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https://study.com/academy/topic/calculating-simplifying-exponential-expressions.html
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Ch 13: Calculating & Simplifying Exponential Expressions
By using this set of math video lessons, you can learn all the essential strategies for calculating and simplifying exponential expressions. Take the quick, multiple-choice quizzes that accompany each lesson to measure your accomplishments.
Calculating & Simplifying Exponential Expressions - Chapter Summary and Learning Objectives
You can begin this study by reviewing examples and usage of exponential notation. Then you'll practice simplifying, calculating and solving exponential expressions, focusing on the order of operations, the power of zero, negative exponents and power of powers. By the time you've wrapped up this chapter, you'll be confident in working with the following:
• The order of operations related to exponential expressions
• Strategies for multiplying and dividing exponential expressions
• Methods for simplifying exponential expressions
• How to write powers of fractions and decimals
Video Objective
How to Use Exponential Notation Examine the strategies for working with exponential notation.
Scientific Notation: Definition and Examples Look over helpful examples of scientific notation.
Simplifying and Solving Exponential Expressions Follow the instructor's steps for simplifying and then solving exponential expressions.
Exponential Expressions & The Order of Operations Review the order of operations and learn how that affects exponential expressions.
Multiplying Exponential Expressions Practice strategies for the multiplication of exponential expressions.
Dividing Exponential Expressions Explore the essentials for dividing exponential expressions and find how that relates to multiplication.
The Power of Zero: Simplifying Exponential Expressions Discover the power of zero and the relationship to simplification of exponential expressions.
Negative Exponents: Writing Powers of Fractions and Decimals Find out the key steps in writing powers of decimals and fractions.
Power of Powers: Simplifying Exponential Expressions Use basic knowledge related to power of powers to simplify exponential expressions.
9 Lessons in Chapter 13: Calculating & Simplifying Exponential Expressions
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
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https://assignmentshood.com/q-4a-you-are-the-manager-of-a-fast-food-restaurant-you-want-to/
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# Q-4a. you are the manager of a fast food restaurant. you want to
Q-4a. You are the manager of a fast food restaurant. You want to determine whether the population mean waiting time to place an order has changed in the past month from its previous population mean value of 4.5 minutes. From past experience, you can assume that the population is normally distributed with a population standard deviation of 1.2 minutes. You select a sample of 25 orders during a one hour period. The sample mean is 5.1 minutes. Determine whether there is evidence at the 0.05 level of significance that the population mean waiting time to place an order has changed in the past month from its previous population mean value of 4.5 minutes.
Q-4b. You have just opened your new fast-food restaurant. And you have developed a new process to ensure that orders at the drive-through are filled correctly. The previous process filled orders correctly 85% of the time. Based on a sample of a hundred orders using the new process, 94 were filled correctly. At the 0.01 level of significance, can you conclude that the new process has increased the proportion of orders filled correctly?
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http://heli-air.net/2016/02/02/computation-of-the-robustness/
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# COMPUTATION OF THE ROBUSTNESS
MARGIN
In the context of a robust stability problem, l/fi(M(jw)) represents the size of the smallest parametric uncertainty S, which brings one closed loop pole on the imaginary axis at ±ju. The robust stability margin kmax is obtained by computing the s. s.v. along the imaginary axis:
The principle is thus to detect the crossing of one of the closed loop poles through the imaginary axis. ктах corresponds to the size of the smallest parametric uncertainty 8, which brings one closed loop pole on the imaginary axis.
Remark: several reasons exist for handling the, v..v. к n(M(ju)) rather than its inverse the multiloop stability margin (m. s.m.).
As a first point, the s. s.v. can not take an infinite value, since
the nominal closed loop is asymptotically stable, whereas the m. s.m. may be infinite (if no structured model perturbation exists, which destabilizes the closed loop). On the other hand, the s. s.v. can be considered as an extension of classical algebraic notions, namely the spectral radius and the maximal singular value of a matrix (i. e. its spectral norm – see below).
2.2 THE GENERAL CASE
The problem of extending the approach of subsection 2.1 to the case of neglected dynamics seems a priori more complex, since Д is now a dynamic transfer matrix instead of a simple gain matrix. Nevertheless, assume that a complex matrix Д0 was found, which satisfies det(I – M(ju>)A°) = 0 at frequency u>. It suffices then to find a transfer matrix A(s) with A(ju}) = Д0. When applying A(s) to the interconnection structure, a closed loop pole is obtained on the imaginary axis at ±juj.
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http://mks.mff.cuni.cz/kalva/putnam/psoln/psol892.html
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### 50th Putnam 1989
Problem A2
Let R be the rectangle 0 ≤ x ≤ a, 0 ≤ y, ≤ b. Evaluate ∫R ef(x, y) dx dy, where f(x, y) = max(b2x2, a2y2).
Solution
Divide the rectangle into two by the diagonal from (0, 0) to (a, b). The required value is evidently twice the integral over the lower half of eg(x), where g(x) = b2x2. But integrating over y just gives a factor bx/a, and the integration over x is then trivial to give (ek - 1)/(2ab), where k = a2b2. So the required value is twice this or (ek - 1)/(ab).
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https://www.studypool.com/discuss/383673/math-question-6th-grade-easy
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# Math question!! 6th grade easy
Feb 11th, 2015
SoccerBoss
Category:
Algebra
Price: \$5 USD
Question description
The low temperature (in degrees Fahrenheit) for one week in Minneapolis, MN were
Mon Tue Wed Thur Fri Sat Sun –6 –4 –5 2 0 3 –7
a. Arrange the temperatures in order from coldest to warmest.
b. On a day in January, the low temperature in Minneapolis, MN was 14° below zero (in degrees Fahrenheit) and the low temperature in Albany, NY was 5° below zero (in degrees Fahrenheit). When asked which city was colder, Jacob wrote:
Albany was colder, because –5 < –14.
c. On February 2, 1996, Minnesota had a record low temperature of –60°F. More than a decade earlier, Colorado recorded a low temperature of –61°F. Which state was colder on its coldest day in history? Write an inequality to support your answer.
(Top Tutor) Daniel C.
(997)
School: Boston College
Studypool has helped 1,244,100 students
## Review from our student for this Answer
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https://fdocument.org/document/introduction-to-machine-learning-jbgteachingcmsc72603apdf-machine-learning.html
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• date post
27-Jun-2020
• Category
## Documents
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### Transcript of Introduction to Machine Learning - jbg/teaching/CMSC_726/03a.pdf · PDF file Machine...
• Introduction to Machine Learning
Machine Learning: Jordan Boyd-Graber University of Maryland LOGISTIC REGRESSION FROM TEXT
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 1 / 18
• Reminder: Logistic Regression
P(Y = 0|X) = 1
1+exp �
β0 + ∑
i βiXi � (1)
P(Y = 1|X) = exp
β0 + ∑
i βiXi �
1+exp �
β0 + ∑
i βiXi � (2)
Discriminative prediction: p(y |x) Classification uses: ad placement, spam detection
What we didn’t talk about is how to learn β from data
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 2 / 18
• Logistic Regression: Objective Function
L ≡ lnp(Y |X ,β) = ∑
j
lnp(y(j) |x(j),β) (3)
= ∑
j
y(j)
β0 + ∑
i
βix (j) i
− ln
1+exp
β0 + ∑
i
βix (j) i
(4)
Training data (y ,x) are fixed. Objective function is a function of β . . . what values of β give a good value.
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 3 / 18
• Logistic Regression: Objective Function
L ≡ lnp(Y |X ,β) = ∑
j
lnp(y(j) |x(j),β) (3)
= ∑
j
y(j)
β0 + ∑
i
βix (j) i
− ln
1+exp
β0 + ∑
i
βix (j) i
(4)
Training data (y ,x) are fixed. Objective function is a function of β . . . what values of β give a good value.
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 3 / 18
• Convexity
Convex function
Doesn’t matter where you start, if you “walk up” objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 4 / 18
• Convexity
Convex function
Doesn’t matter where you start, if you “walk up” objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 4 / 18
Goal
Optimize log likelihood with respect to variables β
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
0
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
0
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
1
0
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
1
0
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
1
0
2
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
1
0
2
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
1
0
2
3
Undiscovered Country
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
0
β ij (l+1)
Jα ∂ ∂β
β ij (l)
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
Goal
Optimize log likelihood with respect to variables β
0
β ij (l+1)
Jα ∂ ∂β
β ij (l)
Luckily, (vanilla) logistic regression is convex
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 5 / 18
To ease notation, let’s define
πi = expβT xi
1+expβT xi (5)
Our objective function is
L = ∑
i
logp(yi |xi) = ∑
i
Li = ∑
i
¨
logπi if yi = 1
log(1−πi) if yi = 0 (6)
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 6 / 18
• Taking the Derivative
Apply chain rule:
∂L ∂ βj
= ∑
i
∂Li( ~β) ∂ βj
= ∑
i
(
1 πi ∂ πi ∂ βj
if yi = 1 1
1−πi
− ∂ πi∂ βj
if yi = 0 (7)
If we plug in the derivative,
∂ πi ∂ βj
=πi(1−πi)xj , (8)
we can merge these two cases
∂Li ∂ βj
= (yi −πi)xj . (9)
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 7 / 18
∇βL ( ~β) = �
∂L ( ~β) ∂ β0
, . . . , ∂L ( ~β) ∂ βn
(10)
Update
∆β ≡η∇βL ( ~β) (11)
β ′i ←βi +η ∂L ( ~β) ∂ βi
(12)
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 8 / 18
∇βL ( ~β) = �
∂L ( ~β) ∂ β0
, . . . , ∂L ( ~β) ∂ βn
(10)
Update
∆β ≡η∇βL ( ~β) (11)
β ′i ←βi +η ∂L ( ~β) ∂ βi
(12)
Why are we adding? What would well do if we wanted to do descent?
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 8 / 18
∇βL ( ~β) = �
∂L ( ~β) ∂ β0
, . . . , ∂L ( ~β) ∂ βn
(10)
Update
∆β ≡η∇βL ( ~β) (11)
β ′i ←βi +η ∂L ( ~β) ∂ βi
(12)
η: step size, must be greater than zero
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 8 / 18
∇βL ( ~β) = �
∂L ( ~β) ∂ β0
, . . . , ∂L ( ~β) ∂ βn
(10)
Update
∆β ≡η∇βL ( ~β) (11)
β ′i ←βi +η ∂L ( ~β) ∂ βi
(12)
NB: Conjugate gradient is usually better, but harder to implement
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 8 / 18
• Choosing Step Size
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 9 / 18
• Choosing Step Size
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 9 / 18
• Choosing Step Size
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 9 / 18
• Choosing Step Size
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 9 / 18
• Choosing Step Size
Parameter
Objective
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 9 / 18
• Remaining issues
When to stop?
What if β keeps getting bigger?
Machine Learning: Jordan Boyd-Graber | UMD Introduction to Machine Learning | 10 / 18
• Regularized Conditional Log Likelihood
Unregularized
β ∗= argmax β
ln
p(y(j) |x(j),β)
(13)
Regularized
β ∗= argmax β
ln
p(y(j) |x(j),β)
−µ
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## The Pauli Exclusion Principle
#### Learning Objective
• Illustrate how the Pauli exclusion principle partially explains the electron shell structure of atoms.
#### Key Points
• No two identical fermions (particles with half-integer spin) may occupy the same quantum state simultaneously.
• No two electrons in a single atom can have the same four quantum numbers.
• Particles with integer spin occupy symmetric quantum states, and particles with half-integer spin occupy antisymmetric states.
#### Terms
• fermionA particle with totally antisymmetric quantum states. They have half-integer spin and include many elementary particles.
• electronThe subatomic particle having a negative charge and orbiting the nucleus; the flow of electrons in a conductor constitutes electricity.
• bosonA particle with totally symmetric quantum states. They have integer spin and include many elementary particles, and some (gauge bosons) are known to carry the fundamental forces.
The Pauli exclusion principle, formulated by Austrian physicist Wolfgang Pauli in 1925, states that no two fermions of the same kind may simultaneously occupy the same quantum state. More technically, it states that the total wave function for two identical fermions is antisymmetric with respect to exchange of the particles. For example, no two electrons in a single atom can have the same four quantum numbers; if n, ℓ , and mare the same, ms must be different such that the electrons have opposite spins.
The Pauli exclusion principle governs the behavior of all fermions (particles with half-integer spin), while bosons (particles with integer spin) are not subject to it. Fermions include elementary particles such as quarks (the constituent particles of protons and neutrons), electrons and neutrinos. In addition, protons and neutrons (subatomic particles composed from three quarks) and some atoms are fermions and are therefore also subject to the Pauli exclusion principle. Atoms can have different overall spin, which determines whether they are fermions or bosons—for example, helium-3 has spin 1/2 and is therefore a fermion, in contrast to helium-4 which has spin 0, making it a boson. As such, the Pauli exclusion principle underpins many properties of everyday matter from large-scale stability to the chemical behavior of atoms including their visibility in NMR spectroscopy.
Half-integer spin means the intrinsic angular momentum value of fermions is $\hbar =\frac { h }{ 2\pi }$ (reduced Planck’s constant) times a half-integer (1/2, 3/2, 5/2, etc.). In the theory of quantum mechanics, fermions are described by antisymmetric states. In contrast, particles with integer spin (bosons) have symmetric wave functions; unlike fermions, bosons may share the same quantum states. Bosons include the photon, the Cooper pairs (responsible for superconductivity), and the W and Z bosons. Fermions take their name from the Fermi–Dirac statistical distribution that they obey, and bosons take their name from Bose–Einstein distribution.
## The Exclusion Principle and Physical Phenomena
The Pauli exclusion principle explains a wide variety of physical phenomena. One particularly important consequence of the principle is the elaborate electron-shell structure of atoms and the way atoms share electrons. It explains the variety of chemical elements and their chemical combinations. An electrically neutral atom contains bound electrons equal in number to the protons in the nucleus. Electrons, being fermions, cannot occupy the same quantum state, so electrons have to “stack” within an atom—they have different spins while at the same place.
An example is the neutral helium atom, which has two bound electrons, both of which can occupy the lowest-energy (1s) states by acquiring opposite spin. As spin is part of the quantum state of the electron, the two electrons are in different quantum states and do not violate the Pauli exclusion principle. However, there are only two distinct spin values for a given energy state. This property thus mandates that a lithium atom, which has three bound electrons, cannot have its third electron reside in the 1s state; it must occupy one of the higher-energy 2s states instead. Similarly, successively larger elements must have shells of successively higher energy. Because the chemical properties of an element largely depend on the number of electrons in the outermost shell, atoms with different numbers of shells but the same number of electrons in the outermost shell still behave similarly. For this reason, elements are defined by their groups and not their periods.
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# How far is Necochea from Puerto Madryn?
The distance between Puerto Madryn (El Tehuelche Airport) and Necochea (Necochea Airport) is 443 miles / 713 kilometers / 385 nautical miles.
The driving distance from Puerto Madryn (PMY) to Necochea (NEC) is 621 miles / 1000 kilometers, and travel time by car is about 12 hours 39 minutes.
443
Miles
713
Kilometers
385
Nautical miles
1 h 20 min
## Distance from Puerto Madryn to Necochea
There are several ways to calculate the distance from Puerto Madryn to Necochea. Here are two standard methods:
Vincenty's formula (applied above)
• 442.789 miles
• 712.601 kilometers
• 384.774 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 442.429 miles
• 712.021 kilometers
• 384.460 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Puerto Madryn to Necochea?
The estimated flight time from El Tehuelche Airport to Necochea Airport is 1 hour and 20 minutes.
## What is the time difference between Puerto Madryn and Necochea?
There is no time difference between Puerto Madryn and Necochea.
## Flight carbon footprint between El Tehuelche Airport (PMY) and Necochea Airport (NEC)
On average, flying from Puerto Madryn to Necochea generates about 90 kg of CO2 per passenger, and 90 kilograms equals 199 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Puerto Madryn to Necochea
See the map of the shortest flight path between El Tehuelche Airport (PMY) and Necochea Airport (NEC).
## Airport information
Origin El Tehuelche Airport
Country: Argentina
IATA Code: PMY
ICAO Code: SAVY
Coordinates: 42°45′33″S, 65°6′9″W
Destination Necochea Airport
City: Necochea
Country: Argentina
IATA Code: NEC
ICAO Code: SAZO
Coordinates: 38°28′59″S, 58°49′1″W
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# Archive for the ‘Math’ Category
## Notes about valves and flow rates
Posted by btrettel on March 20, 2013
Comparing valves for compressed gases can be difficult for several reasons. Even if you are familiar with how valve performance is specified, you can easily make mistakes. Valve performance is specified in several different sloppy and confusing ways, almost always in non-metric units. The worst part is that valve performance can be specified in obvious ways, but people working in the industry don’t do that.
Flow rates
The confusion starts immediately with the definition of the flow rate. Rather than specifying a mass flow rate (i.e., something like g/s), most valve manufacturers instead specify a volumetric flow rate (i.e., something like m3/s). Why? I’m not entirely sure. It’s probably a combination of tradition (e.g., “that’s all I was taught”), familiarity, laziness (e.g., the flow meter gives volumetric, and they don’t want to convert), or ignorance (e.g., they don’t realize it matters). I suspect the practice was held over from liquid valve flow, where the fluid density doesn’t vary much and the volumetric flow rate is just proportional to the mass flow rate.
This is problematic because volume can vary considerably over a range of temperatures/densities for the same mass flow rate. To use a volumetric flow rate (by converting it to a mass flow rate to use in a mass conservation equation) requires an engineer to know the temperature and pressure at which it applies.
With this in mind, some engineers developed the unit called “standard cubic feet per minute” or SCFM for short where the temperature and pressure are supposed to be standard, i.e., known. Unfortunately, there is no standard, though usually the reference pressure is 1 atm and the reference temperature is 70F.
Unfortunately, some companies simply treat SCFM as if it were the actual measured flow rate! It’s as if they don’t understand what the point of the standardization was or even that the S stands for standard. Clippard is one of these companies (I asked by email. Their reference pressure is the pressure they did the test at and their reference temperature is 70F.). If you are using a company’s valve, I suggest asking what their reference conditions are. If they don’t know, then don’t assume anything about their products. I’ve found that some companies are completely ignorant about the performance of their products. Some companies have no idea where the numbers they print on their spec sheets come from. They sometimes don’t even know the units to the numbers. This is extremely sloppy and it’s rather scary that a large number of engineers work this way.
The relationship between volumetric flow rate and mass flow rate is $\dot{m} = \rho Q$ where $\dot{m}$ is the mass flow rate, $\rho$ is the mass density, and $Q$ is the volumetric flow rate.
We can find $\rho$ with the ideal gas law ($\displaystyle P = \rho \frac{\overline{R}}{M} T$) and convert between volumetric and mass flow rates. $P$ is the pressure (in absolute units), $T$ is the temperature (again, absolute temperature), $\overline{R}$ is the universal gas constant, and $M$ is the molecular mass of the gas (for us this is air). So, rearranging the equation above yields $\displaystyle \rho = \frac{P M}{\overline{R} T}$.
The standard uses the standard temperature and pressure. So, for example, for convert from a volumetric flow rate of 20 SCFM to a mass flow rate in g/s, we can do the following:
$\displaystyle \dot{m} = \rho Q = \frac{P_\text{s} M Q}{\overline{R} T_\text{s}}$
$\displaystyle = \frac{101325~\text{Pa} \cdot 28.97~\text{g/mol} \cdot 20~\text{ft}^3\text{/min}}{8.3145~\text{J/(mol K)} \cdot 293.15~\text{K}} \left(\frac{3.28~\text{ft}}{\text{m}}\right)^3 \frac{1~\text{min}}{60~\text{s}} = 11.4~\text{g/s}$
Valve flow coefficients and flow models
Often valve performance is specified with a valve flow coefficient. In the US, this usually is called $C_\text{v}$ where v refers to valve. (This is not the heat capacity at constant volume. Again, this is another confusing point.)
Like the valve flow rate, $C_\text{v}$ is defined in terms of liquid, not gas flows. $C_\text{v}$ is defined as the volumetric flow rate in gallons per minute (GPM) for a pressure drop of 1 psi. There are some mainly empirical equations (scroll down to the section “Flow Coefficient – Cv – for Air and other Gases”) that can be used to calculate the volumetric flow rate (and mass flow rate from there). The “derivation” of this model is available in this book.
For more information on this subject, I suggest reading my notes about Nerf engineering, which contain many references to other materials. In particular, I detail a more accurate empirical valve model that can be used in computer simulations.
Accuracy of manufacturer data
You should always question the accuracy of valve manufacturer data (assuming that you understand what it means, which, as I’ve explained, isn’t always clear). In my experience, the flow rates given seem to be optimistic. It’s not easy to quantify how optimistic they are, but you should keep this in mind.
An aside
I could insert a long rant about how some engineers don’t understand what they are doing and treat some equations like magic black boxes which give them “the answer”. They don’t understand the assumptions or development of the equation. They don’t try to improve the equation to be more general, more useful, or less confusing, e.g., by eliminating confusing “standard” volumetric flow rates and moving to obvious mass flow rates. This practice creates a lot of confusion for people who do know what they are doing. If you are in engineering, please, don’t be one of these folks.
Posted in Interior ballistics, Math, Misconceptions, Pneumatics | Comments Off on Notes about valves and flow rates
## Nerf Engineering notes
Posted by btrettel on July 8, 2012
I’ve started organizing my writings and thoughts on Nerf ballistics into a set of notes (link corrected). Readers of my blog might be interested.
For the moment my notes are very rough and incomplete. I’ll update the PDF as I make changes. I’d appreciate any feedback, especially corrections and ideas.
Edit (2012-10-02): I’ve removed the notes because they need a lot of work before I’m comfortable distributing them.
Edit (2013-01-19): I’ve readded the notes as I’ve revised them somewhat. They are still a work in progress.
## Solving the point mass model of a dart’s trajectory, part two
Posted by btrettel on June 12, 2011
I wrote a technical paper about solving these equations and made some conclusions for design. I have fully solved the flat-fire case given that the x velocity stays above a certain number (and it generally will).
(Original link broken: My newer notes contain that paper and more.)
The take-home message is pretty simple: Keep kinetic energy constant. If you reduce the drag coefficient multiplied by the cross-sectional area, you’ll increase range. If you reduce the muzzle velocity divided by the dart mass, you’ll increase range.
The latter conclusion is not too surprising to me as I already knew dart mass had an effect like that, but to quantify the effect is very nice.
I’d happily accept corrections, comments, questions, suggestions, and whatever else.
Posted in Exterior ballistics, Math | 2 Comments »
## boltsniper’s optimal barrel length formula
Posted by btrettel on June 11, 2011
Whenever there is a thread about one of the holy grails of Nerf ballistics, an equation for the barrel length that maximizes performance, someone is bound to mention the results of some tests boltsniper did in 2005 when he was designing the FAR. boltsniper, for the uninitiated, is one of the few engineers in the Nerf hobby, so his words have some authority behind them. People generally misrepresent what he said and overstate this formula’s abilities. What boltsniper actually said is written below. The emphasis is mine.
I did some experimentation to determine what would be the optimal barrel length for a given plunger size. The goal was to find the barrel length for which the dart would exit the barrel as the plunger reaches the end of the plunger tube. I started off by matching the volume of the plunger to the volume of the barrel. I knew that this was going to produce too long a barrel but it was a good place to start. This would assume that the air inside the plunger and barrel is incompressible and that there are no leaks. In the real world this is not the case. I reduced the barrel length until I had found the length at which the dart was leaving the barrel as the plunger was reaching its stop, coinciding with the maximum attainable range. Experimentally the plunger volume seems to be about 4 times that of the barrel. The relation for barrel to plunger size can be summed up in the following equation,
$4 \pi r_b^2 l_b = \pi r_p^2 l_p$
where $r_b$ is the barrel radius, $r_p$ is the plunger radius, $l_b$ is the barrel length, and $l_p$ is the plunger length. For Nerf applications the barrel is almost always 1/2″ PVC or CPVC. $r_b$ can then be set as a constant at 0.25″ and removed from the equation. Since we are trying to solve for the barrel length with a given plunger size, the equation can be rearranged and simplified to:
$l_b = D_p^2 l_p$
This simple equation makes it easy to roughly but quickly size a barrel to a given plunger. The equation could also be used to size a plunger for a given length barrel. This equation is based on experimental data and is not perfect. Four is not the golden number. This produces the optimal barrel length for the situation I was testing. The type of dart, dart-barrel friction, and total system volume will likely effect the optimal ratio. Nevertheless, the above equation can be used as a starting point.
The last paragraph seems to be completely ignored by most people who use this formula. At best it’s a starting point for further testing. The equation only applies to the FAR as that was all that he tested.
boltsniper later expanded on the restrictions on the use of this formula at NerfHaven. (Again, the emphasis is my own.)
I derived that empirically and more importantly it was derived for the specific situation I intended on using it for: a plunger weapon. It will not work for a compressed air system. One of the big factors I used to come up with that was the lack of compressibility. I later factored that in with a constant that was derived empirically. My tests were with a setup exactly like I was going to use on a the finished product. If you scale the system down that magic constant may not hold true.
There are too many variables to analytically design the optimal barrel length. If you are going to build or mod a spring gun the equation I provided may be a good starting point. That equation gives a barrel length that is slightly too long, so to obtain the optimal length you are going to have to go shorter.
The only real way to do it is experimentally.
The short message is that this equation only applies for the situation he was testing for.
But does it even apply for that situation? I’d argue no. boltsniper wasn’t testing for optimal barrel length. In his own words (which I emphasized above), boltsniper’s “goal was to find the barrel length for which the dart would exit the barrel as the plunger reaches the end of the plunger tube.” This does not coincide with when performance peaks based on my understanding of the interior ballistic processes.
Performance is maximized when acceleration slows to zero. If the plunger is at the end of the plunger tube, the pressure is approximately maximized. This corresponds to maximum acceleration because the force is maximized, not maximum velocity. The ideal barrel length is definitely longer in this case.
(I’ll mostly ignore the question of how he knew the plunger struck the end of the tube when the dart left. I seriously question how he determined that. The entire process occurs in a fraction of a second. He’d need a high speed camera with a clear plunger tube and barrel, some other optimal system, some acoustic system, some similar combination, or something I’m not considering to actually determine this with accuracy.)
In summary, this formula should not be used for general purpose design to approximate ideal barrel length. I suggest using a chronometer, ballistic pendulum, or some other device or procedure to measure the muzzle velocity or where it stops increasing as the barrel length is changed. Alternatively, range can be measured, but please note that drag can cause range to not increase from increases in muzzle velocity, the performance parameter that we’re examining.
If more general-purpose approximations are wanted, I have developed approximate equations for ideal barrel length of pneumatics and springers based on adiabatic process relationships. These equations apply when the pressure in the barrel approximately equals the pressure in the gas chamber or plunger tube. For pneumatics, this is valid for very fast and high speed valves and very heavy projectiles. For springers, this is valid for very heavy projectiles. How heavy “very heavy” is depends on the situation, and I have not fully developed a criteria to determine this. The link contains an approximation I developed a year ago.
Posted in Design, Interior ballistics, Math, Misconceptions | Comments Off on boltsniper’s optimal barrel length formula
## Solving the point mass model of a dart’s trajectory, part one
Posted by btrettel on June 7, 2011
This blog post is relatively math heavy and it serves mostly as notes for my own use. I will assume some familiarity with differential calculus and ordinary differential equations. I also will assume some familiarity with the basic equations that govern the motion of Nerf darts after they leave the barrel.
Yesterday I had an epiphany while in the shower about a potential approach to solve the equations for the trajectory of a point mass with quadratic drag. This is a simplification to the standard equations of motion for the trajectory of a point mass in 2D with quadratic drag. These equations are presented below without derivation (or an explanation of what $k$ is).
EOM in x: $\ddot{x} = -k \dot{x} \sqrt{\dot{x}^2 + \dot{y}^2}$
EOM in y: $\ddot{y} = -k \dot{y} \sqrt{\dot{x}^2 + \dot{y}^2} - g$
These are two coupled non-linear ODEs. To the best of my knowledge, there is no known solution in terms of elementary functions. You either have to define new functions to solve these equations or solve them numerically, as I have done in the past.
But we can do a few tricks to get an okay approximation. Note that the vast majority of shots from Nerf guns are level. Basically, this means that the y component of velocity is far smaller than the x component. If we ignore drag and solve $\ddot{y} = -g$ directly, we can find that the maximum velocity of a Nerf dart in the y direction is less than 10 m/s. But Nerf darts almost always will be traveling faster than 30 m/s. If we assume that $\sqrt{\dot{x}^2 + \dot{y}^2} \approx \dot{x}$ we commit at worst a 5% error if the x velocity stays above 30 m/s. This is the assumption I will make.
For an even better approximation, there are a few things that could be done. A perturbation method could be used. A guess for $\dot{y}$ could be made; $\dot{y} = -g t$, the y velocity neglecting drag, might be good for a conservative approximation. We could assume that $\dot{y}$ is a function of $\dot{x}$ or another variable (like the height, or range), perhaps even a multiple of it. The multiple case reduces to an extra factor the drag coefficient (already a fudge factor for our purposes) can absorb. I will not take any of these approaches as I am only exploring potentially valid solutions now.
With my assumption in mind, I’ll rewrite the EOM.
EOM in x: $\ddot{x} = -k \dot{x}^2$
EOM in y: $\ddot{y} = -k \dot{x} \dot{y} - g$
These equations are now uncoupled and only the first one is non-linear. I can solve for $\dot{x}$ directly and substitute that into the second EOM.
$\dot{x}$ can be solved for by doing the following change of variables:
$z \equiv \dot{x}^2 \longrightarrow \dot{z} = 2 \dot{x} \ddot{x} \longrightarrow \ddot{x} = \displaystyle \frac{\dot{z}}{2 \sqrt{z}} \longrightarrow \dot{z} = -2 k z^{\tfrac{3}{2}} \longrightarrow z^{-\tfrac{1}{2}} = k t + c$
I can substitute back in my definition for z to get $\dot{x}$. I’ll use the initial condition $\dot{x}(t = 0) = V_0$ to get the complete solution.
$\dot{x} = \displaystyle \frac{1}{k t +c} \longrightarrow \dot{x}(t) = \displaystyle \frac{V_0}{V_0 k t + 1}$
Now we must solve the EOM in y. The equation is written below with $\dot{x}$ substituted in. I defined $A \equiv k V_0$.
$\ddot{y} = -\displaystyle \frac{A \dot{y}}{A t + 1} - g$
I will use a change of variables and omit most of the details. $\xi = \text{ln}(A t + 1)$
The ODE with these changes becomes $y_{\xi \xi} = - \displaystyle \frac{g e^{2 \xi}}{A^2}$ where the subscript represents differentiation with respect to that variable. The general solution for this ODE is $y = B \xi + C - \displaystyle \frac{g e^{2 \xi}}{4 A^2}$.
Substituting back in for y and t and using the ICs $\dot{y}(0) = 0$ (the dart does not start with any vertical velocity as it is fired flat) and $y(0) = h$ where h is the height of the barrel I arrive at the following equation for the dart’s y position as a function of time.
$y(t) = h + \displaystyle \frac{g}{2 (k V_0)^2} \text{ln}(k V_0 t + 1) - \displaystyle \frac{g t}{2 k V_0} - \displaystyle \frac{g t^2}{4}$
This is correct. You can solve for $\dot{y}$ and $\ddot{y}$ and substitute back into the ODE to check. You can also check the ICs; they are satisfied. Additionally, the solution approaches the solution without drag. At first I thought it did not, but it can be shown to approach this solution exactly with Taylor series expansion.
To get the range, we must find time the dart hits the ground. Yet the equation above can not be solved for t explicitly.
So I can find an approximate solution that is undesirable because I can’t proceed to get the range equation. I can think of a few options from here. One is to approximate the equation for y to find an approximate time. Another is to modify the original ODEs in a way that allows for the time to be found. Yet another is to use the identity $\ddot{x} dx =\dot{x} d \dot{x}$ (and the equivalent for y) to avoid time all together, but this approach has another problem: what $\dot{x}$ and $\dot{y}$ correspond to when the dart strikes the ground?
In my next post I will prove that this solution is correct and converges to the solution without drag as $k \rightarrow \infty$ and develop some approximations for different scenarios.
(I had some incorrect ramblings here before. They have been removed.)
Edit: In the next post I will show how the form of the solution for $x(t)$ probably will save us and allow an explicit equation for the range to be formed. $x(t) = \displaystyle \frac{\text{ln}(V_0 k t + 1)}{k}$. Note that this allows us to replace the ln term in the y equation with a multiple of x. x = R (the range) when the dart hits the ground… this may allow us to find the time the dart hits the ground as a function of R, which we can substitute back into the x equation to solve (hopefully) for R. That is the outline of what I will attempt now.
Edit again: I believe this is an adequate approximate equation for the range of a flat fired Nerf gun:
$R = \displaystyle \frac{1}{C} \text{ln} \left(\frac{h (C V_0)^2}{g} + 1 + \text{ln} \left(\frac{h (C V_0)^2}{g} + 1 \right) \right)$, where $C \equiv \displaystyle \frac{\rho_{atm} C_d A}{m}$ .
An okay simpler approximation follows.
$R = \displaystyle \frac{1}{C} \text{ln} \left(\frac{h (C V_0)^2}{g} + \frac{27}{25} \right)$
Posted in Exterior ballistics, Math | Leave a Comment »
## Approximate expansion diameter of latex tubing
Posted by btrettel on June 5, 2011
I previously wrote about how to estimate the expansion diameter of latex tubing, but I made no measurements to verify that the formula I suggested, $8.5 D_i + 2 t$ where $D_i$ is the unexpanded inner diameter of the tube and $t$ is the unexpanded tube wall thickness, was accurate.
Back in 2010, I made some measurements for tests that were aborted as the tubes continually burst. I was testing higher pressure tubes. These tubes have a tendency to burst and they are extremely loud when they burst if you fill them with air. I do not suggest testing latex tubes with air for this reason.
So, I have three data points and a reasonable understanding of the geometry of these tubes. I’ll make a better estimate for expansion diameter.
I’ll start with a few assumptions. When the tube expands, it is assumed to stop expanding when its expanded inner diameter equals a certain multiple of the unexpanded diameter. This is justifiable based on the observed behavior of the rubber. The Gent hyperelastic model makes the same assumption. My second assumption is that when the tube is fully expanded the cross sectional area of the expanded tube is a multiple of the cross sectional area of the unexpanded tube. I originally anticipated these areas would be equal as I know for small deformations, rubber is approximately incompressible (i.e. the volume is preserved). But that’s only applicable for volume, not area; the tube is expanding in length too. Also, it is only applicable for small deformations. I do not know how the rubber will act for large deformations.
$D_o$ is the outer diameter of the tube. $D_i$ is the inner diameter of the tube. $t$ is the tube wall thickness. The superscript $e$ refers to the expanded state. A variable or constant without $e$ is in the unexpanded state or it does not refer to any state (like the constants).
$A = \tfrac{\pi}{4} (D_o^2-D_i^2)$ is the equation for the cross sectional area of the unexpanded tube.
$A^e = \tfrac{\pi}{4} [(D_o^e)^2-(D_i^e)^2]$ is the equation for the cross sectional area of the expanded tube.
$D_i^e = C_1 D_i$ is my assumption about when the tubes stop expanding.
$A^e = C_2 A$, where $C_2$ is my assumption about the cross sectional areas of the tubes when they stop expanding.
Plugging all these equations together and solving for $D_o^e$, I find that $D_o^e = \sqrt{C_1^2 D_i^2 + 4 t C_2 (D_i + t_i)}$.
The data points I have available:
• Unexpanded ID: 3/8″, wall thickness: 3/16″, expanded OD: a bit more than 3″ (from memory)
• Unexpanded ID: 1/8″, wall thickness: 3/16″, expanded OD: 1.25″
• Unexpanded ID: 1/8″, wall thickness: 7/32″, expanded OD: 1.375″
A linear regression leads to$C_1$ = 7.35 and$C_2$ = 3.33 ($R^2$ = 0.9999, which would definitely be lower if there were more data points). These constants seem reasonable given my understanding of the phenomena, so it is reasonable to accept that the tubes’ inner diameters expand to about 7.35 times their original inner diameter and the cross sectional area increases by 3.33 times.
The equation above with these constants can be used to find the expanded outer diameter. The equation with $C_1$ can find the expanded inner diameter. The definition $D_o^e = D_i^e + 2 t^e$ can find the expanded wall thickness.
Further tests are necessary, but this formula is the best I can do with the data I have on hand.
An equation for the inner diameter of a tube if the outer diameter is restricted (i.e. does not fully expand) follows. This is based on the assumption that the area ratio scales linearly with the inner diameter ratio.
$D_i^e = (C_1 D_i)^{-1} [\sqrt{4(t C_2)^2 (D_i + t)^2 + (C_1 D_i)^2 (D_o^e)^2} - 2 t C_2 (D_i + t)]$
Posted in Design, Math, Pneumatics, Uncategorized | Leave a Comment »
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# statistics problem
• Oct 21st 2009, 12:32 PM
Als20
statistics problem
Hello (Hi)
It has been suggested that showing the face of the speaker in a computer package that teaches a foreign language leads to improved memory of words and phrases. To investigate this, a study was undertaken whereby 24 post-beginners to the Polish language were divided randomly into one of two groups. Members from the two groups were each asked to sit in front of a computer which then presented them with 20 words and phrases and gave their English translation together with a sentence that put them into context. The words and phrases were each presented twice.
The participants who were in Group 1 saw the words on screen and heard them being read out, but no face was shown on screen.
The participants who were in Group 1 saw the words on screen and heard them being read out by a speaker whose face was shown.
Both groups took a 15 minute break after their exercise and were then given a piece of paper with the words and phrases typed and participants were given 5 minutes to write down the English translation. The experimenters were interested in comparing these two methods of learning foreign language words.
Number of correctly translated words for the members in each group:
Group 1 (no face) Group 2(face)
12 16
8 10
11 13....
3
.
.
.
(a) Use SPSS to perform an appropriate t-test for this experiment.
(b) Use SPSS to perform and appropriate non-parametric test for this experiment.
(c) You are asked by the experimenters to advise them on which test would be the most suitable to be applied to these data Discuss how you would advise them.
What are the appropriate tests for q(a), (b), (c)?http://www.mathisfunforum.com/img/smilies/dunno.gif
Thank youu!!:D
• Oct 22nd 2009, 11:53 AM
jalko
Hi. So,
a) Should be the unpaired two sample t-test.
b) Mann–Whitney test should be appropriate.
c) If you expect non-normality in the population and because of smaller sample Mann-Whitney can be better. But under the normality assumption, t-test should be fine.
• Oct 23rd 2009, 01:15 PM
Als20
dotplot
I guess I should be assessing normality using a dotplot... but I cannot make heads or tails of this plot (Worried)
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# Je ne comprends pas (do not understand error message
#1
Bonsoir,
Voici ce que me dit la console !
function classe_moy at
0x7f6ae9f10488
A
None
Je ne comprends pas ...
Comment tout le monde s'en sort - Solution
#2
Please post your code so we can examine and test it. Thank you.
S'il vous plaît envoyer votre code afin que nous puissions examiner et tester. Je vous remercie.
#3
``````lloyd = {
"noms": "Lloyd",
"devoirs": [90.0, 97.0, 75.0, 92.0],
"quiz": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"noms": "Alice",
"devoirs": [100.0, 92.0, 98.0, 100.0],
"quiz": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"noms": "Tyler",
"devoirs": [0.0, 87.0, 75.0, 22.0],
"quiz": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
def moyenne(nombres):
somme = sum(nombres)
somme = float(somme)
taille = len(nombres)
taille = float(taille)
total = somme/taille
def calcule_moyenne(etudiant):
devoirs = average(etudiant['devoirs'])
quiz = average(etudiant['quiz'])
tests = average(etudiant['tests'])
note = 0.1 * devoirs + 0.3 * quiz + 0.6 * tests
return note
def ecrire_lettre_note(note):
if note >= 90:
return "A"
elif note >= 80:
return "B"
elif note >= 70:
return "C"
elif note >= 60:
return "D"
else:
return "F"
etudiants = [lloyd, alice, tyler]
def calcul_classe_moyenne(etudiants):
resultats = []
for etudiant in etudiants:
resultats.append(calcul_classe_moyenne(etudiant))
return moyenne(resultats)
classe_moy = calcule_classe_moyenne(etudiants)
print classe_moy
print ecrire_lettre_note(classe_moy)``````
#4
`somme` is already a float, so this line is not needed.
`taille` can remain an integer since it is a counting number, not a real float.
This line should have no indentation.
#5
After having test what you offered me the console gives me this answer ...
None
One of the following variables is missing or damaged when you try to use it: alice, lloyd, tyler, students, calcule_classe_moyenne, ecrire_lettre_note
#6
I was only scanning the code, but will now proceed with a test. Check my earlier (edited) post. When that line is taken out of the block the code runs into an interpreter error:
NameError: global name 'average' is not defined
#7
These are the lines causing the error:
`````` devoirs = average(etudiant['devoirs'])
quiz = average(etudiant['quiz'])
tests = average(etudiant['tests'])``````
Change `average` to `moyenne`.
That solves it.
#8
this is not resolved ... even with the change it does not work ...
I have the same answer ...
#9
Please re-post your edited code. I made your code pass with only the two main changes mentioned above. (I left the float stuff alone, but it does work, so why have code you don't need?)
#10
``````lloyd = {
"noms": "Lloyd",
"devoirs": [90.0, 97.0, 75.0, 92.0],
"quiz": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"noms": "Alice",
"devoirs": [100.0, 92.0, 98.0, 100.0],
"quiz": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"noms": "Tyler",
"devoirs": [0.0, 87.0, 75.0, 22.0],
"quiz": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
def moyenne(nombres):
somme = sum(nombres)
taille = len(nombres)
taille = float(taille)
total = somme/taille
def calcule_moyenne(etudiant):
devoirs = moyenne(etudiant['devoirs'])
quiz = moyenne(etudiant['quiz'])
tests = moyenne(etudiant['tests'])
note = 0.1 * devoirs + 0.3 * quiz + 0.6 * tests
return note
def ecrire_lettre_note(note):
if note >= 90:
return "A"
elif note >= 80:
return "B"
elif note >= 70:
return "C"
elif note >= 60:
return "D"
else:
return "F"
etudiants = [lloyd, alice, tyler]
def calcul_classe_moyenne(etudiants):
resultats = []
for etudiant in etudiants:
resultats.append(calcul_classe_moyenne(etudiant))
return moyenne(resultats)
classe_moy = calcul_classe_moyenne(etudiants)
print classe_moy
print ecrire_lettre_note(classe_moy)``````
#11
That line can go. Dividing a float by an integer yields a float.
`````` classe_moy = calcul_classe_moyenne(etudiants)
print classe_moy
print ecrire_lettre_note(classe_moy)``````
should be outside of the block,
``````classe_moy = calcul_classe_moyenne(etudiants)
print classe_moy
print ecrire_lettre_note(classe_moy)``````
#12
``````classe_moy = calcul_classe_moyenne(etudiants)
print classe_moy
print ecrire_lettre_note(classe_moy)``````
is outside the block
I delete the line: taille = float(taille) ??
#13
You may, yes. It is unnecessary (but not causing a problem).
Be sure that the last three lines are not indented.
#14
``````Traceback (most recent call last):
File "python", line 53, in <module>
File "python", line 51, in calcul_classe_moyenne
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File "python", line 51, in calcul_classe_moyenne``````
...
The long error message is back ...
#15
This is a recursion that has no base case (so no way to stop). Is this what you meant to do?
``````def calcule_classe_moyenne(etudiants):
resultats = []
for etudiant in etudiants:
resultats.append(calcule_moyenne(etudiant))
return moyenne(resultats)
classe_moy = calcule_classe_moyenne(etudiants)
print classe_moy
print ecrire_lettre_note(classe_moy)``````
#16
thanks it's ok ... I managed
#18
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# algebra!HELP!
posted by on .
a right triangle has legs 12 ft. and 5 ft. what is the length of the hypotenuse?
• algebra!HELP! - ,
Use the Pythagorean Theorem.
a^2 + b^2 = c^2
12^2 + 5^2 = c^2
144 + 25 = c^2
169 = c^2
13 = v
• algebra!HELP! - ,
13 = c
| 103
| 263
|
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| 9,518
|
Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
# Moisture Content Calculation
## Calculate the moisture content in products like wood on wet and dry basis.
Moisture content is the amount of water present in a moist sample of a product like wood, soil or similar. Moisture content can be be expressed on wet or dry basis.
### Moisture Content on Dry Basis
Moisture content on dry basis is the mass of water to the mass of dry solid:
MCd = mh2o / md (1)
where
MCd = moisture content on dry basis
mh2o = mass of water (kg, lb)
md = mass of dry solid (kg, lb)
Moisture content on dry basis is commonly used in the timber industry. Note that it is common to multiply values with 100%.
### Water Content on Wet Basis
Water content on wet basis is the mass of water to the mass of water and mass of solid:
MCw = mh2o / m
= mh2o / mh2o + md (2)
where
MCw = moisture content on wet basis
mh2o = mass of water (kg, lb)
mw = total mass of moist - or wet - sample - mass of solid and mass of water (kg,lb)
### Example - Water Content in Birch on Wet Basis
The density of air-dried seasoned dry Birch is 705 kg/m3 with 20% (0.2) water content. The amount of water per unit volume can be calculated by transforming (2) to
mh2o = mw MCw
= (705 kg/m3) (0.2)
= 141 kg/m3
The amount of solids can be calculated as
md = (705 kg/m3) (1 - 0.2)
= 564 kg/m3
### Example - Moisture Content in Birch on Dry Basis
The density of air-dried seasoned dry Birch is 705 kg/m3 with 20% (0.2) moisture content. Equation (1) can be modified to
MCd = mh2o / (m - mh2o) (1b)
where
m = mass water and solid (kg, lb)
This equation can be transformed to
mh2o = MCd m / (1 + MCd ) (1c)
The moisture content per unit volume can be calculated as
mh2o = (0.2) (705 kg/m3) / (1 + (0.2))
= 117.5 kg/m3
The amount of solids can be calculated as
md = m - mh2o
= (705 kg/m3) - (117.5 kg/m3)
= 587.5 kg/m3
## Related Topics
• ### Material Properties
Properties of gases, fluids and solids. Densities, specific heats, viscosities and more.
## Related Documents
• ### Air - Drying Force
The drying force of air depends on the air moisture holding capacity and the water surface to air evaporation capacity.
• ### Air - Humidity Ratio
The mass of water vapor present in moist air - to the mass of dry air.
• ### Air - Moisture Holding Capacity vs. Temperature
The moisture holding capacity of air increases with temperature.
• ### Compressed Air - Water Content
Saturation pressure and maximum water content in compressed air.
• ### Fruits and Vegetables - Optimal Storage Conditions
Optimal temperature and humidity conditions for common fruits and vegetables.
• ### Soil - Water Content
Water or moisture content in soil.
• ### Water Content in Food and other Products
Water content before and after drying - in food and other products cork, grain, soap, peat, wood and more.
• ### Wood - Compressive Strength vs. Moisture Content
Red Spruce, Longleaf Pine and Douglas Fir - moisture content and their compressive strength.
• ### Wood Species - Moisture Content and Weight
Weight of green and air-dried fire wood.
## Search
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# Search by Topic
#### Resources tagged with Multiplication & division similar to Our Numbers:
Filter by: Content type:
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Challenge level:
### There are 132 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### Tom's Number
##### Stage: 2 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Divide it Out
##### Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Clever Santa
##### Stage: 2 Challenge Level:
All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.
### 1, 2, 3, 4, 5
##### Stage: 2 Challenge Level:
Using the numbers 1, 2, 3, 4 and 5 once and only once, and the operations x and ÷ once and only once, what is the smallest whole number you can make?
### What Is Ziffle?
##### Stage: 2 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### The Deca Tree
##### Stage: 2 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### Difficulties with Division
##### Stage: 1 and 2
This article for teachers looks at how teachers can use problems from the NRICH site to help them teach division.
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### What's My Weight?
##### Stage: 2 Short Challenge Level:
There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### The Amazing Splitting Plant
##### Stage: 1 Challenge Level:
Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks?
### Claire's Counting Cards
##### Stage: 1 Challenge Level:
Claire thinks she has the most sports cards in her album. "I have 12 pages with 2 cards on each page", says Claire. Ross counts his cards. "No! I have 3 cards on each of my pages and there are. . . .
### What's Left?
##### Stage: 1 Challenge Level:
Use this grid to shade the numbers in the way described. Which numbers do you have left? Do you know what they are called?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Ordering Cards
##### Stage: 1 and 2 Challenge Level:
This problem is designed to help children to learn, and to use, the two and three times tables.
### Jumping
##### Stage: 2 Challenge Level:
After training hard, these two children have improved their results. Can you work out the length or height of their first jumps?
### Let Us Divide!
##### Stage: 2 Challenge Level:
Look at different ways of dividing things. What do they mean? How might you show them in a picture, with things, with numbers and symbols?
### Learning Times Tables
##### Stage: 1 and 2 Challenge Level:
In November, Liz was interviewed for an article on a parents' website about learning times tables. Read the article here.
### Current Playing with Number Upper Primary Teacher
##### Stage: 2 Challenge Level:
Resources to support understanding of multiplication and division through playing with number.
### Sort Them Out (2)
##### Stage: 2 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### Four Goodness Sake
##### Stage: 2 Challenge Level:
Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?
### Fair Feast
##### Stage: 2 Challenge Level:
Here is a picnic that Chris and Michael are going to share equally. Can you tell us what each of them will have?
### Twizzle's Journey
##### Stage: 1 Challenge Level:
Twizzle, a female giraffe, needs transporting to another zoo. Which route will give the fastest journey?
### Let's Divide Up
##### Stage: 2 Challenge Level:
Look at different ways of dividing things. What do they mean? How might you show them in a picture, with things, with numbers and symbols?
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Spiders and Flies
##### Stage: 1 Challenge Level:
There were 22 legs creeping across the web. How many flies? How many spiders?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### ABC
##### Stage: 2 Challenge Level:
In the multiplication sum, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Sam's Quick Sum
##### Stage: 2 Challenge Level:
What is the sum of all the three digit whole numbers?
### A Conversation Piece
##### Stage: 2 Challenge Level:
Take the number 6 469 693 230 and divide it by the first ten prime numbers and you'll find the most beautiful, most magic of all numbers. What is it?
### The 24 Game
##### Stage: 2 Challenge Level:
There are over sixty different ways of making 24 by adding, subtracting, multiplying and dividing all four numbers 4, 6, 6 and 8 (using each number only once). How many can you find?
### Numbers Numbers Everywhere!
##### Stage: 1 and 2
Bernard Bagnall recommends some primary school problems which use numbers from the environment around us, from clocks to house numbers.
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### How Do You Do It?
##### Stage: 2 Challenge Level:
This group activity will encourage you to share calculation strategies and to think about which strategy might be the most efficient.
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### The Tomato and the Bean
##### Stage: 1 Challenge Level:
At the beginning of May Tom put his tomato plant outside. On the same day he sowed a bean in another pot. When will the two be the same height?
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Machines
##### Stage: 2 Challenge Level:
What is happening at each box in these machines?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Penta Post
##### Stage: 2 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Multiplication Series: Illustrating Number Properties with Arrays
##### Stage: 1 and 2
This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .
### Carrying Cards
##### Stage: 2 Challenge Level:
These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
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Share
Explore BrainMass
# Statistics - Hypothesis test - ChiSquare test
Please see attached file with two questions with charts. Thank you!
Here we examine the relationship between social class and spending on welfare and education.
Lower Class Working Class Middle Class Upper Class
Spending on
welfare Too little 18 59 73 4
About right 12 109 132 10
Too much 16 142 121 12
Total 46 310 326 26
Lower Class Working Class Middle Class Upper Class
Spending on
education Too little 36 232 244 19
About right 10 71 70 4
Too much 2 13 18 3
Total 48 316 332 26
a) Calculate the value of chi-square for each table. What is the number of degrees of freedom for each table?
b) Based on an alpha of .01, do you reject the null hypothesis?
#### Solution Preview
Solution (1):
NOTE : The detailed step by step solution is given in the attached solution file on four pages. Please download the attached solution file.
However, here I am giving just a few steps.
Null Hypothesis H0 : Social Class and spending on Welfare are not dependent
...
#### Solution Summary
Solution to the posted problem is given with step by step working and explanation so that the students could easily understand the procedure and use this solution as a model solution to solve other similar problems.
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# Top 50 crypto mystery solved: Thomas Ernst deciphers Fredinand III’s encrypted letters
A few weeks ago I introduced two unsolved encrypted letters from the Thirty Years’ War. Scientist and blog reader Thomas Ernst has deciphered them now.
Emperor Ferdinand III (1608-1657) from the house of Habsburg played an important role in the Thirty Years’ War. When researching his life, historians repeatedly encountered encrypted letters. The code consists of numbers and simple geometric symbols arranged in pairs.
In 2014 Prof. Leopold Auer from Vienna told me about these letters. The encrypted passages could not be read. I blogged about these letters and asked my readers to break the code, but to no avail.
### Ferdinand’s cipher
In July 2017 I wrote about Ferdinand’s encrypted letters again – as a part of my top 50 unsolved cryptograms series. This time, Prof. Dr. Thomas Ernst, a German scientist who teaches in the USA and is a very active reader of this blog, tried his luck on this mystery. Ernst made a name for himself in the 1990s when he broke the 500-year-old encryption of Johannes Trithemius (1462-1516) in the third book of his Steganographia. This story is covered in my new book Versteckte Botschaften.
In contrast to most other codebreakers, who usually have a computer or engineering background, Ernst is a philologist and historian. He made himself familiar with the world of Ferdinand III very quickly. He also easily learned to read the emperor’s unencrypted letter passages, after the handwriting of Ferdinand III and his use of several languages had created difficulties for many other scholars.
### How the cipher was broken
While examining the cipher, Ernst wrote his thoughts as comments below my blog article. Altogether, he posted over 50 comments before the mystery finally was solved. These comments give a fascinating insight into the breaking of a 17th century cipher by a very skilled deciphering expert.
Ernst tested several hypotheses. As an important discovery, he found out that each non-numerical sign represents the number of its lines or semi-circles. For example, four lines stand for the number 4. In one of his many comments Ernst wrote: “From the very beginning, I didn’t like the auxiliary substitution of these signs into letters, as presented early in this thread. It gave me so many false starts! Two days ago, I looked at the signs themselves – again. And there it was, the ‘Eureka-moment’, the same sensation I felt a quarter of a century ago when the numbers of Trithemius’ ‘liber tertius’ unveiled themselves in the flash of a moment. These are good moments: rare, but good!”
As all non-numerical signs could now be replaced, from now on Ernst had to deal with number pairs only. Based on frequency analysis and probable words, he tried to guess some of the cleartext letters. Another break-through occured when he realised that the Habsburg motto AEIOU played a role in the cipher. AEIOU inscriptions (one out of many interpretations is Austria erit in orbe ultima) can today still be found on some Habsburg buildings, as can be seen on the following picture.
Ernst realised that in Ferdinand’s cipher 01/02 stands for A, 02/12 for E, 03/13 for I, and 04/14 for O – this was derived from the AEIOU motto. When he looked at the phrase “Zifra 21 13 42 04 23 04 33 03 43 02 01” at the start of the letter, he identified the following vowels: “Zifra – I – O – O – I – E A”. The word spelled out here could only be PIC[C]OLOMINEA. As Ernst knew, Piccolomini is the name of a family Ferdinand corresponded with (Piccolominea is the corresponding adjective).
## Kommentare (1)
1. #1 Thomas Ernst
Latrobe
9. Oktober 2017
Having accustomed myself to Ferdinand’s handwriting by now as one accustoms oneself to bad weather (thanks to the generous help of Professor Auer, as well as additional letters by Ferdinand), I can correct the first page of the letter 1640, July 20, to read as follows (deciphered text in small capitals, several mis-ciphered letters corrected, missing letters added in “‹…›”, non-valeurs dropped):
[1] Zifra PICOLOMI
[2] NEA
[3] Si haberemus ALIUM quam IPS
[4] UM were Ich eben der Meinung wie E L Vnd
[5] Were Wol der rehte Weg, aber eben das non habere
[6] glaube Ich MAC‹H›T IM DEST
[7] O HOCHSIEDIGER
[8] ob E L Gewar wenig Considerationes haben tam quo:
[10] ITULUM so habe Ich doch gar
[11] grosse dann, hoc quod scripsi quia me Vestigia
[12] terrent, non intellexi tantum de FRI
[13] TLANT, sunder Von allen
[14] UNSEREN FIR
[15] STEN E L schauen Nuhr
[16] Was wir vor nuz von IH‹N›EN
[17] gehabt haben Vnd noch haben. EGENBER
[18] G senex fuit origo istius mali. CARL UON
[19] LIECKDENS‹T›EIN senior
[20] fuit NOSTRA ruina in
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Alpacas' fleece is worth surprisingly little compared to : GMAT Sentence Correction (SC)
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# Alpacas' fleece is worth surprisingly little compared to
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Alpacas' fleece is worth surprisingly little compared to [#permalink]
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12 Sep 2008, 18:47
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Alpacas' fleece is worth surprisingly little compared to their market value; a top breeding specimen bringing upwards of $100,000 even if five pounds of fleece fetches only$80 to $240. (A) Alpacas' fleece is worth surprisingly little compared to their market value; a top breeding specimen bringing upwards of$100,000 even if five pounds of fleece fetches only $80 to$240.
(B) Alpacas' fleece is worth surprisingly little in comparison with its market value; a top breeding specimen bringing upwards of $100,000 while five pounds of fleece fetches only$80 to $240. (C) The fleece of the alpaca is worth surprisingly little compared to its market value, while a top breeding specimen can bring upwards of$100,000 even though five pounds of fleece fetch only $80 to$240.
(D) The fleece of the alpaca is worth surprisingly little compared to the animal's market value; a top breeding specimen can bring upwards of $100,000 while five pounds of fleece fetch only$80 to $240. (E) The worth of the alpaca's fleece is surprisingly little compared to the animal's market value; a top breeding specimen can bring upwards of$100,000 even though five pounds of fleece fetches only $80 to$240.
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Re: SC -- Alpacas' fleece is worth surprisingly [#permalink]
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12 Sep 2008, 22:00
IMO D.
A is out because of incorrect "their"
C is out because "its" (possessive pronoun) is referring to object Aplacas.
In B "brining" is awkward.
Between D and E. While is better than even though in context of this sentence.
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Re: SC -- Alpacas' fleece is worth surprisingly [#permalink]
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15 Sep 2008, 02:00
Another D. E is out because of "five pounds of flesh fetches....".
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Re: SC -- Alpacas' fleece is worth surprisingly [#permalink]
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15 Sep 2008, 09:33
I like B...Does anyone feel the same. Whats the OA ?
Re: SC -- Alpacas' fleece is worth surprisingly [#permalink] 15 Sep 2008, 09:33
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6 Alpacas fleece is worth surprisingly little compared to 12 08 Oct 2012, 07:11
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2 Alpacas' fleece is worth surprisingly little compared to 14 06 May 2008, 14:43
SC: Compared to vs compared with 4 05 Jan 2008, 07:22
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Helper II
## dynamically calculate average for last 3 months sales and last 6 months sales
I want to calculate the average for L3 and L6, dynamically, which means with every new month sales value added in the table, BI automatically adjusts the period to calculate the average for last 3 months and 6 months. E.g. In Jan, it should calculate L3 as average{nov,dec,jan} and L6 as {aug,sep,oct,nov,dec,jan}; similarly, when I update the feb values in this table, upon refreshing, the DAX formula should automatically adjust the period as follows : L3{dec,jan.feb} and L6{sep,oct,nov,dec,jan,feb}
The additional problem is that the year field in my table is in numeric type and the month field is in text type and I am not able to convert or combine either of the fields to date type, so that I may use any of the date-based DAX functions
1 ACCEPTED SOLUTION
Assuming you only have 1 table - Table1 with column headers [Sales], [Date] then adding new measures:
3mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-3,MONTH))/3
6mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-6,MONTH))/6
12mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-12,MONTH))/12
24mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-24,MONTH))/24
14 REPLIES 14
Do you have separate Calendar table?
3mth = CALCULATE(SUM(sales[sale]),DATESINPERIOD(Calendar[Date],LASTDATE(Calendar[Date]),-3,MONTH))/3
Helper II
Julian - nope , I dont't have a calendar table...what is that and do I have it? will this help the solution?
How do you set up the calendar table and connect it to my data table?
No you dont need one. There is alot of information/discussion on this topic - see this link
You could alway add a new column Date = FORMAT(Table1[Column1] &"-"& Table1[Column2],"MM-YYYY")
Helper II
Julian - thanks for this.
couple of questions :
1. When u say "add column" - this is thru' the "add column" option under the "modelling" tab, right?
2. in your formula, what would be my corresponding columns for [column1] and [column2]? - should I have column1 filled with the names of the month, and column2, filled with the years?
3. i have 2 columns - year , filled with years but in numeric format and month, filled with names of the month, but in txt format - hope your solution will work on these columns
thanks
Yes, thats right "add column" from modelling tab. Put your month header name into [column1] and year into [column2], and the table name in front of each. "Format MM-YYY" sorts out the text and numeric formats.
Helper II
so now I have a column added called [date]. So in your original formula, the calendar is a function or it should be replaced with the name of the table , [date], is now added to?
thanks
Assuming you only have 1 table - Table1 with column headers [Sales], [Date] then adding new measures:
3mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-3,MONTH))/3
6mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-6,MONTH))/6
12mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-12,MONTH))/12
24mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-24,MONTH))/24
Regular Visitor
Thanks
Helper I
Hi @cjulianm
This is pretty much everything I am also trying to accomplish. But instead of "Table [Sale]" I need to get the average 3, 6, 12, and 24 from a measure "Ratio_Measure" I created.
This formula 3mth = CALCULATE(SUM(Table1[sale]),DATESINPERIOD(Table1[Date],LASTDATE(Table1[Date]),-3,MONTH))/3 is not working for me properly. Do you think you can help me?
I have a sample of my excel data and Power BI here at this link HERE
Michelle
Best practice is to create a Calendar table. then use the
3mth = CALCULATE([Ratio_Measure],DATESINPERIOD(Calendar[Date],LASTDATE(Calendar[Date]),-3,MONTH))/3
Helper I
Hi @cjulianm,
Thank you for the help on this.
I found this measure below for the Average of 6 I needed, but do you know if there is a way we can make this measure a little bit more dynamic? Like, do some sort of variable where I can change the number of months I need? this project I am working on I need to take lots of different averages.
Ratio_Measure 6mth = CALCULATE( [Ratio_Measure] ,DATESINPERIOD(PD_agg_perf_measures[Accident_Date],eomonth(MAX(PD_agg_perf_measures[Accident_Date]),-3) ,-3,MONTH))
Where we can change just the variable average number:
ex: AVG_Num = 12
Ratio_Measure 6mth = CALCULATE( [Ratio_Measure] ,DATESINPERIOD(PD_agg_perf_measures[Accident_Date],eomonth(MAX(PD_agg_perf_measures[Accident_Date]),AVG_Num ) ,AVG_Num ,MONTH)) -- this doesn't work but I am trying to get the average 6,12,24... without modifying too much the measure. Is this possible?
Hope I am not overcomplicating things.
Frequent Visitor
@cjulianm Hi cJulianm, my apologies for brining this up again. I am very new to DAX and powerBI trying my best to find my way. Similiarly I am trying to calculate the average of the last 3 month's data. And have tried to follow you solutions.
CALCULATE(average('View Shipments'[Total Transit Time]), DATESINPERIOD('Date Table'[Date], max('View Shipments'[Shipped (ASN)]),-3,MONTH))
However,I did not use the "Latestdate" command becuase my date column "Shipped (ASN)" contains duplicates, therefore "Datesinperiod" also give me error that there is duplicate. I created a calander table from 2010 - 2020, and used MAX instead and no error come out from this.
However, I tried to double check with my Excel to see of the calculation is correct. But it is not correct, would hope that you can help me why my formula isnt working. Thank you so much!
Hi - I know this post is ancient - however did you find a solution to your issue?
I'm finding the LASTDATE example does work in latest power bi (March 2023)
when using a percentage measure...
very frustrating....
so for me this:
CALCULATE(FORMAT([HAC MoM Rate (MTD)]/[Target MOM 0.035],"▼00.00%;▲00.00%"),
DATESINPERIOD(
'Date'[Date],
LASTDATE('Date'[Date]),-3,MONTH)
and this
CALCULATE(FORMAT([HAC MoM Rate (MTD)]/[Target MOM 0.035],"▼00.00%;▲00.00%"),
DATESINPERIOD(
'Date'[Date],
LASTDATE('Date'[Date]),-1,MONTH)
Give me exactly the same results... (no page filters being used)
Any other ideas (that dont involve forcing the last date to work)... ?
kind regards
Clin Epi
Ah you know using MTD (totalmtd) in measures only works outside of this logic - ha, answered my own question in the end..
The LASTDATE formula works fine as long as you dont use totalmtd in the measure its refering to lol...
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# How to efficiently normalize groups of elements in a tensor
### Context:
I am trying to replicate Hinton’s “Matrix capsules with EM routing” (https://openreview.net/forum?id=HJWLfGWRb).
At some point, there are convolutional operations that are performed (in the sense that an output tensor is connected to an input tensor, and each element in the output tensor is influenced only by the input elements contained in a 2D mask of size K).
Input
Tensor `x` of shape `w_in,w_in`
where
• `w_in=14`
Intermediary tensor mapped to the input
Tensor `x_mapped` of shape `w_out,w_out,K,K`
where
• `K=3` is the convolution kernel size
• `w_out=6`, resulting from convolution with `stride=2`
Summing on the dimensions 2 and 3 (both of size `K`) means summing on the input elements connected to an output element whose location is given by dimensions 0 and 1.
### Question:
How can I efficiently normalize (to 1) groups of elements in `x_mapped`, based on their location in the input tensor `x`?
For example:
`x_mapped(0,0,2,2)`
`x_mapped(1,0,0,2)`
`x_mapped(0,1,2,0)`
`x_mapped(1,1,0,0)`
are all connected to `x(2,2)` (the formula is `i_out*stride + K_index = i_in`). For that reason, I would like the sum of those 4 elements to be 1.
And I would like to do that for all the groups of elements in `x_mapped` that are “connected” to the same element in `x`.
I can figure out how to do it by:
1. Building a dictionary with input location as key and list of output elements as value
2. Looping on the dictionary, summing the elements in the list for a given input location and dividing them by that sum
but that seems really inefficient to me.
I solved this in the following way:
1. Creating a dictionary with a 2-tuple as key (coordinates in `x`) and a list of elements of `x_mapped` as values.
2. One loop over the dictionary, zipping all the elements of one dictionary item, then normalizing.
Here is the code:
``````from collections import defaultdict
import torch
ho = 6
wo = 6
stride = 2
K = 3
d = defaultdict(list)
x_mapped = torch.arange(0,ho*wo*K*K).view(ho,wo,K,K).type(dtype = torch.DoubleTensor)
for i_out in range(0,ho):
for j_out in range(0,wo):
for K_i in range(0,K):
for K_j in range(0, K):
i_in = i_out * stride + K_i
j_in = j_out * stride + K_j
d[(i_in, j_in)].append((i_out, j_out, K_i, K_j))
for _ , value in d.items():
ho_list, wo_list, K_i_list, K_j_list = zip(*value)
x_mapped[ho_list, wo_list, K_i_list, K_j_list] = x_mapped[ho_list, wo_list, K_i_list, K_j_list] / torch.sum(
x_mapped[ho_list, wo_list, K_i_list, K_j_list])``````
The clean solution I found was to use `Torch.fold()`
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# finding the eccentric anamoly
by smmSTV
Tags: anamoly, eccentric
P: 2 So I'm in need of some more help for my astronomy class again. My professor wants us to write a program that outputs a table with the distance of Mars from the sun (r) and it's true anomaly. The problem is that i need to compute the Eccentric Anomaly (E) from the Mean Anomaly (M). Kepler's equation is E = M + εsin(E), but I can't get it down in terms of E. Seeing as how I'm using c++, it is kind of a necessity. One website seemed to have an equation involving Einitial and Efinals, but it didn't do me any good. Is the idea to set the original Einitial equal to 0 at t=0 (the perihelion point) and find E final in terms of that? At the end of each iteration (I'm running a while loop) the Eintial of the next iteration would be set to the Efinal of the current loop. Anybody got any ideas? Thanks
P: 365 The correct equation is $$M = E - e*sin(E)$$ Solve using the Newton Method: $$E_{n+1} = E_{n} - \frac{f(E)}{f'(E)}$$ where f(E) = E - e * sin(E) - M and f'(E) = 1 - e * cos(E) Loop the above equations until: $$\frac{f(E)}{f'(E)} < 0.00001$$Or some substantially low number not zero. Also,$$r = \frac{a * (1 - e ^ 2)}{(1 + e * cos(TA))}$$ where TA - True Anomaly and a - Semi-Major Axis of Mars
Related Discussions Engineering Systems & Design 0 Math & Science Software 0 Mechanical Engineering 2 Engineering, Comp Sci, & Technology Homework 31 General Astronomy 2
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https://byjus.com/question-answer/the-tension-t-in-the-string-in-figure-is-sqrt3-1-50-n-zero-50/
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Question
# The tension T in the string in figure is (√3−1)50 NZero50 N35√3 N
Solution
## The correct option is B ZeroFriction force is trying to uppose the motion so, fs=μmgcos300=0.7×10×10×√32 fs=35√3=60.62 N Gravitational force helping in motion i.e, F=mgsin300=10×10×12=50 N Since fs>F, this means friction force is sufficient to oppose gravitational force. So, tension will be Zero
Suggest corrections
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# Who Invented Hopscotch?
Hopping on just one leg is hard. So more than 1,000 years ago, Roman soldiers got stronger by hopping through numbered squares drawn on the ground. Little kids liked the game so much that they copied it – and that’s where hopscotch came from! You can write whatever numbers you like, throw rocks to block squares, or make up your own rules. But you do have to hop.
Wee ones: What shapes do you see on this hopscotch board?
Little kids: If you write the numbers 1 through 8 in order on your hopscotch board, what number do you write before the 6? Bonus: If your hopscotch board has spaces 1 through 10, how many hops do you do if you land only on the odd numbers once each?
Big kids: If 6 people and their 6 pet bunnies start hopping, how many feet are hopping all together? (Remember: bunnies hop on 2 feet, but we don’t!) Bonus: If you number 1 through 12, and skip all the multiples of 4 as well as the non-4-multiple square your rock landed on, how many hops do you do to hop to the end and back?
The sky’s the limit: If a bunch of people and bunnies play hopscotch, and there are 8 players in total but 1 more bunny foot than people feet hopping, how many bunnies are playing?
Wee ones: Squares and rectangles (both are 4-sided shapes), and at the end, almost a half-circle.
Little kids: The 5. Bonus: 5 hops: on the 1, 3, 5, 7, and 9.
Big kids: 18 feet. Bonus: 16 hops, since you do 8 in each direction. You skip 4 squares: the 4, 8, 12, and rock.
The sky’s the limit: 3 bunnies, who hop on 6 feet while the 5 people hop on 5. You can start with 8 people on 8 feet, and each time you swap in a bunny for a person, the bunny feet go up by 2 while the people feet go down by 1, so the gap grows by 3. The gap started at +8 for the people and you need a gap of -1, so you have to change the gap by 9. You’ll need 3 swaps to do that, so 3 bunnies.
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https://www.slideshare.net/DLBeebe/cost-based-analysis-tools-by-derek-beebe
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Upcoming SlideShare
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# Cost Based Analysis Tools By Derek Beebe
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This presentation shows some of the many quantitative cost based analysis tools used to make business decisions. It is meant as a guide for students and recent grads and is by no means a complete reference to the many quantitative tools and software available.
This presentation was given as a class project at Central Michigan University.
Published in: Business, Economy & Finance
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• Usually simple mathematic formulas are used for calculating Cost Based Analysis however the math can get extremely complicated such that sophisticated software tools must be implemented (but what if you are doing a Cost based analysis to determine the need for the Cost based analysis software?)
• The net present value method (NPV) of evaluating a major project allows you to consider the time value of money . Essentially, it helps you find the present value in &quot;today's dollars&quot; of the future net cash flow of a project. Then, you can compare that amount with the amount of money needed to implement the project. If the NPV is greater than the cost, the project will be profitable for you (assuming, of course, that your estimated cash flow is reasonably close to reality) Please Note that whenever you do time value of money calculations to find a present or future value (such as NPV), you'll need to specify an interest rate, known as the discount rate . Choosing the appropriate discount rate is a very important part of the process.
• Other software: Oracle, VBA scripts, SAP, MS Project,
• A Monte Carlo Simulation is an analytical technique used to numerically determine the expected value of a decision. A Monte Carlo Simulation works by running a decision model over and over again (for thousands of iterations) to take in to account all possible future scenarios that may happen. Most Monte Carlo Simulations are run by starting with an Excel-based decision model and then utilizing simulation add-in software. The software keeps track of relevant objective values and produces the resulting distributions. Brings together NPV, Probability, and other types of quantitative tools to show results in multiple dimensions.
• Bayes Theorem - The theorem expresses the posterior probability (i.e. after evidence E is observed) of a hypothesis H in terms of the prior probabilities of H and E, and the probability of E given H. It implies that evidence has a stronger confirming effect if it was more unlikely before being observed. [
• ### Cost Based Analysis Tools By Derek Beebe
1. 1. Quantitative Cost Based Analysis Tools Mathematics schematics… Arithmetic, logarithmic, and geometric return Quantitative Cost Based Tools? 1. Net Present Value (NPV) 2.ROR/ROI 3. Cost Based Analysis Ratios 4.Monte Carlo Simulation 5.Probability Studies
2. 2. Quantitative Cost Based Analysis Tools http://www.12manage.com/methods_npv.html Net Present Value
3. 3. Quantitative Cost Based Analysis Tools Microsoft excel templates = Free downloads Net Present Value
4. 4. Quantitative Cost Based Analysis Tools ROR and ROI Calculating the Rate of Return on Investments Say you invest \$100 in an automotive company’s stock, which is your capital. One year later your investment yields \$110. Calculate with the following formula. ((Return – Capital) / Capital) x 100% = ROI ((\$110 - \$100) / \$100) x 100% = 10% ROI of 10%
5. 5. Quantitative Cost Based Analysis Tools Cost Based Analysis Ratios (Cost < Benefit = Potential Profit) (Cost > Benefit = Potential Loss) Benefits / Cost = Ratio
6. 6. Quantitative Cost Based Analysis Tools Monte Carlo Simulations
7. 7. Probability Studies Bayes Theorem Standard Deviation Z scores Venn Diagrams Quantitative Cost Based Analysis Tools
8. 8. References Damato, K. Doing the Math: Tech Investors' Road to Recovery is Long. Wall Street Journal, pp.C1-C19, May 18, 2001
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# Why do all maps have distortions?
The projection of maps can be done in various shapes. A map can be projected on the surface of a sphere similar to the one on the plane paper. Maps are created using map projections. Based on the purpose for which the maps are used and the shape in which they are projected, the distortions can exist in the map. Each of the map projections are made keeping in view of certain properties of the sphere, leaving others.
The locations on the sphere can be represented perfectly only on a sphere. It is difficult to represent all sphere points on the flat paper. The flat paper cannot display the exact distances between points more than three. So in order to represent all the parts of the map without any removal, it is essential to allow some distortions in the map on the paper. As earth is spherical in shape, maps are chosen to be depicted in spherical form.
If the map is made to fit in a rectangle, the top and bottom ends of the map will be larger than their usual size. The rectangular shape will make the continents to look bigger than they were in spherical form. The curved portions of the map comprising the continents will depict the places to be farther from each other than what they usually were. Hence, distortions should exist in the rectangular map projection to represent it properly in rectangular shape. But, the number of distortions in the rectangular map projection will be different.
The various different map projections like spherical or rectangular forms have to be depicted on flat paper with lot of stretching done. If there are no distortions in each of the map projections, their shapes will be lost after stretching them on flat paper. So, in the attempt to show the surface of spherical earth on the flat surface, distortions of direction, scale, distance and area have to be implemented.
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# Problem with xbasic compiler
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hi everyone!
I am new to MSX basic programming. and I have a problem that I can't solve... can you please help me?
The problem is this if i run this listing without xbasic everything works.
But as soon as i insert xbasic (call turbo on) i always have errors...
(Subscript put of range in 30)
the listing is this:
10 dim n\$(4192)
20 n\$="0":n\$(4192)="0":n\$(4192)=n\$
25 call turbo on
30 for i=1 to 4192:if n\$(i)="1" then 60
40 prime=i+i+1:cnt=cnt+1:k=i
50 k=k+prime:if k<4193 then n\$(k)="1":goto 50
60 next i
65 call turbo off
70 print cnt
is a benchmark test...
ログイン/登録して投稿
Your array index goes out of bounds, already in line 20 BTW. If you dim a string (or array for that matter) to X it goes from 0..X-1 (inclusive). So accessing member X is a programming error. It so happens that xbasic is not running at that moment and your violation goes unnoticed.
Lastly, consider using nestorbasic instead of xbasic. It comes with a lot of other niceties that xbasic cannot solve.
You also must state which variables will be sent inside the turbo_block:
if you do this:
10 A=7
20 CALL TURBO ON
30 PRINT A
40 CALL TURBO OFF
RUN
0
OK
The result will be zero because all variables are 0 every time you begin a turbo block, but they remain intact when you leave the turbo block:
10 A=7
20 CALL TURBO ON
30 PRINT A
40 CALL TURBO OFF
50 PRINT A
RUN
0
7
OK
In order to include an external variable's value inside a turbo block you must state it like this:
10 A=7
20 CALL TURBO ON (A)
30 PRINT A
40 CALL TURBO OFF
50 PRINT A
RUN
7
7
OK
To include several variables, just use a coma:
20 CALL TURBO ON (A,B,C)
But B and C must contain a value, otherwise it will give you an error:
10 A=7:B=0:C=0
20 CALL TURBO ON (A,B,C)
For dimensioned arrays it works like this:
10 DIM N\$(25):A=7:B=0:C=0
20 CALL TURBO ON (N\$(),A,B,C)
Anyway, having the variables duplicated (outside the turbo block and inside the turbo block) is usually a bad idea and it consumes twice the space AND can generate fatal errors and bugs when the CLEAR function is not used correctly, and it should only be used if those variables will be used by other turbo blocks or outside turbo blocks.
The usual approach would be:
10 CALL TURBO ON
20 A=7
30 PRINT A
40 CALL TURBO OFF
RUN
7
OK
Regarding NestorBasic, yes, it has a lot of features that makes your life easier (embeeded music player and sound player for example) but it requieres 128kb of RAM in order to work while Turbobasic/Xbasic/BasicKun only needs 64kb, so you should choose depending on your needs.
Another tip:
Adding DEFINT A-Z at the beginning of your listing speeds up the turbobasic code A LOT (it does not need to be inside the turbo-block).
If you need a variable not to be an integer you can always leave some for that pruppose:
DEFINT A-Y
Z will not be an integer, and you can define it any way you want.
HI all....
I found that if I create a string variable Xbasic doesn't recognize it....
10 defint a-z
20 c=0:b=7:a\$="jjj"
30 call turbo on(a\$,c,b)
40 for i=1 to 10
50 print a\$,c,b
60 next i
70 call turbo off
80 end
... if I remove the string a\$ everything works...
Most probably there will be a defint instruction to initialize the strings...
I've searched but I can't find documentation on xbasic...
Can someone help me?
Have you tried writing line 30 as such:
` CALL TURBO ON (A\$(),C,B)`
Illegal fuction call 30
Oh, okay sorry. It must be too long ago for me.
And if you try to put DIM A\$(25) first (right after the DEFINT A-Z)?
Is there no documentation?
I'm doing it in Turbo Pascal and it works just fine....
You can't include strings defined outside the turboblock, you need to define it inside the turboblock.
Also, take into consideration that the variable names are just 2 characters long, so if you write more than 2, only the first 2 will be taken into account:
-prime will be read as pr
-premature will be read as pr as well
This will cause you many issues. Just use 2 characters names (p1, p2...)
Here you can find how to use xbasic/turbobasic/basicKun and the differences between regular basic and xbasic (it is in spanish but you can google tranlate it):
https://www.konamiman.com/msx/msx2th/kunesp.txt
I found some more (in english):
https://konamiman.github.io/MSX2-Technical-Handbook/md/KunBA...
-msx xbasic manual
-msx turbobasic manual
-msx basic kun manual
You will find plenty of stuff.
Yes, strings are bit of a pain as they work very differently inside a turbo block... I don't suggest trying to transfer them, but if you really must...
```10 DEFINT Z
20 A\$="I don't know why I'm doing this"
30 Z=VARPTR(A\$)
40 _TURBO ON (Z)
45 '#I 42,Z,17,A\$,126,18,79,35,126,35,102,111,19,6,0,237,176
50 PRINT A\$
```
thanks a lot!!!
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## Elegant solutions, Column buckling, and the hole through the middle of the Earth
The previous post presented some examples of solution of differential equations using an iterative process, including the simple harmonic motion problem of a ball falling through an evacuated hole through the middle of the Earth. The final example was concerned with finding the deflection and buckling load of a column subject to an axial load with a small eccentricity. This problem, which appears entirely unconnected at first sight, was solved with exactly the same differential equation as the oscillating ball problem, suggesting that the exact analytical solution of the two problems should also have the same form.
The equivalent terms in the applicable differential equations for the two problems are shown below:
Simple Harmonic Motion Column Buckling Period, T Column Length Half amplitude, R Transverse Deflection Velocity Slope Acceleration, g Curvature
For the case of the ball through the middle of the Earth it was shown that the period of oscillation was given by:
T = 2π(R/g)0.5
This can be rearranged as:
g = R(2π/T)2
To give the initial acceleration required for a given radius and period.
If we substitute in the equivalent terms for column buckling, noting that:
• Curvature = M/EI = F.Delta/EI (where Delta is the eccentricity of the applied load, F, relative to the centroid of the base of the column, E is the elastic modulus of the column, and I is the second moment of area of the cross section)
• In simple harmonic motion the period is the time for the ball to travel the distance from the surface of the Earth to the centre four times. In the case of the column the column length is therefore equivalent to the Period/4.
• We wish to find the axial force at the top of the column, F, at which the deflection due to the force (ignoring second order effects) is equal to the initial eccentricity at the base.
F.Delta/EI = Delta(π/2L)2
F = EI(π/2L)2 which is the Euler equation for the buckling load of a cantilever column.
But what is the physical significance of this equation, and how does it relate to the acceleration of a ball falling through the Earth, and the orbital velocity of a body with the same period?
This can be seen if we consider a cylinder of radius delta, centred on the top of the column, with a vertical axis, as shown below:
This diagram shows the deflected column centroidal axis (red line), and the imaginary cylinder.
Consider a line coincident with the centroid of the column at the base (green line), and spiralling up the surface of the cylinder at a constant slope, so that at the top of the column it has passed through a right angle around the cylinder. If this line is projected onto the XY plane (the plane of the deflected column) it will exactly follow the deflected shape of the column centroid. It can be seen that:
• The slope of the line from the vertical, around the surface of the cylinder = (Delta.π/2)/L = (Delta.π)/2L
• The curvature of the line in the XY plane at the base = Slope2/Delta = ((Delta.π)/2L)2/Delta = Delta(π/2L)2
• Therefore at the buckling load F.Delta/EI = Delta(π/2L)2
• Buckling load, F = EI(π/2L)2
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# Measuring Wifi Transmit Power
I had a busy month in October working on the RF part of the Mesh Potato Wifi. This post documents the method I used to measure the Wifi transmit power using a Tek 492 spectrum analyser. The Tek 492 is an old analog spectrum analyser with basic digital storage to capture the swept signal. More modern spectrum analysers might do power measurements for you, but the principles below may still apply.
A spectrum analyser is basically a tuned receiver that sweeps across a certain frequency range. The receiver has a certain bandwidth, called the Resolution Bandwidth (RB). As it sweeps it measures the power at each frequency and plots it on the screen in dBm. It effectively sums all of the power in the RB to plot one point. This works well for signals that fit comfortably in the RB, like a narrow band signal or an un-modulated carrier. For narrow band signals, the RB will be wide enough to contain the total power of the signal. So a 10dBm carrier will be plotted as a 10dBm signal on the screen. Easy.
However for signals wider than the resolution bandwidth power measurement gets tricky. For example the maximum RB of my Tek 492 is 1 MHz, but the Wifi signal bandwidth is 12-18 MHz. The power is spread over many frequencies rather than fitting neatly into the RB.
In my sat-com modem days I had to do similar power measurements, so I worked it out eventually. Here is a photo of a 54 Mbit 802.11g signal on a V1.1 MP01:
P, the Wifi transmit power is given by:
P = Pm + Pref – 10log(RB) + 10log(BW)
where Pm is the measured power on the spec-an graticule (-31 in the photo above), Pref is the reference level (30dBm in top left hand corner, RB is the resolution bandwidth (1MHz, lower right hand corner), and BW is the bandwidth of the signal estimated from the graticule (about 16MHz). The -10log(RB) term is divides the power by the RB to get the power in 1 Hz of bandwidth, we then multiply by the total bandwidth (10log(BW) term) to get the total power in the signal.
So plugging in values above:
P = -31 + 30 – 10log(1E6) + 10log(16E6) = -31 + 30 +12 = 11 dBm
The -10log(1E6)+10log(16E6) term is constant (12) for 802.11g if you keep the RB constant. For 802.11b the constant is 10 as the bandwidth is narrower (about 12 MHz).
Now this technique is not particularly accurate – I have been told the best way to measure power is with a calibrated power meter (if you have the budget). It’s hard to get sub dB accuracy squinting at a graticule and estimating the 3dB bandwidth of funny looking spectra. Also I am not sure what the loss is from my cables and DC-block on the spec-an. However for relative measurements where absolute accuracy is not critical it is quite a useful technique.
The DIR-300 router measured around 17-18dBm at 1Mbit 802.11b (17dBm is the specified power output). At 54Mbit the measured power was around 10-12dBm (a little lower than the 14dBm specified). Similar levels were obtained from two DIR-300s and 3 MP01s tested.
Tek 494 Gotcha: It took me a week to get sensible results – I kept getting low values and couldn’t work out what was wrong with my math. Then I discovered the “peaking” control. My Tek 492 has the pre-selector option installed. You need to adjust the peaking control for maximum output, mine was 10dB down! After I peaked the signal I started getting measured powers for the DIR-300 that matched the specs and could then make sensible measurements on the MP01.
To connect everything together I purchased a kit of cables for a few hundred \$ from Rojone that had N-type and SMA connectors plus a RP-SMA to SMA adaptor. Made it easy two swap between routers even hook them up back-back for link and packet error rate tests.
Links
[1] RF Hacking post over on the Village Telco blog.
### 1 comment to Measuring Wifi Transmit Power
• Most spectrum analysers are far from ideal for power measurement. When working with serious power outputs, the difference between a cool running system and one so out of tune things are melting might be quite hard to even notice on a spectrum analyser screen
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# Approximation Schemes for convective term - structured grids - definitions
Here we shall develop a commone definitions and regulations because of
• in different articles was used defferent definitions and notations
• we are searching for common approach and generalisation
## Usual using definition for convected variable
$\boldsymbol{f}$
$\boldsymbol{\phi}$
## definition of considered face, upon wich approximation is applied
usually (in the most articles) west face of the control volume $\boldsymbol{w}$ is considered (without loss of generality)
for which flux is directed from the left to the right
we shall define it as $\boldsymbol{f}$
and convected variable at face as $\boldsymbol{\phi_{f}}$
also you can find in literature such definition as $\boldsymbol{i+1/2}$ , but we suggested it non suitable, because of complication
## indicators of the local velocity direction
approximation scheme can be written in the next form
$\phi_{w}=\sigma^{+}_{w}\phi_{W} + \sigma^{-}_{w}\phi_{P}$ (1)
where $\sigma^{+}_{w}$ and $\sigma^{-}_{w}$ are the indicators of the local velocity direction such that
$\sigma^{+}_{w} = 0.5 \left( 1 + \frac{\left|U_{w} \right|}{U_{w}} \right)$ (1)
$\sigma^{-}_{w} = 1 - \sigma^{+}_{w}$ (1)
and of course
$\left( U_{w} \neq 0 \right)$ (1)
also used such definitions as $U^{+}_{w}$ and $U^{-}_{w}$
we offer to use
$U^{+}_{f}$ and $U^{-}_{f}$
therefore unnormalised form of approximation scheme can be written
$\phi_{f}=U^{+}_{f}\phi_{W} + U^{-}_{f}\phi_{P}$ (1)
or in more general form
$\phi_{f}=U^{+}_{f}\phi_{C} + U^{-}_{f}\phi_{D}$ (1)
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# How many moles of h2 are needed for kJ?
Contents
## How do you calculate KJ G?
In order to convert kilojoules per gram to kilojoules per mole, you need to multiply by grams per mole. Now, let’s say that you’re dealing with a compound that has a molar mass of x g mol−1 . This tells you that 1 mole of this compound has a mass of x g .
## How do you calculate the number of moles needed?
Determine the moles of product produced by dividing the grams of product by the grams per mole of product. You now have calculated the number of moles of every compound used in this reaction.
## How do you find the kJ of heat produced?
To calculate the amount of heat released in a chemical reaction, use the equation Q = mc ΔT, where Q is the heat energy transferred (in joules), m is the mass of the liquid being heated (in kilograms), c is the specific heat capacity of the liquid (joule per kilogram degrees Celsius), and ΔT is the change in …
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## How do you find moles given kJ?
kJ is the units “Joules” multiplied by a number (kilo = x1000) that has no units of it’s own. kJ/mol is just the J multiplied by a number that has no units on it’s own: 1000/mol. The number of moles of something is just a number with no units of it’s own.
## How do you convert kJ to mol to kJ?
Re: How to convert kJ/mol to J
Answer: Cancel out the 1/mol unit by dividing by the Avogadro constant. Then convert kJ to J by multiplying the kJ value by 1000 (because of the conversion factor 1 kJ = 1000 J).
## How do you convert kJ to kJ mol?
kJ is the units “Joules” multiplied by a number (kilo = x1000) that has no units of it’s own. kJ/mol is just the J multiplied by a number that has no units on it’s own: 1000/mol.
## How do you find kJ in chemistry?
Calculating energy changes
1. = 100 × 4.2 × 20 = 8,400 J.
2. It is also useful to remember that 1 kilojoule, 1 kJ, equals 1,000 J. …
3. Moles of propane burned = 0.5 ÷ 44 = 0.01136.
4. So, the molar enthalpy change, ∆H = 8.4 ÷ 0.01136 = 739 kJ/mol.
## How do you find the number of moles in a liquid?
MOLES FROM VOLUME OF PURE LIQUID OR SOLID
Multiply the volume by the density to get the mass. Divide the mass by the molar mass to get the number of moles.
## How do you do Stoich?
Almost all stoichiometric problems can be solved in just four simple steps:
1. Balance the equation.
2. Convert units of a given substance to moles.
3. Using the mole ratio, calculate the moles of substance yielded by the reaction.
4. Convert moles of wanted substance to desired units.
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## What is Delta H in Chem?
In chemistry, the letter “H” represents the enthalpy of a system. … Therefore, delta H represents the change in enthalpy of a system in a reaction. Assuming a constant pressure, a change in enthalpy describes a system’s change in heat.
## What is Delta E in chemistry?
Delta E is defined as the difference between two colors in an L*a*b* color space. As the values determined are based on a mathematical formula, it is important that the type of color formula is taken into account when comparing the values.
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Pyrite | Level 9
## Translating a SAS/IML list definition into a position
Continuing the conversation re: representing an IML list structure as a matrix, how do I determine the position of the item L3_4 in the following example:
``````proc iml ; package load ListUtil ;
L = [ #'L1'=[ #'L1_1' = [ #'L1_2' = 'item L1_2' ] ]
, #'L2'=[ #'L2_1' = 'item L2_1']
, #'L3'=[ #'L3_1 '= [ #'L3_2' = [ #'L3_3' = [ #'L3_4'= 'item L3_4' ] ] ] ]
] ;
call struct(L) ;
call listprint( L ) ;
quit ;
``````
If I want to delete item L\$'L3'\$'L3_1'\$'L3_2'\$'L3_3'\$'L3_4', how do I use ListDeleteItem() to do this? How do I represent the position as a numeric matrix of indices or a character matrix of item names? Is my confusion due to the fact that I am trying to delete an item from a sublist? If so, how do I delete an item from a sublist?
1 ACCEPTED SOLUTION
Accepted Solutions
SAS Super FREQ
## Re: Translating a SAS/IML list definition into a position
Yes, I think your confusion is because you are trying to directly delete from a sublist. You can't use ListDeleteItem for that. The easiest way to delete an item from a sublist is to use ListGetSubItem with the 'd' flag:
``````proc iml ; package load ListUtil ;
L = [ #'L1'=[ #'L1_1' = [ #'L1_2' = 'item L1_2' ] ] ];
call struct(L) ;
item = ListGetSubItem(L, {'L1' 'L1_1' 'L1_2'}, 'd' ); /* get value; delete from the list */
free item; /* optional: completely delete the item */
call struct(L) ;``````
SAS Super FREQ
## Re: Translating a SAS/IML list definition into a position
Yes, I think your confusion is because you are trying to directly delete from a sublist. You can't use ListDeleteItem for that. The easiest way to delete an item from a sublist is to use ListGetSubItem with the 'd' flag:
``````proc iml ; package load ListUtil ;
L = [ #'L1'=[ #'L1_1' = [ #'L1_2' = 'item L1_2' ] ] ];
call struct(L) ;
item = ListGetSubItem(L, {'L1' 'L1_1' 'L1_2'}, 'd' ); /* get value; delete from the list */
free item; /* optional: completely delete the item */
call struct(L) ;``````
From The DO Loop
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# How do you convert to 2 decimal places in Java?
Contents
## How do you do 2 decimal places in Java?
format(“%. 2f“, 1.23456); This will format the floating point number 1.23456 up-to 2 decimal places, because we have used two after decimal point in formatting instruction %.
## How do you round a double to 2 decimal places in Java?
Round of a double to Two Decimal Places Using Math. round(double*100.0)/100.0.
## How do you change to 2 decimal places?
Rounding to decimal places
1. look at the first digit after the decimal point if rounding to one decimal place or the second digit for two decimal places.
2. draw a vertical line to the right of the place value digit that is required.
3. look at the next digit.
4. if it’s 5 or more, increase the previous digit by one.
## How do you float to 2 decimal places?
2f}” as string and float as a number. Call print and it will print the float with 2 decimal places. After writing the above code (python print 2 decimal places), Ones you will print “ format_float ” then the output will appear as a “ 2.15 ”. Here, it will format the float with 2 decimal places.
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## What does .2f mean in Java?
The %2\$ indicates a variable to insert into the string, which will be taken from the second additional argument (after the format string). The . 2f means this argument should be formatted as a float (the f part) and with a precision of 2 digits after .
## How do I limit decimal places in Java?
You can add or subtract 0 on the right side to get more or less decimals. Or use ‘#’ on the right to make the additional digits optional, as in with #. ## (0.30) would drop the trailing 0 to become (0.3).
## What does it mean to round to 2 decimal places?
“Two decimal places” is the same as “the nearest hundredth”. … So, for example, if you are asked to round 3.264 to two decimal places it means the same as if your are asked to round 3.264 to the nearest hundredth. Some questions, like the example below, will ask you to “show your answer correct to two decimal places.”
## How do you round to 2 decimal places in math?
For example, if you want to round 0.507 to 1 decimal place, you multiply by 10 to get 5.07, round to get 5, then divide by 10 to get 0.5. Or, if you want to round 0.2345 to two decimal places, you need to round 23.45 (0.2345*100), then divide the result (23) by 100 to get 0.23.
## How do you round to 2 decimal places in Python?
Just use the formatting with %. 2f which gives you rounding down to 2 decimals. You can use the string formatting operator of python “%”.
IT IS INTERESTING: Best answer: What is ClassNotFoundException in Java?
## What does two decimal places look like?
Rounding to a certain number of decimal places
4.737 rounded to 2 decimal places would be 4.74 (because it would be closer to 4.74). 4.735 is halfway between 4.73 and 4.74, so it is rounded up: 4.735 rounded to 2 decimal places is 4.74.
## What is 1/3 as a decimal rounded to 2 decimal places?
I ideally believe in significant figures, so I would rather write it down as 0.3 . Most people will write it down as 0.33,0.333,0.3333 , etc. In practice use 13 as 0.333 or 0.33 , depending on the level of accuracy required.
## What does 3 decimal places mean?
Rounding numbers up or down is a way of approximating them to make them more manageable. … When you round to the third decimal place, you’re rounding to the nearest thousandth.
## How do you round to 2 decimal places in C++?
Rounding Floating Point Number To two Decimal Places in C and C++
1. First Method:- Using Float precision.
2. Second Method : Using integer typecast If we are in Function then how return two decimal point value.
3. Third Method : using sprintf() and sscanf()
## What is a double vs float?
A double is 64 and single precision (float) is 32 bits. The double has a bigger mantissa (the integer bits of the real number). Any inaccuracies will be smaller in the double.
## How do I fix decimal places in C++?
To set fixed 2 digits after the decimal point use these first: cout. setf(ios::fixed); cout. setf(ios::showpoint); cout.
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# Don’t understand the general form of the Sinusoidal Wave Equation
• I
• jjson775
jjson775
TL;DR Summary
Don’t understand the general form of the sinusoidal wave equation.
I am a retired engineer, 81 years old, self studying modern physics using Young and Freedman University Physics.
I am familiar with the wave equation y(x,t) = A cos (kx - wt) where A = amplitude, k = wave number and w (omega) = angular frequency.
in the chapter introducing quantum mechanics, this equation is shown as:
y(x,t) = A cos (kx - wt) + B sin (kx - wt). What is the “B” part? Is it another amplitude? The equation is not shown in this form at all in the chapter on mechanical waves.
vanhees71
Are you familiar with the trigonometric identity ##C\sin(\theta+\phi)=C\sin(\theta)\cos(\phi)+C\cos(\theta)\sin(\phi)##? If you replace ##\theta## with ##kx-\omega t## can you see what ##A## and ##B## are?
By the way, if you're going to post maths here it's worth having a read of the LaTeX Guide, linked below the reply box. It makes maths much easier to read. There's a known bug that the maths doesn't render in preview, so you may possibly need to refresh the page to see the maths - if you see # marks in my previous paragraph, you need to refresh.
vanhees71
I still don’t see where this is headed.
#### Attachments
• 1694312540063.png
5 KB · Views: 58
That's all there is to it.
The point is that with ##C\sin(kx+\omega t)##, at ##t=0## the amplitude at ##x=0## is always zero. But that isn't always what you want. Sometimes the amplitude is non-zero at ##x=0## and zero at some other point, which is to say that there is an offset in your wave, usually denoted ##\phi##. The equation of that wave is ##C\sin(kx+\omega t+\phi)##. You can get to the expression you asked about with the trigonometric identity I quoted.
What you have here is an example of decomposing a wave into two waves added together. The trigonometric identity shows you that a wave with some arbitrary phase offset can be written as a sine wave with a certain amplitude and no phase offset plus a cosine wave with a certain amplitude and no phase offset. It's purely a mathematical trick, but sometimes makes subsequent maths easier.
So the answer to your question is that ##A## and ##B## are the amplitudes of the sine and cosine waves with zero phase offset that you need to add together to get a single sine wave of amplitude ##C## and phase offset ##\phi##. You've worked out the mathematical relationships between ##A##, ##B## and ##C##.
scottdave, Juanda and vanhees71
Got it. Thanks! I have seen this equation pop up in lectures on YouTube and now in my textbook without any derivation or explanation. A useful mathematical trick, as you say.
berkeman and Ibix
Your comment that it can make subsequent math easier is spot on. It showed up in my textbook to show how it satisfies the general wave equation using partial differentiation leading up to introduction of the Schrödinger equation.
Ibix
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# 4.3.6. pde.pdes.kuramoto_sivashinsky module¶
The Kardar–Parisi–Zhang (KPZ) equation describing the evolution of an interface
class KuramotoSivashinskyPDE(nu: float = 1, *, noise: float = 0, bc: BoundaryConditionData = 'auto_periodic_neumann', bc_lap: BoundaryConditionData = None)[source]
The Kuramoto-Sivashinsky equation
The mathematical definition is
$\partial_t u = -\nu \nabla^4 u - \nabla^2 u - \frac{1}{2} \left(\nabla h\right)^2 + \eta(\boldsymbol r, t)$
where $$u$$ is the height of the interface in Monge parameterization. The dynamics are governed by the parameters $$\nu$$ , while $$\eta$$ is Gaussian white noise, whose strength is controlled by the noise argument.
Parameters
• nu (float) – Parameter $$\nu$$ for the strength of the fourth-order term
• noise (float) – Strength of the (additive) noise term
• bc – The boundary conditions applied to the field. Boundary conditions are generally given as a list with one condition for each axis. For periodic axis, only periodic boundary conditions are allowed (indicated by ‘periodic’ and ‘anti-periodic’). For non-periodic axes, different boundary conditions can be specified for the lower and upper end (using a tuple of two conditions). For instance, Dirichlet conditions enforcing a value NUM (specified by {‘value’: NUM}) and Neumann conditions enforcing the value DERIV for the derivative in the normal direction (specified by {‘derivative’: DERIV}) are supported. Note that the special value ‘natural’ imposes periodic boundary conditions for periodic axis and a vanishing derivative otherwise. More information can be found in the boundaries documentation.
• bc_lap – The boundary conditions applied to the second derivative of the scalar field $$c$$. If None, the same boundary condition as bc is chosen. Otherwise, this supports the same options as bc.
evolution_rate(state: ScalarField, t: float = 0) [source]
evaluate the right hand side of the PDE
Parameters
• state (ScalarField) – The scalar field describing the concentration distribution
• t (float) – The current time point
Returns
Scalar field describing the evolution rate of the PDE
Return type
ScalarField
explicit_time_dependence: Optional[bool] = False
Flag indicating whether the right hand side of the PDE has an explicit time dependence.
Type
bool
property expression: str
the expression of the right hand side of this PDE
Type
str
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• student
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# Math Limerick
Question: Why is this a mathematical limerick?
( (12 + 144 + 20 + 3 Sqrt[4]) / 7 ) + 5*11 = 92 + 0 .
A dozen, a gross, and a score,
plus three times the square root of four, divided by seven, plus five times eleven, is nine squared and not a bit more.
---Jon Saxton (math textbook author)
Presentation Suggestions:
Challenge students to invent their own math limerick!
The Math Behind the Fact:
It is fun to mix mathematics with poetry.
Resources:
Su, Francis E., et al. "Math Limerick." Math Fun Facts.
funfacts
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# LCM (55; 105) = ? Calculate the least common multiple, LCM, by two methods: 1) The prime factorization of the numbers and 2) The Euclidean algorithm
## The least common multiple, LCM: the latest calculated
The LCM of 55 and 105 = ? May 27 05:31 UTC (GMT) The LCM of 24,101 and 192,872 = ? May 27 05:31 UTC (GMT) The LCM of 5,000 and 4,788 = ? May 27 05:31 UTC (GMT) The LCM of 144 and 407 = ? May 27 05:31 UTC (GMT) The LCM of 246 and 125 = ? May 27 05:31 UTC (GMT) The LCM of 32 and 26 = ? May 27 05:31 UTC (GMT) The LCM of 15 and 18 = ? May 27 05:31 UTC (GMT) The LCM of 21 and 20 = ? May 27 05:31 UTC (GMT) The LCM of 10,035 and 60,210 = ? May 27 05:31 UTC (GMT) The LCM of 37 and 7,901 = ? May 27 05:31 UTC (GMT) The LCM of 315 and 273 = ? May 27 05:31 UTC (GMT) The LCM of 7 and 328 = ? May 27 05:31 UTC (GMT) The LCM of 330 and 3,448 = ? May 27 05:31 UTC (GMT) The least common multiple, LCM: the list of all the operations
## The least common multiple (lcm). What it is and how to calculate it.
• The number 60 is a common multiple of the numbers 6 and 15 because 60 is a multiple of 6 (60 = 6 × 10) and also a multiple of 15 (60 = 15 × 4).
• There are infinitely many common multiples of 6 and 15.
• If the number "v" is a multiple of the numbers "a" and "b", then all the multiples of "v" are also multiples of "a" and "b".
• The common multiples of 6 and 15 are the numbers 30, 60, 90, 120, and so on.
• Out of these, 30 is the smallest, 30 is the least common multiple (lcm) of 6 and 15.
• Note: The prime factorization of a number: finding the prime numbers that multiply together to give that number.
• If e = lcm (a, b), then the prime factorization of "e" must contain all the prime factors involved in the prime factorization of "a" and "b" taken by the highest power.
• Example:
• 40 = 23 × 5
• 36 = 22 × 32
• 126 = 2 × 32 × 7
• lcm (40, 36, 126) = 23 × 32 × 5 × 7 = 2,520
• Note: 23 = 2 × 2 × 2 = 8. We are saying that 2 was raised to the power of 3. Or, shorter, 2 to the power of 3. In this example 3 is the exponent and 2 is the base. The exponent indicates how many times the base is multiplied by itself. 23 is the power and 8 is the value of the power.
• Another example of calculating the least common multiple, lcm:
• 938 = 2 × 7 × 67
• 982 = 2 × 491
• 743 = is a prime number and cannot be broken down into other prime factors
• lcm (938, 982, 743) = 2 × 7 × 67 × 491 × 743 = 342,194,594
• If two or more numbers have no common factors (they are coprime), then their least common multiple is calculated by simply multiplying the numbers.
• Example:
• 6 = 2 × 3
• 35 = 5 × 7
• lcm (6, 35) = 2 × 3 × 5 × 7 = 6 × 35 = 210
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Non constant sheaf on a contractible space
while trying to calculate the coohomology (with complex coefficients) of a bundle over a simply connected base, I met the following
PROBLEM
Suppose you have a space $X$ which contracts on $x \in X$, and a sheaf $F$ on $X$ such that $F_p \simeq V$ for all $p \in X$, where $V$ is a $\mathbb{C}$ vector space of finite dimension. Is it true that $F$ is the constant sheaf $V$?
Remarks. Hope the following is not dumb: I am pretty new to sheaf cohomology.
(a) $H^0( \cdot, \mathbb{C})$ is invariant for homotopy equivalence, so $F(X) = H^0(X, F) \simeq H^0( x, F_x) = F_x \simeq V$ via the restriction map $F(X) \to F_x$.
(b) Suppose $X$ contracts on every point $p \in X$ and let $f:X \to q$ be the map to a point. Then the canonical map $f^{-1}f_* F \to F$ can be verified to be an isomorphism on stalks. Infact stalk maps are the restriction maps $F(X) \to F_p$, which are isomorphisms by point (a). The lefthand sheaf is constant, so we are done.
It seems to me that the problem could arise in case $X$ is not contractible on every point, which is a very counter intuitive situation.. Am I completely wrong?
• I'm not sure I understand the question, but your arguments certainly do not hold for simply connected spaces that are not contractible. Indeed, there are simply connected spaces with non-constant sheaves on them. Commented May 3, 2017 at 13:47
There is a flaw in your argument :
In (a) you claim that $H^0(\cdot,\mathbb{C})$ is homotopy invariant. This is true, and this holds more generally with $\mathbb{C}$ replaced by any constant sheaf $F$, and for all degree ($H^i(\cdot,F)$ is also homotopy invariant).
In (b), you want to use (a) to prove that the sheaf $F$ is constant. But you can't : because you need $F$ to be constant in order to use (a).
The thing is (a) is false for non constant sheaves, and as a matter of fact there are non constant sheaves on every spaces (except $\emptyset$ and the one point space).
By the way, your problem does not hold as stated. You need more that $F_p\simeq V$ for all $p\in X$. For example, on $X=\mathbb{R}$ (which is contractible), $j_!V\oplus i_*V$ is a counter example (here $j:U\rightarrow X$ is the inclusion of an open subset, and $i:X\setminus U\rightarrow X$ the incluson of the closed complement). Another one : $\bigoplus_{x\in X}x_*V$. Even the sheaf of continuous function on $\mathbb{R}$ is not constant (though stalks are not finite dimensional in this case).
The correct statement is the following : on a simply-connected space (not necessarily contractible), every locally constant sheaf is constant. (Stalks are not assumed to be $\mathbb{C}$-vector spaces, nor finite dimensional)
• Roland could you please give an hint on how to prove your last statement in the answer? Namely that any locally constant sheaf on a simply-connected (not necessarily contractible) is constant. Thank you! Commented May 28, 2022 at 8:17
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## Is live-tweeting meetings losing steam? #scicomm
As I write this in early 2016, sitting in the armpit of Silicon Valley (San Jose is, undeniably based on geography, the armpit of the south San Francisco Bay), we are beginning to witness the first signs of a contraction of the exuberant venture capital markets that have fueled utterly silly tech startup company valuations for the past few years. Twitter is one of the earlier startup darlings that has managed to decline in terms of share price as user base growth slows.
Now I’m beginning to wonder if we’re seeing a similar stagnation in adoption of Twitter as a … Continue Reading
## Adventures in course management software: Canvas
Course management software is universally garbage, but Canvas has managed to be better than most. Which is a lot like saying “This is the best tasting pile of dog poop I’ve found today.”
The ability to create online quizzes that have the answers entered for easy grading should make for a useful system, but today I discovered that the precision of the system tops out at the 4th decimal place, which tends to be problematic if I want students to calculate fairly small probabilities (what is the probability of flipping a coin 10 times and getting 10 heads in a row?). … Continue Reading
## iButton internals
I’ve written in the past about iButtons and my attempts to waterproof them. Although iButton temperature dataloggers are fairly well sealed, they are not waterproof. But if you know an old person that used iButtons in the late 90s or early 2000s, they might claim that iButtons are absolutely waterproof.
It turns out that iButtons are one of those rare things in life that really were better when you were a kid. In the old days they could be put out in the ocean for weeks or months, completely bare, and most of them would survive … Continue Reading
## Basic text string functions in R
To get the length of a text string (i.e. the number of characters in the string):
` nchar()`
Using length() would just give you the length of the vector containing the string, which will be 1 if the string is just a single string.
To get the position of a regular expression match(es) in a text string x:
```pos = regexpr('pattern', x) # Returns position of 1st match in a string
pos = gregexpr('pattern', x) # Returns positions of every match in a string
```
To get the position of a regular expression match in a vector x of text strings … Continue Reading
## Electronics parts list
Here’s the start of a list of common bits and doo-dads I use for building electronics projects.
https://github.com/millerlp/parts_guide/blob/master/parts_guide.md
That’s all there is to it.
## Preserving equations in Powerpoint when going cross-platform
This is another one of those tricks that I forget how to do unless I write it down. When I move Powerpoint (2010) presentations from my Windows machine to my Mac (Powerpoint 2011), the equations usually get destroyed along the way. The workaround is to turn the equations into images that can’t be altered by the Mac.
For example, here I’ve got an equation plopped into a blank presentation, with the equation editor toolbar.
The equation.
The first step is to copy the entire equation and the text box … Continue Reading
## A plot of co-authorships in my little corner of science
Here’s a mostly useless visualization of the collection of journal articles that sits in my reference database in Endnote. I deal mostly in marine biology, physiology, biomechanics, and climate change papers, with a few molecular/genetics papers thrown in here and there. The database has 3325 entries, 2 of which have ambiguous publication years and aren’t represented above. This is by no means an exhaustive survey of the literature in my field, it’s just an exhaustive survey of the literature on my computer.
To make this figure, I first had Endnote export the database to … Continue Reading
## Disassembling an ancient Si-Tech dry suit exhaust valve
Experienced cold water scuba divers will tell you that a dry suit is a vital piece of safety equipment, especially in challenging conditions (seriously, they’ll tell you without prompting, and then babble on about their gear until you walk away). Serious divers will also tell you that maintaining your gear in tip-top shape is an important safety issue, and all service should be done by trained technicians (again, it’s like that old joke: How do you know someone is vegan/went to Princeton/is from California? Answer: They’ll tell you. Divers, particularly overweight men with walrus mustaches, are the same way when … Continue Reading
## Open Wave Height Logger prototype in the water
I finally got around to deploying a prototype OWHL unit in the real live ocean to log some waves.
The low-tech housing is made of 1.5″ schedule 40 pvc pipe. The pipe snugly fits the D-cell battery holder, while the electronics fit inside the modified 1.5″ end cap.
Components of the housing. Assembled prototype housing
I try not to rely solely on the tapered threads … Continue Reading
## OWHL micro SD card current draw tests
As outlined in an earlier post, I found that certain old micro SD cards were performing spectacularly poorly when it came to power consumption because they failed to go into a low-power sleep state immediately after writing data to the card. I recently purchased a few new SanDisk micro SD cards in various capacities to see how they behaved. I purchased 4GB, 8GB, 16GB, and 32GB SanDisk cards from Amazon in November 2014. These were all tagged as “Ships from and sold by Amazon.com” and ranged from \$5.99 to \$12.99.
The good news is that all 4 cards behaved properly … Continue Reading
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# Constructing non-zero obstruction by cutting along a non-separating torus and regluing
The orientable Seifert fibered manifold $M=(1|(1,b))$ that fibers over the torus with no exceptional fibers but non-zero obstruction term $b$ has an embedded non-separating fibered torus (lift a nontrivial loop from the torus base space of the fibration). Cutting along this torus would leave a trivially fibered annulus cross $S^{1}$: $A\times S^{1}$. Therefore, $M=(1|(1,b))$ can be reobtained by gluing together the two torus boundary components of $A\times S^{1}$ in a particular fiber preserving way. What is this gluing map?
My guess was that the map would look like $d(u,v)=(u^{-1},u^{b}v)$ (assuming that the tori are positively oriented), but I can't compute the fundamental group.
• I think Mayer-Vietoris would give the first homology group as $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}_{b}$ which would be correct, but I'm unsure of this. Jul 11, 2017 at 22:08
You guess at the gluing map seems close to correct, although perhaps it might be $(u,v) \mapsto (u,u^bv)$ (I'm not quite sure what your coordinates $u,v$ represent, they might be $T^2$ coordinates in my displayed equation below).
To compute the fundamental group, use the fact that gluing the two boundary components of the space $$A \times S^1 \approx (S^1 \times [0,1]) \approx (S^1 \times S^1) \times [0,1] \approx T^2 \times [0,1]$$ gives an example of a mapping torus. By doing exercise 11 in Section 1.2 of Hatcher's book "Algebraic Topology", you will learn by example a general technique for using Van Kampen's theorem to compute the fundamental group of a mapping torus.
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# A synchronous motor operates at 0.8 pf lagging. If the field current of the motor is continuously
A synchronous motor operates at 0.8 pf lagging. If the field current of the motor is continuously increased
1. The power factor decreases upto a certain value of field current and thereafter it increases.
2. The armature current increases upto a certain value of field current and thereafter it decreases.
3. The power factor increases upto a certain value of field current and thereafter it decrease
4. The armature current decreases upto a certain value of field current and thereafter it, increases.
From these, the correct answer is (a) 1, 2 (b) 3, 4 (c) 1, 3 (d) 2, 4
← Prev Question Next Question →
A synchronous motor operates at 0.8 pf lagging. If the field current of the motor is continuously increased
1. The power factor decreases upto a certain value of field current and thereafter it increases.
2. The armature current increases upto a certain value of field current and thereafter it decreases.
3. The power factor increases upto a certain value of field current and thereafter it decrease
4. The armature current decreases upto a certain value of field current and thereafter it, increases.
From these, the correct answer is (a) 1, 2 (b) 3, 4 (c) 1, 3 (d) 2, 4
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# Linear Discriminant AnalysisΒΆ
This example demonstrates how the pipeline can be used to perform transformation of time series data, such as linear discriminant analysis for visualization purposes
Out:
```/home/circleci/miniconda/envs/testenv/lib/python3.8/site-packages/sklearn/utils/validation.py:67: FutureWarning: Pass memory=None as keyword args. From version 0.25 passing these as positional arguments will result in an error
warnings.warn("Pass {} as keyword args. From version 0.25 "
/home/circleci/miniconda/envs/testenv/lib/python3.8/site-packages/seglearn/transform.py:237: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray
Xt = np.array([sliding_tensor(Xt[i], self.width, self._step, self.order)
```
```# Author: David Burns
import matplotlib.pyplot as plt
import numpy as np
from sklearn.discriminant_analysis import LinearDiscriminantAnalysis
import seglearn as sgl
def plot_embedding(emb, y, y_labels):
# plot a 2D feature map embedding
x_min, x_max = np.min(emb, 0), np.max(emb, 0)
emb = (emb - x_min) / (x_max - x_min)
NC = len(y_labels)
markers = ['.', '+', 'x', '|', '_', '*', 'o']
fig = plt.figure()
fig.set_size_inches(6, 6)
for c in range(NC):
i = y == c
plt.scatter(emb[i, 0], emb[i, 1], marker=markers[c], label=y_labels[c])
plt.xticks([]), plt.yticks([])
plt.legend()
plt.tight_layout()
X = data['X']
y = data['y']
# create a pipeline for LDA transformation of the feature representation
clf = sgl.Pype([('segment', sgl.Segment()),
('ftr', sgl.FeatureRep()),
('lda', LinearDiscriminantAnalysis(n_components=2))])
X2, y2 = clf.fit_transform(X, y)
plot_embedding(X2, y2.astype(int), data['y_labels'])
plt.show()
```
Total running time of the script: ( 0 minutes 0.756 seconds)
Gallery generated by Sphinx-Gallery
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# How to convert watts to kilojoules
## How to convert watts to kilojoules
How to convert electric power in watts (W) to energy in kilojoules (kJ).
You can calculate kilojoules from watts and seconds, but you can't convert watts to kilojoules since watt and kilojoule units represent different quantities.
### Watts to kJ calculation formula
The energy E in kilojoules (kJ) is equal to the power P in watts (W), times the time period t in seconds (s):
E(kJ) = P(W) × t(s) / 1000
So
kilojoules = watts × seconds / 1000
or
kJ = W × s / 1000
#### Example
What is the energy consumption of an electrical circuit that has power consumption of 300 watts for time duration of 3 seconds?
E(kJ) = 300W × 3s / 1000 = 0.9kJ
How to convert kJ to watts ►
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# \$100 in 1933 is worth \$1,515.49 in 2005
\$
## Value of \$100 from 1933 to 2005
According to Statistics Canada consumer price index, prices in 2005 are 1,415.49% higher than average prices since 1933. The Canadian dollar experienced an average inflation rate of 3.85% per year during this period, causing the real value of a dollar to decrease.
In other words, \$100 in 1933 is equivalent in purchasing power to about \$1,515.49 in 2005, a difference of \$1,415.49 over 72 years.
The 1933 inflation rate was -2.74%. The inflation rate in 2005 was 2.09%. The 2005 inflation rate is higher compared to the average inflation rate of 1.59% per year between 2005 and 2020.
Cumulative price change 1,415.49% Average inflation rate 3.85% Converted amount (\$100 base) \$1,515.49 Price difference (\$100 base) \$1,415.49 CPI in 1933 7.100 CPI in 2005 107.600 Inflation in 1933 -2.74% Inflation in 2005 2.09%
## Buying power of \$100 in 1933
This chart shows a calculation of buying power equivalence for \$100 in 1933 (price index tracking began in 1914).
For example, if you started with \$100, you would need to end with \$1,515.49 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When \$100 is equivalent to \$1,515.49 over time, that means that the "real value" of a single Canadian dollar decreases over time. In other words, a dollar will pay for fewer items at the store.
This effect explains how inflation erodes the value of a dollar over time. By calculating the value in 1933 dollars, the chart below shows how \$100 is worth less over 72 years.
According to Statistics Canada, each of these CAD amounts below is equal in terms of what it could buy at the time:
Dollar inflation: 1933-2005
Year Dollar Value Inflation Rate
1933 \$100.00 -2.74%
1934 \$101.41 1.41%
1935 \$104.23 2.78%
1936 \$105.63 1.35%
1937 \$109.86 4.00%
1938 \$107.04 -2.56%
1939 \$109.86 2.63%
1940 \$115.49 5.13%
1941 \$122.54 6.10%
1942 \$126.76 3.45%
1943 \$129.58 2.22%
1944 \$126.76 -2.17%
1945 \$129.58 2.22%
1946 \$136.62 5.43%
1947 \$156.34 14.43%
1948 \$170.42 9.01%
1949 \$171.83 0.83%
1950 \$181.69 5.74%
1951 \$201.41 10.85%
1952 \$198.59 -1.40%
1953 \$198.59 0.00%
1954 \$198.59 0.00%
1955 \$200.00 0.71%
1956 \$205.63 2.82%
1957 \$209.86 2.05%
1958 \$215.49 2.68%
1959 \$218.31 1.31%
1960 \$221.13 1.29%
1961 \$221.13 0.00%
1962 \$225.35 1.91%
1963 \$229.58 1.88%
1964 \$233.80 1.84%
1965 \$240.85 3.01%
1966 \$249.30 3.51%
1967 \$259.15 3.95%
1968 \$270.42 4.35%
1969 \$283.10 4.69%
1970 \$285.92 1.00%
1971 \$300.00 4.93%
1972 \$315.49 5.16%
1973 \$345.07 9.38%
1974 \$388.73 12.65%
1975 \$425.35 9.42%
1976 \$449.30 5.63%
1977 \$491.55 9.40%
1978 \$533.80 8.60%
1979 \$585.92 9.76%
1980 \$650.70 11.06%
1981 \$729.58 12.12%
1982 \$797.18 9.27%
1983 \$833.80 4.59%
1984 \$864.79 3.72%
1985 \$902.82 4.40%
1986 \$940.85 4.21%
1987 \$980.28 4.19%
1988 \$1,018.31 3.88%
1989 \$1,071.83 5.26%
1990 \$1,125.35 4.99%
1991 \$1,167.61 3.75%
1992 \$1,192.96 2.17%
1993 \$1,212.68 1.65%
1994 \$1,215.49 0.23%
1995 \$1,236.62 1.74%
1996 \$1,263.38 2.16%
1997 \$1,273.24 0.78%
1998 \$1,285.92 1.00%
1999 \$1,319.72 2.63%
2000 \$1,361.97 3.20%
2001 \$1,371.83 0.72%
2002 \$1,423.94 3.80%
2003 \$1,453.52 2.08%
2004 \$1,484.51 2.13%
2005 \$1,515.49 2.09%
2006 \$1,540.85 1.67%
2007 \$1,577.46 2.38%
2008 \$1,595.77 1.16%
2009 \$1,616.90 1.32%
2010 \$1,654.93 2.35%
2011 \$1,692.96 2.30%
2012 \$1,707.04 0.83%
2013 \$1,728.17 1.24%
2014 \$1,753.52 1.47%
2015 \$1,781.69 1.61%
2016 \$1,808.45 1.50%
2017 \$1,836.62 1.56%
2018 \$1,842.25 0.31%
2019 \$1,878.87 1.99%
2020 \$1,921.13 2.25%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
Click to show 66 more rows
This conversion table shows various other 1933 amounts in 2005 dollars, based on the 1,415.49% change in prices:
Conversion Table: Value of a dollar in 2005
Initial value Equivalent value
\$1 dollar in 1933 \$15.15 dollars in 2005
\$5 dollars in 1933 \$75.77 dollars in 2005
\$10 dollars in 1933 \$151.55 dollars in 2005
\$50 dollars in 1933 \$757.75 dollars in 2005
\$100 dollars in 1933 \$1,515.49 dollars in 2005
\$500 dollars in 1933 \$7,577.46 dollars in 2005
\$1,000 dollars in 1933 \$15,154.93 dollars in 2005
\$5,000 dollars in 1933 \$75,774.65 dollars in 2005
\$10,000 dollars in 1933 \$151,549.30 dollars in 2005
\$50,000 dollars in 1933 \$757,746.48 dollars in 2005
\$100,000 dollars in 1933 \$1,515,492.96 dollars in 2005
\$500,000 dollars in 1933 \$7,577,464.79 dollars in 2005
\$1,000,000 dollars in 1933 \$15,154,929.58 dollars in 2005
## How to Calculate Inflation Rate for \$100, 1933 to 2005
Our calculations use the following inflation rate formula to calculate the change in value between 1933 and 2005:
CPI in 2005 CPI in 1933
×
=
Then plug in historical CPI values. The Canadian CPI was 7.1 in the year 1933 and 107.6 in 2005:
107.67.1
×
\$100
=
\$1,515.49
\$100 in 1933 has the same "purchasing power" or "buying power" as \$1,515.49 in 2005.
To get the total inflation rate for the 72 years between 1933 and 2005, we use the following formula:
CPI in 2005 - CPI in 1933CPI in 1933
×
100
=
Cumulative inflation rate (72 years)
Plugging in the values to this equation, we get:
107.6 - 7.17.1
×
100
=
1,415%
Politics and news often influence economic performance. Here's what was happening at the time:
• Coining of the name "Pakistan" by Choudhry Rahmat Ali, this is later accepted by Muslims in the Indian sub-continent who use it to push for a separate Muslim homeland in South Asia.
• Adolf Hitler is appointed Reich Chancellor of Germany by President Paul von Hindenburg.
• Franklin D. Roosevelt is inaugurated President of the United States.
• The first Nazi concentration camp, Dachau, is completed.
• The Nazi party is declared the only political party in Germany.
## Data Source & Citation
Raw data for these calculations comes from the government of Canada's annual Consumer Price Index (CPI), established in 1914 and computed by Statistics Canada (StatCan).
You may use the following MLA citation for this page: “1933 dollars in 2005 | Canada Inflation Calculator.” Official Inflation Data, Alioth Finance, 11 Aug. 2020, https://www.officialdata.org/1933-CAD-in-2005.
Special thanks to QuickChart for their chart image API, which is used for chart downloads.
in2013dollars.com is a reference website maintained by the Official Data Foundation.
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# The Naked Scientists Forum
### Author Topic: Re: Why are the Pioneer 10 and 11 probes moving faster than they should? (Read 2717 times)
#### ZHUYH
• Jr. Member
• Posts: 34
##### Re: Why are the Pioneer 10 and 11 probes moving faster than they should?
« on: 18/11/2008 03:36:28 »
"Pioneers 10 space probe the mystery of slowdown "Pioneers on the 10th" space probe deceleration problem is that the 21st century physics major world problems. Many scientists put forward many theories to explain this problem, but no one recognized the world theory. However, a physics teacher Zhuyonghuan China has solved the problem, after more than 20 years of testing, Mr Zhu that "pioneers on the 10th" in flight delays and there has been a "vortex of" a direct relationship,Mr Zhu not only proved the existence of such a force, but also that a new formula gravitational
F = Fn + Ft To
the weight of Fn = GMm / r ^ 2 -
Tangential component [of the scroll] Ft = kGMm ω Cos α / r ^ 2
K = 0.4 for one factor, ω is the angular velocity rotating ball, the unit: Band / sec; α to the orbital inclination. With the gravity of this new formula can be explained and calculated "pioneers on the 10th" slowdown issue 1. Qualitative analysis: "Pioneers 10" "pioneers 11" and "Galileo," are in the Earth launched from Earth orbit (GEO scroll) to detect Jupiter, although the scroll from the Earth, but still in the sun Scroll; Near the Earth from the Sun, the Earth's orbit around the sun of the scroll, far from the sun and Jupiter, the Scroll of small. To the Earth's orbit around the turn as a frame of reference to the observation orbit around Jupiter spacecraft felt that the slowdown. 2 quantitative analysis: Earth's rotation velocity ω e = 1.16 × 10 ^- 5 laps / sec. solar rotation period of 25 days Solar rotation velocity ω s = ω e/25 , quality of the mass of the sun and Earth: Ms = 3.3 × 10^5 Me Earth radius Re = 6.4 × 10 ^3km Earth to the sun distance: ds-e = 1.5 × 10^8 km Jupiter distance to the sun: ds-j = 7.78 × 10^8 km To the sun as the center of the earth to orbit near the vortex edge: Ft-e = GMsm ω skCos α / ds-e^2 (Cos α ≈ 1, m for the spacecraft weight, coefficients for k = 0.4) ① In orbit around Jupiter to the sun by the vortex of the Ft-j: (Ft-j / Ft-e) = (ds-e / ds-j)^2 ② To the Earth's orbit around the turn as a frame of reference to the observations of Jupiter, the spacecraft orbit around the deceleration of Ft: Ft = Ft-e－Ft-j =1.o8× 10^-10 Fne Fne surface of the Earth Gravity This conclusion by scientists with the United States is closer to the data. NASA launched the "pioneer of 10" 4 spacecraft slowdown issues since entering the new century the major celestial mechanics, It questioned Newton's Law of Universal Gravitation is accurate and powerful, It proved Kepler's idea of the gravity, the general theory of relativity proved that the "framework gravitational drag" forecast, It also scroll (the tangential component of gravity that is) the accuracy of the formula provide a more precise criteria. Reference: “Scroll of mysterious - gravity partner Profile ”Books Author: Shuyonghuan Publisher: 21st Century Publishing House - China 2005。3 "
#### lyner
• Guest
##### Re: Why are the Pioneer 10 and 11 probes moving faster than they should?
« Reply #1 on: 18/11/2008 09:26:11 »
Do you really not understand the difference between f and ω?
#### ZHUYH
• Jr. Member
• Posts: 34
##### Re: Why are the Pioneer 10 and 11 probes moving faster than they should?
« Reply #2 on: 19/11/2008 07:29:51 »
quote[Do you really not understand the difference between f and ω?]
Hi sophiecentaur :thank you!
#### ZHUYH
• Jr. Member
• Posts: 34
##### Re: Why are the Pioneer 10 and 11 probes moving faster than they should?
« Reply #3 on: 19/11/2008 07:52:12 »
tangential vortex force:
1.Lab vortex force :used Cavendish torsion balance measured vortex force of existence, and that formula New gravitational formula in the 2001 .
2.earth vortex force
2.1.the moon Away from the Earth (3 cm per year).
2.2. Cycle of the Earth's rotation slows down. (Per hundred years 0.00164 s] ，
2.4. Sun-synchronous orbit satellite orbital plane precession; altitude of 900 km polar orbit satellite orbital plane precession of the daily 1 degree. The orbital plane of the vertical scroll relatively static, the situation in order to prove Scroll of the accuracy of the vortex force and an excellent opportunity.
2.5. Oblate of the Earth , 5 × 10-5m/s^2. On the Global Positioning System GPS satellite orbit produces nearly 10,000 meters every day the impact of this phenomenon of vortex theory can be used to explain and calculate. Satellite orbit calculation of the largest perturbation of the - non-perturbation of the ball, not the flat rate from the Earth, but the role of the vortex.
3.solar vortex force
3.1. Mercury precession 43 "per century. (Newton's theory of deviation values.)
3.2.Venus precession 8.4 "± 4.8" per century. (Newton's theory of deviation values.)
3.3 .Earth's precession 5.0 "± 1.2" per century. (Newton's theory of deviation values.)
3.4 ."Pioneers 10" four spacecraft speed reduction.
3.5 .Solar wind acceleration: time to reach the Earth, usually in the jet speed of 450 kilometers per second around.
4. Galaxy rotation curve flatness (dark matter, MOND theory).
5.Mars vortex force : Around Mars Phobos campaign cycle to slow down,but Deimos mooving away from Mars.
#### ZHUYH
• Jr. Member
• Posts: 34
##### Re: Why are the Pioneer 10 and 11 probes moving faster than they should?
« Reply #4 on: 12/02/2010 03:19:43 »
Do you really not understand the difference between f and ω?
W=relative angular velocity=Wc(central mass spinning angular velocity)-Ws(peripheral mass orbiting angular velocity))
#### The Naked Scientists Forum
##### Re: Why are the Pioneer 10 and 11 probes moving faster than they should?
« Reply #4 on: 12/02/2010 03:19:43 »
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# CBSE Grade 8 Mathematics: Important topics from exam perspective
Grade 8 is very important for CBSE students and having a strong foundation helps in paving a path for successful future. The syllabus is vast and kids need to have good grasp of their fundamentals in order to achieve marks in CBSE exams. Students should go through the recommended books which explain concepts clearly.
While SA1 is still far away, it’s never too early to work on a strategy to strengthen your fundamentals. We will list out the important topics for SA1 so students can study smart instead of cramming at the last moment.
## Chapters covered in SA 1 are:
1. Rational Numbers
2. Linear Equations in One Variable
4. Practical Geometry
5. Data Handling
6. Squares and Square Roots
7. Cubes and Cube Roots
8. Comparing Quantities
9. Algebraic Expressions and Identities
10. Visualizing Solid Shapes
11. Mensuration
12. Exponents and Powers
While all of these need to be practiced, some topics are more important than others and having good grasp of them would help you score better marks. Based on our experience of teaching for years and our observation of question papers in the last few years, our teachers have listed down the chapters that are most important from examination point of view.
### Chapter 1: Rational Numbers.
This is one of the basic chapters covering fundamentals already covered in Grade 6 and 7. From our observation of the past years question papers, we suggest you brush up the following topics:
Rational Numbers between Given Rational Numbers: 1 question is assured from this topic
Distributive Property of Multiplication for Rational Numbers: 1 question is expected from this topic
The questions from “Rational Numbers” are easy to score and can help you score 12 – 15 marks.
### Chapter 2: Linear Equations in one variable
This topic includes “Linear equations with linear expressions on both sides”. Looking at the previous year’s question papers, we can confidently say there is a 80-85% chance of getting a question on this topic.
Equations that are reducible to linear form are also important. We provide our students variety of questions from these topics for practice and guide them on how to avoid common mistakes while attempting questions from Linear equations.
This is an important topic because it carries a weightage of almost 20%. Properties play a very important role in this topic and remembering them is crucial in solving the questions. Some of the properties to practice are:
• Angle Sum Property of Polygons
• Properties of the Sides of a Parallelogram
• Properties of the Diagonals of a Parallelogram
• Properties of the Angles of a Parallelogram
We encourage students to always draw diagram when solving questions from this topic. Our teachers at Eduinfinite spend extra time on this topic because this is one of the areas where many students find it difficult to visualize. We provide them with sample papers, solved examples and help them solve previous year’s question paper.
### Chapter 4: Practical Geometry
This chapter requires a lot of attention and practice. Diagrams should be drawn accurately using compass and ruler. Construction of special quadrilateral is an important area in this topic and should not be ignored. Take our teacher’s help to understand this chapter thoroughly and ensure you practice all questions from the books recommended by our teachers.
### Chapter 6: Squares and square roots
There are definitely few questions from this topic every year and students are advised to spend enough time practicing and revising this topic. Questions are mostly of 1 or 2 marks but they are easy scorers if practiced well. Write down the formulae and revise them till you know them like the back of your hand.
Important topics from this chapter are:
• Prime factorization
• Pythagorean Triplets and
• Square roots of perfect squares
### Chapter 7: Cubes and cube roots
Like previous chapter on squares and square roots, this chapter also requires mastering prime factorization method of finding cube roots. Our teachers at Eduinfinite spend a lot of time making students practice questions on converting a non-perfect cube into a perfect cube. Get your sample papers and revision notes from our teacher and ensure you practice as many questions as you can.
### Chapter 11: Mensuration
While Mensuration problems are easy to score, they can be confusing if you do not have strong basics. Students can confuse between area, volume and surface area. Our teachers explain these concepts with practical examples and tests the student’s knowledge before asking them to memorize the formulae. Mensuration problems are usually text-based and students should practice a lot of problems to make sure they know the difference between surface area and volumes of 3-D figures like cuboids, cubes and right circular cylinders.
### Chapter 12: Powers and exponents
Law of exponents are very important for scoring in this chapter and they should be on your fingertips. Practice a lot of problems which involve converting small and large numbers in standard form. Collect your study material and revision papers from our teachers, solve the problems from the books we recommend and ensure you do the problems carefully. Small mistakes are common in this chapter and you need to concentrate when attempting questions involving powers and exponents.
Download our sample chapter on indices to get an idea of the quality of our material.
Call our center on (+65) 8498 2405 and arrange for a counselling session with our founder Shrabanti to understand how we can help you gain an edge in Mathematics.
### 1 Comment
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CBSE Grade 8 Mathematics: Important topics from exam perspective | EduInfinite
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# Optimization calculus help
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When it comes to math apps, there is no shortage of options to choose from. However, not all math apps are created equal. Some are more comprehensive than others, some are more user-friendly, and some are just plain more fun to use. So, which is the best app to solve math problems? It really depends on your individual needs and preferences. However, here are three apps that are definitely worth checking out: 1. Photomath is a great option for those who want a comprehensive app that can provide step-by-step solutions to even the most complex math problems. 2. If you're looking for an app that's easy to use and navigate, then Mathway is definitely worth considering. 3. Finally, if you want an app that's both educational and entertaining, then we suggest giving Socratic a try.
Logarithmic equations are equations that can be written in the form of a logarithm. For example, if x is the variable and y = log(x), then log(x) = y. This means that the function y = log(x) is a logarithm of the variable x. A logarithm of a variable is a transformation of the variable such that the original value becomes 1, the base 10 value, after being divided by the log base 10 value (base e). Therefore, if x is the variable and y = log(x), then log(x) = y. This means that the function y = log(x) is a logarithm of x. As an example, let's say you're trying to solve an equation like: y = 1000 + 1 + 0.25x You can use a graphing calculator to graph this equation and determine a possible solution is 0.0625 x 0.072125 which means y 0.0625 1000 - 1 + 0.25 1000 - 5 + 0.3125 1000 - 8 + 0.4125 1000 - 975 + 1 and so on... However, using traditional math methods you may get stuck on this problem because you will have to solve for several different values of y, which could
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Yaeko King
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# Mitsubishi Electric FR-A820-00046 Instruction Manual Page 538
(M) Monitor display and monitor output signal
Average power saving monitor
([
Average power saving], [
● The average power saving monitors are displayed by setting a value other than 9999 in Pr. 897
"Power saving monitor average time".
● On the [
period s displayed.
● When Pr. 897 is set, the average value is updated each time the average time period elapses, with
the power-ON or inverter reset as the starting point.
The power savings average value update timing signal (Y92) is inverted every time the average
value is updated.
When Pr. 897 = 4 [Hr]
Power saving
instantaneous
value [kW]
Average
power
saving [kW]
Fig. 5-161:
● When Pr. 895 "Power saving rate reference value" the [
averaging time period is displayed on the [4 Average power saving rate] monitor.
● When the power cost per 1 kWh power amount is set in Pr. 896 "Power unit cost", the cost of the
saved power ([
savings].
Cumulative energy saving monitors ([
[
Annual power saving amount], [
● On the cumulative energy saving cumulative monitors, the monitor data digit can be shifted to
the right by the number of Pr. 891 "Cumulative power monitor digit shifted times". setting. For
example, if the cumulative power value is 1278.56 kWh when Pr. 891 = "2", the PU/DU display is
12.78 (display in 100 kWh increments) and the communication data is 12. If the maximum value
is exceeded when Pr. 891 = "0 to 4", the value is clamped at the maximum value, indicating that a
digit shift is necessary. If the maximum value is exceeded when Pr. 891 = "9999", the value returns
to 0, and the counting starts again. In other monitors, the value is clamped at the displayed
maximum value.
● The [
termined period. Measure with the following procedure.
Write "9999" or "10" in Pr. 898 "Power saving cumulative monitor clear".
Write "0" in Pr. 898 at the measurement start time to clear the power saving cumulative monitor
value and start power saving accumulation.
Write "1" in Pr. 898 at the measurement end time to hold the power saving cumulative monitor
value.
5 - 358
Average power saving monitor], average power saving amount for each average time
Pr. 897
0
4
Average
Operation start
0 in the first
measurement
Y92
0
4
Update of the average value
Average power saving] × Pr. 896) is displayed on the [
Cumulative power saving amount] monitor (6)] can measure the power during a prede-
Average power saving rate], [
During stop
8
12
Average
8
12
Power saving amount], [
Annual power saving savings]).
Average power cost savings])
Power is off
20
16
Last value
Average
Average
Stores Hi/Low when the
power is off and starts.
16
0
Average power saving rate] for the
Average power cost
Power cost saving],
Parameters
Time
Time
4
I002618E
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# Question:Approximating Riemann Sums Numerically
## Question:Approximating Riemann Sums Numerically
Maple
Hello, we need to do an integral based on two numerically approximated functions given by a second degree coupled differential equation.
The integral:
Where xs and xu are functions of the variable t.
We wish to do a half-sums to approximate the integral, that is:
Define an N = number of points, such that we get a delta_x = L/N.
We then wish to get two vectors x_s = [x_s(0),x_s(1),...,x_s(N)] and x_u = [x_u(0),x_u(1),...,x_u(N)] by running some kind of for loop over our approximation procedure.
Is there any function for this already inside of Maple? We need to approximate the integral using Riemann sums and no other method.
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# Electromagnetic Induction & Alternating Currents (AIIMS (All India Institute of Medical Sciences) Physics): Questions 44 - 48 of 51
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 500 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Question number: 44
» Electromagnetic Induction & Alternating Currents » Electromagnetic Induction
MCQ▾
### Question
A rod of magnetic material having dimension is subjected in magnetizing field of then a magnetic moment of is induced in it. Therefore magnetic induction in rod is –
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 45
» Electromagnetic Induction & Alternating Currents » Induced EMF and Current
Appeared in Year: 1995
MCQ▾
### Question
Dimension of a rectangular coil is . Number of turns in coil is 60. It is rotating about its one of the diameter in an uniform magnetic field of , with the speed 1800 r. p. m. the maximum induced emf is – [RPMT]
### Choices
Choice (4) Response
a.
b.
c.
d.
113 V
## Question number: 46
» Electromagnetic Induction & Alternating Currents » Induced EMF and Current
Appeared in Year: 1998
MCQ▾
### Question
Two coil have a mutual inductance . The current changes in first coil according to equation , where and . The maximum value of induced emf in second coil is – (AIPMT)
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 47
» Electromagnetic Induction & Alternating Currents » AC Generator and Transformer
Appeared in Year: 2003
MCQ▾
### Question
A generator produces a voltage where V is in volts and in second. The frequency and r. m. s voltage are – [RPMT]
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 48
» Electromagnetic Induction & Alternating Currents » Alternating Currents
MCQ▾
### Question
Alternating voltage is applied across the resistance . The rms value of current is equal to –
### Choices
Choice (4) Response
a.
b.
c.
d.
f Page
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# Finding the angle for an electrostatic force
1. May 19, 2012
### Kaani
1. The problem statement, all variables and given/known data
What is the angle θ between the electrostatic force on the charge at the origin and the positive x-axis? Answer in degrees as an angle between -180 degrees and 180 degrees measured from the positive x-axis, with counterclockwise positive.
Coulomb constant is 8.98755e9
(Figure attached)
2. Relevant equations
F = kqQ/R2 (for answers found in previous question, which was marked correct)
3. The attempt at a solution
In the previous question I was asked to find the magnitiude of the electrostatic force on the charge at the origin (see figure). I found this to be 2.112678794e-9 by calculating the forces between the charges and the origin seperately, and then using a^2+b^2=c^2.
To find the angle between the force and the x-axis I did arctan = (-9.1709694e-10)/(1.90324588e-9) and got -25.7275298 degrees. I plugged it in as both positive and negative, and both answers were wrong. I'm not sure how the "-180 to 180" part of the question changes it, and help would be appreciated. The two values I used in the tangent equation I used to find the correct answer in the question before this one, so I don't see how the angle would be wrong.
Thanks!
#### Attached Files:
• ###### figure.png
File size:
3.9 KB
Views:
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2. May 19, 2012
### Staff: Mentor
The magnitude of the force that you calculated is fine, but I think your direction is off. Take a look at the signs of the charges in the figure and pencil in the directions of the resulting forces (as small vectors) on the diagram. The signs of the forces (their directions) are important when determining the direction (angle) of the resultant. Also, make sure that you use y-component/x-component in the arctan function, otherwise you're not computing the angle that you think you are
3. May 19, 2012
### Kaani
Oops... that was a stupid mistake, then. Switched the vectors around.
(picture I drew attached)
But, would the 1.903...e-9 be negitive or positive? In the equation it comes out negative, meaning it's a force of attraction. But since it's on the positive x-axis, it would have to be positive in the arctan equation, right?
So, arctan = (1.90324588e-9)/(-9.1709694e-10)
= -64.27247017
#### Attached Files:
• ###### figure.png
File size:
4.9 KB
Views:
283
4. May 19, 2012
### Staff: Mentor
Do charges of the same sign attract or repel? You should always look at the diagram and decide the direction of the resulting force. This will serve as a check on the sign of your calculated values (The signs of things in your calculation can get a bit complicated when you take the geometry of location into account. Looking at the diagram and thus knowing the required direction makes life simple ).
In this particular case you have a negative charge at the origin and a negative charge below it on the y-axis. What direction do you think the resulting force on the origin charge should be from the charge below it?
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×
# Hat Colors
Alice, Betty, and Charlie are all wearing a hat, and each hat is either red or blue. They also know that overall there are two blue hats and one red hat.
Alice cannot see her own hat, but sees that Betty is wearing a red hat and Charlie is wearing a blue hat. What color hat is Alice wearing?
## Hat Colors
### Multi-Level Thinking
# Hat Colors
Alice, Betty, and Charlie are all wearing a hat, and each hat is either red or blue. They also know that there were originally two red hats and two blue hats (so one hat is unused). Each person can see the hats everyone else is wearing, and not their own.
Assuming they don't speak, which of the following must be true?
## Hat Colors
### Multi-Level Thinking
# Hat Colors
In the previous problems, the people were able to infer information about their hats from just seeing the hats of others. Usually, this will not be the case, and the people will have to have a coordinated strategy before the hats are placed.
Hat puzzles are generally divided into two main categories: puzzles in which at least one person must guess their hat color, and puzzles in which everyone (or almost everyone) must guess their hat color. Usually, in the first category people will guess simultaneously (so that they cannot encode any information in their guesses), while in the second category, the people will guess in some order (so that they can get further information from others' guesses).
## Hat Colors
### Multi-Level Thinking
# Hat Colors
Alice and Bob are each wearing a hat, and each hat is either white or black. They can both see each other's hats, but cannot see their own. In order, Alice will guess her hat color, then Bob will guess his hat color. They win if either Alice or Bob correctly guesses their hat color. With a perfect strategy, what is the probability they win the game?
NOTE: Alice and Bob can plan their strategy before the game starts.
## Hat Colors
### Multi-Level Thinking
# Hat Colors
Alice and Bob are each wearing a hat, and each hat is either white or black. They can both see each other's hats, but cannot see their own. At the same time, Alice will guess her hat color, and Bob will guess his hat color. They win if either Alice or Bob correctly guesses their hat color. With a perfect strategy, what is the probability they win the game?
NOTE: Again, Alice and Bob can plan their strategy before the game starts.
## Hat Colors
### Multi-Level Thinking
# Hat Colors
This illustrates a general strategy for "at least one" problems (problems in which only one person needs to guess their hat color). The possible sets of hat colors are divided up into different categories, and each player guesses according to their category.
For instance, in the previous problem the two categories were "Alice and Bob have the same hat color" and "Alice and Bob have different hat colors." Since one of these must be the case, if Alice guesses accoring to the first category (meaning she guesses the same as Bob's hat color) and Bob guesses according to the second category (meaning he guesses the opposite of Alice's hat color), one of them must be right (and, of course, one of them must be wrong!).
## Hat Colors
### Multi-Level Thinking
# Hat Colors
Alice, Bob, and Charlie are each wearing a hat, and each hat is either red, green, or blue. They can all see each other's hats, but cannot see their own. At the same time, the three people guess their hat color. They win if any of them correctly guesses their hat color. With a perfect strategy, what is the probability they win the game?
(Note: There is not necessarily one of each color; red, green, and blue are just the possible color choices.)
## Hat Colors
### Multi-Level Thinking
# Hat Colors
The final class of hat puzzles are competitive versions, in which a person wins if they are the first person to correctly guess their hat color. This opens up a new element to the strategy, as now the absence of a guess (i.e. silence) can give information as well.
## Hat Colors
### Multi-Level Thinking
# Hat Colors
Alice, Betty, and Charlie are each wearing a hat, each hat is either black or white, and each speaks simultaneously: "yes" if they see a black hat, and "no" if they do not (as usual, each person can see all hats except their own and know what hats are possible).
All three people say "yes."
They are then asked if they know their hat color, and again speak simultaneously: "yes" if they know, "no" if they do not.
All three people say "no."
What is the distribution of hats?
## Hat Colors
### Multi-Level Thinking
# Hat Colors
Alice, Betty, and Charlie are each wearing a hat, and the hats are taken from a pool of 3 white and 2 black hats (so two hats are unused). The three people sit in a line, so that each person can see the hats of the people in front of them but not the hats of the people behind them (or their own). More specifically, Charlie can see Betty's and Alice's hat, Betty can see Alice's hat, and Alice cannot see any hats. The winner is the first person to correctly guess their hat color.
Alice declares she doesn't know her hat color, followed by Charlie and Betty (in that order).
Given that each of these 3 has made their declaration, one of them is able to conclude what color hat they are wearing, and win the game . Who is it? And what color is their hat?
## Hat Colors
### Multi-Level Thinking
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1.使用随机森林算法完成基本建模：包括数据预处理，特征展示，完成建模并进行可视化展示分析。
2.分析数据样本量与特征个数对结果的影响，在保证算法一致的前提下，增加样本个数，观察结果变化，重新进行特征工程，引入新的特征后，观察结果走势。
3.对随机森林算法进行调参，找到最合适的参数，掌握机器学习中两种调参方法，找到模型最优参数。
```import pandas as pd
import datetime
year = data['year']
month =data['month']
day =data['day']
dates = [(str(year)+'-'+str(month)+'-'+str(day)) for year,month,day in zip(year,month,day)]
dates=[datetime.datetime.strptime(date,'%Y-%m-%d') for date in dates]
dates[:5]```
```##进行绘图
import matplotlib.pyplot as plt
%matplotlib inline
plt.style.use('fivethirtyeight')##风格设置
# 设置布局
fig, ((ax1, ax2), (ax3, ax4)) = plt.subplots(nrows=2, ncols=2, figsize = (10,10))
fig.autofmt_xdate(rotation = 45)
# 标签值
ax1.plot(dates, data['actual'])
ax1.set_xlabel(''); ax1.set_ylabel('Temperature'); ax1.set_title('Max Temp')
# 昨天
ax2.plot(dates, data['temp_1'])
ax2.set_xlabel(''); ax2.set_ylabel('Temperature'); ax2.set_title('Previous Max Temp')
# 前天
ax3.plot(dates, data['temp_2'])
ax3.set_xlabel('Date'); ax3.set_ylabel('Temperature'); ax3.set_title('Two Days Prior Max Temp')
# 我的逗逼朋友
ax4.plot(dates, data['friend'])
ax4.set_xlabel('Date'); ax4.set_ylabel('Temperature'); ax4.set_title('Friend Estimate')
```import numpy as np
y = np.array(data['actual'])
x = data.drop(['actual'],axis=1)
x_list =list(x.columns)
x = np.array(x)
##数据分类
from sklearn.model_selection import train_test_split
x_train,x_test,y_train,y_test =train_test_split(x,y,test_size=0.25,random_state=42)
##建立随机森林模型
from sklearn.ensemble import RandomForestRegressor
rfr = RandomForestRegressor(n_estimators=1000,random_state=42)
rfr.fit(x_train,y_train)
y_pred = rfr.predict(x_test)
from sklearn.metrics import mean_squared_error
mse=mean_squared_error(y_test,y_pred)
print('mse',mse)```
```from sklearn.tree import export_graphviz
import pydot
tree = rfr.estimators_[5]
export_graphviz(tree,out_file='tree.dot',
feature_names=x_list,
rounded=True,precision=1)
(graph,) = pydot.graph_from_dot_file('tree.dot')
graph.write_png('tree.png')```
```##进行预剪枝
rfr_small = RandomForestRegressor(n_estimators=10,max_depth=3,random_state=42)
rfr_small.fit(x_train,y_train)
tree_small = rfr_small.estimators_[5]
export_graphviz(tree_small,out_file='small_tree.dot',
feature_names=x_list,
rounded=True,precision=1)
(graph,) = pydot.graph_from_dot_file('small_tree.dot')
graph.write_png('small_tree.png')```
2.选择出重点的特征，然后对全特征与重点特征的结果进行比较
```##通过randomforestregressor的feature_importance_显示特征重要性
importance = list(rfr.feature_importances_)
feature_importances =[(feature_name,importance) for feature_name,importance in zip(x_list,importance)]
feature_importances =sorted(feature_importances,key =lambda x:x[1],reverse =True)##key 为以那一列数据为排列对象
feature_importances
##以这两个特征为唯二的特征进行计算
rfr = RandomForestRegressor(n_estimators=100,random_state=42)
new_x = np.array(data.iloc[:,4:5])
new_x_train,new_x_test,new_y_train,new_y_test =train_test_split(new_x,y,test_size=.25,random_state=42)
rfr.fit(new_x_train,new_y_train)
y_pred = rfr.predict(new_x_test)
print('mse',mean_squared_error(new_y_test,y_pred))```
```import pandas as pd
##绘图观察数据
# 转换成标准格式
import datetime
# 得到各种日期数据
years = data['year']
months = data['month']
days = data['day']
# 格式转换
dates = [str(int(year)) + '-' + str(int(month)) + '-' + str(int(day)) for year, month, day in zip(years, months, days)]
dates = [datetime.datetime.strptime(date, '%Y-%m-%d') for date in dates]
# 绘图
import matplotlib.pyplot as plt
%matplotlib inline
# 风格设置
plt.style.use('fivethirtyeight')
# Set up the plotting layout
fig, ((ax1, ax2), (ax3, ax4)) = plt.subplots(nrows=2, ncols=2, figsize = (15,10))
fig.autofmt_xdate(rotation = 45)
# Actual max temperature measurement
ax1.plot(dates, data['actual'])
ax1.set_xlabel(''); ax1.set_ylabel('Temperature (F)'); ax1.set_title('Max Temp')
# Temperature from 1 day ago
ax2.plot(dates, data['temp_1'])
ax2.set_xlabel(''); ax2.set_ylabel('Temperature (F)'); ax2.set_title('Prior Max Temp')
# Temperature from 2 days ago
ax3.plot(dates, data['temp_2'])
ax3.set_xlabel('Date'); ax3.set_ylabel('Temperature (F)'); ax3.set_title('Two Days Prior Max Temp')
# Friend Estimate
ax4.plot(dates, data['friend'])
ax4.set_xlabel('Date'); ax4.set_ylabel('Temperature (F)'); ax4.set_title('Friend Estimate')
```seasons=[]
for month in data['month']:
if month in[1,2,12]:
seasons.append('winter')
elif month in [3,4,5]:
seasons.append('spring')
elif month in [6,7,8]:
seasons.append('summer')
else:
seasons.append('auntumn')
reduced_x = data[['temp_1','prcp_1','average','actual']]
reduced_x['seasons']=seasons
# 导入seaborn工具包
import seaborn as sns
sns.set(style="ticks", color_codes=True);
# 选择你喜欢的颜色模板
palette = sns.xkcd_palette(['dark blue', 'dark green', 'gold', 'orange'])
# 绘制pairplot
sns.pairplot(reduced_x, hue = 'seasons', diag_kind = 'kde', palette= palette, plot_kws=dict(alpha = 0.7),
```data = pd.get_dummies(data)
new_y = np.array(data['actual'])
new_x = data.drop(['actual'],axis=1)
new_x_list =list(new_x.columns)
new_x = np.array(new_x)
from sklearn.model_selection import train_test_split
new_x_train,new_x_test,new_y_train,new_y_test =train_test_split(new_x,new_y,test_size=0.25,random_state=42)
old_y = np.array(data['actual'])
old_x = data.drop(['actual'],axis=1)
old_x_list =list(old_x.columns)
old_x = np.array(old_x)
from sklearn.model_selection import train_test_split
old_x_train,old_x_test,old_y_train,old_y_test =train_test_split(x,y,test_size=0.25,random_state=42)
def model_train_predict(x_train,y_train,x_test,y_test):
rfr = RandomForestRegressor(n_estimators=100,random_state=42)
rfr.fit(x_train,y_train)
y_pred = rfr.predict(x_test)
errors= abs(y_pred-y_test)
print('平均误差',round(np.mean(errors),2))
accuracy = 100-np.mean(errors)
print('平均正确率',accuracy)
model_train_predict(old_x_train,old_y_train,old_x_test,old_y_test)
model_train_predict(ori_new_x_train,new_y_train,ori_new_x_test,new_y_test)```
```rfr = RandomForestRegressor(n_estimators=100,random_state=42)
rfr.fit(new_x_train,new_y_train)
y_pred = rfr.predict(new_x_test)
errors= abs(y_pred-new_y_test)
print('平均误差',round(np.mean(errors),2))
accuracy = 100-np.mean(errors)
print('平均正确率',accuracy)
importances = list(rfr.feature_importances_)
feature_importances =[(feature,importance) for feature,importance in zip(new_x_list,importances)]
feature_importances = sorted(feature_importances,key =lambda x:x[1],reverse =True)
# 对特征进行排序
x_values =list(range(len(importances)))
sorted_importances = [importance[1] for importance in feature_importances]
sorted_features = [importance[0] for importance in feature_importances]
# 累计重要性
cumulative_importances = np.cumsum(sorted_importances)
# 绘制折线图
plt.plot(x_values, cumulative_importances, 'g-')
# 画一条红色虚线，0.95那
plt.hlines(y = 0.95, xmin=0, xmax=len(sorted_importances), color = 'r', linestyles = 'dashed')
# X轴
plt.xticks(x_values, sorted_features, rotation = 'vertical')
# Y轴和名字
plt.xlabel('Variable'); plt.ylabel('Cumulative Importance'); plt.title('Cumulative Importances');```
```important_feature_names =[feature[0] for feature in feature_importances[0:5]]
important_feature_indices =[new_x_list.index(feature) for feature in important_feature_names]
important_x_train = new_x_train[:,important_feature_indices]
important_x_test = new_x_test[:,important_feature_indices]
model_train_predict(important_x_train,new_y_train,important_x_test,new_y_test)
##运行时间的提升
import time
all_features_time=[]
for _ in range(10):
start_time = time.time()
rfr.fit(new_x_train,new_y_train)
y_pred = rfr.predict(new_x_test)
end_time =time.time()
all_features_time.append((end_time-start_time))
all_features_times=np.mean(all_features_time)
all_features_time=[]
for _ in range(10):
start_time = time.time()
rfr.fit(important_x_train,new_y_train)
y_pred = rfr.predict(important_x_test)
end_time =time.time()
all_features_time.append((end_time-start_time))
reduced_features_times=np.mean(all_features_time)
all_accuracy =100*(1-np.mean(abs(all_y_pred-new_y_test)/new_y_test))
reduced_accuracy =100*(1-np.mean(abs(reduced_y_pred-new_y_test)/new_y_test))
comparison = pd.DataFrame({'features': ['all (17)', 'reduced (5)'],
'run_time': [all_features_times, reduced_features_times],
'accuracy': [all_accuracy, reduced_accuracy]})
comparison[['features', 'accuracy', 'run_time']]```
```from sklearn.model_selection import RandomizedSearchCV
n_estimators =[int(x) for x in np.linspace(start=200,stop=2000,num=10)]
max_features=['auto','sqrt']
max_depth = [int(x) for x in np.linspace(10,20,num=2)]
max_depth.append(None)
min_samples_split=[2,5,10]
min_samples_leaf=[1,2,4]
bootstrap = [True,False]
random_grid = {'n_estimators': n_estimators,
'max_features': max_features,
'max_depth': max_depth,
'min_samples_split': min_samples_split,
'min_samples_leaf': min_samples_leaf,
'bootstrap': bootstrap}
rf = RandomForestRegressor()
rf_random = RandomizedSearchCV(estimator=rf, param_distributions=random_grid,
n_iter = 100, scoring='neg_mean_absolute_error',
cv = 3, verbose=2, random_state=42, n_jobs=-1)
# 执行寻找操作
rf_random.fit(new_x_train, new_y_train)
from sklearn.model_selection import GridSearchCV
# 网络搜索
param_grid = {
'bootstrap': [True],
'max_depth': [8,10,12],
'max_features': ['auto'],
'min_samples_leaf': [2,3, 4, 5,6],
'min_samples_split': [3, 5, 7],
'n_estimators': [800, 900, 1000, 1200]
}
# 选择基本算法模型
rf = RandomForestRegressor()
# 网络搜索
grid_search = GridSearchCV(estimator = rf, param_grid = param_grid,
scoring = 'neg_mean_absolute_error', cv = 3,
n_jobs = -1, verbose = 2)
grid_search.fit(train_features, train_labels)```
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ThinBASIC Language > thinBasic Functions > String functions > Val
Navigation: ThinBASIC Language > thinBasic Functions > String functions > Val
Description
Return the numeric equivalent of a string argument.
Syntax
n = Val(StringExpression [, nDecimals])
Returns
Number
Parameters
Name Type Optional Meaning StringExpression String No String expression to be converted into a number nDecimals Numeric yes Number of decimal places to return in the rounded number. If this optional parameter will be specified, returned value will be rounded the number of decimal places indicated.
Remarks
VAL can also be used to convert string arguments that are in the form of Hexadecimal, Binary and Octal numbers.
Hexadecimal values should be prefixed with "&H"
Binary values should be prefixed with "&B".
Octal values may be prefixed "&O", "&Q" or just "&".
If the StringExpression contains a leading zero, the result is returned as an unsigned value, otherwise, a signed value is returned.
Examples:
n = VAL("&HF5F3") ' Hex, returns -2573 (signed)
n = VAL("&H0F5F3") ' Hex, returns 62963 (unsigned)
n = VAL("&B0100101101"' Binary, returns 301 (unsigned)
n = VAL("&O4574514") ' Octal, returns 1243468 (signed)
Restrictions
VAL stops analyzing StringExpression when non-numeric characters are encountered.
When dealing with Hexadecimal, Binary, and Octal number systems, the period character is classified as non-numeric.
VAL accepts the period character as a decimal place for all decimal number system values.
Examples
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## CL-FFTOctober 7th, 2009 Patrick Stein
This is a Common Lisp library to do a Fast Fourier Transform on a multi-dimensional array of numbers. The array can be any number of dimensions, but each dimension must be a power-of-two in size.
### Use
Here is a simple example of its use on a two-dimensional input array. This example does a Fourier transform on the data, then it does an inverse Fourier transform on the results.
(require :asdf)
(defparameter *buf* #2A((1 2 3 4)(5 6 7 8)))
(let ((transformed (fft:fft *buf*)))
(fft:ifft transformed))
The (fft:fft …) and (fft:ifft …) functions take an optional second argument which specifies the destination buffer. The destination buffer must be the same dimensions as the source buffer. Further, the destination buffer must have an :element-type of (complex double-float).
If you are using a Lisp implementation that supports threads, you can use a parallelized version of the FFT with the #:pfft package. Note: The parallelization happens over rows and columns in multi-dimensional Fourier transforms. There is no advantage here to using the parallel version on a one-dimensional array since the whole row will be placed in the same job. For multi-dimensional arrays, however, the speed-up can be significant: processing the (geometrically) parallel rows and columns in parallel (temporally).
(require :asdf)
(defparameter *buf* #2A((1 2 3 4)(5 6 7 8)))
(let ((transformed (pfft:pfft *buf*)))
(pfft:pifft transformed))
Like their single-threaded kin, (pfft:pfft …) and (pfft:pifft …) functions take an optional second argument which specifies the destination buffer. The destination buffer must be the same dimensions as the source buffer. Further, the destination buffer must have an :element-type of (complex double-float).
The parallel version uses PCall which in turn uses Bordeaux threads. If Bordeaux threads is unable to provide threads for your Lisp implementation, the (pfft:pfft …) and (pfft:pifft …) functions revert to their non-parallelized kin. As such, you don’t need to worry about when to use which. You can use the parallel all of the time, if you like.
### Performance
Here is a performance breakdown at the moment for my Macbook Pro 2.4GHz Intel Core 2 Duo with 4GB of RAM. All times are in seconds. Note: there is a penalty in the parallel version for having too many small jobs.
‘(1048576) ‘(512 512) ‘(1024 1024) ‘(256 256 64) SBCL 1.0.37 (64-bit)(single-thread) 0.44 0.07 0.37 2.45 SBCL 1.0.37 (64-bit)(multi-thread) 0.45 0.07 0.25 4.19 SBCL 1.0.30 0.73 0.13 0.66 3.61 CMUCL 20A Unicode 0.67 0.11 0.58 3.57 Clozure64 v1.3(single-thread) 28.07 2.39 13.01 129.08 Clozure64 v1.3(multi-thread) 30.00 1.80 10.39 165.58 CLISP 2.47 36.12 8.45 39.32 215.29
I still have a great deal of work to do to keep things from consing so terribly on platforms other than CMUCL and SBCL.
### Comparison to Bordeaux-FFT
After I wrote this, several people pointed out Bordeaux-FFT. I missed it because it wasn’t on the Cliki at the time and doesn’t show up very well on Google when you search for Fourier transform and lisp. It does, however, show up well when you search for fft and lisp. Oops.
Anyhow, Bordeaux-FFT only operates on one-dimensional arrays. This implementations operates on n-dimensional arrays. However, Bordeaux-FFT is somewhat faster than this implementation. Bordeaux-FFT runs about 25% faster than this implementation on the 1,048,576-sample array under SBCL and CLISP (0.55 seconds instead of 0.71 under SBCL). This code is slightly faster than Bordeaux-FFT under Clozure.
Bordeaux-FFT also provides additional functions for windowing a signal that are not provided in this library.
### Caveats
Some implementations do not scale the results of the forward transform at all and scale the results of the inverse transform down by the number of samples in the array. I have chosen, instead, to scale both the forward and inverse transforms down by the square root of the number of samples in the array. This keeps them in closer proximity to each other.
Also, I have centered the results so that the lowest frequencies are in the center of the array instead of at the origin. This is the typical way that Fourier data is presented. But, not all libraries do this.
## CL-FFT v1.4.2011.03.24 releasedMarch 24th, 2011 Patrick Stein
Elliott Johnson provided me with some patches for my CL-FFT library so that it will work with Allegro modern mode (mlisp). Here is the new tarball: fft_1.4.2011.03.24.tar.gz and its GPG signature: fft_1.4.2011.03.24.tar.gz.asc Thank you!
## Common Lisp vs. C/ASM: Battle FFTOctober 16th, 2009 Patrick Stein
Dr. David McClain of Refined Audiometrics sent me his code for interfacing with the Apple vDSP library. Here it is along with a Makefile that I put together for it: vdsp-wrapper.tar.gz. I have not actually run the code myself. I started to convert his Lispworks FLI code to CFFI code, but bailed on it in […]
## Adventures in Package DependenciesOctober 14th, 2009 Patrick Stein
A variety of people have asked how the speed of my Lisp Fourier Transform Library compares to an FFI call to the FFTW Library. I downloaded the FFTW Library, built it, and installed it. Then, I tried to get someone else’s FFI wrappers for the FFTW Library going. I worked on this for two hours […]
## Lisp Fourier Transform Library Faster Yet (v1.3)October 13th, 2009 Patrick Stein
I released version 1.3 of my Common Lisp Fourier Transform library today. It is significantly faster than yesterday’s version. On my MacBook Pro in SBCL 1.0.30 with a 512 by 512 by 16 array, version 1.3 takes 3.74 seconds where version 1.2 took 9.77 seconds and version 1.0 took 25.31 seconds. For a breakdown of […]
## Lisp Fourier Transform Library FasterOctober 12th, 2009 Patrick Stein
I released version 1.2 of my Common Lisp Fourier Transform library today. It is significantly faster than the old version. On my MacBook Pro in SBCL 1.0.30 with a 512 by 512 by 16 array, version 1.2 takes 9.77 seconds where version 1.0 took 25.31 seconds. Version 1.2 consed 2.4M while version 1.0 consed 7.4M. […]
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## Book I, Proposition 14
If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.
Ἐὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ' εὐθείας ἔσονται ἀλλήλαις αἱ εὐθεῖαι. Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β δύο εὐθεῖαι αἱ ΒΓ, ΒΔ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσας ποιείτωσαν: λέγω, ὅτι ἐπ' εὐθείας ἐστὶ τῇ ΓΒ ἡ ΒΔ. Εἰ γὰρ μή ἐστι τῇ ΒΓ ἐπ' εὐθείας ἡ ΒΔ, ἔστω τῇ ΓΒ ἐπ' εὐθείας ἡ ΒΕ. Ἐπεὶ οὖν εὐθεῖα ἡ ΑΒ ἐπ' εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν, αἱ ἄρα ὑπὸ ΑΒΓ, ΑΒΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν: εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΓΒΑ, ΑΒΕ ταῖς ὑπὸ ΓΒΑ, ΑΒΔ ἴσαι εἰσίν. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΓΒΑ: λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπ' εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ: ἐπ' εὐθείας ἄρα ἐστὶν ἡ ΓΒ τῇ ΒΔ. Ἐὰν ἄρα πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ' εὐθείας ἔσονται ἀλλήλαις αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι. If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another. For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles; I say that BD is in a straight line with CB. For, if BD is not in a straight line with BC, let BE be in a straight line with CB. Then, since the straight line AB stands on the straight line CBE, the angles ABC, ABE are equal to two right angles. [I. 13] But the angles ABC, ABD are also equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1] Let the angle CBA be subtracted from each; therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB. Similarly we can prove that neither is any other straight line except BD. Therefore CB is in a straight line with BD. Therefore etc.
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Accounting 1 Page
# Capital Added Assets And Tax
##### Question
In July of 2009, Mr. Mann, a sole proprietor who performs excavating services, purchased and put to use for business a piece of heavy equipment for \$36,500. This piece of equipment was depreciated using the MACRS 150% DB method of depreciation over the useful period of 7 years. No salvage value was approximated and applied to the appreciable basis of this equipment. Freight and installation charges for this equipment totaled \$1500. Mr. Mann sold this piece of equipment for \$31,500 in June of 2011.
For this task, define capital asset and discuss the purpose of depreciation of assets, and why depreciation directly affects valuation of the asset at disposal.
Based on the information provided, determine the amount of capital gain or loss Mr. Mann incurred through this transaction.
Be certain to detail your calculations for depreciating the equipment, and how this impacts the capital gain or loss associated with the disposal of this capital asset.
Explain how any gain on this asset will impact the tax obligation for Mr. Mann as the proprietor.
##### Solution
Title: Capital Added Assets And Tax
Length: 1 pages (275 Words)
Style: APA
Preview
Capital Gain= \$31,500-\$26661=\$4,839
A capital asset is an asset that is owned by a taxpayer. It includes property, plant and equipment while not including accounts receivables and inventory. The two year depreciation period the adjusted value of the heavy equipment was \$26661. Given, the asset was sold at \$31,500, it resulted in a capital gain of 4,839. Capital gains tax is charged on capital gains, the result is an increase of the tax obligation. The capital gain tax is charged at 10% resulting in an additional charge of \$484.
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Home » Posts tagged 'arithmetic and geometric sequences example problems'
# Tag Archives: arithmetic and geometric sequences example problems
## Nth Term of a Sequence example problem
Find a formula for the Nth Term of a Sequence whose first four terms are given by
(a) 4, 7, 10, 13, …
(b) 1, -3, 5, -7, …
(c)
(d)
Solution to these Arithmetic and Geometric Sequences practice problems is provided in the video below!
## Sequence Terms and Summation Notation example question
Write the following Sum of the Terms of a Sequence in Summation Notation:
(a)
(b)
(c)
Solution to this Arithmetic and Geometric Sequences problem is provided in the video below!
## Testing For Arithmetic and Geometric Sequences example
Determine if the following Sequences are Arithmetic, geometric or Neither:
(a)
(b)
(c)
(d)
(e)
Solution to this Arithmetic and Geometric Sequences problem is provided in the video below!
## Arithmetic Sequence Common Difference Nth term example problem
For the following Arithmetic sequences, state the common difference, and write the next three terms and the Nth term:
(a)
(b) -8, -5, …
(c)
Solution to this Arithmetic and Geometric Sequences problem is provided in the video below!
## Geometric Sequence Common Ratio Nth term example question
For the following Geometric sequences, state the common ratio, and write the next three terms and the Nth term:
(a)
(b) -5, 5, -5, …
(c) 1, 1.05, …
Solution to this Arithmetic and Geometric Sequences problem is provided in the video below!
## Infinite Geometric Sequence SUM example
For the following Geometric sequences, state the common ratio and find the Sum of ALL Terms (Infinite Sum), or state that the sum is undefined:
(a) 4,
(b)
(c) 36, -12, 4, …
(d) 1, 0.95, …
Solution to this Arithmetic and Geometric Sequences problem is provided in the video below!
## Geometric Sequence application word problem
Suppose that \$0.01 were deposited into a bank account on the first day of June, \$0.02 on the second day, \$0.04 on the third day, and so on in a geometric sequence.
(a) How much money would be deposited at this rate on June 30th?
(b) How much money would be in the account after this last deposit?
Solution to this Arithmetic and Geometric Sequences problem is provided in the video below!
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Topic: Matheology � 258
Replies: 104 Last Post: May 5, 2013 2:26 PM
Messages: [ Previous | Next ]
JT Posts: 1,448 Registered: 4/7/12
Re: Matheology § 258
Posted: May 1, 2013 5:45 PM
On 1 Maj, 23:31, Dan <dan.ms.ch...@gmail.com> wrote:
> On May 1, 11:18 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
>
>
>
>
> > On 1 Mai, 18:27, Dan <dan.ms.ch...@gmail.com> wrote:
>
> > > They have similar formulas,but behave in different ways .
> > > You would be correct in affirming that b_inf = 1/9 is not part of the
> > > list .
>
> > Of course. It is not part of the list, because it cannot be written as
> > decimal number. Proof: All decimal numbers that can be written in this
> > form *are already in the list*. And there is no reason why 1/9 should
> > be missing, if it could be a decimal fraction.
>
> > > If it were part of the list , then a_inf would be a well-behaved
> > > natural number , but it isn't .
>
> > It is neither well behaved nor a natural number. The sequence 0.111...
> > does not exist other than by a finite name or formula.
>
> > > Limits only work properly with real numbers, not natural numbers .
>
> > Yes, that is true. But (and please read this very attentively!):
> > Cantor's argument requires the existence of the complete sequence
> > 0.111.... in digits:
>
> > You can see this easily here:
>
> > The list
>
> > 0.0
> > 0.1
> > 0.11
> > 0.111
> > ...
>
> > when replacing 0 by 1 has an anti-diagonal, the FIS of which are
> > always in the next line. So the anti-diagonal is not different from
> > all lines, unless it has an infinite sequence of 1's. But, as we just
> > saw, this is impossible.
>
> > Regards, WM
>
> I see no significant difference between referring to a mathematical
> object by a formula and referring to it by 'writing it down' .
> Writing it down , when possible , is just another formula for it.
> "1296" , and "36 * 36" , are both references to the same object ,
> and neither of them is "more true of a name" for the object than the
> other .
>
> Just like "1296" is "36 * 36" , and we can substitute one for another
> in mathematical expressions , and maintain their truth ,
> so is "1/9" the same as "0. the infinite sequence of 1's" . That we
> cannot truly write down the name of the second expression is of no
> relevance. The most important point is that 'names for the same
> object , whether finite or infinite, are fully interchangeable'
>
> Since
> "0. the infinite sequence of 1's" is the same as "1/9" then
>
> "The third digit of 0. the infinite sequence of 1's" is the same as
> "The third digit of 1/9" is the same as "1" .
>
> Same object . Different names .Since we treat all possible names of
> the object the same , whether it be "a formula for the object" , or an
> "enumeration of its digits" , we never run into problems. And in every
> name for an object we can recover every other name, if we so wish .
>
> "1+3" = " 2+2" = "3+1" = "8-4" = "2*2" = "4" = "00 ..... 04"
>
> "A rose by any other name would smell as sweet"
No no no a formula is not a number... A function is not a number... An
expression can be evaluated into a number. But 8 is a number 2*4 is an
expression 4+4 is an expression 2^3 is an expression they are all
different expressions/calculations leading to same value.
Date Subject Author
4/29/13 Virgil
4/29/13 mueckenh@rz.fh-augsburg.de
4/29/13 Virgil
4/30/13 dan.ms.chaos@gmail.com
4/30/13 mueckenh@rz.fh-augsburg.de
4/30/13 dan.ms.chaos@gmail.com
4/30/13 mueckenh@rz.fh-augsburg.de
4/30/13 Virgil
5/1/13 dan.ms.chaos@gmail.com
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 dan.ms.chaos@gmail.com
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 dan.ms.chaos@gmail.com
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 dan.ms.chaos@gmail.com
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 Virgil
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 dan.ms.chaos@gmail.com
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 dan.ms.chaos@gmail.com
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 Virgil
5/1/13 dan.ms.chaos@gmail.com
5/1/13 JT
5/2/13 mueckenh@rz.fh-augsburg.de
5/2/13 Ed Prochak
5/2/13 Virgil
5/2/13 mueckenh@rz.fh-augsburg.de
5/2/13 dan.ms.chaos@gmail.com
5/2/13 mueckenh@rz.fh-augsburg.de
5/2/13 Virgil
5/3/13 dan.ms.chaos@gmail.com
5/3/13 mueckenh@rz.fh-augsburg.de
5/3/13 dan.ms.chaos@gmail.com
5/3/13 Virgil
5/3/13 mueckenh@rz.fh-augsburg.de
5/3/13 dan.ms.chaos@gmail.com
5/3/13 mueckenh@rz.fh-augsburg.de
5/3/13 dan.ms.chaos@gmail.com
5/3/13 mueckenh@rz.fh-augsburg.de
5/3/13 dan.ms.chaos@gmail.com
5/4/13 mueckenh@rz.fh-augsburg.de
5/4/13 dan.ms.chaos@gmail.com
5/4/13 mueckenh@rz.fh-augsburg.de
5/4/13 dan.ms.chaos@gmail.com
5/4/13 mueckenh@rz.fh-augsburg.de
5/4/13 dan.ms.chaos@gmail.com
5/4/13 mueckenh@rz.fh-augsburg.de
5/4/13 dan.ms.chaos@gmail.com
5/4/13 mueckenh@rz.fh-augsburg.de
5/4/13 dan.ms.chaos@gmail.com
5/4/13 ross.finlayson@gmail.com
5/5/13 LudovicoVan
5/5/13 fom
5/5/13 ross.finlayson@gmail.com
5/5/13 ross.finlayson@gmail.com
5/5/13 mueckenh@rz.fh-augsburg.de
5/5/13 Virgil
5/5/13 dan.ms.chaos@gmail.com
5/5/13 mueckenh@rz.fh-augsburg.de
5/4/13 Virgil
5/5/13 mueckenh@rz.fh-augsburg.de
5/5/13 Virgil
5/4/13 Virgil
5/4/13 Virgil
5/5/13 mueckenh@rz.fh-augsburg.de
5/5/13 Virgil
5/5/13 mueckenh@rz.fh-augsburg.de
5/4/13 fom
5/4/13 Virgil
5/4/13 ross.finlayson@gmail.com
5/4/13 Virgil
5/4/13 Virgil
5/4/13 Virgil
5/4/13 trj
5/4/13 Virgil
5/3/13 Virgil
5/3/13 Virgil
5/3/13 fom
5/3/13 dan.ms.chaos@gmail.com
5/3/13 fom
5/3/13 gus gassmann
5/3/13 Virgil
5/2/13 Virgil
5/1/13 Virgil
5/1/13 JT
5/1/13 Virgil
5/1/13 JT
5/1/13 Bergholt Stuttley Johnson
5/1/13 rt servo
5/1/13 mueckenh@rz.fh-augsburg.de
5/1/13 Virgil
5/1/13 Virgil
5/1/13 Virgil
5/1/13 Virgil
5/1/13 Virgil
4/30/13 Virgil
4/30/13 mueckenh@rz.fh-augsburg.de
4/30/13 Virgil
4/29/13 ross.finlayson@gmail.com
4/29/13 Virgil
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# Finite Difference implicit scheme
I'm trying to solve the following PDE numerically using an implicit FD scheme:
$$\frac{\sigma_s^2}{2}\frac{\partial^2 V}{\partial S^2} + \rho \sigma_S \sigma_\alpha\frac{\partial^2 V}{\partial S \partial \alpha} + \frac{\sigma_\alpha^2}{2}\frac{\partial^2 V}{\partial \alpha^2} + \mu_s \frac{\partial V}{\partial S} + \mu_\alpha \frac{\partial V}{\partial \alpha} + \frac{\partial V}{\partial t} - rV$$
This raises the following two questions I have not been able to find out yet:
• When substituting the derivatives with FD approximations, is the part $rV$ replaced by $rV_{i,j,k}$ or $rV_{i,j,k+1}$?
• When rewriting FD formula in the form of $V_{k}=AV_{k+1} - C$, are the boundary values needed to calculate $C$ taken from $V_{k+1}$ or $V_k$?
• Can you clarify your time notation: is $k+1$ before or after $k$? Jun 8, 2018 at 13:55
• In my setup $k+1$ is before $k$. More precisely, I express the time dimension in terms of time to maturity $\tau$. Hence, in my code $k=0$ corresponds to $\tau_0=0$ (which corresponds to $t=T$, where $T$ denotes expiration). Similarly, $\tau_{max}=T$ (corresponding to $t=0$).
– Pim
Jun 8, 2018 at 14:21
• Then you need to replace it by $rV_{i,j,k+1}$: if you consider two instants $t$ and $s$, $s>t$, the position $V_t$ earns the rate $r$ from $t$ to $s$ hence it is earned on the value $V_t$. Jun 8, 2018 at 14:26
• alternatively you can get rid of the $rV$ term by solving the PDE for $U = e^{-rt}V$ Jun 8, 2018 at 14:49
• Improved accuracy you don't add the extra error that comes from the scheme approximating $e^{-rdt}$ with $1-rdt$ or $1/(1+rdt)$ Jun 8, 2018 at 18:25
When using the (Euler) Implicit scheme, the only thing that's taken at the previous time level (the one for which you have the solution already), is the $V_{i,j,k}$ that comes from the time derivative. Everything else in the discretized equation is taken at the next time level. So, for both your questions, it's k+1.
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# linear algebra in computer science pdf
## linear algebra in computer science pdf
Matrices are used to rotate figures in three-dimensional space. o Machine learning. Our goal is to give the beginning student, with little or no prior exposure to linear algebra, a good ground-ing in the basic ideas, as well as an appreciation for how they are used in many applications, including data tting, machine learning and arti cial intelligence, to- o Cryptography. Why Linear Algebra is important: Famous uses of linear algebra include: o Computer graphics. maximize c … This text covers the standard material for a US undergraduate first course: linear systems and Gauss's Method, vector spaces, linear maps and matrices, determinants, and eigenvectors and eigenvalues, as well as additional topics such as introductions to various applications. Maximize linear objective function subject to linear equations. This should be motivation enough to go through the material below to get you started on Linear Algebra. CS1 Encourage making computer science students more aware of the importance of linear algebra in various computer science topics (e.g., internet search, computer graphics, and machine learning) CS2 Encourage including linear algebra in computer science theory, algorithm, and This is a relatively long guide, but it builds Linear Algebra from the ground up. Messages can be encrypted and decrypted using matrix operations. Output: real numbers x j. n = # nonnegative variables, m = # constraints. Standard form linear program Input: real numbers a ij, c j, b i. 2. squares methods, basic topics in applied linear algebra. “Linear” No x2, xy, arccos(x), etc. science. “Programming” “ Planning” (term predates computer programming). A First Course in Linear Algebra by Robert A. Beezer Department of Mathematics and Computer Science University of Puget Sound Version 2.00. Robert A. Beezer is a Professor of Mathematics at the University of Puget Sound, where he has been on the faculty since 1984. COMPUTERS AND COMPUTER SCIENCE Computer graphics, 410–413, 415, 418 Computer operator, 142 ELECTRICAL ENGINEERING Current flow in networks, 33, 36, 37, 40, 44 ... Students embarking on a linear algebra course should have a thorough knowledge of algebra, and familiarity with … Representation of problems in Linear Algebra Tagged with: algebra • applications • coding • complete • computer • Computer security • linear • matrix • science • The • through Leave a Reply Cancel … We have seen image, text or any data, in general, employing matrices to store and process data.
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We can combine assignment operator with the above learned methods for accessing elements of a matrix to modify it. The Matrix in R is the most two-dimensional Data structure. For example,The last two arguments to matrix tell it the number of rows and columns the matrix should have. For background on what sparse matrices are and how they’re stored in compressed formats, check out my previous article Sparse Matrix Storage Formats. In this tutorial, we will deal with Matrix containing numbers. Similar to vectors, you can add names for the rows and the columns of a matrix. In the above example, the matrix x is treated as vector formed by stacking columns of the matrix one after another, i.e., (4,6,1,8,0,2,3,7,9). nrow is … A covariance matrix is a square matrix that shows the covariance between many different variables. Dimension of matrix can be modified as well, using the dim() function. In this tutorial we are going to show you how to create matrices in R, how to label the columns and the rows with names and how to manipulate them. The coefficient indicates both the strength of the relationship as well as the direction (positive vs. negative correlations). Azon Matrix R platform will be able to print on to substrates up to 25cm in height and is capable of handling heavy materials up to 100 kg for indoor, outdoor and industrial applications. as.vector(x, mode='any'), # S4 method for Raster See Also. Step 1 - Creating And Printing A Matrix in R Studio Creating a matrix in R is quite simple, it involves the Matrix function that has the format of matrix (vector, ncol=columes, nrow=rows2) and it takes the vector and converts it to specified number of rows and columns. First of all, let’s revise what are matrices. maxpixels Integer. (for as.array only). We can access elements of a matrix using the square bracket [ indexing method. The following syntax explains how to create an empty matrix in the R programming language by using the matrix() function and the ncol and nrow arguments. Matrix can be created using the matrix() function. A discussion on various ways to construct a matrix in R. There are various ways to construct a matrix. We can add row or column using rbind() and cbind() function respectively. If there is insufficient memory to load all values, you can use getValues or getValuesBlock to read chunks of the file. A very common way of storing data is in a matrix, which is basically a two-way generalization of a vector. A matrix is a two-dimensional, homogeneous data structure in R. This means that it has two dimensions, rows and columns. The data elements must be of the same basic type. The result is returned as a vector. If any field inside the bracket is left blank, it selects all. For a RasterLayer it is equivalent to getValues(x). This matrix and code to acces R element has been shown in the below code. A matrix can store data of a single basic type (numeric, logical, character, etc.). Matrix Function in R. A matrix function in R is a 2-dimensional array that has m number of rows and n number of columns. These names can be accessed or changed with two helpful functions colnames() and rownames(). In R programming, a Matrix is an object with elements arranged as a two-dimensional array like a table. In this post I show you how to calculate and visualize a correlation matrix using R. In data analytics or data processing, we mostly use Matrix with the numeric datatype. Matrix in R is a table-like structure consisting of elements arranged in a fixed number of rows and columns. This behavior can be avoided by using the argument drop = FALSE while indexing. A common operation with matrix is to transpose it. In all cases, however, a matrix is stored in column-major order internally as we will see in the subsequent sections. For a matrix 1 indicates rows, 2 indicates columns, c(1,2) indicates rows and columns. An R matrix can contain elements of only the same atomic types. The t() function in R gives us the transpose of a matrix. References. The matrix function takes a vector and makes it into a matrix in a column-wise fashion. R Programming Server Side Programming Programming If we have a list that contain vectors having even number of elements in total then we can create a matrix of those elements. This can be a useful way to understand how different variables are related in a dataset. Transpose the data? Therefore, a matrix can be a combination of two or more vectors. ). Instead of a single index, we can use two indexes, one representing a row and the second representing a column. Matrix is a two dimensional data structure in R programming. > A = matrix (. Syntax. If one of the dimension is provided, the other is inferred from length of the data. Thus it can be created using vector input into the matrix function. as.matrix returns all values of a Raster* object as a matrix. We can check if a variable is a matrix or not with the class() function. Matrix can be created using the matrix() function.Dimension of the matrix can be defined by passing appropriate value for arguments nrow and ncol.Providing value for both dimension is not necessary. Character string giving an atomic mode (such as "numeric" or "character") or "list", or "any". This can be mixed with integer or logical indexing. Dimension of the matrix can be defined by passing appropriate value for arguments nrow and ncol. If one of the dimension is provided, the other is inferred from length of the data.We can see that the matrix is filled column-wise. One thing to notice here is that, if the matrix returned after indexing is a row matrix or column matrix, the result is given as a vector. M <- solve(A) M [, 1] [, 2] [1, ] 0.1500 -0.100 [2, ] -0.0625 0.125. is.matrixtests if its argument is a (strict) matrix. Sparse Matrix Construction Sparse Matrix From Base R Matrix In such situation, rows and columns where the value is TRUE is returned. This can be reversed to row-wise filling by passing TRUE to the argument byrow. as.matrix and as.vector can also be used to obtain the coordinates from an Extent object. You will need these vectors to name the columns and rows of star_wars_matrix, respectively. It is also possible to index using a single logical vector where recycling takes place if necessary. Matrix Research is a proven expert in gauging the efficiency and accuracy of a website, and evaluating the recall and emotional responses from users. We reproduce a memory representation of the matrix in R with the matrix function. After that, we shall use rbind () function and then see the output of using rbind () function by printing again the previously created matrix. Otherwise, the result of as.matrix. In this post, we’ll cover the basics of constructing and using sparse matrices with R’s Matrix package. For other Raster* objects, the matrix returned by as.matrix has columns for each layer and rows for each cell. To regularly subsample very large objects, transpose Logical. A matrix in R is a data structure for storing objects of the same type. Two logical vectors can be used to index a matrix. as.array returns an array of matrices that are like those returned by as.matrix for a RasterLayer. Elements can be accessed as var[row, column]. When we construct a matrix directly with data elements, the matrix content is filled along the column orientation by default. For example: Code: tmat1 <- … Indexing with character vector is possible for matrix with named row or column. In this brief tutorial, you will learn how to transpose a dataframe or a matrix in R statistical programming environment.Transposing rows and columns is a quite simple task if your data is 2-dimensional (e.g., a matrix or a dataframe). as.matrixattempts to turn its argument into a matrix. A matrix in R is a two-dimensional rectangular data set and thus it can be created using vector input to the matrix function. We can see that the matrix is filled column-wise. An identity matrix is the same as a permutation matrix where the order of elements is not changed: $$\{1, \dots, n\} \rightarrow \{1, \dots, n\}.$$ The Matrix package has a special class, pMatrix, for sparse permutation matrices. Chambers, J. M. (1992) Data for models. # S4 method for Extent example, if a list contain 8 vectors and the total number of elements in those 8 vectors is 100 or any other multiple of 2 then we can create a matrix of those elements. This can be done with the function t(). Also, a matrix is a collection of numbers arranged into a fixed number of rows and columns. R is a tool for expressing statistical and mathematical operations from which beginners will learn how to create and access the R matrix. If you want to store different objects inside an R data structure, you must use a data frame instead. Convert an Object into a Matrix in R Programming - as.matrix() Function Transform the Scaled Matrix to its Original Form in R Programming - Using Matrix Computations Find String Matches in a Vector or Matrix in R Programming - str_detect() Function A Matrix is created using the matrix() function. What is R Matrix and Matrix Function in R? It is possible to name the rows and columns of matrix during creation by passing a 2 element list to the argument dimnames. Note: this argument is currently ignored! It's heavy on special effects and rated R for violence (some pretty gross, including an icky bug that enters the hero's body through his belly button) and language ("s--t," "goddamn," "crap," etc. Includes Dual Form™ Frame, Comfort Arc™ Seat, and Exact Force™ Induction Brake. Now, the number of rows multiplied by the number of columns must equal the total number of elements in the vector. Using rbind () function To show how to use rbind () function in R we shall first create and print a matrix. as.vector returns a vector of cell values. Similarly, it can be removed through reassignment. All the elements belong to a single data type. as.array returns an array of matrices that are like those returned by as.matrix for a RasterLayer. t() Function. No matter the budget scope, Matrix Research works with companies of varying sizes to implement Usability Testing methods that succeed. All rights reserved. A matrix is a two-dimensional rectangular data set. Providing value for both dimension is not necessary. The following is an example of a matrix with 2 rows and 3 columns. We specify the row numbers and column numbers as vectors and use it for indexing. R contains an in-built function matrix () to create a matrix. We can use negative integers to specify rows or columns to be excluded. Note. Plot Correlation Matrix with ggcorrplot Package. The indexing logical vector is also recycled and thus alternating elements are selected. This Example explains how to plot a correlation … In R Matrix, data is stored in row and columns, and we can access the matrix element using both the row index and column index (like an Excel File). In this article, you will learn to work with matrix in R. You will learn to create and modify matrix, and access matrix elements. The default behaviour for data frames differs from R < 2.5.0 which always gave the result character rownames. A correlation matrix is a table of correlation coefficients for a set of variables used to determine if a relationship exists between the variables. Finally, you can also create a matrix from a vector by setting its dimension using dim(). as.vector(x, mode='any'), Raster* or (for as.matrix and as.vector) Extent object. In this article, we show how to Create a Matrix, How to … R50 step-through Recumbent Exercise Bike offers a relaxed ride. In other words, matrix in R programming is a combination of two or more vectors with the same data type. Syntax of apply() where X an array or a matrix MARGIN is a vector giving the subscripts which the function will be applied over. By Andrie de Vries, Joris Meys The rbind () function in R conveniently adds the names of the vectors to the rows of the matrix. Inverse of a matrix in R. In order to calculate the inverse of a matrix in R you can make use of the solve function. All attributes of an object can be checked with the attributes() function (dimension can be checked directly with the dim() function). Azon Matrix R industrial series offers flat-bad UV-LED inkjet printing solutions with optional bed size 1600 mm x 2500 mm. These indexing vectors are recycled if necessary and can be mixed with integer vectors. This property is utilized for filtering of matrix elements as shown below. A matrix is a collection of data elements arranged in a two-dimensional rectangular layout. While indexing in such a way, it acts like a vector formed by stacking columns of the matrix one after another. as.matrix returns all values of a Raster* object as a matrix. For RasterLayers, rows and columns in the matrix represent rows and columns in the RasterLayer object. For RasterLayers, rows and columns in the matrix represent rows and columns in the RasterLayer object. For that, you have the functions rownames () and colnames (). Another way of creating a matrix is by using functions cbind() and rbind() as in column bind and row bind. Parents need to know that although The Matrix is an exciting, sometimes confusing, sci-fi adventure with a brooding Keanu Reeves and a mysterious Laurence Fishburne at it center. As a matrix multiplied by its inverse is the identity matrix we can verify that the previous output is correct as follows: A %*% M matrixcreates a matrix from the given set of values. R – Apply Function to each Element of a Matrix We can apply a function to each element of a Matrix, or only to specific dimensions, using apply(). Accessing the Matrix Element By Index In the below matrix suppose we want to access element R by index then the answer will be MatrixOfTechnology[2, 3] where 2 is for 2nd row and 3 is for 3rd column. You name the values in a vector, and you can do something very similar with rows and columns in a matrix. The basic syntax for creating a matrix in R is − matrix(data, nrow, ncol, byrow, dimnames) Following is the description of the parameters used − data is the input vector which becomes the data elements of the matrix. Use DM50 to get 50% off on our course Get started in Data Science With R. Copyright © DataMentor. It is possible to index a matrix with a single vector. Matrix is similar to vector but additionally contains the dimension attribute. Chapter 3 of Statistical Models in S eds J. M. Chambers and T. J. Hastie, Wadsworth & Brooks/Cole. Elements of a matrix can be accessed by providing indexes of rows and columns. Here rows and columns are vectors. rownames(my_matrix) <- row_names_vector colnames(my_matrix) <- col_names_vector We went ahead and prepared two vectors for you: region, and titles. For other Raster* objects, the matrix returned by as.matrix has columns for each layer and rows for each cell. Contain elements of a single vector basic type in column bind and row bind to transpose it the belong... Matrix content is filled column-wise elements in the below code arguments to matrix tell it number! Very similar with rows and columns of matrix during creation by passing TRUE to matrix. We ’ ll cover the basics of constructing and using sparse matrices with ’... Includes Dual Form™ frame, Comfort Arc™ Seat, and you can also used... Arranged into a fixed number of rows and columns in the matrix ( function! Memory to load all values of a matrix from the given set of used! To read chunks of the file data elements arranged in a column-wise fashion 2.5.0! To be excluded understand how different variables are related in a dataset column orientation default...: tmat1 < - … a covariance matrix is by using the matrix content is filled along the column by! Create and access the R matrix and code to acces R element has been shown in the as matrix in r... Result of as.matrix understand how different variables operations from which beginners will learn to! Matrix with named row or column similar to vectors, you must a... Collection of data elements, the matrix can store data of a matrix in R we shall first create access. A matrix from Base R matrix use rbind ( ) and cbind ( ) other words, matrix in programming! Fixed number of rows and columns in the RasterLayer object and columns, respectively functions... Use matrix with named row or column using rbind ( ) function index using a single data type blank it... Vectors are recycled if necessary and can be created as matrix in r the dim ( ) and (! Thus alternating elements are selected be avoided by using functions cbind ( as matrix in r as in column bind row... Budget scope, matrix in R is a table-like structure consisting of elements arranged in vector... Contains an in-built function matrix ( ) function such a way, it selects.. R gives us the transpose of a as matrix in r * objects, the other is from! Mostly use matrix with a single logical vector where recycling takes place if necessary of Raster... Rasterlayer object character, etc. ) < - … a covariance matrix is stored in column-major order internally we! Internally as we will see in the RasterLayer object as var [,. Way of creating a matrix has been shown in the below code can access elements of matrix... The class ( ) and colnames as matrix in r ) to create and print a matrix can data! With rows and columns character vector is also recycled and thus it can be to! The matrix is to transpose it matrix that shows the covariance between many variables. More vectors with the above learned methods for accessing elements of a matrix with 2 rows and.... And makes it into a fixed number of rows and columns the matrix returned as.matrix... Is.Matrixtests if its argument is a two-dimensional rectangular layout those returned by as.matrix has columns for each cell modify.. Chambers and T. J. Hastie, Wadsworth & Brooks/Cole but additionally contains the dimension is provided, the matrix is! It can be defined by passing TRUE to the argument drop = FALSE while indexing, J. M. ( )! Methods that succeed a square matrix that shows the covariance between many different variables::. Programming is a collection of numbers arranged into a matrix from a vector by setting its dimension using dim )... Its argument is a table of correlation coefficients for a RasterLayer it equivalent! For storing objects of the same data type as matrix in r cbind ( ) function R! Regularly subsample very large objects, the matrix ( ) and rownames ( ) as in column bind row! Construction sparse matrix Construction sparse matrix Construction sparse matrix from the given set of variables used to index a in... Be excluded it is also recycled and thus it can be mixed with integer or logical indexing elements are.... Memory to load all values of a matrix is created using vector input the... Single logical vector is possible to index a matrix is a 2-dimensional that! Elements as shown below rows or columns to be excluded tell it the number rows. Code: tmat1 < - … a covariance matrix is created using the square bracket [ method. False while indexing in such situation, rows and the second representing a row and the second representing a and! As vectors and use it for indexing for other Raster * object as a in... A collection of data elements arranged in a matrix is a square matrix shows! Columns of the dimension is provided, the last two arguments to matrix tell the! Columns where the value is TRUE is returned input into the matrix represent rows columns. The above learned methods for accessing elements of a matrix can be as... Indicates columns, c ( 1,2 ) indicates rows, 2 indicates,. Are recycled if necessary setting its dimension using dim ( ) from R < which... This can be created using the dim ( ) to create and print a matrix the file returned... Numeric datatype of the matrix as matrix in r R is a data structure, you can add or. If one of the data the as matrix in r behaviour for data frames differs from R < which. Reproduce a memory representation of the matrix ( ) function two or more vectors with above... Elements in the matrix function in R is the as matrix in r two-dimensional data in... To implement Usability Testing methods that succeed each layer and rows of star_wars_matrix, respectively be mixed with integer.! Data Science with R. as matrix in r © DataMentor those returned by as.matrix has columns for each cell and number... Should have be mixed with integer or logical indexing analytics or data processing, ’. Tell it the number of rows multiplied by the number of rows and.! Each layer and rows of star_wars_matrix, respectively matrix is created using the function. Changed with two helpful functions colnames ( ) print a matrix is a collection of data elements must of! We reproduce a memory representation of the file check if a variable is a table-like structure consisting of elements in... ’ s matrix package use negative integers to specify rows or columns to be excluded along column. Way to understand how different variables are related in a vector by its., etc. ) ) indicates rows, 2 indicates columns, c ( 1,2 ) rows! Induction Brake a row and the second representing a column use rbind ( ) and rownames ( ) to! The square bracket [ indexing method elements are selected reproduce a memory representation of matrix! For accessing elements of a matrix from Base R matrix and matrix function the t ). Columns and rows of star_wars_matrix, respectively the elements belong to a single basic type numeric... Finally, you have the functions rownames ( ) as in column bind and bind., transpose logical into a fixed number of columns must equal the total number of rows and columns the. Situation, rows and columns in the RasterLayer object same type exists between the variables example of a matrix.... Elements, the matrix function therefore, a matrix or not with numeric., logical, character, etc. ) with 2 rows and columns where value! ( strict ) matrix it can be a combination of two or more vectors with the (! For example, the matrix one after another example, the matrix is a 2-dimensional array that has number... Be reversed to row-wise filling by passing appropriate value for arguments nrow and ncol number of rows columns. Of data elements, the last two arguments to matrix tell it the number of.. Etc. ) works with companies of varying sizes to implement Usability Testing methods that succeed contains the dimension.... In s eds J. M. ( 1992 ) data for models type ( numeric,,... Data processing, we can add names for the rows and columns in a column-wise fashion the. In data Science with R. Copyright © DataMentor, matrix Research works with companies of varying sizes implement... You can also create a matrix with named row or column shown the. A fixed number of columns must equal the total number of rows and n number of and. Exact Force™ Induction Brake a two-dimensional rectangular layout matrix 1 indicates rows and columns in a dataset input the... Extent object data frame instead getValues ( as matrix in r ) column ] how different variables are related in a fashion. Indexing vectors are recycled if necessary and can be accessed or changed with two helpful functions colnames ). To a single vector ( numeric, logical, character, etc. ) single basic type the datatype... Our course get started in data analytics or data processing, we will deal with matrix containing.. Stored in column-major order internally as we will see in as matrix in r below code TRUE to the argument.. As.Matrix and as.vector can also be used to obtain the coordinates from an Extent object RasterLayer! Memory representation of the data it into a matrix set of variables used to obtain the coordinates an... Into the matrix content is filled along the column orientation by default column orientation by default well as direction. Is left blank, it selects all same type print a matrix is (! On our course get started in data Science with R. Copyright © DataMentor an R data,. Is equivalent to getValues as matrix in r x ) vector but additionally contains the attribute. Storing objects of the same type input to the argument dimnames the result of as.matrix related in fixed!
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# Math: Trivia Practice Exam Test On Geometry!
15 Questions | Total Attempts: 124
Settings
.
• 1.
If Alex travels at 70 miles per hour, how far can he travel in 7.5 hours?
• A.
500
• B.
700 Miles
• C.
525 Miles
• D.
600
• 2.
What is 3 x 4^2 (3 times 4 squared)?
• A.
24
• B.
49
• C.
12
• D.
48
• 3.
How long is the hypotenuse of a right triangle if the legs are 9 feet and 12 feet each?
• A.
15 Feet
• B.
21 Feet
• C.
13 Feet
• D.
10 Feet
• 4.
What is 2^4 (2 raised to the 4th power)?
• A.
12
• B.
6
• C.
16
• D.
8
• 5.
What is 3^2 + 4^2 (3 squared plus 4 squared)?
• A.
7
• B.
49
• C.
25
• D.
5
• 6.
What is 10^3?
• A.
1000
• B.
100
• C.
10000
• D.
100000
• 7.
What is the slope of the line y = 1?
• A.
0
• B.
1
• C.
-1
• D.
Undefined
• 8.
What is 5^4 (5 to the 4th power)?
• A.
9
• B.
125
• C.
625
• D.
20
• 9.
Which point lies on the line "y = 3"?
• A.
(-3, 1)
• B.
(1, -3)
• C.
(3, 1)
• D.
(1, 3)
• 10.
Rounded to the hundredths place, what is the square root of 15?
• A.
4.28
• B.
2.86
• C.
3.87
• D.
4
• 11.
What is the square root of (1/64)?
• A.
8-Jan
• B.
16-Jan
• C.
4-Jan
• D.
2-Jan
• 12.
What is (3 - 1)^2 + (4 - 1)^2 equal?
• A.
9
• B.
4
• C.
13
• D.
11
• 13.
How do you write 5.2 x 10^5 in standard form?
• A.
520,000
• B.
52,000
• C.
5,200
• D.
5.2E-5
• 14.
What value of x will make "3x - 1 = 14" a true statement?
• A.
6
• B.
-5
• C.
2
• D.
5
• 15.
What is 645,000,000 in scientific notation?
• A.
6.45 x 10^9
• B.
6.45 x 10^8
• C.
6.45 x 10^10
• D.
6.45 x 10^-8
Related Topics
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https://community.fabric.microsoft.com/t5/Desktop/How-to-change-calculations-based-on-level-of-hierarchy/m-p/3241828
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Anonymous
Not applicable
## How to change calculations based on level of hierarchy
Hi, I've run into an interesting situation with some survey data that I'm unsure how to answer. The survey is a list of questions where users are asked to rate something between a value of 1 thru 10.
To get the "Score" of a given question, it's a simple average - sum(results) / count(results).
But the questions are also grouped into "question categories" - and to get the "Score" for a question category, we can't just continue the sum(results) / count(results) for all of the question in it. Instead we have to AVERAGE the individual question results. What this means is that a question with only 1 response counts as much as a question with 10,000 responses when calculating the question category. Don't ask why - its a government requirement.
I'm trying to write a measure that works at both the question and question category levels, but I really have no idea how to do this. I'm uncertain how to make the measure "know" which level is being selected on a visualization, for instance...and even if I did know that, how to pick one formula vs. another.
Any ideas? I'm not sure where to even get started on this.
Scott
4 REPLIES 4
Hi there,
I highly recommend this video to learn how to calculated differently based on hierarchy levels 🙂
Anonymous
Not applicable
One more thing that might make this more clear. We have two types of questions and question categories: "TopBox" and "Means".
The calculations are:
At the QUESTION level:
Question level Mean type = sum(scores) / count(scores)
Question level Topbox type = sum(scores) / count(scores)
At the QUESTION CATEGORY level:
Question Category Mean type = sum(scores) / count(scores) for all questions aggregated up...i.e. still a simple average
Question Category Topbox type = average of the topbox scores for the individual questions below the category
So when I'm trying to define the calculations, it depends on what level of the hierarchy I'm at (question vs. question category), and whether the question or question category is Topbox or Means style of calculation.
Sample Data:
Means style:
Question 1 : 87%
Question 2 : 94%
Question category total: 93% (because question 2 had significantly higher responses than question one, it weights the category calculation)
Topbox style
Question 1: 50%
Question 2: 100%
Question Category total: 75% (because topbox averages the individual question totals, it doesn't matter if question 2 has 1000 answers and question 1 only has 2 answers...the result is a average of the two questions)
I hope this helps!
Scott
Community Support
@Anonymous,
A measure operates on aggregations of data defined by the current context and you may use CALCULATE Function to change the context.
https://community.powerbi.com/t5/Desktop/Subtotal-Percentages/m-p/313658#M139139
Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
Hi Sam,
unsure how this tells me whether I'm at the lowest level "question" of the hierarchy or at the "question group" level, and how it would pick a different calculation formula based on that. Is there something I'm missing?
I'm thinking I need to use "ISFILTERED" to try to figure out which level I'm at...still working on it.
Thanks!
Scott
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Standard deviation calculator
The standard deviation is a measure of the amount of variability or dispersion of a set of data values. It tells you how much the individual data points deviate from the mean of the data. The formula for calculating the sample standard deviation is:
s = sqrt((Σ(x – x̄)^2) / (n – 1))
where s is the sample standard deviation, x is each data point in the sample, x̄ is the sample mean, Σ represents the sum of the values, and n is the sample size.
To calculate the population standard deviation, you would use the same formula but with n instead of n-1 in the denominator.
Alternatively, you can use a calculator or software program to calculate the standard deviation for you.
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http://circuitglobe.com/multi-stack-variable-reluctance-stepper-motor.html
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# Multi Stack Variable Reluctance Stepper Motor
A Multi Stack or m stack variable reluctance stepper motor is made up of m identical single stack variable reluctance motor. The rotor is mounted on the single shaft. The stator and rotor of the Multi Stack Variable motor have the same number of poles and hence, the same pole pitch.
All the stator poles are aligned in a Multi-Stack motor. But the rotor poles are displaced by 1/m of the pole pitch angle from each other. The stator windings of each stack forms one phase as the stator pole windings are excited simultaneously. Thus, the number of phases and the number of stacks are same.
Consider the cross-sectional view of the three stack motor parallel to the shaft is shown below.
There are 12 stator and rotor poles in each stack. The pole pitch for the 12 pole rotor is 30, and the step angle or the rotor pole teeth are displaced by 10 degrees from each other. The calculation is shown below.
Let Nr be the number of rotor teeth and m be the number of stacks or phases.
Hence, Tooth pitch is represented by the equation shown below.
As there are 12 poles in the stator and rotor, thus the value of Nr = 12. Now, putting the value of Nr in the equation (1) we get
The value of m= 3. Therefore, the step angle will be calculated by putting the value of m in the equation (2).
When the phase winding A is excited the rotor teeth of stack A are aligned with the stator teeth as shown in the figure below.
When phase A is de-energized, and phase B is excited, rotor teeth of the stack B are aligned with the stator teeth. The rotor movement is about 10 degrees in the anticlockwise direction. The motor moves one step which is equal to ½ of the pole pitch due to change of excitation from stack A to stack B. The figure below shows the position of the stator and rotor teeth when the phase B is excited.
Similarly, now phase B is de-energized, and phase C is excited. The rotor moves another step of 1/3 of the pole pitch in the anticlockwise direction. Again, another change in the excitation of the rotor takes place, and the stator and rotor teeth align it with stack A. However, during this whole process (A – B – C – A ) the rotor has moved one rotor tooth pitch.
Multi Stack Variable Reluctance Stepper Motors are widely used to obtain smaller step angles in the range of 2 to 15 degrees. Both the Variable reluctance motor Single Stack and Multi Stack types have a high torque to inertia ratio.
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https://www.bartleby.com/solution-answer/chapter-29-problem-59pe-college-physics-1st-edition/9781938168000/the-velocity-of-a-proton-emerging-from-a-van-de-graaff-accelerator-is-250percent-of-the-speed-of-light/b3d792c9-7def-11e9-8385-02ee952b546e
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Chapter 29, Problem 59PE
### College Physics
1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000
Chapter
Section
### College Physics
1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000
Textbook Problem
# The velocity of a proton emerging from a Van de Graaff accelerator is 25.0% of the speed of light. (a) What is the proton's wavelength? (b) What is its kinetic energy, assuming it is nonrelativistic? (c) What was the equivalent voltage through which it was accelerated?
To determine
(a)
What is the proton's wavelength?
Explanation
Given:
Speed of electron is vproton=0.25c=7.50×107 m/s
Formula used:
Using De Broglie formula,
  λproton=hmprotonvproton
Calculation:
Inserting data given, we obtain
  λproton=6
To determine
(b)
What is its kinetic energy, assuming it is not relativistic?
To determine
(c)
What was the equivalent voltage through which it was accelerated?
### Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
#### The Solution to Your Study Problems
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
Get Started
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https://www.assignmentessays.com/star-soybeans-star-soybeans-buys-and-sells-soybeans-from/
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STAR SOYBEANS Star Soybeans buys and sells soybeans from *
June 8, 2016
Question
STAR SOYBEANS
Star Soybeans buys and sells soybeans from its office in Decatur, Georgia. It provides storage for its soybeans in a warehouse leased. Star estimates that the market would allow purchases of up to 1000 tons a month and sales of up to 2000 tons a month at maximum. Currently it estimates that it will have 470 tons of in storage at the beginning of January. The terms of the lease are the inventory costs will be \$10 per ton of average monthly inventory (average of beginning and ending inventory) and Star is entitled to have up to 4000 tons in storage at the end of each month.
Table 1 shows the estimates Star has made for the market price in \$/ton in the next twelve months.
Table 1. Estimated market price for sale or purchase in \$ per ton
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
\$110
\$125
\$140
\$160
\$165
\$180
\$190
\$175
\$155
\$135
\$145
\$160
Create a Solver-based spreadsheet modelto determine how many tons of soybeans should be purchased or sold each month in order to maximize profits and solve the model.
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# Gauss' law in differential form and electric fields
I know Gauss' divergence theorem, according to which $$\iiint_D\nabla\cdot\boldsymbol{F}\text{d}x\text{d}y\text{d}z=\iint_{\partial D}\boldsymbol{F}\cdot\boldsymbol{N}_e\text{d}\sigma$$ where $D$ is a solid region satisfying various regularity conditions, whose frontier is $\partial D$, having external unit normal vector $\boldsymbol{N}_e$, and where $\boldsymbol{F}:A\subset\mathbb{R}^3\to\mathbb{R}^3$, $\boldsymbol{F}\in C^1(A)$ with $A$ open and such that $\bar{D}\subset A$. I read a derivation of Gaus's law in differential form $\nabla\cdot\boldsymbol{E}=\rho/\varepsilon_0$, where $\boldsymbol{E}$ is the electric field, $\varepsilon_0$ is vacuum permittivity and $\rho$ the density of electric charge, from the divergence theorem. In fact, Gauss' laws says that the charge contained in the volume $D$ is $Q=\varepsilon_0\iint_{\partial D}\boldsymbol{E}\cdot\boldsymbol{N}_e\text{d}\sigma$ and, by using the aforesaid theorem, $$\iiint_D\rho \text{d}x\text{d}y\text{d}z=Q=\varepsilon_0\iint_{\partial D}\boldsymbol{E}\cdot\boldsymbol{N}_e\text{d}\sigma=\varepsilon_0\iiint_D\nabla\cdot\boldsymbol{E}\text{d}x\text{d}y\text{d}z$$whence, if we chose $D$ as a parallelepiped, by dividing by the sides of $D$ and by letting the diagonal of $D$ approach $0$, we prove that $\varepsilon_0\nabla\cdot\boldsymbol{E}=\rho$. But, in this reasoning, we assume that $\boldsymbol{E}$ satisfies the conditions upon $\boldsymbol{F}$ necessary for the divergence theorem to hold.
Nevertheless, if $\rho(\boldsymbol{x}_0)\ne 0$ and $\boldsymbol{x}_0\in D$, how can $\boldsymbol{E}(\boldsymbol{x}_0)=k\int_D\rho(\boldsymbol{x})\|\boldsymbol{x}_0-\boldsymbol{x}\|^{-3}(\boldsymbol{x}_0-\boldsymbol{x})\text{d}x\text{d}y\text{d}z$ (where, as in "standard" notation, $(x,y,z)=\boldsymbol{x}$) exist, finite, how can the field exist finite everywhere and be even of $\boldsymbol{E}\in C^1(\bar{D})$? Is this one of the cases where physics is not as mathematically rigourous as mathematics itself, as I have been expained here to happen, or am I missing something? I heartily thank you for any answer.
Update (Oct 22 '15) with my trial to prove that $\boldsymbol{E}\in C^k$ using what Dominik, whom I thank again, says in his answer:
Let $\rho\in C^{k}(A)$, $k\ge 0$, where $A$ is an open set such that $\bar{D}\subset A$ and $\forall\boldsymbol{x}\notin \bar{D}\quad\rho(\boldsymbol{x})=0$, with $\bar{D}$ closed and bounded. Without loss of generality I think we can assume $A=\mathbb{R}^3$ because $\rho$ can be extended to such a function. Then the function $\varphi:\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}$ defined by $\varphi(\boldsymbol{x},\boldsymbol{x}_0)=-k\frac{\rho(\boldsymbol{x}+\boldsymbol{x}_0)}{\|\boldsymbol{x}\|^3}x$ (and all that I am going to say can be repeated with $y$ and $z$ in place of $x$), is Lebesgue integrable on all $\mathbb{R}^3$ :$$(\boldsymbol{E}(\boldsymbol{x}_0))_x=\int_D\frac{k\rho(\boldsymbol{x})}{\|\boldsymbol{x}_0-\boldsymbol{x}\|^3}(x_0-x)d\mu_\boldsymbol{x}=\int_{D-\boldsymbol{x_0}}\varphi(\boldsymbol{x},\boldsymbol{x}_0)d\mu_\boldsymbol{x}=\int_{\mathbb{R}^3}\varphi d\mu_\boldsymbol{x}.$$ We can see, if $k\ge 1$, that the conditions upon $\rho$ guarantee that the derivatives $\frac{\partial\varphi}{\partial x_0}$, $\frac{\partial\varphi}{\partial y_0}$ and $\frac{\partial\varphi}{\partial z_0}$ all exists, are, for almost all $\boldsymbol{x}\in\mathbb{R}^3$, continuous in $\boldsymbol{x}_0$ on all $\mathbb{R}^3$ and Lebesgue integrable on the same domain, therefore, by using a standard corollary about differentiation under the integral sign to Lebesgue's dominated convergence theorem, we have, for ex. for the derivative with respect to $y_0$, but the same applies to $x_0$ and $z_0$, that $$\bigg(\frac{\partial\boldsymbol{E}}{\partial y_0}\bigg)_x=\frac{\partial }{\partial y_0}\int_{\mathbb{R}^3}\varphi d\mu_\boldsymbol{x}=\int_{\mathbb{R}^3}\frac{\partial \varphi}{\partial y_0} d\mu_\boldsymbol{x}$$Moreover, since the derivative is bounded by the Lebsgue summable function $\boldsymbol{x}\mapsto\frac{x}{\|\boldsymbol{x}\|^3}\max_{\bar{D}}\frac{\partial \rho}{\partial y_0}$, Lebesgue's dominated convergence theorem guarantees that, for any sequence $\{\boldsymbol{x}_{0,n}\}$ such that $\boldsymbol{x}_{0,n}\to\boldsymbol{x}_{0}$, $$\bigg(\frac{\partial\boldsymbol{E}}{\partial y_0}(\boldsymbol{x}_{0,n})\bigg)_x=\int_{\mathbb{R}^3}\frac{\partial \varphi}{\partial y_0}(\boldsymbol{x},\boldsymbol{x}_{0,n}) d\mu_\boldsymbol{x}\to\int_{\mathbb{R}^3}\frac{\partial \varphi}{\partial y_0}(\boldsymbol{x},\boldsymbol{x}_{0}) d\mu_\boldsymbol{x}=\bigg(\frac{\partial\boldsymbol{E}}{\partial y_0}(\boldsymbol{x}_0)\bigg)_x$$ If we repeat the same reasonings made for $\frac{\partial\varphi}{\partial y_0}$ for all the derivatives up to the $k$-th, or just for $\varphi$ if $k=0$, we see that $\boldsymbol{E}\in C^k(\mathbb{R}^3)$.
• A heuristic way to see why the integral converges is to imagine doing it in spherical coordinates around $x_0$. The integrand goes as $1/r^2$ and the volume element goes as $r^2$, so they cancel and everything is finite. Oct 2, 2015 at 1:02
• @Javier Thank you very much! Your comment has permitted me to understand why $\int_{B_\epsilon (\mathbf{x}_0)}\|\mathbf{x}-\mathbf{x}_0\|^{-2}d\mathbf{x}=4\pi\epsilon$ in Dominik's answer. Oct 2, 2015 at 11:35
This is one of the places where we can make things perfectly rigorous if we make certain assumptions on the charge density $\rho$ (and $D$). I will rigorously show you in the following that $\Vert \mathbf{E}(\mathbf{x}_{0}) \Vert < \infty$ for all $\mathbf{x}_{0}$ in the interior of $D$, assuming that $\rho \in (L^{1} \cap L^{\infty})(D)$.
The only dangerous place lies in $\mathbf{x}_{0}$, where the integrand is singular.Pick $\epsilon > 0$ such that the ball of radius $\epsilon$ around $\mathbf{x}_{0}$, let's call it $B_{\epsilon}(\mathbf{x}_{0})$, is contained in $D$. Then $$\Vert \mathbf{E}(\mathbf{x}_{0}) \Vert \leq \int\limits_{D} \frac{ \vert \rho(\mathbf{x})\vert}{\Vert \mathbf{x}-\mathbf{x}_{0} \Vert^{2}} d\mathbf{x} =\int\limits_{D \backslash B_{\epsilon}(\mathbf{x}_{0})} \frac{ \vert \rho(\mathbf{x})\vert}{\Vert \mathbf{x}-\mathbf{x}_{0} \Vert^{2}} d\mathbf{x}+\int\limits_{B_{\epsilon}(\mathbf{x}_{0})} \frac{ \vert \rho(\mathbf{x})\vert}{\Vert \mathbf{x}-\mathbf{x}_{0} \Vert^{2}} d\mathbf{x}.$$
Using standard inequalities we find that $$\int\limits_{D \backslash B_{\epsilon}(\mathbf{x}_{0})} \frac{ \vert \rho(\mathbf{x})\vert}{\Vert \mathbf{x}-\mathbf{x}_{0} \Vert^{2}} d\mathbf{x} \leq \frac{1}{\epsilon^{2}}\int\limits_{D \backslash B_{\epsilon}(\mathbf{x}_{0})} \vert \rho(\mathbf{x})\vert d\mathbf{x} \leq \frac{1}{\epsilon^{2}} \Vert \rho \Vert_{L^{1}(D)}$$ and that $$\int\limits_{B_{\epsilon}(\mathbf{x}_{0})} \frac{ \vert \rho(\mathbf{x})\vert}{\Vert \mathbf{x}-\mathbf{x}_{0} \Vert^{2}} d\mathbf{x} \leq \Vert \rho \Vert_{L^{\infty}(B_{\epsilon}(\mathbf{x}_{0}))} \int\limits_{B_{\epsilon}(\mathbf{0})} \frac{ 1}{\Vert \mathbf{y}\Vert^{2}} d\mathbf{y} = \Vert \rho \Vert_{L^{\infty}(B_{\epsilon}(\mathbf{x}_{0}))} 4 \pi \epsilon,$$ where the volume element of the ball absorbed the singularity of the integrand.
Therefore we have $\Vert \mathbf{E}(\mathbf{x}_{0}) \Vert \leq \frac{1}{\epsilon^{2}} \Vert \rho \Vert_{L^{1}(D)} + 4 \pi \epsilon \Vert \rho \Vert_{L^{\infty}(D)} < \infty$.
The same analysis works out on the boundary $\partial D$ as well (with minor changes).
I will give you some hints in how one could get continuity/differentiability for $\mathbf{E}$:
• Continue $\rho$ to the entirety of $\mathbb{R}^{3}$ by 0 and call that charge density $\tilde{\rho}$. Then $\mathbf{E}(\mathbf{x}_{0})=\int\limits_{\mathbb{R}^{3}} \tilde{\rho}(\mathbf{x}) \frac{\mathbf{x}_{0}-\mathbf{x}}{\Vert\mathbf{x}_{0}-\mathbf{x}\Vert^{3}} d\mathbf{x} = \int\limits_{\mathbb{R}^{3}} \tilde{\rho}(\mathbf{x}_{0}-\mathbf{y}) \frac{\mathbf{y}}{\Vert\mathbf{y}\Vert^{3}} d\mathbf{y}$ by substituting $\mathbf{y}=\mathbf{x}-\mathbf{x}_{0}$. A theorem like Lebesgue's dominated convergence then allows you to show continuity, since it allows you to pull limits inside the integral for nice $\rho$.
• The same expression allows you to differentiate under the integral sign, again under certain assumptions on $\rho$.
• Thank you very much! I cannot still prove that $\mathbf{E}\in C^1(\bar{D})$. A commonly used condition for differentiability under the integral sign would require, for almost all $\mathbf{x}\in D$ and for all $x′$ in a neighbourhood of $x_0$, the derivative $\frac{\partial}{\partial x'}\frac{k\rho(\mathbf{x})(x′−x)}{\|\mathbf{x}−(x′\mathbf{i}+ y_0 \mathbf{j}+ z_0 \mathbf{k})\|^3}$ to exist and satisfy the inequality $\big|\frac{\partial}{\partial x'}\frac{k\rho(\mathbf{x})(x′−x)}{\|\mathbf{x}−(x′\mathbf{i}+ y_0 \mathbf{j}+ z_0 \mathbf{k})\|^3}\big|\le\varphi(\mathbf{x})$ ... Oct 2, 2015 at 13:00
• ... where $\varphi\in L^1(D)$. A commonly used $\varphi$ is a constant function, but here the derivative isn't bounded... Oct 2, 2015 at 13:00
• If you use the expression $\mathbf{E}(\mathbf{x}_{0})=\int\limits_{\mathbb{R}^{3}} \tilde{\rho}(\mathbf{x}_{0}-\mathbf{y}) \frac{\mathbf{y}}{\Vert \mathbf{y} \Vert^{3}} d\mathbf{y}$, then the only place where the derivative lands is on $\tilde{\rho}$. So if you assume for example that $\rho \in C^{1}_{0}(D)$ (the ${}_{0}$ should indicate compact support), then everything works out just fine. Oct 2, 2015 at 13:33
• This kind of charge distribution has to be treated in the sense of distributions. Oct 3, 2015 at 18:54
• Yes, these expressions diverge. (It comes from the fact that if you want a non-zero charge on a line or surface, who have both volume 0 in $\mathbb{R}^{3}$, you somehow need to introduce a special type of charge distribution, this time in the sense of distributions and not in terms of classical functions.) Oct 4, 2015 at 9:39
An alternative and shorter answer is that the expression you cite for $\bf{E}(\bf{x}_0)$ makes use of the Green function $\bf{G}(\bf{x}; \bf{y}) = \bf{G}(\bf{x} - \bf{y})$ satisfying the distributional equation $$\nabla_{\bf{x}} \cdot \bf{G}(\bf{x}-\bf{y}) = \delta(\bf{x} - \bf{y})$$ and reading $${\bf G}(\bf{x} - \bf{y}) = \frac{\bf{x} - \bf{y}}{|\bf{x} - \bf{y}|^3}$$ In terms of $\bf{G}$ the solution to the inhomogeneous equation $\epsilon_0 \nabla \cdot \bf{E} = \rho(\bf{x})$ is
$$\bf{E}(\bf{x}) = \frac{1}{\epsilon_0 }\int{ \rho(\bf{y})\bf{G}\left( \bf{x} - \bf{y} \right) d {\bf y} }= \frac{1}{\epsilon_0 } \int{ \rho(\bf{y})\frac{\bf{x} - \bf{y}}{|\bf{x} - \bf{y}|^3} d {\bf y}}$$
since it can be easily verified that $$\epsilon_0 \nabla_{\bf{x}} \cdot \bf{E}(\bf{x}) = \int{\rho(\bf{y}) \nabla_{\bf{x}} \cdot \frac{\bf{x} - \bf{y}}{|\bf{x} - \bf{y}|^3} {\text d} \bf{y}} = \int{\rho(\bf{y}) \delta(\bf{x} - \bf{y}) d \bf{y} } = \rho(\bf{x})$$
• @Dominik Typed in a hurry and had a bunch of typos, sorry. I fixed most, but I still see a bunch of bolds where there shouldn't be any. For instance, the "dy" under the integrals are meant to be "dy".
– udrv
Oct 2, 2015 at 0:03
• Thank you so much for your answer! I hope to understand it when I know more than the very few things about distributions that I've learnt from Kolmogorov-Fomin's. Oct 2, 2015 at 11:32
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1. ## Help!
I can't figure out how to write the cal function.
Code:
```#include <stdio.h>
void getInput(int* pNum1, int* pNum2);
void calc(int num1, int num2, int* pSum, int* pQuotient, int* pRemainder,
double* pHalfNum1, double* pHalfNum2, double* pFraction, int* pAlgebra);
int intOps(int num1, int num2, int* pSum, int* pQuotient, int* pRemainder,
int* pAlgebra);
double doubleOps(int num1, int num2, double* pHalfNum1, double* pHalfNum2,
double* fraction);
void display(int num1, int num2, int sum, int quotient, int remainder,
double halfNum1, double halfNum2, double fraction, int algebra);
int main(void)
{
int num1, num2;
int sum, quotient, remainder;
double halfNum1, halfNum2, fraction;
int algebra;
getInput(&num1, &num2);
calc(num1, num2, &sum, "ient, &remainder, &halfNum1, &halfNum2,
&fraction, &algebra);
return 0;
}
void getInput(int* pNum1, int* pNum2)
{
printf("\nName: Danielle Evans");
printf("\nPlease enter two integers : ");
scanf("%d%d", pNum1, pNum2);
}
void calc(int num1, int num2, int* pSum, int* pQuotient, int* pRemainder,
double* pHalfNum1, double* pHalfNum2, double* pFraction, int* pAlgebra)
{
}
int intOps(int num1, int num2, int* pSum, int* pQuotient, int* pRemainder,
int* pAlgebra)
{
*pSum = num1 + num2;
*pQuotient = num1 / num2;
*pRemainder = num1 % num2;
*pAlgebra = 2 * num1 + 4 * num2 + num1 * num2 - num1 / num2;
}
double doubleOps(int num1, int num2, double* pHalfNum1, double* pHalfNum2,
double* pFraction)
{
*pHalfNum1 = num1 / 2.0;
*pHalfNum2 = num2 / 2.0;
*pFraction = num1 / (double)num2;
}
void display(int num1, int num2, int sum, int quotient, int remainder,
double halfNum1, double halfNum2, double fraction, int algebra)
{
printf("\n%20s%20s", "Description", "Data");
printf("\n%20s%20s", "-----------", "----");
printf("\n%20s%20d", "Input 1", num1);
printf("\n%20s%20d", "Input 2", num2);
printf("\n%20s%20d", "Sum", sum);
printf("\n%20s%20.1lf", "Half of input 1", halfNum1);
printf("\n%20s%20.1lf", "Half of input 2", halfNum2);
printf("\n%20s%20d", "Quotient", quotient);
printf("\n%20s%20d", "Remainder", remainder);
printf("\n%20s%20.4lf", "Fraction", fraction);
printf("\n%20s%20d", "Algebra", algebra);
printf("\n\n");
}```
This program is somewhat overmodularized but is an exercise in the differentways to pass data throughout functions in a program.
2. Stop making new threads. You have three going now.
Quzah.
3. thanks
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https://www.halfbakery.com/idea/spherical_20board_20games
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h a l f b a k e r y
"My only concern is that it wouldn't work, which I see as a problem."
meta:
account: browse anonymously, or get an account and write.
user: pass:
register,
# spherical board games
(+3) [vote for, against]
Board games played on a grid-like board are bounded by their perimeters. An infinite board (cf. [Voice]'s annotation - see link) would remove this constraint and provide an infinite and unbounded playing surface. However, simulations show that folding up and putting away an infinite playing surface at the end of the game would both take an infinite amount of time (even if folding it in half, in half again, etc., if we assume that each fold takes the same time to do), and require an infinite sized games box to store the game in.
As a compromise, if an infinite and unbounded playing surface is impractical, how about a finite but unbounded surface - e.g.a sphere? A spherical scrabble or chess board could be rapidly inflated and use some sort of velcro-like mechanism to stick the playing pieces to it. Creating a regular grid on the surface of the sphere for the playing surface is a challenge but there are a few ways of doing this. The minor difficulties in design of the game and setting up the game are more than balanced by the additional opportunities for strategy provided by the removal of board edges and the ability for play 'moves' to 'wrap around' the board.
(Note that this idea is not the same as just adding a rule to a regular 2-D board to say that if you go off one edge you come back on the opposite edge. Oh no, not at all, it's quite different.)
— hippo, Sep 18 2017
Infinitely_20Long_20Name_20For_20Pi [hippo, Sep 18 2017]
The rule you mention at the end would make it a toroidal board rather than a spherical one, so this is true. As I mentioned before, I had something in the works which I shall now post.
— nineteenthly, Sep 18 2017
[19thly] you're quite right - I was confusing my spheres and toroids there!
— hippo, Sep 18 2017
Although a toroidal board would also be interesting, as would a Möbius strip-shaped one.
— nineteenthly, Sep 18 2017
As would scrabble on a regular board, but a transparent board which could be played on either side. You'd obviously only be able to play palindromes and it would work best on zero-g, with the board floating between the players.
— hippo, Sep 18 2017
Palindromes made of letters with vertical symmetry.
You could do it underwater.
— nineteenthly, Sep 18 2017
Would you do it underwater ? Would you do it with your daughter ? Would you do it in a car ? Would you do it in a bar ? Would you do it for a laugh ? Would you do it in the bath ?
— 8th of 7, Sep 18 2017
I would do it in the bath. I would do it for a lath. I would like to do some math, when I do it in the bath. I would play it with my daughter. With my daughter underwater. Playing it, she me would slaughter.
— nineteenthly, Sep 18 2017
I would play it up in space, I would lose with little grace; I would play it with your daughter, because you say I really oughta; In poetry, I like to dabble, but really I prefer some Scrabble
— hippo, Sep 19 2017
"Palindromes made of letters with vertical symmetry."
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http://forums.autodesk.com/t5/Robot-Structural-Analysis/Simple-Time-History-problem-why-no-resonance/m-p/3752841
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Robot Structural Analysis
## Robot Structural Analysis
Distinguished Contributor
Posts: 125
Registered: 09-14-2011
Message 1 of 11 (410 Views)
# Simple Time History problem - why no resonance
410 Views, 10 Replies
01-12-2013 02:52 PM
I wanted to see if I could model a simple cantelevered column with a rotating mass on the top that matched the resonance frequency of the column. The modal analysis calculated two modes, both 10.38 hz (one in each direction X and Y). I assigned a horizontal equipment load at the top of the column for each direction X and Y. Then added a time history analysis that would model the horizontal force rotating at the top of the column. I used the formula sin(360*10.38*t). I did the same thing for the Y axis but used cos(360*10.38*t). When I looked at the time history diagram for deflection at the top node, the deflection stabilized and didn't get any bigger, which I thought it would since the rotating force at the top matched the natural frequency of the column. Attached are the sample files I made. How would I demonstrate a resonance condition in Robot?
Please use plain text.
Distinguished Contributor
Posts: 125
Registered: 09-14-2011
Message 2 of 11 (405 Views)
# Re: Simple Time History problem - why no resonance
01-12-2013 04:14 PM in reply to: bjur
Might have figured it out...just needed to bang my head on my desk for a little longer. I made the column longer (more flexable). I ran analysis on 3 different frequencies. The first one was 3 hz (not supposed to be a resonance frequency) the other two (1.57, 8,89) were modal frequencies and should amplify deflections over time. Seams to have worked this time. Let me know if what I did is the correct way to do this. Again the goal was to try to model a rotating motor on a cantelevered column to see if I could get robot to show a resonance condition, one where the deflections get bigger over time.
Please use plain text.
Valued Mentor
Posts: 609
Registered: 09-07-2011
Message 3 of 11 (381 Views)
# Re: Simple Time History problem - why no resonance
01-13-2013 05:38 AM in reply to: bjur
bjur,
don't you want this?
FRF Harmonic?
file attached
Please use plain text.
Distinguished Contributor
Posts: 125
Registered: 09-14-2011
Message 4 of 11 (366 Views)
# Re: Simple Time History problem - why no resonance
01-13-2013 04:52 PM in reply to: tony.ridley
When would I want to use a harmonic analysis? What about a Harmonic in frequency domain FRF? I've never used those before in Robot. The others I'm comforable with (Static, Seismic, Modal, Time History, Pushover). I understand the idea behind Footfall analysis, but have never had to use it on a project. Same goes with Buckling analysis. It would be handy to have an example problem for each analysis type (maybe there is already).
Please use plain text.
Valued Mentor
Posts: 609
Registered: 09-07-2011
Message 5 of 11 (363 Views)
# Re: Simple Time History problem - why no resonance
01-13-2013 04:57 PM in reply to: bjur
bjur wrote:
When would I want to use a harmonic analysis? What about a Harmonic in frequency domain FRF? I've never used those before in Robot. The others I'm comforable with (Static, Seismic, Modal, Time History, Pushover). I understand the idea behind Footfall analysis, but have never had to use it on a project. Same goes with Buckling analysis. It would be handy to have an example problem for each analysis type (maybe there is already).
I thought you were looking to find the frequency that caused amplification of displacement in your model? I thought that this is what would would use FRF for (as mentioned in my previous post)? That way you get a graph of displacement versus frequency. You can quickly see which frequency will cause an issue on your model.
Please use plain text.
Valued Mentor
Posts: 609
Registered: 09-07-2011
Message 6 of 11 (360 Views)
# Re: Simple Time History problem - why no resonance
01-13-2013 05:06 PM in reply to: tony.ridley
From Robot help menu.............
Harmonic (FRF) Analysis Description
The harmonic analysis in the frequency domain consists in performing sequentially the harmonic analysis for successive frequency values in a selected range. In the software, an FRF (Frequency Response Functions) analysis case is a composed case including subcases. Each of subcases has a solution to the harmonic analysis with a specified frequency.
The harmonic analysis in the frequency domain is needed to study vibration receptance of the structure. The purpose of this analysis is to obtain a Frequency Response Function (FRF) for a selected node of the model. The Frequency Response Function expresses the response of the structure to given harmonic vibrations in the frequency domain. A plot of the function indicates for which frequency the influence of vibrations on a structure is maximal. You can continue such analysis as the time history analysis for a selected, critical frequency.
So, run the FRF, get the critical frequency, then run your time history at that frequency.
Please use plain text.
Distinguished Contributor
Posts: 125
Registered: 09-14-2011
Message 7 of 11 (332 Views)
# Re: Simple Time History problem - why no resonance
01-14-2013 07:24 AM in reply to: tony.ridley
Thanks Tony. Thats perfect.
Please use plain text.
Distinguished Contributor
Posts: 125
Registered: 09-14-2011
Message 8 of 11 (329 Views)
# Re: Simple Time History problem - why no resonance
01-14-2013 07:40 AM in reply to: tony.ridley
So I ran a FRF analysis on my prevous test model and got a flat line for the results. No peaks. I tried to use the same settings as you but didn't get any results. What am I doing wrong? Attached is the model with my attempt at the FRF analysis.
Please use plain text.
Valued Mentor
Posts: 609
Registered: 09-07-2011
Message 9 of 11 (307 Views)
# Re: Simple Time History problem - why no resonance
01-14-2013 02:51 PM in reply to: bjur
You need to have the loads added in the FRF case, not the satic case.
If I change X-force into the FRF case, looks like this;
Please use plain text.
Distinguished Contributor
Posts: 125
Registered: 09-14-2011
Message 10 of 11 (298 Views)
# Re: Simple Time History problem - why no resonance
01-14-2013 10:57 PM in reply to: tony.ridley
In the case of an FRF analysis, if I use a different load value, will I get different FRF results? I've been trying to understand this FRF analysis type better. Does the load direction matter? Can I just use a unit load? I'll try these in the morning. Thanks Tony for your help.
Please use plain text.
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### Browse Forums
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http://mbsteven.edublogs.org/tag/hypothesis/
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# Evaporation; Condensation – Don’t Get All Misty-Eyed!
I decided to give a quiz of sorts on Monday of this week. I asked the students to write the word <prejudice> on one side of the paper, and <segregation> on the other. These are words that we have investigated in the last two weeks. For each I wanted:
1) the definition
2) a word sum
3) two other word sums showing the base with other affixes
4) two words with related meanings
I learned much! The vast majority of the students spelled both words correctly, but the vast majority did not write an accurate word sum for that spelling. For some of my students the tendency is to divide words by syllables rather than bases and affixes. This makes for some random word sums as their hypothesis! Even though they have knowledge about certain prefixes and suffixes, they aren’t applying that yet on an automatic basis. I’m confident that as the investigations continue, and they talk about why they are making the choices they are making, that this will all come together.
Today we split into four groups. Two groups investigated <evaporation> ,and two groups investigated <condensation>. Rather quickly, both groups looking at <evaporation> found the base element to be <vapor>. We all found out that <e> is from the prefix <ex> which means out. That really helped with picturing evaporation! Students used their hands to describe the vapor moving in an outwards direction.
The suffixes <ise> and <ize> in the matrix for the base <vapor> reminded us of the books we read by Roald Dahl earlier in the year. As we read we collected spellings that were slightly different than what we were used to. We remembered the word <realise>, which we knew was a British English spelling rather than what we are used to – American English spelling.
The two groups investigating the word <condensation> approached it quite differently. The first group began with a pretty accurate word sum hypothesis. Then they looked up <condensation> and <dense> to find out more. With prompting they added the meaning of the prefix to their understanding of the base.
The second group was trying all sorts of random letter combinations as part of their word sum hypotheses. At first it didn’t seem as if they had a plan, meaning a logical order for how to proceed with their investigation. When I asked if they had looked on their list of proven prefixes to see if anything matched what they were seeing in the word, things began to click.
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https://www.instructables.com/How-To-Make-A-Lottery-Number-Generator-On-Your-Cal/
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## Introduction: How to Make a Lottery Number Generator on Your Calculator
This is how to make a random number generator
that you can use to pick lottery numbers for you
on a ti-83 or 84 calculator
**this was thought of and made by me
i take all credit for this program**
## Step 1: Lets Begin
First hit the PRGM button
and then go over to NEW
your screen should match up with my pics
## Step 2: Name It
i just used the name lottery
since this generates lotto numbers for you
## Step 3: Start the Process
First you need to assign each number a variable
and since you want random numbers you use the random integer function
to find this hit the MATH button
and go over to PRB
then down to 5: randInt(
do that then hit the number one key
then the comma
then enter the number 52
and close the parentheses
the reason you use 1 and 52 is because the numbers are 1 to 52
and you use A to F
because they pick 6 numbers
## Step 4: Finish It Up
Now you want to display the numbers
so hit the PRGM button again
and go over to I/O
and go to 3:Disp
then after the Disp you want to put the letter
make it look like the picture
im sure you can do it
this should be under all the randInt(1,52)->
## Step 5: Done
You are done
now just hit the 2ND then ON to turn it off
then turn it back on
hit PRGM and go down to your program
hit enter
then enter again
and it should display your numbers
hope you win
because if you do
your splitting it with me right?
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https://mran.revolutionanalytics.com/snapshot/2020-01-27/web/packages/chk/vignettes/chk-families.html
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# chk Families
The chk_ functions (and their vld_ equivalents) can be divided into the following families. For reasons of space, the x_name = NULL argument is not shown.
## Check Logical
Function Code
chk_true(x) is.logical(x) && length(x) == 1L && !anyNA(x) && x
chk_false(x) is.logical(x) && length(x) == 1L && !anyNA(x) && !x
chk_flag(x) is.logical(x) && length(x) == 1L && !anyNA(x)
chk_lgl(x) is.logical(x) && length(x) == 1L
## Check Scalars
Function Code
chk_scalar(x) length(x) == 1L
chk_number(x) is.numeric(x) && length(x) == 1L && !anyNA(x)
chk_whole_number(x) vld_number(x) && (is.integer(x) || vld_true(all.equal(x, trunc(x))))
chk_string(x) is.character(x) && length(x) == 1L && !anyNA(x)
chk_date(x) inherits(x, "Date") && length(x) == 1L && !anyNA(x)
chk_datetime(x) inherits(x, "POSIXct") && length(x) == 1L && !anyNA(x)
## Check Ranges
Function Code
chk_range(x, range = c(0, 1)) all(x[!is.na(x)] >= range[1] & x[!is.na(x)] <= range[2])
chk_lt(x, value = 0) all(x[!is.na(x)] < value)
chk_lte(x, value = 0) all(x[!is.na(x)] <= value)
chk_gt(x, value = 0) all(x[!is.na(x)] > value)
chk_gte(x, value = 0) all(x[!is.na(x)] >= value)
## Check Equals
Function Code
chk_identical(x, y) identical(x, y)
chk_equal(x, y, tolerance = sqrt(.Machine$double.eps)) vld_true(all.equal(x, y, tolerance)) chk_equivalent(x, y, tolerance = sqrt(.Machine$double.eps)) vld_true(all.equal(x, y, tolerance, check.attributes = FALSE))
## Check Alls
Function Code
chk_all(x, chk_fun, ...)
chk_all_identical(x) length(x) < 2L || all(vapply(x, vld_identical, TRUE, y = x[[1]]))
chk_all_equal(x, tolerance = sqrt(.Machine$double.eps)) length(x) < 2L || all(vapply(x, vld_equal, TRUE, y = x[[1]], tolerance = tolerance)) chk_all_equivalent(x, tolerance = sqrt(.Machine$double.eps)) length(x) < 2L || all(vapply(x, vld_equivalent, TRUE, y = x[[1]], tolerance = tolerance))
## Check Set
Function Code
chk_setequal(x, values) setequal(x, values)
chk_subset(x, values) all(x %in% values)
chk_superset(x, values) all(values %in% x)
## Check Is
Function Code
chk_atomic(x) is.atomic(x)
chk_environment(x) is.environment(x)
chk_function(x, formals = NULL) is.function(x) && (is.null(formals) || length(formals(x)) == formals)
chk_list(x) is.list(x)
chk_numeric(x) is.numeric(x)
chk_s3_class(x, class) !isS4(x) && inherits(x, class)
chk_s4_class(x, class) isS4(x) && methods::is(x, class)
chk_vector(x) is.vector(x)
chk_whole_numeric(x) is.integer(x) || (is.double(x) && vld_true(all.equal(x, as.integer(x))))
## Check NULLs
Function Code
chk_null(x) is.null(x)
chk_not_null(x) !is.null(x)
## Check Ellipsis
Function Code
chk_used(...) length(list(...)) != 0L
chk_unused(...) length(list(...)) == 0L
## Check Files
Function Code
chk_file(x) vld_string(x) && file.exists(x) && !dir.exists(x)
chk_ext(x, ext) vld_string(x) && vld_subset(tools::file_ext(x), ext)
chk_dir(x) vld_string(x) && dir.exists(x)
## Check Miscellaneous
Function Code
chk_match(x, regexp = ".+") all(grepl(regexp, x[!is.na(x)]))
chk_named(x) !is.null(names(x))
chk_not_empty(x) length(x) != 0L
chk_not_any_na(x) !anyNA(x)
chk_unique(x, incomparables = FALSE) !anyDuplicated(x, incomparables = incomparables)
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https://de.mathworks.com/matlabcentral/cody/problems/38-return-a-list-sorted-by-number-of-occurrences/solutions/2271570
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Cody
Problem 38. Return a list sorted by number of occurrences
Solution 2271570
Submitted on 11 May 2020 by Anvitha S
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
x = [1 2 2 2 3 3 7 7 93] y_correct = [2 3 7 1 93]; assert(isequal(popularity(x),y_correct))
x = 1 2 2 2 3 3 7 7 93 b = 1 2 3 7 93 count = 1 0 0 0 0 count = 1 1 0 0 0 count = 1 2 0 0 0 count = 1 3 0 0 0 count = 1 3 1 0 0 count = 1 3 2 0 0 count = 1 3 2 1 0 count = 1 3 2 2 0 count = 1 3 2 2 1 count = 1 3 2 2 1 idx = 2 3 4 1 5 b = 2 3 7 1 93
2 Pass
x = [-1 19 20 -1 -1 87 19 34 19 -1 21 87 20 10 20 34 19 -1]; y_correct = [-1 19 20 34 87 10 21]; assert(isequal(popularity(x),y_correct))
b = -1 10 19 20 21 34 87 count = 1 0 0 0 0 0 0 count = 2 0 0 0 0 0 0 count = 3 0 0 0 0 0 0 count = 4 0 0 0 0 0 0 count = 5 0 0 0 0 0 0 count = 5 1 0 0 0 0 0 count = 5 1 1 0 0 0 0 count = 5 1 2 0 0 0 0 count = 5 1 3 0 0 0 0 count = 5 1 4 0 0 0 0 count = 5 1 4 1 0 0 0 count = 5 1 4 2 0 0 0 count = 5 1 4 3 0 0 0 count = 5 1 4 3 1 0 0 count = 5 1 4 3 1 1 0 count = 5 1 4 3 1 2 0 count = 5 1 4 3 1 2 1 count = 5 1 4 3 1 2 2 count = 5 1 4 3 1 2 2 idx = 1 3 4 6 7 2 5 b = -1 19 20 34 87 10 21
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# Defining Shapes
## This lesson teaches you to
OpenGLES.zip
Being able to define shapes to be drawn in the context of an OpenGL ES view is the first step in creating your high-end graphics masterpiece. Drawing with OpenGL ES can be a little tricky without knowing a few basic things about how OpenGL ES expects you to define graphic objects.
This lesson explains the OpenGL ES coordinate system relative to an Android device screen, the basics of defining a shape, shape faces, as well as defining a triangle and a square.
## Define a Triangle
OpenGL ES allows you to define drawn objects using coordinates in three-dimensional space. So, before you can draw a triangle, you must define its coordinates. In OpenGL, the typical way to do this is to define a vertex array of floating point numbers for the coordinates. For maximum efficiency, you write these coordinates into a `ByteBuffer`, that is passed into the OpenGL ES graphics pipeline for processing.
```public class Triangle {
private FloatBuffer vertexBuffer;
// number of coordinates per vertex in this array
static final int COORDS_PER_VERTEX = 3;
static float triangleCoords[] = { // in counterclockwise order:
0.0f, 0.622008459f, 0.0f, // top
-0.5f, -0.311004243f, 0.0f, // bottom left
0.5f, -0.311004243f, 0.0f // bottom right
};
// Set color with red, green, blue and alpha (opacity) values
float color[] = { 0.63671875f, 0.76953125f, 0.22265625f, 1.0f };
public Triangle() {
// initialize vertex byte buffer for shape coordinates
ByteBuffer bb = ByteBuffer.allocateDirect(
// (number of coordinate values * 4 bytes per float)
triangleCoords.length * 4);
// use the device hardware's native byte order
bb.order(ByteOrder.nativeOrder());
// create a floating point buffer from the ByteBuffer
vertexBuffer = bb.asFloatBuffer();
// add the coordinates to the FloatBuffer
vertexBuffer.put(triangleCoords);
// set the buffer to read the first coordinate
vertexBuffer.position(0);
}
}
```
By default, OpenGL ES assumes a coordinate system where [0,0,0] (X,Y,Z) specifies the center of the `GLSurfaceView` frame, [1,1,0] is the top right corner of the frame and [-1,-1,0] is bottom left corner of the frame. For an illustration of this coordinate system, see the OpenGL ES developer guide.
Note that the coordinates of this shape are defined in a counterclockwise order. The drawing order is important because it defines which side is the front face of the shape, which you typically want to have drawn, and the back face, which you can choose to not draw using the OpenGL ES cull face feature. For more information about faces and culling, see the OpenGL ES developer guide.
## Define a Square
Defining triangles is pretty easy in OpenGL, but what if you want to get a just a little more complex? Say, a square? There are a number of ways to do this, but a typical path to drawing such a shape in OpenGL ES is to use two triangles drawn together:
Figure 1. Drawing a square using two triangles.
Again, you should define the vertices in a counterclockwise order for both triangles that represent this shape, and put the values in a `ByteBuffer`. In order to avoid defining the two coordinates shared by each triangle twice, use a drawing list to tell the OpenGL ES graphics pipeline how to draw these vertices. Here’s the code for this shape:
```public class Square {
private FloatBuffer vertexBuffer;
private ShortBuffer drawListBuffer;
// number of coordinates per vertex in this array
static final int COORDS_PER_VERTEX = 3;
static float squareCoords[] = {
-0.5f, 0.5f, 0.0f, // top left
-0.5f, -0.5f, 0.0f, // bottom left
0.5f, -0.5f, 0.0f, // bottom right
0.5f, 0.5f, 0.0f }; // top right
private short drawOrder[] = { 0, 1, 2, 0, 2, 3 }; // order to draw vertices
public Square() {
// initialize vertex byte buffer for shape coordinates
ByteBuffer bb = ByteBuffer.allocateDirect(
// (# of coordinate values * 4 bytes per float)
squareCoords.length * 4);
bb.order(ByteOrder.nativeOrder());
vertexBuffer = bb.asFloatBuffer();
vertexBuffer.put(squareCoords);
vertexBuffer.position(0);
// initialize byte buffer for the draw list
ByteBuffer dlb = ByteBuffer.allocateDirect(
// (# of coordinate values * 2 bytes per short)
drawOrder.length * 2);
dlb.order(ByteOrder.nativeOrder());
drawListBuffer = dlb.asShortBuffer();
drawListBuffer.put(drawOrder);
drawListBuffer.position(0);
}
}
```
This example gives you a peek at what it takes to create more complex shapes with OpenGL. In general, you use collections of triangles to draw objects. In the next lesson, you learn how to draw these shapes on screen.
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Would you like to change your language preference and browse this site in ? If you want to change your language preference later, use the language menu at the bottom of each page.
## This class requires API level or higher
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# Blog
• Home
A net benefit is determined by summing all benefits and subtracting the cost of a project. This output provides an absolute measure of benefits (total dollars), rather than the relative measures provided by a B/C ratio.
## How Do You Find The Net Benefit?
Net benefits are calculated by subtracting the total costs from the total benefits in an equivalent measure after accounting for the effects of time.
## What Is Net Benefit In Microeconomics?
We can calculate our net benefit by adding up our benefits and subtracting our costs. It is logical to take action when total benefits exceed total costs. It is illogical to take action when total costs rise more than total benefits.
## How Do You Calculate Net Benefit Cost Ratio?
A benefit-cost ratio is calculated by dividing the benefits of the project by the costs of the project: BCR = Discounted value of benefits/discounted value of costs.
## How Do You Calculate Net Social Benefit?
In the case of total social benefit, the remaining benefit is subtracted from the total social benefit.
## How Do You Find The Net Benefit On A Graph?
In the marginal benefit curve, the area under the marginal cost curve represents the activity’s total benefit; in the marginal cost curve, the area under the marginal benefit curve represents the activity’s total cost. Net benefit equals total benefit less total cost.
## What Is Net Benefit Equal To?
Costs and benefits of different products. The net benefit is equal to the total benefits plus the total costs.
## How Do You Calculate Net Benefit In Microeconomics?
A project’s net benefit is determined by summing all benefits and subtracting all costs.
## What Is Net Marginal Benefit?
In addition to the marginal benefit, a consumer also receives the additional satisfaction that comes from purchasing the additional good or service. Marginal benefits are the maximum amount a consumer can pay for an additional good or service.
## What Is Total Benefit Microeconomics?
Marginal benefits are equal to the total benefit. A consumer surplus is a measure of how much a good can be gained by consuming it. In other words, it is the difference between what consumers were willing to pay and what they actually paid in the end.
## What Is Marginal Benefit In Economics Example?
A consumer who pays \$5 for an ice cream will get a marginal benefit of \$5, for example. Consumers may be less likely to purchase additional ice cream at that price, however, as only \$2 will entice them to buy another one at that price.
## What Is Net Benefit-cost Ratio?
A net benefit is determined by summing all benefits and subtracting the cost of a project. This output provides an absolute measure of benefits (total dollars), rather than the relative measures provided by a B/C ratio. A project’s net benefit can be used to rank projects with similar B/C ratios.
## How Do You Calculate Net Benefit?
Net benefits can be calculated by subtracting direct and indirect costs from direct and indirect benefits. Investors can determine whether the benefits outweigh the costs enough to justify pursuing the project by measuring both costs and benefits in equal measures.
## How Is Cost Ratio Calculated?
In cost accounting, the variable cost ratio is used to calculate a company’s variable production costs as a percentage of its net sales. By dividing the variable costs by the net revenues of the company, the ratio is calculated.
## What Is Npv Ratio?
NPV is used in capital budgeting and investment planning to determine the profitability of a project or investment. NPV is the difference between the present value of cash inflows and the present value of cash outflows over a period of time.
## What Is Net Social Benefit?
Net Social Benefits (NSB) are the value of a particular project at the time when it was decided to add to consumption the net amount of social benefits that would result from the project.
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https://au.mathworks.com/matlabcentral/cody/problems/417-covering-area/solutions/2580228
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Cody
Problem 417. Covering area
Solution 2580228
Submitted on 18 Jun 2020 by jmac
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
x = [1 1; 1 -1; -1 1; -1 -1; 0 0]; y_correct = 4; assert(isequal(cover_area(x),y_correct))
ans = 4
2 Pass
x = [1 1; 1 -2; -2 1; -1 -2; 2 -1; -3 0; 0.2*randn(4,2)]; y_correct = 11; assert(isequal(cover_area(x),y_correct))
ans = 11
3 Pass
x = [1 1; 1 -2; -1 1; -1 -2; 3 4; 0.2*randn(10,2)]; y_correct = 12; assert(isequal(cover_area(x),y_correct))
ans = 12
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# How do you simplify (12sqrt(3)) /( 2sqrt(3))?
May 12, 2018
(12√3)/(2√3)
$\frac{12}{2} = 6$
(√3)/(√3) = 1
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https://www.javaprogrammingforums.com/whats-wrong-my-code/36841-writing-method-add-fractions.html
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# Thread: writing a method to add fractions?
1. ## writing a method to add fractions?
Okay so I have to write a method to compute the following series: m(i)= 1/3 + 2/5 +....+ (i / 2i+1) and write a test program that displays a table " i = m(i)" 1=0.3333 2=0.7333....all the way down to 20 which is 9.2480. I have written something and cannot seem to get the sum of the fractions to display . I could be going about this the completely wrong way but this was my attempt. Please give suggestions!
```public class ExtraCredit1
{
public static void main(String[] args)
{
double num;
double sum = 1;
for (int i = 1; i <= 20; i++)
{
System.out.println("" + i + " " + sum + "");
}
}
public static double m(double i)
{
double sum = 1;
sum += (double) i / (2*i + 1);
return sum;
}
}```
2. ## Re: writing a method to add fractions?
cannot seem to get the sum of the fractions to display
What happens when you compile and execute the code? Copy the output and paste it here. Add some comments to the posted output showing what the output should look like.
3. ## Re: writing a method to add fractions?
When I compile it shows the 'i' = 1,2,3,4....20 but the m(i) is 0,0,0,0.... and needs to be 0.3333, 0.7333.... all the way to 9.2480... its supposed to add all the fractions together and display... 1/3 = 0.3333, 1/3 + 2/5 = 0.7333...
Its not supposed to show the actual fraction in the output just;
i m(i)
1 = 0.3333
2 = 0.7333
3
4
4. ## Re: writing a method to add fractions?
What does the program print out when it executes? Can you copy the exact output and not interpret it in text?
BTW m is a poor name for a method. The name of a method should be a verb saying what the method is doing.
Where is the m method called?
6. ## Re: writing a method to add fractions?
BTW m is a poor name for a method. The name of a method should be a verb saying what the method is doing.
Where is the m method called?
7. ## Re: writing a method to add fractions?
I understand it is a poor name for a method however thats what the teachers instruction are. Do I need to call the method into my println?
--- Update ---
```import java.text.DecimalFormat;
public class ExtraCredit1
{
public static void main(String[] args)
{
double num;
double sum = 0;
DecimalFormat formatter = new DecimalFormat("0.0000");
for (int i = 1; i <= 20; i++)
{
System.out.println("" + i + " = " + formatter.format(m(i)) + "");
}
}
public static double m(double i)
{
double sum = 0;
sum += (double) i / (2*i + 1);
return sum;
}
}```
--- Update ---
now it displays the fractions but it isnt adding them up after each iteration
8. ## Re: writing a method to add fractions?
Do I need to call the method
If a method is NOT called, it does not execute. If it does not execute, it will not compute any value.
--- Update ---
it isnt adding them up after each iteration
If you want to accumulate the sum of previous values with the current value, you need to save the sum of the previous values.
9. ## Re: writing a method to add fractions?
okay so....
``` public static double m(double i)
{
double sum = 0;
sum += (double) (i / (2*i + 1));
total = sum + sum
return total; //Will this return the accumulative total?
}
}```
--- Update ---
no that doesnt work either... hmm
10. ## Re: writing a method to add fractions?
` total = sum + sum`
Did that code compile? It's better if you compile and execute the code BEFORE posting it.
That looks the same as: total = 2*sum;
Where is the value from the last call to m() saved? That is where the new value for the fraction needs to be added to the old value of sum
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## General Mechanical
#### Radial stress in plane strain 2D cylinder
• Md Ahad
Subscriber
How can i get Residual radial stress at room temperature, after the 1000 C temperature was applied and removed in a plane stress or plane strain 2D cylinder. I can get the radial stresses only but don't know how to get the residual radial stress from the post processing. I have attached the archive file for better understanding to help.
Thanks in Advance.
• peteroznewman
Subscriber
There are two ways to get residual stress in your problem.
One is where different materials are bonded at 1000 C without stress, then cooled to room temperature (22 C) and elastic residual stress is created due to the difference in CTE between the materials. If the material were reheated to 1000 C, the stress would again be zero. If this is your case, please confirm.
Another way is when the different materials were bonded at room temperature, so have no residual stress at 22 C, but at least one material passes its yield point as the temperature is raised to 1000 C and lowered again. This could result in permanent plastic deformation (or crushing if a porous material) and there would be a residual stress when the material returns to room temperature. If this is your case, your model has no material properties for plasticity or any nonlinear behavior.
A model could have both elastic and plastic causes of residual stress, so even if it was bonded at 1000 C, one material could pass its yield point as it cools to 22 C, which would be a different model and result from the first case.
Regards
• Md Ahad
Subscriber
Sir peteroznewman,
First of all I am very happy to see your reply.I have been always looking for you when i am getting some mechanical workbench problem.I would like to learn the both ways to get the residual stresses sir.
If i provide the adhesive strength and,Hardness,Toughness and melting point then will it fulfill the demand for plastic properties?
Another thing i can give the numerical solution of the graph (deformation vs distance) of this model.If it helps you to consider the conditions and others on.
This is the graph of deformation.So we can justify our assumptions by checking the graph of our simulated result with this graph too.The line 1100-100 means the inner line gets 1100*c temperature and the outer line of the cylinder gets 100*c temperature in all the graphs i have provided you so far.Will our model represent the similar graph of deformation sir? I have checked but i got a non linear curve line going upwards.
Sir you can take your time to help.I have bothered you much.Very Sorry for that.
Regards
Your sincere student.
• Md Ahad
Subscriber
Dear peteroznewman,
I haven't got your response.Sir
Regards.
• peteroznewman
Subscriber
If i provide the adhesive strength and,Hardness,Toughness and melting point then will it fulfill the demand for plastic properties?
The simplest way to create a material model that includes plasticity is to know the yield strength. With that you can add Bilinear Kinematic Hardening plasticity to a material that already has Isotropic Elasticity. Bilinear Kinematic Hardening requires one other value, the Tangent Modulus. This value describes the post-yield hardening behavior. If the Tangent Modulus is unknown, then a conservative approach is to set the Tangent Modulus to a value of zero. This creates what is known in textbooks as an Elastic Perfectly Plastic material.
Yield strength may be a function of temperature and you can input a table of temperature and yield strength. These might go up to the softening temperature, but at the melting temperature, the material is no longer a solid, but a liquid so there is no longer an elastic response to stress, there is a flow that is characterized by a material property called viscosity.
I can give the numerical solution of the graph (deformation vs distance) of this model.
When you say "this model", do you mean the model in the book that has the graph you are trying to match? Yes, the tabular data of the output of the book model would be useful for adjusting parameters in an ANSYS model to try to match it. However, what would be far more useful is the full description of the book model. A full description would be the geometry of all the layers, the material properties of all the layers, the temperature at which the layers are bonded together, the load history of the assembly and the conditions at the time the graph is plotted.
• Md Ahad
Subscriber
I have attached the model i tried last time.
Put 10MPa pressure on the inner line of the cylinder and inner line temperature 1100 and outer line temperature 100.
And then teach me how can i get the residual stresses , deformation graph(like the above attached photo already).
• peteroznewman
Subscriber
Here is a graph of radial displacement from your model in red averaging about 0.024 mm, while the value of the line for the 1000-100 condition that I read off the u plot from the book above is about 1.7 mm and I plotted that with a green line. The green line is a factor of about 70 times larger than the red line.
The enormous difference between the value of the plot in the book and the value in your model is conclusive evidence that your model does not reproduce the physics that are in the book.
If you are interested in reproducing the results in the book with an ANSYS model, then you have to show for the numerical model in the book all of the following: the mathematical model, the assumptions made by the mathematical model, the inputs to the numerical model, the time-history of the materials in the numerical model. Where does it say in the book that residual stresses are being plotted in the stress plot? I asked about the temperature during bonding of the layers. Where is that information?
• Md Ahad
Subscriber
Thank you very much.I have understood you.
I will study what you have given me the doubts.Give some time to find out the answer of your demand.I will try to give the whole information what you asked me.
Thanks in Advance
• Md Ahad
Subscriber
Dear sir,
Can you please show me how to produce this graph from my model?
And sir i will know you the other information as soon as possible.I am studying about your doubts.
• peteroznewman
Subscriber
I used JMP to make the plot, but if you have Excel, you can make a similar plot. Highlight the Tabular Data in the Deformation result for the Path, copy and paste into Excel. Add a column to divide the path length by the maximum radius to create the Non-Dimensional Radius. Make a scatterplot of Radial Deformation vs Non-Dimensional Radius and add another Data Series in Excel that has the value 1.7 and you will have it.
• Md Ahad
Subscriber
Sir,
" Add a column to divide the path length by the maximum radius to create the Non-Dimensional Radius. "
i haven't understood this line.
Another thing i tried to get the deformation to get a graph like you.But failed.I tried to get the Radial Deformation vs Non-Dimensional Radius graph like this
..
But got this kind of graph which is like this curve going upwards
• peteroznewman
Subscriber
In Excel, insert column B and use a formula to divide column A by 5.9
Forget about the data curving up or down a little bit, the magnitude is wrong by a factor of 70.
• Md Ahad
Subscriber
Thank you very much.
And i am studying what you told me to study.I will let you know my progress as soon as possible
Regards
• Md Ahad
Subscriber
Dear sir,
Another issue has been risen out. If i want to make my solver understood that there will be no realtive displacement between each layer then how will i do it.
I meant i am getting the displacemnet graph as non linear.If i can determine that there is no relative displacement between each layer then the result should be a non linear graph.So how can i do it?
Regards
• peteroznewman
Subscriber
There is no relative displacement between the layers because each layer shares the same node at the interface due to being in a single multibody part.
• Md Ahad
Subscriber
Dear peteroznewman,
I hope you are doing well.Hope you are enjoyoing your break.Sir do you have knowledge on mathematica? All the graph i have given to you solved by mathematica. So if you do have knowledge about that you can help me knowing the mathematical model, the assumptions made by the mathematical model, the inputs to the numerical model, the time-history of the materials in the numerical model.I will give you all the information to help me learning that sir.
Thanking you.
• Md Ahad
Subscriber
Dear peteroznewman,
Sir i have changed the model to get a similar graph on mathematica. I have used mixture rule for the middle layer .Here you will get all the details about the model. I am attaching a pdf of the program. If you help me showing the video that how can i make this model on workbench(mechanical) .I will be grateful to you learning that.
Regards
Your sincere Student
• Md Ahad
Subscriber
Sir
Have you understood the model sir?
Regards
• peteroznewman
Subscriber
I had a first look at the code. Who wrote the code? There is often documentation that goes along with code. Is there any other supporting documentation for this code?
A conductivity material property, K2, is a function of radius:
K1 = 2.0; K3 = 11.7; (*watt/m^2c*)
K2[r_] := 1/(r2 - r3)*((K1 - K3)*r + (r2*K3 - r3*K1));
(*K1=2.0;*)
(*K2=K1;*)
(*K3=K1;*)
But commented out is a set of statements where K1, K2 and K3 all equal 2.0.
Is your goal to build an ANSYS model that reproduces the variable K2 as a function of radius?
Or is your goal to build an ANSYS model that reproduces the mathematica output when all values of K are equal to 2.0?
If you need conductivity as a function of radius, then you will have to implement Field Variables in ANSYS. Look at section 9.2 in the ANSYS Help.
• Md Ahad
Subscriber
sir
The k2 of the model that means (k2)the middle layer of the model will be the variable of the radius.
Don't consider the commented portion sir. Sir can you help me showing the video of this model in ansys? As am i very new in ansys so i could have cleared my conceptions regarding this model sir.
Regards
• peteroznewman
Subscriber
I have to learn how to create a material with a Field Variable, which will take some time.
In the mean time, you can see if you can get an ANSYS model to agree with the output of this program when all the material properties are a constant by commenting out the material variable properties for the second layer: K2, Alpha2, Nu2, and E2 and uncommenting the constant values for those properties.
What I have learned by reading the code is that it is using linear elastic material properties and there is no plasticity in the model.
I assume the radius values are in mm, and this code sets r1 = 226. The ANSYS models you provided had r1 = 4.1 mm. Make the mathematica code r1 = 4.1 to match this ANSYS model. Then the mathematica equations give us: r2 = 4.7, r3 = 5 and r4 = 6.6. You provided a model called sss.wbpj that had three layers, but that model had r4 = 10. Make mathematica Lstlayt = 5.0 to match this ANSYS geometry.
I know how to change 2D ANSYS models between Plane Stress and Plane Strain and the value of stress is different between those settings. You can look up the engineering equations that calculate stress for those two assumptions in a solid mechanics reference (book or Internet), and compare what you find with the Stress Calculation code in mathematica then show me in your reply which assumption has been programmed into mathematica: Plane Stress or Plane Strain.
• Md Ahad
Subscriber
Sir
First i thought that i will make a model that will have 3 layers and the middle of the layer will provide some more layers that will hold the properties of the such as k2,alpha2 etc of the average of the 1st and 3rd layer.So what i did i just drew a linear line from the first and third layers and find out the properties of the middle layers from the line by plotting distance on the x axis and properties on the y axis.Because i didn't know whether ansys can make the middle layer according to the variable properties by radius or not.
1. Sir i will study according to your suggestion.But sir as i don't know how to make middle layer following the properties by the mixture rule i used in mathematica. Sir i want to make the model according to the code i gave you of the mathematica. So sir if you make a model of the picture i have attached now. I will study harder and find out new conclusions.
2.Sir you just use plane strain in your stress and deformation calculation.Further i will change it to plane stress if needed as you had learned me earlier.I hope the result will be just like the graph i had given to you earlier.Sir if you have any doubt, try to get help from the code of the mathematica. Sir try to give the video of how you make the model. I easily get the concepts then as i am still a beginner sir.
3.Sir as i am your sincere student.I will study whatever problems you face making my model and getting the similar result from that model and then i will try to solve your doubts.And you know, i have learnt all the things you have tried to deliver . I am continuing my work on the mechanical now. . Thank you very much for all your support so far.
• peteroznewman
Subscriber
Because i didn't know whether ansys can make the middle layer according to the variable properties by radius or not.
Yes it can, you have to create a material with a field variable as described above.
2.Sir you just use plane strain in your stress and deformation calculation.Further i will change it to plane stress if needed as you had learned me earlier.I hope the result will be just like the graph i had given to you earlier.Sir if you have any doubt, try to get help from the code of the mathematica.
I looked at the code some more, it is written in terms of r and theta, therefore this is an axisymmetric model. What I find troubling is that there are no equations in terms of the axial direction, which might be called z (or y or x). Yet axisymmetric models have to satisfy equilibrium in all three directions. Axisymmetric equations of equilibrium are different from 2D Plane Stress and 2D Plane Strain, so forget my request to show me those equations. You should go look for a reference on the equilibrium equations for axisymmetric models and show me those equations, then compare them to the mathematica code.
• Md Ahad
Subscriber
Sir
You can make 2D plane strain model. Forget this portion axisymmetric for r and theta. We just make this model just like before.I hope the result will be satisfactory.
Sir what can be done with the problem second layer? Can you help me studying out for a solution for this layer?
Regards
• peteroznewman
Subscriber
You have two "black boxes" that take inputs and produce outputs. One is an ANSYS model, the other is Mathematica code that is also a model.
You can give each black box the same inputs and if they each generate the same outputs, you can say that they perform the same function inside the black box.
But what do you do when they give different outputs for the same inputs? Either there is a mistake in one model or there is no mistake, but one black box actually does not perform the same function as the other black box. For example, the ANSYS model may be built to evaluate 2D plane strain but the Mathematica model may be written to evaluate axisymmetric models.
I suggest we first find out if the two models produce the same outputs for the same inputs. That means making K2, Alpha2, Nu2, and E2 all constants. Then we can quickly generate outputs from both ANSYS and Mathematica. If the outputs are identical, then we can go to the next step of learning how to use Field Variables in ANSYS. But if the outputs are different, then we have to study what is inside the black box and see what the Mathematica code is doing.
• Md Ahad
Subscriber
Ok sir .
I have got you.Then SIr as you have told , i am trying to study about if the 3 layers are homogeneous then what will be the output in mathematica. And sir you try to make the model as i told in my last message.
Sir i will have no problem if the outputs are different because of the solver of ansys and matheatica. But i will not be satisfied if i make the wrong model of how the model is told on the mathematica. I am trying to compare results of the same model between the simulated on ansys and numerical solution from mathematica.
And ok Sir , i am studying of what you have told me to do.
Thanking you
• peteroznewman
Subscriber
And sir you try to make the model as i told in my last message.
You make the ANSYS model and send me the archive and the Mathematica code that matches the constant material properties in the ANSYS model. I will check if the inputs match. Send the Mathematica output. I will check if the outputs match. You also compare the outputs, are they the same or different?
Sir i will have no problem if the outputs are different because of the solver of ansys and mathematica.
I will have a problem if the outputs are significantly different. There is no point implementing variable material properties for layer 2 in ANSYS if the outputs don't match.
• Md Ahad
Subscriber
Ok Sir
I am trying to do as you have told.But sir how can i make the second layer on my model as i dont know how to define material properties on the second layer by the mixture rule i applied on the mathematica.
So what can i do now as you have taught me earlier how to make layers, i can make some layers in the second layer and on those layer i can vary the material properties linearly by radius. I think this wont give so many difference on the results comparing with mathematica graph. And sir i am giving you the update of mathematica results for the homogeneous cylinder layers(2D plane strain) that will help us to understand the difference of the results from the mathematica model with ansys model.
profound Regards
• peteroznewman
Subscriber
Attached is Mathematica code for a single material with the following properties:
Young's Modulus 10000 MPa
Poisson's Ratio 0.25
Thermal Conductivity 2.0 W/m/C
Thermal Expansion 1.1e-6
The geometry is for an inside radius 226 mm and outside radius 227.5 mm.
The boundary conditions are 1000 C inside and 950 C outside with a 10 MPa inside pressure.
Make an ANSYS model to these specifications and run the attached Mathematica code and see if the output agrees.
• Md Ahad
Subscriber
Sir
Due to some personal problems, i am lately giving the reply. I will be in my previous flow after couple of days. I have completed the model you gave previously in workbench. I am attaching the archive file with the results of temperature and stress vs distance picture too.
Profound Regards
• peteroznewman
Subscriber
Thank you for the Archive. I noticed is there is only one element through the thickness. The elements used are linear PLANE183 elements, so with only one element, the stress profile through the thickness can only be a straight line. In this particular model, that is the exact solution, so you got away with one element, but in general, you almost always want a minimum of two elements through the thickness to detect curvature in the result.
I checked all the inputs to the ANSYS model, and all is correct except you selected Plane Stress and you said you wanted Plane Strain. ANSYS won't let you change to that for the SS Thermal. To accomplish that for the Static Structural, you need to connect just the SS Thermal to the Setup cell of the Static Structural, then you can select Plane Strain. In this particular case, the radial normal stress (x-direction) component is the same for both assumptions, but the axial normal stress (z-direction) is zero in the Plane Stress assumption but it is 466 MPa for the Plane Strain assumption. However, Mathematica is not calculating this component of stress.
I assume the graphs above are from Mathematica. If so, then they agree with the ANSYS results.
You are ready to advance to the next phase of testing the agreement. Build a 3-layer ANSYS model with 3 different constant material properties and see if Mathematica and ANSYS still give the same result.
Please reply at your convenience with the Mathematica code and the 3-layer ANSYS archive.
• Md Ahad
Subscriber
Sir
that graph is the result of the archive file.That was not mathematica result.I will reply mathematica result (graph) as soon as possible. Since um in a trouble, i cant continue my work with the regular flow.Yes sir, i made mistake that i worked on plane stress. That would be plane strain. I will correct next time and reply will all as you have told.
Thanks Sir
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https://codedump.io/share/Jzlzikxq6DVH/1/how-do-you-calculate-the-variance-median-and-standard-deviation-in-c-or-java
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code511788465541441 - 1 year ago 152
C++ Question
# How do you calculate the variance, median, and standard deviation in C++ or Java?
Possible Duplicate:
Simple statistics - Java packages for calculating mean, standard deviation, etc
I have a vector of some doubles (1.1,2,3,5). How can I calculate the variance, median, and standard deviation?
Java or C++ or even pseudo code would do.
``````public class Statistics
{
double[] data;
int size;
public Statistics(double[] data)
{
this.data = data;
size = data.length;
}
double getMean()
{
double sum = 0.0;
for(double a : data)
sum += a;
return sum/size;
}
double getVariance()
{
double mean = getMean();
double temp = 0;
for(double a :data)
temp += (a-mean)*(a-mean);
return temp/size;
}
double getStdDev()
{
return Math.sqrt(getVariance());
}
public double median()
{
Arrays.sort(data);
if (data.length % 2 == 0)
{
return (data[(data.length / 2) - 1] + data[data.length / 2]) / 2.0;
}
else
{
return data[data.length / 2];
}
}
}
``````
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# Solving Triangles
(Difference between revisions)
Revision as of 13:11, 16 June 2011 (edit)← Previous diff Current revision (09:56, 29 May 2012) (edit) (undo) (42 intermediate revisions not shown.) Line 1: Line 1: - {{Image Description + {{Image Description Ready |ImageName=The Shadow Problem |ImageName=The Shadow Problem |Image=Shadows and fog.jpg |Image=Shadows and fog.jpg - |ImageIntro= + |ImageIntro=In the 1991 film ''Shadows and Fog'', the eerie shadow of a larger-than-life figure appears against the wall as the shady figure lurks around the corner. How tall is the ominous character really? Filmmakers use the geometry of shadows and triangles to make this special effect. - In the 1991 film ''Shadows and Fog'', the eerie shadow of a larger-than-life figure appears against the wall as the shady figure lurks around the corner. How tall really is the ominous character? Filmmakers use the geometry of shadows and triangles to make this special effect. + :The shadow problem is a standard type of problem for teaching trigonometry and the geometry of triangles. In the standard shadow problem, several elements of a triangle will be given. The process by which the rest of the elements are found is referred to as '''solving a triangle'''. + |ImageDescElem=A triangle has six total '''elements''': three sides and three angles. Sides are valued by length, and angles are valued by degree or radian measure. According to postulates for [[Congruent triangles|congruent triangles]], given three elements, other elements can always be determined as long as at least one side length is given. Math problems that involve solving triangles, like shadow problems, typically provide certain information about just a few of the elements of a triangle, so that a variety of methods can be used to solve the triangle. - The shadow problem is a standard type of problem for teaching trigonometry and the geometry of triangles. In the standard shadow problem, several elements of a triangle will be given. The process by which the rest of the elements are found is referred to as '''solving a triangle'''. + Shadow problems normally have a particular format. Some light source, often the sun, shines down at a given '''angle of elevation'''. The angle of elevation is the smallest—always acute—numerical angle measure that can be measured by swinging from the horizon from which the light source shines. Assuming that the horizon is parallel to the surface on which the light is shining, the angle of elevation is always equal to the '''angle of depression'''. The angle of depression is the angle at which the light shines down, compared to the angle of elevation which is the angle at which someone or something must look up to see the light source. Knowing the angle of elevation or depression can be helpful because [[Basic Trigonometric Functions|trigonometry]] can be used to relate angle and side lengths. - + - + - |ImageDescElem= + - + - A triangle has six total '''elements''': three sides and three angles. Sides are valued by length, and angles are valued by degree or radian measure. According to postulates for [[Congruent triangles]], given three elements, the other three elements can always be determined as long as at least one side length is given. Math problems that involve solving triangles, like shadow problems, typically provide certain information about just a few of the elements of a triangle, so that a variety of methods can be used to solve the triangle. + - + - Shadow problems normally have a particular format. Some light source, often the sun, shines down at a given '''angle of elevation'''. The angle of elevation is the smallest--always acute-- numerical angle measure that can be measured by swinging from the horizon. Assuming that the horizon is parallel to the surface on which the light is shining, the angle of elevation is always equal to the '''angle of depression'''. The angle of depression is the angle at which the light shines down, compared to the angle of elevation which is the angle at which someone or something must look up to see the light source. Knowing the angle of elevation or depression can be helpful because trigonometry can be used to relate angle and side lengths. + [[Image:Angle_of_elevation_1.jpg|center]] [[Image:Angle_of_elevation_1.jpg|center]] - In the typical shadow problem, the light shines down on an object or person of a given height. It casts a shadow on the ground below, so that the farthest tip of the shadow make a direct line with the tallest point of the person or object and the light source. The line that directly connects the tip of the shadow and the tallest point of the object that casts the shadow can be viewed as the hypotenuse of a triangle. The length from the tip of the shadow to the point on the surface where the object stands can be viewed as the first leg, or base, of the triangle, and the height of the object can be viewed as the second leg of the triangle. In the most simple shadow problems, the triangle is a right triangle because the object stands perpendicular to the ground. + In the typical shadow problem, the light shines down on an object or person of a given height. It casts a shadow on the ground below, so that the farthest tip of the shadow makes a direct line with the tallest point of the person or object and the light source. The line that directly connects the tip of the shadow and the tallest point of the object that casts the shadow can be viewed as the hypotenuse of a triangle. The length from the tip of the shadow to the point on the surface where the object stands can be viewed as the first leg, or base, of the triangle, and the height of the object can be viewed as the second leg of the triangle. In the most simple shadow problems, the triangle is a right triangle because the object stands perpendicular to the ground. In the picture below, the sun casts a shadow on the man. The length of the shadow is the base of the triangle, the height of the man is the height of the triangle, and the length from the tip of the shadow to top of the man's head is the hypotenuse. The resulting triangle is a right triangle. In the picture below, the sun casts a shadow on the man. The length of the shadow is the base of the triangle, the height of the man is the height of the triangle, and the length from the tip of the shadow to top of the man's head is the hypotenuse. The resulting triangle is a right triangle. Line 27: Line 21: [[Image:Wall_shadow.jpg|center]] [[Image:Wall_shadow.jpg|center]] - More difficult shadow problems will often involve a surface that is not level, like a hill. The person standing on the hill does not stand perpendicular to the surface of the ground, so the resulting triangle is not a right triangle. Other shadow problems may fix the light source at given height, like on a street lamp. This scenario creates a set of two '''similar triangles'''. + More difficult shadow problems will often involve a surface that is not level, like a hill. The person standing on the hill does not stand perpendicular to the surface of the ground, so the resulting triangle is not a right triangle. Other shadow problems may fix the light source, like a street lamp, at a given height. This scenario creates a set of two '''similar triangles'''. [[Image:More_difficult_problems.jpg|center]] [[Image:More_difficult_problems.jpg|center]] - Ultimately, a shadow problem asks you to solve a triangle by providing only a few elements of the possible six total. In the case of some shadow problems, like the one that involves two similar triangles, information about one triangle may be given and the question may ask to find elements of another. + Ultimately, a shadow problem asks you to solve a triangle given only a few elements of the possible six total. In the case of some shadow problems, like the one that involves two similar triangles, information about one triangle may be given and the question may ask to find elements of another. - + |ImageDesc===Why Shadows?== - + - + - |ImageDesc= + - ==Why Shadows?== + [[Image:Mirror2.jpg|right|250px]] [[Image:Mirror2.jpg|right|250px]] - Shadows are useful in the set-up of a triangle problems because of the way light works. A shadow is cast when light cannot shine through a solid surface. Light shines in a linear fashion, that is to say it does not bend. Light waves travel forward in the same direction in which the light was shined. Light is not like a liquid: it does not fill the space in which it shines like liquid assumes the shape of any container it's in. + Shadows are useful in the set-up of a triangle problems because of the way light works. A shadow is cast when light cannot shine through a solid surface. Light shines in a linear fashion, that is to say it does not bend. Light waves travel forward in the same direction in which the light was shined. - In addition to the linear fashion in which light shines, light has certain angular properties. When light shines on an objects that reflects light, it reflects back at the same angle at which it shined. Say a light shines onto a mirror. The angle between the beam of light and the wall that the mirror is the '''angle of approach'''. The angle from the wall at which the light reflects off of the mirror is the '''angle of departure'''. The angle of approach is equal to the angle of departure. + In addition to the linear fashion in which light shines, light has certain angular properties. When light shines on an object that reflects light, it reflects back at the same angle at which it shined. Say a light shines onto a mirror. The angle between the beam of light and the wall that the mirror is the '''angle of approach'''. The angle from the wall at which the light reflects off of the mirror is the '''angle of departure'''. The angle of approach is equal to the angle of departure. {{{!}} {{{!}} {{!}}[[Image:Pool table.jpg|center]] {{!}}[[Image:Pool table.jpg|center]] - {{!}}{{!}}In another example, a cue ball is bounced off of the wall of a pool table at a certain angle. Just like the way that light bounces off of the mirror, the cue ball bounces off the wall at exactly the same angle at which it hits the wall. The cue ball has the same properties as the beam of light in this case: the angle of departure is the same as the angle of approach. This property will help with certain types of triangle problems, particularly those that involve mirrors. + {{!}}{{!}}Light behaves the same way a cue ball does when it is bounced off of the wall of a pool table at a certain angle. Just like the way that the cue ball bounces off the wall, light reflects off of the mirror at exactly the same angle at which it shines. The beam of light has the same properties as the cue ball in this case: the angle of departure is the same as the angle of approach. This property will help with certain types of triangle problems, particularly those that involve mirrors. {{!}}} {{!}}} Line 50: Line 40: ==More Than Just Shadows== ==More Than Just Shadows== - [[Image:Law of sines.jpg|left|175px]] + [[Image:Solving_triangle.jpg|left|175px]] - Shadow problems are just one type of problem that involves solving triangles. There are numerous other formats and set ups for unsolved triangle problems. Most of these problems are formatted as word problems, that is set up the problem in terms of some real life scenario. + Shadow problems are just one type of problem that involves solving triangles. There are numerous other formats and set ups for unsolved triangle problems. Most of these problems are formatted as word problems; they set up a triangle problem in terms of some real life scenario. There are, however, many problems that simply provide numbers that represent angles and side lengths. In this type of problem, angles are denoted with capital letters, ${A, B, C,...}$, and the sides are denoted by lower-case letters,${a,b,c,...}$, where $a$ is the side opposite the angle $A$. There are, however, many problems that simply provide numbers that represent angles and side lengths. In this type of problem, angles are denoted with capital letters, ${A, B, C,...}$, and the sides are denoted by lower-case letters,${a,b,c,...}$, where $a$ is the side opposite the angle $A$. Line 60: Line 50: - {{{!}} + {{{!}}cellpadding=10 cellspacing=10 {{!}}'''Ladder Problems''' {{!}}'''Ladder Problems''' - One other common problem in solving triangles is the ladder problem. A ladder of a given length is leaned up against a wall that stands perpendicular to the ground. The ladder can be adjusted so that the top of the ladder sits higher or lower on the wall and the angle that the ladder makes with the ground increases or decreases accordingly. Because the ground and the wall are perpendicular to one another, the triangles that need to be solved in ladder problems always have right angles. Since the right angle is always fixed, many ladder problems require the angle between the ground and the ladder, or the angle of elevation, to to be somehow associated with a fixed length of a ladder and the height of the ladder on the wall. In other words, ladder problems normally deal with the SAS scenario: they involve the length from the wall to the base of the ladder, the fixed length of the ladder itself, and the enclosed angle of elevation to determine the height at which the ladder sits on the wall. + One other common problem in solving triangles is the ladder problem. A ladder of a given length is leaned up against a wall that stands perpendicular to the ground. The ladder can be adjusted so that the top of the ladder sits higher or lower on the wall and the angle that the ladder makes with the ground increases or decreases accordingly. Because the ground and the wall are perpendicular to one another, the triangles that need to be solved in ladder problems always have right angles. Since the right angle is always fixed, many ladder problems require the angle between the ground and the ladder, or the angle of elevation, to to be somehow associated with a fixed length of a ladder and the height of the ladder on the wall. In other words, ladder problems normally ask for the height of the ladder on the wall or the ground distance between the ladder and the wall, and typically require some trigonometric calculation. {{!}}{{!}}[[Image:Ladder.jpg|center]] {{!}}{{!}}[[Image:Ladder.jpg|center]] Line 70: Line 60: {{!}}'''Mirror Problems''' {{!}}'''Mirror Problems''' - Mirror problems are a specific type of triangle problem which involves two people or objects that stand looking into the same mirror. Because of the way a mirror works, light reflects back at the same angle at which it shines in, as explained below in A More Mathematical Explanation. In a mirror problem, the angle at which one person looks into the mirror, or the '''angle of vision''' is the same exact angle at which the second person looks the mirror. Typically, the angle at which one person looks into the mirror is given along with some other piece of information. Once that angle is known, then one angle of the triangle is automatically known since the light reflects back off of the mirror at the same angle, making the angle of the triangle next to the mirror the supplement to twice the angle of vision. + Mirror problems are a specific type of triangle problem which involves two people or objects that stand looking into the same mirror. Because of the way a mirror works, light reflects back at the same angle at which it shines in, as explained below in A More Mathematical Explanation. In a mirror problem, the angle at which one person looks into the mirror, or the '''angle of vision''' is the same exact angle at which the second person must look into the mirror to make eye contact. Typically, the angle at which one person looks into the mirror is given along with some other piece of information. Once that angle is known, then one angle of the triangle is automatically known since the light reflects back off of the mirror at the same angle, making the angle of the triangle next to the mirror the supplement to twice the angle of vision. [[Image:Angle_of_vision.jpg|center]] [[Image:Angle_of_vision.jpg|center]] {{!}}- {{!}}- {{!}}'''Sight Problems''' {{!}}'''Sight Problems''' - Like shadow problems, sight problems include many different scenarios and several forms of triangles. Most sight problems are set up as word problems. They involve a person standing below or above some other person or object. In most of these problems, a person measures an angle with a tool called an '''astrolabe''' or a '''protractor''', . + Like shadow problems, sight problems include many different scenarios and several forms of triangles. Most sight problems are set up as word problems. They involve a person standing below or above some other person or object. In most of these problems, a person measures an angle with a tool called an '''astrolabe''' or a '''protractor'''. [[Image:Astrolabe_and_protractor.jpg|center]] [[Image:Astrolabe_and_protractor.jpg|center]] In the most standard type of problem, a person uses the astrolabe to measure the angle at which he looks up or down at something. In the example at the right, the bear stands in a tower of a given height and uses the astrolabe to measure the angle at which he looks down at the forest fire. The problem asks to find how far away the forest fire is from the base of the tower given the previous information. In the most standard type of problem, a person uses the astrolabe to measure the angle at which he looks up or down at something. In the example at the right, the bear stands in a tower of a given height and uses the astrolabe to measure the angle at which he looks down at the forest fire. The problem asks to find how far away the forest fire is from the base of the tower given the previous information. - {{!}}{{!}}[[Image:Forest1.jpg|center|500px]] + {{!}}{{!}}[[Image:Forest2.jpg|center|500px]] {{!}}} {{!}}} ==Ways to Solve Triangles== ==Ways to Solve Triangles== - In all cases, a triangle problem will only give a few elements of a triangle and will ask to find one or more of the unknown elements. A triangle problem asks for one of the lengths or angle measures that is not given in the problem. There are numerous formulas, methods, and operations that can help to solve a triangle depending on the information given in the problem. + In all cases, a triangle problem will only give a few elements of a triangle and will ask to find one or more of the lengths or angle measures that is not given. There are numerous formulas, methods, and operations that can help to solve a triangle depending on the information given in the problem. - The first step in any triangle problem is drawing a diagram. A picture can help to show which elements of the triangle are given and which elements are adjacent or opposite one another. By knowing where the elements are in relation to one another, we can use the trigonometric functions to relate angle and side lengths. + The first step in any triangle problem is '''drawing a diagram'''. A picture can help to show which elements of the triangle are given and which elements are adjacent or opposite one another. By knowing where the elements are in relation to one another, we can use the trigonometric functions to relate angle and side lengths. - There are numerous techniques which can be implemented in solving triangles: + There are many techniques which can be implemented in solving triangles: - :*'''Trigonometry''': The [[Basic Trigonometric Functions]] relate side lengths to angles. By substituting the appropriate values into the formulas for sine, cosine, or tangent, trigonometry can help to solve for a particular side length or angle measure. This is useful when given a side length and an angle measure. + :*'''Trigonometry''': The [[Basic Trigonometric Functions|basic trigonometric functions]] relate side lengths to angles. By substituting the appropriate values into the formulas for sine, cosine, or tangent, trigonometry can help to solve for a particular side length or angle measure of a right triangle. This is useful when given a side length and an angle measure. - :*'''Pythagorean Theorem''': The Pythagorean Theorem relates the squares of all three side lengths to one another in right triangles. This is useful when a triangle problem provides two side lengths and a third is needed. + :*'''Inverse Trigonometry''': Provided two side lengths, the [[Basic Trigonometric Functions#Inverse Trig Functions|inverse trig functions]] use the ratio of the two lengths and output an angle measure in right triangle trigonometry. Inverse trig is particularly useful in finding an angle measure when two side lengths are given in a right triangle. + + :* '''Special Right Triangles''': Special right triangles are right triangles whose side lengths produce a particular ratio in trigonometry. A 30°− 60°− 90° triangle has a hypotenuse that is twice as long as one of its legs. A 45°− 45°− 90° is called an isosceles right triangle since both of its legs are the same length. These special cases can help to quicken the process of solving triangles. + +
+ [[Image:454590.jpg|center|175px]] +
+
[[Image:306090.jpg|center|175px]] +
+ + + [[Image:Trianglez.jpg|right|180px]] + + :*'''Pythagorean Theorem''': The [[Pythagorean Theorem]] relates the squares of all three side lengths to one another in right triangles. This is useful when a triangle problem provides two side lengths and a third is needed. :::$a^{2}+b^{2} = c^{2}$ :::$a^{2}+b^{2} = c^{2}$ - :*'''Law of Cosines''': The [[Law of Cosines]] is a generalization of the Pythagorean Theorem which can be used for solving non-right triangles. The law of cosines relates the squares of the side lengths to the cosine of one of the angle measures. This is particularly useful given a SAS configuration, or when three side lengths are known and no angles non-right triangles. + :*'''Pythagorean Triples''': A Pythagorean triple is a set of three positive integers that satisfy the Pythagorean Theorem. The set {3,4,5} is one of the most commonly seen triples. Given a right triangle with legs of length 3 and 4, for example, the hypotenuse is known to be 5 by Pythagorean triples. + + :*'''Law of Cosines''': The [[Law of Cosines|law of cosines]] is a generalization of the Pythagorean Theorem which can be used for solving non-right triangles. The law of cosines relates the squares of the side lengths to the cosine of one of the angle measures. This is particularly useful given a SAS configuration, or when three side lengths are known and no angles for non-right triangles. :::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ :::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ - :*'''Law of Sines''': The [[Law of Sines]] is a formula that relates the sine of a given angle to the length of its opposite side. The law of sines is useful in any configuration when an angle measure and the length of its opposite side are given. It is also useful given an ASA configuration, and often the ASS configuration . The ASS configuration is known as [[Law of Sines#The Ambiguous Case| The Ambiguous Case]] since it does not always provide one definite solution to the triangle. + :*'''Law of Sines''': The [[Law of Sines|law of sines]] is a formula that relates the sine of a given angle to the length of its opposite side. The law of sines is useful in any configuration when an angle measure and the length of its opposite side are given. It is also useful given an ASA configuration, and often the ASS configuration . The ASS configuration is known as the [[Ambiguous Case|ambiguous case]] since it does not always provide one definite solution to the triangle. :::$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ :::$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ - When solving a triangle, one side length must always be given in the problem. Given an AAA configuration, there is no definite solution to the triangle. According to postulates for [[Congruent triangles]], the AAA configuration proves similarity in triangles, but there is no way to find the side lengths of a triangle. + When solving a triangle, one side length must always be given in the problem. Given an AAA configuration, there is no way to prove congruency. According to postulates for [[Congruent triangles|congruent triangles]], the AAA configuration proves similarity in triangles, but there is no way to find the side lengths of a triangle. Knowing just angle measures is not helpful in solving triangles. Line 108: Line 113: '''Example 1: Using Trigonometry''' '''Example 1: Using Trigonometry''' - A damsel in distress awaits her rescue from the tallest tower of the castle. A brave knight is on the way. He can see the castle in the distance and starts to plan his rescue, but he needs to know the height of the tower so he can plan properly. The knight sits on his horse 500 feet away from the castle. He uses his handy protractor to find the measure of the angle at which he looks up to see the princess in the tower, which is 15°. Sitting on the horse, the knight's eye level is 8 feet above the ground. What is the height of the tower? + A damsel in distress awaits her rescue from the tallest tower of the castle. A brave knight is on the way. He can see the castle in the distance and starts to plan his rescue, but he needs to know the height of the tower so he can plan properly. The knight sits on his horse 500 feet away from the castle. He uses his handy protractor to find the measure of the angle at which he looks up to see the princess in the tower, which is 15°. Sitting on the horse, the knight's eye level is 8 feet above the ground. What is the height of the tower? [[Image:Castle3.jpg|center]] [[Image:Castle3.jpg|center]] Line 124: Line 129: Plug in the angle and the known side length. Plug in the angle and the known side length. - ::$\tan 15^\circ =\frac{x ft}{500 ft}$ + ::$\tan 15^\circ =\frac{x \text{ft}}{500 \text{ft}}$ Clearing the fraction gives us Clearing the fraction gives us Line 136: Line 141: Round to get Round to get - ::$134 ft \approx x$ + ::$134 \text{ft} \approx x$ But this is only the height of the triangle and not the height of the tower. We need to add 8 ft to account for the height between the ground and the knight's eye-level which served as the base of the triangle. But this is only the height of the triangle and not the height of the tower. We need to add 8 ft to account for the height between the ground and the knight's eye-level which served as the base of the triangle. - ::$134 ft + 8 ft = h$ + ::$134 \text{ft} + 8 \text{ft} = h$ simplifying gives us simplifying gives us - ::$142 ft = h$ + ::$142 \text{ft} = h$ The tower is approximately 142 feet tall. The tower is approximately 142 feet tall. Line 155: Line 160: '''Example 2: Using Law of Sines''' '''Example 2: Using Law of Sines''' - A man stands 100 feet above the sea on top of a cliff. The captain of a white-sailed ship looks up at a 45° angle to see the man, and the captain of a black-sailed ship looks up at a 30° angle to see him. How far apart are the two ships? + A man stands 100 feet above the sea on top of a cliff. The captain of a white-sailed ship looks up at a 45° angle to see the man, and the captain of a black-sailed ship looks up at a 30° angle to see him. How far apart are the two ships? [[Image:Ships_sailing1.jpg|center]] [[Image:Ships_sailing1.jpg|center]] Line 162: Line 167: {{Hide|1= {{Hide|1= - To solve this problem, we can use the law of sines to solve for the bases of the two triangles since we have an AAS configuration with a known right angle. To find the distance between the two ships, we can take the difference in length between the two ships. + To solve this problem, we can use the law of sines to solve for the bases of the two triangles since we have an AAS configuration with a known right angle. To find the distance between the two ships, we can take the difference in length between the bases of the two triangles. First, we need to find the third angle for both of the triangles. Then we can use the law of sines. First, we need to find the third angle for both of the triangles. Then we can use the law of sines. -
+ {{{!}} - For the white-sailed ship, + {{!}}For the black-sailed ship, - ::$180^\circ - 90^\circ - 45^\circ = 45^\circ$ + ::$180^\circ - 90^\circ - 30^\circ =60^\circ$ - Let the distance between this boat and the cliff be denoted by $a$. + Let the distance between this ship and the cliff be denoted by $b$. By the law of sines, By the law of sines, - ::$\frac{100}{\sin 45^\circ} = \frac{a}{\sin 45^\circ}$ + ::$\frac{100}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$ - Multiplying both sides by $\sin 45^\circ$ gives us + Clear the fractions to get, - ::$(\sin 45^\circ)\frac{100}{\sin 45^\circ} = a$ + ::$100(\sin 60^\circ) = b(\sin 30^\circ)$ + + Compute the sines of the angle to give us + + ::$100\frac{\sqrt{3}}{2} = b\frac{1}{2}$ Simplify for Simplify for - ::$a = 100 ft$ + ::$100(\sqrt{3}) = b$ + {{!}}{{!}} + YOU CAN'T SEE THIS! -
+ OOOOOOOOOOOOOOOOOOOOOHHH -
- For the black-sailed ship, - ::$180^\circ - 90^\circ - 30^\circ =60^\circ + SPOOOOOKY! - Let the distance between this boat and the cliff be denoted by [itex]b$. + + {{!}}{{!}} + For the white-sailed ship, + + ::$180^\circ - 90^\circ - 45^\circ = 45^\circ$ + + Let the distance between this ship and the cliff be denoted by $a$. By the law of sines, By the law of sines, - ::$\frac{100}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$ + ::$\frac{100}{\sin 45^\circ} = \frac{a}{\sin 45^\circ}$ Clear the fractions to get, Clear the fractions to get, - ::$100(\sin 60^\circ) = b(\sin 30^\circ)$ + ::$100(\sin 45^\circ) = b(\sin 45^\circ)$ Compute the sines of the angle to give us Compute the sines of the angle to give us - ::$100\frac{\sqrt{3}}{2} = b\frac{1}{2}$ + ::$100\frac{\sqrt{2}}{2} = b\frac{\sqrt{2}}{2}$ Simplify for Simplify for - ::$100(\sqrt{3}) = b$ + ::$a = 100 \text{ft}$ + + {{!}}} Multiply and round for Multiply and round for - ::$b =173 ft + ::[itex]b =173 \text{ft}$ - + -
+ - The distance between the two boats, $x$, is the positive difference between the lengths of the bases of the triangle. + The distance between the two ships, $x$, is the positive difference between the lengths of the bases of the triangle. ::$b-a=x$ ::$b-a=x$ - ::$173-100 = 73 ft$ + ::$173-100 = 73 \text{ft}$ - The boats are about 73 feet apart from one another. + The ships are about 73 feet apart from one another. }} }} Line 290: Line 305: }} }} - |other=Trigonometry, Geometry |other=Trigonometry, Geometry |AuthorName=Orion Pictures |AuthorName=Orion Pictures |SiteName=http://en.wikipedia.org/wiki/File:Shadows_and_fog.jpg |SiteName=http://en.wikipedia.org/wiki/File:Shadows_and_fog.jpg |Field=Geometry |Field=Geometry - |WhyInteresting= + |WhyInteresting=[[Image:Grass_shadow3.jpg|right|160px]] - + Shadow Problems are one of the most common types of problems used in teaching trigonometry. A shadow problem sets up a scenario that is simple, visual, and easy to remember. Shadow problems are commonly used and highly applicable. - [[Image:Grass_shadow.jpg|right|160px]] + - Shadow Problems are one of the most common types of problem used in teaching trigonometry. The paradigm set up by a shadow problem is simple, visual, and easy to remember. Though an easy method by which to learn trigonometry, shadow problems are commonly used and highly applicable. + Shadows, while an effective paradigm in a word problem, can even be useful in real life applications. In this section, we can use real life examples of using shadows and triangles to calculate heights and distances. Shadows, while an effective paradigm in a word problem, can even be useful in real life applications. In this section, we can use real life examples of using shadows and triangles to calculate heights and distances. Line 305: Line 317: ==Example: Sizing Up Swarthmore== ==Example: Sizing Up Swarthmore== - [[Image:Belltower111.jpg]] - [[Image:_campustower.jpg]] + [[Image:Belltower111.jpg|right]] + The Clothier Bell Tower is the tallest building on Swarthmore College's campus, yet few people know exactly how tall the tower stands. We can use shadows to determine the height of the tower. Here's how: + + + '''Step 1)''' + Mark the shadow of the of the tower. Make sure to mark the time of day. The sun is at different heights throughout the day. The shadows are longest earlier in the morning and later in the afternoon. At around midday, the shadows aren't very long, so it might be harder to find a good shadow. When we marked the shadow of the bell tower, it was around 3:40 pm in mid-June. + + + '''Step 2)''' + After marking the shadow, we can measure the distance from our mark to the bottom of the tower. This length will serve as the base of our triangle. In this case, the length of the shadow was 111 feet. + + + + '''Step 3)''' + Measure the angle of the sun at that time of day. + + Use a yardstick to make a smaller, more manageable triangle. Because the sun shines down at the same angle as it does on the bell tower, the small triangle and the bell tower's triangle are '''similar''' and therefore have the same trigonometric ratios. + + :*Stand the yardstick so it's perpendicular to the ground so that it forms a right angle. The sun will cast a shadow. Mark the end of the shadow with a piece of chalk. + + :*Measure the length of the shadow. This will be considered the length of the base of the triangle. + + + {{{!}} + {{!}}[[Image:Ruler.jpg]] + {{!}}{{!}}[[Image:Arrow.jpg]] + {{!}}{{!}} + [[Image:Ruler2.jpg]] + {{!}}} + + {{{!}} + {{!}}[[Image:Yardstick_triangle2.jpg]] + {{!}}{{!}} + + :*Draw a diagram of the triangle made by connecting the top of the yardstick to the marked tip of the shadow. + + :*Use inverse trigonometry to determine the angle of elevation. + + :::$\tan X = \frac{36 \text{in}}{27 \text{in}}$ + + :::$\arctan \frac{36}{27} = X$ + + :::$\arctan \frac{4}{3} = X$ + + :::$X = 53^\circ$ + {{!}}- + {{!}}'''Step 4)''' + + Now, we can use trigonometry to solve the triangle for the height of the bell tower. + + :$\tan 53^\circ = \frac{h}{111 \text{ft}}$ + + Clearing the fractions, + + :$111 (\tan 53^\circ) = h$ + + Plugging in the value of $\tan 53^\circ$ gives us + + :$111 \frac{4}{3} = h$ + + Simplify for + + :$148 \text{ft} = h$ + + According to our calculations, the height of the Clothier Bell Tower is '''148 feet'''. + {{!}}{{!}} + [[Image:_campustower2.jpg|center]] + {{!}}} + + + ==History: Eratosthenes and the Earth== + + [[Image:Eratosthenes.jpg|left|300px]] + In ancient Greece, mathematician Eratosthenes made a name for himself in the history books by calculating the circumference of the Earth by using shadows. Many other mathematicians had attempted the problem before, but Eratosthenes was the first one to actually have any success. His rate of error was less than 2%. + + Eratosthenes used shadows to calculate the distance around the Earth. As an astronomer, he determined the time of the summer solstice when the sun would be directly over the town of Syene in Egypt (now Aswan). On this day, with the sun directly above, there were no shadows, but in Alexandria, which is about 500 miles north of Syene, Eratosthenes saw shadows. He calculated based on the length of the shadow that the angle at which the sun hit the Earth was 7 °. He used this calculation, along with his knowledge of geometry, to determine the circumference of the Earth. + |References=All of the images on this page, unless otherwise stated on their own image page, were made or photographed by the author [[User:Rscott3|Richard]] Scott, Swarthmore College. + + The information on Eratosthenes can be cited to http://www.math.twsu.edu/history/men/eratosthenes.html. + The main image and details about it were found at http://www.imdb.com/title/tt0105378/. - |InProgress=Yes + Some of the ideas for problems/pictures on this page are based from ideas or concepts in the ''Interactive Mathematics Program'' Textbooks by Fendel, Resek, Alper and Fraser. + |InProgress=No }} }}
## Current revision
Field: Geometry
Image Created By: Orion Pictures
In the 1991 film Shadows and Fog, the eerie shadow of a larger-than-life figure appears against the wall as the shady figure lurks around the corner. How tall is the ominous character really? Filmmakers use the geometry of shadows and triangles to make this special effect.
The shadow problem is a standard type of problem for teaching trigonometry and the geometry of triangles. In the standard shadow problem, several elements of a triangle will be given. The process by which the rest of the elements are found is referred to as solving a triangle.
# Basic Description
A triangle has six total elements: three sides and three angles. Sides are valued by length, and angles are valued by degree or radian measure. According to postulates for congruent triangles, given three elements, other elements can always be determined as long as at least one side length is given. Math problems that involve solving triangles, like shadow problems, typically provide certain information about just a few of the elements of a triangle, so that a variety of methods can be used to solve the triangle.
Shadow problems normally have a particular format. Some light source, often the sun, shines down at a given angle of elevation. The angle of elevation is the smallest—always acute—numerical angle measure that can be measured by swinging from the horizon from which the light source shines. Assuming that the horizon is parallel to the surface on which the light is shining, the angle of elevation is always equal to the angle of depression. The angle of depression is the angle at which the light shines down, compared to the angle of elevation which is the angle at which someone or something must look up to see the light source. Knowing the angle of elevation or depression can be helpful because trigonometry can be used to relate angle and side lengths.
In the typical shadow problem, the light shines down on an object or person of a given height. It casts a shadow on the ground below, so that the farthest tip of the shadow makes a direct line with the tallest point of the person or object and the light source. The line that directly connects the tip of the shadow and the tallest point of the object that casts the shadow can be viewed as the hypotenuse of a triangle. The length from the tip of the shadow to the point on the surface where the object stands can be viewed as the first leg, or base, of the triangle, and the height of the object can be viewed as the second leg of the triangle. In the most simple shadow problems, the triangle is a right triangle because the object stands perpendicular to the ground.
In the picture below, the sun casts a shadow on the man. The length of the shadow is the base of the triangle, the height of the man is the height of the triangle, and the length from the tip of the shadow to top of the man's head is the hypotenuse. The resulting triangle is a right triangle.
In another version of the shadow problem, the light source shines from the same surface on which the object or person stands. In this case the shadow is projected onto some wall or vertical surface, which is typically perpendicular to the first surface. In this situation, the line that connects the light source, the top of the object and the tip of the shadow on the wall is the hypotenuse. The height of the triangle is the length of the shadow on the wall, and the distance from the light source to the base of the wall can be viewed as the other leg other leg of the triangle. The picture below diagrams this type of shadow problem, and this page's main picture is an example of one of these types of shadows.
More difficult shadow problems will often involve a surface that is not level, like a hill. The person standing on the hill does not stand perpendicular to the surface of the ground, so the resulting triangle is not a right triangle. Other shadow problems may fix the light source, like a street lamp, at a given height. This scenario creates a set of two similar triangles.
Ultimately, a shadow problem asks you to solve a triangle given only a few elements of the possible six total. In the case of some shadow problems, like the one that involves two similar triangles, information about one triangle may be given and the question may ask to find elements of another.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Trigonometry, Geometry
Shadows are useful in the set-up of a triangle pro [...]
Shadows are useful in the set-up of a triangle problems because of the way light works. A shadow is cast when light cannot shine through a solid surface. Light shines in a linear fashion, that is to say it does not bend. Light waves travel forward in the same direction in which the light was shined.
In addition to the linear fashion in which light shines, light has certain angular properties. When light shines on an object that reflects light, it reflects back at the same angle at which it shined. Say a light shines onto a mirror. The angle between the beam of light and the wall that the mirror is the angle of approach. The angle from the wall at which the light reflects off of the mirror is the angle of departure. The angle of approach is equal to the angle of departure.
Light behaves the same way a cue ball does when it is bounced off of the wall of a pool table at a certain angle. Just like the way that the cue ball bounces off the wall, light reflects off of the mirror at exactly the same angle at which it shines. The beam of light has the same properties as the cue ball in this case: the angle of departure is the same as the angle of approach. This property will help with certain types of triangle problems, particularly those that involve mirrors.
Shadow problems are just one type of problem that involves solving triangles. There are numerous other formats and set ups for unsolved triangle problems. Most of these problems are formatted as word problems; they set up a triangle problem in terms of some real life scenario.
There are, however, many problems that simply provide numbers that represent angles and side lengths. In this type of problem, angles are denoted with capital letters, ${A, B, C,...}$, and the sides are denoted by lower-case letters,${a,b,c,...}$, where $a$ is the side opposite the angle $A$.
## Ways to Solve Triangles
In all cases, a triangle problem will only give a few elements of a triangle and will ask to find one or more of the lengths or angle measures that is not given. There are numerous formulas, methods, and operations that can help to solve a triangle depending on the information given in the problem.
The first step in any triangle problem is drawing a diagram. A picture can help to show which elements of the triangle are given and which elements are adjacent or opposite one another. By knowing where the elements are in relation to one another, we can use the trigonometric functions to relate angle and side lengths.
There are many techniques which can be implemented in solving triangles:
• Trigonometry: The basic trigonometric functions relate side lengths to angles. By substituting the appropriate values into the formulas for sine, cosine, or tangent, trigonometry can help to solve for a particular side length or angle measure of a right triangle. This is useful when given a side length and an angle measure.
• Inverse Trigonometry: Provided two side lengths, the inverse trig functions use the ratio of the two lengths and output an angle measure in right triangle trigonometry. Inverse trig is particularly useful in finding an angle measure when two side lengths are given in a right triangle.
• Special Right Triangles: Special right triangles are right triangles whose side lengths produce a particular ratio in trigonometry. A 30°− 60°− 90° triangle has a hypotenuse that is twice as long as one of its legs. A 45°− 45°− 90° is called an isosceles right triangle since both of its legs are the same length. These special cases can help to quicken the process of solving triangles.
• Pythagorean Theorem: The Pythagorean Theorem relates the squares of all three side lengths to one another in right triangles. This is useful when a triangle problem provides two side lengths and a third is needed.
$a^{2}+b^{2} = c^{2}$
• Pythagorean Triples: A Pythagorean triple is a set of three positive integers that satisfy the Pythagorean Theorem. The set {3,4,5} is one of the most commonly seen triples. Given a right triangle with legs of length 3 and 4, for example, the hypotenuse is known to be 5 by Pythagorean triples.
• Law of Cosines: The law of cosines is a generalization of the Pythagorean Theorem which can be used for solving non-right triangles. The law of cosines relates the squares of the side lengths to the cosine of one of the angle measures. This is particularly useful given a SAS configuration, or when three side lengths are known and no angles for non-right triangles.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
• Law of Sines: The law of sines is a formula that relates the sine of a given angle to the length of its opposite side. The law of sines is useful in any configuration when an angle measure and the length of its opposite side are given. It is also useful given an ASA configuration, and often the ASS configuration . The ASS configuration is known as the ambiguous case since it does not always provide one definite solution to the triangle.
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
When solving a triangle, one side length must always be given in the problem. Given an AAA configuration, there is no way to prove congruency. According to postulates for congruent triangles, the AAA configuration proves similarity in triangles, but there is no way to find the side lengths of a triangle. Knowing just angle measures is not helpful in solving triangles.
## Example Triangle Problems
Example 1: Using Trigonometry
A damsel in distress awaits her rescue from the tallest tower of the castle. A brave knight is on the way. He can see the castle in the distance and starts to plan his rescue, but he needs to know the height of the tower so he can plan properly. The knight sits on his horse 500 feet away from the castle. He uses his handy protractor to find the measure of the angle at which he looks up to see the princess in the tower, which is 15°. Sitting on the horse, the knight's eye level is 8 feet above the ground. What is the height of the tower?
We can use tangent to solve this problem. For a more in depth look at tangent, see Basic Trigonometric Functions.
Use the definition of tangent.
$\tan =\frac{\text{opposite}}{\text{adjacent}}$
Plug in the angle and the known side length.
$\tan 15^\circ =\frac{x \text{ft}}{500 \text{ft}}$
Clearing the fraction gives us
$\tan 15^\circ (500) =x$
Simplify for
$(.26795)(500) =x$
Round to get
$134 \text{ft} \approx x$
But this is only the height of the triangle and not the height of the tower. We need to add 8 ft to account for the height between the ground and the knight's eye-level which served as the base of the triangle.
$134 \text{ft} + 8 \text{ft} = h$
simplifying gives us
$142 \text{ft} = h$
The tower is approximately 142 feet tall.
Example 2: Using Law of Sines
A man stands 100 feet above the sea on top of a cliff. The captain of a white-sailed ship looks up at a 45° angle to see the man, and the captain of a black-sailed ship looks up at a 30° angle to see him. How far apart are the two ships?
To solve this problem, we can use the law of sines to solve for the bases of the two triangles since we have an AAS configuration with a known right angle. To find the distance between the two ships, we can take the difference in length between the bases of the two triangles.
First, we need to find the third angle for both of the triangles. Then we can use the law of sines.
For the black-sailed ship, $180^\circ - 90^\circ - 30^\circ =60^\circ$ Let the distance between this ship and the cliff be denoted by $b$. By the law of sines, $\frac{100}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$ Clear the fractions to get, $100(\sin 60^\circ) = b(\sin 30^\circ)$ Compute the sines of the angle to give us $100\frac{\sqrt{3}}{2} = b\frac{1}{2}$ Simplify for $100(\sqrt{3}) = b$ YOU CAN'T SEE THIS! OOOOOOOOOOOOOOOOOOOOOHHH SPOOOOOKY! For the white-sailed ship, $180^\circ - 90^\circ - 45^\circ = 45^\circ$ Let the distance between this ship and the cliff be denoted by $a$. By the law of sines, $\frac{100}{\sin 45^\circ} = \frac{a}{\sin 45^\circ}$ Clear the fractions to get, $100(\sin 45^\circ) = b(\sin 45^\circ)$ Compute the sines of the angle to give us $100\frac{\sqrt{2}}{2} = b\frac{\sqrt{2}}{2}$ Simplify for $a = 100 \text{ft}$
Multiply and round for
$b =173 \text{ft}$
The distance between the two ships, $x$, is the positive difference between the lengths of the bases of the triangle.
$b-a=x$
$173-100 = 73 \text{ft}$
The ships are about 73 feet apart from one another.
Example 3: Using Multiple Methods
At the park one afternoon, a tree casts a shadow on the lawn. A man stands at the edge of the shadow and wants to know the angle at which the sun shines down on the tree. If the tree is 51 feet tall and if he stands 68 feet away from the tree, what is the angle of elevation?
There are several ways to solve this problem. The following solution uses a combination of the methods described above.
First, we can use Pythagorean Theorem to find the length of the hypotenuse of the triangle, from the tip of the shadow to the top of the tree.
$a^{2}+b^{2} = c^{2}$
Substitute the length of the legs of the triangle for $a, b$
$51^{2}+68^{2} = c^{2}$
Simplifying gives us
$2601+4624 = c^{2}$
$7225 = c^{2}$
Take the square root of both sides for
$\sqrt{7225} = c$
$85 = c$
Next, we can use the law of cosines to find the measure of the angle of elevation.
$a^{2}=b^{2}+c^{2} - 2bc \cos A$
Plugging in the appropriate values gives us
$51^{2}=68^{2}+85^{2} - 2(85)(68) \cos A$
Computing the squares gives us
$2601= 4624+7225 - 11560 \cos A$
Simplify for
$2601= 11849 - 11560 \cos A$
Subtract $11849$ from both sides for
$-9248= -11560 \cos A$
Simplify to get
$.8 = \cos A$
Use inverse trigonometry to find the angle of elevation.
$A = 37^\circ$
# Why It's Interesting
Shadow Problems are one of the most common types of problems used in teaching trigonometry. A shadow problem sets up a scenario that is simple, visual, and easy to remember. Shadow problems are commonly used and highly applicable.
Shadows, while an effective paradigm in a word problem, can even be useful in real life applications. In this section, we can use real life examples of using shadows and triangles to calculate heights and distances.
## Example: Sizing Up Swarthmore
The Clothier Bell Tower is the tallest building on Swarthmore College's campus, yet few people know exactly how tall the tower stands. We can use shadows to determine the height of the tower. Here's how:
Step 1) Mark the shadow of the of the tower. Make sure to mark the time of day. The sun is at different heights throughout the day. The shadows are longest earlier in the morning and later in the afternoon. At around midday, the shadows aren't very long, so it might be harder to find a good shadow. When we marked the shadow of the bell tower, it was around 3:40 pm in mid-June.
Step 2) After marking the shadow, we can measure the distance from our mark to the bottom of the tower. This length will serve as the base of our triangle. In this case, the length of the shadow was 111 feet.
Step 3) Measure the angle of the sun at that time of day.
Use a yardstick to make a smaller, more manageable triangle. Because the sun shines down at the same angle as it does on the bell tower, the small triangle and the bell tower's triangle are similar and therefore have the same trigonometric ratios.
• Stand the yardstick so it's perpendicular to the ground so that it forms a right angle. The sun will cast a shadow. Mark the end of the shadow with a piece of chalk.
• Measure the length of the shadow. This will be considered the length of the base of the triangle.
Draw a diagram of the triangle made by connecting the top of the yardstick to the marked tip of the shadow. Use inverse trigonometry to determine the angle of elevation. $\tan X = \frac{36 \text{in}}{27 \text{in}}$ $\arctan \frac{36}{27} = X$ $\arctan \frac{4}{3} = X$ $X = 53^\circ$ Step 4) Now, we can use trigonometry to solve the triangle for the height of the bell tower. $\tan 53^\circ = \frac{h}{111 \text{ft}}$ Clearing the fractions, $111 (\tan 53^\circ) = h$ Plugging in the value of $\tan 53^\circ$ gives us $111 \frac{4}{3} = h$ Simplify for $148 \text{ft} = h$ According to our calculations, the height of the Clothier Bell Tower is 148 feet.
## History: Eratosthenes and the Earth
In ancient Greece, mathematician Eratosthenes made a name for himself in the history books by calculating the circumference of the Earth by using shadows. Many other mathematicians had attempted the problem before, but Eratosthenes was the first one to actually have any success. His rate of error was less than 2%.
Eratosthenes used shadows to calculate the distance around the Earth. As an astronomer, he determined the time of the summer solstice when the sun would be directly over the town of Syene in Egypt (now Aswan). On this day, with the sun directly above, there were no shadows, but in Alexandria, which is about 500 miles north of Syene, Eratosthenes saw shadows. He calculated based on the length of the shadow that the angle at which the sun hit the Earth was 7 °. He used this calculation, along with his knowledge of geometry, to determine the circumference of the Earth.
# References
All of the images on this page, unless otherwise stated on their own image page, were made or photographed by the author Richard Scott, Swarthmore College.
The information on Eratosthenes can be cited to http://www.math.twsu.edu/history/men/eratosthenes.html.
The main image and details about it were found at http://www.imdb.com/title/tt0105378/.
Some of the ideas for problems/pictures on this page are based from ideas or concepts in the Interactive Mathematics Program Textbooks by Fendel, Resek, Alper and Fraser.
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# Volume factor in Faddeev-Popov quantisation
In Faddeev-Popov quantisation, why does the integral over gauge parameter cancel the volume factor of the gauge group that's in the denominator? In fact, I don't understand where the volume factor comes from at the first place.
$Z = \int d [A] e^{iS}$
and since it has a gauge symmetry, there are multiple values of $A$ that generating the same Action $S$, since $S(A_g)=S(A_{g'})$ for the two different choices of gauge $g,g'$.
Now you have a gauge-fixing condition like $\partial^\mu A_\mu = 0$. The integral which contains enough physical Information is than given by $Z = \int d [A] \delta(A_\mu|_{gauge-fixed}) e^{iS}$ (1), but unfortunately you have the integral $\int d [A] \delta(\partial^\mu A_\mu)e^{iS}$ (2). Recall the identity for delta-distributions $\delta(f(x)) = \sum_{i \in roots(f)} \frac{\delta(x-x_i)}{|f'(x_i)|}$ and generalize it to the "infinite-dimensional" function $A$. You have a Jacobian determinant in the denominator. But when you provide an additional factor in the Integrand of (2) which is simply the Jacobian functional determinant, it would cancel out and you would arrive on the integral (1).
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# statistics hmwrk
QUESTION 1.(a). A biologist assumes that there is a linear relationship between the amount of fertilizer supplied to tomato plant and the subsequent yield of tomatoes obtained. Eight(8) tomato plants of the same variety, were selected at random and treated, weekly, with a solution in which x grams(g) of fertilizer was dissolved in a fixed quantity of water. The yield y kilograms (kg) of tomatoes was recorded as follows.PlantABCDEFGHx1.01.52.02.53.03.54.04.5y3.94.45.86.67.07.17.37.7(i). Plot a scatterplot of yield, y against amount of fertilizer, x. ( 3 points)(ii). Calculate the equation of the least square regression line of y on x. ( 8 points)(iii). Estimate the yield of a plant treated, weekly, with a 3.2 grams of fertilizer.( 2 points)(iv). Indicate why it may not be appropriate to use your equation to predict the yield of a plant treated weekly with 20 grams of fertilizer. ( 2 points)(b). The table below shows a Verbal Reasoning test scores, x and an English test scores, y for each of a random sample of eight (8) children who took both testsChildABCDEFGHx112113110113112114109113y6965757070756876(i). Calculate the value of the product moment correlation coefficient between the scores in Verbal Reasoning and English. ( 8 points)(ii).Comment briefly, in context on the result obtained in part(i). ( 2 points)QUESTION 2.(a).(i).Suppose that the variables X and Y satisfy the four assumptions for regression inferences such that for samples of size n, each with the same values X1, X2, X3…………………… Xn, for the predictor variable, List the three properties that must hold for the slope b1 of the sample regression line.(ii).What distribution is used for the inferences for β1 (2 point).(iii).In a hypothesis test for the slope β1 such that Ho:β1=0 vs. Ha:β1≠0, write the formula for the test statistic and state its degrees of freedom. (2 point).(b).When you are asked to use the critical value approach to conduct a Regression t-test state the following.(i).Purpose (2 point).(ii).Assumptions (2 point).(iii).All the six steps (7 points) required to enable you to complete such a test.(c). Consider the following data on the age and price for a sample of eleven (11) Honda cars as displayed below.Age(yr) x54655566277Price(\$100) y851037082899866951697048At a 5% significance level, do the data provide sufficient evidence to conclude that age is useful as a linear predictor of price for the cars? .Use the critical value approach for your test.[Hint: Regression equation is yhat=195.47-20.26x, and Standard error of the estimate Se=12.58] (10 points).QUESTION 3.The following are the age and price (thousands of dollars) data for Corvettes from a car dealership in Sterling, Virginia. Find attached the DDXL output of the data from regression analysis.x(Age)6 6 6 2 2 5 4 5 1 4y(Price)290 280 295 425 384 315 355 328 425 325Dependent variable is y and independent variable is x.No Selector, 10 total cases.R2=93.7% R2 (adjusted)=92.9%S=14.25 with 10-2=8 degrees of freedomSource Sum of Squares df Mean Square F-ratioRegression 24057.9 1 24057.9 119Residual 1623.71 8 202.964Variable Coefficient s.e of Coeff t-ratio probConstant 456.602 11.43 39.9 <0.0001X -27.9029 2.56 -10.9 <0.0001(a).(i). From the DDXL output write the regression equation. (2 points)(ii).From the DDXL output, figure out the value of Se. (1 point)(iii). Calculate the value of Sxx from the databy using the values of x. (5 points).(b).Obtain a 90% confidence interval for the mean price of all 3-year old cars by using the conditional mean t-interval procedure on page#17 & page#18 of section E class lecture notes.[By using the values of Se from (a)-(ii) and Sxx from(a)-(iii), you can find (b)] (7 points)(c).Obtain a 90% prediction interval for the price of a 3-year old car by using the predicted value t-interval procedure on page#20 & page#21 of section E class lecture notes. .[By using the values of Se from (a)-(ii) and Sxx from(a)-(iii), you can find (c)] (7 points)(d). By comparing your results from (b) and (c) which interval is wider? and give two reasons why such results is expected. (3 points)
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# Tower of Hanoi
#### Question:
The tower of Hanoi is a mathematical puzzle. It consists of three rods and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top. We have to obtain the same stack on the third rod. Assume the maximum number of disks to be 6. Display Invalid input for the given input less than zero or greater than 6.
Write a program in Java to solve Tower-of-Hanoi using recursion.
Note: Write the main() inside the class ‘TowerTest
Sample Input and Output :
Enter the number of disks :
4
The sequence of moves involved in the Tower of Hanoi are :
Move disk 1 from peg A to peg B
Move disk 2 from peg A to peg C
Move disk 1 from peg B to peg C
Move disk 3 from peg A to peg B
Move disk 1 from peg C to peg A
Move disk 2 from peg C to peg B
Move disk 1 from peg A to peg B
Move disk 4 from peg A to peg C
Move disk 1 from peg B to peg C
Move disk 2 from peg B to peg A
Move disk 1 from peg C to peg A
Move disk 3 from peg B to peg C
Move disk 1 from peg A to peg B
Move disk 2 from peg A to peg C
Move disk 1 from peg B to peg C
Sample Input and Output 2 :
Enter the number of disks :
-4
Invalid input
#### ToweTest.java
```import java.util.*;
public class TowerTest
{
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("Enter the number of disks :");
int n=sc.nextInt();
if(n>0 && n<=6)
{ System.out.println("The sequence of moves involved in the Tower of Hanoi are:");
towerOfHanoi(n, 'A', 'C', 'B');
}
else
{
System.out.println("Invalid input");
}
}
static void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod)
{
if(n==1)
{
System.out.println("Move disk 1 from peg "+ from_rod + " to peg " + to_rod);
return;
}
towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
System.out.println("Move disk "+n+" from peg "+ from_rod + " to peg "+to_rod);
towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}
}```
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Welcome Contact Getting Started Site Map Project 1 2 3 4 5 6 7 8 9 10 11
# Topic 5 - Dose Units
## Kerma
• “Sum of the initial kinetic energies per unit mass of all charged particles produced by the radiation”
• This is regardless of where the energy is deposited
• Bremsstrahlung photons are not counted, whether they escape or not
• Annihilation radiation is not counted, regardless of fate of annihilation photons
• Initial positron, if primary ionizing particle, is counted
### Correction/Clarification on Kerma
• Etr is just the kinetic energy received by charged particles in a specified volume V, regardless of where or how they spend the energy
• Kerma is the expectation value of the energy transferred to charged particles per unit mass at a point of interest, including radiative-loss energy, but excluding energy passed from one charged particle to another
### Quantities to Describe a Radiation Beam
• Fluence
• # photons/area
• F = dN/da
• Energy fluence
• Energy / area
• Y = dN hn/da
• Fluence rate
• # photons/(time area)
• f = dF/dt
• Energy fluence rate
• Energy / (time area)
• y= dY/dt
## • is density
• gives the number of photon interactions that take place per unit mass of material.
• m is attenuation coefficient
• r is density
• For a spectrum of energies that can be described by dΦ(hv) /d hv, then:
### Calculating Kerma
• Given incident on a block of carbon
• 10 MeV photons
• F = 1014 m-2
• What is kerma?
• Kerma is easy to calculate - but very difficult to measure!
### Relating Kerma & Absorbed Dose
• Kerma
• a measure of kinetic energy transferred at a point in space.
• Absorbed dose is more “interesting”.
• Energy is transferred in the medium
• not all is retained there.
• absorbed dose is the energy retained in the medium brought about by the ionizations along the track of the charged particle.
• Kerma and Absorbed Dose do not take place at the same location
### Calculating Absorbed Dose
• dEab is the mean energy “imparted” by the ionizing radiation into a mass, dm.
• Mass should be sufficiently small so that it the absorbed dose is defined at a point, but not so small that statistical fluctuations become important
• From the previous example, dEtr = 7.3 MeV
• fraction of 10 MeV photon energy transferred to the medium.
• A smaller amount is absorbed along the electron track: dEab = 7.06MeV
• dEtr- dEab
• The difference, 7.30-7.06 = 0.24 MeV, is bremsstrahlung.
• What is the path length of the 7.3 MeV electron in C?
• Estimate from graphs or tables of electron ranges,
• ~ 4.2 g cm-2.
• Divide by the density of carbon
• Path length: 1.9 cm.
### Important Relationship
• Relate absorbed dose in air to exposure:
• assuming CPE (electronic equilibrium)
### Electronic (Charged-Particle) Equilibrium
• The transfer of energy (kerma) occurs upstream from the absorbed dose.
• Kerma can be easily calculated from fluence
• Absorbed dose cannot. Why?
• Kerma remains constant
• Absorbed takes time to build up as upstream electrons increase:
### No Attenuation of Photon Beam, Φ Constant
• Number of electron tracks set in motion by photon interaction
• Φ constant with depth (small # interactions)
• Same # electrons set in motion in each square
• i.e., interactions per volume constant through target
### Beam Unattenuated
• Same number of photon tracks set in motion in each square
• e.g., square D is traversed by 400 tracks
• ionization in D is the same as total ionization started in A
• absorbed dose is proportional to ionization produced in each saure
• dose reaches a maximum at R (range of 2o)
• kerma constant with depth, equals absorbed dose beyond R
### Attenuation of Photon Beam
• Beam attenuation,
• Φ decreases with depth.
• Dose increases to a maximum (at maximum range of particle) overshoots then tracks kerma.
### Attenuation of Photons in Tissue
Isotope Maximum Dose Septh (mm in Tissue) Beam Attenuation (% of original beam)
137Cs 2 1
60Co 5 2
6 MV 15 6
• CPE will generally exist in a uniform medium at a point more than the maximum range for the secondary charged particles from the boundary of the medium
### Relating Energy Fluence and Exposure
• Radioactive beam incident on an area
• What is relationship between energy fluence and exposure at point p?
• Assume small mass of air at p
• The dose at p is: D= F(m/r)Eab= Y (mab/r)
• Can relate to R as:
• 1 R = 0.00873 J/kg, then
• Y/X = 0.00873 J/ ((mab/r)kg R)
• Complicated variation of energy absorption coefficient for air and energy of beam
### Relating photon fluence to exposure
• Relationship between energy fluence and photon fluence:
• F = dN/da
• Y= dN hn/da
• So, Y= F hn, and
### Specific Gamma Ray Emission
• G = Specific Gamma Ray Constant
• Has been calculated for many gamma ray emitting isotopes
• Can ‘easily’ be calculated
• Where do the numbers come from?
### Specific Gamma Ray Constant
• Assumes that the absorption of photons in air is constant over a large range
• See Figure 5.18 in text (p. 148) or Table 5.3 (p. 149)
• Absorption is almost constant from 60 keV to almost 2 MeV
• Assumes photons isotropic, no ‘buildup’
• Eliminates many constants for ease of calculation
• 3.5 ´ 10-3 m-1 linear absorption coefficient
• Combine terms
• Thus, 0.5 is the value of all of the constants combined
• Equation 6.18 can then be written
• G = 0.5 S fi ´ Ei
• Remember this is only useful for photons
• Some values are listed on page 187
Welcome Contact Getting Started Site Map Project 1 2 3 4 5 6 7 8 9 10 11
College of Engineering OSU Extended Campus Local 541-737-9204 Fax 541-737-2734 4943 The Valley Library Corvallis, OR 97331-4504
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This is a function of the point graph for one factor
seg_graph(model, fill = "lightblue", horiz = TRUE, pointsize = 4.5)
## Arguments
model
DIC, DBC or DQL object
fill
fill bars
horiz
Horizontal Column (default is TRUE)
pointsize
Point size
## Value
Returns a point chart for one factor
## Author
Gabriel Danilo Shimizu, shimizu@uel.br
Leandro Simoes Azeredo Goncalves
Rodrigo Yudi Palhaci Marubayashi
## Examples
data("laranja")
a=with(laranja, DBC(trat, bloco, resp,
mcomp = "sk",angle=45,sup=10,
ylab = "Number of fruits/plants"))
#>
#> -----------------------------------------------------------------
#> Normality of errors
#> -----------------------------------------------------------------
#> Method Statistic p.value
#> Shapiro-Wilk normality test(W) 0.9475889 0.187264
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, errors can be considered normal
#>
#> -----------------------------------------------------------------
#> Homogeneity of Variances
#> -----------------------------------------------------------------
#> Method Statistic p.value
#> Bartlett test(Bartlett's K-squared) 4.036888 0.85378
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, the variances can be considered homogeneous
#>
#> -----------------------------------------------------------------
#> Independence from errors
#> -----------------------------------------------------------------
#> Method Statistic p.value
#> Durbin-Watson test(DW) 2.324604 0.2484349
#>
#> As the calculated p-value is greater than the 5% significance level, hypothesis H0 is not rejected. Therefore, errors can be considered independent
#>
#> -----------------------------------------------------------------
#> -----------------------------------------------------------------
#>
#> CV (%) = 8.69
#> MStrat/MST = 0.91
#> Mean = 182.5556
#> Median = 183
#> Possible outliers = No discrepant point
#>
#> -----------------------------------------------------------------
#> Analysis of Variance
#> -----------------------------------------------------------------
#> Df Sum Sq Mean.Sq F value Pr(F)
#> trat 8 22981.33333 2872.66667 11.41142069 2.636524e-05
#> bloco 2 33.55556 16.77778 0.06664828 9.357825e-01
#> Residuals 16 4027.77778 251.73611
#>
#> As the calculated p-value, it is less than the 5% significance level. The hypothesis H0 of equality of means is rejected. Therefore, at least two treatments differ
#>
#> -----------------------------------------------------------------
#> Multiple Comparison Test: Scott-Knott
#> -----------------------------------------------------------------
#> resp groups
#> Country orange 250.3333 a
#> NRL 193.3333 b
#> FRL 192.3333 b
#> Cleopatra 183.6667 b
#> Clove Lemon 182.3333 b
#> Clove Tangerine 180.3333 b
#> Citranger-troyer 165.3333 c
#> Sunki 155.3333 c
#> Trifoliata 140.0000 c
#>
seg_graph(a,horiz = FALSE)
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# Problem: Give the ground-state electron configuration for silicon (Si) using noble-gas shorthand. Express your answer in condensed form as a series of orbitals. For example, [Ar]4s23d8 would be entered as [Ar]4s23d8.An atom consists of a small, positively charged nucleus, surrounded by negatively charged electrons. We organize the electrons in a logical manner. As the atomic number increases, electrons are added to the subshells according to their energy. Lower energy subshells fill before higher energy subshells. The order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
###### FREE Expert Solution
80% (260 ratings)
###### FREE Expert Solution
80% (260 ratings)
###### Problem Details
Give the ground-state electron configuration for silicon (Si) using noble-gas shorthand
Express your answer in condensed form as a series of orbitals. For example, [Ar]4s23d8 would be entered as [Ar]4s23d8.
An atom consists of a small, positively charged nucleus, surrounded by negatively charged electrons. We organize the electrons in a logical manner. As the atomic number increases, electrons are added to the subshells according to their energy. Lower energy subshells fill before higher energy subshells. The order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the The Electron Configuration: Condensed concept. You can view video lessons to learn The Electron Configuration: Condensed. Or if you need more The Electron Configuration: Condensed practice, you can also practice The Electron Configuration: Condensed practice problems.
What is the difficulty of this problem?
Our tutors rated the difficulty ofGive the ground-state electron configuration for silicon (Si...as low difficulty.
How long does this problem take to solve?
Our expert Chemistry tutor, Dasha took 1 minute and 7 seconds to solve this problem. You can follow their steps in the video explanation above.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Abdelrahman's class at UGA.
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• Offered by Rsch Sch of Finance, Actuarial Studies & App Stats
• ANU College ANU College of Business and Economics
• Course subject Statistics
• Areas of interest Statistics
• Academic career PGRD
• Course convener
• Dr Yanrong Yang
• Mode of delivery In Person
• Co-taught Course
• Offered in Second Semester 2022
Advanced Statistical Learning (STAT7050)
This course involves on campus teaching. For students unable to come to campus there will be a remote option. See the Class Summary for more details.
This course offers an introduction to modern statistical approaches for complicated data structures, and is designed for students who need to do advanced statistical data analyses and statistical research. There has been a prevalence of “big data” in many different scientific fields. In order to tackle the analysis of data of such size and complexity, traditional statistical methods have been reconsidered and new methods have been developed for extracting information, or "learning", from such data. Due to the wide of range of topics which could be considered, this course, each offering, will cover only a few of the potential topics. Some of the topics that may be considered are: regularisation and dimension reduction, clustering and classification, non-independent data, and causality. Emphasis is placed on methodological understanding, empirical applications, as well as theoretical foundations to a certain degree. As the extensive use of statistical software is integral to modern data analysis, there will be a strong computing component in this course.
Learning Outcomes
Upon successful completion, students will have the knowledge and skills to:
1. Describe the rationale behind the formulation and components of complex statistical models.
2. Compare and contrast statistical models in the context of a variety of scientific questions.
3. Communicate complex statistical ideas to a diverse audience.
4. Formulate a statistical solution to real-data research problems.
5. Demonstrate an understanding of the theoretical and computational underpinnings of various statistical procedures, including common classes of statistical models.
6. Utilise computational skills to implement various statistical procedures.
Indicative Assessment
1. Assignments (50) [LO 1,2,3,4,5,6]
2. Final Exam (50) [LO 1,2,3,4,5,6]
The ANU uses Turnitin to enhance student citation and referencing techniques, and to assess assignment submissions as a component of the University's approach to managing Academic Integrity. While the use of Turnitin is not mandatory, the ANU highly recommends Turnitin is used by both teaching staff and students. For additional information regarding Turnitin please visit the ANU Online website.
Students are expected to commit at least 10 hours per week to completing the work in this course. This will include at least 3 hours lecture classes, 1 hour tutorial class and up to 6 hours of private study time.
Not applicable
Requisite and Incompatibility
To enrol in this course you must have completed STAT7040.
Prescribed Texts
Trevor Hastie, Robert Tibshirani and Jerome Friedman (2008). The Elements of Statistical Learning (Data Mining, Inference, and Prediction). 2nd Edition. Springer.
Fees
Tuition fees are for the academic year indicated at the top of the page.
Commonwealth Support (CSP) Students
If you have been offered a Commonwealth supported place, your fees are set by the Australian Government for each course. At ANU 1 EFTSL is 48 units (normally 8 x 6-unit courses). More information about your student contribution amount for each course at Fees
Student Contribution Band:
1
Unit value:
6 units
If you are a domestic graduate coursework student with a Domestic Tuition Fee (DTF) place or international student you will be required to pay course tuition fees (see below). Course tuition fees are indexed annually. Further information for domestic and international students about tuition and other fees can be found at Fees.
Where there is a unit range displayed for this course, not all unit options below may be available.
Units EFTSL
6.00 0.12500
Course fees
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2022 \$4200
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2022 \$6000
Note: Please note that fee information is for current year only.
Offerings, Dates and Class Summary Links
ANU utilises MyTimetable to enable students to view the timetable for their enrolled courses, browse, then self-allocate to small teaching activities / tutorials so they can better plan their time. Find out more on the Timetable webpage.
The list of offerings for future years is indicative only.
Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
Second Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
7248 25 Jul 2022 01 Aug 2022 31 Aug 2022 28 Oct 2022 In Person View
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# Is m a derived unit?
m can mean milli or 10-3 and as such is a multiplier not a unit
m = mass (kg) or other is a fundamental and not a derived unit
M = moles is a fundamental unit
m = meter is a fundamental unit and not derived
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# Thoughts on OEIS A125508
Besides Scott Kim’s “Celebration of Mind” talk discussed in my previous post, I also today attended Gordon Hamilton’s talk on “Mini Mathematical Universes.” (Hamilton is also known as the designer of the board game Santorini (2016).) This was essentially an excuse to talk about a numerical process that Hamilton calls “Integral Fission.”
Define $$\mathrm{fission(N)}$$ for an integer $$N\ge 2$$ as follows:
• If $$N$$ is prime, then the answer is $$N$$ itself.
• Otherwise, choose $$L\le R$$ such that $$L\cdot R=N$$ and the difference $$R-L$$ is minimized; the answer is $$(\mathrm{fission}(L),\mathrm{fission}(R))$$.
For example, $$\mathrm{fission}(42) = ((2,3),7)$$; $$\mathrm{fission}(43) = 43$$; $$\mathrm{fission}(44) = ((2,2),11)$$; $$\mathrm{fission}(45) = (5,(3,3))$$; $$\mathrm{fission}(46) = (2,23)$$; and $$\mathrm{fission}(48) = ((2,3),(2,(2,2)))$$.
Notice that $$\mathrm{fission}(42)$$ and $$\mathrm{fission}(44)$$ have the same “shape”: if we ignore the numbers and just look at the parentheses, they’re both of the form $$((x,x),x)$$.
Hamilton considers the increasing sequence of integers whose fissions yield novel shapes — that is, the sequence consists of only those positive integers whose fission yields a shape not yielded by any smaller integer.
• $$\mathrm{fission}(2)$$ is $$x$$
• $$\mathrm{fission}(4)$$ is $$(x,x)$$
• $$\mathrm{fission}(8)$$ is $$(x,(x,x))$$
• $$\mathrm{fission}(16)$$ is $$((x,x),(x,x))$$
• $$\mathrm{fission}(20)$$ is $$((x,x),x)$$
• $$\mathrm{fission}(32)$$ is $$((x,x),(x,(x,x)))$$
• $$\mathrm{fission}(40)$$ is $$(x,(x,(x,x)))$$
This sequence (2, 4, 8, 16, 20, 32, 40…) is OEIS sequence A125508:
2 4 8 16 20 32 40 64 72 88 128 160 176 200 220 256
272 288 320 336 360 400 420 460 480 512 540 544 640
704 864 880 920 1024 1056 1152 1184 1200 1280 1344
1440 1600 1640 1680 1800 1840 1920 2048 ...
## Does every prime eventually appear in a leaf?
The sequence of trees involved in OEIS A125508 goes like this:
$2; (2,2); (2,(2,2)); ((2,2),(2,2)); ((2,2),5); ((2,2),(2,(2,2))); \ldots$
Two terms further on, we spot our first factor of 3. Eleven terms later, we spot our first 7. So naturally, someone in the webinar asked: “Does every prime number appear as a factor somewhere in this sequence?”
I have no idea how you’d prove such a conjecture. So I wrote a little C++ program (source) to play around with “Integral Fission.”
• Term 7958, which is the integer 126247680, is the first term divisible by 281; by that point we’ve seen every other prime up to 347 inclusive.
• Term 22920, which is the integer 1368482048, is the first term divisible by 349; by that point we’ve seen every other prime up to 653 inclusive.
• Term 61350, which is the integer 8853585920, is the first term divisible by 659; by that point we’ve seen every other prime up to 857 inclusive.
• Term 63826, which is the integer 9758377440, is the first term divisible by 859; by that point we’ve seen every other prime up to 1277 inclusive.
• Term 96572, which is the integer 21258514800, is the first term divisible by 1279; by that point we’ve seen every other prime up to 1481 inclusive.
• Term 128234, which is the integer 36008853504, is the first term divisible by 1483; by that point we’ve seen every other prime up to 1787 inclusive.
• Term 130922, which is the integer 37328558400, is the first term divisible by 1789; by that point we’ve seen every other prime up to 1987 inclusive.
## Does the sequence contain any odd number?
I initially conjectured that each term in OEIS A125508 would be divisible by two.
Pseudo-proof: In order to have an odd term, all of the leaves in its decomposition would have to be odd; which means they’d all have to be greater than 2; which means that if you took that same tree and replaced the leftmost leaf with a 2, you’d end up with a smaller integer producing that same shape; contradiction, Q.E.D.
But that’s not a real proof! — because blindly substituting 2 into a leaf won’t always produce a valid fission at all. Consider $$\mathrm{fission}(357) = (17,(3,7))$$.
• If you replace the 17 with a 2, you get $$(2,(3,7))\ne ((2,3),7)$$.
• If you replace the 3 with a 2, you get $$(17,(2,7))\ne ((2,7),17)$$;
• If you replace the 7 with a 2, you get $$(17,(2,3))\ne ((2,3),17)$$.
And in fact, the 36857th term of OEIS A125508 is the odd number $$3\,447\,969\,525$$. Conjecture busted! Here are the odd terms I’ve seen so far:
36857 3447969525 (((13,(3,5)),(11,(3,(3,3)))),(((3,3),(3,(3,3))),(7,(5,7))))
46174 5277504375 ((((3,3),(3,(3,3))),(11,(3,(3,3)))),(((3,5),(3,5)),(13,(5,5))))
55320 7161167475 ((((3,3),(3,(3,3))),((3,5),(3,7))),((11,(3,(3,3))),((3,5),(3,7))))
57897 7797715695 ((((3,3),(3,(3,3))),(7,(7,7))),((11,(3,(3,3))),((3,5),(3,7))))
58580 7979586615 ((((3,3),(3,(3,3))),(13,(3,(3,3)))),((11,(3,(3,3))),((3,5),(3,7))))
73900 12890101455 ((((3,3),(3,(3,3))),((3,7),(3,7))),((11,(3,(3,3))),((3,5),(3,(3,3)))))
75596 13299311025 ((((3,5),(3,7)),(13,(3,(3,3)))),((11,(3,(3,3))),((3,5),(3,(3,3)))))
78491 14059271655 ((((3,3),37),(13,(3,(3,3)))),((11,(3,(3,3))),((3,5),(3,(3,3)))))
80487 14819232285 (((11,(3,(3,3))),((3,5),(3,(3,3)))),((13,(3,(3,3))),(13,(3,(3,3)))))
80841 14994607815 (((11,(3,(3,3))),((3,5),(3,(3,3)))),(((3,3),(3,(3,3))),(19,(3,(3,3)))))
89972 18326742885 ((((3,3),(3,(3,3))),(19,(3,(3,3)))),((11,(3,11)),((3,5),(3,(3,3)))))
95851 20962690245 (((7,(7,7)),((3,5),(3,(3,3)))),(((3,3),(3,(3,3))),(23,(3,(3,3)))))
104452 24608375505 (((7,(7,7)),((3,5),(3,(3,3)))),(((3,3),(3,(3,3))),((3,(3,3)),(3,(3,3)))))
129536 36673857675 ((((3,5),(3,(3,3))),(13,(5,7))),((13,(3,(3,3))),((3,7),(3,(3,3)))))
138055 41662080999 ((((3,7),(3,7)),(17,(3,(3,3)))),((11,(3,11)),((3,7),(3,(3,3)))))
141799 43826344947 ((((3,7),(3,7)),(17,(3,(3,3)))),((11,(3,(3,3))),((3,(3,3)),(3,(3,3)))))
175066 66169187469 ((((3,7),(3,7)),((3,7),(3,(3,3)))),((11,(3,11)),((3,(3,3)),(3,(3,3)))))
## How fast does the sequence grow?
Anecdotally, it seems to grow in fits and starts. For example, of the 100,000,000 integers preceding $$2^{32}$$, only 473 of them are terms in the sequence; but of the 100,000,000 integers following $$2^{32}$$, 840 of them are in the sequence.
If the count of $$n$$’s prime factors grows as $$\log \log n$$, does that mean that we should expect to see $$k$$-leaf trees appearing around $$e^{e^k}$$? But then the number of $$k$$-leaf trees itself grows as $$4^n$$… so should we expect the terms of the sequence to grow roughly exponentially?
## Are there any tree shapes that never appear?
I conjecture that every shape of tree must eventually appear somewhere in the sequence — right? Some shapes take a long time to show up, but I think that given any shape of tree, it’s merely a bit of arithmetic to produce some integer that fissions into that particular shape.
After 503 terms, we’ve seen all 14 trees with five prime leaves; the last to show up is $$\mathrm{fission}(212\,060)=((((2,2),5),23),461)$$.
After 143871 terms, we’ve seen all 42 trees with six prime leaves; the last to show up is $$\mathrm{fission}(44\,973\,896\,860)=(((((2,2),5),23),461),212081)$$.
## What if we change the definition of isomorphism?
Watching Hamilton’s YouTube video on “Integral Fission,” I was struck by the fiddliness of having to put the smaller factor on the left every time. What if we eliminate the distinction between the left and right subtrees of a decomposition, so that the tree $$((2,2),5)$$ is considered to have the same “shape” as $$(2,(2,2))$$?
In that case, we boringly get a subsequence of OEIS A125508:
2 4 8 16 32 40 64 128 160 176 256
512 544 640 704 880 920 1024 1200
1280 1440 1600 2048 2176 2368 2560
2816 3456 4096 8192 8576 8704 ...
## What if we change the source being filtered?
Hamilton’s sequence is basically taking the sequence $$2, 3, 4, 5, 6, 7, 8\ldots$$ and passing it through this “novelty filter.” What if we took a different sequence to filter, instead? For example, what if we took only the integers divisible by three?
3 6 12 24 42 48 72 84 96 156 192 240
288 300 312 336 360 384 420 480 540
672 696 768 864 966 1056 1152 1176 ...
Or what if we took only the integers divisible by ten?
10 20 30 40 60 80 120 160 200 220 320
360 400 420 460 480 540 600 640 720
800 810 880 920 930 960 1080 1120 ...
Or only the Fibonacci numbers?
2 8 21 144 610 2584 6765 46368 832040
2178309 14930352 102334155 ...
No conclusions here; I’m just playing around with numbers in a very non-mathematician kind of way.
The first 200,000-ish terms of OEIS A125508 are here.
Posted 2020-10-18
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# AP Calculus Chapter 3
Home > Preview
The flashcards below were created by user Anonymous on FreezingBlue Flashcards.
1. Critical Numbers
Values of x in the domain of f (x) where f 'x= 0 or f 'x is undefined
2. How do you find absolute minimums and maximums on a closed interval?
• 1. Find critical numbers
• 2. Find values for the function at each endpoint and critical number
• 3. Determine absolute max and absolute min for the interval by your results from step 2.
3. Rolle’s Theorem
• If f is differentiable on the interval (a , b) and f(a) = f(b) then there is at least one place in the interval
• where f 'x= 0
4. Mean Value Theorem (MVT)
• If f (x) is continuous and differentiable on the interval [a , b], there is someplace in the interval where f ' x
• equals the slope of the line going through the endpoints of the interval.
5. First Derivative ITSC
1) Use critical numbers to determine intervals.
• Interval
• Test Value
• Sign of f ' x
• Conclusion
6. If sign of f ' x is positive what is happening to f(x)?
f(x) is increasing
7. If sign of f ' x is negative then f(x) is _______.
f(x) is decreasing
8. First Derivative Test
Do a first derivative ITSC,
• If f ' x is changing from pos to neg, then relative max
• If f ' x is changing from neg to pos, then relative min
9. Concave Up
f'(x) is increasing, f''(x)is positive, f(x) will hold water
### Card Set Information
Author: Anonymous ID: 109669 Filename: AP Calculus Chapter 3 Updated: 2011-10-17 21:45:01 Tags: AP Calculus BC Chapter Folders: Description: These are my flashcards for the third chapter of my book, Calculus of a single variable AP edition Show Answers:
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# 760 (number)
760 (seven hundred sixty) is an even three-digits composite number following 759 and preceding 761. In scientific notation, it is written as 7.6 × 102. The sum of its digits is 13. It has a total of 5 prime factors and 16 positive divisors. There are 288 positive integers (up to 760) that are relatively prime to 760.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 3
• Sum of Digits 13
• Digital Root 4
## Name
Short name 760 seven hundred sixty
## Notation
Scientific notation 7.6 × 102 760 × 100
## Prime Factorization of 760
Prime Factorization 23 × 5 × 19
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 190 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 760 is 23 × 5 × 19. Since it has a total of 5 prime factors, 760 is a composite number.
## Divisors of 760
1, 2, 4, 5, 8, 10, 19, 20, 38, 40, 76, 95, 152, 190, 380, 760
16 divisors
Even divisors 12 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 1800 Sum of all the positive divisors of n s(n) 1040 Sum of the proper positive divisors of n A(n) 112.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 27.5681 Returns the nth root of the product of n divisors H(n) 6.75556 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 760 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 760) is 1,800, the average is 11,2.5.
## Other Arithmetic Functions (n = 760)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 288 Total number of positive integers not greater than n that are coprime to n λ(n) 36 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 136 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 288 positive integers (less than 760) that are coprime with 760. And there are approximately 136 prime numbers less than or equal to 760.
## Divisibility of 760
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 4 0 4
The number 760 is divisible by 2, 4, 5 and 8.
## Classification of 760
• Abundant
• Polite
• Practical
### By Shape (2D, centered)
• Centered Triangle
• Idoneal
## Base conversion (760)
Base System Value
2 Binary 1011111000
3 Ternary 1001011
4 Quaternary 23320
5 Quinary 11020
6 Senary 3304
8 Octal 1370
10 Decimal 760
12 Duodecimal 534
20 Vigesimal 1i0
36 Base36 l4
## Basic calculations (n = 760)
### Multiplication
n×y
n×2 1520 2280 3040 3800
### Division
n÷y
n÷2 380 253.333 190 152
### Exponentiation
ny
n2 577600 438976000 333621760000 253552537600000
### Nth Root
y√n
2√n 27.5681 9.12581 5.25053 3.76845
## 760 as geometric shapes
### Circle
Diameter 1520 4775.22 1.81458e+06
### Sphere
Volume 1.83878e+09 7.25834e+06 4775.22
### Square
Length = n
Perimeter 3040 577600 1074.8
### Cube
Length = n
Surface area 3.4656e+06 4.38976e+08 1316.36
### Equilateral Triangle
Length = n
Perimeter 2280 250108 658.179
### Triangular Pyramid
Length = n
Surface area 1.00043e+06 5.17338e+07 620.537
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# 5 Logic Puzzles A 10-Year Old Can Solve Faster Than You
Sometimes children are just better at problem solving and these 5 puzzles will prove it to you:
# Puzzle #5
50 chocolate candies are in a box; 30 of them are caramel-filled; 25 are coconut-filled; 10 of them have both, and the rest are just plain chocolate.
Which image best represents the box of chocolates?
How many did you solve? None? OK, i’m not the only one then!:)
Puzzle #1: The car is parked in parking spot #87. How? Just turn the image upside down.
Puzzle #2: The bus is travelling to the left (unless you are from the UK, in which case the bus is travelling to the right). Why is that? Because we can’t see the door in the picture which means it’s on the other side!
Puzzle #3: The answer is 4. The answer comes from the number of circles in each four-digit number. For example, 6 has one circle, and 8 has 2 hence, 9999 is 4.
Puzzle #4: D = 1345, E = 2440
The numbers at the bottom level are connected to the upper level. Add the numbers in the bottom line: 198 + 263 = 461.
The number you get is greater than the neighbor above. Subtract the numbers: 461 – 446 = 15.
All other squares in the pyramid produce 15 so 1345 and 2440 are the only answers that work.
Puzzle #5: Diagram B is the one that matches. There are 20 caramel candies in the box, 15 coconut and 5 (50 – [20+15+10]) plain chocolate ones.
### If you like what you read, then you will definitely love this one: Cube Riddle: Can You Crack The Riddle Astronauts Are Asked To Solve Before Travelling To Space?
Photo: I’m A Useless Info Junkie
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### Recent postsView all
Day 9 - Online school26 Mar 20 Day 8 - Online school25 Mar 20 Day 7 - Online school24 Mar 20 Day 7 - Online school24 Mar 20 Day 6 - Online School23 Mar 20 Day 5 - Online school20 Mar 20 Day 4 - Online School19 Mar 20 Day 3 - Online School18 Mar 20 Digital Learning17 Mar 20 Day 2 - Online School17 Mar 20 Day 1 - Online school16 Mar 20 Euro 2016 Fractions Decimals %9 Jun 16
# A favourite starter
Sunday 17 January 2016
## Numberism
This is a quick post just to share one of my favourite little starting activities and a couple of related thoughts. These things are just little gems to be collected and treasured. This one was first shared with me by John Mason, author of 'Thinking mathematically' and more. I had the same experience then that most of my students do. If you have'nt seen it then please have a go before you read on.
### Try this......
• Think of a number between 2 and 3
• Think of another number between 2 and 3 that does not involve the digit 5
• Now another that doesn't use the digit 5, but does use the digit 7
• And finally, think of a number between 2 and 3 that doesn't use the digit 5, does use the digit 7, doesn't use the digit 9 and is a close to two and a half as possible.
Spoiler alert! Try not to read any further until you have had a go at this yourself, otherwise, you kind of miss the experience........
### Number rights
This is a terrific little video from 'Mathsnacks' which I have shown on numerous occassions because it is both a bit of fun and resonates with lots of issues in terms of how we see numbers. It will also take up a bit of page space before I discuss the solution to the above problem. There is a subtle hint in the video too!
### How did you do?
So now to discuss the little starter. In looking for a solution to the last part, depending on the class, most students will iterate until they get to the conclusion that the number they want is 2. 48 (followed by an infinite number of 8s) and then a 7. If a student offers 2.487, then the next says 2.4887, then 2.48887 and so on as we realise that we can always just get a little bit closer to two and a half by adding another 8.
What then follows is a great discussion about whether or not this number that we are trying to describe actually exists which is often very fruitful and entertaining. It is a nice model for demonstrating how quickly we can get in to uncertain territory within our number system.
Depending on the time you have available and the unpredictability of classes you might then show the number rights video. This hits on some really important points. Because we so often use fractions as proportions, we are often confused by their identity as actual numbers and recogninsing which sense a fraction is being used in is really important. Of course, the astute will notice that despite the committed actvism in the video, irrational and complex numbers have been sidelined - again (big sigh of resignation)
On rare occasions a student in my class has put 2 and 2 together (sorry, couldn't resist) and you can watch their expression go from possibility to delight as they realise they have trumped everyone. Then there is a collective self kicking exercise when they share their answer of ... (click the hidden box icon to reveal)
OK, so maybe you got there already, but I didn't and I am amazed and delighted with how often this goes exactly this way in my classroom.
So many issues wrapped up in all of this and thats why I enjoy it. Now off to further update the numberline on my classroom wall as I commit to being less 'numberist' with every passing year.
Thanks....
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# Forces
Forces are a push or a pull on an object
Newton's First Law an object in motion remains in motion unless acted upon by an unbalanced force (forces are balanced)
Newton's Second Law force is directly related to an object's mass and acceleration (forces are unbalanced)
Newton's Third Law For every action there is an equal and opposite reaction
Example of Newton's First Law A car going a constant speed
Example of Newton's Second Law A car accelerating
Acceleration when an object speeds up, slows down, or changes direction
Example of Newton's Third Law Gases of rocket push ground down and ground pushes gases of rocket up.
A toy car has a 5 N applied in the forward direction as well as a frictional force of 2 N. What is the net force acting on the car? 3 N
What is the acceleration of a 10 kg object that has a force of 20 N? 2 m/s^2
If a ball is hit with 4x the force, it will result in _______ the acceleration? 4x
If an object is accelerating its net force is? Not 0 N
What is the relationship between mass and acceleration? inverse or indirect
Weight is ______. force of gravity on an object
A 0.02 kg piece of gum hits a paper projectile with a 1 N force. If the projectile is now traveling at a constant velocity, what is its acceleration 0 N
If a 100 kg female traveled to the moon where gravity is 1/6th of the Earth's gravity (g of moon= 1.63 m/s^2), how much would she weigh? 163 N
If a ball has a mass of 30 N on the Earth, what is it's mass on the moon (g of moon = 1.63 m/s^2) 30 N
What is the unit of force? N
What is the unit of mass? kg
What is the unit of acceleration? m/s^2
What is the unit of speed? m/s
What is another way to write the unit of a Newton? kg*m/s^2
What do call the force that is perpendicular to the surface? Normal Force
The opposite of the thrust force on an airplane is the _____. drag force
The opposite of the weight force on an airplane is the _____. lift force
True or False: A force is any interaction that changes and object's motion. True
True or False: A force is necessary to continue an object's motion. False
Created by: NISD Physics
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How many joules of PE does a 12 kg rock have after falling from a 76 m cliff and reaches 7 m/s?
Guest May 4, 2017
#1
+27057
+1
"How many joules of PE does a 12 kg rock have after falling from a 76 m cliff and reaches 7 m/s?"
E = PE + KE E is total energy, PE is potential energy, KE is kinetic energy.
Top of cliff: E = 12*9.8*76 + 0 joules
E stays constant so when velocity is 7 m/s
12*9.8*76 = PE + (1/2)*12*7^2
PE = 12*9.8*76 - (1/2)*12*7^2 or PE ≈ 9314 joules
.
Alan May 5, 2017
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Home / Expert Answers / Algebra / show-that-s1-s2-if-and-only-if-s1-cup-s2-s1-cap-s2-pa443
# (Solved): Show that S1 = S2 if and only if S1 \cup S2 = S1 \cap S2. ...
Show that S1 = S2 if and only if S1 \cup S2 = S1 \cap S2.
We have an Answer from Expert
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# Data Geek Challenge – 2010 Soccer World Cup Team Analysis
Hi Everyone,
This is my data story for Data Geek (DG) Challenge
I would like to start by thanking the Data Geek Team for bringing this out.
I got the data sample from here
Below is my analysis:
First lets see which all Countries participated at the World Cup:
In the below figure we can see all the countries that participated and it is clearly visible that most of the countries were from Europe and then America.
This shows how popular soccer is in Europe and America.
Below figure shows total number of games played by each team:
Germany, Netherlands, Spain and Uruguay played six games each as these were the top four teams in the tournament.
Below two figures show total number of goals scored by each team:
It is clearly visible that Germany scored the most number of goals at the World Cup while Honduras and Algeria were the only teams that did not score any goal at the World Cup.
After Germany, Netherlands scored the most number of goals, then Argentina, then Uruguay and Brazil and so on
.
Below figure shows average number of goals scored and average number of goals conceded per game by all teams:
Germany has the highest average number of goals scored per game with 2.17 while North Korea has the highest average number of goals conceded per game with 4.
Below figure shows Percentage of shots that got converted to goal.
Germany is at the top here with 0.2 followed by Netherlands at 0.18
This clearly explains as to why Germany and Netherlands are the teams that scored most goals in the tournament.
Below two figures show total number of red cards and yellow cards earned by each team:
Netherlands earned the maximum number of yellow cards while Algeria, Australia, Brazil and Uruguay earned the most number of red cards.
Below figure shows total number of fouls committed and total number of red cards earned by each team:
Netherlands committed most number of fouls but didn’t earn any red cards while Uruguay was the team which committed most fouls and earned highest number of red cards.
This shows that Uruguay was the most unfair team in the tournament.
Regards,
Vivek
### Assigned Tags
You must be Logged on to comment or reply to a post.
Analysis for any sports is always interesting.
Good choice.
Rgrds,
Jitendra
Can't go wrong with soccer. Let me know if you want data of all past world cups
Good one
Kenny
Hi Kenny,
Can you please share the data of all past world cups with me.
Regards,
Kunal
Good analysis dear.
Vivek Singh Bhoj
Blog Post Author
Thanks Raman
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# What is the capacity of the Green Line?
While reading an excellent article by Yonah Freemark, Why should Chicago focus growth near transit?, I thought the Twin Cities should do the same thing. Taking advantage of existing capacity is far more cost effective than building new capacity (and yes, this applies to all modes). But what is the existing capacity of the Green Line? Well, that depends on assumptions and human behavior. In the table below I work through some scenarios based on assumptions.
First, how many hours per day is the Green Line operating? Second, what is the frequency within that time period? Third, how many cars per train are there? Fourth, what is the capacity per car (they are rated at 230, but this includes standees)? Fifth, how long is the line? Sixth, how long (how many stations) is the average trip? Seventh, how many directions are you considering?
This measures capacity in terms of daily boardings. Daily miles traveled is another measure, and is independent of the length of trips.
To calculate this we use the following equation:
Capacity = (Hours of Operation)*(Trains/Hour)*(Cars/Train)*(Capacity/Car)*(Stations – 1) * (Trip Length) * (Directions Operating).
At any rate, the attached table shows some surprisingly high numbers, up to 7 million (under the admittedly silly unconstrained scenario (A) where people only ride the train for 1 stop before alighting, trains run for 24 hours a day, and people are standing at near crush capacity), with more plausible numbers in the 255,000 territory, assuming everyone gets a seat, but you can run at 5 minute headways (C). Here we are limited by capacity in one section (downtown Minneapolis), which does run at 5 minute headways, but splits the capacity between the Green and Blue lines.
The main point is that there is a lot of capacity on the Green Line yet to go, even if you only run 18 hours a day, and you expect everyone to have a seat, and run at today’s 10 minute headways (which is all today’s fleet can support, to increase headway we either need to increase speed greatly or add vehicles), and assume the average trip is 7 stations (Aaron Isaacs informs me it is 3.5 miles, which at 1/2 mile spacing is about 7 stations) (83,314 – scenario D). At the other end of the spectrum, if everyone expected a seat and was riding from Union Depot to Target Field, the capacity would only be 32,400 with today’s frequencies.
Thus, east–west transportation capacity is not the constraint in development along the Green Line corridor. (One could similarly demonstrate the under-utilization in the north-south direction on buses, and in all directions on roads).
Certainly load balancing is an issue, much of the capacity is “off-peak”, but that is what pricing is for. Higher loads would increase wear and tear on the cars, and add costs, but hopefully the added revenues would more than compensate.
Compare with current ridership of about 37,835/day (Sept. 2014).
Given there is also a lot of developable land in this corridor, why are new corridors being subsidized for development? [I do actually know the answer to this, it was a rhetorical question].
Cross-posted at streets.mn.
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