y,z) {
var result = x*y*z;
return result;
}
</script>
// or //
<script>
var result = addNumbers (2,3,4);
function addNumbers (a, b, c) {
return (a + b + c);
}
</script> ```
;
### Object Initialisers
``` <!DOCTYPE html>
<html lang="en">
<title>STANDARD TEMPLATE</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, user-scalable=no, minimum-scale=1.0, maximum-scale=1.0">
<style>
</style>
<script>
athletes = {
type: "runners",
country: "American"
//0bject = {property:value, property:value}//
//can have unlimited properties//
};
food = {
type: "dairy",
engergy: "high"
};
</script>
<body>
<div>Standard Index Template</div>
<script>
/*object initializer is faster than a constructor function because it
only uses one line of code*/
/*use when creating one or two objects of the same type. If creating 500
person objects for example, then use the constructor function*/
document.write(athletes.country + athletes.type + " eat " + food.type + " for breakfast. ");
</script>
</body>
</html>
```
;
### Create New Instance of an Object Using This.
```<!DOCTYPE html>
<html lang="en">
<title>STANDARD TEMPLATE</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, user-scalable=no, minimum-scale=1.0, maximum-scale=1.0">
<style>
</style>
<script>
function trucks(brand, color) {
this.brand = brand;
this.color = color;
//we are working with this curent object//
//constructor function, is a blueprint for a function//
}
var mechanics = new trucks("Mac", "Blue");
var performance = new trucks("Mercedes Benz", "Silver");
//creating a new instance of an object (unlimited) with two properties//
//variable can be an object too//
</script>
<body>
<div>Standard Index Template</div>
<script>
document.write(mechanics.brand);
document.write(performance.color);
//name of the object and a property that we want to access//
</script>
</body>
</html> ```
; | 838 | 3,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-51 | longest | en | 0.411761 |
http://www.qsstudy.com/chemistry/thermo-chemical-equations | 1,518,975,685,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812247.50/warc/CC-MAIN-20180218173208-20180218193208-00462.warc.gz | 519,571,708 | 8,173 | Thermo-chemical Equations - QS Study
A thermo-chemical equation is a chemical equation for a reaction in which the enthalpy of reaction for the molar amounts stated is written directly after the equation.
In a thermo-chemical equation, it is important to note state symbols (solid, liquid or gas) because the enthalpy change, ΔH, depends on the phase of the substances.
Manipulating Thermo-chemical Equations: The following are two important rules for manipulating thermo-chemical equations:
i. When a chemical equation is reversed, the value of ΔH is reversed in sign. For example,
Sr(s) + ½ O2 (g) → SrO(s) ΔH = -592 kJmol-1
SrO(s) → Sr(s) + ½ O2 (g) ΔH = +592 kJmol-1
ii. When a thermo-chemical equation is multiplied by any factor, the value of AH for the new equation is obtained by multiplying the AH in the original equation by that same factor. For example,
Sr(s) + ½ O2 (g) → SrO(s) ΔH = -592 kJmol-1
2 Sr(s) + O2 (g) → 2 SrO(s) →H = -1184 kJmo1-1 | 277 | 965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-09 | longest | en | 0.870434 |
https://goprep.co/ex-6.3-q24-find-the-value-of-5x-6-x-1-5x-2y-3-x-12xy-2-when-i-1njmrn | 1,618,639,468,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00596.warc.gz | 391,631,947 | 33,685 | # Express each of t
5 × -1.5 × -12 × x6 × x2 × x × y3 × y2
= 90 × x9 × y5
= 90x9y5
Verification:
x = 1 and y = 0.5
R.H.S = 90x9y5
= 90 (1)9 (05)5
= 2.8125
L.H.S = 2.8125
Therefore,
L.H.S = R.H.S
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Find each of the RS Aggarwal - Mathematics | 293 | 979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-17 | latest | en | 0.748627 |
http://mathhelpforum.com/pre-calculus/2993-someone-help.html | 1,526,998,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864790.28/warc/CC-MAIN-20180522131652-20180522151652-00599.warc.gz | 185,814,752 | 8,464 | 1. ## someone help!
The equation 2k³ - 4k^(3/2) + 1 = 0
Subsitute p² for k³ to find the value of k.
i got 2p²-4p + 1, and it doesnt factorise to give k=0.44, which is at the back of the book!
i even used the discriminant thing.
2. Originally Posted by ify00
The equation 2k³ - 4k^(3/2) + 1 = 0
Subsitute p² for k³ to find the value of k.
i got 2p²-4p + 1, and it doesnt factorise to give k=0.44, which is at the back of the book!
i even used the discriminant thing.
Hello,
keep in mind:
$\displaystyle p^2=k^3 \Longrightarrow k=\sqrt[3]{p^2}$
$\displaystyle 2p^2-4p+1=0\Longleftrightarrow p=1+{1\over2}\sqrt{2}\ \vee \ p=1-{1\over2}\sqrt{2}$
Now plug in these values of p into the equation to calculate k and you'll get:
$\displaystyle k=\sqrt[3]{\left( 1+{1\over2}\sqrt{2} \right)^2} \approx 1.4283...$ or
$\displaystyle k=\sqrt[3]{\left( 1-{1\over2}\sqrt{2} \right)^2} \approx .4410...$
Greetings
EB | 352 | 913 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-22 | latest | en | 0.768649 |
http://stackoverflow.com/questions/6385700/calculating-a-histogram-on-a-streaming-data-online-histogram-calculation?answertab=votes | 1,417,045,166,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931007607.9/warc/CC-MAIN-20141125155647-00138-ip-10-235-23-156.ec2.internal.warc.gz | 283,177,957 | 18,596 | # Calculating a histogram on a streaming data - Online histogram calculation
I am looking for an algorithm to generate a histogram over a large amount of streaming data, the max and min are not known in advance but standard deviation and mean are in a particular range.
Cheers,
-
what's an approximate histogram? – CharlesB Jun 17 '11 at 12:33
I meant I don't want to have the exact histogram (the number of elements in each bucket doesn't need to be exact). – Ali Jun 18 '11 at 3:07
– mtrw Jun 18 '11 at 5:16
The propose python solution, requires knowing min/max in advance, or one pass through the data, if min/max changes the whole histogram has to be re calculated as bins change, this is not useful for online histogram calculation. Thanks for the suggestion anyway. Please see my answer below. – Ali Jun 19 '11 at 1:04
Doesn't this question belong in cs.stackexchange.com ? – einpoklum Dec 14 '13 at 11:58
Standard deviation and mean do not matter for a histogram. Simply choose your resolution and draw a bar as high as you have hits for its range. This will, of course, get more expensive with a higher resolution. You can try adjusting the resolution by trying to fit the existing data into a normal curve (or whatever model you like) and finding the standard deviation to choose a reasonable granularity.
Edit: Read it wrong the first time around. If you know the approximate standard deviation, you can choose reasonable sizes for your histogram groups from the get-go. Just compare every new entry to your current min and max and adjust your range accordingly.
-
I just found one solution. Sec. 2.2 of "On-line histogram building from A streaming parallel decision tree algorithm" paper. The algo is implemented by NumericHistogram class in Hive project :
A generic, re-usable histogram class that supports partial aggregations. The algorithm is a heuristic adapted from the following paper: Yael Ben-Haim and Elad Tom-Tov, "A streaming parallel decision tree algorithm", J. Machine Learning Research 11 (2010), pp. 849--872. Although there are no approximation guarantees, it appears to work well with adequate data and a large (e.g., 20-80) number of histogram bins.
-
I use a package called "GoHistogram" which provides two streaming approximattion histograms (NumericHistogram and Weighted Numeric Histogram). It is implemented in Golang (https://code.google.com). Here is the link:
https://github.com/VividCortex/gohistogram
- | 553 | 2,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-49 | longest | en | 0.903619 |
https://metanumbers.com/559020 | 1,624,314,019,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504838.98/warc/CC-MAIN-20210621212241-20210622002241-00316.warc.gz | 334,110,598 | 11,433 | ## 559020
559,020 (five hundred fifty-nine thousand twenty) is an even six-digits composite number following 559019 and preceding 559021. In scientific notation, it is written as 5.5902 × 105. The sum of its digits is 21. It has a total of 8 prime factors and 96 positive divisors. There are 116,160 positive integers (up to 559020) that are relatively prime to 559020.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 21
• Digital Root 3
## Name
Short name 559 thousand 20 five hundred fifty-nine thousand twenty
## Notation
Scientific notation 5.5902 × 105 559.02 × 103
## Prime Factorization of 559020
Prime Factorization 22 × 3 × 5 × 7 × 113
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 8 Total number of prime factors rad(n) 2310 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 559,020 is 22 × 3 × 5 × 7 × 113. Since it has a total of 8 prime factors, 559,020 is a composite number.
## Divisors of 559020
96 divisors
Even divisors 64 32 16 16
Total Divisors Sum of Divisors Aliquot Sum τ(n) 96 Total number of the positive divisors of n σ(n) 1.96762e+06 Sum of all the positive divisors of n s(n) 1.4086e+06 Sum of the proper positive divisors of n A(n) 20496 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 747.676 Returns the nth root of the product of n divisors H(n) 27.2746 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 559,020 can be divided by 96 positive divisors (out of which 64 are even, and 32 are odd). The sum of these divisors (counting 559,020) is 1,967,616, the average is 20,496.
## Other Arithmetic Functions (n = 559020)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 116160 Total number of positive integers not greater than n that are coprime to n λ(n) 7260 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45892 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 116,160 positive integers (less than 559,020) that are coprime with 559,020. And there are approximately 45,892 prime numbers less than or equal to 559,020.
## Divisibility of 559020
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 0 4 3
The number 559,020 is divisible by 2, 3, 4, 5, 6 and 7.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (559020)
Base System Value
2 Binary 10001000011110101100
3 Ternary 1001101211110
4 Quaternary 2020132230
5 Quinary 120342040
6 Senary 15552020
8 Octal 2103654
10 Decimal 559020
12 Duodecimal 22b610
20 Vigesimal 39hb0
36 Base36 bzcc
## Basic calculations (n = 559020)
### Multiplication
n×i
n×2 1118040 1677060 2236080 2795100
### Division
ni
n⁄2 279510 186340 139755 111804
### Exponentiation
ni
n2 312503360400 174695628530808000 97658350261292288160000 54592970963067614927203200000
### Nth Root
i√n
2√n 747.676 82.3776 27.3437 14.1087
## 559020 as geometric shapes
### Circle
Diameter 1.11804e+06 3.51243e+06 9.81758e+11
### Sphere
Volume 7.31763e+17 3.92703e+12 3.51243e+06
### Square
Length = n
Perimeter 2.23608e+06 3.12503e+11 790574
### Cube
Length = n
Surface area 1.87502e+12 1.74696e+17 968251
### Equilateral Triangle
Length = n
Perimeter 1.67706e+06 1.35318e+11 484126
### Triangular Pyramid
Length = n
Surface area 5.41272e+11 2.05881e+16 456438 | 1,293 | 3,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-25 | latest | en | 0.807742 |
https://www.assignmentexpert.com/homework-answers/mathematics/algebra/question-8421 | 1,585,718,099,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505366.8/warc/CC-MAIN-20200401034127-20200401064127-00313.warc.gz | 830,780,062 | 84,842 | 84 505
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# Answer to Question #8421 in Algebra for Shreyah mishra
Question #8421
If (x + 3) whole squared is a factor of f(x) = 3 * x cubed + k * x + 6, then find the remainder obtained when f(x) is divided by x - 6
1
2012-04-13T11:58:47-0400
Let's use the method of undetermined coefficients to find k:
3x³ + kx + 6 = (ax + b)(x + 3)²
3x³ + kx + 6 = (ax + b)(x² + 6x + 9)
3x³ + kx + 6 = ax³ + (6a + b)x² + (9a + 6b)x + 9b
a = 3
0 = 6a + b
k = 9a + 6b
6 = 9b
a = 3
0 = 6*3 + b ==> b = -18
k = 9a + 6b
6 = 9b ==> b = 2/3
Statements b = -18 and b = 2/3 are inconsistent, so (x + 3)² cannot be a factor of 3x³ + kx + 6.
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for any assignment or question with DETAILED EXPLANATIONS! | 378 | 879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-16 | latest | en | 0.820174 |
https://www.codespeedy.com/check-for-symmetric-binary-tree-iterative-approach-in-java/ | 1,674,996,054,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00626.warc.gz | 733,440,316 | 15,136 | Check for Symmetric Binary Tree (Iterative Approach) in Java
In this tutorial, will see how we can check for a Symmetric Binary Tree in Java. A Symmetric Binary Tree is a tree in which a tree is a mirror of itself. We will be doing this code without recursion.
For Example: –
``` 9
/ \
5 5
/ \ / \
2 1 1 2
```
This is an example of a symmetric binary tree.
In this, we will be using a queue. So, in this we see each level behaves like a palindrome. As we can see, the left child of our left subtree is equal to the right child of the right subtree and vice-versa. So we will be inserting these into the queue and check whether they are equal or not.
Code: –
```import java.util.* ;
public class Main
{
Node r;
static class Node
{
int data;
Node left;
Node right;
Node(int data)
{
this.data = data;
left = null;
right = null;
}
}
Main(Node r){
this.r = r;
}
public boolean symmetric(Node root)
{
//Add null elements to the queue
while (!queue.isEmpty())
{
// Remove and check last 2 nodes
Node tleft = queue.remove();
Node tright = queue.remove();
// both null check more
if (tleft==null && tright==null) {
continue;
}
// If any of the one is null then false
if ((tleft==null && tright!=null) || (tleft!=null && tright==null)) {
return false;
}
// If both not equal false
if (tleft.data != tright.data) {
return false;
}
queue.add(tleft.right); // Than right child left subtree
queue.add(tright.left); // At last left child right subtree
}
return true;
}
public static void main(String[] args)
{
Node node = new Node(9);
Main tree = new Main(node);
tree.r.left = new Node(5);
tree.r.right = new Node(5);
tree.r.left.left = new Node(2);
tree.r.left.right = new Node(1);
tree.r.right.left = new Node(1);
tree.r.right.right = new Node(2);
if (tree.symmetric(tree.r)) {
System.out.println("Symmetric");
}
else{
System.out.println("Not Symmetric");
}
}
}```
Output: –
`Symmetric`
This is how we check for a symmetric binary tree with Java. | 527 | 1,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-06 | latest | en | 0.497364 |
https://community.ptc.com/t5/PTC-Mathcad/Using-a-range-variable-to-repeat-the-same-calc-many-times/m-p/123739 | 1,632,227,592,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00403.warc.gz | 235,337,168 | 37,958 | cancel
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1-Newbie
Using a range variable to repeat the same calc many times
The code that I am working with is shown below.
I want to eventually make the matrix for the variable "Reactions(Conc)", extremely large (~1400 rows, 1 column), but I'm working with only 2 rows for now to learn how to code in the array mathematics. The calcs below are working fine, and are reporting the correct answers, but I don't want to type the Greek capital PI and the arguements for it 1400 times, so I'm trying to use the range variable i2 in place of the 0 and the 1 in the first and second rows of "Reactions(Conc)", and then I want to only enter the capital PI statement once, and just have the range variable repeat it for all ~1400 rows. This isn't working for me though. When I sub in the i2 where the 0 or 1 numbers are right no, it does not continue to give me the nice 1 X 2 matrix with the Conc variables that is shown here. It just shows up in red, doesn't symbolically evaluate it, and says "pattern match exception." Is there a different way to use this range variable so that it will take care of repeating the rows of "Reactions(Conc)" for me?
5 REPLIES 5
24-Ruby I
(To:ptc-3031221)
I think it will be better use programming tool (the for loop) for this task.
21-Topaz I
(To:ptc-3031221)
You can't really use an externally defined range variable inside a function definition (a type of program).
A range variable simply holds the range itself (ie, a,b..c). The main worksheet itself sees a range variable as an instruction to iterate over the range definition's expansion. A program, however, sees it simply as the value 'a,b..c', which is not a number and, therefore, use it as an exponent. However, a function (or program) can deal with it in a for loop, as can be seen from the example at the top of the attached worksheet. It's usually good programming practice to pass a range variable as an argument rather than as global, so I've shown two alternative choices for dealing with what I think(?) may be your need, and I'd recommend using the latter.
Stuart
PS. Is there some particular reason you want to use the symbolics? If not, then I'd probably stick to the standard numerics.
1-Newbie
(To:StuartBruff)
Thanks for the advice. The reason why I'm using symbolics is because I'm using these lines of code to generate a long list of equations that I'm going to feed into a differential equation solver. This is going to work now!
15-Moonstone
(To:ptc-3031221)
Perhaps you're looking for something like this:
Alan
1-Newbie
(To:AlanStevens)
That code works for me. Thanks for your help!
Announcements
Check out the latest | 671 | 2,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.937052 |
https://www.freelists.org/post/lit-ideas/Try-a-Logic-Problem | 1,582,558,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00225.warc.gz | 720,366,588 | 12,721 | [lit-ideas] Re: Try a Logic Problem
• From: wokshevs@xxxxxx
• To: lit-ideas@xxxxxxxxxxxxx
• Date: Thu, 22 Jun 2006 17:45:52 -0230
```I had a hell of a time getting this message through to the List due to our
university spam filter. I think the words "bored and lonely" caught it's eye.
Trying once again. Cheers,
> > Here, as promised, for all the bored and lonely souls, is a problem I
> > offer my students. I often get e-mail
> > requests years after the course from students who have forgotten the
> > solution and desperately need it for their Saturday evening escapades and
> > reflections. I am informed that some have made quick money in pubs with it
> > and/or gotten lucky in some other manner. However, I ask for no
> > commission.
> >
> >
> > A guard tells his 3 prisoners: "Gentlemen, I have here 3 white hats and 2
> > red hats. I shall place one hat on each of your heads, but you won't know
> > which hat is on your own head. Whoever among you can tell me the colour of
> > the hat he's wearing, will be granted an amnesty." The prisoners
> > immediately agree. So the guard places one hat on each of the 3
> > prisoners' heads and puts the other 2 away. None of the inmates know
> > which hats were put away and none of the inmates has seen which colored
> > hat was placed on his own head.
> >
> > The guard now asks prisoner #1 "Can you tell me the colour of the hat
> > you're wearing. The prisoner responds "No, I cannot."
> >
> > The guard then asks prisoner #2 the same question and he receives the same
> >
> > Finally, the guard asks prisoner #3 and prisoner #3 replies that he does
> > indeed know the colour of the hat he's wearing.
> >
> > The problem, should you decide to accept it, is: what is the colour of the
> > hat #3 is wearing, and more importantly, in logic as in love, how did he
> > figure it out?
> >
> > I forgot to mention one small detail here. Prisoner #3 is completely
> > blind. This is a problem in strict logic, nothing up my sleeves, Rocky.
> >
> > Enjoy!
> >
> > We of coure need a prize for the winning solution, should there be one.
> > Suggestions? Solutions may be judged communally (i.e., David tells John
> > his solution fails at step 5 and why.) This is beginning to look like a
> > Web-CT course.
> >
> > Walter O.
> > Department of Logic and Long Gun Registery
> > Ministry of Tourism and Recreation
> > Ottawa, Ontario
> >
>
------------------------------------------------------------------
To change your Lit-Ideas settings (subscribe/unsub, vacation on/off,
digest on/off), visit www.andreas.com/faq-lit-ideas.html
``` | 685 | 2,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-10 | latest | en | 0.958394 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=76&t=26703 | 1,604,110,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00086.warc.gz | 358,130,213 | 11,801 | ## Question 8.101 [ENDORSED]
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am
### Question 8.101
A technician carries out the reaction 2SO2(g) + O2(g) 2SO3(g) at 25°C
and 1.00 atm in a constant-pressure cylinder fitted with a piston. Initially,
0.0300 mol SO2 and 0.0300 mol O2 are present in the cylinder. The
technician then adds a catalyst to initiate the reaction.
a. calculate the volume of the cylinder containing the reactant gases before the reactions begin
b. what is the limiting reactant
c. assuming that the reaction goes to completion and that the temperature and pressure of the reaction remain constant, what is the final volume of the cylinder
d. how much work takes place and is it done by the system or on the system
e. how much enthalpy is exchanged and does it leave or enter the system
f. calculate the change in internal energy for the reaction
Ive worked through parts a-c and specifically need help?/and explanation for the last three parts
Chem_Mod
Posts: 18739
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 635 times
### Re: Question 8.101 [ENDORSED]
D. Because this is a constant external pressure problem, you will use the equation
$w=-P\Delta V$.
The problem says it occurs at 1 atm of pressure, and our change in volume was determined in parts a and c.
$\Delta V=$=1.1L - 1.5 L = -0.4L.
Plugging it into the equation, you get:
$w=-P\Delta V=-(1.00 \textup{atm})(-0.4\textup{L})(101.325 J\cdot L^{-1}\cdot atm^{-1})=40J$
Work here is positive because it is done on the system.
E. The enthalpy of the reaction requires you to use the standard enthalpies of formation, which can be found in the back of the textbook, and the balanced equation. When you calculate it, you'll find it is -197.78 kJ/mole.
The problem states that we are using 0.030 mol of SO2, so you would use the enthalpy of the reaction (remember, the balanced equation says that we use 2 moles of SO2, meaning 2 moles produces -197.78 kJ) to determine the enthalpy exchanged.
F. To calculate the internal energy, you would use the equation:
$\Delta U=q+w$
where q is calculated in part E and w is calculated in part D. | 599 | 2,118 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-45 | latest | en | 0.895103 |
https://rustgym.com/leetcode/631 | 1,632,869,167,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00687.warc.gz | 520,286,438 | 5,002 | ## 631. Design Excel Sum Formula
Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions:
`Excel(int H, char W):` This is the constructor. The inputs represents the height and width of the Excel form. H is a positive integer, range from 1 to 26. It represents the height. W is a character range from 'A' to 'Z'. It represents that the width is the number of characters from 'A' to W. The Excel form content is represented by a height * width 2D integer array `C`, it should be initialized to zero. You should assume that the first row of `C` starts from 1, and the first column of `C` starts from 'A'.
`void Set(int row, char column, int val):` Change the value at `C(row, column)` to be val.
`int Get(int row, char column):` Return the value at `C(row, column)`.
`int Sum(int row, char column, List of Strings : numbers):` This function calculate and set the value at `C(row, column)`, where the value should be the sum of cells represented by `numbers`. This function return the sum result at `C(row, column)`. This sum formula should exist until this cell is overlapped by another value or another sum formula.
`numbers` is a list of strings that each string represent a cell or a range of cells. If the string represent a single cell, then it has the following format : `ColRow`. For example, "F7" represents the cell at (7, F).
If the string represent a range of cells, then it has the following format : `ColRow1:ColRow2`. The range will always be a rectangle, and ColRow1 represent the position of the top-left cell, and ColRow2 represents the position of the bottom-right cell.
Example 1:
```Excel(3,"C");
// construct a 3*3 2D array with all zero.
// A B C
// 1 0 0 0
// 2 0 0 0
// 3 0 0 0
Set(1, "A", 2);
// set C(1,"A") to be 2.
// A B C
// 1 2 0 0
// 2 0 0 0
// 3 0 0 0
Sum(3, "C", ["A1", "A1:B2"]);
// set C(3,"C") to be the sum of value at C(1,"A") and the values sum of the rectangle range whose top-left cell is C(1,"A") and bottom-right cell is C(2,"B"). Return 4.
// A B C
// 1 2 0 0
// 2 0 0 0
// 3 0 0 4
Set(2, "B", 2);
// set C(2,"B") to be 2. Note C(3, "C") should also be changed.
// A B C
// 1 2 0 0
// 2 0 2 0
// 3 0 0 6
```
Note:
1. You could assume that there won't be any circular sum reference. For example, A1 = sum(B1) and B1 = sum(A1).
2. The test cases are using double-quotes to represent a character.
3. Please remember to RESET your class variables declared in class Excel, as static/class variables are persisted across multiple test cases. Please see here for more details.
## Rust Solution
``````#[derive(Debug, Clone)]
enum Cell {
Value(i32),
Sum(Vec<(usize, usize)>),
}
struct Excel {
mat: Vec<Vec<Cell>>,
}
impl Excel {
fn new(h: i32, w: char) -> Self {
let n = h as usize;
let m = (w as u8 - b'A' + 1) as usize;
let mat = vec![vec![Cell::Value(0); m]; n];
Excel { mat }
}
fn set(&mut self, r: i32, c: char, v: i32) {
let i = Self::row(r);
let j = Self::col(c);
self.mat[i][j] = Cell::Value(v);
}
fn get(&self, r: i32, c: char) -> i32 {
let i = Self::row(r);
let j = Self::col(c);
self.get_by_index(i, j)
}
fn get_by_index(&self, i: usize, j: usize) -> i32 {
match &self.mat[i][j] {
Cell::Value(v) => *v,
Cell::Sum(nums) => nums.iter().map(|&(i, j)| self.get_by_index(i, j)).sum(),
}
}
fn sum(&mut self, r: i32, c: char, strs: Vec<String>) -> i32 {
let i = Self::row(r);
let j = Self::col(c);
let mut nums = vec![];
for s in strs {
let parts: Vec<&str> = s.split(':').collect();
if parts.len() == 1 {
nums.push(Self::cell(parts[0]));
} else {
let top_left = Self::cell(parts[0]);
let bottom_right = Self::cell(parts[1]);
for i in top_left.0..=bottom_right.0 {
for j in top_left.1..=bottom_right.1 {
nums.push((i, j));
}
}
}
}
self.mat[i][j] = Cell::Sum(nums);
self.get_by_index(i, j)
}
fn row(r: i32) -> usize {
(r - 1) as usize
}
fn col(c: char) -> usize {
(c as u8 - b'A') as usize
}
fn cell(s: &str) -> (usize, usize) {
let mut col = 0;
let mut row = 0;
for (i, c) in s.char_indices() {
if i == 0 {
col = (c as u8 - b'A') as usize;
} else {
row *= 10;
row += (c as u8 - b'0') as usize;
}
}
row -= 1;
(row, col)
}
}
#[test]
fn test() {
let mut obj = Excel::new(3, 'C');
obj.set(1, 'A', 2);
assert_eq!(obj.get(1, 'A'), 2);
assert_eq!(obj.sum(3, 'C', vec_string!["A1", "A1:B2"]), 4);
obj.set(2, 'B', 2);
assert_eq!(obj.get(3, 'C'), 6);
}
``````
Having problems with this solution? Click here to submit an issue on github. | 1,476 | 4,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | longest | en | 0.780672 |
https://de.mathworks.com/matlabcentral/profile/authors/6317359-michael-jarboe | 1,568,734,828,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573080.8/warc/CC-MAIN-20190917141045-20190917163045-00095.warc.gz | 435,223,303 | 20,234 | Community Profile
# Michael Jarboe
354 total contributions since 2015
Automation/Autonomous Development Engineer
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6 Monate ago | 1,336 | 5,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-39 | latest | en | 0.858942 |
http://www.mathworks.com/matlabcentral/answers/76992-how-to-iterate-it-please-help-me?requestedDomain=www.mathworks.com&nocookie=true | 1,480,856,146,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541321.31/warc/CC-MAIN-20161202170901-00268-ip-10-31-129-80.ec2.internal.warc.gz | 584,278,178 | 15,777 | on 26 May 2013
### Walter Roberson (view profile)
Hello!! I need help!! For Iteration
Here is my first iteration:
```S12t=6+2*1i;
S01t=10+5*1i;
U0=110;
U1=110;
U2=110;
R12 = 16.2+24*1i;
U0i=115;
R01= 15.75+31.2*1i;
dx01=106.125;
dx12=81;
U20=10;
e=0.01;
```
```dQ12=dx12*U2.^2*10.^-6;
S12i=S12t-complex(0,dQ12);
DS12 =((real(S12i).^2+imag(S12i).^2)/U2.^2)*R12;
S12ii=DS12+S12i;
S12e=S12ii-complex(0,dQ12);
```
```S01i=S12e+S01t;
dQ01=dx01*U1.^2*10.^-6;
S01ii=S01i-complex(0,dQ01);
DS01=((real(S01ii).^2+imag(S01ii).^2)/U1.^2)*R01;
dQ01i=dx01*U0i.^2*10.^-6;
S01ii2=S01ii+DS01;
S01iii=S01ii2-complex(0,dQ01i);
```
```dU01i=(real(S01ii2)*real(R01)+imag(S01ii2)*imag(R01))./U0i;
dU01ii=(real(S01ii2)*imag(R01)-imag(S01ii2)*real(R01))./U0i;
U1=sqrt((U0i-dU01i).^2+dU01ii.^2);
```
```dU12i=(real(S12ii)*real(R12)+imag(S12ii)*imag(R12))./U1;
dU12ii=(real(S12ii)*imag(R12)-imag(S12ii)*real(R12))./U1;
U2=sqrt((U1-dU12i).^2+dU12ii.^2);
```
And i want it continue to next iteration.
continue it if U0-(max of U1 U2)>e
U1 U2 will be changed last 2 value in next iteration
end it if U0-(max of U1 U2)<e
How can i put the if continue (while, return, end )
I can't understand I tried so many times, but still error message "usage might invalid Matlab syntax" appeared.
Just help me tell where can i put that statements If, while, continue in Long Function. Please help me!!
Thank You!!
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### Walter Roberson (view profile)
on 26 May 2013
I have not studied the code thoroughly but it appears to me that you do not assign new values to any variables you used previously. Unless at least one input variable gets a new value, then when you iterate the results would have to come out exactly the same and you would iterate forever. Something needs to change inside the loop before you can hope to have your iteration terminate.
Your general code structure will look something like this:
```Initialize your variables
```
```while true %"true" is literal here, not an example variable name
```
``` Do some calculation
break; %"break" is literal, a MATLAB command
end```
``` %you only get here if you did not break out, so if the termination
%condition did not hold```
` end %of while loop. If the termination condition did not hold, loop back`
Walter Roberson
### Walter Roberson (view profile)
on 26 May 2013
You start out with U1 and U2 being the same, so their difference is 0, which is less than e, so your loop would not run if you use that "while".
```S12t=6+2*1i;
S01t=10+5*1i;
U0=110;
U1=110;
U2=110;
R12 = 16.2+24*1i;
U0i=115;
R01= 15.75+31.2*1i;
dx01=106.125;
dx12=81;
U20=10;
e=0.01;
```
```iteration_number = 0;
```
```while true
```
``` iteration_number = iteration_number + 1;
fprintf('Starting iteration #%d\n', iteration_number);```
``` dQ12=dx12*U2.^2*10.^-6;
S12i=S12t-complex(0,dQ12);
DS12 =((real(S12i).^2+imag(S12i).^2)/U2.^2)*R12;
S12ii=DS12+S12i;
S12e=S12ii-complex(0,dQ12);```
` .... and so on ...`
``` if U0-max(U1,U2) <= e
break;
end```
```end
```
Light
### Light (view profile)
on 26 May 2013
Thank you very much!! Almost done.
But it stopped at iteration 1 unless prove that U0-U2 <= e.
It must reach iteration 3
How can i give iteration number?
And can i model it in Simulink by whileloop block?
It is embarrassing asking so many thing Sorry :-)
Walter Roberson
### Walter Roberson (view profile)
on 26 May 2013
No, the code will stop because it finds that U0-U2 <= e . Remember what I wrote earlier about you starting out U1 and U2 with the same value. Any algorithm that tries to iterate to find something between the two of them is going to stop because they are the same already.
I do not know what you mean about "How can I give iteration number" ?
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MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi | 1,333 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-50 | latest | en | 0.718519 |
https://timov.la/blog/how/how-is-a-raven-like-a-writing-desk.html | 1,696,022,017,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00787.warc.gz | 609,329,995 | 52,457 | ## How Is A Raven Like A Writing Desk?
Jul 26, 2023
Stationary – ln relativity, a situation or spacetime is stationary if there is no change over time. One has to take into account that, in general reletivity, time can be defined in many different ways, all of them equally valid for formulating the laws of physics.
1. This leads to a modified definition: A situation or a spacetime is stationary if it is possible to define time in a way so that there is no change of its properties over time – if you follow the properties of a given region of space over time, they will not change.
2. Did Carroll have similar thinking on the properties of time being relative and potentially stationary even before Einsteins Special Theory of Relativity and did Einstein understand Wonderland? On seeing the simpler answer ‘one is nevaR backwards and one is forWords’ for the first time, Jenny Woolf said, “I have read many attempts to solve the riddle but yours jumped out because it’s exactly the kind of answer he would give.
AND it works when spoken but not when written. (He drew on what he had said to the children in the Alice books, and, as he was very much a story teller rather than primarily a writer I would have expected it to be a riddle which worked when spoken.) Thirdly, his own written comment on the riddle was phrased in a way which instantly conveyed to me that it was not the solution but a clue.” We may ‘nevaR’ find a more logical and reasoned argument in discovering an answer to the riddle and if this was always Carroll’s intension that someday the answer to the riddle would be explained, some 150 years in TIME after it was first set out in the story of Alice in Wonderland.
1. Whether this answer to the riddle was originally intended or not, it seems clear that Dodgson created a convoluted riddle answer that not only answered the original riddle, but gave us a second riddle to solve to provide a much simpler intended answer.
2. Such is the genius of the man.
3. Lewis Carroll wrote thousands of replies to fans that were full of puns, riddles, acrostic verses, charades, word play and anagrams.
He sent letters that could only be read by looking in a mirror. Some even had words in the proper order but spelled backwards and others read entirely backwards, letter by letter. He even signed his name backwards in one letter. This is more evidence of him providing riddles within riddles.
1. I’d like to think that Charles Dodgson could not inform us directly of the riddles answer because of the answer itself.
2. I would suggest that after all this explanation, the riddle answer put into its simplest form of ‘Why is a raven is like a writing desk? is Because it takes ‘TIME’ and ‘SPACE’ to solve.
Or even; Everything has a time and a place! `Your hair wants cutting,’ said the Hatter. He had been looking at Alice for some time with great curiosity, and this was his first speech. Why SPACE? To find out why space is mentioned you’ll need to continue to read the next section of the article but first a bit more evidence on time is below, if you have the time! What is amazing is that this was not the first time that this answer had been proposed independently.
This answer was first proposed by Mary Hammond, author of The Mad Hatter, the Role of Mercury in the Life of Lewis Carroll and she said in an online article about the riddle on 20th August 2014; “Why is a raven like a writing desk? Because that which is never backwards is always forwards, and a raven is nevar backwards, and a writing desk is always for words.
If you knew a little bit about Carroll’s habit of feeding out teasing clues bit by bit, you’d know that Carroll’s statement that the riddle was not intended to have an answer was very similar in form to other little games he played. For those of you who are not familiar with the statement, it’s in the form of “no meaning was originally intended, but if it was, hint, hint, and hint”.
I’m paraphrasing here, but I truly believe that if Carroll hadn’t died so soon after making the statement, he would have given us more clues.” An extract from Mary Hammonds book published earlier on 29th May 2014, the first time ever that this riddle answer was ever proposed concludes, “I persist in the belief that Lewis Carroll had an answer in mind when he wrote the raven-writing desk riddle, and that its genesis had something to do with Poe’s poem ‘The Raven’ (as so many have suspected), and his Philosophy of Composition (which Carroll took so closely to heart).
It may be an answer as cerebral as “because Poe wrote on both backwards”, which is very close to one of the solutions ventured by the American puzzle and chess genius Sam Loyd, but realistically, as Carroll was writing for children, it is probably something pun-tastic like, because a raven is nevar backwards, and a writing desk is always for words.” Mary Hammond tried to get her thoughts acknowledged by the North American Lewis Carroll Society (see below) and wrote an article on her thoughts for submission for the Rectory Umbrella but she was denied publication.
She then as Mary Hibbs, published a youtube video in October 2014 which argues that this answer is related to Poes poem on the Raven and how he wrote that backwards. See her short video on the idea here: http://www.youtube.com/channel/UCEI9NuFzz_1oR4S6TyXmVvA The hidden charade that Neil Bant spotted in Carroll’s answer of ‘Letter’ and ‘Backwards’ in 2017, a year after his initial thoughts on the idea were published in Bandersnatch, is a final piece of the puzzle that provides further evidence for some to accept this as the intended answer by Charles Dodgson and the idea of time and a few flat notes being stationery (stationary) in 2023, concludes that the riddle has Time related properties that preceed Einsteins thoughts on the Special Theory of Relativity! It would not surprise me at all, if one day someone discovers that Alices’ sisters daydream conclusion was also editted to a ‘never-ending meal’ from an intended ‘nevaR-ending meal’, which would provide the absolute satisfaction that the riddle was intended to be solved in TIME.
“Your hair wants cutting,” said the Hatter. He had been looking at Alice for some TIME. Again The Hatter wanted ‘some time’ related answer to the riddle he then asked her! The word time is mentioned 18 times which is curious as the Tea Party is also stuck at 18:00, 6 o’clock tea-time for such a long time together! This is the final clue to indicate the riddle answer has Time related properties.
#### Why is the Raven like the writing desk?
Because it can produce a few notes. Particularly if its name is Lewis Carroll. The answer lies in the quill: both may be penned, but they can never truly be captive.
## Who said how is a raven like a writing desk?
Quote by Lewis Carroll : ‘Mad Hatter: ‘Why is a raven like a writing-desk’
## Where did Alice ask why is a raven like a writing desk?
Why is a Raven Like a Writing Desk? A couple of weeks ago, I spent time with a group of leaders in Singapore working on how to form better questions as part of a workshop on leading with agility. I returned home through Tokyo, which meant that I arrived in Dallas two hours earlier on the same day than when I departed Japan.
You would think that after years of international travel, I would no longer be entertained by the idea of arriving earlier than I departed. “What happened to those two hours?” I thought when I landed in Dallas. Of course the question can’t be answered because it rests on a fundamental misunderstanding.
However, asking myself the question got me thinking about nonsense, which in turn got me thinking about Lewis Carroll. In of Lewis Carroll’s Alice in Wonderland, after some back and forth about whether there is room at the table for Alice to join the Mad Hatter’s tea party, the Hatter poses the question, “Why is a raven like a writing desk?” The precocious Alice is eager to work out the riddle, but gets caught up in the chaotic tea party conversation.
Later, when the Hatter asks Alice about the riddle, she admits that she has not worked it out and asks the Hatter for the answer. He tells her that he does not have the slightest idea*. Some nonsense questions amuse us in the same way we might be amused by the charming innocence of a child’s question.
Decades before Bill Cosby shocked and disappointed a whole generation, my friends and I spent hours memorizing his routines. I can still picture the cover of his album, Cosby’s question is elegant, simple and nonsensical. Asking, “Why is there Air?” and “What happened to the two hours I lost during my twelve hour flight?” indicate that the person asking the question is either confused or trying to be funny.
1. Like Lewis Carroll, I’m a fan of wordplay, puns and riddles.
2. I pay close attention to how people express themselves looking for interesting or clever ways to interpret a turn of phrase.
3. It turns out, not everyone delights in my attempts at wit.
4. What I imagine to be an endearing habit quickly becomes obnoxious if I’m not careful.
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The is based on the idea that you can tell a lot about how people think by paying attention to the questions they ask. The key to helping people explore the thinking behind their choice of question is not to place too much emphasis on their choice of words.
• Consultants should not engage with a philosophical or lawyerly mindset.
• Philosophers worship clarity.
• Lawyers weaponize clarity.
• Consultants and coaches should focus on constructing meaning, not deconstructing meaning.
• Don’t focus on what the question means, focus on what the person means by asking it.
• As an example, when a client frames a consulting request as, “How do we get people to be more accountable?” I need to let go of my reflex to dismiss the question as nonsense and instead, help my client clarify the unexpressed need.
I might take an appreciative approach and say, “Tell me a story about someone acting with accountability to help me picture what you want more people to do.” Or, I might offer options to get the conversation moving, “When you say ‘accountable,’ is it more about keeping commitments or not blaming others or maybe it’s simply about complying with directives?” I don’t ask questions to hear answers.
### What is the Mad Hatter’s riddle?
Why is a Raven Like a Writing Desk? – The Mad Hatter, for his part, finds her frustration hilarious. When Alice said she couldn’t answer it, the Mad Hatter finally admits the riddle has no solution. Arguably the craziest of the characters at the table, the Mad Hatter, asks her the now infamous riddle. “Why is a Raven like a Writing Desk?” Out of all of the Lewis Carroll Riddles, this one is by far the wackiest. Alice ponders the question. She rolls it around in her mind like hot tea sipped from in a porcelain cup that is clutched by a shaking hand.
She figures, she fidgets. Much to her frustration, she cannot produce an answer. He was just asking for no reason at all. Alice sighed at the Mad Hatter’s response and says, “I think you might do something better with the time, then waste it in asking riddles that have no answers.” The Mad Hatter replies, “If you knew Time as well as I do, you wouldn’t talk about wasting it.
It’s him.” For Alice’s part, she becomes insulted and tired of being bombarded with riddles and she leaves, saying it is the stupidest tea party that she has ever been to. The riddle was so wildly insane that Carroll felt obliged years later to answer it himself, via an updated version.
He himself proposed the answer in the 1897 final revision of Alice’s Adventures. A raven is like a writing desk because, “Because it can produce a few notes, though they are very flat; and it is never put with the wrong end in front!” The early issues of the revision spell “never” as “nevar”, ie “raven.” Some conjecture that a overzealous proofreader, while typesetting, fixed the intended error.
Many were introduced to the unanswered riddle in their formative years. It got under people’s skin and itched and bothered them like a bed bug, just as it did to Alice, and still does to this day.
## Why does the Mad Hatter’s hat say 10 6?
Since 1986, October 6 is marked as the Mad Hatter Day — a famous character in Lewis Carroll’s classic Alice’s Adventures in Wonderland. Here are some interesting facts about the iconic comic character: English illustrator John enniel depicted Hatter wearing a hat with 10/6 written on it. The 10/6 refers to the cost of a hat — 10 shillings and 6 pence, and later became the date and month to celebrate Mad Hatter Day. The idiom “mad as a hatter” was around long before Carroll started writing. Colloquially used to describe an eccentric person, “mad as a hatter” is based on a problem that arose in the 1800s when hat companies used lead in the hat-making process. The lead got into their systems and they went insane, hence the term “mad as a hatter”. Rumour has it that Carroll intended the character of the Mad Hatter to be an outlandish caricature of a man named Theophilus Carter — an eccentric British furniture dealer from Oxford. Even though Hatter is popularly known as the Mad Hatter, Lewis Carroll never refers to the character as the Mad Hatter. The Mad Hatter comic book character made his debut in Batman #49 in 1948. He is the supervillain who keeps his Wonderland counterpart’s costume and personality, with a lot of his gadgets stored in his hat. In the world of Batman, he is a scientist who uses mind-controlling devices to manipulate his victims.
### What does Raven symbolize writing?
Symbolism of Raven in Literature – Raven, as symbolism in literature, is mostly depicted as a sign of death, supernatural, and evil. However, these intelligent birds are also powerful representations in cultures like Native America, symbolizing a connection to the spirit world and teachers of magical studies.
## What is the Mad Hatter famous for saying?
If I had a world of my own, everything would be nonsense.
#### What is the meaning of writing desk?
: a desk that often has a sloping top for writing on also : a portable case that contains writing materials and has a surface for writing
### What’s the difference between a raven and a crow?
Common Raven ( Corvus corax ) American Crow ( Corvus brachyrhynchos )
What is the difference between a raven and a crow?
Tail Feathers: Ravens have wedge-shaped tails and crows have fan-shaped tails. When you see the bird flying overhead, you can often get a good look at the shape of the tail. (Drawing by Jenifer Rees. Courtesy of WDFW.)
Ravens differ from crows in appearance by their larger bill, tail shape, flight pattern and by their large size. Ravens are as big as Red-tailed Hawks, and crows are about the size of pigeons. The raven is all black, has a 3.5-4 ft wingspan and is around 24-27 inches from head to tail.
1. The crow is also black, has a 2.5 ft wingspan and is about 17 inches long.
2. The raven weighs around 40 oz while the crow is 20 oz – half the weight of a raven.
3. The raven has highly glossed plumage showing iridescent greens, blues, and purples.
4. Sometimes the feathers have an oily or wet sheen.
5. Crows also have feathers with iridescent purple and blue, but with less sheen than the raven.
Ravens are uncommon in populated urban areas. If you see a “really big crow” in the city, the chances are good that it really is a crow and not a raven. Ravens have wedge-shaped tails and crows have fan-shaped tails ( view drawing ). Ravens are longer necked in flight than crows. The larger bill of the raven can be seen in flight, but it is less apparent than the long neck. Raven wings are shaped differently than are crow wings, with longer primaries (“fingers”) with more slotting between them.
Ravens have pointed wings, while crows have a more blunt and splayed wing tip. A raven’s wing sometimes makes a prominent “swish, swish” sound, while a crow’s wingbeat is usually silient. Ravens soar more than crows. If you see a “crow” soaring for more than a few seconds, take another look – it might be a raven.
Common Ravens can do a somersault in flight and even fly upside down. Ravens are longer necked in flight than crows. The most familiar call of a raven is a deep, reverberating croaking or “gronk-gronk.” Crows make the familiar “caw-caw,” but also have a large repertoire of rattles, clicks, and bell-like notes.
## How is Alice related to Raven?
” I’ve got a 180 IQ. I’m always the smartest kid in my class. ” — Alice to Ivy,
Alice is a main character on the Disney Channel series, Raven’s Home, She is the daughter of Raven’s country cousin, Betty Jane and the granddaughter of Victor’s country cousin, Delroy Baxter. She lives with Victor to attend Katherine Johnson Tech, a special school for geniuses. Alice now attends Johnson Tech Middle School, as a Sixth grader. She is portrayed by Mykal-Michelle Harris,
#### Who does the Mad Hatter represent in Alice’s life?
Since 1986, October 6 is marked as the Mad Hatter Day — a famous character in Lewis Carroll’s classic Alice’s Adventures in Wonderland. Here are some interesting facts about the iconic comic character: English illustrator John enniel depicted Hatter wearing a hat with 10/6 written on it. The 10/6 refers to the cost of a hat — 10 shillings and 6 pence, and later became the date and month to celebrate Mad Hatter Day. The idiom “mad as a hatter” was around long before Carroll started writing. Colloquially used to describe an eccentric person, “mad as a hatter” is based on a problem that arose in the 1800s when hat companies used lead in the hat-making process. The lead got into their systems and they went insane, hence the term “mad as a hatter”. Rumour has it that Carroll intended the character of the Mad Hatter to be an outlandish caricature of a man named Theophilus Carter — an eccentric British furniture dealer from Oxford. Even though Hatter is popularly known as the Mad Hatter, Lewis Carroll never refers to the character as the Mad Hatter. The Mad Hatter comic book character made his debut in Batman #49 in 1948. He is the supervillain who keeps his Wonderland counterpart’s costume and personality, with a lot of his gadgets stored in his hat. In the world of Batman, he is a scientist who uses mind-controlling devices to manipulate his victims.
## Why did Hatter go mad?
Mad hatter’s disease is a form of mercury poisoning that affects the brain and nervous system. People can develop mercury poisoning by inhaling mercury vapors. Mad hatter’s disease is caused by chronic mercury poisoning. It is characterized by emotional, mental, and behavioral changes, among other symptoms. Share on Pinterest Early symptoms of mercury poisoning may include sleep disturbances, a wet cough, and muscle pain. Mercury is a metal that can turn to vapor at room temperatures. The lungs can easily absorb this vapor, and once mercury is in the body, it can pass through cell membranes and the blood-brain barrier.
When chronic mercury poisoning affects the brain and nervous system, a person might be said to have mad hatter’s disease. The doctor may instead refer to the neurological changes as erethism. In medieval Europe, mercury was used in medicine and manufacturing.
• Later, hatmakers commonly cured felt using a form of mercury called mercurous nitrate.
• As the hatmakers inhaled mercury vapors over time, many experienced neurological symptoms of mercury poisoning.
• By 1837, “mad as a hatter” was a common saying.
• Nearly 30 years later, Lewis Carroll published Alice in Wonderland, which contained the now-famous Mad Hatter character.
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In the United States, hatmakers continued to use mercury until 1941, There are early and late symptoms of mercury poisoning, depending on the level of exposure. The neurological changes that characterize Mad hatter’s disease occur after long-term exposure.
a rashskin itchinessmuscle paina metallic taste in the mouthsores or inflammation in the mouthvomitingstomach pain diarrhea sleep disturbancesa wet cough
Later symptoms of mercury poisoning may include :
irritability and a lack of patienceshyness and the desire to avoid people anxiety insomnia a tremor that begins in the hands and affects the face and headtrouble thinking or concentratingmemory losschanges in movement, which may become coarse or jerky
The World Health Organization (WHO) explain that exposure to mercury may be:
Inorganic: A person may be exposed through their job or through contact with mercury in dental fillings or cosmetics, for example. Organic: A person can be exposed to mercury in their diet.
The three most common sources of exposure to mercury are:
certain dental fillingscontaminated fishworkplaces
## What does the Mad Hatter symbolize in Alice in Wonderland?
Through the Mad Hatter, Carroll is seen by some observers as critiquing England’s mistreatment of its workers and its mentally ill. During the Victorian era, workers in the textile industries were subjected to hazardous conditions, including exposure to lead and mercury.
### What does the Cheshire Cat symbolism?
Introduction – The Cheshire Cat was another addition for the printed version of Alice and is now one of the best known and best loved of all the Wonderland creatures. Almost every illustrator chooses to represent it and it is present in almost every film version, beginning with the very first Alice in Wonderland film in 1903,
The Cheshire Cat is another character that employs the ” flawed logic ” of Wonderland, especially in his argument to prove that he is “mad,” “And how do you know that you’re mad?” “To begin with,” said the Cat, “a dog’s not mad. You grant that?” “I suppose so,” said Alice. “Well, then,” the Cat went on, “you see a dog growls when it’s angry, and wags its tail when it’s pleased.
Now I growl when I’m pleased, and wag my tail when I’m angry. Therefore I’m mad.” The Cat draws bad conclusions from faulty assumptions, but when Alice tries to call him out, he changes the subject. The result is once again a frustrated Alice. However, when the Cat appears again on the Queen’s croquet ground, Alice is actually pleased to see it.
“How are you getting on?” said the Cat, as soon as there was mouth enough for it to speak with.Alice put down her flamingo, and began an account of the game, feeling very glad she had someone to listen to her.” The Cheshire Cat is sometimes interpreted as a guiding spirit for Alice, as it is he who directs her toward the March Hare ‘s house and the mad tea party, which eventually leads her to her final destination, the garden.
The Cat also seems to have some sort of privileged knowledge of the workings of Wonderland, which combined with its ability to immaterialize is certainly spirit-like. It is also through the Cheshire Cat that we learn the essential secret of Wonderland: it’s mad !
### Is the Hatter good or bad?
Answer and Explanation: No, the Mad Hatter is not a villain in Alice’s Adventures in Wonderland. He is an ultimately benign individual, despite his odd behavior. Like Wonderland’s other inhabitants, he wishes to go about his day without incurring the wrath of the Queen of Hearts, the temperamental ruler of Wonderland.
### Does Alice fall in love with the Mad Hatter?
Alice falls in love with the Mad Hatter and they both want to get married. The moment captures their sparking love and happiness. Alice wants to stay forever in Wonderland with the Mad Hatter. But what is White Rabbit doing there and does not seem happy.
#### What is Hatter full name?
Tarrant Hightopp
Alice in Wonderland character
Johnny Depp as Tarrant Hightopp in Alice in Wonderland (2010)
First appearance Alice in Wonderland (2010)
Last appearance Alice Through the Looking Glass (2016)
Based on Hatter by Lewis Carroll
Adapted by Tim Burton Linda Woolverton
Designed by Colleen Atwood
Portrayed by
Johnny Depp
Other:
In-universe information
Full name Tarrant Hightopp
Occupation Milliner
Family Zanik Hightopp (father) Tyva Hightopp (mother) Bim Hightopp (unspecified), Bumalig Hightopp and Poomally Hightopp (aunt and uncle), Baloo Hightopp and Pimlick Hightopp (cousins)
Home Wonderland
Tarrant Hightopp, also known as The Mad Hatter, is a fictional character in the 2010 film Alice in Wonderland and its 2016 sequel Alice Through the Looking Glass, based upon the original character from Lewis Carroll ‘s Alice novels. He is portrayed by actor Johnny Depp, He serves as the films’ male protagonist, Audience reception of the character was positive.
#### What are the 3 symbols in the raven?
There are three primary symbols in “The Raven”: the raven, the bust of Pallas, and the speaker’s chamber. All of these symbols work together to form a portrait of the speaker’s grief.
#### Why did the raven keep saying nevermore?
Answer and Explanation: In ‘The Raven,’ the raven says ‘nevermore’ because it appears to be the only word that the bird knows how to say. No matter what question the narrator asks the raven, his response is always the same.
### What is the meaning of writing desk?
: a desk that often has a sloping top for writing on also : a portable case that contains writing materials and has a surface for writing
#### Why is the narrator reading books in the raven?
The Raven | Los Angeles Public Library Review: Poe’s narrative poem follows the development of an unnamed speaker/narrator whose interaction with a raven causes him to go mad. The story implies that the narrator lost his lover Lenore and is greatly devastated due to her death. He decides to cope by reading, hoping it will drive his mind off the terrible circumstances.
• On a cold December night, the narrator hears knocking on his chamber door and wonders who it could be at this time.
• He opens it only to find no one, which causes him to question more.
• He looks around into the darkness and sees no one.
• Upon closing the door, a raven enters.
• The narrator humorously asks the bird for its name and hears it speak.
Shocked by the raven’s ability for speech, the narrator begins to question it as if it were a person to see if it can give him answers and solutions. Like many of Poe’s other pieces, this one inspired the feeling of grim due to the poem’s darker tone.
1. He accomplishes and achieves this tone by using words in his diction associated with feelings of despair and sorrow, such as “haunted” and “grim,” in addition to depicting the setting with a cold and unsettling vibe.
2. I would give the book a rating of 4 due to the elements it possesses.
3. I enjoyed how Poe uses the raven as a symbolic figure for the narrator’s pain and sorrow to further display its message to the reader by portraying the raven as a dark figure who casts a shadow upon the narrator.
I also liked how the book sounded like a poem. Throughout the dialogue, it connects the narrator’s thoughts well, allowing the reader to comprehend his thinking without disruptions, such as abstract or vague language, as seen in other poems that need to be read more than once to better understand them.
## What is the narrator obsessed with in the raven?
Edgar Allen Poe is famous for eerie literary works preoccupied with terror and littered with mystery. The Raven, arguably his most famous text, is no exception. The Gothic and Fantastic elements that permeate the poem are most evident in the second, thirteenth, fourteenth, and eighteenth stanzas.
These passages are the most explicit in terms of the division between the real and unreal. These two concepts are juxtaposed in a concrete way through the narrator (real) and the raven (unreal). The raven serves as the representation of the unreal because it is nothing more than an anthropomorphized version of the narrators subconscious despair.
In this way, the poem consists of a pseudo-dialogue between the narrator and his own psychological echo. Every time he addresses the empty night, the raven reciprocates by reminding him of his pain using a single word that is the embodiment of the narrators despair and anguish.
This repetition of nevermore posits the raven as a figment of the narrators imagination, thus resolving the mystery between reality and fantasy. To understand the nature of the relationship between the narrator and the raven, it is helpful to examine Tzvetan Todorovs concept of the Fantastic. First and foremost, for a work to be considered fantastic, there must be a hesitation on the part of the reader in distinguishing reality from unreality.
Furthermore, a character in the text must share this uncertainty. Most would agree that a talking raven achieves this on both ends. Eventually, a decision must be made between the two, while still accepting the strangeness as literal rather than metaphorical or allegorical.
When this choice occurs, the idea of the fantastic breaks down into one of two subcategories. If the events in the work are deemed strange, yet explicable in the natural world, the fantastic becomes the uncanny. If the events are supernatural and inexplicable, the fantastic becomes the marvelous. To simplify, the uncanny is real but unfamiliar; the marvelous is impossible.
By replacing reality and unreality with uncanny and marvelous respectively, it is easier for the reader to discern between the two and obtain a more complete understanding of The Raven. In order for conflict between the uncanny and marvelous to exist, the reader must first be able to share in a characters hesitant thoughts and ambiguous frame of mind.
• The narrators purposeful detachment from reality makes him an unreliable source of information, thus causing hesitation on the part of the reader regarding what is real.
• In the second stanza, Poe writes, Eagerly I wished the morrow;- vainly I had sought to borrow / From my books surcease of sorrow- sorrow for the lost Lenore-
(lines 2-3).
This is an early warning sign that the narrator is losing or possibly relinquishing his grasp on reality. In order to quell the very real pain of his loss, he attempts a mental escape into the very unreal world of books. This expatriation from his own sense of rationality creates a rift in the narrators consciousness that will only increase as the poem continues.
When the raven enters, the uncanny and marvelous reach maximum conflict capacity. To determine how the fantastic will break down and thus resolve the conflict between reality and unreality the raven must be assigned a certain role. He is either a mystical talking bird in a marvelous literary world, or he is a psychotic manifestation of the narrators subconscious desire to be reunited with his lost lover in an uncanny literary world.
Based on the narrators tenuous grip on reality, the latter seems to be more plausible. The pivotal moment of decision occurs in the thirteenth stanza, after the raven has answered every question or assertion posed by the narrator with nevermore. As he is contemplating why the bird continues to reply as such, he says, This and more I sat divining, with my head at ease reclining / On the cushions velvet lining that the lamp-light gloated oer, / But whose velvet violet lining with the lamp-light gloating oer, / She shall press, ah, nevermore
(lines 2-5).
By answering his own question in the very same fashion as the raven, he creates a mirror between the two. Because each nevermore draws an anxious response from the narrator, it is easy to see that the word has been constantly planted in the readers head in order to draw attention to the narrators suffering.
This repetition conveys the narrators obsession with the raven and reflects his repressed grief over the loss of Lenore. The habitual association of a single word with anguish and despondency creates a conspicuous connection between the raven and the narrator, revealing each character as a fractured part of a single psyche.
Further evidence of the narrators neuroticism and mental split is given as he suddenly smells incense and begins to praise God for granting him relief: Wretch, I cried, thy God hath lent thee- by these angels he hath sent thee / Respite- respite and nepenthe, from thy memories of Lenore! / Quaff, oh quaff this kind nepenthe and forget this lost Lenore! / Quoth the Raven, Nevermore
(lines 3-6).
This immediate, unprompted switch from depression to joy, then back to angst in the next stanza reinforces the idea that this man is not mentally stable and is absolutely ruled by a despair so intense that it has developed a mind of its own. The final stanza of the poem marks the strongest confirmation that the raven exists only in the narrators mind.
1. Poe writes, And the Raven, never flitting, still is sitting, still is sitting
And my soul from out that shadow that lies floating on the floor / Shall be lifted- nevermore! (lines 1&5-6).
2. The poem starts just as the narrator is beginning to fall asleep.
3. Considering the events that unfold, it could be argued that he did fall asleep and the entire poem should be read as a dream sequence.
By this token the work would fall into the category of marvelous, based on the fact that the occurrences are more readily categorized as inexplicable in a dream world. However, in these lines the narrator points out that the bird has been there for some time, thus nixing the idea that the whole poem has been a dream.
The way he refers to the ravens shadow in the final line seems to reinforce the idea that he is incapable of clawing his way out of the all-consuming despair that has plagued him from the beginning. In Poes The Raven, the repeated use of the word nevermore provides a key for deciphering the true meaning of the relationship between the narrator and the raven.
It is important to see the word in the overall context of the narrators desperate, paranoid, and delusional psyche. Keeping this in mind while using Todorovs fantastic as a reference point for the conflict between the uncanny and marvelous, it becomes clear that the raven is an uncanny manifestation of the narrators subconscious, thus resolving the conflict between reality and unreality.
### Why is the narrator reading in the raven?
Poe’s unnamed narrator is a scholar who is mourning the death of his beloved, Lenore. He is alone in his house on a cold December midnight, trying to distract himself from his thoughts of her by reading old books. | 7,694 | 34,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.979314 |
http://www.realtimerendering.com/resources/GraphicsGems/gemsiii/fillet.c | 1,500,742,790,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-30/segments/1500549424088.27/warc/CC-MAIN-20170722162708-20170722182708-00212.warc.gz | 552,767,198 | 2,571 | /* Joining Two Lines with a Circular Arc Fillet Robert D. Miller */ #include "math.h" #include "stdio.h" #include "stdlib.h" #define pi 3.14159265358974 #define halfpi 1.5707963267949 #define threepi2 4.71238898038469 #define twopi 6.28318530717959 #define degree 57.29577951308232088 #define radian 0.01745329251994329577 typedef struct point { float x; float y;} point; point gv1, gv2; float xx, yy, sa, sb, qx, qy, rx, ry; int x1, yy1, x2, y2; float arccos(); float arcsin(); float cross2(); float dot2(); void moveto(); /* External draw routine */ void lineto(); /* External draw routine */ void drawarc(); float linetopoint(); void pointperp(); void fillet(); float cross2(v1,v2) point v1, v2; { return (v1.x*v2.y - v2.x*v1.y); } float dot2(v1,v2) /* Return angle subtended by two vectors. */ point v1, v2; { float d, t; d= (float) sqrt(((v1.x*v1.x)+(v1.y*v1.y)) * ((v2.x*v2.x)+(v2.y*v2.y))); if (d != (float) 0.0) { t= (v1.x*v2.x+v1.y*v2.y)/d; return (arccos(t)); } else return ((float) 0.0); } /* Draw circular arc in one degree increments. Center is (xc,yc) with radius r, beginning at starting angle, startang through angle ang. If ang < 0 arc is draw clockwise. */ void drawarc(xc, yc, r, startang, ang) float xc, yc, r, startang, ang; { #define sindt 0.017452406 #define cosdt 0.999847695 float a, x, y, sr; int k; a= (float) startang*radian; x= (float) r*cos(a); y= (float) r*sin(a); moveto(xc+x, yc+y, 3); if (ang >= (float) 0.0) sr= (float) sindt; else sr= (float) -sindt; for (k= 1; k <= (int) floor(fabs(ang)); k++) { x= x*cosdt-y*sr; y= x*sr+y*cosdt; lineto(xc+x,yc+y); } return; } /* Find a,b,c in Ax + By + C = 0 for line p1,p2. */ void linecoef(a,b,c,p1,p2) float *a, *b, *c; point p1, p2; { *c= (p2.x*p1.y)-(p1.x*p2.y); *a= p2.y-p1.y; *b= p1.x-p2.x; return; } /* Return signed distance from line Ax + By + C = 0 to point P. */ float linetopoint(a,b,c,p) float a, b, c; point p; { float d, lp; d= sqrt((a*a)+(b*b)); if (d == 0.0) lp = 0.0; else lp= (a*p.x+b*p.y+c)/d; return ((float) lp); } /* Given line l = ax + by + c = 0 and point p, compute x,y so p(x,y) is perpendicular to l. */ void pointperp(x, y, a, b, c, p) float *x, *y, a, b, c; point p; { float d, cp; *x= 0.0; *y= 0.0; d= a*a +b*b; cp= a*p.y-b*p.x; if (d != 0.0) { *x= (-a*c-b*cp)/d; *y= (a*cp-b*c)/d; } return; } /* Compute a circular arc fillet between lines L1 (p1 to p2) and L2 (p3 to p4) with radius R. The circle center is xc,yc. */ void fillet(p1, p2, p3, p4, r, xc, yc, pa, aa) point *p1, *p2, *p3, *p4; float r, *xc, *yc, *pa, *aa; { float a1, b1, c1, a2, b2, c2, c1p, c2p, d1, d2, xa, xb, ya, yb, d, rr; point mp, pc; linecoef(&a1,&b1,&c1,*p1,*p2); linecoef(&a2,&b2,&c2,*p3,*p4); if ((a1*b2) == (a2*b1)) /* Parallel or coincident lines */ goto xit; mp.x= ((*p3).x + (*p4).x)/2.0; mp.y= ((*p3).y + (*p4).y)/2.0; d1= linetopoint(a1,b1,c1,mp); /* Find distance p1p2 to p3 */ if (d1 == 0.0) goto xit; mp.x= ((*p1).x + (*p2).x)/2.0; mp.y= ((*p1).y + (*p2).y)/2.0; d2= linetopoint(a2,b2,c2,mp); /* Find distance p3p4 to p2 */ if (d2 == 0.0) goto xit; rr= r; if (d1 <= 0.0) rr= -rr; c1p= c1-rr*sqrt((a1*a1)+(b1*b1)); /* Line parallel l1 at d */ rr= r; if (d2 <= 0.0) rr= -rr; c2p= c2-rr*sqrt((a2*a2)+(b2*b2)); /* Line parallel l2 at d */ d= a1*b2-a2*b1; *xc= (c2p*b1-c1p*b2)/d; /* Intersect constructed lines */ *yc= (c1p*a2-c2p*a1)/d; /* to find center of arc */ pc.x= *xc; pc.y= *yc; pointperp(&xa,&ya,a1,b1,c1,pc); /* Clip or extend lines as required */ pointperp(&xb,&yb,a2,b2,c2,pc); (*p2).x= xa; (*p2).y= ya; (*p3).x= xb; (*p3).y= yb; gv1.x= xa-*xc; gv1.y= ya-*yc; gv2.x= xb-*xc; gv2.y= yb-*yc; *pa= (float) atan2(gv1.y,gv1.x); /* Beginning angle for arc */ *aa= dot2(gv1,gv2); if (cross2(gv1,gv2) < 0.0) *aa= -*aa; /* Angle subtended by arc */ xit: return; } | 1,625 | 3,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-30 | longest | en | 0.315938 |
https://www.kodytools.com/units/angularacc/from/gradpns2/to/degph2 | 1,713,147,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00466.warc.gz | 775,373,776 | 17,589 | # Gradian/Square Nanosecond to Degree/Square Hour Converter
1 Gradian/Square Nanosecond = 1.1664e+25 Degree/Square Hour
## One Gradian/Square Nanosecond is Equal to How Many Degree/Square Hour?
The answer is one Gradian/Square Nanosecond is equal to 1.1664e+25 Degree/Square Hour and that means we can also write it as 1 Gradian/Square Nanosecond = 1.1664e+25 Degree/Square Hour. Feel free to use our online unit conversion calculator to convert the unit from Gradian/Square Nanosecond to Degree/Square Hour. Just simply enter value 1 in Gradian/Square Nanosecond and see the result in Degree/Square Hour.
Manually converting Gradian/Square Nanosecond to Degree/Square Hour can be time-consuming,especially when you don’t have enough knowledge about Angular Acceleration units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Gradian/Square Nanosecond to Degree/Square Hour converter tool to get the job done as soon as possible.
We have so many online tools available to convert Gradian/Square Nanosecond to Degree/Square Hour, but not every online tool gives an accurate result and that is why we have created this online Gradian/Square Nanosecond to Degree/Square Hour converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Gradian/Square Nanosecond to Degree/Square Hour (grad/ns2 to deg/h2)
By using our Gradian/Square Nanosecond to Degree/Square Hour conversion tool, you know that one Gradian/Square Nanosecond is equivalent to 1.1664e+25 Degree/Square Hour. Hence, to convert Gradian/Square Nanosecond to Degree/Square Hour, we just need to multiply the number by 1.1664e+25. We are going to use very simple Gradian/Square Nanosecond to Degree/Square Hour conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gradian/Square Nanosecond} = 1 \times 1.1664e+25 = \text{1.1664e+25 Degree/Square Hour}$$
## What Unit of Measure is Gradian/Square Nanosecond?
Gradian per square nanosecond is a unit of measurement for angular acceleration. By definition, if an object accelerates at one gradian per square nanosecond, its angular velocity is increasing by one gradian per nanosecond every nanosecond.
## What Unit of Measure is Degree/Square Hour?
Degree per square hour is a unit of measurement for angular acceleration. By definition, if an object accelerates at one degree per square hour, its angular velocity is increasing by one degree per hour every hour.
## What is the Symbol of Degree/Square Hour?
The symbol of Degree/Square Hour is deg/h2. This means you can also write one Degree/Square Hour as 1 deg/h2.
## How to Use Gradian/Square Nanosecond to Degree/Square Hour Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Gradian/Square Nanosecond and in the first input field, enter a value.
• From the second dropdown, select Degree/Square Hour.
• Instantly, the tool will convert the value from Gradian/Square Nanosecond to Degree/Square Hour and display the result in the second input field.
## Example of Gradian/Square Nanosecond to Degree/Square Hour Converter Tool
1
Degree/Square Hour
1.1664e+25
# Gradian/Square Nanosecond to Other Units Conversion Table
ConversionDescription
1 Gradian/Square Nanosecond = 900000000000000000 Degree/Square Second1 Gradian/Square Nanosecond in Degree/Square Second is equal to 900000000000000000
1 Gradian/Square Nanosecond = 900000000000 Degree/Square Millisecond1 Gradian/Square Nanosecond in Degree/Square Millisecond is equal to 900000000000
1 Gradian/Square Nanosecond = 900000 Degree/Square Microsecond1 Gradian/Square Nanosecond in Degree/Square Microsecond is equal to 900000
1 Gradian/Square Nanosecond = 0.9 Degree/Square Nanosecond1 Gradian/Square Nanosecond in Degree/Square Nanosecond is equal to 0.9
1 Gradian/Square Nanosecond = 3.24e+21 Degree/Square Minute1 Gradian/Square Nanosecond in Degree/Square Minute is equal to 3.24e+21
1 Gradian/Square Nanosecond = 1.1664e+25 Degree/Square Hour1 Gradian/Square Nanosecond in Degree/Square Hour is equal to 1.1664e+25
1 Gradian/Square Nanosecond = 6.718464e+27 Degree/Square Day1 Gradian/Square Nanosecond in Degree/Square Day is equal to 6.718464e+27
1 Gradian/Square Nanosecond = 3.29204736e+29 Degree/Square Week1 Gradian/Square Nanosecond in Degree/Square Week is equal to 3.29204736e+29
1 Gradian/Square Nanosecond = 6.224263236e+30 Degree/Square Month1 Gradian/Square Nanosecond in Degree/Square Month is equal to 6.224263236e+30
1 Gradian/Square Nanosecond = 8.96293905984e+32 Degree/Square Year1 Gradian/Square Nanosecond in Degree/Square Year is equal to 8.96293905984e+32
1 Gradian/Square Nanosecond = 54000000000000000000 Arcmin/Square Second1 Gradian/Square Nanosecond in Arcmin/Square Second is equal to 54000000000000000000
1 Gradian/Square Nanosecond = 54000000000000 Arcmin/Square Millisecond1 Gradian/Square Nanosecond in Arcmin/Square Millisecond is equal to 54000000000000
1 Gradian/Square Nanosecond = 54000000 Arcmin/Square Microsecond1 Gradian/Square Nanosecond in Arcmin/Square Microsecond is equal to 54000000
1 Gradian/Square Nanosecond = 54 Arcmin/Square Nanosecond1 Gradian/Square Nanosecond in Arcmin/Square Nanosecond is equal to 54
1 Gradian/Square Nanosecond = 1.944e+23 Arcmin/Square Minute1 Gradian/Square Nanosecond in Arcmin/Square Minute is equal to 1.944e+23
1 Gradian/Square Nanosecond = 6.9984e+26 Arcmin/Square Hour1 Gradian/Square Nanosecond in Arcmin/Square Hour is equal to 6.9984e+26
1 Gradian/Square Nanosecond = 4.0310784e+29 Arcmin/Square Day1 Gradian/Square Nanosecond in Arcmin/Square Day is equal to 4.0310784e+29
1 Gradian/Square Nanosecond = 1.975228416e+31 Arcmin/Square Week1 Gradian/Square Nanosecond in Arcmin/Square Week is equal to 1.975228416e+31
1 Gradian/Square Nanosecond = 3.7345579416e+32 Arcmin/Square Month1 Gradian/Square Nanosecond in Arcmin/Square Month is equal to 3.7345579416e+32
1 Gradian/Square Nanosecond = 5.377763435904e+34 Arcmin/Square Year1 Gradian/Square Nanosecond in Arcmin/Square Year is equal to 5.377763435904e+34
1 Gradian/Square Nanosecond = 3.24e+21 Arcsec/Square Second1 Gradian/Square Nanosecond in Arcsec/Square Second is equal to 3.24e+21
1 Gradian/Square Nanosecond = 3240000000000000 Arcsec/Square Millisecond1 Gradian/Square Nanosecond in Arcsec/Square Millisecond is equal to 3240000000000000
1 Gradian/Square Nanosecond = 3240000000 Arcsec/Square Microsecond1 Gradian/Square Nanosecond in Arcsec/Square Microsecond is equal to 3240000000
1 Gradian/Square Nanosecond = 3240 Arcsec/Square Nanosecond1 Gradian/Square Nanosecond in Arcsec/Square Nanosecond is equal to 3240
1 Gradian/Square Nanosecond = 1.1664e+25 Arcsec/Square Minute1 Gradian/Square Nanosecond in Arcsec/Square Minute is equal to 1.1664e+25
1 Gradian/Square Nanosecond = 4.19904e+28 Arcsec/Square Hour1 Gradian/Square Nanosecond in Arcsec/Square Hour is equal to 4.19904e+28
1 Gradian/Square Nanosecond = 2.41864704e+31 Arcsec/Square Day1 Gradian/Square Nanosecond in Arcsec/Square Day is equal to 2.41864704e+31
1 Gradian/Square Nanosecond = 1.1851370496e+33 Arcsec/Square Week1 Gradian/Square Nanosecond in Arcsec/Square Week is equal to 1.1851370496e+33
1 Gradian/Square Nanosecond = 2.24073476496e+34 Arcsec/Square Month1 Gradian/Square Nanosecond in Arcsec/Square Month is equal to 2.24073476496e+34
1 Gradian/Square Nanosecond = 3.2266580615424e+36 Arcsec/Square Year1 Gradian/Square Nanosecond in Arcsec/Square Year is equal to 3.2266580615424e+36
1 Gradian/Square Nanosecond = 30000000000000000 Sign/Square Second1 Gradian/Square Nanosecond in Sign/Square Second is equal to 30000000000000000
1 Gradian/Square Nanosecond = 30000000000 Sign/Square Millisecond1 Gradian/Square Nanosecond in Sign/Square Millisecond is equal to 30000000000
1 Gradian/Square Nanosecond = 30000 Sign/Square Microsecond1 Gradian/Square Nanosecond in Sign/Square Microsecond is equal to 30000
1 Gradian/Square Nanosecond = 0.03 Sign/Square Nanosecond1 Gradian/Square Nanosecond in Sign/Square Nanosecond is equal to 0.03
1 Gradian/Square Nanosecond = 108000000000000000000 Sign/Square Minute1 Gradian/Square Nanosecond in Sign/Square Minute is equal to 108000000000000000000
1 Gradian/Square Nanosecond = 3.888e+23 Sign/Square Hour1 Gradian/Square Nanosecond in Sign/Square Hour is equal to 3.888e+23
1 Gradian/Square Nanosecond = 2.239488e+26 Sign/Square Day1 Gradian/Square Nanosecond in Sign/Square Day is equal to 2.239488e+26
1 Gradian/Square Nanosecond = 1.09734912e+28 Sign/Square Week1 Gradian/Square Nanosecond in Sign/Square Week is equal to 1.09734912e+28
1 Gradian/Square Nanosecond = 2.074754412e+29 Sign/Square Month1 Gradian/Square Nanosecond in Sign/Square Month is equal to 2.074754412e+29
1 Gradian/Square Nanosecond = 2.98764635328e+31 Sign/Square Year1 Gradian/Square Nanosecond in Sign/Square Year is equal to 2.98764635328e+31
1 Gradian/Square Nanosecond = 2500000000000000 Turn/Square Second1 Gradian/Square Nanosecond in Turn/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Turn/Square Millisecond1 Gradian/Square Nanosecond in Turn/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Turn/Square Microsecond1 Gradian/Square Nanosecond in Turn/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Turn/Square Nanosecond1 Gradian/Square Nanosecond in Turn/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Turn/Square Minute1 Gradian/Square Nanosecond in Turn/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Turn/Square Hour1 Gradian/Square Nanosecond in Turn/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Turn/Square Day1 Gradian/Square Nanosecond in Turn/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Turn/Square Week1 Gradian/Square Nanosecond in Turn/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Turn/Square Month1 Gradian/Square Nanosecond in Turn/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Turn/Square Year1 Gradian/Square Nanosecond in Turn/Square Year is equal to 2.4897052944e+30
1 Gradian/Square Nanosecond = 2500000000000000 Circle/Square Second1 Gradian/Square Nanosecond in Circle/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Circle/Square Millisecond1 Gradian/Square Nanosecond in Circle/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Circle/Square Microsecond1 Gradian/Square Nanosecond in Circle/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Circle/Square Nanosecond1 Gradian/Square Nanosecond in Circle/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Circle/Square Minute1 Gradian/Square Nanosecond in Circle/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Circle/Square Hour1 Gradian/Square Nanosecond in Circle/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Circle/Square Day1 Gradian/Square Nanosecond in Circle/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Circle/Square Week1 Gradian/Square Nanosecond in Circle/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Circle/Square Month1 Gradian/Square Nanosecond in Circle/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Circle/Square Year1 Gradian/Square Nanosecond in Circle/Square Year is equal to 2.4897052944e+30
1 Gradian/Square Nanosecond = 16000000000000000000 Mil/Square Second1 Gradian/Square Nanosecond in Mil/Square Second is equal to 16000000000000000000
1 Gradian/Square Nanosecond = 16000000000000 Mil/Square Millisecond1 Gradian/Square Nanosecond in Mil/Square Millisecond is equal to 16000000000000
1 Gradian/Square Nanosecond = 16000000 Mil/Square Microsecond1 Gradian/Square Nanosecond in Mil/Square Microsecond is equal to 16000000
1 Gradian/Square Nanosecond = 16 Mil/Square Nanosecond1 Gradian/Square Nanosecond in Mil/Square Nanosecond is equal to 16
1 Gradian/Square Nanosecond = 5.76e+22 Mil/Square Minute1 Gradian/Square Nanosecond in Mil/Square Minute is equal to 5.76e+22
1 Gradian/Square Nanosecond = 2.0736e+26 Mil/Square Hour1 Gradian/Square Nanosecond in Mil/Square Hour is equal to 2.0736e+26
1 Gradian/Square Nanosecond = 1.1943936e+29 Mil/Square Day1 Gradian/Square Nanosecond in Mil/Square Day is equal to 1.1943936e+29
1 Gradian/Square Nanosecond = 5.85252864e+30 Mil/Square Week1 Gradian/Square Nanosecond in Mil/Square Week is equal to 5.85252864e+30
1 Gradian/Square Nanosecond = 1.1065356864e+32 Mil/Square Month1 Gradian/Square Nanosecond in Mil/Square Month is equal to 1.1065356864e+32
1 Gradian/Square Nanosecond = 1.593411388416e+34 Mil/Square Year1 Gradian/Square Nanosecond in Mil/Square Year is equal to 1.593411388416e+34
1 Gradian/Square Nanosecond = 2500000000000000 Revolution/Square Second1 Gradian/Square Nanosecond in Revolution/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Revolution/Square Millisecond1 Gradian/Square Nanosecond in Revolution/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Revolution/Square Microsecond1 Gradian/Square Nanosecond in Revolution/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Revolution/Square Nanosecond1 Gradian/Square Nanosecond in Revolution/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Revolution/Square Minute1 Gradian/Square Nanosecond in Revolution/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Revolution/Square Hour1 Gradian/Square Nanosecond in Revolution/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Revolution/Square Day1 Gradian/Square Nanosecond in Revolution/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Revolution/Square Week1 Gradian/Square Nanosecond in Revolution/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Revolution/Square Month1 Gradian/Square Nanosecond in Revolution/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Revolution/Square Year1 Gradian/Square Nanosecond in Revolution/Square Year is equal to 2.4897052944e+30 | 4,474 | 14,573 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.895827 |
http://mathworks.com/matlabcentral/contest/contests/24/submissions/53486 | 1,386,865,368,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164645800/warc/CC-MAIN-20131204134405-00050-ip-10-33-133-15.ec2.internal.warc.gz | 113,101,935 | 13,189 | Winner Paulo Uribe (dm1)
2004-04-28 09:00:00 UTC
# Imagine that
by cyclist
Status: Passed
Results: 0.219%
CPU Time: 9.894
Score: 4.5487
Submitted at: 2004-04-28 15:49:35 UTC
Scored at: 2004-04-28 15:50:12 UTC
Current Rank: 24th
Based on: chkIntMap (diff)
cyclist
28 Apr 2004
Inspired by vague memories that "i" and "j" are not the best variable names to use, because Matlab needs to check that the user does not mean sqrt(-1).
Code
```function d = solver(map,n)
goal = sum(map(:))/n;
map = map/goal;
d = solver0(map,n,goal);
s = myscore(map,d,n)*goal;
if s>801 && s<1999
d1=crystallise6(map,n);
s1=myscore(map,d1,n)*goal;
if s1<s
d=d1;
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = solver0(map,n,goal)
[s1,s2] = size(map);
d = solver1(map,n,s1,s2,goal);
s=myscore(map,d,n)*goal;
if s<21
return
end
if ~any(map(1,:)) || ~any(map(s1,:))
notemptyrows = find(any(map,2));
row1=notemptyrows(1);
rowe=notemptyrows(end);
trim=1;
else
trim=0;
row1=1;
rowe=s1;
end
if ~any(map(:,1)) || ~any(map(:,s2))
notemptycols = find(any(map));
col1=notemptycols(1);
cole=notemptycols(end);
trim=1;
else
trim=0;
col1=1;
cole=s2;
end
if trim
d1=d;
d1(row1:rowe,col1:cole)=solver1(map(row1:rowe,col1:cole),n,rowe-row1+1,cole-col1+1,goal);
sc1=myscore(map,d1,n)*goal;
if sc1<s
s=s1;
d1(1:row1-1,:)=d1(row1,col1);
d1(rowe+1:s1,:)=d1(rowe,cole);
d1(:,1:col1-1)=d1(row1,col1);
d1(:,cole+1:s2)=d1(rowe,cole);
d=d1;
end
end
if s<80|| (s>185&&s<6000)||s>8000
return
end
% Use prince of Darkness
if s1*s2 == 20
d_2 = ozzy(map,n);
if myscore(map,d_2,n)*goal<s
d = d_2;
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = solver1(map,n,s1,s2,goal)
is1=s1:-1:1;
is2=s2:-1:1;
for ndivs = [23,19,17,13,11,7,5,3,2]
if (rem(n,ndivs) == 0) && (rem(s1,ndivs) == 0)
k = s1/ndivs; mm = map(1:k,:);
idivs = (1:k)'; idivs = idivs(:,ones(1,ndivs));
MM = mm(idivs,:);
if isequal(map,MM)
n1 = n/ndivs;
d1 = solver1(mm,n1,k,s2,goal);
d = map;
for ccc = 1:ndivs
d((ccc-1)*k+1:ccc*k,:) = d1 + (ccc-1)*n1;
end
return
end
end
if (rem(n,ndivs) == 0) && (rem(s2,ndivs) == 0)
k = s2/ndivs; mm = map(:,1:k);
idivs = (1:k)'; idivs = idivs(:,ones(1,ndivs));
MM = mm(:,idivs);
if isequal(map,MM)
n1 = n/ndivs;
d1 = solver1(mm,n1,s1,k,goal);
d = map;
for ccc = 1:ndivs
jk=ccc*k;
d(:,jk-k+1:jk) = d1 + (ccc-1)*n1;
end
return
end
end
end
% is this an all-integer map?
if any( any( rem(map*goal,1)>0 ) )
minscore=0;
else
minscore = n*abs(goal-round(goal))/goal + 1e-13;
end
len = s1*s2;
limiet=n/570;
limiet = max(limiet,minscore);
np = 11;
D = cell(1,8);
siz = [s1 s2];
siz2= [s2 s1];
D{1} = map;
D{2} = map';
D{3} = map(is1,:);
D{4} = D{2}(is2,:);
D{5} = map(:,is2);
D{6} = D{2}(:,is1);
D{7} = D{3}(:,is2);
D{8} = D{4}(:,is1);
Score = inf;
dec =0;
ind = cell(1,np*8);
stop = 0;
pat3x3a = [1 8 9; ...
2 7 6; ...
3 4 5];
pat3x3b = [1 2 9; ...
4 3 8; ...
5 6 7];
pat4x4 = [1 2 15 16; ...
4 3 14 13; ...
5 8 9 12; ...
6 7 10 11];
pat5x5 = [1 20 21 22 25; ...
2 19 18 23 24; ...
3 16 17 12 11; ...
4 15 14 13 10; ...
5 6 7 8 9];
%mj = 1;
init_idx=zeros(22);
for bbb = 1:8
nmap = D{bbb}*0+n;
% mj=~mj;
mj = rem(bbb-1,2);
for ccc=[0:5 7:np-2]
idx=ccc+mj*np+1;
if ~init_idx(idx)
if idx==9
ind{idx} = refind1(siz);
elseif idx==11
ind{idx} = refind2(siz);
elseif idx==4
ind{idx} = refind3(siz,2);
elseif idx==3
ind{idx} = refind3(siz,3);
elseif idx==8
ind{idx}= refind4(siz);
elseif idx==1
ind{idx} = refind5(siz,pat3x3a);
elseif idx==2
ind{idx} = refind5(siz,pat3x3b);
elseif idx==7
ind{idx} = refind5(siz,pat5x5);
elseif idx==5
ind{idx} = refind5(siz,pat4x4);
elseif idx==6
ind{idx} = refind3(siz,4);
elseif idx==10
ind{idx} = refind3(siz,5);
end
if s1 ~= s2
if idx==20
ind{idx} = refind1(siz2);
elseif idx==22
ind{idx} = refind2(siz2);
elseif idx==15
ind{idx} = refind3(siz2,2);
elseif idx==14
ind{idx} = refind3(siz2,3);
elseif idx==19
ind{idx} = refind4(siz2);
elseif idx==12
ind{idx} = refind5(siz2,pat3x3a);
elseif idx==13
ind{idx} = refind5(siz2,pat3x3b);
elseif idx==18
ind{idx} = refind5(siz2,pat5x5);
elseif idx==16
ind{idx} = refind5(siz2,pat4x4);
elseif idx==17
ind{idx} = refind3(siz2,4);
elseif idx==21
ind{idx} = refind3(siz2,5);
end
else
for jj = 1:np
ind{np+jj} = ind{jj};
init_idx(np+jj)=1;
end
end
init_idx(idx)=1;
end
d1 = pattern(D{bbb},n,len,ind{idx},nmap);
Score1 = myscore(D{bbb},d1,n);
if Score1<Score
Score=Score1;
d=d1;
dec=bbb;
if Score1 < limiet
dec = bbb;
stop = 1;
break
end
end
end
if stop
break;
end
end
switch dec;
case 2
d = d';
case 3
d = d(is1,:);
case 4
d = d(is2,:)';
case 5
d = d(:,is2);
case 6
d = d(:,is1)';
case 7
d = d(is1,is2);
case 8
d = d(is2,is1)';
end
if Score*n < 5e-2, return; end
d = mysolver(map, n, d, siz,goal);
%s=myscore(map,d,n);
%if (s>80&&s<185)
% if s1*s2==20
% d = ozzy(map,n);
% end
% return
%end
%if s>6000&&s<8000
% if s1*s2==20
% d = ozzy(map,n);
% end
%
%end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function X = refind1(s)
nel = s(1)*s(2);
X = reshape(1:nel,s(1),s(2));
X(:,2:2:s(2)) = X(s(1):-1:1,2:2:s(2));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function X = refind2(siz)
a=siz(1);b=siz(2);
d = zeros(a,b);
ss = a*b;
upper = 1;
lower = a;
jj = 1;
kk = b;
ii = 1;
while ii <= ss,
dd = lower-upper;
d(upper:lower,jj) = (ii:ii+dd)';
ii = ii + dd + 1;
jj = jj + 1;
if ii > ss, break; end
dd = kk-jj;
d(lower,jj:kk) = (ii:ii+dd);
ii = ii + dd + 1;
lower = lower - 1;
if ii > ss, break; end
dd = lower-upper;
d(lower:-1:upper,kk) = (ii:ii+dd)';
ii = ii + dd + 1;
kk = kk - 1;
if ii > ss, break; end
dd = kk-jj;
d(upper,kk:-1:jj) = (ii:ii+dd);
ii = ii + dd + 1;
upper = upper + 1;
if ii > ss, break; end
end
X = zeros(1,ss);
for dec = 1:ss,
X(d(dec)) = dec;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function X = refind3(siz,nr)
a=siz(1);b=siz(2);
d = zeros(a,b);
len = a*b;
bbb = 1;
dir = 1;
row = 1;
col = 1;
c = 1;
if mod(b,2) == 0,
d(:,1) = (a:-1:1)';
bbb = bbb + a;
c = 2;
col = 2;
row = 1;
end
q = mod(a,nr);
while q
if dir > 0
d(row,c:b) = bbb:bbb+b-c;
else
d(row,b:-1:c) = bbb:bbb+b-c;
end
bbb = bbb+b-c+1;
col = col + (b-c)*dir;
dir = -dir;
row = row + 1;
q = q - 1;
end
while bbb <= len,
while bbb <= len
for g = 0:nr-1
d(row+g,col) = bbb+g;
end
bbb = bbb + nr;
if (col == c && dir == -1) || (col == b && dir == 1)
break;
end
col = col + dir;
for g = 0:nr-1
d(row+nr-1-g,col) = bbb+g;
end
bbb = bbb + nr;
col = col + dir;
end
row = row + nr;
dir = -dir;
end
X = zeros(1,len);
for dec = 1:len,
X(d(dec)) = dec;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Indices = refind4(siz)
s1 = siz(1);
s2 = siz(2);
bbb = 1;
d = zeros(s1,s2);
r = s1;
c = s2;
x = 1;
y = 1;
while r >= 3 && c >= 2
d(y,x:end) = bbb:bbb+c-1;
bbb = bbb + c;
d(y+1,end:-1:x) = bbb:bbb+c-1;
bbb = bbb + c;
y = y + 2;
r = r - 2;
d(y:end,x) = (bbb:bbb+r-1)';
bbb = bbb + r;
d(end:-1:y,x+1) = (bbb:bbb+r-1)';
bbb = bbb + r;
x = x + 2;
c = c - 2;
end
d(y:end,x:end) = reshape((1:r*c)+bbb-1,c,r)';
d(y+1:2:end,end:-1:x) = d(y+1:2:end,x:end);
l = s1*s2;
Indices = zeros(1,l);
for bbb = 1:l
Indices(d(bbb)) = bbb;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Indices = refind5(siz,p1)
s1 = siz(1);
s2 = siz(2);
d = zeros(s1,s2);
[gx,gy]=size(p1);
p2 = p1(:,gy:-1:1);
ny = floor((s1-1)/(2*gy))*2;
nx = floor((s2-1)/gx);
bbb = 1;
r = 1;
c = 1;
dir = gx;
for y = 1:ny
for x = 1:nx
if dir > 0
d(r:r+gy-1,c:c+gx-1) = p1+(bbb-1);
else
d(r:r+gy-1,c:c+gx-1) = p2+(bbb-1);
end
bbb = bbb + gx*gy;
c = c + dir;
end
c = c - dir*(gx-1)/gx;
if dir < 0
d(r:r+gy-1,c) = (bbb:bbb+gy-1)';
else
d(r:r+gy-1,c+gx-1) = (bbb:bbb+gy-1)';
end
bbb = bbb + gy;
dir = -dir;
r = r + gy;
end
dr = s1 - r + 1;
dc = gx*nx+1;
d(r:end,1:dc) = reshape(bbb:bbb+dr*dc-1,dr,dc);
d(r:end,2:2:dc) = d(end:-1:r,2:2:dc);
if dc < s2
c = dc + 1;
bbb = bbb + dr*dc;
dc = s2 - c + 1;
d(1:end,c:s2) = reshape(bbb:bbb+s1*dc-1,s1,dc);
d(1:end,c:2:end) = d(end:-1:1,c:2:end);
end
l = s1*s2;
Indices = zeros(1,l);
for bbb = 1:l
Indices(d(bbb)) = bbb;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = pattern(map,n,l,ind,d)
% function d = pattern(map,n,l,ind)
% d = map*0+n;
w = 1;
ij = 1;
ji = l;
md = max(1,floor((n-1)/2));
k = l+1;
for bbb = 1:l,
ccc = ind(bbb);
m = map(ccc);
if ji<=n-ij || w+w < m ,
if ij == md,
ij = ij + 1;
k = bbb;
break;
end
ij = ij + 1;
w = 1-m;
else
w = w - m;
end
ji = ji-1;
d(ccc) = ij;
end
w = 1;
for bbb = l:-1:k
ccc = ind(bbb);
m = map(ccc);
if ji<=n-ij || w+w < m ,
if ij == n-1,
break;
end
ij = ij + 1;
w = 1-m;
else
w = w - m;
end
ji = ji-1;
d(ccc) = ij;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = mysolver(map,n,d,siz,goal)
s1=siz(1);s2=siz(2);
rand('state',101);
w = zeros(1,n);
for bbb = 1:s1*s2
w(d(bbb)) = w(d(bbb)) + map(bbb);
end
Score = sum(abs(w-1)) / n;
nl = 3000;
if Score > 0.007241
nl = 5000;
end
pop = rand(3,nl);
pop(1,:)=pop(1,:)*(s1-2)+2;
pop(2,:)=pop(2,:)*(s2-2)+2;
pop(3,:)=pop(3,:)*4;
pop=double(uint16(pop));
p=uint16(d);
lastcheck = 0;
bestscore = Score;
for kk=1:nl
y1 = pop(1,kk);
x1 = pop(2,kk);
y2 = y1;
x2 = x1;
switch pop(3,kk)
case 0
y2 = y2 + 1;
case 1
y2 = y2 - 1;
case 2
x2 = x2 + 1;
case 3
x2 = x2 - 1;
end
row1 = p(y1,x1);
rowe = p(y2,x2);
if row1 == rowe
continue;
end
a = w(row1);
b = w(rowe);
da = b-a;
if da>0
p(y2,x2) = row1;
m = map(y2,x2);
if da <= abs(da-2*m) || isNCont(p==rowe,siz)
p(y2,x2) = rowe;
else
w(rowe) = b - m;
w(row1) = a + m;
end
else
p(y1,x1) = rowe;
m = map(y1,x1);
if da >= -abs(da+2*m) || isNCont(p==row1,siz)
p(y1,x1) = row1;
else
w(rowe) = b + m;
w(row1) = a - m;
end
end
if kk-lastcheck > 500
s = sum(abs(w-1)) / n;
if s - bestscore > -0.000001
break;
else
bestscore = s;
lastcheck = kk;
end
end
end
d=double(p);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function tf = isNCont(a,siz)
s1=siz(1);s2=siz(2);
[ii,jj] = find(a);
tf = length(ii);
if tf==0;return;end
dec = 1;
a(ii(1),jj(1)) = 0;
tf = tf - 1;
while dec
r = ii(dec);
c = jj(dec);
dec = dec-1;
if (r > 1) && a(r-1,c)
dec = dec+1;
ii(dec) = r-1;
jj(dec) = c;
a(r-1,c) = 0;
tf = tf - 1;
end
if (r < s1) && a(r+1,c)
dec = dec+1;
ii(dec) = r+1;
jj(dec) = c;
a(r+1,c) = 0;
tf = tf - 1;
end
if (c > 1) && a(r,c-1)
dec = dec+1;
ii(dec) = r;
jj(dec) = c-1;
a(r,c-1) = 0;
tf = tf - 1;
end
if (c < s2) && a(r,c+1)
dec = dec+1;
ii(dec) = r;
jj(dec) = c+1;
a(r,c+1) = 0;
tf = tf - 1;
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Score = myscore(ma,in,n)
Total = zeros(1,n);
for bbb = 1:numel(ma),
Total(in(bbb)) = Total(in(bbb)) + ma(bbb);
end
Score = abs(Total(1)-1);
for k=2:n
Score = Score + abs(Total(k)-1);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = ozzy(map,n)
[xn,yn] = size(map);
num = xn*yn;
d=zeros(xn,yn);
[e,w]=sort(-map(:));
d(w(1:n))=1:n;
liberties = zeros(n,num);
libnz=zeros(n,1);
groupsize(1:n) = map(w(1:n));
ym=1:yn;
ymesh=ym(ones(xn,1),:);
xm=(1:xn)';
xmesh=xm(:,ones(1,yn));
for bbb=1:n
wi=w(bbb);
nzi=libnz(bbb);
x = xmesh(wi); y = ymesh(wi);
if x>1 && ~d(x-1,y),
nzi=nzi+1;
liberties(bbb,nzi)=wi-1;
end
if x<xn && ~d(x+1,y)
nzi=nzi+1;
liberties(bbb,nzi)=wi+1;
end
if y>1 && ~d(x,y-1)
nzi=nzi+1;
liberties(bbb,nzi)=wi-xn;
end
if y<yn && ~d(x,y+1)
nzi=nzi+1;
liberties(bbb,nzi)=wi+xn;
end
libnz(bbb)=nzi;
end
num=num-n;
while num
if all(groupsize==inf)
break
end
[smallest,bbb]=min(groupsize);
nzi = libnz(bbb);
notzero = (liberties(bbb,1:nzi));
test=~d(notzero);
if ~any(test) || ~nzi
groupsize(bbb)=inf;
continue;
end
if ~all(test),
nzi=sum(test);
notzero = notzero(test);
end
notzero = sort(notzero);
choice = map(notzero);
[tmp,move]=max(choice);
nzm = notzero(move);
liberties(bbb,1:nzi)=notzero;
if ~d(nzm)
x = xmesh(nzm); y = ymesh(nzm);
if x>1 && ~d(x-1,y)
nzi=nzi+1;
liberties(bbb,nzi)=nzm-1;
end
if x<xn && ~d(x+1,y)
nzi=nzi+1;
liberties(bbb,nzi)=nzm+1;
end
if y>1 && ~d(x,y-1)
nzi=nzi+1;
liberties(bbb,nzi)=nzm-xn;
end
if y<yn && ~d(x,y+1)
nzi=nzi+1;
liberties(bbb,nzi)=nzm+xn;
end
num = num - 1;
d(nzm) = bbb;
groupsize(bbb) = groupsize(bbb)+ tmp; %map(nzm);
liberties(bbb,move:nzi-1)=liberties(bbb,move+1:nzi);
libnz(bbb)=nzi-1;
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = crystallise6(map,n)
[m1,m2] = size(map);
m=m1*m2;
m11=m1+2;
m22=m2+2;
targ = 1;
dpop=zeros(n,1); % population of district
dsq = zeros(n,1); % number of squares in district
ihalos = zeros(n,m);
nhalos = zeros(n,1);
% pad d matrix with ones
e=ones(m11,m22);
e(2:m1+1,2:m2+1)=0;
d=e;
e(:)=10000;
e(2:m1+1,2:m2+1)=map;
map=e;
hmap=e;
apop=0; %assigned population to date
for idis=1:n
% choose starting pixel - highest unassigned pixel
[mmax,imax]=max(~d(:).*map(:));
i1=imax;
d(i1)=idis;
dpop(idis)=map(i1);
apop = apop+dpop(idis);
dsq(idis)=1;
% make the halo list
nhalo=0;
halo=zeros(1,m);
notstuck=1;
hmap(:)=d(:);
% make the halo for the first point
if ~hmap(i1-1)
nhalo=nhalo+1;
halo(nhalo)=i1-1;
hmap(i1-1) = 1;
end
if ~hmap(i1+1)
nhalo=nhalo+1;
halo(nhalo)=i1+1;
hmap(i1+1) = 1;
end
if ~hmap(i1-m11)
nhalo=nhalo+1;
halo(nhalo)=i1-m11;
hmap(i1-m11)=1;
end
if ~hmap(i1+m11)
nhalo=nhalo+1;
halo(nhalo)=i1+m11;
hmap(i1+m11)=1;
end
if nhalo>0
halo=halo(1:nhalo);
end
oldhalo=halo;
idtarg=idis*targ;
updatehalo=0; halo_up=1;
while notstuck && apop < idtarg
if nhalo>0 % if there is the possibility of adding a point
halo=halo(1:nhalo);
notstuck=1;
% choose the best halo point to add, and add it
best = Inf;
for ih=1:nhalo
%testval = abs(dpop(idis) + map(halo(ih)) - targ);
testval = abs( apop + map(halo(ih)) - idtarg);
if testval < best
best = testval;
end
end
if abs(dpop(idis)+map(i1)-targ) < abs(dpop(idis)-targ) % will this point make things better?
d(i1)=idis; % add the pixel
halo=halo(2:nhalo);
halo=halo(1:nhalo-1);
else
end
halo_up=0;
nhalo=nhalo-1;
dpop(idis) = dpop(idis) + map(i1);
apop = apop + map(i1);
dsq(idis)=dsq(idis)+1;
updatehalo=0;
elseif ~halo_up % if the halo is not up to date, update it and try again
updatehalo=1;
else
notstuck=0; % go no further
end
elseif ~halo_up % if the halo is not up to date, update it and try again
updatehalo=1;
else
notstuck=0;
end % if nhalo>0
% constuct the halo of the last added point, [i1]
if updatehalo
for ia = 1:length(iadded)
if ~hmap(i1-1)
nhalo=nhalo+1;
halo(nhalo)=i1-1;
hmap(i1-1) = 1;
end
if ~hmap(i1+1)
nhalo=nhalo+1;
halo(nhalo)=i1+1;
hmap(i1+1) = 1;
end
if ~hmap(i1-m11)
nhalo=nhalo+1;
halo(nhalo)=i1-m11;
hmap(i1-m11)=1;
end
if ~hmap(i1+m11)
nhalo=nhalo+1;
halo(nhalo)=i1+m11;
hmap(i1+m11)=1;
end
if nhalo>0
halo=halo(1:nhalo);
end
end
oldhalo=halo;
halo_up=1;
end
end %while - working through this district and its halo
% Finish with this district:
% constuct the halo of the last added point, [i1,j2] - it won't
% have been updated inside the while loop
% update teh halo before leaving this district
if ~halo_up
for ia = 1:length(iadded)
if ~hmap(i1-1)
nhalo=nhalo+1;
halo(nhalo)=i1-1;
hmap(i1-1) = 1;
end
if ~hmap(i1+1)
nhalo=nhalo+1;
halo(nhalo)=i1+1;
hmap(i1+1) = 1;
end
if ~hmap(i1-m11)
nhalo=nhalo+1;
halo(nhalo)=i1-m11;
hmap(i1-m11)=1;
end
if ~hmap(i1+m11)
nhalo=nhalo+1;
halo(nhalo)=i1+m11;
hmap(i1+m11)=1;
end
if nhalo>0
halo=halo(1:nhalo);
end
end
end
nhalos(idis)=nhalo;
if nhalo>0
ihalos(idis,1:nhalo)=halo;
end
end % loop over districts
% UNASSIGNED SQUARES - GROW DISTRICTS THAT ARE TOO SMALL
% unassigned squares? grow all districts as far as possible, starting with
% the smallest
if sum(dsq)<m
[sss,idsort] = sort(dpop-targ);
idsort=idsort(sss<0);
for k=1:length(idsort);
hmap=d;
idis=idsort(k);
% Retrieve the halo for district idis, and delete entries that have
% become occupied since halo was constructed.
halo = ihalos(idis,:);
nhalo = nhalos(idis);
for ih = 1:nhalo
if d(halo(ih))
halo(ih)=0;
end
end
halo(halo==0)=[];
nhalo = length(halo);
nhalos(idis)=nhalo;
if nhalos(idis)>0
ihalos(idis,1:nhalos(idis))=halo;
end
for ih = 1:nhalo
hmap(halo(ih))=1;
end
% halo is now updated. Start growing the district as before, this time
% until it gets stuck.
notstuck=1; updatehalo=0;
if nhalo==0
notstuck=0;
end
while dpop(idis)<targ&¬stuck
if updatehalo % construct the halo of the last added point, [i1,j2]
if ~hmap(i1-1)
nhalo=nhalo+1;
halo(nhalo)=i1-1;
hmap(i1-1) = 1;
end
if ~hmap(i1+1)
nhalo=nhalo+1;
halo(nhalo)=i1+1;
hmap(i1+1) = 1;
end
if ~hmap(i1-m11)
nhalo=nhalo+1;
halo(nhalo)=i1-m11;
hmap(i1-m11)=1;
end
if ~hmap(i1+m11)
nhalo=nhalo+1;
halo(nhalo)=i1+m11;
hmap(i1+m11)=1;
end
end
if nhalo>0
notstuck=1;
nhalo=length(halo);
% choose the best halo point to add, and add it
mmax = -1;
for ih=1:nhalo
if map(halo(ih))>mmax
mmax=map(halo(ih));
end
end
d(i1)=idis;
nhalo=nhalo-1;
apop = apop + map(i1);
dpop(idis)=dpop(idis)+map(i1);
dsq(idis)=dsq(idis)+1;
updatehalo=1; % a square has been added - next iteration will need halo to be reconstructed
else
notstuck=0;
end
end % while growing district in second pass
% Finish with this district:
% construct the halo of the last added point, [i1,j2] - it won't
% have been updated inside the while loop
if updatehalo
if ~hmap(i1-1)
nhalo=nhalo+1;
halo(nhalo)=i1-1;
hmap(i1-1) = 1;
end
if ~hmap(i1+1)
nhalo=nhalo+1;
halo(nhalo)=i1+1;
hmap(i1+1) = 1;
end
if ~hmap(i1-m11)
nhalo=nhalo+1;
halo(nhalo)=i1-m11;
hmap(i1-m11)=1;
end
if ~hmap(i1+m11)
nhalo=nhalo+1;
halo(nhalo)=i1+m11;
hmap(i1+m11)=1;
end
if nhalo>0
halo=halo(1:nhalo);
end
end
% store the halo
nhalos(idis)=nhalo;
if nhalo>0
ihalos(idis,1:nhalo)=halo;
end
end % phase 2 loop over districts
end
% UNASSIGNED SQUARES - GROW ANY DISTRICTS
% unassigned squares? grow all districts as far as possible, starting with
% the smallest
if sum(dsq)<m
[sss,idsort] = sort(dpop-targ);
for k=1:n
hmap=d;
idis=idsort(k);
% Retrieve the halo for district idis, and delete entries that have
% become occupied since halo was constructed.
nhalo = nhalos(idis);
halo = ihalos(idis,1:nhalo);
for ih = 1:nhalo
if d(halo(ih))
halo(ih)=0;
end
end
halo(halo==0)=[];
nhalo = length(halo);
nhalos(idis)=nhalo;
if nhalos(idis)>0
ihalos(idis,1:nhalo)=halo;
end
for ih = 1:nhalo
hmap(halo(ih))=1;
end
% halo is now updated. Start growing the district as before, this time
% until it gets stuck.
notstuck=1; updatehalo=0;
if nhalo==0
notstuck=0;
end
while notstuck
if updatehalo % construct the halo of the last added point, [i1,j2]
if ~hmap(i1-1)
nhalo=nhalo+1;
halo(nhalo)=i1-1;
hmap(i1-1) = 1;
end
if ~hmap(i1+1)
nhalo=nhalo+1;
halo(nhalo)=i1+1;
hmap(i1+1) = 1;
end
if ~hmap(i1-m11)
nhalo=nhalo+1;
halo(nhalo)=i1-m11;
hmap(i1-m11)=1;
end
if ~hmap(i1+m11)
nhalo=nhalo+1;
halo(nhalo)=i1+m11;
hmap(i1+m11)=1;
end
end
if nhalo>0
notstuck=1;
nhalo=length(halo); %?????????
% choose the best halo point to add, and add it
mmax = -1;
for ih=1:nhalo
if map(halo(ih))>mmax
mmax=map(halo(ih));
end
end
d(i1)=idis;
nhalo=nhalo-1;
dpop(idis)=dpop(idis)+map(i1);
dsq(idis)=dsq(idis)+1;
updatehalo=1; % a square has been added - next iteration will need halo to be reconstructed
else
notstuck=0;
end
end % while growing district in second pass
end % loop over districts
end
d = d(2:m1+1,2:m2+1);
% Empty districts: try to create single-square districts that don't cause
% contiguity problems
emptydis = find(~dsq);
for k = 1:length(emptydis)
idis = emptydis(k);
for i1=1:m1
for j1=1:m2
if ~critcon(d,i1,j1,m1,m2)
id = d(i1,j1);
d(i1,j1) = idis;
dpop(id) = dpop(id)-map(i1,j1);
dpop(idis) = dpop(idis)+map(i1,j1);
idis=idis+1;
if idis>n
break
end
end
end
if idis>n
break
end
end
end % if
return
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = trade(d, map, n)
% Init some vars
popcount = full(sparse(d(:),1,map(:))');
totalpop = sum(popcount);
targetpop = totalpop / n;
err = popcount - targetpop;
[r,c] = size(map);
% Trades squares between districts to improve score
notdone = 1;
while notdone
% Find district boundaries and get indices to those points
ddiff = diff(d); % Down boundary
[dy,dx] = find(ddiff);
dndx1 = (dx - 1)*r + dy; % above the boundry
dndx2 = dndx1 + 1; % below the boundry
rdiff = diff(d,1,2); %diff(d')'; % Right boundary
[ry,rx] = find(rdiff);
rndx1 = (rx - 1)*r + ry;
rndx2 = rndx1 + r;
ndx1 = [dndx1; rndx1]'; % Combine down and right
ndx2 = [dndx2; rndx2]';
if isempty(ndx1)
notdone = 0;
continue
end
% Get district of each square
dist1 = d(ndx1);
dist2 = d(ndx2);
% Check if population errors are in opposite sense
err1 = err(dist1);
err2 = err(dist2);
ndx = sign(err1) ~= sign(err2);
ndx1 = ndx1(ndx);
if isempty(ndx1)
notdone = 0;
continue
end
ndx2 = ndx2(ndx);
dist1 = dist1(ndx);
dist2 = dist2(ndx);
err1 = err1(ndx);
err2 = err2(ndx);
% Swap vars so dist1 (the giver) always gives a square to dist2 (the taker)
ndx = err1 < 0;
ndx1(ndx) = ndx2(ndx);
%ndx2(ndx) = dummy; % never used after this, so dropped
dummy = dist1(ndx);
dist1(ndx) = dist2(ndx);
dist2(ndx) = dummy;
dummy = err1(ndx);
err1(ndx) = err2(ndx);
err2(ndx) = dummy;
% Check if score will be improved
ndx = tradevalue < max(err1,abs(err2)); % do examples
ndx1 = ndx1(ndx);
if isempty(ndx1)
notdone = 0;
continue
end
dist1 = dist1(ndx);
dist2 = dist2(ndx);
err1 = err1(ndx);
err2 = err2(ndx);
% Check if trade benefit is large enough to be worthwhile
%ndx = [tradeben > 0.00001*totalpop]; % Limit amount of trading
ndx = tradeben > 0.0000011*totalpop; % Limit amount of trading
%ndx = [tradeben > 0]; % Make all trades - way too slow!
ndx1 = ndx1(ndx);
if isempty(ndx1)
notdone = 0;
continue
end
dist1 = dist1(ndx);
dist2 = dist2(ndx);
% Make as many trades as possible but don't affect a single district more
% than once during this loop
%ndx = flplr(ndx); % Max benefit first
k = 0;
for bbb = ndx(end:-1:1)
continue % One or both districts already traded in this iteration
end
if traded(ndx1(bbb)) % can only trade each square once ever
k = k + 1; % This square has been traded too many times
continue
end
% Consider this square traded
% Make the trade
d(ndx1(bbb)) = dist2(bbb);
% Check if districts are still contiguous
% if ~isconnected(d,dist1(bbb))
if isNCont(d==dist1(bbb),[r c])
d(ndx1(bbb)) = dist1(bbb); % Undo trade
continue
end
% Update vars
err(dist1(bbb)) = err(dist1(bbb)) - tradevalue(bbb); % the giver
err(dist2(bbb)) = err(dist2(bbb)) + tradevalue(bbb); % the taker | 9,499 | 22,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2013-48 | latest | en | 0.532413 |
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Introduction to Stochastic Gradient Descent Our previous articles investigated the importance of linear regression and batch gradient descent in machine learning modeling, as well as the intricacies of Gradient Descent models. With respect to gradient descent, here we investigate linear regression as it relates to stochastic gradient descent machine learning. Before embarking on this discussion,Continue reading “Stochastic Gradient Descent Algorithms: Topics of Machine Learning”
## Batch Gradient Descent Algorithms: Topics of Machine Learning
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## Multi-Class Classification In Machine Learning: Topics of Machine Learning
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## Performance Measurement in Machine Learning: 100 Days of Code (1/100)
The Motivation The Art of Better Programming has grown beyond what we had ever anticipated. With several collaborators now working together to bring programmers unique material to follow along in their training endeavors, the sky’s the limit right now. With that being said, for those who are just now joining us on this journey, weContinue reading “Performance Measurement in Machine Learning: 100 Days of Code (1/100)”
## Confusion Matrix and Measurement of Algorithmic Recall: Topics of Machine Learning
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## Unsupervised Machine Learning Models: Topics of Machine Learning
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## Supervised Machine Learning Algorithms: Topics of Machine Learning
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## An Overview Of Machine Learning Systems: Topics of Machine Learning
An Introduction to Machine Learning Systems This article marks the commencement on our series pertaining to machine learning and the learning mechanisms thereof. In order to make discussion of those learning mechanisms, we must first supply reason and logic to the topic at hand. Particularly, we must answer the question: “What exactly is machine learning?”Continue reading “An Overview Of Machine Learning Systems: Topics of Machine Learning” | 928 | 5,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-50 | latest | en | 0.864981 |
https://studysoup.com/note/75105/wfu-phy-712-fall-2015 | 1,477,220,010,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719215.16/warc/CC-MAIN-20161020183839-00355-ip-10-171-6-4.ec2.internal.warc.gz | 873,746,918 | 16,391 | ×
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Electromagnetism
by: Miss Lucienne Hamill
24
0
2
Electromagnetism PHY 712
Miss Lucienne Hamill
WFU
GPA 3.91
Natalie Holzwarth
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This 2 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 712 at Wake Forest University taught by Natalie Holzwarth in Fall. Since its upload, it has received 24 views. For similar materials see /class/230730/phy-712-wake-forest-university in Physics 2 at Wake Forest University.
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Date Created: 10/28/15
January 22 2002 Notes for Lecture 2 Examples of solutions of the onedimensional Poisson equation Consider the following one dimensional charge distribution 0 formlt a pu for altalt0 pu for 0 lt lt a 1 for gt a We want to nd the electrostatic potential such that 12 p 12 l E l 2 J 1 with the boundary condition oo 0 The solution to the differential equation is given by U for lt a PittHIV for 1ltFIIltU 25 12 pua2su for 0 lt lt a 39 7311 for gt a The electrostatic eld is given by U for lt a E for altatlt0 4 for0ltarlta U formgta The electrostatic potential can be determined by piecewise solution within each of the four regions or by use of the Green s function Cm 47rarlt where 1 47mg ltIgtltargt ammar39wmt a In the expression for Garar arlt should be taken the smaller of and ar It can be shown that Eq 5 gives the identical result for given in Eq 3 Notes on the onedimensional Green s functions The Green s function for the Poisson equation can be de ned a solution to the equation V2Gat 47r6a7 6 Here the factor of 47r is not really necessary but ensures consistency with your text s treat ment of the 3dimensional case The meaning of this expression is that 37 is held xed while taking the derivative with respect to It is easily shown that with this de nition of the Green s function 6 Eq 5 nds the electrostatic potential for an arbitrary charge density In order to nd the Green s function which satis es Eq 6 we notice that we can use two independent solutions to the homogeneous equation Wad o 739 where i 1 or 2 to form 4 GW39 73 1ltmltgt 2ltxgtgt 8 This notation means that a7lt should be taken the smaller of and 37 and a7gt should be taken the larger In this expression IV is the VVronskian 03 2 IV E 96 1at We can check that this recipe works by noting that for 35 37 Eq 8 satis es the de ning equation 6 by virtue of the fact that it is equal to a product of solutions to the homogeneous equation 7 The de ning equation is singular at ar but integrating 6 over in the neighborhood of 37 6 lt lt 37 6 gives the result dGm dGm I 7 1 l b 6 l e 4 0 In our present case we can choose 1at and 12 1 so that W 1 and the Green s function is given above For this piecewise continuous form of the Green s function the integration 5 can be evaluated 1 47161 cm Gm m pm dm x Ga m pm dm 11 which becomes 1 at x j m39pm dat pat da 12 cu 730 L Evaluating this expression we nd that we obtain the same result given in Eq In general the Green s function Cm solution 5 depends upon the boundary conditions of the problem well as on the charge density In this example the solution is valid for all neutral charge densities that is 13 0
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# reg1-67 - STA 3024 Section 7454/7459 Name Homework 1 for...
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STA 3024 Homework 1 for Regression Section 7454/7459 Form C Name: UFID: This form is only for students whose UFID number begins with 6 or 7. Answer all problems neatly . If you turn in more than one piece of paper, they must be stapled . Work that isn’t neat or stapled may lose points. You don’t need to turn in this problem listing, but you do need to include your name , UF-ID number , and form code on your homework. 1. When paleontologists identify a new species of dinosaur, it’s often the case that they find only a few bones, rather than an entire skeleton. Still, they are usually able to figure out some basic facts, such as how big their new species of dinosaur was. They do this by comparing their new bones to the corresponding bones of related species whose sizes are already known. Suppose paleontologists discover a femur (thigh bone) for a new species of dinosaur. This femur is 0.44 m long. Shown below are the femur lengths and overall lengths of ten species that the paleontologists believe are closely related to their new species. ( X ) ( Y ) Species Femur (m) Overall (m) 1 0.63 9.1 2 0.46 7.6 3 0.24 4.2 4 0.50 7.7 5 0.39 6.5 6 0.22 3.1 7 0.47 6.6 8 0.58 7.7 9 0.60 7.6 10 0.35 5.1 0.3 0.4 0.5 0.6 3 4 5 6 7 8 9 Femur Length (m) Overall Length (m) (a) i. Calculate the regression equation using the following information: ¯ X = 0 . 444 , S X = 0 . 1434 , ¯ Y = 6 . 52 , S Y = 1 . 85 , r = 0 . 949 . (So that you can check your work, part of your answer should be that a = 1 . 08 . ) ii. Calculate the predicted overall length for Species 4 in the data set. iii. Calculate the residual for Species 4 in the data set. iv. Calculate the predicted overall length for the paleontologists’ new species. (Recall that its femur is 0.44 m long.) v. Suppose that some other paleontologists discover another femur with a length of 1.22 m from another new species that is also related to the species in this data set. Explain why it would
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This section has the following videos:
• Lesson Video
• Example Set A Practice Problems (Video Solutions)
• Example Set B Practice Problems (Video Solutions)
• DOWNLOAD the Practice Problem Worksheet Here (the videos below explain the solutions to the problems)
One of the top mistakes that cause algebra students to fail is not mastering positive and negative numbers. The rules are easy however many students tend to confuse and mix the rules up- resulting in poor algebra scores. Don’t think this problem of mastering the basics in only contained in middle and high school students even the best college students make these algebra errors. The best way to remember the rules for positive and negative numbers is to group all the rules into two major rules. One rule is for multiplying and dividing and another rule for adding and subtracting. Another words if you can multiply positive and negative numbers than you can divide them because the rule is the same. Likewise if you can add positive and negative numbers than you can subtract them. Take a look at the notes for these two rules and do whatever it takes to master them- your algebra grades or college grades will be helped big time! | 250 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-47 | longest | en | 0.901658 |
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# Check Your Understanding Direction: Read and analyze each item. Identify the what is being asked in the following. Choose your answer inside the box 1. It is done by a conservative force in a closed path is zero 2. It is the capacity of an object to do work by the virtue of its motion or configuration (position) 3. The sum total of an object's kinetic and potential energy at any given point in time is its total mechanical energy 4. It is the ability of an object to do work by the virtue of its configuration or position 5. It is the ability of an object to do work by the virtue of its motion 6. It is the kinetic energy of the object in joules (J). 7. Term means that energy has been stored and can be used at another time 8. As the car descends the first slope, its PE is converted to and t 9. As the car moves up the next slope, some of the KE is transformed back into car slows down. 10. It is expressed as, where, K is the kinetic energy of the object in joules (J), m is the of the object in kilograms and v is the velocity of the object: POTENTIAL ENERGY POTENTIAL CONSERVATION OF ENERGY OF MECHANICAL ENERGY KINETIC ENERGY CONSERVATIVE FORCE MECHANICAL ENERGY KE PE K
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If you were not able to experience the above listed changes, what might have caused difference?
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Who propose the negative and positive charges | 363 | 1,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.934103 |
http://www.netlib.org/lapack/explore-html/df/dd1/group___o_t_h_e_rauxiliary_ga6680aa4ad62702cd9a00d7530c8ab53a.html | 1,643,455,235,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304883.8/warc/CC-MAIN-20220129092458-20220129122458-00020.warc.gz | 116,549,245 | 17,744 | LAPACK 3.10.0 LAPACK: Linear Algebra PACKage
## ◆ slasd4()
subroutine slasd4 ( integer N, integer I, real, dimension( * ) D, real, dimension( * ) Z, real, dimension( * ) DELTA, real RHO, real SIGMA, real, dimension( * ) WORK, integer INFO )
SLASD4 computes the square root of the i-th updated eigenvalue of a positive symmetric rank-one modification to a positive diagonal matrix. Used by sbdsdc.
Purpose:
``` This subroutine computes the square root of the I-th updated
eigenvalue of a positive symmetric rank-one modification to
a positive diagonal matrix whose entries are given as the squares
of the corresponding entries in the array d, and that
0 <= D(i) < D(j) for i < j
and that RHO > 0. This is arranged by the calling routine, and is
no loss in generality. The rank-one modified system is thus
diag( D ) * diag( D ) + RHO * Z * Z_transpose.
where we assume the Euclidean norm of Z is 1.
The method consists of approximating the rational functions in the
secular equation by simpler interpolating rational functions.```
Parameters
[in] N ``` N is INTEGER The length of all arrays.``` [in] I ``` I is INTEGER The index of the eigenvalue to be computed. 1 <= I <= N.``` [in] D ``` D is REAL array, dimension ( N ) The original eigenvalues. It is assumed that they are in order, 0 <= D(I) < D(J) for I < J.``` [in] Z ``` Z is REAL array, dimension ( N ) The components of the updating vector.``` [out] DELTA ``` DELTA is REAL array, dimension ( N ) If N .ne. 1, DELTA contains (D(j) - sigma_I) in its j-th component. If N = 1, then DELTA(1) = 1. The vector DELTA contains the information necessary to construct the (singular) eigenvectors.``` [in] RHO ``` RHO is REAL The scalar in the symmetric updating formula.``` [out] SIGMA ``` SIGMA is REAL The computed sigma_I, the I-th updated eigenvalue.``` [out] WORK ``` WORK is REAL array, dimension ( N ) If N .ne. 1, WORK contains (D(j) + sigma_I) in its j-th component. If N = 1, then WORK( 1 ) = 1.``` [out] INFO ``` INFO is INTEGER = 0: successful exit > 0: if INFO = 1, the updating process failed.```
Internal Parameters:
``` Logical variable ORGATI (origin-at-i?) is used for distinguishing
whether D(i) or D(i+1) is treated as the origin.
ORGATI = .true. origin at i
ORGATI = .false. origin at i+1
Logical variable SWTCH3 (switch-for-3-poles?) is for noting
if we are working with THREE poles!
MAXIT is the maximum number of iterations allowed for each
eigenvalue.```
Contributors:
Ren-Cang Li, Computer Science Division, University of California at Berkeley, USA
Definition at line 152 of file slasd4.f.
153 *
154 * -- LAPACK auxiliary routine --
155 * -- LAPACK is a software package provided by Univ. of Tennessee, --
156 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
157 *
158 * .. Scalar Arguments ..
159 INTEGER I, INFO, N
160 REAL RHO, SIGMA
161 * ..
162 * .. Array Arguments ..
163 REAL D( * ), DELTA( * ), WORK( * ), Z( * )
164 * ..
165 *
166 * =====================================================================
167 *
168 * .. Parameters ..
169 INTEGER MAXIT
170 parameter( maxit = 400 )
171 REAL ZERO, ONE, TWO, THREE, FOUR, EIGHT, TEN
172 parameter( zero = 0.0e+0, one = 1.0e+0, two = 2.0e+0,
173 \$ three = 3.0e+0, four = 4.0e+0, eight = 8.0e+0,
174 \$ ten = 10.0e+0 )
175 * ..
176 * .. Local Scalars ..
177 LOGICAL ORGATI, SWTCH, SWTCH3, GEOMAVG
178 INTEGER II, IIM1, IIP1, IP1, ITER, J, NITER
179 REAL A, B, C, DELSQ, DELSQ2, SQ2, DPHI, DPSI, DTIIM,
180 \$ DTIIP, DTIPSQ, DTISQ, DTNSQ, DTNSQ1, DW, EPS,
181 \$ ERRETM, ETA, PHI, PREW, PSI, RHOINV, SGLB,
182 \$ SGUB, TAU, TAU2, TEMP, TEMP1, TEMP2, W
183 * ..
184 * .. Local Arrays ..
185 REAL DD( 3 ), ZZ( 3 )
186 * ..
187 * .. External Subroutines ..
188 EXTERNAL slaed6, slasd5
189 * ..
190 * .. External Functions ..
191 REAL SLAMCH
192 EXTERNAL slamch
193 * ..
194 * .. Intrinsic Functions ..
195 INTRINSIC abs, max, min, sqrt
196 * ..
197 * .. Executable Statements ..
198 *
199 * Since this routine is called in an inner loop, we do no argument
200 * checking.
201 *
202 * Quick return for N=1 and 2.
203 *
204 info = 0
205 IF( n.EQ.1 ) THEN
206 *
207 * Presumably, I=1 upon entry
208 *
209 sigma = sqrt( d( 1 )*d( 1 )+rho*z( 1 )*z( 1 ) )
210 delta( 1 ) = one
211 work( 1 ) = one
212 RETURN
213 END IF
214 IF( n.EQ.2 ) THEN
215 CALL slasd5( i, d, z, delta, rho, sigma, work )
216 RETURN
217 END IF
218 *
219 * Compute machine epsilon
220 *
221 eps = slamch( 'Epsilon' )
222 rhoinv = one / rho
223 tau2= zero
224 *
225 * The case I = N
226 *
227 IF( i.EQ.n ) THEN
228 *
229 * Initialize some basic variables
230 *
231 ii = n - 1
232 niter = 1
233 *
234 * Calculate initial guess
235 *
236 temp = rho / two
237 *
238 * If ||Z||_2 is not one, then TEMP should be set to
239 * RHO * ||Z||_2^2 / TWO
240 *
241 temp1 = temp / ( d( n )+sqrt( d( n )*d( n )+temp ) )
242 DO 10 j = 1, n
243 work( j ) = d( j ) + d( n ) + temp1
244 delta( j ) = ( d( j )-d( n ) ) - temp1
245 10 CONTINUE
246 *
247 psi = zero
248 DO 20 j = 1, n - 2
249 psi = psi + z( j )*z( j ) / ( delta( j )*work( j ) )
250 20 CONTINUE
251 *
252 c = rhoinv + psi
253 w = c + z( ii )*z( ii ) / ( delta( ii )*work( ii ) ) +
254 \$ z( n )*z( n ) / ( delta( n )*work( n ) )
255 *
256 IF( w.LE.zero ) THEN
257 temp1 = sqrt( d( n )*d( n )+rho )
258 temp = z( n-1 )*z( n-1 ) / ( ( d( n-1 )+temp1 )*
259 \$ ( d( n )-d( n-1 )+rho / ( d( n )+temp1 ) ) ) +
260 \$ z( n )*z( n ) / rho
261 *
262 * The following TAU2 is to approximate
263 * SIGMA_n^2 - D( N )*D( N )
264 *
265 IF( c.LE.temp ) THEN
266 tau = rho
267 ELSE
268 delsq = ( d( n )-d( n-1 ) )*( d( n )+d( n-1 ) )
269 a = -c*delsq + z( n-1 )*z( n-1 ) + z( n )*z( n )
270 b = z( n )*z( n )*delsq
271 IF( a.LT.zero ) THEN
272 tau2 = two*b / ( sqrt( a*a+four*b*c )-a )
273 ELSE
274 tau2 = ( a+sqrt( a*a+four*b*c ) ) / ( two*c )
275 END IF
276 tau = tau2 / ( d( n )+sqrt( d( n )*d( n )+tau2 ) )
277 END IF
278 *
279 * It can be proved that
280 * D(N)^2+RHO/2 <= SIGMA_n^2 < D(N)^2+TAU2 <= D(N)^2+RHO
281 *
282 ELSE
283 delsq = ( d( n )-d( n-1 ) )*( d( n )+d( n-1 ) )
284 a = -c*delsq + z( n-1 )*z( n-1 ) + z( n )*z( n )
285 b = z( n )*z( n )*delsq
286 *
287 * The following TAU2 is to approximate
288 * SIGMA_n^2 - D( N )*D( N )
289 *
290 IF( a.LT.zero ) THEN
291 tau2 = two*b / ( sqrt( a*a+four*b*c )-a )
292 ELSE
293 tau2 = ( a+sqrt( a*a+four*b*c ) ) / ( two*c )
294 END IF
295 tau = tau2 / ( d( n )+sqrt( d( n )*d( n )+tau2 ) )
296
297 *
298 * It can be proved that
299 * D(N)^2 < D(N)^2+TAU2 < SIGMA(N)^2 < D(N)^2+RHO/2
300 *
301 END IF
302 *
303 * The following TAU is to approximate SIGMA_n - D( N )
304 *
305 * TAU = TAU2 / ( D( N )+SQRT( D( N )*D( N )+TAU2 ) )
306 *
307 sigma = d( n ) + tau
308 DO 30 j = 1, n
309 delta( j ) = ( d( j )-d( n ) ) - tau
310 work( j ) = d( j ) + d( n ) + tau
311 30 CONTINUE
312 *
313 * Evaluate PSI and the derivative DPSI
314 *
315 dpsi = zero
316 psi = zero
317 erretm = zero
318 DO 40 j = 1, ii
319 temp = z( j ) / ( delta( j )*work( j ) )
320 psi = psi + z( j )*temp
321 dpsi = dpsi + temp*temp
322 erretm = erretm + psi
323 40 CONTINUE
324 erretm = abs( erretm )
325 *
326 * Evaluate PHI and the derivative DPHI
327 *
328 temp = z( n ) / ( delta( n )*work( n ) )
329 phi = z( n )*temp
330 dphi = temp*temp
331 erretm = eight*( -phi-psi ) + erretm - phi + rhoinv
332 * \$ + ABS( TAU2 )*( DPSI+DPHI )
333 *
334 w = rhoinv + phi + psi
335 *
336 * Test for convergence
337 *
338 IF( abs( w ).LE.eps*erretm ) THEN
339 GO TO 240
340 END IF
341 *
342 * Calculate the new step
343 *
344 niter = niter + 1
345 dtnsq1 = work( n-1 )*delta( n-1 )
346 dtnsq = work( n )*delta( n )
347 c = w - dtnsq1*dpsi - dtnsq*dphi
348 a = ( dtnsq+dtnsq1 )*w - dtnsq*dtnsq1*( dpsi+dphi )
349 b = dtnsq*dtnsq1*w
350 IF( c.LT.zero )
351 \$ c = abs( c )
352 IF( c.EQ.zero ) THEN
353 eta = rho - sigma*sigma
354 ELSE IF( a.GE.zero ) THEN
355 eta = ( a+sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
356 ELSE
357 eta = two*b / ( a-sqrt( abs( a*a-four*b*c ) ) )
358 END IF
359 *
360 * Note, eta should be positive if w is negative, and
361 * eta should be negative otherwise. However,
362 * if for some reason caused by roundoff, eta*w > 0,
363 * we simply use one Newton step instead. This way
364 * will guarantee eta*w < 0.
365 *
366 IF( w*eta.GT.zero )
367 \$ eta = -w / ( dpsi+dphi )
368 temp = eta - dtnsq
369 IF( temp.GT.rho )
370 \$ eta = rho + dtnsq
371 *
372 eta = eta / ( sigma+sqrt( eta+sigma*sigma ) )
373 tau = tau + eta
374 sigma = sigma + eta
375 *
376 DO 50 j = 1, n
377 delta( j ) = delta( j ) - eta
378 work( j ) = work( j ) + eta
379 50 CONTINUE
380 *
381 * Evaluate PSI and the derivative DPSI
382 *
383 dpsi = zero
384 psi = zero
385 erretm = zero
386 DO 60 j = 1, ii
387 temp = z( j ) / ( work( j )*delta( j ) )
388 psi = psi + z( j )*temp
389 dpsi = dpsi + temp*temp
390 erretm = erretm + psi
391 60 CONTINUE
392 erretm = abs( erretm )
393 *
394 * Evaluate PHI and the derivative DPHI
395 *
396 tau2 = work( n )*delta( n )
397 temp = z( n ) / tau2
398 phi = z( n )*temp
399 dphi = temp*temp
400 erretm = eight*( -phi-psi ) + erretm - phi + rhoinv
401 * \$ + ABS( TAU2 )*( DPSI+DPHI )
402 *
403 w = rhoinv + phi + psi
404 *
405 * Main loop to update the values of the array DELTA
406 *
407 iter = niter + 1
408 *
409 DO 90 niter = iter, maxit
410 *
411 * Test for convergence
412 *
413 IF( abs( w ).LE.eps*erretm ) THEN
414 GO TO 240
415 END IF
416 *
417 * Calculate the new step
418 *
419 dtnsq1 = work( n-1 )*delta( n-1 )
420 dtnsq = work( n )*delta( n )
421 c = w - dtnsq1*dpsi - dtnsq*dphi
422 a = ( dtnsq+dtnsq1 )*w - dtnsq1*dtnsq*( dpsi+dphi )
423 b = dtnsq1*dtnsq*w
424 IF( a.GE.zero ) THEN
425 eta = ( a+sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
426 ELSE
427 eta = two*b / ( a-sqrt( abs( a*a-four*b*c ) ) )
428 END IF
429 *
430 * Note, eta should be positive if w is negative, and
431 * eta should be negative otherwise. However,
432 * if for some reason caused by roundoff, eta*w > 0,
433 * we simply use one Newton step instead. This way
434 * will guarantee eta*w < 0.
435 *
436 IF( w*eta.GT.zero )
437 \$ eta = -w / ( dpsi+dphi )
438 temp = eta - dtnsq
439 IF( temp.LE.zero )
440 \$ eta = eta / two
441 *
442 eta = eta / ( sigma+sqrt( eta+sigma*sigma ) )
443 tau = tau + eta
444 sigma = sigma + eta
445 *
446 DO 70 j = 1, n
447 delta( j ) = delta( j ) - eta
448 work( j ) = work( j ) + eta
449 70 CONTINUE
450 *
451 * Evaluate PSI and the derivative DPSI
452 *
453 dpsi = zero
454 psi = zero
455 erretm = zero
456 DO 80 j = 1, ii
457 temp = z( j ) / ( work( j )*delta( j ) )
458 psi = psi + z( j )*temp
459 dpsi = dpsi + temp*temp
460 erretm = erretm + psi
461 80 CONTINUE
462 erretm = abs( erretm )
463 *
464 * Evaluate PHI and the derivative DPHI
465 *
466 tau2 = work( n )*delta( n )
467 temp = z( n ) / tau2
468 phi = z( n )*temp
469 dphi = temp*temp
470 erretm = eight*( -phi-psi ) + erretm - phi + rhoinv
471 * \$ + ABS( TAU2 )*( DPSI+DPHI )
472 *
473 w = rhoinv + phi + psi
474 90 CONTINUE
475 *
476 * Return with INFO = 1, NITER = MAXIT and not converged
477 *
478 info = 1
479 GO TO 240
480 *
481 * End for the case I = N
482 *
483 ELSE
484 *
485 * The case for I < N
486 *
487 niter = 1
488 ip1 = i + 1
489 *
490 * Calculate initial guess
491 *
492 delsq = ( d( ip1 )-d( i ) )*( d( ip1 )+d( i ) )
493 delsq2 = delsq / two
494 sq2=sqrt( ( d( i )*d( i )+d( ip1 )*d( ip1 ) ) / two )
495 temp = delsq2 / ( d( i )+sq2 )
496 DO 100 j = 1, n
497 work( j ) = d( j ) + d( i ) + temp
498 delta( j ) = ( d( j )-d( i ) ) - temp
499 100 CONTINUE
500 *
501 psi = zero
502 DO 110 j = 1, i - 1
503 psi = psi + z( j )*z( j ) / ( work( j )*delta( j ) )
504 110 CONTINUE
505 *
506 phi = zero
507 DO 120 j = n, i + 2, -1
508 phi = phi + z( j )*z( j ) / ( work( j )*delta( j ) )
509 120 CONTINUE
510 c = rhoinv + psi + phi
511 w = c + z( i )*z( i ) / ( work( i )*delta( i ) ) +
512 \$ z( ip1 )*z( ip1 ) / ( work( ip1 )*delta( ip1 ) )
513 *
514 geomavg = .false.
515 IF( w.GT.zero ) THEN
516 *
517 * d(i)^2 < the ith sigma^2 < (d(i)^2+d(i+1)^2)/2
518 *
519 * We choose d(i) as origin.
520 *
521 orgati = .true.
522 ii = i
523 sglb = zero
524 sgub = delsq2 / ( d( i )+sq2 )
525 a = c*delsq + z( i )*z( i ) + z( ip1 )*z( ip1 )
526 b = z( i )*z( i )*delsq
527 IF( a.GT.zero ) THEN
528 tau2 = two*b / ( a+sqrt( abs( a*a-four*b*c ) ) )
529 ELSE
530 tau2 = ( a-sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
531 END IF
532 *
533 * TAU2 now is an estimation of SIGMA^2 - D( I )^2. The
534 * following, however, is the corresponding estimation of
535 * SIGMA - D( I ).
536 *
537 tau = tau2 / ( d( i )+sqrt( d( i )*d( i )+tau2 ) )
538 temp = sqrt(eps)
539 IF( (d(i).LE.temp*d(ip1)).AND.(abs(z(i)).LE.temp)
540 \$ .AND.(d(i).GT.zero) ) THEN
541 tau = min( ten*d(i), sgub )
542 geomavg = .true.
543 END IF
544 ELSE
545 *
546 * (d(i)^2+d(i+1)^2)/2 <= the ith sigma^2 < d(i+1)^2/2
547 *
548 * We choose d(i+1) as origin.
549 *
550 orgati = .false.
551 ii = ip1
552 sglb = -delsq2 / ( d( ii )+sq2 )
553 sgub = zero
554 a = c*delsq - z( i )*z( i ) - z( ip1 )*z( ip1 )
555 b = z( ip1 )*z( ip1 )*delsq
556 IF( a.LT.zero ) THEN
557 tau2 = two*b / ( a-sqrt( abs( a*a+four*b*c ) ) )
558 ELSE
559 tau2 = -( a+sqrt( abs( a*a+four*b*c ) ) ) / ( two*c )
560 END IF
561 *
562 * TAU2 now is an estimation of SIGMA^2 - D( IP1 )^2. The
563 * following, however, is the corresponding estimation of
564 * SIGMA - D( IP1 ).
565 *
566 tau = tau2 / ( d( ip1 )+sqrt( abs( d( ip1 )*d( ip1 )+
567 \$ tau2 ) ) )
568 END IF
569 *
570 sigma = d( ii ) + tau
571 DO 130 j = 1, n
572 work( j ) = d( j ) + d( ii ) + tau
573 delta( j ) = ( d( j )-d( ii ) ) - tau
574 130 CONTINUE
575 iim1 = ii - 1
576 iip1 = ii + 1
577 *
578 * Evaluate PSI and the derivative DPSI
579 *
580 dpsi = zero
581 psi = zero
582 erretm = zero
583 DO 150 j = 1, iim1
584 temp = z( j ) / ( work( j )*delta( j ) )
585 psi = psi + z( j )*temp
586 dpsi = dpsi + temp*temp
587 erretm = erretm + psi
588 150 CONTINUE
589 erretm = abs( erretm )
590 *
591 * Evaluate PHI and the derivative DPHI
592 *
593 dphi = zero
594 phi = zero
595 DO 160 j = n, iip1, -1
596 temp = z( j ) / ( work( j )*delta( j ) )
597 phi = phi + z( j )*temp
598 dphi = dphi + temp*temp
599 erretm = erretm + phi
600 160 CONTINUE
601 *
602 w = rhoinv + phi + psi
603 *
604 * W is the value of the secular function with
605 * its ii-th element removed.
606 *
607 swtch3 = .false.
608 IF( orgati ) THEN
609 IF( w.LT.zero )
610 \$ swtch3 = .true.
611 ELSE
612 IF( w.GT.zero )
613 \$ swtch3 = .true.
614 END IF
615 IF( ii.EQ.1 .OR. ii.EQ.n )
616 \$ swtch3 = .false.
617 *
618 temp = z( ii ) / ( work( ii )*delta( ii ) )
619 dw = dpsi + dphi + temp*temp
620 temp = z( ii )*temp
621 w = w + temp
622 erretm = eight*( phi-psi ) + erretm + two*rhoinv
623 \$ + three*abs( temp )
624 * \$ + ABS( TAU2 )*DW
625 *
626 * Test for convergence
627 *
628 IF( abs( w ).LE.eps*erretm ) THEN
629 GO TO 240
630 END IF
631 *
632 IF( w.LE.zero ) THEN
633 sglb = max( sglb, tau )
634 ELSE
635 sgub = min( sgub, tau )
636 END IF
637 *
638 * Calculate the new step
639 *
640 niter = niter + 1
641 IF( .NOT.swtch3 ) THEN
642 dtipsq = work( ip1 )*delta( ip1 )
643 dtisq = work( i )*delta( i )
644 IF( orgati ) THEN
645 c = w - dtipsq*dw + delsq*( z( i ) / dtisq )**2
646 ELSE
647 c = w - dtisq*dw - delsq*( z( ip1 ) / dtipsq )**2
648 END IF
649 a = ( dtipsq+dtisq )*w - dtipsq*dtisq*dw
650 b = dtipsq*dtisq*w
651 IF( c.EQ.zero ) THEN
652 IF( a.EQ.zero ) THEN
653 IF( orgati ) THEN
654 a = z( i )*z( i ) + dtipsq*dtipsq*( dpsi+dphi )
655 ELSE
656 a = z( ip1 )*z( ip1 ) + dtisq*dtisq*( dpsi+dphi )
657 END IF
658 END IF
659 eta = b / a
660 ELSE IF( a.LE.zero ) THEN
661 eta = ( a-sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
662 ELSE
663 eta = two*b / ( a+sqrt( abs( a*a-four*b*c ) ) )
664 END IF
665 ELSE
666 *
667 * Interpolation using THREE most relevant poles
668 *
669 dtiim = work( iim1 )*delta( iim1 )
670 dtiip = work( iip1 )*delta( iip1 )
671 temp = rhoinv + psi + phi
672 IF( orgati ) THEN
673 temp1 = z( iim1 ) / dtiim
674 temp1 = temp1*temp1
675 c = ( temp - dtiip*( dpsi+dphi ) ) -
676 \$ ( d( iim1 )-d( iip1 ) )*( d( iim1 )+d( iip1 ) )*temp1
677 zz( 1 ) = z( iim1 )*z( iim1 )
678 IF( dpsi.LT.temp1 ) THEN
679 zz( 3 ) = dtiip*dtiip*dphi
680 ELSE
681 zz( 3 ) = dtiip*dtiip*( ( dpsi-temp1 )+dphi )
682 END IF
683 ELSE
684 temp1 = z( iip1 ) / dtiip
685 temp1 = temp1*temp1
686 c = ( temp - dtiim*( dpsi+dphi ) ) -
687 \$ ( d( iip1 )-d( iim1 ) )*( d( iim1 )+d( iip1 ) )*temp1
688 IF( dphi.LT.temp1 ) THEN
689 zz( 1 ) = dtiim*dtiim*dpsi
690 ELSE
691 zz( 1 ) = dtiim*dtiim*( dpsi+( dphi-temp1 ) )
692 END IF
693 zz( 3 ) = z( iip1 )*z( iip1 )
694 END IF
695 zz( 2 ) = z( ii )*z( ii )
696 dd( 1 ) = dtiim
697 dd( 2 ) = delta( ii )*work( ii )
698 dd( 3 ) = dtiip
699 CALL slaed6( niter, orgati, c, dd, zz, w, eta, info )
700 *
701 IF( info.NE.0 ) THEN
702 *
703 * If INFO is not 0, i.e., SLAED6 failed, switch back
704 * to 2 pole interpolation.
705 *
706 swtch3 = .false.
707 info = 0
708 dtipsq = work( ip1 )*delta( ip1 )
709 dtisq = work( i )*delta( i )
710 IF( orgati ) THEN
711 c = w - dtipsq*dw + delsq*( z( i ) / dtisq )**2
712 ELSE
713 c = w - dtisq*dw - delsq*( z( ip1 ) / dtipsq )**2
714 END IF
715 a = ( dtipsq+dtisq )*w - dtipsq*dtisq*dw
716 b = dtipsq*dtisq*w
717 IF( c.EQ.zero ) THEN
718 IF( a.EQ.zero ) THEN
719 IF( orgati ) THEN
720 a = z( i )*z( i ) + dtipsq*dtipsq*( dpsi+dphi )
721 ELSE
722 a = z( ip1 )*z( ip1 ) + dtisq*dtisq*( dpsi+dphi)
723 END IF
724 END IF
725 eta = b / a
726 ELSE IF( a.LE.zero ) THEN
727 eta = ( a-sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
728 ELSE
729 eta = two*b / ( a+sqrt( abs( a*a-four*b*c ) ) )
730 END IF
731 END IF
732 END IF
733 *
734 * Note, eta should be positive if w is negative, and
735 * eta should be negative otherwise. However,
736 * if for some reason caused by roundoff, eta*w > 0,
737 * we simply use one Newton step instead. This way
738 * will guarantee eta*w < 0.
739 *
740 IF( w*eta.GE.zero )
741 \$ eta = -w / dw
742 *
743 eta = eta / ( sigma+sqrt( sigma*sigma+eta ) )
744 temp = tau + eta
745 IF( temp.GT.sgub .OR. temp.LT.sglb ) THEN
746 IF( w.LT.zero ) THEN
747 eta = ( sgub-tau ) / two
748 ELSE
749 eta = ( sglb-tau ) / two
750 END IF
751 IF( geomavg ) THEN
752 IF( w .LT. zero ) THEN
753 IF( tau .GT. zero ) THEN
754 eta = sqrt(sgub*tau)-tau
755 END IF
756 ELSE
757 IF( sglb .GT. zero ) THEN
758 eta = sqrt(sglb*tau)-tau
759 END IF
760 END IF
761 END IF
762 END IF
763 *
764 prew = w
765 *
766 tau = tau + eta
767 sigma = sigma + eta
768 *
769 DO 170 j = 1, n
770 work( j ) = work( j ) + eta
771 delta( j ) = delta( j ) - eta
772 170 CONTINUE
773 *
774 * Evaluate PSI and the derivative DPSI
775 *
776 dpsi = zero
777 psi = zero
778 erretm = zero
779 DO 180 j = 1, iim1
780 temp = z( j ) / ( work( j )*delta( j ) )
781 psi = psi + z( j )*temp
782 dpsi = dpsi + temp*temp
783 erretm = erretm + psi
784 180 CONTINUE
785 erretm = abs( erretm )
786 *
787 * Evaluate PHI and the derivative DPHI
788 *
789 dphi = zero
790 phi = zero
791 DO 190 j = n, iip1, -1
792 temp = z( j ) / ( work( j )*delta( j ) )
793 phi = phi + z( j )*temp
794 dphi = dphi + temp*temp
795 erretm = erretm + phi
796 190 CONTINUE
797 *
798 tau2 = work( ii )*delta( ii )
799 temp = z( ii ) / tau2
800 dw = dpsi + dphi + temp*temp
801 temp = z( ii )*temp
802 w = rhoinv + phi + psi + temp
803 erretm = eight*( phi-psi ) + erretm + two*rhoinv
804 \$ + three*abs( temp )
805 * \$ + ABS( TAU2 )*DW
806 *
807 swtch = .false.
808 IF( orgati ) THEN
809 IF( -w.GT.abs( prew ) / ten )
810 \$ swtch = .true.
811 ELSE
812 IF( w.GT.abs( prew ) / ten )
813 \$ swtch = .true.
814 END IF
815 *
816 * Main loop to update the values of the array DELTA and WORK
817 *
818 iter = niter + 1
819 *
820 DO 230 niter = iter, maxit
821 *
822 * Test for convergence
823 *
824 IF( abs( w ).LE.eps*erretm ) THEN
825 * \$ .OR. (SGUB-SGLB).LE.EIGHT*ABS(SGUB+SGLB) ) THEN
826 GO TO 240
827 END IF
828 *
829 IF( w.LE.zero ) THEN
830 sglb = max( sglb, tau )
831 ELSE
832 sgub = min( sgub, tau )
833 END IF
834 *
835 * Calculate the new step
836 *
837 IF( .NOT.swtch3 ) THEN
838 dtipsq = work( ip1 )*delta( ip1 )
839 dtisq = work( i )*delta( i )
840 IF( .NOT.swtch ) THEN
841 IF( orgati ) THEN
842 c = w - dtipsq*dw + delsq*( z( i ) / dtisq )**2
843 ELSE
844 c = w - dtisq*dw - delsq*( z( ip1 ) / dtipsq )**2
845 END IF
846 ELSE
847 temp = z( ii ) / ( work( ii )*delta( ii ) )
848 IF( orgati ) THEN
849 dpsi = dpsi + temp*temp
850 ELSE
851 dphi = dphi + temp*temp
852 END IF
853 c = w - dtisq*dpsi - dtipsq*dphi
854 END IF
855 a = ( dtipsq+dtisq )*w - dtipsq*dtisq*dw
856 b = dtipsq*dtisq*w
857 IF( c.EQ.zero ) THEN
858 IF( a.EQ.zero ) THEN
859 IF( .NOT.swtch ) THEN
860 IF( orgati ) THEN
861 a = z( i )*z( i ) + dtipsq*dtipsq*
862 \$ ( dpsi+dphi )
863 ELSE
864 a = z( ip1 )*z( ip1 ) +
865 \$ dtisq*dtisq*( dpsi+dphi )
866 END IF
867 ELSE
868 a = dtisq*dtisq*dpsi + dtipsq*dtipsq*dphi
869 END IF
870 END IF
871 eta = b / a
872 ELSE IF( a.LE.zero ) THEN
873 eta = ( a-sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
874 ELSE
875 eta = two*b / ( a+sqrt( abs( a*a-four*b*c ) ) )
876 END IF
877 ELSE
878 *
879 * Interpolation using THREE most relevant poles
880 *
881 dtiim = work( iim1 )*delta( iim1 )
882 dtiip = work( iip1 )*delta( iip1 )
883 temp = rhoinv + psi + phi
884 IF( swtch ) THEN
885 c = temp - dtiim*dpsi - dtiip*dphi
886 zz( 1 ) = dtiim*dtiim*dpsi
887 zz( 3 ) = dtiip*dtiip*dphi
888 ELSE
889 IF( orgati ) THEN
890 temp1 = z( iim1 ) / dtiim
891 temp1 = temp1*temp1
892 temp2 = ( d( iim1 )-d( iip1 ) )*
893 \$ ( d( iim1 )+d( iip1 ) )*temp1
894 c = temp - dtiip*( dpsi+dphi ) - temp2
895 zz( 1 ) = z( iim1 )*z( iim1 )
896 IF( dpsi.LT.temp1 ) THEN
897 zz( 3 ) = dtiip*dtiip*dphi
898 ELSE
899 zz( 3 ) = dtiip*dtiip*( ( dpsi-temp1 )+dphi )
900 END IF
901 ELSE
902 temp1 = z( iip1 ) / dtiip
903 temp1 = temp1*temp1
904 temp2 = ( d( iip1 )-d( iim1 ) )*
905 \$ ( d( iim1 )+d( iip1 ) )*temp1
906 c = temp - dtiim*( dpsi+dphi ) - temp2
907 IF( dphi.LT.temp1 ) THEN
908 zz( 1 ) = dtiim*dtiim*dpsi
909 ELSE
910 zz( 1 ) = dtiim*dtiim*( dpsi+( dphi-temp1 ) )
911 END IF
912 zz( 3 ) = z( iip1 )*z( iip1 )
913 END IF
914 END IF
915 dd( 1 ) = dtiim
916 dd( 2 ) = delta( ii )*work( ii )
917 dd( 3 ) = dtiip
918 CALL slaed6( niter, orgati, c, dd, zz, w, eta, info )
919 *
920 IF( info.NE.0 ) THEN
921 *
922 * If INFO is not 0, i.e., SLAED6 failed, switch
923 * back to two pole interpolation
924 *
925 swtch3 = .false.
926 info = 0
927 dtipsq = work( ip1 )*delta( ip1 )
928 dtisq = work( i )*delta( i )
929 IF( .NOT.swtch ) THEN
930 IF( orgati ) THEN
931 c = w - dtipsq*dw + delsq*( z( i )/dtisq )**2
932 ELSE
933 c = w - dtisq*dw - delsq*( z( ip1 )/dtipsq )**2
934 END IF
935 ELSE
936 temp = z( ii ) / ( work( ii )*delta( ii ) )
937 IF( orgati ) THEN
938 dpsi = dpsi + temp*temp
939 ELSE
940 dphi = dphi + temp*temp
941 END IF
942 c = w - dtisq*dpsi - dtipsq*dphi
943 END IF
944 a = ( dtipsq+dtisq )*w - dtipsq*dtisq*dw
945 b = dtipsq*dtisq*w
946 IF( c.EQ.zero ) THEN
947 IF( a.EQ.zero ) THEN
948 IF( .NOT.swtch ) THEN
949 IF( orgati ) THEN
950 a = z( i )*z( i ) + dtipsq*dtipsq*
951 \$ ( dpsi+dphi )
952 ELSE
953 a = z( ip1 )*z( ip1 ) +
954 \$ dtisq*dtisq*( dpsi+dphi )
955 END IF
956 ELSE
957 a = dtisq*dtisq*dpsi + dtipsq*dtipsq*dphi
958 END IF
959 END IF
960 eta = b / a
961 ELSE IF( a.LE.zero ) THEN
962 eta = ( a-sqrt( abs( a*a-four*b*c ) ) ) / ( two*c )
963 ELSE
964 eta = two*b / ( a+sqrt( abs( a*a-four*b*c ) ) )
965 END IF
966 END IF
967 END IF
968 *
969 * Note, eta should be positive if w is negative, and
970 * eta should be negative otherwise. However,
971 * if for some reason caused by roundoff, eta*w > 0,
972 * we simply use one Newton step instead. This way
973 * will guarantee eta*w < 0.
974 *
975 IF( w*eta.GE.zero )
976 \$ eta = -w / dw
977 *
978 eta = eta / ( sigma+sqrt( sigma*sigma+eta ) )
979 temp=tau+eta
980 IF( temp.GT.sgub .OR. temp.LT.sglb ) THEN
981 IF( w.LT.zero ) THEN
982 eta = ( sgub-tau ) / two
983 ELSE
984 eta = ( sglb-tau ) / two
985 END IF
986 IF( geomavg ) THEN
987 IF( w .LT. zero ) THEN
988 IF( tau .GT. zero ) THEN
989 eta = sqrt(sgub*tau)-tau
990 END IF
991 ELSE
992 IF( sglb .GT. zero ) THEN
993 eta = sqrt(sglb*tau)-tau
994 END IF
995 END IF
996 END IF
997 END IF
998 *
999 prew = w
1000 *
1001 tau = tau + eta
1002 sigma = sigma + eta
1003 *
1004 DO 200 j = 1, n
1005 work( j ) = work( j ) + eta
1006 delta( j ) = delta( j ) - eta
1007 200 CONTINUE
1008 *
1009 * Evaluate PSI and the derivative DPSI
1010 *
1011 dpsi = zero
1012 psi = zero
1013 erretm = zero
1014 DO 210 j = 1, iim1
1015 temp = z( j ) / ( work( j )*delta( j ) )
1016 psi = psi + z( j )*temp
1017 dpsi = dpsi + temp*temp
1018 erretm = erretm + psi
1019 210 CONTINUE
1020 erretm = abs( erretm )
1021 *
1022 * Evaluate PHI and the derivative DPHI
1023 *
1024 dphi = zero
1025 phi = zero
1026 DO 220 j = n, iip1, -1
1027 temp = z( j ) / ( work( j )*delta( j ) )
1028 phi = phi + z( j )*temp
1029 dphi = dphi + temp*temp
1030 erretm = erretm + phi
1031 220 CONTINUE
1032 *
1033 tau2 = work( ii )*delta( ii )
1034 temp = z( ii ) / tau2
1035 dw = dpsi + dphi + temp*temp
1036 temp = z( ii )*temp
1037 w = rhoinv + phi + psi + temp
1038 erretm = eight*( phi-psi ) + erretm + two*rhoinv
1039 \$ + three*abs( temp )
1040 * \$ + ABS( TAU2 )*DW
1041 *
1042 IF( w*prew.GT.zero .AND. abs( w ).GT.abs( prew ) / ten )
1043 \$ swtch = .NOT.swtch
1044 *
1045 230 CONTINUE
1046 *
1047 * Return with INFO = 1, NITER = MAXIT and not converged
1048 *
1049 info = 1
1050 *
1051 END IF
1052 *
1053 240 CONTINUE
1054 RETURN
1055 *
1056 * End of SLASD4
1057 *
subroutine slasd5(I, D, Z, DELTA, RHO, DSIGMA, WORK)
SLASD5 computes the square root of the i-th eigenvalue of a positive symmetric rank-one modification ...
Definition: slasd5.f:116
subroutine slaed6(KNITER, ORGATI, RHO, D, Z, FINIT, TAU, INFO)
SLAED6 used by SSTEDC. Computes one Newton step in solution of the secular equation.
Definition: slaed6.f:140
real function slamch(CMACH)
SLAMCH
Definition: slamch.f:68
Here is the call graph for this function:
Here is the caller graph for this function: | 10,887 | 26,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-05 | latest | en | 0.779327 |
http://www.algebra-answer.com/algebra-help/graphing-lines/algebra-2-answer.html | 1,513,026,537,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00330.warc.gz | 310,653,312 | 7,296 | Related topics:
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• graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
• graphing general functions
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ORDER NOW! | 948 | 4,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-51 | longest | en | 0.948256 |
http://caspoc.com/help/userguide/modeling_topics/electrical_machines/permanent_magnet_synchronous_machines/ | 1,544,907,269,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827097.43/warc/CC-MAIN-20181215200626-20181215222626-00042.warc.gz | 41,467,795 | 25,928 | • Introduction
• What is in this manual
• What is Caspoc
• User interface
• Introduction
• Starting
• Simulation
• Editing
• Viewing and printing
• Getting Started
• Basic editing
• Simulation in the time domain
• Basic User Interface Topics
• Editing
• Simulation
• Viewing
• Library
• Reports
• Project management
• Circuit and Block Diagram Components
• Introduction
• Cscript and user defined functions
• Component parameters
• Modeling Topics
• Introduction
• Power Electronics
• Semiconductors
• Electrical Machines
• Electrical drives
• Power Systems
• Mechanical Systems
• Thermal Systems
• Magnetic Circuits
• Green Energy
• Coupling to FEM
• Experimenter
• Analog hardware description language
• Embedded C code Export
• Coupling to Spice
• Small Signal Analysis
• Matlab coupling
• Tips and tricks
• Appendices
Permanent Magnet Synchronous Machines
Click to Close this View Usually, a brushless dc motor is commutated. One stator pole is turned-off and the next pole is energized. The windings of a brushless dc motor are distinct separate poles. This arrangement is the most simple to wind and control, but produces some torque ripple.
A variation of a brushless dc motor uses a slightly different winding. The windings are distributed around the stator to produce a roughly sinusoidal distribution. The stator winding is similar to an ac induction motor. This type of motor is called a brushless ac motor or Permanent Magnet Synchronous Machine (PMSM).
A PMSM is driven with ac sinewave voltages. The permanent magnet rotor rotates synchronous to the rotating magnetic field. The rotating magnetic field is illustrated using a red and green gradient. An actual simulation of the magnetic field would show a far more complex magnetic field.
The basic block BLDCM from the block diagram models the Permanent Magnet Synchronous Machine. The output equals the electric torque produced by the machine. The inputs are the three phase currents, the angular speed of the rotor and the machine constant for the calculation of the torque and the produced EMF.
u1:=K*sin(angle);u2:=K*sin(angle-(2*PI/3));u3:=K*sin(angle-(4*PI/3));if(p1=0)then begin if(u1 < 0) then u1:=K else u1:=-K; if(u2 < 0) then u2:=K else u2:=-K; if(u3 < 0) then u3:=K else u3:=-K; end;T=u1*i1+u2*i2+u3*i3
Inside the basic BLDCM block from the block diagram the type of the machine can be defined.
p1=0 Brushless DC machine BLDCM
p1=1 Permanent Magnet Synchronous Machine
Use an INT_VAR block to get the EMF voltages and position of the rotor:
INT_VAR Output p1=1 u1 p1=2 u2 p1=3 u3 p1=4 Position of the rotor (in Radians) p1=5 Rotor position polar x-coordinate p1=6 Rotor position polar y-coordinate
Equivalent circuit model
The library blocks for the Permanent Magnet Synchronous Machine have electrical and mechanical connections. The electrical part of the Permanent Magnet Synchronous Machine is modeled by a series connection of the stator inductance, resistance and the above mentioned EMF.
The connections R, S and T are the electrical terminals. The stator current is depends on the produced EMF:
us=Rs · is+Lsdis/dt+uEMF)
where uEMFr*ui
The torque produced by the machine equals:
Te=u1 · i1+u2 · i2+u3 · i3
Mechanical part:
The mechanical shaft is modeled by a rigid shaft model. The inertia J [kgm2]of the shaft can be modeled in most of the Permanent Magnet Synchronous Machine library blocks. Most machine models have internal parameters for modeling the inertia and friction of the shaft. The Permanent Magnet Synchronous Machine can either run without load or can be connected to a mechanical rotational model. Use the models from components/circuit/rotational and from the various mechanical libraries for modeling the mechanical model.
DQ-model
There are also models for the Permanent Magnet Synchronous Machine in the library that are based on the dq model. The dq model for the PMSM models the dynamics of a three-phase permanent magnet synchronous machine with sinusoidal flux distribution.
The electrical and mechanical parts of the machine are each represented by a second-order state-space model. The model assumes that the flux established by the permanent magnets in the stator is sinusoidal, which implies that the electromotive forces are sinusoidal.
The block implements the following equations expressed in the rotor reference frame (qd frame).
Electrical System
where (all quantities in the rotor reference frame are referred to the stator)
Lq, Ld q and d axis inductance's R Resistance of the stator windings iq, id q and d axis currents vq, vd q and d axis voltages ωr Angular velocity of the rotor λ Amplitude of the flux induced by the permanent magnets of the rotor in the stator phases p Number of pole pairs Te Electromagnetic torque
Mechanical System
where
J Rotor inertia F Rotor friction θ Rotor angular position Tm Shaft mechanical torque
With excitation winding
Here the parameter Lm defines the coupling between Ld and the exitation winding Le
Ψd=Ld*id+Lm*ie
Ψe=Lm*id+Le*ie
Le Excitation winding inductance Re Excitation winding resistance Lm Mutual inductance between the excitation winding and Ld | 1,222 | 5,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-51 | latest | en | 0.747376 |
https://www.crikey.com.au/2015/05/20/what-the-phenomenal-small-business-write-off-is-really-worth/ | 1,660,426,035,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00662.warc.gz | 608,147,278 | 32,645 | The government’s small business budget has been a big success, it seems. It got positive headlines about the budget being stimulatory, and now some polls show the Prime Minister’s approval rating on the rise.
It’s a PR success. And a lot of that is due to attention lavished on the \$20,000 tax write-off for small business.
It’s “phenomenal”, apparently.
Get ready for a lot more asset write-off announcements in future. Because they buy the government a lot more headlines than they deserve.
You wouldn’t know this from reading about it, but the “\$20,000″ asset write-off is worth only about \$1000.
Here’s why:
The \$20,000 is not taken off the tax bill of a small business. Instead it’s a deduction from income — same as when an individual gives to charity. After a small business takes \$20,000 off its income, it saves the 28.5% of tax it would have paid on it. 28.5% of \$20,000 is \$5700.
That’s the actual value of being able to instantly write off a car or machine from your tax this year.
But here’s the thing. Businesses could always write off asset purchases against their income. They just had to do it more slowly.
Using a depreciation schedule from the ATO website, I calculated how much a small business would have been able to save off its tax under the old rules.
It’s \$5700.
The only advantage is that under the new policy, a business can claim all that \$5700 in this tax year, instead of claiming it in dribs and drabs over the next decade.
Here’s a graph for how your depreciation works under instant write-off versus slow depreciation.
We can measure how much benefit instant access to the write-off provides. All we need to do is make an assumption about how small business values money over time. We do that with a discount rate. Let’s assume a discount rate of 8% (a hypothetical figure at the higher end of the realistic discount rate spectrum).
If that is the case, the net present value (NPV) of the flow of money is \$4660. Only \$1040 less than the value of the money right now.
(If you assume small business is even more patient, the value of the instant write-off is even less. At a 2% discount rate — a figure much lower than the average discount rate for small business — the NPV is \$5350, and the net value of the new policy is a mere \$350.)
In summary, the government is getting great value from this policy in media coverage terms.
Compare it to another tax break the Coalition gave small business in the budget — a 5% tax cut for unincorporated businesses. You probably haven’t seen mention of that anywhere.
But this 5% tax cut (full disclosure, I run an unincorporated business!) is worth even more to the budget bottom line. That’s it in the blue bubble on the right side — worth \$1.8 billion. This graphic was in the glossy brochures journalists got in the budget lock-up.
It’s one of the most expensive measures in the budget. And it has barely got a headline. The government will not make that mistake again.
Expect asset write-off thresholds to be even higher in the next budget as governments seek a headline that says something like \$100,000 Asset Tax Bonus for Small Business.
*This article was originally published at Jason Murphy’s blog, Thomas the Think Engine | 743 | 3,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-33 | latest | en | 0.954611 |
http://www.haskell.org/pipermail/haskell-cafe/2011-March/090404.html | 1,371,533,479,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706933615/warc/CC-MAIN-20130516122213-00085-ip-10-60-113-184.ec2.internal.warc.gz | 500,008,599 | 2,426 | # [Haskell-cafe] Computing the multiplication table of a group using the GHC inliner ; )
Daniel Schüssler anotheraddress at gmx.de
Wed Mar 23 00:45:49 CET 2011
```Hello,
turns out that you can define the group operation of the symmetric group on 3
elements in this abstract way (via the isomorphism to the group of bijective
functions from a three-element type to itself):
s3mult g2 g1 = fromFun (toFun g2 . toFun g1)
and convince GHC to compile it down to a nested case statement. It even
somehow made the left multiplication with the identity non-strict. Just
thought it's neat ;)
\$ ghc-core S3.hs -funfolding-use-threshold=64
...
-- identifiers manually un-qualified for readability
s3mult =
\ (g2_ahA :: S3) (g1_ahB :: S3) ->
case g2_ahA of _ {
S3abc -> g1_ahB;
S3bca ->
case g1_ahB of _ {
S3abc -> S3bca;
S3bca -> S3cab;
S3cab -> S3abc;
S3acb -> S3bac;
S3bac -> S3cba;
S3cba -> S3acb
};
S3cab ->
case g1_ahB of _ {
S3abc -> S3cab;
S3bca -> S3abc;
S3cab -> S3bca;
S3acb -> S3cba;
S3bac -> S3acb;
S3cba -> S3bac
};
S3acb ->
case g1_ahB of _ {
S3abc -> S3acb;
S3bca -> S3cba;
S3cab -> S3bac;
S3acb -> S3abc;
S3bac -> S3cab;
S3cba -> S3bca
};
S3bac ->
case g1_ahB of _ {
S3abc -> S3bac;
S3bca -> S3acb;
S3cab -> S3cba;
S3acb -> S3bca;
S3bac -> S3abc;
S3cba -> S3cab
};
S3cba ->
case g1_ahB of _ {
S3abc -> S3cba;
S3bca -> S3bac;
S3cab -> S3acb;
S3acb -> S3cab;
S3bac -> S3bca;
S3cba -> S3abc
}
}
-- inverse
s3inv =
\ (g_ahC :: S3) ->
case g_ahC of _ {
S3abc -> S3abc;
S3bca -> S3cab;
S3cab -> S3bca;
S3acb -> S3acb;
S3bac -> S3bac;
S3cba -> S3cba
}
--- end core ---
--- source ---
module S3 where
-- | Symmetric group / permutation group on 3 elements
data S3 = S3abc | S3bca | S3cab | S3acb | S3bac | S3cba deriving(Eq)
-- | Returns an element of S3 satisfying the given predicate
s3the :: (S3 -> Bool) -> S3
s3the p
| p S3abc = S3abc
| p S3acb = S3acb
| p S3bac = S3bac
| p S3bca = S3bca
| p S3cba = S3cba
| p S3cab = S3cab
| otherwise = error "s3the: no element satisfies the predicate"
data ABC = A | B | C deriving(Eq)
toFun :: S3 -> ABC -> ABC
toFun g = case g of
S3abc -> mkFun A B C
S3bca -> mkFun B C A
S3cab -> mkFun C A B
S3acb -> mkFun A C B
S3bac -> mkFun B A C
S3cba -> mkFun C B A
where
mkFun imA _ _ A = imA
mkFun _ imB _ B = imB
mkFun _ _ imC _ = imC
fromFun :: (ABC -> ABC) -> S3
fromFun f = s3the (\g ->
toFun g A == f A
&& toFun g B == f B)
s3mult :: S3 -> S3 -> S3
s3mult g2 g1 = fromFun (toFun g2 . toFun g1)
s3inv g = s3the (\g' -> s3mult g' g == S3abc)
``` | 1,019 | 2,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-20 | longest | en | 0.526992 |
https://math.answers.com/math-and-arithmetic/Brian_bought_a_used_bike_for_25_less_than_the_original_price_he_paid_88_for_the_bike_what_ws_the_original_price_of_the_bike | 1,726,133,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00247.warc.gz | 345,580,281 | 49,798 | 0
# Brian bought a used bike for 25 less than the original price he paid 88 for the bike what ws the original price of the bike?
Updated: 10/31/2022
Wiki User
14y ago
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Q: Brian bought a used bike for 25 less than the original price he paid 88 for the bike what ws the original price of the bike?
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### John bought a used bike for 6000 and was given a 15 discount. What was the original price?
If it was a 15 dollar discount, the original price was 60.00+15=75.00 dollars.If it was a 15 percent discount, the original price was \$70.59 as followsoriginalprice - originalprice *.15 = salepriceoriginalprice * (1-.15) = saleproceoriginalprice = saleprice / (1-.15)originalprice = 60.00 / .85originalprice is 70.588 or 70.59 dollars
143.10
250
### How do you find out if a bike you bought is a stolen bike?
I you consult that bike to the nearest police station and you will find out that your bike is stolen or not.
### Where can bike toys be bought?
Bike toys can be bought on eBay, amazon and other online shopping site. Bike toys can also be bought from the stores like Argos who sell toys from their physical as well as online stores .
### Where can a Marin mountain bike be bought?
A Marin mountain bike can be bought from Marin's website, which can be contacted by email or phone. It can also be bought at your local sports shop for a fee.
### What is the average price for a french bike?
the average price for a french bike is about 462 us dollars
### Where can a Razor MX500 be bought?
The Razor MX500 Dirt Rocket Bike can currently only be purchased online and shipped from America. Websites offering this bike at a competitive price include Walmart, Toys R Us and Best Buy.
### What is the price range for Kawasaki dirt bikes?
The price range for Kawasaki dirt bikes are roughly between 1800 dollars and 8000 dollars. And can be bought used or new at any bike shop from across the nation.
### What bike is better for its price?
It is hard to determine which bike is better for the price, you must first decide what features of the bike are the most important to you. Then choose the bike that best fits your needs.
### What is the value of an American chopper?
To me an American Chopper. It is about showing who you are and what you are about. It shows pride in your beliefs and what you have built. There is no value of a chopper. There is always a price tag on everything. But the value is in the rider, the one who is proud to show off his bike. The bike that is not store bought, the one no one else has. It is custom, one of a kind. To me an American Chopper. It is about showing who you are and what you are about. It shows pride in your beliefs and what you have built. There is no value of a chopper. There is always a price tag on everything. But the value is in the rider, the one who is proud to show off his bike. The bike that is not store bought, the one no one else has. It is custom, one of a kind. | 734 | 3,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.950899 |
https://www.scribd.com/document/11437348/Partial-Differential-Equation | 1,537,593,367,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158045.57/warc/CC-MAIN-20180922044853-20180922065253-00010.warc.gz | 864,446,630 | 28,160 | # Partial Differential Equation
Let us assume that z will always represent a function of x and y. i.e., z = f(x,y) where x and y are two independent variables and z is a dependent variable. Notations
p= ∂z∂x, q= ∂z∂y, r= ∂2z∂x2, s= ∂2z∂x ∂y, t= ∂2z∂y2
www.MathsHomeWork123.com
If the number of constants to be eliminated is equal to the number of independent variables then the required Partial Differential Equation will be of First Order. If the number of constants to be eliminated is more than the number of independent variables then the required Partial Differential Equation will be of Second Order or higher order. If the number of functions to be eliminated is one, then the required Partial Differential Equation will be of first order otherwise it will be of second order or higher order. Eliminating ф from фu, v=0 gives a Partial Differential Equation
uxuyvxvy=0
To Solve f(p, q) = 0 Let z = ax + by + c be the solution. Then p = a, q = b, we get f(a,b)=0 Solving, we get b= фa The Complete Integral is z •
= ax + ф(a) y + c. There is no singular integral.
The complete integral of Partial Differential Equation of the type Z = px + qy + f(p, q) is z= ax + by + f(a,b)
To Solve f(z, p, q) = 0 Let z=f(x + ay) be the solution.
www.MathsHomeWork123.com / Partial Differential Equation
www.MathsHomeWork123.com
Put u = x + ay, then z = f(u).
Solving fz, dzdu, adzdu=0 which is an ordinary differential equation, we get the required solution. To Solve f1(x, p) = f2(y, q) Let f1(x, p) = f2(y, q) = k p = F1(x, k) q = F2(y, k) Then z= F1x,kdx+ F2y,kdy To Solve F(xmp, ynq) = 0 and F(z, xmp, ynq) = 0 If m≠1, n≠1, then Xmp = (1 – m)P Ynq = (1 – n)Q, where P= ∂z∂X, Q= ∂z∂Y Solution is F ( P, Q ) = 0 F (z, P, Q) = 0 If m=1, n=1, then put log x = X and log y = Y xp = P and yq = Q Solution is F ( P, Q ) = 0 F (z, P, Q) = 0
www.MathsHomeWork123.com / Partial Differential Equation
www.MathsHomeWork123.com
Lagranges Linear Equation The standard form is Pp and z.
+ Qq = R
where P, Q and R are functions of x, y
The subsidiary equation is
dxP= dyQ= dzR
Choose any three multiplier l, m, n such that
dxP= dyQ= dzR= l dx+m dy+n dzlP+mQ+nR, where lP+mQ+nR=0
i.e., l dx+m dy+n dz = 0 Solving, we get u(x, y, z ) = c1
Similarly choose another set of three multipliers l/, m/, n/ such that
dxP= dyQ= dzR= l/ dx+m/ dy+n/ dzl/P+m/Q+n/R, where l/P+m/Q+n/R=0
Solving, we get v(x, y, z ) = c2
The Solution is given by ф(u, v) = 0
www.MathsHomeWork123.com / Partial Differential Equation
www.MathsHomeWork123.com
HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATION
The general form of Linear Partial Differential Equation is
a0∂nz∂xn+a1∂nz∂xn-1∂y+……+an∂nz∂yn=F (x, y)
The Solution is z = Complementary Function (C.F) + Particular Integral (P.I)
To find Complementary Function (C.F) Auxillary Equation is a0mn+a1mn-1+ ……+ an=0 This equation has n roots say m1,m2,m3,…..mn Case (i) If the roots are real ( or imaginary) and different say m1≠m2≠m3≠…..≠mn, then the C.F is
z= f1y+m1x+f2y+m2x+f3y+m3x+……+ fny+mnx
Case (ii) If any two roots are equal say m1=m2=m and m3≠m4≠……then the C.F is
z= f1y+mx+f2y+mx+f3y+m3x+f4y+m4x……+ fny+mnx
Case (ii) If any three roots are equal say m1=m2=m3=m and m4≠m5≠……then the C.F is
z= f1y+mx+x f2y+mx+x2f3y+m3x+f4y+m4x……+ fny+mnx
To find Particular Integral ( P. I ) If F(x, y) = eax+by, then
www.MathsHomeWork123.com / Partial Differential Equation
www.MathsHomeWork123.com
Rule 1
P.I= 1ф(D,D/) eax+by= 1ф(a,b)eax+by provided фa, b≠0
If фa, b= 0, then refer to caseiv. Rule 2 If F(x, y) = sin (mx+ny) or cos(mx+ny), then
P.I= 1ф(D,D/) sin mx+ny or cos(mx+ny)
Replace D2 by –m2, D/2 by –n2 and DD/ by –mn in фa, b provided the denominator is not equal to zero. If the denominator is zero, then refer to case (iv). Rule 3 If F(x, y) = xmyn then
P.I= 1ф(D,D/) xmyn=[ф(D, D/)]-1xmyn
Expand =[ф(D, D/)]-1 by using Binomial theorem and then operate on xmyn. Note 1
1Df(x, y) means integrate f(x,y) with respect to ‘x’ one time assuming ‘y’ as a constant. 1D/f(x, y) means integrate f(x, y) with respect to ‘y’ one time
assuming ‘x’ as a constant. Note 2
In xmyn, if m < n, then try to write ф(D, D/) as ф(DD/) and if n < m, write ф(D, D/) as ф(D/D)
Rule 4 If F(x, y) is any other function, resolve ф(D, D/) into linear factors say
(D- m1D/)D-m2D/…..etc, then the P.I= 1(D- m1D/)D-m2D/F(x,y)
Now, 1(D- mD)Fx,y= Fx, c-mxdx where y=c-mx Note : If the denominator is zero in Rule 1 and Rule 2, then apply rule 4 to find Particular Integral.
www.MathsHomeWork123.com / Partial Differential Equation
www.MathsHomeWork123.com
www.MathsHomeWork123.com / Partial Differential Equation | 1,668 | 4,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-39 | latest | en | 0.816313 |
http://spotidoc.com/doc/1287523/manova-%C3%A2%C2%80%C2%93-multivariate-analysis-of-variance | 1,603,791,513,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00391.warc.gz | 95,829,635 | 8,074 | # MANOVA â Multivariate analysis of variance
```MANOVA – Multivariate analysis of variance
•
•
•
Multivariate analysis of variance (MANOVA) is simply an ANOVA with several
dependent variables.
o ANOVA tests for the difference in means between two or more groups,
while MANOVA tests for the difference in two or more vectors of means.
Can involve 1 IV or more than 1
Requires parametric DVs.
Why do you need MANOVA?
Can use it when there are multiple DVs and IVs in the model to be tested
•
•
•
•
Type I error probability increases with number of variables (i.e. falsely rejecting
H0 – finding an effect when there isn’t one)
More than 1 DV may lead to multiple correlated responses
o i.e. examines relationships between DVs, which ANOVA doesn’t do.
MANOVA provides a joint test for any significant effects among a set of variables
o i.e. has a greater power to detect any effects on a group of a combination
of variables, rather than just one
Also can be used instead of a repeated measures ANOVA when assumptions of
sphericity are violated (i.e. equal variances among the different levels of the
groups of IVs – tested with Mauchly's sphericity test).
Assumptions in MANOVA
Similar to ANOVA, but extended for multivariate case
1. Independence – observations should be statistically independent
2. Random sampling – data should be randomly sampled from the population of
interest and measured at the interval level.
3. Multivariate normality – in ANOVA we assume the DV is normally distributed
within each group; in MANOVA, we assume that the DVs (collectively) have
multivariate normality within groups.
a. F-test is robust to non-normality, if it’s caused by skewness rather than
outliers
b. Run tests for, and remove or transform any outliers before doing a MANOVA
4. Homogeneity of covariance matrices – in an ANOVA we assume homogeneity of
variance (the variances in each group are roughly equal). In MANOVA we assume
it’s true for
a. Each DV and
b. The correlation between any 2 DVs is the same in all groups (i.e. level of the
IV(s)).
c. Test whether the population variance-covariance matrices of the different
groups of analysis are equal
i. Test univariate equality of variances between the groups – Levene’s
test – shouldn’t be significant for any of the DVs
ii. Compare variance-covariance matrices between groups using Box’s
test – should be non-significant if the matrices are the same. Data must
show multivariate normality, or lead to erroneous results in Box’s test.
Example
•
•
•
•
•
Effect of 2 different text books (IV – 2 levels) on students’ improvements in maths
and physics (2 DVs)
Hypothesis – both DVs will be affected by difference in text book
o Use MANOVA to test this
Calculate a multivariate F-value (Wilks’ λ) based on the comparison of the error
variance / covariance matrix and the effect variance / covariance matrix, instead of
univariate F.
o Can also use Hotelling's trace and Pillai's criterion (robust to violations of
assumptions).
The "covariance" here is included because the two measures are probably correlated and
we must take this correlation into account when performing the significance test.
Test the multiple DVs by creating new DVs that maximize group differences. These
artificial DVs are linear combinations of the measured DVs.
Aim of a MANOVA
•
To establish if response variables (e.g. the students’ maths and physics results) are
altered by manipulation of the IVs.
1.
2.
3.
4.
5.
What are the main effects of each IV?
What is the interaction between the IVs?
What is the importance of the DVs?
How are the DVs related?
What are the effects of the covariates and how can they be used?
e.g.
Limitations
1. Outliers –
a. MANOVA’s very sensitive to outliers, which may produce Type I or Type II
errors, but not give an indication as to which is occurring.
b. Can test for outliers or examine plots.
2. Multicollinearity and Singularity:
a. High correlation between DVs, results in one DV becoming a near-linear
combination of the other DVs.
b. It would then become statistically redundant and suspect to include both
combinations.
``` | 1,004 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-45 | latest | en | 0.87918 |
https://answers-ph.com/math/question2387742 | 1,656,585,273,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00740.warc.gz | 150,615,393 | 41,721 | • Accueil
• Math
• 4. given x=3(raise to the power of 8). find the va...
# 4. given x=3(raise to the power of 8). find the value of x.
• Réponse publiée par: JUMAIRAHtheOTAKU
It easy
Step-by-step explanation:
x= 3^8
so x= 6561
• Réponse publiée par: HaHannah
yes, josh's solution is correct.
step-by-step explanation:
this is josh’s solution for the equation x^2 + 12x + 32 = 0:
x^2 + 12x +32 = 0
x^2 + 12x = -32 (32 is move to the other side that is why it became negative -32 and that is correct)
x^2 + 12x +36 = -32 + 36 (both sides are added with 36 so that the left side could be a perfect square: the process is completing the square: adding any the same number for both sides will result to the same original equation)
〖(x+6)〗^2 = 4 (the left side is a perfect square so it is right)
x + 6 = ±4 (square of any value is ± so it is right)
x=-2 (get the possible values of x)
x=-10
• Réponse publiée par: cyrishlayno
the learner is able to communicate mathematical thinking with coherence and clarity in formulating, investigating formulate the inverse, converse, and contrapositive of an implication.
Connaissez-vous la bonne réponse?
4. given x=3(raise to the power of 8). find the value of x.... | 386 | 1,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-27 | latest | en | 0.714698 |
https://www.expii.com/t/asa-angle-side-angle-congruence-5522 | 1,675,457,393,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500074.73/warc/CC-MAIN-20230203185547-20230203215547-00467.warc.gz | 778,758,501 | 2,377 | Expii
# ASA (Angle-Side-Angle) Congruence - Expii
If two angles and the side between them (ASA) are congruent between two triangles, the triangles have to be congruent. Since the angles coming off the side determine the direction of the other sides, there's only one point where it's possible for them to meet!. | 75 | 313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.914047 |
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