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https://politics.stackexchange.com/questions/13176/could-the-us-president-be-elected-by-kevin-costner | 1,722,705,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00382.warc.gz | 384,755,715 | 39,564 | # Could the US President be elected by Kevin Costner?
In the 2008 movie "Swing Vote", the presidential election is essentially tied. The only state that hasn't reported in is New Mexico, because it's waiting for a single precinct, which is waiting for a single voter, played by Kevin Costner.
I've always wondered if the first part of that scenario could happen: could New Mexico decide the electoral vote?
The country currently has 538 electoral votes, of which New Mexico has 5. I think this means that New Mexico could decide the election if Candidate A had at least 265, Candidate B had 268 at most, and their total would have to be 533.
My questions are:
1) Is my math correct?
2) Since each state has a specific, non-random number of electoral votes, could this situation arise?
• In the movie, is Costner a member of the Electoral College casting his electoral vote or a normal citizen voting on Election Day? And what is the reason that the nation is waiting for him? Commented Nov 10, 2016 at 3:23
• He's a normal citizen. He's the only uncounted vote in his small precinct, and I guess the rest of the state is completely tied. Its a stretch, but it's a comedy-drama. That's partly why my question is about the more realistic issue of whether a state with a small population could swing the electoral vote. Commented Nov 10, 2016 at 3:30
• So the rest of the votes are counted before he voted? And nobody has called for a recount? The short answer is that there is no way that could happen :-) If you'd like to focus on a certain part of the situation, it would be helpful to edit the question to indicate what parts we should suspend our disbelief on. Commented Nov 10, 2016 at 3:45
• I think I already addressed that when I asked "if the first part of that scenario [New Mexico being the remaining swing state] could happen" Commented Nov 10, 2016 at 3:50
• Yeah, I guess you did. Perhaps you could edit your question to include that so future readers don't have to go through the comment chain. Commented Nov 10, 2016 at 3:52
1) Is my math correct?
In so far as it goes, yes, it is. However, see the next point:
2) Since each state has a specific, non-random number of electoral votes, could this situation arise?
The number of electoral votes each state gets changes every 10 years. Each state has a minimum of 3, so in theory there would be some combinations that could never happen (such as 537-1). However, two states (Maine and Nebraska) award an electoral vote to the winner of each district in the state in addition to one for the overall winner, so it should be possible for every numeric combination to happen.
However, the vote of a state is largely influenced by its demographics. So there are certain combinations of votes which are exceedingly unlikely to occur in reality (within any given election - demographics change between elections). That said, not having a winner declared with one state left to report is a very feasible situation - that's what happened in 2000. From Wikipedia:
As the final national results were tallied the following morning, Bush had clearly won a total of 246 electoral votes, while Gore had won 255 votes. 270 votes were needed to win. Two smaller states—New Mexico (5 electoral votes) and Oregon (7 electoral votes)—were still too close to call. It was Florida (25 electoral votes), however, that the news media focused their attention on. Mathematically, Florida's 25 electoral votes became the key to an election win for either candidate. Although both New Mexico and Oregon were declared in favor of Gore over the next few days, Florida's statewide vote took center stage because that state's winner would ultimately win the election. The outcome of the election was not known for more than a month after the balloting ended because of the extended process of counting and then recounting Florida's presidential ballots.
• +1 for the point about Maine and Nebraska, meaning almost any combination of electoral votes could occur, which almost completely answers my question. Commented Jan 19, 2017 at 19:19
Your maths are correct. They exclude the case where a tie would appear (I am not sure what would happen then).
There may be plenty of configurations for the other states where this would occur. To determine one is a purely mathematical problem (known as the subset sum problem). I can display one if you wish. I did not see the movie. Do they say what happens in each of the other states ?
• In the event of an electoral college tie, or indeed any scenario in which no candidate receives a majority of electoral college votes, the election is decided by the houseof representatives. Commented Dec 24, 2016 at 5:30
• I'd like to see your example. (In the movie, I don't recall what happened in the other states, but I think they said the electoral vote was tied otherwise.) Commented Dec 27, 2016 at 19:59 | 1,095 | 4,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-33 | latest | en | 0.983179 |
https://discourse.llvm.org/t/missed-optimizations/11225 | 1,669,825,890,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00178.warc.gz | 248,451,759 | 7,044 | missed optimizations
Hi,
I have two questions about optimizations performed by llvm.
Consider these simple functions:
int x(int b) { return b?4:6; }
int y() { return x(0); }
int x2() { return 5; }
int y2() { return x2(); }
the optimized bitcode (with clang + opt -std-compiler-opts) is:
define i32 @y(...) nounwind {
entry:
ret i32 6
}
define i32 @y2(...) nounwind {
entry:
%call = call i32 (...)* @x2( ) ; <i32> [#uses=0]
ret i32 5
}
So why does LLVM optimizes the more difficult case, but leaves behind the function call in the easiest case?
Second question:
int f();
int g() {
if (f() == 5) return 3;
return 4;
}
int h() {
if (g() == 6) return 2;
return 1;
}
gives the following optimized bc:
define i32 @g(...) nounwind {
entry:
%call = call i32 (...)* @f( ) ; <i32> [#uses=1]
%cmp = icmp eq i32 %call, 5 ; <i1> [#uses=1]
%retval = select i1 %cmp, i32 3, i32 4 ; <i32> [#uses=1]
ret i32 %retval
}
define i32 @h(...) nounwind {
entry:
%call = call i32 (...)* @g( ) ; <i32> [#uses=1]
%cmp = icmp eq i32 %call, 6 ; <i1> [#uses=1]
%retval = select i1 %cmp, i32 2, i32 1 ; <i32> [#uses=1]
ret i32 %retval
}
In function h(), llvm doesn't realize that g() never returns 6, and thus it could reduce that function to { g(); return 1; }. Doesn't llvm produce some sort of summary for functions with e.g. their possible return values? If not, is there any pass where I could implement this? (I have some code that has many opportunities for this kind of optimization)
Also, shouldn't the function g() get inlined in the h() function? It is inlined only if I change the g() function to be static. So isn't llvm being too conservative when inlining global functions?
Thanks,
Nuno
Nuno Lopes a écrit :
Hi,
I have two questions about optimizations performed by llvm.
Consider these simple functions:
int x(int b) { return b?4:6; }
int y() { return x(0); }
int x2() { return 5; }
int y2() { return x2(); }
the optimized bitcode (with clang + opt -std-compiler-opts) is:
define i32 @y(...) nounwind {
entry:
ret i32 6
}
define i32 @y2(...) nounwind {
entry:
%call = call i32 (...)* @x2( ) ; <i32> [#uses=0]
ret i32 5
}
So why does LLVM optimizes the more difficult case, but leaves behind the function call in the easiest case?
I don't know why vararg would inhibit optimisation but maybe it is the problem.
You should try with:
int x2(void) { return 5; }
int y2(void) { return x2(); }
it may solve the problem (or use C++ as langage)
Cédric
The LLVM browser test optimizes both examples as expected... What
optimization options are different? Although that optimization could
be gcc-llvm level and not actual llvm level...
Hi,
I have two questions about optimizations performed by llvm.
Consider these simple functions:
int x(int b) { return b?4:6; }
int y() { return x(0); }
int x2() { return 5; }
int y2() { return x2(); }
the optimized bitcode (with clang + opt -std-compiler-opts) is:
define i32 @y(...) nounwind {
entry:
ret i32 6
}
define i32 @y2(...) nounwind {
entry:
%call = call i32 (...)* @x2( ) ; <i32> [#uses=0]
ret i32 5
}
So why does LLVM optimizes the more difficult case, but leaves behind the
function call in the easiest case?
Pretty simple: LLVM doesn't know how to eliminate the call.
There are a couple of ways it could be eliminated: DCE or inlining.
The current DCE infrastructure isn't clever enough to do
interprocedural analysis, so that doesn't eliminate it. And the
inliner immediately gives up on varargs functions, so that doesn't
eliminate it. And
That said, clang really should be turning int x2() { return x(0); }
into "define i32 @x2()" rather than "define i32 @x2(...)"; the
function isn't varargs, and marking it as such could lead to wrong
code for exotic calling conventions.
Second question:
int f();
int g() {
if (f() == 5) return 3;
return 4;
}
int h() {
if (g() == 6) return 2;
return 1;
}
gives the following optimized bc:
define i32 @g(...) nounwind {
entry:
%call = call i32 (...)* @f( ) ; <i32> [#uses=1]
%cmp = icmp eq i32 %call, 5 ; <i1> [#uses=1]
%retval = select i1 %cmp, i32 3, i32 4 ; <i32> [#uses=1]
ret i32 %retval
}
define i32 @h(...) nounwind {
entry:
%call = call i32 (...)* @g( ) ; <i32> [#uses=1]
%cmp = icmp eq i32 %call, 6 ; <i1> [#uses=1]
%retval = select i1 %cmp, i32 2, i32 1 ; <i32> [#uses=1]
ret i32 %retval
}
In function h(), llvm doesn't realize that g() never returns 6, and thus it
could reduce that function to { g(); return 1; }. Doesn't llvm produce some
sort of summary for functions with e.g. their possible return values? If
not, is there any pass where I could implement this? (I have some code that
has many opportunities for this kind of optimization)
No, there isn't any such infrastructure at the moment. The LLVM pass
that might do that sort of thing is predsimplify, but it isn't an
interprocedural pass. It's not necessarily difficult to implement,
depending on how precise you want it to be, but nobody's implemented
it; as far as I know, it doesn't provide significant benefits for
normal C/C++ code. If there's some specific pattern in your code, you
could probably throw together a specialized pass pretty quickly.
Also, shouldn't the function g() get inlined in the h() function? It is
inlined only if I change the g() function to be static. So isn't llvm being
too conservative when inlining global functions?
The inliner doesn't like varargs; see above.
-Eli
Hi Eli,
That said, clang really should be turning int x2() { return x(0); }
into "define i32 @x2()" rather than "define i32 @x2(...)"; the
function isn't varargs, and marking it as such could lead to wrong
code for exotic calling conventions.
I always understood that this is correct per C language specification. For
functions that are internal (static), the DeadArgElim pass removes the vararg
spec if it's unused, but that's not applicable in this case.
Any particular reason the inliner doesn't touch varargs? I think that when you
call a varargs function without any varargs, it should be able to inline it
just fine? Or at least when the function in question also doesn't call the
va_* functions.
> Also, shouldn't the function g() get inlined in the h() function? It is
> inlined only if I change the g() function to be static. So isn't llvm being
> too conservative when inlining global functions?
The inliner doesn't like varargs; see above.
And the fact that it is inlined when it's static, is probably due to the
Gr.
Matthijs
Hi Eli,
That said, clang really should be turning int x2() { return x(0); }
into "define i32 @x2()" rather than "define i32 @x2(...)"; the
function isn't varargs, and marking it as such could lead to wrong
code for exotic calling conventions.
I always understood that this is correct per C language specification.
What's the question here? The C standard says that int(*)() is not a
varargs function, but rather a function without an explicitly declared
argument list, and is actually incompatible with any varargs function.
As far as I can tell, though, the LLVM varargs specifier is more
forgiving; llvm-gcc seems to be willing to call int(*)() using an llvm
varargs type.
Any particular reason the inliner doesn't touch varargs? I think that when you
call a varargs function without any varargs, it should be able to inline it
just fine? Or at least when the function in question also doesn't call the
va_* functions.
I think it's just unimplemented.
-Eli
Hi,
As a follow up of this thread I've made a patch that implements a simple approach to propagate the function return values as described previously.
It can transform e.g. the following program:
define i32 @f(...) nounwind {
(...)
%cond = select i1 %tobool, i32 2, i32 3 ; <i32> [#uses=1]
ret i32 %cond
}
define i32 @g(...) nounwind {
entry:
%call = call i32 (...)* @f() ; <i32> [#uses=1]
switch i32 %call, label %sw.default [
i32 5, label %sw.bb
i32 2, label %sw.bb1
i32 3, label %sw.bb2
]
sw.bb: ; preds = %entry
ret i32 -1
sw.bb1: ; preds = %entry
ret i32 3
sw.bb2: ; preds = %entry
ret i32 55
sw.default: ; preds = %entry
ret i32 0
}
into:
define i32 @g(...) nounwind {
entry:
%call = call i32 (...)* @f() ; <i32> [#uses=1]
%cond = icmp eq i32 %call, 2 ; <i1> [#uses=1]
%retval = select i1 %cond, i32 3, i32 55 ; <i32> [#uses=1]
ret i32 %retval
}
This kind of transformation isn't currently done by LLVM (note that here this pass is only removing case statements. other transformations are not of my responsibility :).
The patch is available at http://web.ist.utl.pt/nuno.lopes/llvm_function_ret_infer.txt
I would love to ear some feedback, so that this pass can eventually be merged to the LLVM tree. In particular I'm not sure if there's a better way to track the possible return values of each function (e.g. is there any code that I can reuse?)
Thanks,
Nuno
AnalyzeFunction can use BB->getTerminator() rather than iterating over
all instructions to find the terminator.
After you Analyze all the functions, why not just iterator on the uses
of the functions you found you could transform rather than on ever
instruction in the module?
It seems you should be able to handle any comparison with a constant
by using the min or max of your set as appropriate to the cmp.
Andrew
Hi,
Thank you for your review. I've updated the patch to reflect your suggestions.
I've also completed the ICmp instruction folding.
Regards,
Nuno | 2,629 | 9,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.657496 |
https://physics.stackexchange.com/questions/141210/about-completeness-relation-from-discrete-to-continuous-limit | 1,725,816,864,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00804.warc.gz | 437,708,280 | 41,444 | # About completeness relation from discrete to continuous limit
In quantum mechanics, the completeness relation for discrete and continuous basis are \begin{align} \sum_n \lvert n \rangle \langle n\rvert &= 1 \tag{1} \\ \int \lvert x \rangle \langle x \rvert \mathrm{d}x = 1 \tag{2} \end{align} respectively. The integral form can be written as $$\lim_{n \rightarrow \infty} \sum_n \lvert x_n \rangle \langle x_n\rvert \Delta x_n=1 \tag{3}$$ such that $$\langle \psi \lvert \psi \rangle= \int \langle \psi \lvert x \rangle \langle x \lvert \psi \rangle \mathrm{d}x = \lim_{n \rightarrow \infty} \sum_n \langle \psi \lvert x_n \rangle \langle x_n\lvert \psi \rangle \Delta x_n \tag{4}$$ like the definition of definite integral.
My question is, suppose space is discrete, Eq. (2) may be written as $$\sum_n \lvert x_n \rangle \langle x_n \rvert =1 \tag{6}$$ If I approach the continuous limit (2) from discrete basis (6), how the $\Delta x_n$ in Eq. (3) should appear?
To add a slightly different angle to PhotonicBoom's sound answer, the link between the two entities - discrete sum and integral - is the concept of measure, not of limit. You can think of your sum as a Lebesgue integral if you choose a discrete measure for the real line with the measure's "anchors" at a countable set of "allowed values". Discrete and continuous measures are highly analogous insofar that they both have all the "Real MacCoy" properties of measures: non-negativity, positivity and countable ($\sigma$-) additivity. Ultimately, though, the two are as different as are $\aleph_0$ and $\aleph_1$, dramatically illustrated by the Cantor Slash argument: a quantum observable which can in principle yield any real number in an interval as a measurement and one which can only have discrete values as measurements are very different beasts.
In quantum mechanics, or at least all the QM I've seen(see footnote), one makes an assumption of a separable or a first countable Hilbert space for the state space. This means, in effect, that there exists a countable basis for the space of states: for example, the quantum harmonic oscillator's state can be expressed as a superposition of the countable set of energy eigenstates. So in "normal QM", there is always a co-ordinate transformation which will turn an integral completeness relation into a discrete one, although, at the same time, you are changing the observable whose eigenstates span the state space.
Footnote: Separability is part of the Wightman axioms. But sometimes the quantum field theorists drop even this assumption, although I understand that they still assume a separable subspace containing physical fields embedded in a non-separable state space of potential fields.
The integral form is not an alternative way of writing the completeness relation. The integral form is the definition of the completeness relation ONLY when a continuous variable is involved.
What you have included in your question, equation (2) only shows how to compute the completeness relation when the number of states are infinite i.e the spatial configuration x can be infinitely divisible therefore a sum is useless and integration is the solution to the problem.
Answering your actual question now. If space is discrete, the it has a discrete basis and (6) IS the definition of the completeness theorem.
• My question is, is there any way to approach the continuous limit (3) from the discrete basis (6) and vice versa? If the answer is yes, then how should $\Delta x$ come? Commented Oct 14, 2014 at 3:33
• The answer is no. (3) is only the formal definition, or derivation if you like, of (2). Therefore only the discrete basis can be used. Commented Oct 14, 2014 at 3:35 | 890 | 3,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.865081 |
https://www.coursehero.com/file/153048/Spring-2005-Nagys-Class-Quiz-3/ | 1,498,204,880,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00479.warc.gz | 848,925,838 | 24,321 | Spring 2005 - Nagy's Class - Quiz 3
# Spring 2005 - Nagy's Class - Quiz 3 - Math 20F Quiz...
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Unformatted text preview: Math 20F Quiz 3 (version 1) May 13, 2005 1. (3.2.29) Compute det B 5 , where B = 1 1 1 1 2 1 2 1 . det B = 1 1 1 1 2 1 2 2 = 1 1 1 1 1 2 , by subtracting the first column from the third column of B . Thus, det B = 1 · 1 1 2 =- 2. Since the determinant is multiplicative, det B 5 = (det B ) 5 = (- 2) 5 =- 32. 2. (4.4.27) Use coordinate vectors to test the linear independence of the set 1 + t 3 , 3 + t- 2 t 2 ,- t + 3 t 2- t 3 of polynomials. Explain your work. The standard basis E for 3 is { 1 , t, t 2 , t 3 } . Hence, 1 + t 3 E = 1 1 , 3 + t- 2 t 2 E = 3 1- 2 ,- t + 3 t 2- t 3 E = - 1 3- 1 The set 1 1 , 3 1- 2 , - 1 3- 1 is linearly independent since none of the vectors is a linear combination of the other two, as can be seen by observing the location of the zero entries.combination of the other two, as can be seen by observing the location of the zero entries....
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## This note was uploaded on 04/23/2008 for the course MATH 20F taught by Professor Buss during the Spring '03 term at UCSD.
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Spring 2005 - Nagy's Class - Quiz 3 - Math 20F Quiz...
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Ask a homework question - tutors are online | 554 | 1,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-26 | longest | en | 0.845818 |
https://www.mathworks.com/matlabcentral/answers/16-how-to-extract-leading-non-zero-digit | 1,675,505,680,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00130.warc.gz | 928,392,851 | 29,302 | How to extract leading non-zero digit?
24 views (last 30 days)
John Booker on 4 Jan 2011
Commented: Asif Newaz on 22 Nov 2019
I'm working on a research problem related to Benford's Law which states that the distribution of leading digits is not random. This is probably because many things grow logarithmically. I am trying to extract the leading digit from these vectors below:
10 --> 1
13 --> 1
0.3 --> 3
-4 --> 4
-5 --> 5
-0.006 --> 6
Input will be a vector
x = [1 0.3 -2 0.001 -0.0006, 582398, 3020];
Output should be
y = [1 3 2 1 6 5 3];
Any help?
Walter Roberson on 20 Jan 2011
The answer is complicated because numbers such as 0.0006 have no exact representation in binary floating point number, and the closest number that can be represented might not have the same leading binary digit.
If one does NOT take that factor in to account, then:
y = floor(abs(x) ./ 10.^floor(log10(abs(x))));
John Booker on 1 Aug 2013
Sorry it took me so long to accept. I stole this question from Ned before Answers launched and forgot about it =).
Ned Gulley on 11 Jan 2011
You'll probably need to do some kind of textual manipulation. Here's one way to do it.
s = sprintf('%1.2e\n',abs(x));
y = s(1:(length(s)/length(x)):end)-48;
end
Daniel Shub on 28 Jun 2012
Is this really the first question on Answers? I was going to use our new magic power and start accepting answers. The only problem is I think there is now a better answer to this question on Loren's blog.
Jan on 28 Jun 2012
I thought of accepting this answer, because it contains a link to good solutions. But actually the method shown by Walter hits the point (and is found in this blog also). Ned's method is more efficient than STR2DOUBLE, but a clean numerical approach is moire direct for a numerical question.
Oleg Komarov on 5 Sep 2012
Another solution based on regexp (from this question):
regexp(num2str(x), '(?<=(^|\s+)[\-\.0]*)[1-9](?=[\d\.]*)', 'match')
Asif Newaz on 22 Nov 2019
can u explain the 'expression'... i've found regexp quite complicated but very useful
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https://www.gamedev.net/forums/topic/592824-calculating-speed-based-on-travel-distance/ | 1,544,479,723,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00611.warc.gz | 930,542,451 | 27,781 | # Calculating speed based on travel distance
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Hello,
I have the following problem. There are two(or more) objects that follow a path through a level starting at different positions. Now those objects should theoretically be able to hit each other when they cross paths, no matter at what position in the level they started. There is a calculation i'am missing here to accurately set the speed based on the overall length of the path. Just setting the speed to pathLength*scale doesn't give correct results and the objects are missing each other.
Any help or pointer in the right direction would be appreciated.
Edit: Sorry i think i postet this in the wrong forum, could someone move this to Math and Physics ?
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I don't know if I understand you correctly, but I'll try to explain the answer to what I believe you are asking. It is quite simple actually.
Take a look at HelpfulPic.png, it names a few things of interest. First there is POI, a shorthand for Point Of Intersection. You must find the point of intersection between path 1 and path 2. Then there is len1, the length from the start of path 1 to the POI. Same with len2, the length from the start of path 2 to the POI. There is one variable missing though, time. You must set an arbitrary amount of time that you would like to pass before the two objects collide, for example 2.0 seconds or 4 minutes, etc. With these variables you can use the following equations:
Object1_Speed = len1 / t Object2_Speed = len2 / t
These two speeds will make it so object 1 and object 2 collide.
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Yea time it is what i was missing. What i forgot to say is that those two objects might start at different times(1-2 seconds difference), so i need to take this into account.
Thank you very much!
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t in the formula doesn't have to be the same for all objects it needs to be (tcollided - tstarted) aka how long this specific object was traveling.
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https://solvedlib.com/n/what-is-the-average-acceleration-of-the-passengers-on-this,3051158 | 1,660,367,698,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00799.warc.gz | 480,270,834 | 18,025 | # What Is the average acceleration of the passengers on this ride? Express your answer using two significant figures
###### Question:
What Is the average acceleration of the passengers on this ride? Express your answer using two significant figures
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# MECHANICAL ENGINEERING INTERVIEW QUESTIONS
1. Ans. 2. Why Entropy decreases with increase in temperature? ds=dQ/T Entropy is inversely proportional to the temperature so, as temp. Increases, entropy decreases. Why different types of sound are produced in different bikes, though they run on SI Engines?
Ans. Engine specifications are different in different manufactures like as Bore Diameter (CC), Ignition timing. Also the exhaust passage takes more responsible for sound. 3. Ans. 4. Ans. 5. How much Watt means 1Hp? 746.2 Watt Explain Bicycle Rear Wheel Sprocket working? Rear wheel sprocket works under the principle of ratchet and pawl. Definition of Octane Number and Cetane Number?
Ans. Octane No.- Octane number is defined as the percentage, by volume, of iso octane in the mixture of iso octane and hheptane. It is the measure of rating of SI engine. Cetane No.- Cetane number is defined as the percentage, by volume, of n-cetane in the mixture of n-cetane and alpha methyl naphthalene. It is the measure of rating of CI engine. 6. Ans. 7. Ans. 8. Which Mechanism is used in Automobile gearing System? Differential mechanism When Crude Oil is Heated, Which Hydro Carbon comes first? Natural gas (Gasoline) at 20 Celsius How to calculate Bearing number ti Diameter of the inner and outer?
Ans. Divide the shaft diameter size by 5, it will give last two digit of the bearing no. and according to type of load we have to choose the type of bearing and that will give prior number of the bearing. 9. Ans. 10. The Fatigue life of a part can be improved by? Improving the surface finish by Polishing & providing residual stress by Shot peening. What happens if gasoline is used in a Diesel Engine, Siesel Engine will work?
Ans. No, It will not work, as the Compression ratio of Petrol engine is 6 to 10 & that of Diesel engine is 15 to 22. Thus on such high compression, gasoline gets highly compressed & it may blast. 11. Poissons Ratio is Higher in, Rubber/Steel/Wood?
Ans. When a material is compressed in one direction, it usuallytends to expand in the other two directions perpendicular tothe direction of compression. This phenomenon is called thePoisson effect. Poissons ratio is a measure of the Poisson effect. For rubber = 0.5 For steel = 0.288 For wood < 0.2 Thus Poissons ratio is higher in RUBBER. 12. Why the Centrifugal Pump is called High Discharge pump?
Ans. Centrifugal pump is a kinetic device. The centrifugal pump uses the centrifugal force to push out the fluid. So the liquid entering the pump receives kinetic energy from the rotating impeller. The centrifugal action of the impeller accelerates the liquid to a high velocity, transferring mechanical (rotational) energy to the liquid. So it discharges the liquid in high rate. It is given in the following formulae: Centrifugal force F= (M * V )/R. Where, M-Mass V-Velocity R-Radius
2
13. How Cavitation can be eliminated by Pump? Ans. 14. Cavitation means bubbles are forming in the liquid. To avoid Cavitation, we have to increase the Pump size to One or Two Inch; To increase the pressure of the Suction Head, or Decrease the Pump Speed.
Why Cavitation will occur in Centrifugal Pump and not in Displacement Pump?
Ans. The formation of cavities (or bubbles) is induced by flow separation, or non-uniform flow velocities, inside a pump casing. In centrifugal pumps the eye of the pump impeller is smaller than the flow area of pipe. This decrease in flow area of pump results in increase in flow rate. So pressure drop happened between pump suction and the vanes of the impeller. Here air bubbles or cavities are formed because of liquid vapour due to increase in temperature in impeller. This air bubbles are transmitted to pump which forms cavitation. 15. Which Pump is more Efficient Centrifugal Pump or Reciprocating Pump?
Ans. Centrifugal pump. Because flow rate is higher compared to reciprocating pump. Flow is smooth and it requires less space to install. Lower initial cost and lower maintenance cost. 16. Why Centrifugal Pump is not called as a Positive Displacement Type of Pump?
Ans. The centrifugal has varying flow depending on pressure or head, whereas the Positive Displacement pump has more or less constant flow regardless of pressure. Likewise viscosity is constant for positive displacement pump where centrifugal pump have up and down value because the higher viscosity liquids fill the clearances of the pump causing a higher volumetric efficiency. When there is a viscosity change in supply there is also greater loss in the system. This means change in pump flow affected by the pressure change. One more example is, positive displacement pump has more or less constant efficiency, where centrifugal pump has varying efficiency rate. 17. What is the difference between Critical Speed and Whirling Speed?
Ans. In Solid mechanics, in the field of rotor dynamics, the critical speed is the theoretical angular velocity which excites the natural frequency of a rotating object, such as a shaft, propeller or gear. As the speed of rotation approaches the objects natural frequency, the object begins to resonate which dramatically increases system vibration. The resulting resonance occurs regardless of orientation.Whirling Speed is due to the unbalanced forces acting on a rotating shaft. 18. How a Diesel Engine Works as Generator?
Ans. Diesel engine is a prime mover, for a generator, pump,and for vehicles etc. generator is connected to engine by shaft. mostly in thermal power plat ,there is an engine is used to drive generator to generate power. 19. Ans. 20. Explain Second Law of Thermodynamics? The entropy of the universe increases over time and moves towards a maximum value. Compare Brayton Cycle and Otto Cycle?
Ans. The heat addition and rejection processes in Otto cycle are of constant volume, whereas in Brayton cycle, they are of constant pressure. -Otto cycle is the ideal cycle for spark ignition engines. -Brayton cycle is the ideal cycle for gas power turbines. 21. Ans. 22. What is the purpose of Scrapper Ring? scrap the excess lube oil from the cylinder walls. there by preventing oil from entering combustion zone. What is DTSI Technology?
Ans. DTSI stands for Digital Twin Spark Plug Ignition. The vehicles with DTSI Technology use 2 spark plugs which are controlled by digital circuit. It results in efficient combustion of air fuel mixture.
Digital - Since the spark generation will be initiated by a microchip. Twin - Since two spark plugs will be used. Spark ignition - Since the ignition will be done via a spark.
23.
## How to Find, Ductile-Brittle Transition Temperature in Metals?
Ans. The point at which the fracture energy passes below a pre-determined point for a standard Impact tests. DBTT is important since, once a material is cooled below the DBTT, it has a much greater tendency to shatter on impact instead of bending or deforming. 24. What is the importance of Thermodynamics?
Ans. All the mechanical engineering systems are studied with the help of thermodynamics. Hence it is very important for the mechanical engineers. 25. Ans. What is the difference between P11 and P12 Pipes? P11 the chromium molybdenum composition that is 1% ofchromium and 1/4% of molybdenum
P12 the chromium molybdenum composition that is 1% ofchromium and 2% of molybdenum 26. State difference between AnitiFriction Bearing and Journal Bearing?
Ans. Generally, journal bearings have higher friction force, consume higher energy and release more heat, but they have larger contact surface, so normally used in low speed high load applications. In anti friction bearings friction is less. One object just rolls over each other. 27. What is the difference between Fan and Blower?
Ans. Fan is an air pushing device. Either Axial or Centrifugal type systems are used to move the air in low pressure. It is rotated by a motor separately. When the fan is a housing of blades and motor, then it called as Blower. It directs the air in a single path with high pressure. 28. What is Heat Rate of Power Plant?
Ans. Heat rate is a measure of the turbine efficiency. It is determined from the total energy input supplied to the Turbine divided by the electrical energy output 29. Ans. What is Hydrodynamic Cavitation? Hydrodynamic cavitation describes the process of vaporization in a constrained channel at a specific velocity.
Bubble generation and Bubble implosion which occurs in a flowing liquid as a result of a decrease and subsequent increase in pressure. 30. Ans. What is Difference between Hardness and Toughness? Toughness is the ability of a material to absorb energy.
Hardness is the ability of a material to withstand wear. 31. Ans. 32. What is the hottest part of Refrigerant in Refrigeration? Compressor What does F.O.F Stand for Piping Design?
Ans. Face of Flange, The F.O.F (Raised face and Flat face) is used to know the accurate dimension of the flange in order to avoid the minute errors in measurement in case of vertical or horizontal pipe lines. 33. Where Multi stage Pump is Used?
Ans. Pressure washing of Aircraft, Trains, Boats and Road vehicles as well as Spray washing of industrial parts and Electronic components. 34. Ans. 35. What does pump develop:- (a) Flow (b) Pressure A pump does not create pressure, it only creates flow. Pressure is a measurement of the resistance to flow. How to measure temperature of Bearing?
Ans. Normally the temperature range from 80 degree to 110 degree Celsius. So the following temperature measurement devices are used. Dial Type Bearing Temperature Detector Stator Winding Temperature Detector Resistance Bearing Temperature Detector (Bearing RTD)
Specialized in Pin type RTD 35. Difference between Performance and Efficiency?
Ans. The accomplishment of a given task measured against preset known standards of accuracy, completeness, cost, and speed is called as Performance. Efficiency is defined as the input given and the work obtained from that input like money, time, labour etc. Its the main factor of productivity. 36. What is Sentinel Relief Valve?
Ans. Its a special type valve system. The valve will open when exhaust casing pressure is excessive (high). The valve warns the operator only; it is not intended to relieve the casing pressure. 37. Ans. What is the difference between Specification,Codes, Standards? Specification is describing properties of any type of materials.
Code is procedure of acceptance and rejection criteria. Standard is accepted values and compare other with it. 38. Ans. 39. Which is heavier 1kg Cotton or 1kg Iron? Both of them have same weight. What is Auto Dosing?
Ans. Auto dosing is an automated system of feeding the equipment with liquid products. It is the ideal way to ensure the correct calibrated dose at the right time every time in auto. 40. Ans. What is the difference between Sudden Force and Impact Force? An impact is a high force or shock applied over a short time period when two or more bodies collide.
A force which applies on the body (material) suddenly is known as sudden force. 41. Ans. 42. What is Geyser Pressure Valve? To release the pressure created inside due to evaporation of water. What is difference between Corrective actions and Preventive actions?
Ans. Corrective actions are taken on discrepancies noticed during inspection of products/documents/process whereas preventive actions are taken to eliminate the possibility ofdiscrepancy in future. 43. Ans. 44. How do you know air is fully saturated? In its Dew point. (100% relative humidity condition) What is the function of scoop in BFP (Boiler Feed water pump) in Thermal Power Station?
Ans. The Function of Scoop tube is regulating the varying amount of oil level in the coupling during operation of infinite variable speed. 45. What is Operating Pressure?
Ans. The amount of pressure nearest the point of performing work at the output end of a pneumatic system. The system operating pressure is used to specify the capability of valves and actuators. 46. What is the difference between Technology and Engineering?
Ans. Engineering is application of science. Technology shows various methods of Engineering. A bridge can be made by using beams to bear the load,by an arc or by hanging in a cable; all shows different technology but comes under civil engineering and science applied is laws of force/load distribution. 47. How to Measure Temperature in Wet Bulb Thermometer?
Ans. Wet bulb temperature is measured in a wet bulb thermometer by covering the bulb with a wick and wetting it with water. It corresponds to the dew point temperature and relative humidity.
48. Ans.
What are the Advantages and Disadvantages of using LPG in Car? Advantages
1. Complete combustion 2. Fuel saving 3. Homogenous combustion Disadvantages 1. As complete combustion is occurring ,more heat liberated,not advised for long journey, engine will be over heated 2. Installation is difficult 3. Reduce engine life efficiency 49. What is the difference between Speed and Economic Speed?
Ans. The rated speed tells us about the maximum speed which can be achieved by a vehicle or some other machine but the economical speed means the speed limit at which the machine works efficiently with least consumption of fuel.eg-in normal bikes(not racing),the max.speed limit shown on speedometer is upto 120 kmph but companies always advice their customers to drive such bikes at around 60 kmph to have maximum mileage. 50. What is Powder Technology?
Ans. Powder technology is one of the ways of making bearing material. In this method metals like bronze, Al, Fe are mixed and compressed to make an alloy. 51. Ans. State all the laws of Thermodynamics? There are three laws of the thermodynamics.
First Law: Energy can be neither created nor destroyed. It can only change forms. In any process in an isolated system, the total energy remains the same. Second Law: When two isolated systems in separate but nearby regions of space, each in thermodynamic equilibrium in itself, but not in equilibrium with each other at first, are at some time allowed to interact, breaking the isolation that separates the two systems, and they exchange matter or energy, they will eventually reach a mutual thermodynamic equilibrium. The sum of the entropies of the initial, isolated systems is less than or equal to the entropy of the final exchanging systems. In the process of reaching a new thermodynamic equilibrium, entropy has increased, or at least has not decreased. Third Law: As temperature approaches absolute zero, the entropy of a system approaches a minimum. 52. State the difference between Unilateral and Bilateral Tolerance?
Ans. A unilateral tolerance is tolerance in which variation is permitted only in one direction from the specified direction.e.g. 1800 +0.000/-0.060 Bilateral tolerance is tolerance in which variation is permitted in both direction from the specified direction.e.g. 1800 +0.060/-0.060 53. Ans. What is the abbreviation of welding rod 7018? 7018 =
70=tensile strength 70000psi 1= welding position 8=current flux 54. Ans. What is difference between Welding and Brazing? In Welding concentrated heat (high temperature) is applied at the joint of metal and fuse together.
In Brazing involves significantly lower temperatures and does not entail the melting of base metals. Instead, a filler metal is melted and forced to flow into the joint through capillary action. 55. Ans. Which has more Efficiency Diesel Engine or Petrol Engine? Diesel engine has the better efficiency out of two.
## 1. What is the difference between isotropic and anisotropic materials ?
Ans: If a material exhibits same mechanical properties regardless of loading direction, it is isotropic, e.g., homogeneous cast materials. Materials lacking this property are anisotropic. 2. What are orthotropic materials ? Ans: It is a special class of anisotropic materials which can be described by giving their properties in three perpendicular directions e.g. wood; composites. 3. What is view factor ? Ans: View factor is dependent upon geometry of the two surfaces exchanging radiation. 4. What properties need to be considered for applications calling for following requirements : (i) rigidity (ii) strength for no plastic deformation under static load (iii) strength to withstand overload without fracture. (iv) wear resistance (v) reliability and safety. Ans: (i) RigidityElastic modulus and yield strength (ii) Strength (for no plastic deformation under static loading)yield point (iii) Strength (overload)Toughness and impact resistance (iv) Wear resistanceHardness (v) Reliability and safetyEndurance limit and yield point. 5. Explain the effects of alloying chromium and nickel in stainless steel. Ans: Addition of nickel and chromium increases the tensile strength and increase in resistance to corrosion takes place. 6. Mention two types of dislocations. Ans: Dislocation refers to a break in the continuity of the lattice. In edge dislocation, one plane of atoms gets squeezed out. In screw dislocation the lattice atoms move fom their regular ideal positions. 7. What are the principal constituents of brass? Ans: Principal constituents of brass are copper and zinc. 8. What is Curie point ? Ans: Curie point is the temperature at which ferromagnetic materials can no longer be magnetised by outside forces. 9. Specific strength of materials is very high when they are in fibre size but lower when they are in bar form Why ? Ans: Crystal structure has ordered, repeating arrangement of atoms. Fibres are liable to maintain this and thus have high specific strength. As size increases, the condition of ordered and repeating arrangements can't be guaranteed because of several types of defects and dislocations and thus the specific strength gets lower. 10. What is the percentage of carbon in cast iron ? Ans: 2.5%. 11. Which element is added in steel to increase resistance to corrosion ? Ans: Chromium. 12. Whether individual components in composite materials retain their characteristics or not? Ans: yes. 13. An elastomer is a polymer when its percentage elongation rate is ? Ans: Greater than 100%. 14. If percentage elongation of a material is more than 200%, it is classed as ? Ans: Rubber. 15. Why is it that the maximum value which the residual stress can reach is the elastic limit of the material ? Ans: A stress in excess of elastic limit, with no external force to oppose it, will relieve itself by plastic deformation until it reaches the value of the yield stress. 16. Why fatigue strength decreases as size of a part increases beyond around 10 mm? Ans: Perfection of material conditions is possible at lower sizes and as size increases, it is not possible to attain uniform structure of the material. 17. Distinguish between creep and fatigue. Ans: Creep is low and progressive deformation of a material with time under a constant stress at high temperature applications. Fatigue is the reduced tendency of material to offer resistance to applied stress under repeated or fluctuating loading condition. 18. While normal carburising and nitriding surface treatments increase fatigue strength, excessive treatment may decrease the fatigue strength. Why ? .Ans: Normal carburising/nitriding treatments increase volume due to phase transformation at Surface and introduce residual compressive surface stress and thus increase the fatigue strength. By excessive treatment the high compressive stresses are introduced but these are balanced by high internal tensile stresses of equal value and the subsurface fatigue cracks may develop in the regions of high tensile stress and lead to early fatigue failure. 19. List at least two factors that promote transition from ductile to brittle fracture. Ans: Manner of loading, and the rate of loading promote transition from ductile to brittle fracture. A machine member may have ductile failure under static loading but may fail in brittle fashion when the load is fluctuating. Similarly a material may evidence ductile failure under tensile loading at ordinary testing speed but if load is applied at a high velocity then failure may be brittle.
20. Which theories of failure are used for (a) ductile materials, and (b) brittle materials ? Ans: For ductile materials, theories of failure used are maximum shear stress theory, and maximum energy of distortion theory; while for brittle materials, theory of maximum principal stress, and maximum strain are used. 21. What does thermal diffusivity of metals signify. Ans: Thermal diffusivity is associated with the speed of propagation of heat into solids during changes in temperature with time. 22. For conduction of heat, the instantaneous rate of heat flow is product of three factors. What are these ? Ans: (i) Area of the section of the heat flow path, perpendicular to the direction of heat flow. (ii) temperature gradient, i.e. change of temperature w.r.t. length of path. (ii) Thermal conductivity of material. 23. How convective heat transfer is effected and on what factors it depends ? Ans: Convective heat transfer is effected between a solid and fluid by a combination of molecular conduction within the fluid in combination with energy transport resulting from the motion of fluid particles. It depends on boundary layer configuration, fluid properties and temperature difference. 24. Which is the common element between brass and bronze ? Ans: Copper. 25. What does following alloy designation indicate FG 250 ? Ans: Grey cast iron with tensile strength of 250 MPa. 26. How is ceramic defined ? Ans: It is a solid formed by combination of metallic and non-metallic elements. 27. Give one example of metal classified as per structure as BCC, FCC, HCP and CCP. Ans: BCC (body centred cubic) structureMolybdenum FCC (face centred cubic) structureAluminium HCP (hexagonal closed packed) structureZinc CCP (cubic dosed packed) structure-Copper. 28. What is the name of solid solution of carbon in alpha iron and delta iron ? Ans: Ferrite and austenite respectively. 29. Explain the difference between pearlite and cementile ? Ans: Pearlite is eutectoid mixture of ferrite and cementile. Cementite is chemical compound of iron and carbon. 30. Give one example each of the following proportion of materials dimensional, physical, technological and mechanical. Ans: Roughness, enthalpy, toughness, and hardness respectively. 31. For which parts the Wahl factor and Lewis form factor used ? Ans: For springs and gears respectively. 32. How oxygen can be removed from steel during melting? What are fully killed steels ? Ans: Oxygen can be removed by adding elements such as manganese, silicon or aluminium which, because of their high affinity for oxygen, react with it to form non-metallic oxides which rise into the slag. Steels which have had most of their dissolved oxygen removed are called "fully killed steels". 33. Hydrogen cannot be removed easily from molten steel. What harm hydrogen has on property of steel ? Ans: Execessive hydrogen results in the formation of small fissures often described as hairline cracks or flakes in the steel. Large forgings in alloy steel are particularly sensitive to this phenomenon. 34. What is allotrope ? In what forms of cubic pattern, iron exists ? Ans: Some elements exist in more than one crystalline form. Each form is known as "allotrope". Iron exists in two forms of cubic pattern, namely body centered cubic (bcc) and face-centered cubic (fee). 35. What is the difference between alpha iron, delta iron and gamma iron ? Ans: The bcc form of iron exists between room temperature and 910C, and between 1400C and the melting point at 1539C. The lower temperature form is known as "alpha"-iron and the higher temperature form as "delta"-iron. The face-centered cubic form existing between 910C and 1400C is referred to as "gamma-iron". 36. Metals, in general are of low strength and do not possess required physio-chemical and technological properties for a definite purpose. Alloys are therefore more than metals alone. Discuss the arrangement of atoms and structures of alloys. Ans: Alloys are produced by melting or sintering two ore more metals, or metals and a nonmetal, together. Alloys possess typical properties inherent in the metallic state. The chemical elements that make up an alloy are called its components. An alloy can consist of two or more components. The phase and structures of alloys describe the constitution, transformations and properties of metals and alloys. A combination of phases in a state of equilibrium is called a system. A phase is a homogeneous portion of a system having the same composition and the same state of aggregation throughout its volume, and separated from the other portions of the system by interfaces. For instance, a homogeneous pure metal or alloy is a singlephase system. A state in which a liquid alloy (or metal) coexists with its crystals is a two-phase system. Structure refers to the shape, size or the mutual arrangement of the corresponding
phases in metals or alloys. The structural components of an alloy are its individual portions, each having a single structure with its characteristic features. 37. What is the difference between isotropic material and homogeneous material ? Ans: In homogeneous material the composition is same throughout and in isotropic material the elastic constants are same in all directions. 38. Explain the difference between the points of inflexion and contraflexure. Ans: At points of inflexion in a loaded beam the bending moment is zero and at points of contraflexure in loaded beam the bending moment changes sign from increasing to decreasing. 39. What is the difference between proof resilience and modulus of resilience ? Ans: Proof resilience is the maximum strain energy that can be stored in a material without permanent deformation. Modulus of resilience is the maximum strain energy stored in a material per unit volume. 40. What is the difference between column and strut ? Ans: Both column and strut carry compressive load. Column is always vertical but strut as member of structure could carry axial compressive load in any direction. 41. Explain the difference between ferrite, austenite and graphite ? Ans: Ferrite is the solid solution of carbon and other constituents in alpha-iron. It is soft, ductile and relatively weak. Austenite is the solid solution of carbon and other constituents in gamma-iron. It exists in ordinary steels at elevated temperatures, but it is also found at ordinary temperatures in some stainless steels. Graphite has a hexagonal layer lattice. ' 42. Explain the terms solid solution, eutectic, eutectoid and peritectic. Ans: Solid Solution. When a homogeneous mixture of two (or more) atomic forms exists in solid state, it is known as solid solution. Eutectic. A mixture of two (or more) phases which solidify simultaneously from the liquid alloy is called an eutectic. Alloys in which the components solidify simultaneously at a constant temperature the lowest for the given system, are called eutectic alloys. Eutectoid. Eutectoid alloys are the alloys for which two solid phases which are completely soluble become completely insoluble on cooling before a certain temperature called eutectoid temperature. Peritectic. A peritectic transformation involves a reaction between a solid and liquid that form a different and new solid phase. This three phase transformation occurs at a point called peritectic point. 43. What do you understand by critical points in iron, iron-carbide diagram ? Ans: The temperatures at which the phase changes occur are called critical points (or temperatures). 45. Why PERT is preferred over CPM for evaluation of project ? Ans: PERT is based on the approach of multiple time estimates for each activity. 46. What is the percentage of chromium in 18 : 4 : 1 IISS ? Ans: 4%. 47. What is stellite ? Ans: It is a non-ferrous cast alloy containing cobalt, chromium and tungsten. 48. Which rays are produced by cobalt-60 in industrial radiography ? Ans: Gamma rays. 49. What are killed steels and what for these are used ? Ans: Killed steels are deoxidised in the ladle with silicon and aluminium. On solidification no gas evolution occurs in these steels because they are free from oxygen. 50. What is critical temperature in metals ? Ans: It is the temperature at which the phase change occurs in metals. 51. Car tyres are usually made of ? Ans: Styrene-butadine rubber. 52. What is the structure of pure iron and whether it is soft or hard ? Ans: Ferrite and it is soft. 53. Which elements increase the corrosion resistance of steel ? Ans: Chromium and nickel. 54. What causes hardness in steel ? How heat treatment alters properties of steel ? Ans: The shape and distribution of the carbides in the iron determines the hardness of the steel. Carbides can be dissolved in austenite is the basis of the heat treatment of steel. If steel is heated above the A critical temperature to dissolve all the carbides, and then cooled, suitable cooling through the cooling range will produce the desired size and distribution of carbides in the ferrite, imparting different properties. 55. Explain the formation of microstructures of pearlite, bainite and martensite in steel. Ans: If austenite containing about 0.80 percent carbon is slowly cooled through the critical temperature, ferrite and cementite are rejected simultaneously, forming alternate plates or lamellae. This microstructure is called pearlite. At temperatures just belot the A1, the transformation from austenite.to pearlite may take an appreciable time to initiate and complete, but the product will be lameller pearlite. As the transformation temperature is lowered, the time to initiate transformation shortens but the product is pearlite of increasing
fineness, and at temperatures approaching 550C it cannot be resolved into its lamellar constituents. Further deerease in transformation temperature causes a lengthening of the ncubation period and a change in structure of the product to a form known as "bainite". If the temperature is lowered sufficiently, the diffusion controlled nucleation and growth modes of transformation are suppressed completely and the austenite transforms by a diffusionless process in which the crystal lattice effectively shears to a new crystallographic configuration known as "martensite". This phase has a tetragonal crystal structure and contains carbon in supersaturated solid solution. 56. How with alloying of steel it is possible to a achieve properties which can not be achieved with heat treatment ? Ans: A prerequisite to the hardening of steels is that martensite should be formed on cooling, but this can only be achieved if the rate of cooling is great enough to suppress the formation of pearlite or bainite and in plain carbon steels this can be achieved by quenching relatively small specimens 57. What are the major effects of alloying elements? Ans: (1) To alter the transformation temperatures and times (2) To modify the room temperature and elevated temperature strengths of given structures by (a) stiffening the crystals and (b) introducing complex precipitates which tend to harden the steel. (3) To modify the type of oxide film formed on the surface of the steel and thereby affect its corrosion resistance. 58. What is the difference between austenite stabilisers and ferrite stabilisers ? Ans: Austenite stabilisers have the effect of extending the temperature range overwhich austenite is formed. Such elements are carbon, manganese, nickel, copper and cobalt. Ferrite stabilisers have the effect of extending the temperature range over which alpha and delta ferrite are formed, which consequently reduces temperature range over which austenite is formed. Such elements are silicon, chromium, molybdenum, tungsten, titanium and niobium. 59. What are the effects of carbon on the properties of steel. Ans: In general, an increase in carbon content produces higher ultimate strength and hardness but lowers ductility and toughness of steel alloys. Carbon also increases airhardening tendencies and weld hardness, especially in the presence of chromium. In low-alloy steel for high-temperature applications, the carbon content is usually restricted to a maximum of about 0.15% in order to assure optimum ductility for welding, expanding, and bending operations. To minimize intergranular corrosion caused by carbide precipitation, the carbon content of austenitic (18-8 type) alloys is limited in commercial specifications to a maximum of 0.08%, or even less, i.e. 0.03% in the extremely low-carbon grades used in certain corrosionresistant applications. In plain carbon steels in the normalised condition, the resistance to creep at temperatures below 440C appears to increase with carbon content up to 0.4% carbon, at higher temperatures there is but little variation of creep properties with carbon content. An increase in carbon content lessens the thermal and electrical conductivities of steel and increases its hardness on quenching. 60. What is the role of silicon as alloying element in steels ? Ans: Silicon contributes greatly to the production of sound steel because of its deoxidizing and degasifying properties. When added in amounts up to 2.5%, the ultimate strength of the steel is increased without loss in ductility. Silicon in excess of 2.5% causes brittleness, and amounts higher than 5% make the steel non-malleable. Resistance to oxidation and surface stability of steel are increased by the addition of silicon. These desirable effects partially compensate for the tendency of silicon to lower the creep properties of steel. Silicon increases the electrical resistivity of steel and decreases hysteresis losses. 61. Discuss the role of manganese in alloying steels. Ans: Manganese is an excellent deoxidizer and sulfur neutralizer, and improves the mechanical properties of steel, notably the ratio of yield strength to tensile strength at normal temperatures. As an alloying element, manganese serves as an inexpensive means of preventing "hot shortness". It improves rolling properties, hardenability, and resistance to wear. However manganese increases the crack sensitivity of weldments, particularly with steels of higher carbon content. 62. Define buckling factor. Ans: It is the ratio of the equivalent length of column to the minimum radius of gyration. 63. What do you understand by catenary cable ? Ans: A cable attached to the supports and carrying its own weight. 64. What is coaxing ? Ans: It is the process of improving fatigue properties by first under-stressing and then increasing the stress in small increments. 65. What is difference between conjugate beam and continuous beam ? Ans: A conjugate beam is an imaginary beam of same size as original beam and carrying a distributed load in accordance with the bending moment diagram. A continuous beam is one which is resting on more than two supports.
66. What is isotropic material ? Ans: It is a material having same elastic constants in all directions. 67. Explain difference between modulus of resilience and modulus of rigidity ? Ans: Modulus of resilience is the maximum strain energy stored in a material per unit volume and modulus of rigidity is the ratio of shearing stress to the shearing strain within the elastic limit. 68. What is the difference between basic hole and basic shaft ? Ans: A basic hole is one whose lower deviation is zero and in case of basic shaft the upper deviation is zero. 69. What for pyranometer is used ? Ans: It is used to measure the total hemispherical solar radiation. 70. Describe transfer machines in brief. Ans: It is an automatic machine in which workpiece alongwith fixture is transferred from one station to other automatically and several operation on workpiece are performed at each station. 71. What is burnt-out point ? Ans: It corresponds to maximum heat flux at which transition occurs from nucleate boiling to film boiling. 72. What do you understand by eutectic ? Ans: It is mechanical mixture of two or more phases which solidify simultaneously from the liquid alloy. 72. Explain the difference between grey iron and white iron. What is mottled iron ? Ans: The carbon in cast iron could exist at room temperature as either iron carbide, or as graphite which is the more stable form. Irons containing carbon as graphite are soft, easily machinable and are called "grey irons". Irons with carbon present as iron carbide are extremely hard, difficult to machine and are called "white" irons. Irons with fairly equal proportions of graphite and iron carbide have intermediate hardness and are called "mottled" irons. 73. The graphite in grey irons exists in the form of flakes which act as stress-raisers under tensile loading and consequently grey irons have relatively low tensile strength and ductility. Still grey iron is extensively used in engineering. Why ? Ans: Grey iron is extensively used in engineering because of following characteristics. (a) Cheapness. (b) Low melting point and high fluidity making it suitable for castings of intricate shape. (c) Relatively good erosion and corrosion resistance. (d) High damping capacity, with respect to vibration. (e) Relatively good mechanical properties under compressive loading. 74. Under what condition a convergent divergent nozzle required ? Ans: When pressure ratio is greater than critical pressure ratio. 75. What is endurance limit and what is its value for steel ? Ans: Endurance limit is the maximum level of fluctuating stress which can be tolerated indefinitely. In most steels this stress is approximately 50% of the ultimate tensile strength and it is defined as the stress which can be endured for ten million reversals of stress. 76. How the net work to drive a compressor and its volumetric efficiency behave with increase in clearance volume ? Ans: Work remains unaltered and volumetric efficiency decreases. 77. What do you understand by sulphur print ? Ans: Sulphides, when attached with dilute acid, evolve hydrogen sulphide gas which stains bromide paper and therefore can be readily detected in ordinary steels and cast irons. While sulphur is not always as harmful as is sometimes supposed, a sulphur print is a ready guide to the distribution of segregated impurities in general. 78. What is the different between brass and bronze ? Ans: Brass is an alloy of copper with zinc; and bronze is alloy of copper with tin. 79. What is the effect of addition of zinc in copper? What is the use of 70/30 brass ? Ans: By addition of zinc in copper, both tensile strength and elongation increases. The 70/30 brass has excellent deep drawing property and is used for making radiator fins. 80, What for admirality brass used ? Ans: Admirality brass with 29% zinc and 1% tin has good corrosion resistance and is used for condenser and feed heater tubes. Aluminium is also added to brass to improve corrosion resistance. 81. What is the maximum use of magnesium ? Ans: Magnesium is used to alloy with aluminium and as an additive for making SG (Spheroidal Graphite) iron. 82. What for zinc finds applications ? Ans: Galvanizing consumes the largest proportion of zinc. Zinc is resistant to corrosion but is attacked by acids and alkalies. Zinc alloy.s are suited for making die casting since the melting point is reasonably low. 83. Which factors influence the type of fracture in failure of a material ? Ans: Seven factors influencing type of failure are : (i) Type of material (inherent structure properties),
(ii) Manner of loading (Static versus dynamic), (iii) Range of imposed stress, (iv) Strain rate (static, dynamic, impact), (v) Stress distribution (discontinuity in material/shape), (vi) temperature, and (vii) surface treatment. 84. What is the name given to ratio of actual cycle efficiency and ideal cycle efficiency. Ans: Efficiency ratio. 85. List two effects of manganese in plain carbon steels. , Ans: Manganese increases tensile strength and hardness. It decreases weldability. 86. Name the strongest and weakest type of atomic bonds. Ans: Metallic bond is strongest and molecular bond also known as Vander Waals bond is weakest. 87. In which process internal energy remains constant ? Ans: Isothermal process. 88. What is temper embrittlement in alloy steels and what are its effects ? Ans: Embrittlement attack is usually intergranular in metals, i.e. cracks progress between the grains of the polycrystalline material. It imparts a tendency to fail under a static load after a given period of time in those alloy steels which are susceptible to embrittlement. 89. What are whiskers ? Ans: Whiskers are very small crystals which are virtually free from imperfections and dislocations. 90. What is Bauschinger effect ? Ans: According to Bauschinger, the limit of proportionality of material does not remain constant but varies according to the direction of stress under cyclic stresses. 91. What is the difference between heat capacity and specific heat of a material ? Ans: The heat capacity of a material is the amount of heat transformed to raise unit mass of a material 1 degree in temperature. The specific heat of a material is the ratio of the amount of heat transferred to raise unit mass of a material 1 degree in temperature to that required to raise unit mass of water 1 degree of temperature at some specified temperature. For most engineering purposes, heat capacities may be assumed numerically equal to;specific heats. 92. Explain the rule to find specific heat of aqueous solutions. Ans: For aqueous solutions of salts, the specific heat can be estimated by assuming the specific heat of the solution equal to that of the water alone. Thus, for a 15% by weight solution of sodium chloride in water, the specific heat would be approximately 0.85. 93. What do you understand by latent heat ? Give four examples of latent heats. Ans: For pure substances, the heat effects accompanying changes in state at constant pressure (no temperature change being evident) are known as latent heats. Examples of latent heats are : heat of fusion, vaporisation, sublimation, and change in crystal form. 94. Define the terms free energy and free enthalpy. What is their significance and importance ? Ans: Free energy (or Helmholtz function) is defined as/= u -Ts. It is equal to the work during a constant-volume isothermal reversible nonflow process. Free enthalpy (or Gibbs function) is defined as g = h - Ts (where u = internal energy, h = enthalpy, T = temperature, s = entropy) Gibbs function is of particular importance in processes where chemical changes occur. For reversible isothermal steady-flow processes or for reversible constant-pressure isothermal nonflow processes, change in free energy is equal to net work. 95.Which parameter remains constant in isochoric process ? Ans: Volume. 96. What is polytropic process ? Under what conditions it approaches isobaric, isothermal, and isometric process ? In which reversible process no work is done ? Ans: A polytropic process is one that follows the equation pun = constant (index n may have values from - oc to + oo. This process approaches isobaric when n = 0, isothermal when n = 1, and isometric when n = <x>. No work is done in isometric process. 97. Whether superheated steam can be treated like ideal gas ? Ans: Yes. 98. Out of constant pressure and constant volume lines on TS diagram which line has higher slope ? And whether slope is constant or variable ? Ans: Constant volume line. Slope is variable. 99. Whether entropy is intensive property or extensive property ? Ans: Entropy is extensive property. 100. In which process fluid expands but does no work ? Ans: Throttling process. 1. How shear forces are incurred in fluid flow ? Ans: Shear forces between fluid particles and boundary walls and between the fluid particles themselves, result from the viscosity of the real fluid.
2. In laminar flow, how discharge is related to viscosity ? Ans: Discharge varies inversely as viscosity. 3. What is best hydraulic section ? Ans: Best hydraulic section is one in which for a given cross-section area, the channel section has least wetted perimeter. 4. What is best hydraulic trapezoidal section ? Ans: It is a half-hexagon in shape. 5. What is specific energy in a channel section ? Ans: It is defined as the energy head measured with respect to the channel bottom at the section. At any section, it is equal to sum of the velocity head and the water depth at the section. 6. What is critical depth and critical flow ? Ans: Critical depth is the depth, at which the discharge may be delivered through the section at minimum energy. Critical flow is the flow in open channel corresponding to critical depth. 7. What causes boundary layer separation ? Ans: An adverse pressure gradient. 8. What is the improtance of Nose radius ? Ans: Nose radius is favourable to long tool life and good surface finish. A sharp point on the end of a tool is highly stressed, short lived and leaves a groove in the path of cut. 9. At what point below free surface in a uniform laminar flow in a channel the point velocity is equal to mean velocity of flow ? Ans: At 0.577 depth of channel. 10. What is the value of friction factor for smooth pile when Reynolds number is approximately = 10G ? Ans: 0.01. 11. How hydraulic grade line and free surface of open channel flow related ? Ans: They coincide. 12.What are the first and last elements of tool signatures ? Ans: Back rake angle and nose radius. 14. What is open channel flow ? Explain 4 types of open channel flows. Ans: Open channel flow has a free water surface which is normally subject to the atmospheric pressure. Steady open channel flow : In this case the discharge and water depth at any section in the reach do not change with time during the period of interest. Unsteady open channel flow : In this case the discharge and the water depth at any section in the reach change with time. Uniform open channel flow : In this case the discharge and the water depth remain the same in energy section in the channel reach. It is mostly steady. Varied open channel {low : In this case the water depth and/or the discharge change along the length of the channel. It may be steady (such as flow over a spillway crest) or unstgeady (as in the case of flood waves or tidal surge). 15. When does wake occur ? Ans: It occurs after a separation point. 16. In straight polarity, to which terminal the electrode is connected ? Ans: Negative. 17. What is hydraulic radius ? Ans: It is the ratio of fluid flow area/shear perimeter. 18. Define flow work. Ans: Flow work is the amount of mechanical energy required to push or force a flowing fluid across a section boundary. 19. What do you understand by dimensional analysis ? Ans: Dimensional analysis is the mathematics of dimensions and quantities and provides procedural techniques whereby the variables that are assumed to be significant in a problem can be formed into dimensionless parameters, the number of parameters being less than the number of variables. 20. What is hydraulic depth for open channel flow ? Ans: It is the ratio of area and the top width of the channel section. 21. Which welding set will you select for welding both ferrous and non ferrous materials. Ans: D.C. generator set. 22. Which welding process is a combination of arc welding and gas welding processes? Ans: Atomic hydrogen welding. 23. What is hydraulic jump ? Ans: It is a natural phenomenon in open channel. It is an abrupt reduction in flow velocity by means of a sudden increase of water depth in the down stream direction. Through a hydraulic jump, and high-velocity supercritical flow is changed to a low-velocity subcritical flow. 24. What is gradually varied flow ? Ans: The changes in water depth in the open channel take place very gradually with
distance. Depending on the channel slope, the surface conditions, the sectional geometry, and the discharge, these may be classified into five categories : steep channel, critical channel, mild channel, horizontal channel, and adverse channel. 25. Explain 5 types of channels. Ans: Steep channel : In this case normal depth < critical depth. Critical channel : In this case normal depth = critical depth. Mild channel : In this case, normal depth > critical depth. Adverse channel : In this case, the slope of channel bed < 0 Horizontal channel : In this case, the slope of channel bed = 0. 26. Mention two major differences between shaper and planer ? Ans: Shaper Planer (a) The cutting tool reciprocates (a) Work reciprocates (b) The work is stationary (b) Tool is stationary (c) Meant for small work pieces (c) Meant for large work pieces (d) Only lighter cut can be taken (d) Heavy cuts can be taken (e) Tools are smaller in size (e) Tools are bigger in size 27. Differentiate between drilling and reaming ? Ans: Drilling Reaming (a) It is the operation of initiating hole (a) Reaming is the operation of finishing and sizing a drilled/bored hole (b) It can be done at relatively high speeds (b) It should be done only at slow speeds (c) Larger chips are produced (c) Less chips are produced (d) Drill has only two flutes (d) Reamer has more than 4 flutes (e) Drill has chisel edge (e) Reamer has a bevel edge 28. What is similarity between normal shock wave and hydraulic jump ? Ans: Both are analogous and irreversible. 29. For what purpose chills are used in moulds ? Ans: Chills are used to achieve directional solidification. 30. What is tumbler gear mechanism ? Ans: Tumbler gear mechanism is a mechanism consisting of number of different sized gears keyed to the driving shaft in the form of cone and is used to give the desired direction of motion of the lathe spindle. 31. What is relative roughness of pipe ? Ans: It is the ratio of the size of the surface imperfections to the inside diameter of the pipe. 32. What is conicity ? Ans: The ratio of the difference in diameter of the taper to its length is termed conicity. 33. Under what condition the effect of compressibility of fluid can be neglected ? Ans: When Mach number is less than 0.4. 34. For which type of flow the velocity distribution in a pipe is parabolic. Ans: For uniform laminar flow. 35. What is the relationship between center line velocity and average velocity for a laminar flow in a pipe ? Ans: Average velocity = 1/2 centre line velocity. 36. On which factor the friction factor for a smooth pipe in turbulent flow depends ? Ans: Reynolds number. 37. What do you understand by pipe branchings ? Ans: When water is brought by pipes to a junction where more than two pipes meet, them the total amount of water brought by pipes to a junction must always be equal to that carried away from the junction by other pipes, and all pipes that meet at the junction must share the same pressure at the junction. 38. What is pipe networks and what are the condition for network ? Ans: Pipe network comprises a number of pipes connected together to form loops and branches. For a network, (i) at any junction, Y.Q = 0 and (ii) between any two junctions the total head loss is independent of the path taken. 39. What is Hardy-Cross Method ? Ans: It is a commonly used computer program for a pipe network. 40. Explain water hammer. Ans: The force resulting from changing the speed of the water mass (say in a pipe due to sudden change of flow rate) may cause a pressure rise in the pipe with a magnitude several times greater than the normal static pressure in the pipe. Pressure head caused by water hammer = C (V velocity in pipe, C = celerity). 41. What happens to impurities in centrifugal casting ? Ans: They get collected at the center of the casting. 42. Define the terms hydraulic similitude and dimensional analysis. Ans: Hydraulic similitude is the principle on which the model studies are based. Dimensional analysis is the analysis of the basic relationship of the various physical quantities involved in the static and dynamic behaviors of water flow in a hydraulic structure.
43. Explain the difference between geometric similarity, kinematic similarity and dynamic similarity. Ans: Geometric similarity : It implies similarity of form. The model is a geometric reduction of the prototype and is accomplished by maintaining a fixed ratio for all homologous lengths between the model and the prototype. Kinematic similarity : It implies similarity in motion. Kinematic similarity between a model and the prototype is attained if the homologous moving particles have the same velocity ratio along geometrically similar paths. It involves the scale of time as well as length. Dynamic similarity : It implies similarity in forces involved in motion. 44. Define Reynolds law ? Ans: When the inertial force and the viscous force are considered to be the only forces governing the motion of the water, the Reynolds number of the model and the prototype must be kept at the same value. 45. When a built up edge is formed while machining ? Ans: While machining ductile materials at high speed. 46. On what factor the friction factor for a rough pipe in turbulent flow depends ? Ans: Relative roughness. 47. What is unit power of a turbine ? Ans: PHP12. 48. How cavitation causes damage ? Ans: In a a closed system (pipelines, or pumps), water vaporises rapidly in regions where the pressure drops below the vapour pressure. This phenomenon is called cavitation. The vapour bubbles formed in cavitation usually collapse in a violent manner, which may cause considerable damage to the system. 49. In milling process, how the arbor torque can be smoothend ? Ans: Arbor torque can be smoothened by increasing number of teeth, using higher spiral angle of teeth, using higher cutting depth. 49. What are webs in a twist drill ? Ans: Webs are the metal column in the drill which separates the flutes. 50. What is bluff body ? Ans: It is a body with such shape that the flow is separated much ahead of its rear end resulting in a large wake, the pressure drag being much greater than the friction drag. 51. What is bulk modulus ? Ans: It is the ratio of hydrostatic stress to volumetric strain within the elastic limits. 52. Define boundary layer. Ans: It is the fluid layer in the neighborhood of a solid boundary where the effects of fluid friction are predominant. 53. What is creep flow ? Ans: It is flow at very low Reynolds number where viscous forces are larger than the inertia forces. 54. What is Torricellis theorem ? Ans: According to it velocity of jet flowing out of a small opening is proportional to the square root of head of liquid above it. 55. How does cavitation manifest itself in a centrifugal pump ? Ans: Usual symptoms of cavitation of pump are noise, vibration, a drop in head and capacity with a decrease in efficiency, accompanied by pitting and corrosion of the impeller vanes. 56. What happens if centrifugal pump is operated at excessive speeds ? Ans: When a pump is operated at excessive speeds, the differential pressure developed increases and causes a powerful pulsating vacuum on the underside of the blade tips and each particle of water is pulled away from the blade, taking away with it a small particle of metal, producing pitting/grooved effect. Repeated erosive action results in complete honey combing and total destruction of blade, with resultant loss in pump performance. 57. On what parameters the head developed by a centrifugal pump depends ? Ans: It depends upon the impeller diameter and its rotative speed. 58. Which factors determine the selection of electrolyte for electrochemical machining process ? Ans: Electrolyte should be chemically stable and have high electrical conductivity. 59. What is the disadvantages of centrifugal pump having very low specific speed ? Ans: For a.centrifugal pump having very low specific speed, the impeller diameter is large and narrow having excessively high disc friction and excessive hydraulic losses. 60. On what factors the efficiency of pump depends ? Ans: Efficiency of a pump (centrifugal) depends on the size, speed, and proportions of the impeller and casing. 61. On what factors the cavitation in centrifugal pumps is dependent ? Ans: Cavitation in centrifugal pumps depends on the velocity of water entering tne impeller and on the relatiave velocity of the impeller blades where the water is picked up. 62. What are the losses in a centrifugal pump ? Ans: The losses in a centrifugal pump are : Friction losses due to eddies in flow, leakage, friction loss due to rotation of impeller in chamber of water, gland and bearing power losses. 63. What should be done to avoid cavitation in pumps ?
Ans: In order to avoid cavitation, the suction lift and the operating speed must be carefully chosen. 64. What will happen if speed of reciprocating pump is increased ? Ans: If speed of a reciprocating pump is increased beyond a limit, then atmospheric pressure would be insufficient to force water into the pump at the same speed as the piston. This would a break in the continuity of the water in suction pipe (cavitation) and give rise to vibration, noise, and chemical attack by any dissolved gases which may be released from the water owing to the high vacuum and the breaking of the water column. 65. What is the order of efficiency of cenetrifugal and reciprocating pumps ? Ans: The efficiency of centrifugal pump is of the order of 45% whereas that of reciprocting pump is around %. 66. What are the basic operations performed in a slotter ? Ans: The different operations done in slotter are : 1. Machining grooves 2. Machining cylindrical surface 3. Machining irregular profiles. 67. How the positive displacement is obtained in rotary pumps ? Ans: Positive displacement in rotary pumps is achieved in two ways : (i) A rotor carries radially adjustable vanes, the outer tips of which are constrained by a circular casing, whose centre is remote from that of the rotor. (ii) Various combinations of gears, lobes, helices, etc. within a casing arranged so that the lobes or teeth attached to each rotor pass in sequence through the same pumping space. 68. How a slotter differs from shaper ? Ans: Slotter Shaper (a) Ram reciprocates vertically (a) Ram reciprocates horizontally (b) It has a circular table (b) It has a rectangular table (c) Suited for machining internal surfaces (c) Suited for machining external surfaces (d) Table can take less load (d) Table can take much load 69. Under what condition the flow and power of pump vary as square of the size ? Ans: The flow and power vary as size2 when the geometric size changes as inverse of change of speed. 70. How the incidence of cavitation detected ? Ans: The incidence of cavitation is detected by the following in order of appearance. (i) stroboscopic observation of bubble appearance and size (ii) sensitive acoustic tests of cavitation noise (iii) reduction of efficiency and/or head of 3%. 71. How cavitation erosion is assessed ? Ans: Cavitation erosion is assessed by depth of attack or weight of metal removed. 72. At which temperature the cavitation erosion is negligible and why ? Ans: Cavitation erosion is negligible at above 300C because the water and steam have same densities at critical temperature of 374C and thus there is no volume change or bubble collapse in the vicinity of this range. 73. What is an arbor ? Ans: An arbor is an accurately machined shaft for holding and driving the arbor type cutter. It is tapered at one end to fit the spindle nose and has two slots to fit the nose keys for locating and driving it. 74. For a body to float in stable equilibrium where the e.g. should be located w.r.t. metacentre ? Ans: e.g. should be below metacentre. 75. What is the condition for a flow to be fully developed through a pipe ? Ans: Ratio of maximum velocity to average velocity should be 2. 76. When a fluid is moving with uniform velocity, whether the pressure of fluid will depend on its depth and orientation ? Ans: No. 77. How much is the vertical component of force on a curved surface submerged in a static liquid ? Ans: It is equal to weight of the liquid above the curved surface. 78. On which factors depends the friction factor in a rough turbulent flow in a pipe? Ans: It depends on pipe diameter and the condition of the pipe. 79. What does the realisation of velocity potential in fluid flow indicate ? Ans: It indicates that flow is irrotational. 80. What is the difference between potential flow and creep flow ? Ans: Potential flow assumes viscous forces to be zero but same is considered larger than inertia forces in creep flow. 81. For which application Mach number and Froude numbers are significant ? Ans: Mach no. is significant in the study of projectiles and jet propulsion. Froude number is significant in the study of ship hulls. 82. What is the difference between geometric, kinematic, and dynamic similarities between model and prototype ? Ans: Geometric similarity is mainly for same shape for model and prototype, kinematic
similarity is concerned with stream line pattern, whereas dynamic similarity concerns the ratio of forces. 83. A body is floating in water. If it is displaced slightly then about which point it will oscillate. Ans: It will oscillate about centre of buoyancy. 84. Why uniform ramming is considered desirable in green sand moulding process ? Ans: Uniform ramming results in greater dimensional stability of a casting. 85. On what account the friction drag is experienced ? Ans: Friction drag is experienced on separation of boundary layer. 86. What is the characteristic of centrifugally cast components ? Ans: These have fine grain structure with high density. 87, Under what condition the separation of flow occurs ? Ans: When pressure gradient changes abruptly. 88. What is the criterion for shaping of streamlined body ? Ans: It is shaped to suppress the flow separation about the body and to shift the boundary layer separation to near most part of body to reduce the wake size. 89. What do you understand by entrance length and what is its value for turbulent flow in smooth pipe ? Ans: Entrance length of flow is the initial length in which the flow develops fully such that the velocity profile remains unchanged downstream. For turbulent flow in smooth pipe, entrance length is taken as 50 x diameter of pipe. 90. Explain the difference between friction drag and pressure drag. Ans: Friction drag is the drag force exerted by a fluid on the surface due to friction action when the flow occurs past a flat surface at zero incidence. Pressure drag is the additional drag force on account of the differences of pressure over the body surface when flow occurs past a surface which is not everywhere parallel to the fluid stream. Pressure drag depends upon separation of bondary layer and the size of wake. Friction drag is due to shear stresses generated due to viscous action. 91. What do you understand by choking in pipe line ? Ans: When specified mass flow is not able to take place in a pipe line. 92. What is the difference between streamline body and bluff body ? Ans: In streamline body the shape is such that separation in flow occurs past the nearmost part of the body so that wake formed is small and thus friction drag is much greater than pressure drag. In bluff body the flow gets separated much ahead of its rear resulting in large wake and thus pressure drag is much greater than the friction drag. 93. What is the difference between dressing and trueing of a grinding wheel ? Ans: Dressing is the process used to clear the cutting surface of the grinding wheel of any dull grits and embedded swarf in order to improve the cutting action. Trueing is the process employed to bring the wheel to the required geometric shop and also to restore the cutting action of a worn wheel. 94. Whether hard grade or softer grade is required for internal grinding than external grinding? Ans: Softer graders of wheel are required for internal grinding than for external grinding. 95. How is the velocity profile and stress distribution due to laminar flow of an incompressible flow under steady conditions in a circular pipe. Ans: Velocity profile is parabolic with zero velocity at boundary and maximum at the centre. Shear stress distribution is linear, being maximum at boundary and zero at centre. 96. Where the maximum velocity occurs in open channels ? Ans: Near the channel bottom. 97. Define hydraulically efficient channel cross section. Ans: The shape of such section is that which produces minimum wetted perimeter for a given area of flow and carries maximum flow. 9. What is follower rest ? Ans .For slender work a travelling or follower rest is used. This fits on the lathe - saddle and travels to and fro with it. By this means support is constantly provided at the position of cut. Thus the work piece being turn is rigidly held against the tool. 99. What is the meaning of the term sensitive drill press ? Ans: A sensitive drilling press is a light, simple, bench type machine for light duty working with infinite speed ratio. 100. Why carburised machine components have high endurance limit ? Ans: In carburised machine components, the process of carburisation introduces a compressive layer on the surface and thus endurance limit is increased.
1. What is anisotropy ? Ans: The phenomenon of different properties in different directions is called anisotropy. 2. What is vapour pressure and when it becomes important ?
Ans: Vapour pressure is the partial pressure created by the vapour molecules when evaporation takes place within an enclosed space. It becomes important when negative pressures are involved. 3. Explain absolute viscosity ? Ans: It is the ratio of shear stress and the gradient of velocity with distance between a fixed plate and moving plate (Its unit is Pa . s). 4. Define surface tension ? Ans: Surface tension of a liquid is the work that must be done to bring enough molecules from inside the liquid to the surface to form one new unit area of that surface. (Its unit is Nm/m2 or N/m). 5. What is capillarity and when liquid rises or falls ? Ans: Capillarity : The rise or fall of a liquid in a capillary tube caused by surface tension. The magnitude of rise or fall depends on the relative magnitudes of the cohesion of the liquid and adhesion of the liquid to the walls of the containing vessels. Capillarity is of importance in tubes smaller than 10 mm diameter. Rise of liquid : Liquids rise (in capillary tube) when they wet and adhesion of the liquid to the walls is greater than the cohesion of the liquid. Fall of liquid : Liquids fall (in capillary tube) when they do not wet and the cohesion of the liquid is greater than the adhesion of the liquid to the walls. 6. What is the difference between perfect and real fluids ? Ans: Perfect fluids are treated as if all tangential forces created by friction can be ignored. Real fluids refer to the cases in which friction must properly be taken into account. 7. Which factors affect tool size ? Ans: Following factors influence tool size : 1. Process variable (speed, feed and depth of cut) 2. Tool material 3. Tool geometry 4. Work piece material, its hardness, microstructure and surface condition 5. Cutting conditions. 8. When the motion of fluid is fully known ? Ans: The motion of a fluid is fully known when the velocity of each of its particles can be specified. 9. Define fluid. Ans: A fluid is a substance that can't remain at rest under the action of any shear force. 10. How is the size of a vertical boring machine designated ? Ans: The specifications of vertical boring machine are : (i) Column height (ii) Table size (iii) Floor area (iv) Weight of the job (v) Spindle of the motor. 11. What do you mean by drill sleeve and a drill socket ? Ans: The drill sleeve is suitable for holding only one size of shank. If the taper shank of the tool is smaller than the taper in the spindle hole, a taper sleeve is used. When the tapered tool shank is larger than the spindle taper, drill sockets are used to hold the tools. Drill sockets are much longer in size than the drill sleeves. 12. What is nonbarotropic fluid ? Ans: Fluid such as air for which the density is not a single-valued function of the pressure (aerostatics). 13. On account of which property, the falling drops of rain acquire superical shape. Ans: Surface tension. 14. How you can definte standard atmospheric pressure ? Ans: By international agreement, the standard atmospheric pressure is defined as 101.325 kN/m2. 15. What do you understand by acoustic velocity ? Ans: Accoustic velocity is the speed of a small pressure (sound) wave in a fluid. 16. Explain the difference between centipoise and centistoke ? Ans: Centipoise is the unit of dynamic viscosity. It is equal to 1/100 of poise, and poise = 0.1 Pa . s. Value of dynamic viscosity of water at 20C is approximately equal to 1 Centipoise. Centistoke is the unit of kinematic viscosity and is equal to 1/100 x stoke. Stoke is defined as 1 square centimeter per second. 17. What is saybolt seconds universal ? Ans: Saybolt Seconds Universal (SSU) is the unit of viscosity and is equal to the time required for a gravity flow of 60 cc through saybolt universal viscometer. 18. Explain difference between cohesion, adhesion and capillarity ? Ans: Cohesion is the attraction of like molecules. Adhesion is the attraction of unlike molecules for each other. Capillarity is the elevation or depression of a liquid surface in contact with a solid. 19. Differentiate between gear hobbing and gear shaping with reference to various relative
motions and applications. Ans: Gear hobbing is a continuous indexing process in which both cutting tool (a hob) and work piece rotate in a constant relationship while the hob is being fed into work. Hob is also imparted a radial feed. Gear shaping uses a pinion type of cutter which is reciprocated with required cutting speed along the face of workpiece and is gradually fed radially to plunge. Continuous generation motion is obtained by feeding cutter to full depth and rotating the cutter and workpiece slowly. 20. What is bulk modulus of elasticity ? Ans: It represents the compressibility of a fluid. It is the ratio of the change in unit pressure to the corresponding volume change per unit of volume. 21. What is is entropic exponent ? Ans: It is the ratio of the specific heat of a gas at constant pressure to the specific heat at constant volume. 22. The best shape of a runner in sand casting is an inverted frustum of a cone why ? Ans: Tapered shape with bigger diameter at top and smaller at bottom ensures avoidance of entrainment or absorption of air/gases into the metal while passing through the runner. 23. Explain difference between ideal fluid and non-newtonian fluid. Ans: Ideal fluid is one for which resistance to shearing deformation is zero. Non-netwtonian fluids deform in such a way that shear stress is not proportional to the rate of shearing deformation. 24. Explain the difference between poise and stoke. Ans: Poise is the unit of viscosity in CGS unit and is measured in dyne sec/cm2.1 poise = 10_1 Pa s. Stoke is the unit of kinematic viscosity in CGS units and is measured in cm2/s. 25. How does the pressure in an isothermal atmosphere behave ? Ans: It increases exponentially with elevation. 26. The centre of pressure for a plane surface immersed vertically in a static liquid compared to centroid of area is always ... ? Ans: Below. 27. What is buoyant force equal to ? Define centroid of the area. Ans: Volume of liquid displaced. Centroid of the area : The point at which the area might be concentrated and still leave unchanged the first moment of the area around any axis. It is also the centre of gravity. 28. Define vapour pressure and on what parameters it depends ? Ans: The pressure exerted when a solid or liquid is in equilibrium with its own vapour is called vapour pressure. It is a function of the substance and its temperature. 29. Explain the difference between steady flow, one dimensional flow, two dimensional flow and three dimensional flow ? Ans: If at every point in the continuum, the local velocity and other fluid property, remains unchanged with time, it is referred to as steady flow. One dimensional flow is one in which a line is necessary to describe the velocity profile. Two dimensional flow is one in which an area is necessary to describe the velocity profile. Three dimensional flow is one in which a volume is necessary to describe the velocity profile. 30. What is aquifer and explain difference between confined and unconfined aquifers. Ans: Groundwater occurs in permeable, water-bearing geologic formations known as aquifers. Confined aquifer : It is a relatively high-permeable, water-bearing formation. Unconfined aquifer : It is a water-bearing formation with a free water table, below which the soil is saturated. 31. What do you understand by flurial hydraulics ? Ans: The channels may be classified as rigid boundary or mobile boundary channels. Mobile boundary channels include rivers and unlined alluvial canals, the boundaries of which are made of loose soil which can be easily eroded and transported by flowing water. The study of flow of water in mobile boundary channels is dealt in flurial hydraulics. 32. By which instruments the shear stress in fluids can be measured directly ? Ans: By Stanton tube or Preston tube. 33. On what factors does the pressure at a point as a static mass of liquid depends upon? Ans: Specific weight of liquid and the depth below the free liquid surface. 34. What is the difference between hoop or longitudinal tension and circum-ferencial tension ? I Ans: Hoop tension is created in the walls of a cylinder subjected to internal pressure. For thin walled cylinder (t < 0.1 d), hoop stress = pressure x radius/thickness. Longitudinal tension in thin-walled cylinders closed at the ends is equal to half the hoop tension. 35. How much force is exterted by liquid ? Ans: Force exerted by a liquid on a plane area A is equal to the product of the unit weight pg of the liquid, the depth hcg of the centre of gravity of the area, and the area. 36. State standard point angle and helix angle of a twist drill. Ans: Standard point angle of twist drill is 118 and helix angle varies from 16 to 30 depending as diameter of hole.
37. Explain the difference between horizontal and vertical components of hydrostatic force ? Ans: The horizontal component of the hydrostatic force on any surface is equal to the normal force on the vertical projection of the surface and acts through the center of pressure for the vertical projection. Vertical component of the hydrostatic force on any surface is equal to the weight of the volume of liquid abaove the area and passes through the e.g. of the volume. 38. What should be diameter of capillary tube to avoid correction for effect of capillarity in manometer ? Ans: Greater than 6 mm. 39. How much hydrostatic pressure acts on a curved surface ? Ans: The horizontal component of the total hydrostatic pressure force on any surface is always equal to the total pressure on the vertical projection of the surface, and can be located through the center of pressure of this projection. The vertical components of the total hydrostatic pressure force on any surface is always equal to the weight of the entire water column above the surface extending vertically to the free surface, and cvan be located through the cemntroid of this column. 40. How much is the horizontal component of force on a curved surface ? Ans: Force on a vertical projection of the curved surface. 41. State Archimedes principle. Ans: Any weight, floating or immersed in a liquid, is acted upon by a buoyant force equal to the weight of the liquid displaced. This force acts through the center of buoyancy, i.e. the e.g. of the displaced liquid. 42. What do you understand by center of buoyancy ? Ans: Center of buoyancy is the center of gravity of the displaced liquid and buoyant force acts through it. 43. How one can establish stability of submerged body and floating body ? Ans: A submerged body floats in stable equilibrium when the center of gravity of the body lies directly below the center of buoyancy. 44. What is neutral equilibrium ? Ans: A submerged body is in neutral equilibrium for all positions when the center of gravity of body coincides with the center of buoyancy. 45. On what factors the stability of floating objects depends ? Ans: Stability of floating objects depends upon whether righting or overturning moment is developed when the center of gravity and center of buoyancy move out of vertical alignment due to the shifting of position of center of buoyancy. 46. State two products each produced by forward extrusion and reverse impact extrusion. Ans: Two examples of products made by direct extrusion are : gear profile, solder wire. Two examples of reverse impact extrusion are : Short tubes of soft alloys, tooth paste containers. 47. What are the best geometrical shapes suggested for sheet metal drawing and spinning? Ans: Cup shape is best suited by drawing operation. Conical shape is easiest to produce by spinning operation. 48. What are the conditions for stability of buoyant bodies ? Ans: A floating body is stable if the center of gravity is below the metacentre. A submerged body is stable if the center of gravity is below the center of buoyancy. 49. Define pipe flow ? Ans: It refers to full water flow in closed conduits of circular cross section under a certain pressure gradient. 50. How much is vertical component of pressure force on a submerged curved surface ? Ans: The weight of liquid vertically above the curved surface. 51. Multiple coated, disposable carbide tips have more-or-less replaced brazed carbide tipped tools in CNC applications. Why ? Ans: Disposable tips are usually indexable type and have a number of cutting edges. Very convenient gripping devices for holding such inserts on tool holders are available. A wide variety of ready made inserts are available which require no grinding or adjustment. 52. State, sequentially elements of a canned CNC drilling cycle. Ans: Canned cycles are fixed cycles for general sequences of operations. For drilling operation, sequence of operations will be movement of drill or workpiece to position of actual drilling, movement of drill downwards till it is in the proximity of workpiece, then movement of drill at desired feed rate to sufficient depth of hole, retracting the drill above the workpiece. 54. Explain the difference between one and two dimensional flow ? Ans: True one-dimensional flow occurs when the direction and magnitude of the velocity at all points are identical. Velocities and accelerations normal to the streamline are negligible. Two dimensional flow occurs when the fluid particles, move in planes or parallel planes and the stream-line patterns are identical in each plane. 55. What is the difference between irrotational and rotational flow ? Ans: Irrotational flow : An ideal fluid flow in which no shear stresses occur and hence no torques exist. Rotational motion of fluid particles about their own mass centers can not exist. It can be represented by a flow net. Rotational flow occurs when the velocity of each particle varies directly as the distance from
the center of rotation. 56. Explain the difference between steady and unsteady flows. Ans: In steady flow, at any point, the velocity of successive fluid particles is the same at successive periods of time. Flow is unsteady when conditions at any point in a fluid change with time, 57. What is the difference between uniform and non-uniform flows ? Ans: Uniform flow occurs when the magnitude and direction of the velocity do not change from point to point in the fluid. Non uniform flow occurs when velocity, depth, pressure, etc. change from point to point in the fluid flow. 58. Explain the difference between stream lines and stream tube. Ans: Stream lines are imaginary curves drawn through a fluid to indicate the direction of motion in various sections of the flow of the fluid system. There can be no flow across a streamline at any point. Stream tube represents elementary portions of a flowing fluid bounded by a group of streamlines which confine the flow. 59. What is the advantage of independent jaw chuck ? Ans: Irregular job can be fixed. 60. Define vorticity ? Ans: It equals twice the angular velocity. 61. What is irrational How ? Ans: If a flow is uniform in one region (without rotation), this property is conserved as the fluid moves into a region of space where the motion is not uniform, and the flow is termed irrotational flow. 62. What is one-seventh power law ? Ans: According to one-seventh power law of turbulent, velocity variation corresponds to a much favoured velocity profile and the mean velocity increases as the one-seventh power of the distance from the boundary wall. 63. Mention the major difference between drilling and boring ? Ans: Drilling Boring (a) It is the operation of initiating a hole Boring is the operation of enlarging a drilled hole (b) It can produce only standard holes We can produce non-standard hole (c) Drill is a multi point tool A single point tool is used 64. What is the magnitude of buoyant force and where does the line of action of buoyant force act ? Ans: It is equal to the volume of the liquid displaced. The line of action of buoyant force acts through the centroid of the displaced volume of the fluid. 65. What is metacentre ? Ans: Metacentre is the point at the intersection of the buoyancy with the vertical axis of the body. 66. Define Prandtl's pitot-static tube. Ans: It is a combination of pressure probe and pitot tube. It is widely used for pressure measurements in wind-tunnel installations. 67. Define the term fluid friction. Ans: The deformation of real fluids is resisted by forces caused by internal friction or viscosity. Viscosity is that property of a real fluid which creates shear forces (fluid friction) between two fluid selements. 68. Why are hydrodynamic bearings used for grinding machines in preference to ball or roller bearings ? Ans: Hydrodynamic bearings are more suitable at high speed, better tolerances, quiet operation, higher capacity to withstand shock, lesser space requirement, better life under fatigue conditions etc. 69. Distinguish between a jig and fixture. Ans: Jig clamps and locates parts in positive manner and guides cutting tools (drills, reamers, taps etc). Fixture is bolted or fixed securely to the machine table so that work is in correct location ship to the cutter. Fixtures are used for mass milling, turning and grinding, etc. 70. State the similarities and differences between a fixture and a jig. Ans: Both jig and fixture positively locate, hold and clamp the workpiece. Jigs guide the tool for drilling etc but fixture has no facility to guide the cutting tools. Jigs are used for drilling etc and fixture for mass milling, turning, grinding, etc. 71.Explain the difference between relative density, viscosity and kinematic viscosity ? Ans: Relative density is that pure number which denotes the ratio of the mass of a body to the mass of an equal volume of a substance taken as a standard. This standard is usually water (at 4C) for solids and liquids, and air free or C02 or hydrogen (at 0C and 1 atmosphere = 1.013 x 105 Pa pressure) for gases. Viscosity of a fluid is that property which determines the amount of its resistance to a shearing force. It is due primarily to interaction between fluid molecules. Kinematic viscosity is the ratio of absolute viscosity and mass density. (Its unit is m2/s). 72. A right hand helical gear is being cut on a milling machine. What changes in machine
settings have to be made to cut a left hand helical gear of same pitch and number of teeth ? Ans: For cutting right hand helical gear right side of milling table is tilted up from normal position by the helix angle. For cutting left hand helical gear, right side of table needs to be tilted down at helix angle, all other settings remaining unchanged. 73. Explain what you understand by the term hydrostatic paradox ? Ans: Hydrostatic paradox is the phenomenon that the floor load of water in vessels of different shapes carrying water at same level depends only on the area of the bottom surface and not on the shape of the vessel. 74. Define buoyancy in brief. Ans: Buoyancy is the vertical force exerted on the body by the fluid at rest. The buoyancy of a submerged body in a fluid at rest is equal to the weight of the fluid displaced by the body and it acts upward through the center of gravity of the displaced volume (the center of buoyancy). 75. What is the necessary condition for a body to float in stable equilibrium ? Ans: Meta center should be above the center of gravity. 76. If liquid rotates at constant angular velocity about a vertical axis as a rigid body, then how its pressure varies at various radial distances ? Ans: As square of the radial distance. 77. How does the velocity vary along radius in a few vortex ? . Ans: It decreases with radius. 78. What type of flow will occur when liquid discharges at constant rate through a long, straight tapering pipe ? Ans: Steady non-uniform flow. 79. What does continuity equation represent ? Ans: It relates mass rate of flow along a stream tube. 80. Under what condition steady flow occurs ? Ans: When conditions do not change with time at any point. 81. Why a drill can not drill deeper than its flute length ? Ans: If flute submerges in a drilled hole, the chips disposal passage is blocked and the drill will get jammed in the hole. 82. When do you recommend the use of straight fluted drill ? Ans: Drills used for brass and other soft materials and thin sheets need not be provided with any helix angle, to prevent the lips digging into the workpiece. 83. Explain velocity distribution for laminar and turbulent flows. Ans: In case of laminar flow, the velocity distribution follows a parabolic law of variation. The maximum velocity at center of pipe is twice the average velocity. For turbulent flow, more uniform velocity distribution results. 84. In which type of flow the stream lines, streak lines and path lines are identical ? Ans: In steady flow. 85. Under what condition a drill may not cut ? Ans: When the cutting lips are not provided with clearance angles. It will cut if 12 clearance angle is ground. 86. Some drills have straight shanks and some have tapered shank. Why ? Ans: Small drills (less than 12 mm diameter) are provided with straight shanks which can be held in Dniren chucks. Bigger drills have tapered shank using self holding Morse tapers which prevent gravitational fall of drill. 87. What are the main factors that are responsible for the formation of built up edge ? Ans: The main factors that are responsible for the formation of built up edge are : 1. Extreme pressure in the cutting zone 2. High friction in the tool chip interface 3. High local temperature. 88. In which type of flow the Navier-strokes equation is useful ? Ans: Viscous flow. 89. Explain the difference between energy line and hydraulic grade line ? Ans: The energy line is a graphical representation of the energy at each section. Hydraulic grade line lies below the energy line by an amount equal to the velocity head at the section. 90. What is the characteristic of equipotential line ? Ans: It has no velocity component tangent to it. 91. How specific cutting pressure behaves with feed rate and what conclusion can be drawn from this ? Ans: As feed rate increases, the specific cutting pressure decreases. It leads to conclude that feed rates be maximised in metal cutting process. 92. What is hydraulically rough pipe ? Ans: When Reynolds number is very small, friction factor f becomes independent of the Reynolds number and depends only on the relative roughness height. Such a pipe behaves as hydraulically rough pipe. 93. What are the qualities of ideal tool material ? Ans: It should be hard to resist flank wear and deformation, have high toughness to resist fracture, be chemically inert to the workpiece, be chemically stable to resist oxidation and
dissolution, and have good resistance to thermal shocks. 94. What are the factors to be considered while evaluating the machinability ? Ans: 1. Tool life 2. Rate of metal removal 3. Power required 4. Surface texture and size of component 5. Temperature of tool (or) chip. 95. Which equation is satisfied by ideal or real, laminar or turbulent flow ? Ans: Continuity equation. 96. How to judge the unsatisfactory performance of cutting tool ? Ans: Unsatisfactory performance of cutting tool leads to loss of dimensional accuracy, increase in surface roughness and increase in power requirements. 97. What are the criteria for judging machinability ? Ans: Cutting speed, tool life, surface finish, cutting force/energy required, teasperature rise at cutting point. 99. What is critical velocity ? Ans: It is the velocity below which all turbulence is damped out by the viscosity of the fluid. 100. What is laminar flow ? Ans: In laminar flow the fluid particles move along straight, parallel paths in layers or laminae. Laminar flow is governed by the law relating shear stress to rate of angular deformation i.e. the product of the viscosity of the fluid and velocity gradient.
1. What parameters influence the tool life ? Ans.- 1. Tool material 2. Work material 3. Speed, feed and depth of cut 4. Tool geometry work system 5. Cutting fluid 6. Built up edge 7. Vibration behaviour of the machine tool. 2. Mention the function of intermediate stage in a generalised measurement system. (i) Amplify signal without affecting its waveform (ii) Remove unwanted noise signals that tend to obscure the input (iii) Capable of doing other conditioning like differentiation/integration, A/D conversion etc. 3. What is a signal flow graph ? A signal flow graph of a system is a graphical model in which nodes represent the system variables and directed branches between the nodes represent relationship between the variables. In a signal flow graph, a forward path originates from the input mode and terminates at output mode without encountering any mode more than once. 4. Explain the difference between rotational and irrotational flow. In rotational flow the vorticity is non zero and in irrotational flow it is zero. In irrotational flow the net rotation of a fluid element about its own mass centre is zero as it moves from one place to another. For it, dv/dx = du/dy. 5. Define the term stability of a feed back control system ? A system is said to be stable if the output of a system after fluctuations, variations or oscillations, settles at a reasonable value for any change in input. 6. What is meant by precision ? Precision is defined as the ability of instrument to reproduce a certain set of readings within a given accuracy. 7. What is boundary layer ? Boundary layer is the fluid layer in the neighbourhood of a solid boundary where the effects of fluid friction are predominant. 8. On what account the boundry layer exists ? Boundary layer m fluid flow exists on account of fluid density. 9. Explain the difference between laminar flow and turbulent flow. In laminar flow the fluid particles move along smooth, regular paths which can be predicted in well advance. Turbulent flow is characterised by random and erratic movements of fluid particles resulting in formation of eddies. For flow to be laminar in a pipe, the Reynold's number is less than 2000 and for flow to be turbulent it is greater than 4000. 10. What is the difference between path line and stream line ? A path line is a curve traced by a single fluid particle during its motion. A stream line is an imaginary line drawn in a flow field such that a tangent drawn at any point on this line represents the direction of the velocity vector. There is no flow across the stream line. 11. What is the difference between steady and unsteady flow ? In steady flow the velocity at a given point does not change with time whereas in unsteady flow it changes with time. 12. What is stagnation point and stagnation pressure ?
A stagnation point is one at which velocity is zero. Stagnation pressure is the sum of static pressure and dynamic pressure. 13. What is the hydraulic-grade line ? It is the line which shows tne variation of piezometric head. 14. Under what condition the thickness of thermal boundary layer and hydrodynamic boundary layer is same ? When Prandtl number is equal to unity. 15. What keeps the shaft in floating condition in hydrodynamic journal bearing ? Hydrodynamic pressure developed in the lubricant. 16. Why the elements of higher pairs must be force closed ? This is necessary to provide completely constrained motion. 17. What is the value of coefficient of friction for ball bearing ? Ans: 0.1 to 0.25. 18. Why belts are subject to creep ? Due to uneven extension and contraction of belt. 19. What is the ratio of damping coefficient to critical damping coefficient called ? Damping factor. 20. In which type of lubrication the starting friction is low ? In hydrostatic lubrication. 21. Whether Bernoulli's equation is applicable for streamline straight and rotational flows ? Ans: Yes. 22. When a plate slides over a parallel plate, what is the relationship between viscous force and the lubricant thickness ? They are directly proportional. 23. How the stability of empty ship increased. By adding ballasts at the bottom level. 24. Define Froude number. It is the square root of ratio of inertia force to gravity force. 25. What for Maning's formula used ? It is used to determine velocity of flow in pipe. 26. What is the maximum efficiency of power transmission through a pipe ? Ans: 66.67%. 27. What is the necessary and sufficient condition for bodies in floatation to be in stable equilibrium ? Centre of gravity of body should be located below metacentre. 28. Under what condition the thermal boundary layer thickness will be equal to the hydrodynamic boundary layer thickness ? When Prandtl number is unity. 29. How laminar boundary layer thickness varies on a flat plate ? It is proportional to the square root of the distance from the leading edge. 30. Name the axial flow reaction turbine with adjustable vanes. Kaplan turbine. 31. Define the degree of reaction of a turbomachine. It is the ratio of total pressure change in the rotor to that in the stage. 32. What is meant by relative stability ? If the system is found to be stable, then it is necessary to know the stable strength or degree of stability which is called relative stability. Phase margin and gain margins are the measures of relative stability. 33. On what factors the efficiency of a centrifugal pump depends ? The efficiency of a centrifugal pump depends on size, speed and type number. 34. Define transducer ? Transducer is an element which converts the signal from one physical form to another without changing the information content of the signal. 35. A 200 mm impeller of centrifugal pump develops discharge head of 2 kg/cm2. If discharge head is required as 8 kg/cm2, what should be diameter of impeller. Ans: 400 mm. 36. Why thin-plate orifice is frequently recommended for flow measurement in most of the cases ? It is recommended because of its high accuracy, low cost, and extreme flexibility. 37. How the water flow from a hydrant in a city water system measured ? It can be measured by allowing the flow of water from a hydrant and noting down the horizontal distance (x) where jet strikes the ground and the vertical distance (y) of outlet from ground. Then flow = area of hydrant x 38. For the same size, length and condition of pipe, how the friction head loss changes if pipe size is varied. Under same flow conditions the friction head loss increases as the size change to the 4.8th power. 39. How does the capacity changes in pipe flow with increase in pipe diameter if head is constant ? At constant head, capacity is proportional to d25. . 40. How does the head change in pipe flow with change in diameter, if capacity is
constant. At constant capacity, head is proportional to 41. How head varies in pipe flow with change in flow, if pipe diameter is same. At constant diameter, head is proportional to (flow)2. 42. Give some properties of errors ? (i) Errors give the difference between the measured-value and the true value (ii) Accuracy of a measurement system is measured in terms of error (iii) A study of error helps in finding the reliability of the results. 43. Define the term contraction coefficient. The ratio of area of jet (contracted) and the cross section of the orifice. 44. What for pycnometer is used ? Pycnometer is used to measure specific gravity of the liquid. 45. How may types of bonded strain gauges used ? (a) Fine wire strain gauge (b) Metal foil strain gauge (c) Piezo resistive gauge. 46. Name various functional elements of an instrumental system ? (1) Primary sensing element (2) Variable conversion (or) transducer element (3) Variable manipulation element (4) Data transmission element (5) Data processing element (6) Data presentation element (7) Data storage and playback element. 47. Name the three stages of a generalised measuring system. (i) Sensing and converting the input to a convenient and practicable form. (ii) Processing/manipulating the measured variable. (iii) Presenting the processed measured variable in quantitative form. 48. How is a Pelton turbine braked ? A pelton wheel is braked by a nozzle directing a jet on the back of the buckets. 49. What is static pressure ? How is it measured. The pressure caused on the walls of a pipe due to a fluid at rest inside the pipe or due to the flow of a fluid parallel to walls of the pipe is called static pressure. This static pressure is measured by inserting a pressure measuring tube into the pipe carrying the fluid so that the tube is at right angle to the fluid flow path. 50. What are the requirements of a control system ? Stability, accuracy and speed of response are the three requirements of control system. 51. What is the difference between weir, and submerged weir. Any obstruction of a streamflow over which water flows is weir. When the downstream water level rises over the weir crest, it is called submerged weir. 52. What is the difference between spillway and siphon spillway ? Spillway is an essential part of a large dam and provides an efficient and safe means of releasing flood water that exceeds the design capacity of the reservoir. Siphon spillway is spillway designed to discharge water in a closed conduit under negative pressure. 53. What is the difference between culverts, stilling basin, and standpipe ? Culverts are built at the points of lowest valley to pass water across the embankments of highways or railroads. Stilling basin is transitional energy dissipating structure to avoid the damaging process by a highly accelerated spilled water. Standpipe is used principally for alleviating the transient pressures in large pipeline system. It also works as a pressure relief valve for the upstream pipe during the turbine shut off. 54. What do you understand by run-of-river scheme in hydro power generation ? A scheme in which the discharge is varying and only that much water can be utilised for conversion as available in the river. 55. What is the difference between firm power and secondary power in hydro power plants ? The approximate constant and continuous power which is assured at power station and would be available throughout the year is called firm power. In run-of-the river schemes which are generally designed for % availability of water, the firm power would be available for at least % of the years during the life of the scheme. However in some years discharge in the river may be more than % availability discharge and some extra power known as secondary power can be generated. 56. What is the criterion for determining economic diameter of power tunnel ? The economic diameter of a power tunnel is determined such that the total value of following two factors is least (i) annual loss of revenue on account of power head lost due to friction. (ii) recurring annual expenditure. 57. What are the three functions served by a surge tank in a hydro plant ? The three functions served by a surge tank are (i) flow regulation - act as a reservoir for acceptance or delivery of water to meet
requirements of load changes. (ii) water-hammer relief or pressure regulation. (iii) improvement in speed regulation. 58. What is the criterion to determine the economic diameter of penstock ? The economic diameter of a penstock required to carry a discharge is the one at which annual costs due to the greater investment do not exceed the annual value of resulting increment energy output. 5. If jet ratio for a pelton wheel is 12 then number of buckets should be .... ? 0.5 x 12+ 15 = 21. 60. What is the difference between anti-friction and hydrostatic bearings ? In anti-friction bearings rolling components are introduced between the sliding surfaces. In hydrostatic bearings lubricant is supplied at a high pressure to a pocket in the bearing. 61. What is inviscid fluid ? It is the frictionless fluid. 62. State law of conservation of momentum ? The ratio of change of momentum of a mass of fluid is equal to the vector sum of all external forces acting on it. 63. Define turbulent flow ? Flow in which adjacent layers mix continuously, so that the flow pattern is unsteady, full of eddies, and apparently without any mathematically expressible regularity. 64. Define Reynold's law of similitude ? Two geometrically similar flow systems subject only to friction and inertia forces are dynamically similar if both have the same 'Reynolds' number. 65. What is the difference between boundary lubrication and fluid film lubrication ? In boundary lubrication condition the ratio of thickness of lubrication film to surface roughness (CLA value) is less than 1 and for fluid film lubrication condition this ratio is greater than 5 and less than 100. 66. A bearing in which hydrodynamic pressure is generated due to rotation of journal is called... ? Journal bearing. 67. For a journal running in a bearing clockwise at steady state, where will be the minimum clearance ? To the left of the load line. 68. Out of impulse and reaction type hydraulic turbines, which has higher specific speed ? Reaction turbine. 69. What is order of the system. The order of the system is the order of the highest derivative of the ordinary differential equation with constant coefficients which defines the system mathematically. 70. What is Bode plot ? Bode plot is a logarithmic plot used to represent transfer functions. The Bode plot consists of two plots namely. (i) The plot of magnitude in db (on linear scale) vs. frequency to (on log scale) (ii) The plot of phase angle in degrees (on linear scale) vs. frequency co (on log scale) 71. Navier stokes equation represents the conservation of which quantity ? Momentum. 72. What are the characteristics of precision ? (i) If a number of measurements are made on same true value, the degree of closeness of these measurements gives precision (ii) It is a measure of the degree of repeatability or reproducibility of the measuring system. 73. What is meant by loading ? During the act of measurement the measuring instrument takes energy from the signal source or measured medium and thus the signal source is altered. This effect is called loading. 74. Give two uses of Bode plots. (i) They are the logarithmic plots used to represent transfer functions (ii) They are used in the calculation of gain margin and phase margin. 75. What is the name of nondimensional group for ratio of inertia force to gravity force. Froude number. 76. The head loss in a fully developed laminar flow in a circular pipe due to friction is directly proportional to ... ? Square of mean velocity. 77. For what type of flow, the streamlines, pathlines and streaklines are virtually identical ? Steady flow. 78. How the streamlines and equipotential lines are related in a flow field ? They are orthogoial everywhere in the flow field. 79. Under what kind and condition of a fluid the velocity potential exists ? Irrotational flow. 80. Where should centre of gravity of a body lie for it to float in stable equilibrium ? e.g. should be located below the metacentre. 81. For fully developed flow through a pipe, what is the ratio of maximum to average velocity ?
Ans: 2. 82. Out of centrifugal, axial and positive displacement pumps, which has maximum specific speed and which has minimum ? Axial pump has maximum specific speed and reciprocating pump has minimum. 83. Specific speed of a turbine is 800. What type of turbine is this ? Kaplan turbine. 84. In which type of lubrication system the starting friction is low ? Hydrostatic lubrication. 85. At which point the cavitation is most likely to occur in a hydel plant ? At turbine rotor exit. 86. What is the difference between lower pair and higher pair ? In a kinematic pair, if the elements have surface contact when in motion, the pair is called lower pair and if elements have line or point contact the pair is called higher pair. 87. What do you understrand by inversions and how many inversions are possible in a kinematic chain having 'n' links ? Inversions are different mechanisms obtained by fixing different links in a kinematic chain but keeping relative motions of links unchanged with respecat to one another. A kinematic chain with 'n' links can have 'n' inversions. 88. What is the difference between basic kinematic chain and compound kinematic chain ? What is the criteria for a chain to be constrained ? A basic kinematic chain consists of four kinematic links and a compound kinematic chain is a constrained chain of more than four links. A chain is constrained if no. of binary joints and half of number of higher pairs is equal to 3/2 of no. of links minus 2. 89. What is the difference between a machine and a mechanism ? A mechanism consists of links forming a constrained kinematic chain. Its function is basically to transmit or modify motion. A machine is a mechanism but here we are concerned with forces to be transmitted and it is used to modify mechanical work. 90. In which type of motion, acaceleration is directly proportional to displacement ? SHM. 91. What is the function of a transducer element ? The function of a transducer element is to sense and convert the desired input into a more convenient and practicable form to be handled by measurement system. 92. What is the direction of tangential acceleration ? The direction of tangential acceleration may be same or opposite to that of angular velocity. 93. What is the number of degrees of freedom constrained in each case by the kinematic pairs. (i) Itevolute pair, (ii) cylindrical pair, (iii) Screw pair, (iv) Spherical pair. Ans: (i) Revolute pairsingle degree of freedom. (ii) cylindrical pairtwo degrees of freedom. (iii) Screw pairone degree of freedom. (iv) spherical pairthree degrees of freedom. 94. What is the purpose of D-slide valve in steam engines ? It controls the piston position for cut off of steam, for release of steam and compression position for steam in cylinder. 95. What is the purpose of Corliss valve in place of D'-slide valve in steam engine ? Corliss valve reduces condensation, provides independent control on admission, cut off, release and compression, and reduces driving power requirements. 96. On what factor the size of cam depends ? Size of cam depends on base circle. 97. If base circle diameter of cam increases, what happens to pressure angle ? It decreases. 98. If angular speed of cam is increased two times, how much jerk will be increased ? Eight times. 99. Define primary and secondary transducers ? A primary transducer senses a physical phenomena and converts it to an analogous output. The analogous output is then converted into an electrical signal by secondary transducer. 100. What is the relationship between (i) axes of spin, precession, and applied gyroscopic torque and (ii) in between their planes ? The axes of spin, precession, and applied gyroscopic torque are contained in two planes perpendicular to each other, and planes of spin, precession, and applied gyroscopic torque in the three planes perpendicular to one another.
1. Why excess air is required to burn a fuel completely ? Ans: Excess air is required to ensure adequate mixing of fuel and air, avoid smoke, minimize slagging in coal burning, and to ensure maximum steam output. 2. What is factor of evaporation ? Ans: It is the ratio of heat change from feed water to steam and the heat of vaporization of steam at atmospheric pressure.
3. Materials which absorb neutrons and produce fissionable material are called ? Ans: Fertile materials, such as U28 and Th22. 4. What is the process which produces fissionable material from a fertile material ? Ans: Breeding. 5. What is burn up in connection with fissionable material ? Ans: Burn up corresponds to destruction of fissionable material when it undergoes nuclear fission. 6. What kind of nuclear fuel is used in Narora, Kalapakkam and Rana Pratap Sagar nuclear plants ? Ans: Natural uranium. 7. Which type of plant will you recommend for remote location if power is required in six to twelve months time ? Ans: Diesel engine power plant. 8. Under what condition a nuclear reaction is said to be critical ? Ans: For critical condition, the reaction should continue at a steady rate which is possible when rate of production of neutrons is same as the combined rate of absorption of neutrons and the rate of leakage of neutrons. 9. If a nuclear reactor generates more fuel that it consumes, it is called ______ ? Ans: Fast breeder reactor. 10. Natural uranium contains only 0.7% U25 which is capable of switching chain nuclear reaction. How this percentage is increased ? Ans: Percentage of U25 in natural uranium is increased by a process called uranium enrichment. 11. Why the vane passages are gradually increased in size in successive wheels in steam turbine ? Ans: The velocity of steam decreases in successive stages and to accommodate same flow per unit time, the area of flow must increase. 12. At what pressure and temperature the density of water and steam is same ? Ans: At 225 kg/cm2 and 74.6 C. 13. What is the latent heat of evaporation at critical point ? Ans: Zero. 14. How much space steam will occupy at atmospheric pressure corresponding to 1 kg of water ? Ans: 100 times the space occupied by 1 kg of water. 15. What is meant by quality of steam ? Ans: It refers to amount of unevaporated moisture in steam. If steam is perfectly dry, it's quality is 100%. 16. What is diagram efficiency in steam turbine ? Ans: Diagram efficiency = Change in K.E./kg / Energy suppled/kg. 17. What do you understand by effectiveness of heat exchanger ? Ans: Effectiveness of heat exchanger = actual heat transfer/maximum possible heat transfer. 18. If superheat and condenser temperatures are unchanged and if pressure of steam is increased, then what happens to dryness fraction of steam after isentropic expansion ? Ans: It will decrease. 19. As pressure of steam is increased, why reheating is necessary? Ans: As pressure is increased, the dryness fraction of steam lowers on isentropic expansion. Therefore, steam has to be reheated after partial expansion so that dryness fraction remains within limits after expansion. 20. What is boundary layer thickness in free convection case ? Ans: Velocity of fluid flow at solid surface is zero and it increases rapidly as distance from solid surface becomes greater, reaches a maximum value and then decreases to that of undisturbed fluid. The distance between the solid surface and point where boundary layer velocity equals undisturbed flow (edge) is called boundary layer thickness. 21. What do you understand by forced convection ? Ans: When convection heat transfer occurs between a solid body and a fluid and where circulation of fluid is caused and controlled by some mechanical. 22. In radiative heat transfer, a gray surface is one whose emissivity is _____ ? Ans: Independent of wavelength. 23. What is the effect of water injection and steam injection in gas turbine of heat rate and power output ? Ans: Water injection results in higher mass flow rate through turbine section and as rich electrical output is increased. However it increases gas turbine heat rate because of the additional heat consumption required to vaporise the water. Steam injection increases power output and decreases heat rate because of its higher energy entering the combustion zone. 24. Water/steam injection in gas turbines increases power output. What is the limit and reasons for same ? Ans: Water/steam injection is not allowed beyond 5% of compressor flow to avoid flame out in combustion and to minimise operating cost and impact on inspection intervals. 25. What are the advantages and drawbacks of centrifugal compressor and axial flow
compressors in gas turbines? Ans: Centrifugal compressors have following advantages : short length due to high pressure ratio attained in a single stage. simple in design, easy to manufacture, rugged in construction, less costly. more reliable in operation under all operating condition. less susceptible to the effects of deposits left on flow path. less sensitive to the fouling of the flow path. Good efficiency over a wide range of operation. Drawback is larger cross-section area for same capacity and its unsuitability for high pressure ratios. Axial flow compressor is more popular due to high delivery capacity, high compression ratio due to many stages, high efficiency (85-90%), low cross sectional area. Its drawback is sophisticated design and appreciable length due to large number of stages, narrow operating range for good efficiency, higher weight and cost, high starting power. 26. What is the difference between pinch point and approach point in connection with heat recovery steam generator (IIRSG) ? How they affect performance of IIRSG ? Ans: Pinch point is the temperature difference between the gas turbine exhaust leaving the evaporator section and the saturation temperature of steam at corresponding pressure (drum operating pressure). Approach point is the temperature difference between the temperature of steam corresponding to drum operating pressure and water temperature leaving the economiser. Both these variables affect the steam production and the cost and effectiveness of HRSG. If pinch point is lower, total heat recovered in HRSG in higher and steam generation is also high. However lowering pinch point requires more heat exchange surface and increase in cost and draft side loss. Its optimum value is 8-10C. Lower is approach temperature, higher is steam production in economiser due to flashing which needs to be avoided for long life of tubes. Higher approach temperature increases the surface in evaporator section and assures higher stability. 27. How the efficiency of heat recovery in IIRSG increased ? Ans: It can be increased by generating steam at more than one pressure. Further improvement is possible by heating make up water/condensate in HRSG, generating low pressure saturated steam or dearation steam in HRSG using a low pressure evaporator, preheating the make up water in a heat exchanger before it enters deaerator, and circulate more water than necessary through the economiser and recirculate excess to the deaerator. 28. NO.,, reduction in gas turbines is challenge to meet environmental regulations. What measures are taken in this regard ? Ans: NO* in gas turbines is controlled by (a) injecting water/steam into combustor, (6) using selective catalytic redaction (SCR) and non-catalytic reduction system (NRS), and (c) better combustor design. 29. What is the difference between fuel NOx and thermal NOx ? Ans: Fuel NO* is formed by the gas phase oxidation of char nitrogen (CN~ compounds) in the fuel and thermal NO* is formed by high temperature reaction between the nitrogen and oxygen in the combustion air. 30. How steam/water injection in combustion zone lowers NOx formation. Ans: It lowers the flame and gas temperatures. 31. What are disadvantages of injecting steam/water in combustion zone in gas turbine ? Ans: Increased CO at lower loads Increased fuel consumption Loss of extensively treated water Increased complexity of engine control Short life of major turbine components. 32. What for water/steam is injected in combustion zones of a gas turbine ? Ans: Water/steam is injected to limit the amount of NOx formed by lowering the flame and gas temperatures. 33. What are improved low NOx combustors ? Ans: Dry low NOx combustors result in improved air/fuel mixing and reduced flame temperature. The main type are lean premixed combustors, rich/quench lean combustors and catalytic combustors. Variable guide vanes are used to control the fuel-air mixture. In one design combustion chamber is designed into a series of small, lean premixed chambers. 34. What is selective catalytic reduction (SCR) method for NO* control in gas turbines ? Ans: SCR is a post combustion method in which a catalyst is installed in flue gas where temperature could be of the order of 15-400C. It enables vaporised NH to react with NO* in exhaust gas to form N2 and H20. Vaporised NH is injected into exhaust gases before it passes through the catalyst bed. This process can achieve in excess of 90% NOx reduction. 35. What catalyst is used in SCR process for control of NOx in gas turbines ? Ans: V2O5 supported by titanium or a mixture of titanium and silica is generally used as catalyst. Oxides of Ti and Mo are often incorporated as moderators. 36. Measure to minimise formation of NOx inhibit complete combustion and this results in formation of CO and voltatile organic compound (VOCs). How these are controlled ? Ans: Catalytic oxidation, a post combustion treatment at 470-600 temperature is used to
reduce emission of CO and VOCs. 37. It is usual practice to report performance characteristics of gas turbine at ISO conditions. What are these ? Ans: ISO conditions for GT performance are : Ambient dry bulb temperature : 15C, Relative humidity - 60%. Ambient berometric pressure 1.0 kg/cm2. 38. Which parameters influence the heat rate of gas turbines and how ? Ans: Various parameters affecting heat rate of GT are : (i) Lower the gas turbine inlet temperature, lower the heat rate. (ii) Higher the compression ratio, lower the heat rate. (iii) Higher the turbine firing temperature, lower the heat rate. (iv) Lower the exhaust temperature, lower the heat rate. (v) Exhaust gas flow, inlet pressure loss, exhaust pressure loss also affect heat rate. 39. Which parameters influence the performance of gas turbines ? Ans: Ambient conditions, inlet/exhaust pressure losses, fuels, and water/steam injection flow rates'influence the performance of gas turbines. 40. Why the efficiency of superheat Rankine cycle is higher than that of simple Rankine cycle ? Ans: Because the mean temperature of heat addition is higher for superheat cycle. 41. A Carnot cycle is to be designed to attain efficiency of 0.75. If temperature of high temperature reservoir is 727C, then low temperature reservoir will have to be maintained at______C ? Ans: - 2C. 42. A gas having a negative Joule-Thompson coefficient, when throttled, will become .... ? Ans: hotter. 43. What is boundary layer thickness in turbulent boundary layer ? Ans: In ease of turbulent boundary layer, boundary layer thickness is defined as the distance from solid surface at which the fluid velocity is 99% of the undisturbed free stream velocity. Near the surface of solid, there is laminar flow which changes to buffer zone and finally turbulent region. 44. Stirling cycle with regenerative arrangement and Carnot cycle operate within same temperature limits. What can be said about thermal efficiencies of these two ? Ans: Thermal n of both will be same. 45. How does the stagnation temperature behave along a streamline in adiabatic flow with friction ? Ans: It remains constant. 46. What is the frequency of secondary imbalance in four-stroke engine ? Ans: Two times the engine speed. 47. What would be the order of boundary layer in pipe flow (i) laminar (ii) transition, and (iii) fully turbulent ? Ans: In laminar flow, boundary layer extends right from surface upto centre. In fully turbulent it may extend 5-10% of radius only whereas in transition flow, boundary layer may be from 5060% of radius of pipe. 48. In general how the boundary layer and local film coefficient of heat transfer are related ? Ans: In general, the thinner the boundary layer, higher the value of local heat transfer film coefficient. 49. Out of constant volume and constant pressure line, which line on T-S diagram has higher slope ? Ans: Constant volume. 50. A heat engine and a refrigeration cycle operate between two temperature limits T1 and T2(T1 > T2). The product of efficiency of heat engine and COP of refrigeration cycle will be ? Ans: T2/TX. 51. A refrigerator and a heat pump are working on the reversed Carnot cycle between the same temperature limits. How COP of refrigerator and heat pump related ? Ans: COP of refrigerator = COP of heat pump -1. 52. For a given set of operating pressure limits of a Rankine cycle, for which cycle the efficiency will be highest ? Ans: For regenerative cycle which approaches Carnot cycle. 1 53. A steam pipe is to be insulated by two different insulating materials of same thickness. What arrangement is preferred ? Ans: Material with lower thermal conductivity should be used for inner layer and material with higher thermal conductivity for the outer layer. 54. Why steam is in open space and water inside tube in condensers used in power plants ? Ans: Overall heat transfer coefficient can be increased by increasing velocity of water in tube. Further steam needs more space due to higher specific volume. 55. A condenser of a refrigeration system rejects heat at a rate of 60 kW, while its compressor consumes a power of 15 kW. What will be the coefficient of performance of this system ? Ans: . 56. Why a refrigeration compressor designed to operate with R22 refrigerant can't be
operated with R12 ? Ans: Condensing pressure of R22 at any given temperature is higher than that of R12. 57. Leakage of Freon-12 refrigerant can not be easly detected, still it is preferred for airconditioning. Why ? Ans: Leakage of Freon-12 can't attain total concentration but it has all desirable characteristics as good refrigerant. 58. In which season the process of humidification and in which season dehumidifi-cation is used in air-conditioning ? Ans: Humidification in summer and dehumidification in winter. 59. What is the name given to combined process of cooling and humidifying ? Ans: Evaporative Cooling. 60. Which refrigerant is used for ice plant and transport refrigeration ? Ans: Ammonia and CO2 respectively. 61. Which cycle is used in vapour compression and gas cycle refrigerantion system ? Ans: Brayton cycle, and Bell-Coleman cycle. 62. In what connection Wilson line and Willan's line used ? Ans: Wilson line represents saturation line on the Mollier diagram. Willan's line is connected with determination of frictional power in IC engines. 63. How the characteristics of fuel like front end volatility, mid-range volatility and tail end volatility affect S.I. engines ? Ans: These characteristics affect SI engines in cold starting, icing of carburettor and dilution in crankcase respectively. 64. What would be the approximate air fuel ratio at time of cold start, part load operation, full load, and idling ? Ans: 10, 12.5 and 16 respectively. 65. Name pressure compounded and velocity compounded turbine. Ans: Rateau and Curtis respectively. 66. What is the nature of curve in case of transfer of heat by conduction through a cylindrical wall and through a spherical wall ? Ans: Logarithmic and hyperbolic. 67. What is radiosity ? Ans: It is the rate at which radiation leaves a surface. 68. When the compressibility effects become significant in heat transfer problems in high speed flow ? Ans: Compressibility effects usually become significant at gas flow velocities greater than 0.5 times the velocity of sound. 69. What is the name given to process of removing non condensables in steam and other vapour cycles ? Ans: Deaeration process. 70. We often come across use of jet pumps in process industry ? What is the reason for this ? Ans: Jet pumps are easy to maintain and can be used to transport gases, liquids and mixtures of both. 71. When it can be said that a fluid is Newtonian ? 9/24/13 Mechanical Engineering Interview Questions and Answers Set3 - Preparation for Engineering www.engineeringkings.com/2012/11/mechanical-engineering-interview_4878.html 7/9 Ans: When the shear stress is direcly proportional to velocity gradient. 72. Why alcohol can't be used in diesel engine ? Ans: Alcohol has low cetane number and thus its ignition is prevented by compression. 73. Out of 2 stroke SI engine, 4 stroke SI engine, and 4 stroke CI engine, how the brake thermal efficiency behaves if these are to be used for road vehicles ? Ans: 4 stroke CI engine has highest brake thermal efficiency followed by 4 stroke SI engine and the 2 stroke SI has lowest brake thermal efficiency. 74. A jet engine is used to drive a vehicle. When its mechanical efficiency will be maximum ? Ans: It will be maximum when vehicle speed approaches the relative velocity of gases at nozzle exit. 75. The knocking tendency in a S.I. engine can be reduced by increasing ? Ans: Speed 76. How you define the emissivity of a body ? Ans: Emissivity of a body is the ratio of the emitted radiant energy flux density to the emitted radiant energy flux density of a blackbody at the same temperature. 77. What is gray body ? Ans: When the emissivity of material does not change with temperature, it is called gray body. 78. Define fin efficiency for extended heat transfer surfaces. Ans: Fin efficiency is defined as the ratio of the mean temperature difference from surface to fluid divided by the temperature difference from fin to fluid at the base or root of the fin. 79. What are the harmful effects of sulphur in coal used for power generation ? Ans: Sulphur in coal gives rise to number of ecological problems ranging from acid mine drainage to ash waste problems and air pollution due to SO2 emission. Emission of S02 is the cause of acid rain and is responsible for severe environmental damage.
80. What do you understand by microbial desulphurisation ? Ans: Sulphur from coal can be removed by physical, chemical and microbial meAns: Microbial means has a number of advantages over other methods. Thiobalilius ferroxidans is the most widely used micro organism studied for coal desulphurisation. This bacteria can remove 90-98% of pyritic sulphur from coal. The reaction, of course, is slow and may take several days to several weeks to complete the reaction. 81. What causes rotation of steam turbine rotor ? Ans: Rotation of turbine rotor is caused by centrifugal force created due to change of direction of fast flowing jet of steam. 82. What do you understand by the term coagulation ? Ans: Impurities in water are usually in finely divided state which takes long time to settle and these usually pass through the filtering media. By adding certain chemicals (coagulants), gelatinous substances are formed which cause small particles to coalesce into groups large enough to be checked during filtering. 83. What is equivalent evaporation in a boiler plant ? Ans: It is the amount of water that would be evaporated from water at 100C to steam at 100C by the same amount of heat which was actually absorbed by water and steam under operating conditions. 84. What is meant by balanced draft in boiler ? Ans: Balanced draft refers to combination of forced and induced draft fans controlled to keep the furnace under slightly vacuum condition. 85. How is dryness fraction of steam measured ? Ans: By using throttling calorimeter. 86. What is the difference between isentropic process and throttlinglprocess ? Ans: In isentropic process, heat transfer takes place and in throttling process, enthalpy before and after the process is same. 87. What is the difference between pyrometer, pyranometer and pyrheliometer. Ans: Pyrometer measures temperature by radiation principle. Pyranometer is used to measure total hemispherical solar radiation. Pyrheliometer is used to measure beam intensity by collimating the radiation. 88. What is the name given to useful part of available energy which is available for conversion to useful work ? Ans: Exergy. 89. Does the area on PV diagram for a process represent work for both reversible and nonreversible process ? Ans: No. Same is true for only reversible process. 90. Why work is considered as high grade of energy and heat as low grade energy ? Ans: While work can be completely converted to heat, heat can't be converted fully to work. 91. When coluomb friction comes into existence between two surfaces ? Ans: When there is relative motion between them. 92. Under what condition a number of forces acting at a point are in equilibrium ? Ans: When sum of the resolved parts in any two perpendicular directions is zero. 93. What happens to two equal and opposite coplanar couples ? Ans: They balance each other. 94. To slide a heavy block over a rough floor by a rope with minimum force by a man, at what angle the rope should be inclined with the level of floor ? Ans: He should incline the rope with horizontal at angle of friction between floor and block. 95. A body is dropped from a certain height. If same body were to move down a smooth inclined plane from same height, which parameters will be same on reaching the ground ? Ans: Velocity, kinetic energy and momentum. 96. A test specimen is stressed slightly beyond the yield point and then unloaded. What happens to its yield strength ? Ans: It will increase. 97. Out of circular, square, channel, I sections, which is most economical for a component subjected to bending ? Ans: I section. 98. What does the area under the stress-strain curve represent ? Ans: Energy required to cause failure. 99. A tension member of certain diameter is to be replaced by a square bar of the same material. Side of square compared to diameter will be ? Ans: Smaller. 100. A circular bar subjected to tension is designed for a particular safety factor. If both load and diameter are doubled, then factor of safety will be _____? Ans: doubled1. What is the difference between scavenging and supercharging ? Ans: Scavenging is process of flushing out burnt gases from engine cylinder by introducing supplying higher mass of air by compressing the atmospheric air. 2. What are the names given to constant temperature, constant pressure, constant volume, constant internal energy, constant enthalpy, and constant entropy processes.Ans: Isothermal, isochroic, isobaric, free expression, throttling and adiabatic processes respectively.
3. In a Rankine cycle if maximum steam pressure is increased keeping steam temperature and condenser pressure same, what will happen to dryness fraction of steam after expansion ?Ans: It will decrease. 4. Why entropy change for a reversible adiabatic process is zero ? Ans: Because there is no heat transfer in this process. 5. What are two essential conditions of perfect gas ? Ans: It satisfies equation of state and its specific heats are constant. 6. Enthalpy and entropy are functions of one single parameter. Which is that ? Ans: Temperature. 7. Why rate of condensation is higher on a polished surface compared to rusty surface ? Ans: Polished surface promotes drop wise condensation and does not wet the surface. 8. How much resistance is offered to heat flow by drop wise condensation ? Ans: Nil 9. What is the relationship between COP of heating and cooling ? Ans: COP of heating is one(unity) more than COP of cooling. 10. How much is the work done in isochoric process ? Ans: Zero. 11. When maximum discharge is obtained in nozzle ? Ans: At the critical pressure ratio. 1. Under what condition the work done in reciprocating compressor will be least ? Ans: It is least when compression process approaches isothermal. For this purpose, attempts are made to cool the air during compression. 13. What is the difference between stalling and surging in rotary compressions ? Ans: Stalling is a local phenomenon and it occurs when How breaks away from the blades. Surging causes complete breakdown of flow and as such it affects the whole machine. 14. Why the electric motor of a fan with backward curved blades is never got overloaded under any condition ? Ans: The maximum power is consumed at about 70% of maximum flow in case'of fan with backward blades. For higher flow, power consumption gets lower. 15. Why the work per kg of air flow in axial flow compressor is less compared to centrifugal compressor for same pressure ratio ? Ans: Isentropic efficiency of axial flow compressor is higher. 16. What is the name given to portion of thermal energy to be necessarily rejected to environment ? Ans: Anergy. 17. What is pitting ? How it is caused ? Ans: Non uniform corrosion over the entire metal surface, but occuring only in small pits is called pitting. It is caused by lack of uniformity in metal. 18. What is caustic embrittlement ? Ans: It is the actual physical change in metal that makes it extremely brittle and filled with minute cracks. It occurs particularly in the seams of rivetted joints and around the rivet holes. 19. Which impurities form hard scale and which impurities soft scale ? Ans: Sulphates and chlorides of lime and magnesium form hard scale, and carbonates of lime and magnesium form soft scale. 20. What is the difference between hard water and soft water ? Ans: Hard water contains excess of scale forming impurities and soft water contains very little or no scale forming substances. 21. Which two elements in feed water can cause corrosion of tubes and plates in boiler ? ' Ans: Acid and oxygen in feed water lead to corrosion. 22. What should be done to prevent a safety valve to stick to its seat ? Ans: Safety valve should be blown off periodically so that no corrosion can take place on valve and valve seat. 23. Why large boilers are water tube type ? Ans: Water tube boilers raise steam fast because of large heat transfer area and positive water circulation. Thus they respond faster to fluctuations in demand. Further single tube failure does not lead to catastrophy. 24. What type of boiler does not need a steam drum ? Ans: Super-critical pressure boiler. 25. Why manholes in vessels are usually elliptical in shape ? Ans: Elliptical shape has minimum area of opening and thus plate is weakened the least. Further it is very convenient to insert and take out the cover plate from elliptical opening. 26. Low water in boiler drum is unsafe because it may result in overheating of water tubes in furnace. Why it is unsafe to have high water condition in boiler drum ? Ans: High drum level does not allow steam separation to be effective and some water can be carried over with steam which is not desirable for steam turbine. 27. Why boiler is purged everytime before starting firing of fuel ? Ans: Purging ensures that any unburnt fuel in furnace is removed, otherwise it may lead to explosion. 28. What is the principle of mechanical refrigeration ? Axis. A volatile liquid will boil under the proper conditions and in so doing will absorb heat from
surrounding objects. 29. Why high latent heat of vaporisation is desirable in a refrigerant ? Ans: A high latent heat of vaporisation of refrigerant results in small amount of refrigerant and thus lesser circulation system of refrigerant for same tonnage. 30. What is the critical temperature of a refrigerant ? Ans: Critical temperature is the maximum temperature of a refrigerantrat which it can be condensed into liquid and beyond this it remains gas irrespective of pressure applied. 31. Maximum combustion temperature in gas turbines is of the order of 1100 to 10C whereas same is around 00C in I.C. engine ? Why ? Ans: High temperature in I.C. engine can be tolerated because it lasts for a fraction of second but gas turbines have to face it continuously which metals can't withstand. 32. Why efficiency of gas turbines is lower compared to I.C. engines ? Ans: In gas turbines, 70% of the output of gas turbine is consumed by compressor. I.C. engines have much lower auxiliary consumption. Further combustion temperature of I.C. engines is much higher compared to gas turbine. 33. What do you understand by timed cylinder lubrication ? Ans: For effective lubrication, lub oil needs to be injected between two piston rings when piston is at bottom of stroke so that piston rides in oi during upward movement. This way lot of lub oil can be saved and used properly. 34. What is IIUCR in relation to petrol engine ? Ans: HUCR is highest useful compression ratio at which the fuel can be used in a specific test engine, under specified operating conditions, without knocking. 35. In some engines glycerine is used in place of water for cooling of engine. Why ? Ans: Glycerine has boiling point of 90C which increases its heat carrying capacity. Thus weight of coolant gets reduced and smaller riadiator can be used. 36. Why consumption of lubricating oil is more in two-stroke cycle petrol engine than fourstroke cycle petrol engine ? Ans: In two-stroke engine lub oil is mixed with petrol and thus some lub oil is blown out through the exhaust valves by scavenging and charging air. There is no such wastage in four stroke petrol engine. 37. As compression ratio increases, thermal n increases. How is thermal n affected by weak and rich mixture strength ? Ans: Thermal n is high for weak mixture and it decreases as mixture strength becomes rich. 38. How engine design needs to be changed to burn lean mixture ? Ans: Engine to burn lean mixture uses high compression ratio and the highly turbulent movement of the charge is produced by the geometry of the combustion chamber. 39. Horse power of I.C. engines can be expressed as RAC rating, SAE rating, or DIN rating. To which countries these standards belong ? Ans: U.K., USA and Germany respectively. 40. What is the use of flash chamber in a vapour compression refrigeration cycle to improve the COP of refrigeration cycle ? Ans: When liquid refrigerant as obtained from condenser is throttled, there are some vapours. These vapours if carried through the evaporator will not contribute to refrigerating effect. Using a flash chamber at some intermediate pressure, the flash vapour at this pressure can be bled off and fed back to the compression process. The throttling process is then carried out in stages. Similarly compression process is also done in two separate compressor stages. 41. Why pistons are usually dished at top ? Ans: Pistons are usually hollowed at top to (i) provide greater spa'e for combustion, (ii) increase surface for flue gases to act upon, and (iii) better distribution of stresses. 42. What is the function of thermostat in cooling system of an engine ? Ans: Thermostat ensures optimum cooling because excessive cooling decreases the overall efficiency. It allows cooling water to go to radiator beyond a predetermined temperature. 43. What are the causes of failure of boiler tubes ? Ans: Boiler tubes, usually are made from carbon steel and are subject to (a) high rates of heat transfer, (b) bending stresses due to uneven heating, especially at expanded or welded joints into headers or drums, (c) external erosion from burners and flue gas, (d) possible corrosion on the boiler side, and (e) occasional manufacturing defects. Failure may occur due to following reasons : (a) High thermal ratings may lead to rapid failure if the internal fluid flow is reduced for any reason. The resultant overheating leads to a failure by creep, characterised by the bulging of the tube with the eventual development of a longitudinal split. (b) Fatigue cracking due to bending stresses occur. These are associated with change of section and/or weld undercut, where tubes are expanded or welded into headers. (c) Failure may arise due to overstressing of a reduced section of metal. (d) Sudden failure of the boiler tube due to corrosion arises from embrittlement of the carbon steel due to interaction between atomic hydrogen from the corrosion process and the iron carbide present in the steel. (e) Defects in tube manufacture, although far from being a regular occurrence, can be a cause of serious trouble. Lamination in boiler tubes or score marks arising from the cold
drawing of tubes, give rise to premature failure and may promote corrosion at these regions. 44. What are the causes of failure of superheater tubes ? Ans: Superheater tubes are subjected to the most severe combination of stress, temperature and corrosive environment. In addition to high-temperature strength, resistance to corrosion is also important. For example, low-alloy ferritic steel such as -1/% Cr, 1% Mo would not be used at metal temperatures above 580C because of inadequate resistance to corrosion and oxidation over a full service life of 100,000/150,000 hr. Failures in superheater tubes may arise from : (a) Prior fabrication history (b) Faulty heat treatment (c) Consequences of welding (d) Overheating of the tube metal (e) Gas-side corrosion (f) Stress corrosion (austenitic steels). 45. Why supercritical boilers use less amount of steel compared to non-supercritical boilers ? Ans: Supercritical boilers do not head heavy drum for separation of steam from mixture of water and steam. 46. Out of electric heater and heat pump, which is economical in operation ? Ans: Heat pump. 47. Which furnace burns low-ash fusion coal and retains most of the coal ash in the slag? Ans: Cyclone furnace. 48. How the thickness of thermal boundary layer and thickness of hydrodynamic boundary layer related ? Ans: Ratio of their thickness = (Prandtl number)-1/3. 49. What is the effect of friction on flow of steam through a nozzle ? Ans: To decrease both mass flow rate and wetness of steam. 50. Why gas turbine power plant needs efficient compressor ? Ans: Because a large portion of turbine work is eaten away by compressor and its inefficiency will affect net power output and cost of generation. 51. Why rockets using liquid hydrogen have higher specific impulse compared to liquid hydrocarbon ? Ans: Liquid hydrogen has higher burning velocity. 52. Why axial flow compressor is preferred for gas turbines for aeroplanes ? Ans: Because it has low frontal area. 53. What is the effect of inter cooling in gas turbines ? Ans: It decreases thermal efficiency but increases net output. 54. Why iso-octane is chosen as reference fuel for S.I. engines and allotted 100 value for its octane number ? Ans: Iso-octane permits highest compression without causing knocking. 55. Why thermal efficiency of I.C. engines is more than that of gas turbine plant ? Ans: In I.C. engine maximum temperature attained is higher than in gas turbine. 56. Which are the reference fuels for knock rating of S.I. engines ? Ans: n-heptane and ISO-octane. 57. When effect of variations in specific heats is considered then how do maximum temperature and pressure vary compared to air standard cycle ? Ans: Temperature increases and pressure decreases. 58. Quantities like pressure, temperature, density, viscosity, etc. are independent of mass. What are these called ? Ans: Intensive properties. 59. The amount of radiation emitted per scm per sec is called .... ? Ans: Emissive power. 60. In convection heat transfer, if heat flux intensity is doubled then temperature difference between solid surface and fluid will ? Ans: Get doubled. 61. How you can define coal ? Ans: Coal is a naturally occurring hydrocarbon that consists of the fossilised remains of buried plant debris that have undergone progressive physical and chemical alteration, called coalification, in the course of geologic time. 62. Which pollutant is major greenhouse gas and what is its effect ? Ans: CO is major greenhouse gas and it traps the radiation of heat from the sun within earth's atmosphere. 63. In order to increase efficiency and reduce CO emissions and other emissions, clear coal technologies are receiving major attention. What are these ? Ans: (i) Advanced pulverised and pressurised pulverised fuel combustion. (ii) Atmospheric fluidised bed combustion and pressurised fluidised bed combustion. (iii) Supercritical boilers. (iv) Integrated gasification combined cycle systems. (v) Advanced integrated gasification, including fuel cell systems. (vi) Magneto hydrodynamic electricity generation. 64. What are the important operational performance parameters in design of fuel firing equipment ? Ans: Fuel flexibility, electrical load following capability, reliability, availability, and maintenance ease.
65. What is the differenc between total moisture and inherent moisture in coal ? Ans: The moisture content of the bulk as sampled is referred to as total moisture, and that of the air dried sample is called inherent moisture. 66. Proximity analysis of coal provides data for a first, general assessment of a coal's quality and type. What elements it reports ? Ans: Moisture, volatile matter, ash and fixed carbon. 67. Ultimate analysis of coal is elementary analysis. What it is concerned with ? Ans: Carbon, hydrogen, nitrogen, and sulphur in coal on a weight percentage basis. 68. Explain the difference between AFBC, BFBC, PFBC and PCFB in regard to fluidised bed technologies. Ans: AFBC (Atmospheric fluidised bed combustion) process consists of forming a bed of inert materials like finely sized ash or ash mixed with sand, limestone (for sulphur removal), and solid fuel particles in a combustor and fluidising it by forcing combustion air up through the bed mixture. The gas flows thorugh bed without disturbing particles significantly but gas velocity is high enough to support the total weight of bed (fluidisation). At slightly higher velocity excess gas passes through the bed as bubbles (fluidised bed) and gives the bed the appearance of a boiling liquid. Bubbling fluidised bed combustion (BFBC) has a defined height of bed material and operates at or near atmospheric pressure in the furnace. Pressurised fluidised bed combustion (PFBC) system operates the bed at elevated pressure. Exhaust gases have sufficient energy to power a gas turbine, of course, gases need to be cleaned. In fluidised combustion, as ash is removed some unburned carbon is also removed resulting in lower efficiency. In circulating fluidised bed combustion (CFBC) system, bed is operated at higher pressure leading to high heat transfer, higher combustion efficiency, and better fuel feed. Circulating fluidised beds operate with relatively high gas velocities and fine particle sizes. The maintenance of steady state conditions in a fast fluidised bed requires the continuous recycle of particles removed by the gas stream (circulating bed). The term circulating bed is often used to include fluidised bed systems containing multiple conventional bubbling beds between which bed material is exchanged. 69. What for Schmidt plot for is used in heat transfer problems ? Ans: Schmidt plot is a graphical method for determining the temperature at any point in a body at a specified time during the transient heating or cooling period. 70. In which reactor the coolant and moderator are the same ? Ans: Pressurised water reactor. 71. Which reactor has no moderator ? Ans: Fast breeder reactor. 72. What are thermal neutrons ? Ans: Thermal neutrons are slow neutrons (having energy below 1 eV) which are in thermal equilibrium with their surroundings. 73. What is big advantage of fast breeder reactor ? Ans: It has rapid self breeding of fissile fuel during the operation of the reactor, and thus, it offers about sixty times the output with same natural uranium resources through ordinary nonbreeder nuclear reactor. 74. What is the purpose of biological shield in nuclear plants ? Ans: Biological shield of heavy concrete prevents exposure to neutrons, beta rays and gamma rays which kill living things. 75. Which two elements have same percentage in proximate and ultimate analysis of coal? Ans: Moisture and ash. 76. On which analysis is based the Dulong's formula for the heating value of fuel ? Ans: On ultimate analysis. 77. Which element causes difference in higher and lower heating values of fuel ? Ans: Hydrogen. 78. Which heating value is indicated by a calorimeter and why ? Ans: Gross heating value because steam is condensed and heat of vapour formed is recovered. 79. State the difference between ultimate and proximate analysis of coal ? Ans: In ultimate analysis, chemical determination of following elements is made by weight: Fixed and combined carbon, H, O, N, S, water and ash. Heating value is due to C, H and S. In proximate analysis following constituents are mechanically determined by weight. Moisture, volatile matter, fixed carbon and ash. Heating value is due to fixed carbon and volatile matter. 80. What is fuel ratio ? Ans: Fuel ratio is the ratio of its % age of fixed carbon to volatile matter. 81. How the analyses and calorific values of fuels can be reported ? Ans: It may be reported as (a) as received or fired (wet) basis (b) dry or moisture free basis (c) combustible or ash and moisture free basis 82. What is the difference between nuclear fission and fission chain reaction. Ans: The process of splitting of nucleus into two almost equal fragments accompanied by
release of heat is nuclear fission. Self sustained, continuing, sequence of fission reactions in a controlled manner is fission chain reaction. 83. Explain difference between fissile and fertile materials. Ans: The materials which can give nuclear fission e.g. U 35, Pu 39, U 33 are fissile materials. Fertile material itself is not fissionable, but it can be converted to a fissionable material by irradiation of neutrons in a nuclear reactor. 84. What do you understand by fuel cycle in nuclear plants ? Ans: Fuel cycle a series of sequential steps involved in supplying fuel to a nuclear power reactor. The steps include : Mining, refining uranium, fabrication of fuel elements, their use in nuclear reactor, chemical processing to recover remaining fissionable material, re-enrichment of fuel from recovered material, refabrication of new fuel elements, waste storage etc. 85. What is heavy water and what is its use in nuclear plants ? Ans: Water containing heavy isotopes of hydrogen (Deuterium) is known as heavy water. Heavy water is used as a moderator. Heavy water has low cross section for absorption of neutrons than ordinary water. Heavy water slows down the fast neutrons and thus moderates the chain reaction. 86. What is a converter reactor ? Ans: A reactor plant which is designed to produce more fuel than it consumes. The breeding is obtained by converting fertile material to fissile material. 87. Explain nuclear reactor in brief. Ans: A plant which initiates, sustains, controls and maintains nuclear fission chain reaction and provides shielding against radioactive radiation is nuclear reactor. 88. What is the difference between conversion and enrichment ? Ans: The process of converting the non fissile U 38 to fissile U-35 is also called "Conversion". The material like U 38 which can be converted to a fissile material by the neutron flux is called "fertile material". The conversion is obtained within the nuclear reactor during the chain reaction. Enrichment is the process by which the proportion of fissile uranium isotope (U-35) is increased above 0.7% (original % in natural uranium). The concentration of U-35 in the uranium hexafluoride is increased from the 0.7% in natural uranium to to 4%. This is called enrichment and is accomplished in an enrichment plant. 89. Disposal of radioactive waste materials and spent fuel is a major and important technology. How the waste radioactive material is disposed off ? Ans: Nonusable fission products are radioactive and take short/medium/long time for radioactive decay to reach safe level of radioactivity. Accordingly three methods of disposal are : (a) Zero or low radioactivity material is dispersed or stored without elaborate shielding. (b) Medium radioactivity material is stored for short duration of about 5 years to allow decay of radioactivity. (c) High radioactive material. They are stored in water for several months to permit radioactive decay to an accepetable low level. 90. Which nuclear reactor uses water as a coolant, moderator and reflector ? Ans: Pressurised water reactor. 91. Which reactor produces more fissionable material than it consumes ? Ans: Breeder reactor. 92. Which reactor uses natural uranium as fuel ? Ans: Gas cooled reacator. 93. Which reactor uses heavy water as moderator ? Ans: CANDU. 94. Which reactor requires no moderator ? Ans: Breeder reactor. 95. Which reactor uses primary coolant as fluoride salts of lithium, beryllium, thorium and uranium ? Ans: Molten salt breeder reactor. 96. Why an increase in area is required to produce an increase of velocity in case of supersonic flow ? Ans: Increase in area for increase in velocity for supersonic flow is required because the density decreases faster than velocity increases at supersonic speeds and to maintain continuity of mass, area must increase. 97. Under what circumstances would there be an increase in pressure in a divergent nozzle ? Ans: For subsonic flow at inlet section of a diffuser a lower velocity and higher pressure will exist at the exit section. For supersonic isentropic flow at the inlet section a higher velocity and lower pressure will exist at the exit but if a shock wave occurs in the diffuser then a higher pressure will exist at the exit. 98. Why water can't be used as refrigerant for small refrigerating equipment ? Ans: The refrigerant should be such that vapour volume is low so that pumping work will be low. Water vapour volume is around 4000 times compared to R- for a given mass. 99. Which parameter remains constant in a throttling process ? Ans: Enthalpy. 100. What is the difference between isentropic process and throttlinglprocess ?
Ans: In isentropic process, heat transfer takes place and in throttling process, enthalpy before and after the process is same. | 33,182 | 154,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-39 | latest | en | 0.910335 |
http://jwilson.coe.uga.edu/EMAT6680Su12/Carreras/HW_03/HW_3.html | 1,542,415,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743247.22/warc/CC-MAIN-20181116235534-20181117021534-00327.warc.gz | 171,552,858 | 3,054 | Assignment 3:
Exploring the Graphs of the Equations x2 + bx + c = 0 in the Plane xb for Various Values of c
by
Ángel M. Carreras Jusino
Goal:
• Find a relationship between the graphs of the equations of the form x2 + bx + c = 0 in the plane xb for various values of c and the number and types of roots of such equations.
Consider the graph of the equation x2 + bx + 1 = 0 in the plane xb.
If we overlay the graph of b = constant to the graph above, their intersections are the roots of the equation x2 + bx + 1 = 0.
Example for b = -4
Note that for |b| = 2 there is one real solution, |b| > 2 there are two real solutions, and for |b| < 2 there is no real solution. See the following animation.
Now lets explore the graphs of the equations of the form x2 + bx + c = 0 in the plane xb for various values of c.
From the animation above can be noticed that:
Let explore each case.
Case 1. c < 0
When c < 0, we always have two real roots no matter what is the value of b.
Case 2. c = 0
When c = 0, according to this graph the equation has only one real root for any value of b, specifically the solution x = -b.
But if we look to the equation x2 + bx + c = 0 with c = 0,
x2 + bx = 0
x (x + b) = 0
x = 0 and x + b = 0
x = 0 and x = -b
Now, why the solution x = 0 doesn't appear in the graph?
This happen because what is graph in the graphing software was an algebraic manipulation of the original equation which exclude 0 from its domain.
x2 + bx = 0
bx = -x2
b = -x2x where x ≠ 0
b = -x
Therefore for c = 0, we always have two real roots x = 0 and x = -b.
Case 3. c > 0
Here for different values of b we have 0, 1, or 2 real roots for the equation.
Now lets explore for which values of b we have the different number of real roots.
Looking at the animation we can note that the equation have one real root when the graph of the equations x2 + bx + c = 0 for c > 0 in the xb plane have slope 0.
Differentiating.
Equating to zero.
Evaluating the equation in this values of x we get:
So when c > 0 the equation x2 + bx + c = 0 has:
no real solution if b < |2√c|
one real solution if b = |2√c|
two real solution if b > |2√c|
Note that this is equivalent to the discriminant b2 - 4ac, which gave us information about the number and type of solutions of a quadratic equation. | 668 | 2,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2018-47 | latest | en | 0.930128 |
https://www.maplesoft.com/support/help/view.aspx?path=MultivariatePowerSeries/UpdatePrecision&L=E | 1,675,772,274,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00183.warc.gz | 873,111,909 | 34,006 | MultivariatePowerSeries/UpdatePrecision - Maple Help
Home : Support : Online Help : MultivariatePowerSeries/UpdatePrecision
MultivariatePowerSeries
UpdatePrecision
Update the precision to which a power series or univariate polynomial over power series is known
Calling Sequence UpdatePrecision(p,d) UpdatePrecision(u,d)
Parameters
p - power series generated by this package u - univariate polynomial over power series generated by this package d - non-negative integer
Description
• The command UpdatePrecision(p,d) increases the precision of p so that this precision equals at least d. This is achieved by repeated calls to the generator of p.
• The command UpdatePrecision(u,d) increases the precision of each power series coefficient of u so that this precision equals at least d.
• When using the MultivariatePowerSeries package, do not assign anything to the variables occurring in the power series and univariate polynomials over power series. If you do, you may see invalid results.
Examples
> $\mathrm{with}\left(\mathrm{MultivariatePowerSeries}\right):$
We create a new power series object. Initially, it is known only to low precision.
> $p≔\frac{1}{\mathrm{PowerSeries}\left(1+x\right)}$
${p}{≔}\left[{PowⅇrSⅇriⅇs of}\frac{{1}}{{1}{+}{x}}{:}{1}{+}{\dots }\right]$ (1)
> $\mathrm{Precision}\left(p\right)$
${0}$ (2)
If we update the precision, more terms are known.
> $\mathrm{UpdatePrecision}\left(p,10\right)$
$\left[{PowⅇrSⅇriⅇs of}\frac{{1}}{{1}{+}{x}}{:}{1}{-}{x}{+}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{x}}^{{4}}{-}{{x}}^{{5}}{+}{{x}}^{{6}}{-}{{x}}^{{7}}{+}{{x}}^{{8}}{-}{{x}}^{{9}}{+}{{x}}^{{10}}{+}{\dots }\right]$ (3)
> $\mathrm{Precision}\left(p\right)$
${10}$ (4)
We create a new polynomial over power series involving the power series defined above, $p$, and a new power series, $q$.
> $q≔\mathrm{GeometricSeries}\left(\left[x,y\right]\right)$
${q}{≔}\left[{PowⅇrSⅇriⅇs of}\frac{{1}}{{1}{-}{x}{-}{y}}{:}{1}{+}{x}{+}{y}{+}{\dots }\right]$ (5)
> $f≔\mathrm{UnivariatePolynomialOverPowerSeries}\left(\left[q,p\right],z\right)$
${f}{≔}\left[{UnivariatⅇPolynomialOvⅇrPowⅇrSⅇriⅇs:}\left({1}{+}{x}{+}{y}{+}{\dots }\right){+}\left({1}{-}{x}{+}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{x}}^{{4}}{-}{{x}}^{{5}}{+}{{x}}^{{6}}{-}{{x}}^{{7}}{+}{{x}}^{{8}}{-}{{x}}^{{9}}{+}{{x}}^{{10}}{+}{\dots }\right){}{z}\right]$ (6)
Because $q$ is new, it is again known only to low precision.
> $\mathrm{Precision}\left(q\right)$
${1}$ (7)
Updating the precision of $f$ updates the precision of each of its coefficients, $p$ and $q$, if necessary. In this case, $p$ was already known to higher precision, but $q$'s precision is updated.
> $\mathrm{UpdatePrecision}\left(f,8\right)$
$\left[{UnivariatⅇPolynomialOvⅇrPowⅇrSⅇriⅇs:}\left({1}{+}{x}{+}{y}{+}{{x}}^{{2}}{+}{2}{}{x}{}{y}{+}{{y}}^{{2}}{+}{{x}}^{{3}}{+}{3}{}{{x}}^{{2}}{}{y}{+}{3}{}{x}{}{{y}}^{{2}}{+}{{y}}^{{3}}{+}{{x}}^{{4}}{+}{4}{}{{x}}^{{3}}{}{y}{+}{6}{}{{x}}^{{2}}{}{{y}}^{{2}}{+}{4}{}{x}{}{{y}}^{{3}}{+}{{y}}^{{4}}{+}{{x}}^{{5}}{+}{5}{}{{x}}^{{4}}{}{y}{+}{10}{}{{x}}^{{3}}{}{{y}}^{{2}}{+}{10}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{5}{}{x}{}{{y}}^{{4}}{+}{{y}}^{{5}}{+}{{x}}^{{6}}{+}{6}{}{{x}}^{{5}}{}{y}{+}{15}{}{{x}}^{{4}}{}{{y}}^{{2}}{+}{20}{}{{x}}^{{3}}{}{{y}}^{{3}}{+}{15}{}{{x}}^{{2}}{}{{y}}^{{4}}{+}{6}{}{x}{}{{y}}^{{5}}{+}{{y}}^{{6}}{+}{{x}}^{{7}}{+}{7}{}{{x}}^{{6}}{}{y}{+}{21}{}{{x}}^{{5}}{}{{y}}^{{2}}{+}{35}{}{{x}}^{{4}}{}{{y}}^{{3}}{+}{35}{}{{x}}^{{3}}{}{{y}}^{{4}}{+}{21}{}{{x}}^{{2}}{}{{y}}^{{5}}{+}{7}{}{x}{}{{y}}^{{6}}{+}{{y}}^{{7}}{+}{{x}}^{{8}}{+}{8}{}{{x}}^{{7}}{}{y}{+}{28}{}{{x}}^{{6}}{}{{y}}^{{2}}{+}{56}{}{{x}}^{{5}}{}{{y}}^{{3}}{+}{70}{}{{x}}^{{4}}{}{{y}}^{{4}}{+}{56}{}{{x}}^{{3}}{}{{y}}^{{5}}{+}{28}{}{{x}}^{{2}}{}{{y}}^{{6}}{+}{8}{}{x}{}{{y}}^{{7}}{+}{{y}}^{{8}}{+}{\dots }\right){+}\left({1}{-}{x}{+}{{x}}^{{2}}{-}{{x}}^{{3}}{+}{{x}}^{{4}}{+}{\dots }\right){}{z}\right]$ (8)
> $\mathrm{Precision}\left(q\right)$
${8}$ (9)
> $\mathrm{Precision}\left(p\right)$
${10}$ (10)
Compatibility
• The MultivariatePowerSeries[UpdatePrecision] command was introduced in Maple 2021.
• For more information on Maple 2021 changes, see Updates in Maple 2021. | 1,751 | 4,153 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-06 | latest | en | 0.568126 |
humphreytalks.com | 1,701,975,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00700.warc.gz | 354,165,532 | 11,025 | # How To Go From \$0 to \$100,000 in 2023 (in a year)
51% of Americans have \$5,000 or less in savings, while 1 in 3 (35%) have \$1,000 or less in savings.
And according to a study in 2022, 56% of Americans cant cover a \$1000 dollar emergency expense.
Now, having \$100K can be life changing - but getting there is a whole nother story. According to Charlie Munger - the right hand man of Warren Buffett with a net worth of over \$2.3 billion dollars, quote:
The first \$100,000 is a bitch, but you gotta do it. I don’t care what you have to do – if it means walking everywhere and not eating anything that wasn’t purchased with a coupon, find a way to get your hands on \$100,000. After that, you can ease off the gas a little bit.*
So let’s break this down - the quote hints at the fact that we need to save and be frugal in order to reach \$100K as quickly as possible — he literally says to “walk everywhere” and “not eat anything”. Notice how he doesn’t say to “invest in Gold” or “yolo into gamestop stock”.
And this is what’s fascinating….
Contrary to what people believe, the first \$100k is typically not derived from just investment returns in the market. Take a look at this calculator from the Four Pillar Freedom blog:
https://fourpillarfreedom.com/four-visuals-volume-5-comparing-savings-vs-investment-returns/
If you save \$15,000 a year for 6 years and you’re also able to attain a modest 4.5% return on your savings - you would be at \$105k after 6 years but 85% of that would have been comprised of savings, and 15% is interest.
In another example, let’s say you save a lot less, \$10k a year, but you average a 10% return. It will still take you 7 years and at a phenomenal 10% return, savings composes 67% of the total \$104k here.
The point is - rarely is your first 100K just from investment gains. It’s actually from just saving, being smart with money, and earning more money throughout the years. The faster you want \$100K, the more you need to save and less you need to invest. That’s why Charlie Munger’s quote really applies here - you want to amass it by ANY MEANS NECESSARY.
BUT the beauty is, after you get your first 100K, the next 100K is easier to make, and it’ll come faster.
Pretend you save \$12000 a year, or \$1000 a month - and you’re able to get 8% in the market.
Your first 100K will take you 6.25 years. Your next 100K assuming you keep investing and saving \$12k a year would only take you 4.25 years. That’s 2 years faster because of the compounding effect of having \$100K already.
Breakdown:
Ok so if I wanted to get to \$100K asap. I would break down this number in a bit more digestible of a manner.
\$100K in a year is \$273.97 cents a day. But if we only count business days, there are 260 working days in a year so that comes out to roughly \$384 dollars a day. This is your target. You want to be able to make about \$380-400 dollars a working day. Assuming you work 40 hour weeks, that’s an hourly wage of \$48 dollars an hour. If you want to work 60 hours per week, its about \$32 dollars an hour.
Ok so is this actually possible within a year? I would say… statistically no its not likely, but the idea is that I’m gonna try for it - because there are certainly people making this kind of money within a year. Why not me?
Irreplaceability
Let’s talk about how I would make \$380 in a day - or roughly \$48 dollars an hour. First I would start off by reading Hump Days, my free newsletter where we break down business and economy news twice a week - link is in the description and its completely free to sign up - we are just about to cross 10,000 readers.
Alright but for real, to make \$380 a day - we need to talk about the concept of being irreplaceable.
The more irreplaceable you are, the more money you will make. It’s a reason why so many athletes make exorbitant amounts of money. Steph Curry, my favorite NBA player, signed a \$215 million dollar contract back in 2017, why? Because you can argue that his skills in basketball are irreplaceable. He literally can’t be replaced. Who else is going to shoot 3 pointers from half court?
So Steph Curry is a huge asset to any NBA team, and is the reason why he was signed to a supermax contract. He drives value to the Golden State Warriors which has been valued at \$7 billion dollars - the most valuable team in the NBA.
Contrast Steph Curry to someone selling concessions in the stadium. That person selling popcorn and drinks is not going to be paid as high as Steph Curry, thats an extreme example, but there are many hundreds if not thousands of people that could do the job of selling concessions.
In the eyes of the market and the economy this person just naturally won’t make that much because the skill is easily replaced and the pay will reflect that skill.
So in this way, we need to play to our strengths. What are our skills? What are our natural advantages?
Your High Income Skill - how you would do it
Whether we know it or not, we all have advantages and expertise in SOMETHING over another person. Think about your friends or family - there’s going to be things that you are naturally better at than they are, simply because you’ve spent more time doing them than your peers.
So if I took myself for example, at this point in my career I know a lot about a few things more than a majority of people.
1) gaming companies, since I used to monetize one of the highest grossing games in the world
2) ecommerce since I started my own poster business
3) finance and financial advisory since I used to be a FA,
and
#4) is social media and YouTube.
Does it mean I’m the single greatest person at these? Not even close. However, I do know more than someone starting out - for example. If I wanted to go out and consult YouTubers or Tik Tok’ers on how to grow their channels, I could conservatively charge \$100-200 an hour for my knowledge, and same thing with helping people with their finances.
Could I charge people for my basketball skills? Hell no - I’ve probably played basketball for less than 20 hours in my entire life. Not even a 5th grader would pay me for that.
{Journal B Roll} Now if you’re watching this, what you can do is take out a piece of paper and write down what you are good at - seriously, pause the video and try this right now.
For example, if you know a lot about building muscle, working out, and creating workout programs - think about the different ways you could monetize that skill.
You could do 1 on 1 personal training with people — {B roll of me worknig out getting a training session}
You could become an affiliate of fitness related products like supplements and clothing products.
You could start a website that offers free knowledge and sell a course on your workout programs.
You could start a Private Facebook Group where you have a community of like minded fitness people that you drive to your links.
But it all starts with knowing what you’re good at.
Now if you’re watching this and you really don’t think you’re good at anything - I doubt thats true, but for the fun of it lets say its true- what should you do?
This is where I would spend the next year or two building out a high income skill. There are so many skills and types of jobs that I know I can do from home and earn over \$50 an hour in. I could become a graphic designer and help people with logos and branding, I could become a videographer, a copywriter, an online writer, a social media marketing manager, a web designer, etc.
You start small with these skills, build them day by day, and eventually you can scale them to 6 figure + businesses. Learning these skills can all be done online for free on websites like Hubspot Academy. For example they have a email marketing course you could take that teaches you the basics of email marketing that you could then use to apply for gigs online.
I remember watching a CNBC Make It video from a year ago where this Fiverr freelancer was making \$375 thousand dollars a year just writing. Her main income was derived from ghostwriting e-books and blog posts. https://youtu.be/hHhYtZV_T9M
While she is probably in the top 1% of all writers online, it doesn’t mean we need to be in the top 1% of anything to get us to our \$100k goal.
We just need to get to a place where we’re earning more money, ideally above \$50 an hour, and then combine it with cutting costs dramatically.
If I didn’t want to learn a new skill, I’d probably start a service based business that had the potential to scale. Some service based businesses are like dog-sitting, window washing, lawn mowing, cleaning gutters - stuff everyone can do. In all of these service based businesses, you start small to prove the concept, get to know customers and get some referrals. Eventually if I got to a point where I was booked like crazy for lawn services, I would then start to scale that business by adding an employee, and so on. But I want to point out that in this process, you are building the skill of starting a business, even if the action you are taking is not a new skill. so thats pretty cool.
Cutting Costs - behavioral change
Ok so you are figuring out how to be irreplaceable and earning more of an income. Now Let’s talk about the other side of the equation - costs.
This is where you need to live as frugal as possible on your way to 100K.
The dictionary definition of frugal is to be "careful about spending money or using things when you do not need to", in other words - we want to make sure we're getting the most value out of our dollar, pound, euro, or whatever currency it may be ESPECIALLY in the beginning because the beginning is the hardest.
You don’t have to be frugal forever, but I would say in the beginning you need to be thinking critically about all your decisions about money. No matter how small they seem.
Here’s a recent example that my friend thought I was crazy for.
I have the McDonalds app on my phone. I can get an iced coffee using the McDonalds app for \$0.99 cents, no matter the size. The normal price is \$2.29 cents or something like that, that’s more than DOUBLE the price. So while i’m only saving like \$1.29, YES I will go out of my way to use the app to save that money. While that \$1 and change isn’t going to change my life dramatically, nor would I even notice it, I find that its a microcosm of my overall financial behavior. I’m always trying to get the best possible deal if I can. That behavior will likely translate to the rest of my life. And that’s how you should be thinking when trying to amass the first \$100K.
It’s the same reason why you see super rich frugal people still trying to save pennies on the dollar. That’s how they got there in the first place.
Now does this mean I’m cheap? Or that you need to be cheap? No. I think there is still enough room for select times where you can be generous and giving to others, but still frugal with your own personal spending.
So I want you to do this. Whenever you're about to spend money, think about if there's a less costly alternative to doing so. If you get a starbucks everyday, why don't you try making coffee at home? If you like eating out every meal, perhaps switch it up and meal prep for a week to see how much you could save? If you take an Uber to work daily, why don’t you try public transport? And my favorite… when you DO go out for lunch, keep it simple. Get the sandwich, but maybe skip the chips and the drink - this small decision will save you money over long periods of time.
It’s usually the tiny lifestyle changes that start to compound and help you save a ton of money.
# Hump 🐪 Days
Financial news can be overwhelming. Hump Days delivers twice a week so you're never left behind. We're your one-stop-shop for all things remotely finance related. | 2,748 | 11,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.942793 |
https://exactlyhowlong.com/fahrenheit-to-kelvin-converter/ | 1,719,153,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00784.warc.gz | 217,693,848 | 20,457 | # Fahrenheit to Kelvin Converter
Instructions:
• Enter the temperature in Fahrenheit.
• Click "Convert" to get the temperature in Kelvin.
• Click "Clear Results" to clear the input and result.
• Click "Copy Result" to copy the result to the clipboard.
• Your calculation history will be displayed below.
Detailed Calculation
Calculation History
Temperature, a fundamental physical property of matter, is expressed in degrees Fahrenheit (°F) or Kelvin (K). While the Fahrenheit scale is commonly used in the United States, the Kelvin scale is the primary temperature scale in the International System of Units (SI) and is widely used in scientific and engineering applications. Therefore, the need arises for accurate and efficient Fahrenheit to Kelvin conversion tools.
## Temperature Scales: A Conceptual Overview
Temperature scales provide a standardized way to quantify the hotness or coldness of an object or substance. Two commonly used temperature scales are the Fahrenheit and Kelvin scales.
### Fahrenheit Scale (°F): A Widely Used Scale
The Fahrenheit scale, developed by Daniel Gabriel Fahrenheit in 1727, is a widely used temperature scale in the United States and a few other countries. It defines 32°F as the freezing point of water and 212°F as the boiling point of water at standard atmospheric pressure.
### Kelvin Scale (K): The SI Standard for Temperature
The Kelvin scale, named after William Thomson, 1st Baron Kelvin, is the SI base unit of temperature. It is an absolute temperature scale, meaning its zero point corresponds to absolute zero, where molecular motion ceases. Absolute zero is defined as 0 K, which is equivalent to -273.15°C or -459.67°F.
Also Read: Celsius to Rankine Converter
## Converting Fahrenheit to Kelvin: Unveiling the Formula
The formula for converting Fahrenheit to Kelvin is:
``````K = (°F + 459.67) × 5/9
``````
This formula essentially adds 459.67 to the Fahrenheit temperature to shift the origin of the scale to absolute zero and then multiplies by 5/9 to convert from Fahrenheit degrees to Kelvin degrees.
## Advantages of Fahrenheit to Kelvin Converter
Utilizing a Fahrenheit to Kelvin converter offers several advantages, including:
1. Accuracy and Efficiency: Conversion tools can perform calculations quickly and precisely, eliminating the need for manual conversions and reducing the likelihood of errors.
2. Convenience and Accessibility: Fahrenheit to Kelvin converters are readily available online or as mobile applications, making them easily accessible for quick conversions on the go.
3. Standardization and Clarity in Communication: Converting temperatures ensures that temperature measurements are expressed in a universally recognized format, facilitating clear and unambiguous communication and record-keeping across different scientific and engineering disciplines.
## Facts about Fahrenheit to Kelvin Conversion
1. A Shared Origin: A Historical Perspective: While the Fahrenheit and Kelvin scales have different origins, they are both based on the concept of temperature scales defined by fixed points.
2. The Role of Absolute Temperature in Scientific Applications: In scientific and engineering calculations, absolute temperature scales like Kelvin are preferred due to their consistent relationship with thermodynamic properties and their role in defining fundamental physical laws.
3. Versatility across Diverse Fields: The need for temperature conversion arises in various fields, including meteorology, climate science, physics, chemistry, and engineering.
## References
1. “Introduction to Thermodynamics” by Y. V. C. Rao (2005)
This comprehensive textbook provides a detailed introduction to thermodynamics, including concepts of temperature scales and their conversions.
1. “Fundamentals of Engineering Thermodynamics” by Michael J. Moran and Howard N. Shapiro (2006)
This essential engineering thermodynamics reference book covers the fundamental principles of temperature measurement and conversion between different temperature scales.
1. “A Dictionary of Physics” by David J. Acheson (2004)
This comprehensive physics dictionary provides clear and concise definitions of various physical terms, including temperature scales and their conversions.
## Conclusion
Fahrenheit to Kelvin conversion tools play a crucial role in accurately translating temperature measurements between different scales, ensuring consistency and standardization in scientific and engineering applications. These tools not only facilitate communication and record-keeping but also contribute to a deeper understanding of temperature as a fundamental physical property. By employing these tools effectively, scientists and engineers can enhance the precision and clarity of their work, leading to advancements in various fields. | 889 | 4,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.781461 |
https://forum.piedao.org/answers/1158085-how-much-liquid-volume-can-a-soup-spoon-hold | 1,708,597,496,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00350.warc.gz | 275,446,982 | 5,681 | How much liquid volume can a soup spoon hold?
Answer: The most a soup spoon could hold is 3-7 mL.
Related Questions
180 miles
---------------
2 hours Find the unit rate in feet per second
If I did the did heheheudjdjdjdjdjdjdjd
12 is a common denominator of 1/4 and 1/3
True... both 3 and 4 are factors of 12 so it can be used as a common denominator.
1/3 × 4/4 = 4/12
1/4 × 3/3 = 3/12
Yes because 4 times 3 equals 12 and 3 times 4 equals 12
What is the area of a triangle with a base of 5cm and height with 3cm
(5x3)/2 = 7.5
Ron wants to deliver a package to his friend. He is considering two courier services to do the job. The first courier service is charging a flat fee of \$15 plus \$3 per pound of the package’s weight.The second courier service is charging a flat fee of \$25 plus \$2 per pound of the package’s weight.
The costs of the two courier services are shown in the graph below.
What must the package weigh for the costs of the two courier services to be the same?
A.
8
B.
10
C.
14
D.
12
Reset Next Question
B. 10
15 + 3(10) = 45
25 + 2(10) = 45
Freddy is helping buy ingredients for salads for the school spaghetti dinner. He bought 10 pounds of onions at \$0.69 per pound, 100 pounds of tomateos at \$0.99 pounds, 1,000 pounds of bread crumbs at \$0.09 per pound, and 100 pounds of lettuce at \$0.69 per pound. Which of the things he bought cost the most?
Hey there,
If you multiply 10 x 0.69 (for onions) it would be 6.9
If you multiply 100 x 0.99 (for tomatoes) it would be 99
Lastly, if you multiply 100 x 0.69 (for lettuce) if would be 69
Which are the most? Tomatoes.
~ Hope this helps, and if you have any questions or concerns, feel free to pm me or comment down below :)
By multiplying the quantity of each item by its cost, you can see that the tomatoes were the most expensive item that Freddy bought for the school spaghetti dinner.
Explanation:
To calculate which item Freddy spent the most on, multiply the quantity he bought by the cost per unit. For onions, he spent 10 pounds x \$0.69/pound = \$6.90. For tomatoes, his cost was 100 pounds x \$0.99/pound = \$99.00. For bread crumbs, he spent 1000 pounds x \$0.09/pound = \$90.00. Lastly, for lettuce the cost was 100 pounds x \$0.69/pound = \$69.00. Therefore, the most expensive item he bought was tomatoes.
brainly.com/question/33521342
#SPJ2
Given the line 5x – 2y = 10, Find the equation of a parallel line that passes through the point (2, 3). Find the x- and y-intercepts of both lines. Plot the intercepts and use them to graph both lines on the same set of axes. | 740 | 2,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.934735 |
https://www.esaral.com/q/a-and-b-can-do-a-piece-of-work-in-6-days-and-4-days-respectively-35973 | 1,726,155,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00033.warc.gz | 700,505,307 | 11,640 | A and B can do a piece of work in 6 days and 4 days respectively.
Question:
A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.
Solution:
A can do a work in 6 days, and B can do the same work in 4 days.
$\therefore$ Work done by A in 2 days $=\frac{2}{6}=\frac{1}{3}$
Remaining work $=1-\frac{1}{3}=\frac{2}{3}$
$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 1 day $=\left(\frac{1}{6}+\frac{1}{4}\right)$
$=\frac{2+3}{12}=\frac{5}{12}$
$\because \frac{5}{12}$ th work is done by A and B in 1 day.
$\therefore \frac{2}{3}$ rd work will be done by A and B in $\left(\frac{12}{5} \times \frac{2}{3}\right)$ days or $\frac{8}{5}$ days.
$\therefore$ Total time taken $=\left(\frac{8}{5}+2\right)$ days $=\frac{18}{5}$ days $=3 \frac{3}{5}$ days | 321 | 890 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-38 | latest | en | 0.841535 |
https://socratic.org/questions/if-a-rectangle-is-17-ft-long-and-8-ft-wide-what-is-its-perimeter | 1,713,610,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817576.41/warc/CC-MAIN-20240420091126-20240420121126-00355.warc.gz | 492,382,670 | 6,254 | # If a rectangle is 17 ft long and 8 ft wide, what is its perimeter?
Jun 13, 2018
50 ft
#### Explanation:
Length = 17ft
Rectangle has 2 lengths and 2 breadths
So,
$17 \times 2 + 8 \times 2$
$= 50$
Jun 13, 2018
Perimeter $= 50 f t$
#### Explanation:
Length of rectangle $= 17 f t$
Width of rectangle $= 8 f t$
Perimeter $= 2 \left(l + b\right) = 2 \left(17 + 8\right) = 2 \left(25\right) = 50 f t$ | 156 | 404 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | latest | en | 0.619988 |
http://en.wikipedia.org/wiki/Aubin%e2%80%93Lions_lemma | 1,386,423,450,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054548/warc/CC-MAIN-20131204131734-00058-ip-10-33-133-15.ec2.internal.warc.gz | 67,677,069 | 9,018 | # Aubin–Lions lemma
In mathematics, the Aubin–Lions lemma is a result in the theory of Sobolev spaces of Banach space-valued functions. More precisely, it is a compactness criterion that is very useful in the study of nonlinear evolutionary partial differential equations. The result is named after the French mathematicians Thierry Aubin and Jacques-Louis Lions.
## Statement of the lemma
Let X0, X and X1 be three Banach spaces with X0 ⊆ X ⊆ X1. Suppose that X0 is compactly embedded in X and that X is continuously embedded in X1; suppose also that X0 and X1 are reflexive spaces. For 1 < pq < +∞, let
$W = \{ u \in L^p ([0, T]; X_0) | \dot{u} \in L^q ([0, T]; X_1) \}.$
Then the embedding of W into Lp([0, T]; X) is also compact
## References
• Showalter, Ralph E. (1997). Monotone operators in Banach space and nonlinear partial differential equations. Mathematical Surveys and Monographs 49. Providence, RI: American Mathematical Society. p. 106. ISBN 0-8218-0500-2. MR 1422252. (Proposition III.1.3) | 304 | 1,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-48 | latest | en | 0.819326 |
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Math.libretexts.org keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
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### NSF Award Search: Award # 1523786
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ABSTRACT The purpose of this collaborative project from Monroe Community College, Robert Morris University, and the State University of New Your (SUNY) at Buffalo is to use visual exploration to help students develop a deeper understanding of multivariable calculus concepts and to conduct and publish research on the effects of visualization on student understanding …
### Polynomial Congruences
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• The second stage of the process, then, is to deal with polynomial congruences of the form € f(x)≡0 (modpe) where p is prime
• For this we use a powerful result, known as The Lifting Theorem Suppose x≡a (modpe) is a solution to the polynomial congruence
### Solving a system of equations Future Starr
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• (Source: math.libretexts.org) Solve
• While systems of three or four equations can be readily solved by hand (see Cracovian), computers are often used for larger systems
• The standard algorithm for solving a system of linear equations is based on Gaussian elimination with some modifications
• Firstly, it is essential to avoid division by small
### Figure 14 The horizontal component of the force is the
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• The horizontal component of the force is the projection of 𝑭
• Photo credit: math.libretexts.org
• Given : F = 10N d = 60 m θ = 55 ° Photo credit : math.libretexts.org Required : W = ? Formula : W = Fd cosθ Solution : W = Fd cosθ = ( 10N ) x ( 60 m ) cos ( 55 ° ) = ( 10N ) x ( 60 m ) x 0.5736
### Calculus Problem Solving: Step by Step Examples
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• Step 2: Write an equation for the horizontal motion of the baseball as a function of time: x (t) = v * Cos (a) * t
• Step 3: Write an equation to describe the vertical motion of the baseball as a function of time: y (t) = h + v * Sin (a) * t – 16 * t2
• In this formula, t 2 is the square of the variable ‘t
### Probability of Poker Hands
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Probability of Poker Hands Drew Armstrong [email protected] November 1, 2006 In a standard deck of cards, there are 4 possible suits (clubs, diamonds, hearts, spades),
### Βιβλία Μαθηματικών
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### Laplace Transform Of Constant
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Laplace Transform can be used in three cases: applying Laplace Transform to governing equations of lumped parameter model to change ordinary differential equation system into algebraic equations; applying Laplace Transform to governing equations of distributed parameter model for parallel - flow and counterflow heat exchangers to change partial …
### Explanation of Simplex Method for Minimization. by
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The Simplex method is an approach to solving linear programming models by hand using slack variables, tableaus, and pivot variables as a means to finding the optimal solution of an optimization…
### What is the Logic of Inference
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• Peregrin dispute with those theoreticians who claim that it is intuitionistic logic that is the logic most intimately related to inference
• However, my aim here is to present what I believe is a deeper and a more general substantiation for
### Circumscribed and inscribed circles of triangles — Krista
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• When a circle circumscribes a triangle, the triangle is inside the circle and the triangle touches the circle with each vertex
• You use the perpendicular bisectors of each side of the triangle to find the the center of the circle that will circumscribe the triangle
• So for example, given G H I \triangle GHI G H I,
### Www2.karlin.mff.cuni.cz
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Gilbert Strang & Edwin “Jed” Herman 2/24/2021 14.5.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/2604 The left-hand side of this equation is equal to …
### How to find the exact values of the six circular function
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• In finding the six circular functions, we base it on a unit circle whose center is always on the origin (0, 0) and its radius is equal to 1 unit
• Suppose θ is an angle plotted in standard position and P (x, y) is the point on the terminal side of θ which lies on the Unit Circle
• The cosine of θ, denoted cos (θ), is defined by
### NSF Award Search: Award # 1524968
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ABSTRACT The purpose of this collaborative project from Monroe Community College, Robert Morris University, and the State University of New York (SUNY) at Buffalo is to use visual exploration to help students develop a deeper understanding of multivariable calculus concepts and to conduct and publish research on the effects of visualization on student understanding …
### How to Graph an Ellipse: 11 Steps (with Pictures)
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• Know about the two foci of the ellipse
• The foci (plural for "focus") are two points inside the ellipse
• Since an ellipse is the curve formed by all the points such that the sum of its distances from each of the two foci is constant, the foci are one of the main defining inputs for an ellipse, along with the constant distance sum
• When both foci are at the same place, the …
### Differential Calculus Word Problems with Solutions
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• Problem 1: A missile fired ground level rises x meters vertically upwards in t seconds and x = 100t - (25/2)t 2
• (i) the initial velocity of the missile, (ii) the time when the height of the missile is a maximum
• (iii) the maximum height reached and
### What is the hardest part of pre-calculus
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• Answer (1 of 6): For me it was partial fractional decomposition
• Some of the identities in trigonometry are difficult and the conics can be hard at first
• Like everything else in math, just keep practicing and you will be fine.
### LCM Calculator with Steps (Least Common Multiple Calculator)
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• Example: Find LCM (15, 18, 21) Step 1: Write down prime factors of all integers
• Step 2: Multiply the highest number of primary factors for each integer
• A number would be considered one time in multiplication if it occurred two or more times in factoring
• 2 × 3 × 3 × 5 × 7 = 630.
### Math 3527 (Number Theory 1)
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Math 3527 (Number Theory 1) Lecture #29 Polynomial Congruences: Polynomial Congruences Modulo m Polynomial Congruences Modulo pn and Hensel’s Lemma This material represents x5.1 from the course notes.
### Mathematical Logic and Set Theory
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• Larger than the set z, so “absolute infinity” demands that there be a stage slater than all the s x.Then V s works as u
• By absolute infinity, there is an infinite stage s.
### Cylindrical Triple Integral Calculator
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• Three figures stand out for M as the position of a point xyz in cylindrical coordinates in cylindrical coordinates: ρ, φ, z, is the projection of the radius vector of the point ρ to the M axis; xy is the distance calculated by three different angles: φ is the projection of the radius vector of the point x axis
• 1 is the representation of the radius vector on the z EQ z The ellipsoid volume | 3,464 | 12,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-05 | latest | en | 0.725985 |
https://universolinux.es/shaking-table/4img5hy5-shaking.html | 1,632,826,649,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060677.55/warc/CC-MAIN-20210928092646-20210928122646-00610.warc.gz | 622,114,208 | 4,998 | # shaking table z normal distribution
Therefore: Z score = (700-600) / 150 = 0.67 Now, in order to figure out how well George did on the test we need to determine the percentage of his peers who go higher and lower scores. That’s where z-table (i.e. standard normal distribution table) comes handy. If you noticed there are two z-tables …
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### standardnormal distribution tableorz-score
View Standard Normal Distribution Table or z-Score Table1-2.pdf from ACCT 3045 at Fanshawe College. Standard Normal Distribution Table or z-Score
### z-score table| formula,distribution table, chart & example
Solution: The z score for the given data is, z= (85-70)/12=1.25. From the z score table, the fraction of the data within this score is 0.8944. This means 89.44 % of the students are within the test scores of 85 and hence the percentage of students who are above the test scores of 85 = (100-89.44)% = 10.56 %
### normal distribution| gaussiandistribution-z table|z
According to the Z-Score table, we get Therefore P (x<46) = P (Z<-2.37) = 0.00889, which indicates only 0.88 % (0.00889 X 100) of students score less than 46. Example 2: The test score of 50 students in a class is normally distributed with mean 65 and a standard deviation of 8
### what isnormal distribution(z)
Normal distribution is a continuous probability distribution. It is also called Gaussian distribution. The normal distribution density function f (z) is called the Bell Curve because it has the shape that resembles a bell. Standard normal distribution table is used to find the area under the f ( z) function in order to find the probability of a specified range of distribution
### zscores (zvalue) &z table| six sigma study guide
In other words, p (Z<-1.53) = 0.0630. Standard normal table also used to determine the area to the right of any Z-value by subtracting the area on the left from 1. Simply, 1-Area Left = Area right. For example Z-score of 0.83 has an area of 0.7967 to the left of it. So, Area of right is 1-0.7967 = 0.2033
### normal distribution tableforz-test
by Using Normal-Distribution Table Z-scores generally ranges from -3.99 to 0 on the left side and 0 to 3.99 on the right side of the mean. Refer the column & row values for z-score. The point where the row & column meets for the corresponding z-score value is the critical value of Z or the rejection area of one or two tailed z-distribution. For example the -2.95 Z is the left tailed distribution
### z table(standardnormal distribution)
This test has a standard deviation (σ) of 25 and a mean (μ) of 150. Also assuming that you are dealing with a normal distribution, you would need to: z = (x – μ) / σ. z = (190 – 150) / 25. z = 1.6. As you already know, the z score lets you know how many standard deviations from the mean your score is
### zscoretable(standard normal table) |simply psychology
May 30, 2019 · A z-score table shows the percentage of values (usually a decimal figure) to the left of a given z-score on a standard normal distribution. For example, imagine our Z-score value is 1.09. First, look at the left side column of the z-table to find the value corresponding to one decimal place of the z-score (e.g. whole number and the first digit after the decimal point)
### 12. standardnormal distribution| professor mccarthy
The z-table method was the only practical way to find normal distribution probabilities until computers were invented. A table which could easily be converted to our z-table was published in 1799 1 by Christian Kramp
### z score percentile distribution table
Related Statistical Tables Terms Used in Stats. Std normal distribution Z table. Z Score Positive Negative table. F Distribution for α = 0.025. F Distribution for α = 0.01. Chi Square Distribution table. Negative Z Scores table. Z Score percentile table. F Distribution for α = 0.10. Wilcoxon Rank Sum table. Xbar Rchart table…
### zscore calculator: standardnormal distribution
The calculator allows area look up with out the use of tables or charts. In addition it provide a graph of the curve with shaded and filled area. The z-score is the number of standard deviations from the mean. The standard normal distribution is a normal distribution with a standard deviation on 1 and a mean of 0 | 1,054 | 4,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-39 | latest | en | 0.842189 |
https://calcresult.com/temporal/hex-clock.html | 1,632,095,289,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00672.warc.gz | 212,330,997 | 5,449 | Show/Hide:Instructions
## Introduction:
This clock takes inspiration from several sources: mainly the 'Flip Clocks' that were the first truly digital clocks, though long before the days of Quartz chrystals and LED's; from stacked binary clocks which admit: "Yes, It's A Clock. No, Your Mom Can't Read It"; and from Simon Hey's Rotary Word Clock
## Caveats:
If you are reading this paragraph, then you probably don't have JavaScript installed/enabled on your computer.
At this point in time the tools on this site are entirely dependant on JavaScript. Given enough time and money it is hoped to develop a version in the future which will be able to operate independently. Since JavaScript runs on your own computer, rather than on this server, it has been possible to set up this site much more quickly than if a server-side language was used, and it is possible to serve far more pages with the available bandwidth.
If you think that you can assist with this project in any way, then please visit the Support section and leave a message.
## Instructions
This clock takes a moment to synchronise each column, somewhere between 1 and 6 seconds, depending on the particular time and date.
This clock is based a demonstration of what time would look like if it was designed today, in the era of binary computers. While binary is the basis for all digital communication, binary is rather unweildy for humans, or for large numbers. The time-system that we are all familiar with is based on two different number-bases - base 12 (base 24) for hours, and base 60 for minutes and seconds. Instead, this clock is based on base 16, commonly known as Hexadecimal (often abbreviated to 'Hex') for all four units of measurement: hexHours; heximes; hexMinutes &amsp; hexSeconds.
A hexHour is exactly one and a half 'normal' hours long, and similarly a hexSecond is approximately 1.3 'normal' seconds in length.
This clock uses the system time of the computer that it is displayed on, so if will drift if your device isn't properly synchronised.
1. Sunday
2. Monday
3. Tuesday
4. Wednesday
5. Thursday
6. Friday
7. Saturday
1. 1st
2. 2nd
3. 3rd
4. 4th
5. 5th
6. 6th
7. 7th
8. 8th
9. 9th
10. 10th
11. 11th
12. 12th
13. 13th
14. 14th
15. 15th
16. 16th
17. 17th
18. 18th
19. 19th
20. 20th
21. 21st
22. 22nd
23. 23rd
24. 24th
25. 25th
26. 26th
27. 27th
28. 28th
29. 29th
30. 30th
31. 31st
1. Jan
2. Feb
3. Mar
4. Apr
5. May
6. Jun
7. Jul
8. Aug
9. Sep
10. Oct
11. Nov
12. Dec
1. 2000
2. 2001
3. 2002
4. 2003
5. 2004
6. 2005
7. 2006
8. 2007
9. 2008
10. 2009
11. 2010
12. 2011
13. 2012
14. 2013
15. 2014
16. 2015
17. 2016
18. 2017
19. 2018
20. 2019
21. 2020
22. 2021
23. 2022
24. 2023
25. 2024
26. 2025
27. 2026
28. 2027
29. 2028
30. 2029
31. 2030
32. 2031
33. 2032
34. 2033
35. 2034
36. 2035
37. 2036
38. 2037
39. 2038
40. 2039
41. 2040
42. 2041
43. 2042
44. 2043
45. 2044
46. 2045
47. 2046
48. 2047
49. 2048
50. 2049
51. 2050
52. 2051
53. 2052
54. 2053
55. 2054
56. 2055
57. 2056
58. 2057
59. 2058
60. 2059
1. 0
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12. B
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14. D
15. E
16. F
1. 0
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1. 0
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16. F
1. 0
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12. B
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14. D
15. E
16. F | 1,360 | 3,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.954838 |
http://www.primidi.com/configuration_geometry | 1,566,520,744,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317688.48/warc/CC-MAIN-20190822235908-20190823021908-00487.warc.gz | 295,664,062 | 3,414 | # Configuration (geometry)
Configuration (geometry)
In mathematics, specifically projective geometry, a configuration in the plane consists of a finite set of points, and a finite arrangement of lines, such that each point is incident to the same number of lines and each line is incident to the same number of points.
Although certain specific configurations had been studied earlier (for instance by Thomas Kirkman in 1849), the formal study of configurations was first introduced by Theodor Reye in 1876, in the second edition of his book Geometrie der Lage, in the context of a discussion of Desargues' theorem. Ernst Steinitz wrote his dissertation on the subject in 1894, and they were popularized by Hilbert and Cohn-Vossen's 1932 book Anschauliche Geometrie (reprinted in English as Geometry and the Imagination).
Configurations may be studied either as concrete sets of points and lines in a specific geometry, such as the Euclidean or projective planes (these are said to be realizable in that geometry), or as abstract incidence structures. In the latter case they are closely related to regular hypergraphs and biregular bipartite graphs, but with some additional restrictions: every two points of the incidence structure can be associated with at most one line, and every two lines can be associated with at most one point. That is, the girth of the corresponding bipartite graph (the Levi graph of the configuration) must be at least six.
Read more about Configuration (geometry): Notation, Examples, Duality of Configurations, configurations">The Number of (n3) Configurations, Constructions of Symmetric Configurations, Higher Dimensions
### Other articles related to "configuration, configurations":
Configuration (geometry) - Higher Dimensions
... The concept of a configurationmay be generalized to higher dimensions, for instance to points and lines or planes in space ... Notable three-dimensional configurationsare the Möbius configuration consisting of two mutually inscribed tetrahedra, Reye's configuration consisting of twelve points and twelve planes ... a given configurationmay be represented by the matrix The principle extends generally to n dimensions, where 0 ≤ j < n ... | 451 | 2,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-35 | latest | en | 0.941437 |
https://betterlesson.com/lesson/resource/1743813/57360/activating-students-as-resources-for-one-another-mov | 1,526,942,933,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00032.warc.gz | 501,280,482 | 18,873 | ## Activating students as resources for one another.mov - Section 2: Beginning the Activity + Homework
Activating students as resources for one another.mov
# Laws of Exponents - Power Raised to a Power
Lesson 7 of 19
## Big Idea: Memorization is short term knowledge but understanding endures time. Create the rules for exponents by expanding the bases to understand.
Print Lesson
Standards:
51 minutes
### Christa Lemily
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Environment: Urban | 309 | 1,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-22 | latest | en | 0.839242 |
http://www.cs.columbia.edu/~verma/classes/fa17/coms3261/lec/L22-Introduction_to_PAC_Learning.html | 1,512,951,318,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948511435.4/warc/CC-MAIN-20171210235516-20171211015516-00420.warc.gz | 332,909,258 | 3,447 | COMS W3261 CS Theory Lecture 22: Introduction to PAC Learning
The promise of modern Artificial Intelligence (AI) is that computers can learn novel concepts. How do we formalize the process of learning? Doing so will help us answer fundamental questions like
• What does it mean to learn a concept?
• What concepts can computers learn?
• What concepts can computers learn efficiently?
• What concepts can computers never learn?
1. Towards Formalizing the Process of Learning
Dictionary defines learning as "the acquisition of knowledge or skills through experience or study". Thus to learn a concept a learning agent should be able to
• Observe a phenomenon from nature (for experiencing or studying it)
• Construct a model from observations (for knowledge acquisition)
• Use the constructed model for making decision/predictions about similar future observations (for demonstrating that knowledge acquisition, ie learning, was successful)
How can we make this precise? We can use the Probably Approximately Correct (PAC) Learning model proposed by Valiant in 1984.
• S -- a sample space (eg. the set of all possible observations of interest)
• D -- a sample distribution (a probability distribution over points in the sample space)
• c: S → {0,1} -- a concept, which accepts or rejects each point in the sample space
• C -- a concept class, or a collection of concepts
There's some "true concept" c ∈ C that the learning agent trying to find (learn). Given m sample observations x1,...,xm which are drawn independently from D, together with their true classifications c(x1),...,c(xm), the agent's goal is to find a hypothesis h ∈ C such that
Pr[h(x) = c(x)] ≥ 1 − ε
for a future observations x. Furthermore, the learning agent wants to succeed at this goal with probability at least 1 − δ over the choice of xi's. In other words, with high probability we want to output a hypothesis that's approximately correct (hence "Probably Approximately Correct").
2. PAC Learning Finite Concept Classes
First question we can ask concerns sample complexity: how many samples do we need to have seen to learn a concept effectively? It's not hard to show that
m = O( 1ε · log |C|δ )
samples drawn independently from D are sufficient. Any hypothesis h ∈ C we can find that agrees with all of these m samples (i.e., such that h(xi) = c(xi) for all i) will satisfy Pr[h(x) = c(x)] ≥ 1 − ε with probability at least 1 − δ over the choice of m samples.
Proof: (by contrapositive) Let h ∈ C be any "bad" hypothesis: that is, such that Pr[h(x) = c(x)] < 1 − ε. Then if we independently pick m points from the sample distribution D, the hypothesis h will be correct on all of these points with probability at most (1 − ε)m. So the probability that there exists a bad hypothesis in C that nevertheless agrees with all our sample data is at most |C| (1 − ε)m (the total number of hypotheses, good or bad, times the maximum probability of each bad hypothesis agreeing with the sample data).
Note that the above situation is a bad event (we picked a bad hypothesis because it agreed with all m samples). We want to limit the chance of this happening by, say, no more than δ. So we want |C| (1 − ε)m ≤ δ, or equivalently
m ≥ O( 1ε · log |C|δ ).
Note that the true concept c is in C. So the learning agent can use the following algorithm to learn the concept:
1. Find any hypothesis in h ∈ C that agrees with all the sample data (i.e., such that h(xi) = c(xi) for all x1,...,xm).
2. Output h.
Such an h will always exist, and by the theorem above it will probably be a good hypothesis. All we need is to see enough sample observations!
3. PAC Learning Infinite Concept Classes
What happens in the case when |C| is infinite? The sample complexity bound derived in previous section becomes useless (it states that we should observe proportional to log |C| observations to find a good hypothesis).
We need to capture the true complexity of a concept class (not just the number of elements in it). The key idea that fully characterizes the complexity of a concept class is called VC-dimension (after two of its inventors, Vapnik and Chervonenkis). We say the set of points x1,...,xm is shattered by a concept class C if for all 2m possible settings of c(x1),...,c(xm) to 0 or 1 (reject or accept), there is some concept c ∈ C that agrees with those values. Then the VC-dimension of C, denoted Dim(C), is the size of the largest set of points shattered by C. (If we can find arbitrarily large finite-size sets of points that can be shattered, then Dim(C) = ∞).
One can show that a concept class C is PAC-learnable if and only if its VC-dimension is finite, and that the number of samples
m = O( Dim(C)ε · log 1εδ )
are sufficent. The learning algorithm again can simply output any hypothesis h ∈ C that is consistent with the m observations.
4. Computational Complexity of Learning
So far we have only discussed the sample complexity of learning (ie, how many samples suffices to learn a concept). But how hard is it as a computational problem to find a hypothesis that is consistent with the m observations? The learning problem is certainly in NP, since given a hypothesis it is easy to check whether it is consistent with the given m samples.
If the learner's goal is to output a hypothesis in some fixed format (such as DNF expressions), it is possible to show that in some cases finding a hypothesis that is consistent with the observations is NP-complete. But if the learner's output hypothesis can in any general format (ie, output a hypothesis that is some polynomial-time algorithm that predicts on data), we don't know whether finding such a hypothesis is NP-complete.
References: Scott Aaronson's lecture notes on Introduction to PAC learning.
aho@cs.columbia.edu
verma@cs.columbia.edu | 1,318 | 5,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-51 | latest | en | 0.910769 |
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The equation of the line that has a slope of -1/3 and a y-intercept of 5/2 is y = -1/3x + 5/2.
Question
Updated 3/12/2015 12:35:18 AM
Rating
3
The equation of the line that has a slope of -1/3 and a y-intercept of 5/2 is y = -1/3x + 5/2.
Confirmed by Andrew. [3/12/2015 12:44:20 AM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 633 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-09 | latest | en | 0.878603 |
https://barisalcity.org/milwaukee-to-grand-rapids-drive-time/ | 1,657,169,317,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00777.warc.gz | 161,194,772 | 5,028 | Flight distance from grand Rapids come Milwaukee (Gerald R. Ford worldwide Airport – Milwaukee Mitchell international Airport) is 121 mile / 194 kilometers / 105 nautical miles. Approximated flight time is 43 minutes.
You are watching: Milwaukee to grand rapids drive time
Driving street from grand Rapids (GRR) come Milwaukee (MKE) is 143 mile / 230 kilometers and also travel time by vehicle is about 3 hours 51 minutes.
How much is Milwaukee from grand Rapids?
There space several methods to calculation distances between Los Angeles and Chicago. Here are two usual methods:
Vincenty"s formula (applied above)120.525 miles193.966 kilometers104.733 nautical miles
Vincenty"s formula calculates the distance in between latitude/longitude points on the earth surface, utilizing an ellipsoidal version of the earth.
Haversine formula120.204 miles193.449 kilometers104.454 nautical miles
The haversine formula calculates the distance in between latitude/longitude point out assuming a spherical earth (great-circle distance – the shortest distance between two points).
Time difference and also current neighborhood times
The time difference between Grand Rapids and also Milwaukee is 1 hour. Milwaukee is 1 hour behind cool Rapids.
Grand Rapids time come Milwaukee time converter
Time in cool Rapids: 00:00⇆Time in Milwaukee: 00:00
Carbon dioxide emissions
On typical flying from cool Rapids come Milwaukee generates around 43 kg of CO2 every passenger, 43 kilograms is equal to 94 pounds (lbs). The numbers are estimates and include just the CO2 produced by burn jet fuel.
See more: My Gramma And Your Grandma Sitting By The Fire Lyrics, Lyrics For Iko Iko By The Dixie Cups
Map of trip path and also driving directions from grand Rapids to Milwaukee
Shortest flight path between Gerald R. Ford global Airport (GRR) and also Milwaukee Mitchell international Airport (MKE).
Airport information
OriginGerald R. Ford worldwide Airport
City:Grand Rapids, MI
Country:United States
IATA Code:GRR
ICAO Code:KGRR
Coordinates:42°52′50″N, 85°31′22″W
DestinationMilwaukee Mitchell international Airport
City:Milwaukee, WI
Country:United States
IATA Code:MKE
ICAO Code:KMKE
Coordinates:42°56′49″N, 87°53′47″W
Related distances from cool Rapids (GRR)
Grand Rapids to Chicago street (GRR to ORD) Grand Rapids to Madison distance (GRR to MSN) Grand Rapids to Rockford street (GRR to RFD)
Popular ranges from grand Rapids (GRR)
Grand Rapids come Chicago street (GRR to ORD) Grand Rapids come Detroit distance (GRR to DTW) Grand Rapids come Denver distance (GRR to DEN) Grand Rapids come Dallas distance (GRR to DFW) Grand Rapids come Atlanta street (GRR to ATL) Grand Rapids come Orlando street (GRR come MCO) Grand Rapids to Minneapolis distance (GRR come MSP) Grand Rapids come Charlotte street (GRR come CLT) Grand Rapids to Philadelphia distance (GRR to PHL) Grand Rapids to las Vegas distance (GRR to LAS) Grand Rapids to Nashville street (GRR come BNA) Grand Rapids to Tampa distance (GRR to TPA) Grand Rapids come Traverse City distance (GRR come TVC) Grand Rapids to ft Lauderdale street (GRR come FLL) Grand Rapids come Miami distance (GRR come MIA) Grand Rapids come Milwaukee street (GRR to MKE) Grand Rapids to Honolulu street (GRR to HNL) Grand Rapids come Reykjavik street (GRR come KEF) Grand Rapids to Amsterdam distance (GRR come AMS) Grand Rapids to Los Angeles distance (GRR to LAX) | 804 | 3,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-27 | longest | en | 0.874969 |
guile-user.gnu.narkive.com | 1,720,801,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00570.warc.gz | 228,220,359 | 11,478 | Discussion:
I wrote a 3+1 guile program, feedback welcome
Joshua Branson
2018-11-07 15:14:06 UTC
Hello,
I spent a week or two working on the 3+1 problem. The 3+1 problem is a
fun number theory conjecture. It states that there exists a procedure
for turning all positive whole integers into 1. The procedure is:
If the number is even, then divide it by two.
If the number is odd, multiply by 3 and add 1.
repeat.
I decided to write a program to graph the number of iterations it takes
for the set of numbers 2-n to reach 1. It uses ice-9 getopt-long, so
the usual --help and --version work. It uses gnuplot, because I
couldn't get guile-charting to work. If anyone has enough free time to
comment on the state of this program, or any tips about how to make it
better, etc. that would be most helpful.
Thanks,
Joshua
Jérémy Korwin-Zmijowski
2018-11-08 09:37:37 UTC
Hey ! Sounds great I will look at it !
Cheers
Post by Joshua Branson
Hello,
I spent a week or two working on the 3+1 problem. The 3+1 problem is a
fun number theory conjecture. It states that there exists a procedure
If the number is even, then divide it by two.
If the number is odd, multiply by 3 and add 1.
repeat.
I decided to write a program to graph the number of iterations it takes
for the set of numbers 2-n to reach 1. It uses ice-9 getopt-long, so
the usual --help and --version work. It uses gnuplot, because I
couldn't get guile-charting to work. If anyone has enough free time to
comment on the state of this program, or any tips about how to make it
better, etc. that would be most helpful.
Thanks,
Joshua
Jérémy | 445 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.89467 |
http://stackoverflow.com/questions/9497524/displaying-3-histograms-on-1-axis-in-a-legible-way-matplotlib | 1,394,862,382,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678695896/warc/CC-MAIN-20140313024455-00096-ip-10-183-142-35.ec2.internal.warc.gz | 148,301,847 | 15,389 | # Displaying 3 histograms on 1 axis in a legible way - matplotlib
I have produced 3 sets of data which are organised in numpy arrays. I'm interested in plotting the probability distribution of these three sets of data as normed histograms. All three distributions should look almost identical so it seems sensible to plot all three on the same axis for ease of comparison.
By default matplotlib histograms are plotted as bars which makes the image I want look very messy. Hence, my question is whether it is possible to force pyplot.hist to only draw a box/circle/triangle where the top of the bar would be in the default form so I can cleanly display all three distributions on the same graph or whether I have to calculate the histogram data and then plot it separately as a scatter graph.
-
Since the data might cross multiple times, it might makes sense to have an `alpha` parameter (transparency) for whatever method you end up using. – Hooked Feb 29 '12 at 16:49
There are two ways to plot three histograms simultaniously, but both are not what you've asked for. To do what you ask, you must calculate the histogram, e.g. by using `numpy.histogram`, then plot using the `plot` method. Use `scatter` only if you want to associate other information with your points by setting a size for each point.
The first alternative approach to using `hist` involves passing all three data sets at once to the `hist` method. The `hist` method then adjusts the widths and placements of each bar so that all three sets are clearly presented.
The second alternative is to use the `histtype='step'` option, which makes clear plots for each set.
Here is a script demonstrating this:
``````import numpy as np
import matplotlib.pyplot as plt
np.random.seed(101)
a = np.random.normal(size=1000)
b = np.random.normal(size=1000)
c = np.random.normal(size=1000)
common_params = dict(bins=20,
range=(-5, 5),
normed=True)
plt.subplot(311)
plt.title('Default')
plt.hist(a, **common_params)
plt.hist(b, **common_params)
plt.hist(c, **common_params)
plt.subplot(312)
plt.title('Skinny shift - 3 at a time')
plt.hist((a, b, c), **common_params)
plt.subplot(313)
common_params['histtype'] = 'step'
plt.title('With steps')
plt.hist(a, **common_params)
plt.hist(b, **common_params)
plt.hist(c, **common_params)
plt.savefig('3hist.png')
plt.show()
``````
And here is the resulting plot:
Keep in mind you could do all this with the object oriented interface as well, e.g. make individual subplots, etc.
-
Thanks a lot. I ended up discovering the numpy.histogram as you suggested and it did the job perfectly! Very much appreciate the help. – Zykx Mar 1 '12 at 0:35 | 630 | 2,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2014-10 | latest | en | 0.841834 |
https://msdn.microsoft.com/en-us/library/office/gg264147(v=office.14) | 1,490,701,229,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189686.56/warc/CC-MAIN-20170322212949-00324-ip-10-233-31-227.ec2.internal.warc.gz | 845,686,877 | 12,480 | This documentation is archived and is not being maintained.
Int, Fix Functions
Office 2010
This page is specific to the Visual Basic for Applications (VBA) Language Reference for Office 2010.
Returns the integer portion of a number.
Syntax
Int(number)
Fix(number)
The required number argument is a Double or any valid numeric expression. If number contains Null, Null is returned.
Remarks
Both Int and Fix remove the fractional part of number and return the resulting integer value.
The difference between Int and Fix is that if number is negative, Int returns the first negative integer less than or equal to number, whereas Fix returns the first negative integer greater than or equal to number. For example, Int converts -8.4 to -9, and Fix converts -8.4 to -8.
Fix(number) is equivalent to:
```Sgn(number) * Int(Abs(number))
```
Example
This example illustrates how the Int and Fix functions return integer portions of numbers. In the case of a negative number argument, the Int function returns the first negative integer less than or equal to the number; the Fix function returns the first negative integer greater than or equal to the number.
```Dim MyNumber
MyNumber = Int(99.8) ' Returns 99.
MyNumber = Fix(99.2) ' Returns 99.
MyNumber = Int(-99.8) ' Returns -100.
MyNumber = Fix(-99.8) ' Returns -99.
MyNumber = Int(-99.2) ' Returns -100.
MyNumber = Fix(-99.2) ' Returns -99.
```
Show: | 341 | 1,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-13 | latest | en | 0.705549 |
https://www.physicsforums.com/threads/simple-pendulum-and-acceleration.299354/ | 1,590,817,968,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00431.warc.gz | 857,573,389 | 17,490 | # Simple Pendulum and acceleration
A simple pendulum consists of a mass m tied to a string. It swings freely within the limits indicated in the figure below. On a swing from left to right, the string is cut suddenly just as the pendulum is vertical. Select the letters that show the path of m, the direction of the force right after the cut, and the direction of the acceleration right after the cut.
http://http://nicodemus.letu.edu/res/msu/kashy/physicsLib02/09_Force_and_Motion/CutPendulumCpt.problem [Broken]
Last edited by a moderator:
Related Introductory Physics Homework Help News on Phys.org
tiny-tim
Homework Helper
Welcome to PF!
Hi leslie1016! Welcome to PF!
Tell us what you were going to write, and why you're not sure, and then we'll know how to help!
Here is the problem and I have attached the picture:
A simple pendulum consists of a mass m tied to a string. It swings freely within the limits indicated in the figure below. On a swing from left to right, the string is cut suddenly just as the pendulum is vertical. Select the letters that show the path of m, the direction of the force right after the cut, and the direction of the acceleration right after the cut.
#### Attachments
• 2.5 KB Views: 766
LowlyPion
Homework Helper
Here is the problem and I have attached the picture:
A simple pendulum consists of a mass m tied to a string. It swings freely within the limits indicated in the figure below. On a swing from left to right, the string is cut suddenly just as the pendulum is vertical. Select the letters that show the path of m, the direction of the force right after the cut, and the direction of the acceleration right after the cut.
Well... what are your thoughts on what happens?
If it's right at the bottom, then what is the initial direction of motion?
What force is acting on that motion? Gravity of course, but what happens to m after that?
Well, I think that the force is moving in the direction of T and then the acceleration is moving in the direction of X?
LowlyPion
Homework Helper
Draw a diagram of the bob at the moment before it is cut.
What forces are acting on it?
Then imagine that one of the forces (tension) is removed. Which forces are still acting on it?
If tension is removed then it's only gravity acting on it, correct?
LowlyPion
Homework Helper
If tension is removed then it's only gravity acting on it, correct?
Yes, so that can be your only force after T=0 then.
The force right after the cut is U. The path of M is O and the acceleration is X?
LowlyPion
Homework Helper
The force right after the cut is U. The path of M is O and the acceleration is X?
Well if the force is U, what is your theory for the direction of the acceleration being X?
well, gravity is pulling down on it and the acceleration is from when it was swinging, so it's swining back to the right.
LowlyPion
Homework Helper
well, gravity is pulling down on it and the acceleration is from when it was swinging, so it's swining back to the right.
I'm just wondering when F stopped being m*a. | 718 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-24 | latest | en | 0.931927 |
https://www.lonegunman.co.uk/2010/11/16/statistical-significance-explained/ | 1,653,594,497,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00110.warc.gz | 1,003,892,686 | 8,454 | # Statistical Significance Explained
If you didn’t read the House of Commons Library’s statistical literacy guides recently (or you need a refresher on what, exactly, statistical significance means), then you can do much worse than student Warren Davies’ short rundown on the meaning of statistical significance:
In science we’re always testing hypotheses. We never conduct a study to ‘see what happens’, because there’s always at least one way to make any useless set of data look important. We take a risk; we put our idea on the line and expose it to potential refutation. Therefore, all statistical tests in psychology test the probability of obtaining your given set of results (and all those that are even more extreme) if the hypothesis were incorrect – i.e. the null hypothesis were true. […]
This is what statistical significance testing tells you – the probability that the result (and all those that are even more extreme) would have come about if the null hypothesis were true. […] It’s given as a value between 0 and 1, and labelled p. So p = .01 means a 1% chance of getting the results if the null hypothesis were true; p = .5 means 50% chance, p = .99 means 99%, and so on.
In psychology we usually look for p values lower than .05, or 5%. That’s what you should look out for when reading journal papers. If there’s less than a 5% chance of getting the result if the null hypothesis were true, a psychologist will be happy with that, and the result is more likely to get published.
Significance testing is not perfect, though. Remember this: ‘Statistical significance is not psychological significance.’ You must look at other things too; the effect size, the power, the theoretical underpinnings. Combined, they tell a story about how important the results are, and with time you’ll get better and better at interpreting this story.
To get a real feel for this, Davies provides a simple-to-follow example (a loaded die) in the post.
via @sandygautam | 434 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-21 | latest | en | 0.936068 |
http://www.jiskha.com/display.cgi?id=1334884806 | 1,495,703,322,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608058.13/warc/CC-MAIN-20170525083009-20170525103009-00562.warc.gz | 546,838,734 | 4,095 | # Per-calc Helppppp
posted by on .
keep in mind answers should be pi radians or if necessary 3 dec.
csc^2-2=0
all angles
4cos^2x-4cosx+1=0
all angles and name 8 angles
7sin^2x-22sinx+3=0
all angles and name 8 angles
2cos^2x-7cosx=-3
all angles
sinx(2sinx+1)=0
all angles
• Per-calc Helppppp - ,
1. csc^2 x = 2
sin^2 x = 1/2
sin x = ±1/√2
x = π/4, 3π/4 , 5π/4 , 7π/4
2. 4cos^2x - 4cosx + 1 = 0
(2cos x - 1)^2 = 0
(2cosx - 1)= 0
cosx = 1/2
x = π/3 or x = 5π/3
3.
7sin^2 x - 22sinx + 3 = 0
(7sinx - 1)(sinx - 3) = 0
sinx = 1/7 or sinx = 3 , which is not possible
for sinx = 1/7
make sure calculator is set to radians
x = .1433 (in I) or x = π - .1433 = 2.998 ( in II)
do the rest the same way
• Per-calc Helppppp - ,
yup
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Post a New Question | 388 | 864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-22 | latest | en | 0.6844 |
https://maa.org/book/export/html/3184732 | 1,709,485,193,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00485.warc.gz | 371,823,949 | 3,034 | # Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 4
Author(s):
If she had understood the meaning of these theorems, she could have applied theorem (2) to $$u=x^2 (a+x)^3 (b-x)^4$$, where $$v_1=x^2$$, $$v_2= (a+x)^3$$, and $$v_3= (b-x)^4$$. This would have given her
$\begin{array}{ccl}du&=&x^2 (a+x)^3 (b-x)^4 \left \{\frac{2x}{x^2} + 3\frac{(a+x)^2}{(a+x)^3} - \frac{4(b-x)^3}{ (b-x)^4} \right \}dx \\ &=&x^2 (a+x)^3 (b-x)^4 \left \{\frac{2}{x}+\frac{3}{a+x}-\frac{4}{b-x}\right \}dx\\ &=&x^2 (a+x)^3 (b-x)^4 \left \{\frac{(2(a+x)(b-x)+3x(b-x)-4x(a+x)}{ x(a+x)(b-x)}\right \}dx\\ &=&x^2 (a+x)^3 (b-x)^4 \left \{\frac{2ab-6ax+5bx- 9x^2}{x(a+x)(b-x)}\right \}dx\\ &=&x (a+x)^2 (b-x)^3 (2ab-(6a-5b)x-9x^2 )dx.\end{array}$
Lovelace’s mistake was that she did not apply the chain rule to $$(b-x)^4$$ correctly, and obtained $$4 (b-x)^3$$instead of $$-4 (b-x)^3$$. As De Morgan explained to her: ‘It is very common to suppose that if $$\phi(x)$$ differentiated gives $$\psi(x)$$, then $$\phi(-x)$$ gives $$\psi(-x)$$, but this should be $$\psi(-x)\times \mbox{diff. co.}(-x)$$ or $$\psi(-x)\times -1$$’ [LB 170, 14 Nov. 1840, f. 20r]. Sure enough, once Lovelace rectified her mistake she obtained the correct answer. | 534 | 1,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-10 | latest | en | 0.705201 |
https://www.exactlywhatistime.com/days-from-date/january-7/24-days | 1,718,254,960,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00375.warc.gz | 714,731,406 | 6,294 | # Friday January 31, 2025
Adding 24 days from Tuesday January 07, 2025 is Friday January 31, 2025 which is day number 031 of 2024. This page is designed to help you the steps to count 24, but understand how to convert and add time correctly.
• Specific Date: Tuesday January 07, 2025
• Days from Tuesday January 07, 2025: Friday January 31, 2025
• Day of the year: 031
• Day of the week: Friday
• Month: January
• Year: 2024
## Calculating 24 days from Tuesday January 07, 2025 by hand
Attempting to add 24 days from Tuesday January 07, 2025 by hand can be quite difficult and time-consuming. A more convenient method is to use a calendar, whether it's a physical one or a digital application, to count the days from the given date. However, our days from specific date calculatoris the easiest and most efficient way to solve this problem.
If you want to modify the question on this page, you have two options: you can either change the URL in your browser's address bar or go to our days from specific date calculator to enter a new query. Keep in mind that doing these types of calculations in your head can be quite challenging, so our calculator was developed to assist you in this task and make it much simpler.
## Friday January 31, 2025 Stats
• Day of the week: Friday
• Month: January
• Day of the year: 031
## Counting 24 days forward from Tuesday January 07, 2025
Counting forward from today, Friday January 31, 2025 is 24 from now using our current calendar. 24 days is equivalent to:
24 days is also 576 hours. Friday January 31, 2025 is 8% of the year completed.
## Within 24 days there are 576 hours, 34560 minutes, or 2073600 seconds
Friday Friday January 31, 2025 is the 031 day of the year. At that time, we will be 8% through 2025.
## In 24 days, the Average Person Spent...
• 5155.2 hours Sleeping
• 685.44 hours Eating and drinking
• 1123.2 hours Household activities
• 334.08 hours Housework
• 368.64 hours Food preparation and cleanup
• 115.2 hours Lawn and garden care
• 2016.0 hours Working and work-related activities
• 1854.72 hours Working
• 3035.52 hours Leisure and sports
• 1647.36 hours Watching television
## Famous Sporting and Music Events on January 31
• 1960 Songwriter Adolph Green marries actress and singer Phyllis Newman in NYC
• 2015 17 year old Lydia Ko of New Zealand becomes the youngest golfer in men's or women's golf history to be ranked No. 1 in the world | 646 | 2,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.938074 |
https://www.eduzip.com/ask/question/fill-in-the-blanks-to-make-the-following-statements-then-the-oppo-520996 | 1,642,995,062,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00653.warc.gz | 737,443,814 | 8,768 | Mathematics
# Fill in the blanks to make the following statements then the opposite interior angles.Sum of the angles of a triangle is ______
##### SOLUTION
Sum of the angles of a triangle is $180^{o}$
You're just one step away
Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
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1 Verified Answer | Published on 09th 09, 2020 | 490 | 1,551 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-05 | latest | en | 0.745558 |
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# Creating forecast data based on historical data
Hi
I am new to Qlikview and have the following requirement and need some guidance on how to approach it
The user wants to take historical data (Orders for last 6 months), get the total by day of week (Mon, Tues etc) and then get a daily % of the Weekly Total
example:
Historical dataSunMonTueWedThurFriSatWeekly total
Data by resource for last 6 months600500400200507803002830
% of weekly total21%18%14%7%2%28%11%100%
The next step in the process will then be to take a weekly forecast number (Note that this is not available by day) and apply the % calculated above to this number to get a weekly forecast number
Eg Forcast for week 2800
ForecastSunMonTueWedThurFriSat
Forecast for week (2800)59449539619849772297
This will then be used to calculate a daily variance.
Thanks in advance for help on this
Soreen
• ###### Re: Creating forecast data based on historical data
Allocation:
DayOfWeek
,sum(Something) as DaySomething
RESIDENT SomeTable
WHERE conditions that narrow you down to the 6 months you want
GROUP BY DayOfWeek
;
LEFT JOIN (Allocation)
RESIDENT Allocation
;
LEFT JOIN (Allocation)
DayOfWeek
,DaySomething / TotalSometing as DayAllocationPercent
RESIDENT Allocation
;
Now I'm not sure what the forecast data looks like, but it's possible you'll be able to do something like this:
LEFT JOIN (Forecast)
DayOfWeek
,DayAllocationPercent
RESIDENT Allocation
;
LEFT JOIN (Forecast)
ForecastKeyField
,DayOfWeek
,WeeklyForecastAmount * DayAllocationPercent as DailyForecastAmount
RESIDENT Forecast
;
DROP TABLE Allocation
;
• ###### Re: Creating forecast data based on historical data
Hi John
Thanks for your quick feedback. Will let you know if this solved the problem.
Soreen | 453 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-43 | latest | en | 0.83447 |
http://gfs.sourceforge.net/wiki/index.php?title=RiverVortexTest&diff=3799&oldid=3798 | 1,454,897,454,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701152097.59/warc/CC-MAIN-20160205193912-00299-ip-10-236-182-209.ec2.internal.warc.gz | 92,186,623 | 8,839 | # RiverVortexTest
(Difference between revisions)
Revision as of 02:46, 18 May 2011Gjrickard (Talk | contribs) (→River Tidal Turbine 2D Problem)← Previous diff Revision as of 02:58, 18 May 2011Gjrickard (Talk | contribs) (→River Tidal Turbine 2D Problem)Next diff → Line 217: Line 217: - Top frame compares space-time solution from Gerris2D (the red dashes) and from the - analytic Laplace transform solution. - Green crosses indicate x-positions of sampled output. + == River Tidal Turbine 2D Sample Output == - Lower frame plots changes in concentration as a function of time at the indicated + Solutions at 3500 seconds into each run using the default and - x-positions. The dashed lines are from Gerris2D on top of the Laplace Transform + Sweby advection schemes. - solutions. + {| align="center" {| align="center" - |+ Comparison of 1D Heat Flux solutions. + |+ Comparison of River 2D Turbine solutions. |- |- | [[Image:Default_3500_0.png|thumb|500px|center|Gerris]] | [[Image:Default_3500_0.png|thumb|500px|center|Gerris]] + | [[Image:Sweby_3500_0.png|thumb|500px|center|Gerris]] + | [[Image:Default_3500_1.png|thumb|500px|center|Gerris]] + | [[Image:Sweby_3500_1.png|thumb|500px|center|Gerris]] + | [[Image:Default_3500_2.png|thumb|500px|center|Gerris]] + | [[Image:Sweby_3500_2.png|thumb|500px|center|Gerris]] |} |}
## Revision as of 02:58, 18 May 2011
Testing the River version of Gerris. In particular interested in the possibility of tidal turbine like simulations in support of Tim Divett's thesis.
The basic issue is that we note in both the Linear Free Surface and Nonlinear Free Surface versions of Gerris that vortices generated from wakes caused by localised regions of high bottom friction (mimicing the presence of a turbine, say) can sometimes become unstable, i.e., the vortex continues to spin up as it moves downstream leading to (presumably) unphysical magnitudes of flow and fluid height. In the linear free surface version, such vortices persist and can be seen moving with the background flow. In the nonlinear free surface version, these vortices become tighter and more intense, and eventually grind the simulation to a halt via excessive magnitudes of flow and amplitude.
This issue is also relevant to a whole series of linear free surface tidal simulations around headlands and bays around New Zealand. In these simulations an increase in spatial resolution reveals the generation of many eddies that appear to be shed in a physical way via interactions with the coastline. As above, these eddies persist in the background tidal flow. They do not appear to be unstable (ie the runs keep going), but the question remains about the fidelity and actuality of these eddies (and to test this we need more field observations, and perhaps some more thought about the model).
Both models have been tested against the cases present by Signell and Geyer (1991) and both seem to do a good job of reproducing many of the features seen. Thus the underlying schemes seem to be correct, at least for the scales of interest in these tests.
Perhaps it's not surprising that these features arise. After all the models in question here are solving shallow water variants of reality, so that the likely possibility of 3D processes in the water column influencing the vortex behaviour is absent. For the shallow water the only explicit damping is via a bottom friction term, and the only implicit damping is through truncation terms in the advection scheme; these may be missing important physics that allow for a more complete representation of the vortex behaviour.
So, here show some examples of the unstable vortex behaviour in the River code. This can perhaps lead to a better understanding of the limitations of the shallow water models, and how far we should be pushing them to look at the (presumably) complex tidal interactions with coastline and in-situ structures.
## River Tidal Turbine 2D Problem
A typical 2D run set up with constant bathymetry of 50 metres depth, with an increased region of bottom friction defined by TT(x,y) in the centre of the domain. M2U(t) drives the tidal flow at the boundary, anticipating a maximum input speed of 1 m/s.
The change of relevance here is the choice of advection scheme. The default (Minmod) produces a stable simulation, with a smooth wake behind the turbine. Change that to Sweby, and the wake becomes unstable, leading to vortices that then continue to deepen and amplify.
The question is which is "correct"?
Run out to around t=5000, taking about the same number of cpu seconds on 6 processor elements.
# run me using: gerris2D -m
Define RXMID1 (1.0)
Define RYMID1 (0.0)
Define rmag1(rx,ry) ( sqrt((rx-RXMID1)*(rx-RXMID1)+(ry-RYMID1)*(ry-RYMID1)) )
Define RBIT1V(rx,ry) (rmag1(rx,ry) > 0.0 && rmag1(rx,ry) < 0.15 ? 1 : 0.0)
Define RBIT2V(rx,ry) ( (fabs(ry) < 0.4 && rx > -0.4 && rx < 2.4) ? 1 : 0.0)
Define RBIT1S(rx,ry) ( exp(-rmag1(rx,ry)/0.01) )
Define RBIT0V(rx,ry) (rmag1(rx,ry) > 0.0 && rmag1(rx,ry) < 0.35 ? 1 : 0.0)
Define LEVMAX 8
Define LEVMID 8
Define LEVMIN 5
Define LEVBIG 9
Define LEVBIGGER 11
Define R2 (RBIT2V(rx,ry))
Define R3 (RBIT1V(rx,ry))
Define LEVEL1 ( R3 ? LEVMAX : LEVMIN )
Define LEVEL2 ( R2 ? LEVMAX+1 : LEVEL1)
# Turbine parameters, diameter is in y direction, width in x direction,
Define DIAMETER 20.
Define WIDTH 10.
Define Cdbig 12.
#Spacing distance between turbines
Define XSPACING DIAMETER*5.
Define YSPACING DIAMETER*1.5
# M2 tidal frequency. The period is 12h25 (44700 seconds) and M2 tidal elevation.
Define M2F (2.*M_PI/44700.)
Define M2(t) (0.0*cos (M2F*t)+0.01*sin (M2F*t))
# Variable U is velocity*depth here, hence set
# M2U to be velocity (a number) by depth (P)...
#
Define M2U(t) (1.0*(P)*sin (M2F*t))
# Define tau (-0.8e-4)
Define tau (0.0)
Define cd (3.e-3 + TT(x,y)*Cdbig)
Define theta (1.0)
Define onemth (1.0 - theta)
Define abfac (dt*cd*Velocity/P)
Define bffac (tau*dt)
Define onepa (1.+abfac)
Define bth (bffac*theta)
Define bonemt (bffac*onemth)
# Use the "GfsRiver" model
3 2 GfsRiver GfsBox GfsGEdge {} {
Global {
/* Dimensional initial potential temperature profile */
#define TT(x,y) (fabs(x) < WIDTH/2. && fabs(y) < DIAMETER/2. ? 1. : 0.0)
}
# Box size (metres)
PhysicalParams { L = 250.0 g = 9.81 }
# I want the centre of the 2nd box at x = 0 so translate by x = 1000
GfsMapTransform { tx = -250 ty = 0 tz = 0 }
# End the simulation
Time { end = 8640 dtmax = 10 }
# Use terrain module
GModule terrain
Refine LEVEL1
VariableTerrain Zb {
basename = bathy50Wide
path = /data01/marine-physics/rickard/FROMTIM/COSB:/data01/marine-physics/rickard/FROMTIM
} { reconstruct = 1 }
Init {} {
P = MAX (0., -Zb)
}
## remember the initial grid Level...
Init{} {MYMINL = Level
RSEE1V = RBIT1V(rx,ry)
RSEE2V = RBIT2V(rx,ry)
CTT = TT(x,y)
}
#AdvectionParams { cfl = 1.0 }
}
ProjectionParams { tolerance = 1e-6 }
ApproxProjectionParams { tolerance = 1e-6 }
cmax = 0.06
cfactor = 2
maxlevel = LEVEL2
minlevel = MYMINL
} (P < 1e-4 ? 0. : P + Zb)
# Bottom friction parameterisation semi-implict, plus
# plus Coriolis terms added with implicitness parameter theta...
#
Init { start=360 istep = 1 } {
U = ( P > 1e-6 ? (onepa*(U-bonemt*V)-bth*(V+bonemt*U))/(onepa*onepa + bth*bth) : 0.)
V = ( P > 1e-6 ? (onepa*(V+bonemt*U)+bth*(U-bonemt*V))/(onepa*onepa + bth*bth) : 0.)
}
# Try to construct a Vorticity by dividing fluxes by local
# depth and applying gradient operators...here in 2D of course...
#
Init { istep = 1 } {
RU = ( P < 1e-4 ? 0.0 : (U/P))
RV = ( P < 1e-4 ? 0.0 : (V/P))
VORT = (P < 1e-4 ? 0. : (dx("RV") - dy("RU")))
}
# Vorticity criterion
# Takes 1 m/s as normalising maximum velocity
AdaptFunction { start = 360 istep = 1 } {
cmax = 5e-2
cfactor= 2
maxlevel = LEVEL2
minlevel = MYMINL
} (fabs(VORT*dL/1.0))
EventBalance { istep = 10 } 0.1
OutputTime { istep = 100 } stdout
OutputSimulation { step = 100 } boi_wiki_3-%g.gfs
OutputSimulation { start = end } boi_wiki_3_end.gfs
}GfsBox {
left = Boundary {
BcDirichlet U M2U(t)
BcNeumann P 0
}
top = Boundary
bottom = Boundary
}
GfsBox { top = Boundary
bottom = Boundary
}
GfsBox {
right = Boundary {
BcDirichlet U M2U(t)
BcNeumann P 0
}
top = Boundary
bottom = Boundary
}
1 2 right
2 3 right
## River Tidal Turbine 2D Sample Output
Solutions at 3500 seconds into each run using the default and Sweby advection schemes.
Gerris Gerris Gerris Gerris Gerris Gerris | 2,600 | 8,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-07 | longest | en | 0.85452 |
http://mymathforum.com/real-analysis/16241-proof-recurrence-sequence-convergence.html | 1,566,241,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314904.26/warc/CC-MAIN-20190819180710-20190819202710-00247.warc.gz | 134,398,593 | 9,252 | My Math Forum Proof of recurrence sequence convergence
Real Analysis Real Analysis Math Forum
December 6th, 2010, 07:25 AM #1 Newbie Joined: Dec 2010 Posts: 7 Thanks: 0 Proof of recurrence sequence convergence Let $(x_n)_{n\in\mathbb{N}}$ be a recurrence sequecnce such that: (1) $x_0$ and $x_1$ are positive reals (2) For $n\geq 2$ we have $x_n=\frac{2}{x_{n-1}+x_{n-2}}$. Proof that $(x_n)_{n\in\mathbb{N}}$ is convergent.
December 6th, 2010, 08:51 AM #2 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Proof of recurrence sequence convergence What happens when the sum is > 1? What happens when the sum is < 1?
December 6th, 2010, 01:20 PM #3 Newbie Joined: Dec 2010 Posts: 7 Thanks: 0 Re: Proof of recurrence sequence convergence From such cases (if you are talking about denominator) follows estimation for the next element of our sequence and I dont see anything more. I tried to divide it into two subsequences such that 1st is composed of elements greater then one, and 2nd consists of numbers lesser then one. After that if 1st was nonincreasingand 2nd nondecreasing (actually it requiers their convergance only) with some easy calculus I can show that they have a common limit. My prove attempt distinguished eight cases (there cant be three elements in a row that belong into on of these subsequances) but some of them are troublesome or simply false (for example they may be monotonic only for enought big values).
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Topic: Interpretation of coefficients in multiple regressions which model
linear dependence on an IV
Replies: 146 Last Post: Dec 15, 2012 6:44 PM
Messages: [ Previous | Next ]
Halitsky Posts: 600 Registered: 2/3/09
Holy Cow! Look at your "average a1" slope regressed on Len Int
Posted: Nov 27, 2012 3:16 PM
If you have a moment, regress your "average a1" on LenInt:
LenInt "average a1"
1 -0.487393513
2 -0.178292604
3 -0.164174277
4 -0.088771216
5 -0.104084177
6 -0.043135845
7 0.02423492
8 -0.021704118
9 0.097815884
10 -0.03499833
11 -0.01778355
12 -0.013260234
These data are for a1_N_1_S (fold a1, nonrandom method, dicodon set 1,
dicodon subset S),
and underlying regression is c on (u', u'^2), where u' = u/(1+u).
I have to believe that you're picking up on something quite strongly
here, but of course,
will have to wait to see what happens when all folds, sets, subsets
are done.
Date Subject Author
11/21/12 Halitsky
11/21/12 Halitsky
11/22/12 Ray Koopman
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12/9/12 gimpeltf@hotmail.com
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12/4/12 Ray Koopman | 1,714 | 3,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-09 | longest | en | 0.645716 |
https://www.claymath.org/euclid_index/triangles-parallels-and-area/?chapter=24 | 1,718,496,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00315.warc.gz | 654,039,217 | 28,030 | Index ← Previous Next →
Translations
Ἐκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις [εὐθείαις], τρίγωνον συστήσασθαι: δεῖ δὲ τὰς δύο τῆς λοιπῆς μείζονας εἶναι πάντῃ μεταλαμβανομένας [διὰ τὸ καὶ παντὸς τριγώνου τὰς δύο πλευρὰς τῆς λοιπῆς μείζονας εἶναι πάντῃ μεταλαμβανομένας]. Ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ, ὧν αἱ δύο τῆς λοιπῆς μείζονες ἔστωσαν πάντῃ μεταλαμβανόμεναι, αἱ μὲν Α, Β τῆς Γ, αἱ δὲ Α, Γ τῆς Β, καὶ ἔτι αἱ Β, Γ τῆς Α: δεῖ δὴ ἐκ τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συστήσασθαι. Ἐκκείσθω τις εὐθεῖα ἡ ΔΕ πεπερασμένη μὲν κατὰ τὸ Δ ἄπειρος δὲ κατὰ τὸ Ε, καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΖ, τῇ δὲ Β ἴση ἡ ΖΗ, τῇ δὲ Γ ἴση ἡ ΗΘ: καὶ κέντρῳ μὲν τῷ Ζ, διαστήματι δὲ τῷ ΖΔ κύκλος γεγράφθω ὁ ΔΚΛ: πάλιν κέντρῳ μὲν τῷ Η, διαστήματι δὲ τῷ ΗΘ κύκλος γεγράφθω ὁ ΚΛΘ, καὶ ἐπεζεύχθωσαν αἱ ΚΖ, ΚΗ: λέγω, ὅτι ἐκ τριῶν εὐθειῶν τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συνέσταται τὸ ΚΖΗ. Ἐπεὶ γὰρ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου, ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ: ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση. καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου, ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ: ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση: καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση. ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση: αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΚΖ, ΖΗ, ΗΚ τρισὶ ταῖς Α, Β, Γ ἴσαι εἰσίν. Ἐκ τριῶν ἄρα εὐθειῶν τῶν ΚΖ, ΖΗ, ΗΚ, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις εὐθείαις ταῖς Α, Β, Γ, τρίγωνον συνέσταται τὸ ΚΖΗ: ὅπερ ἔδει ποιῆσαι.
Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [I. 20] Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C. Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3] With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined; I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C. For, since the point F is the centre of the circle DKL, FD is equal to FK. But FD is equal to A; therefore KF is also equal to A. Again, since the point G is the centre of the circle LKH, GH is equal to GK. But GH is equal to C; therefore KG is also equal to C. And FG is also equal to B; therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C. Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed. | 1,259 | 2,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.479329 |
https://u-next.com/blogs/cyber-security/what-is-cryptography/ | 1,722,658,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00361.warc.gz | 476,950,674 | 55,188 | # What is Cryptography: A Complete Guide In 6 Easy Points
Ajay Ohri
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## Introduction
All of us know that the tech- giant WhatsApp has ‘end-to-end encryption’. But do we know what the word encrypted means? Let us understand what it means. Encryption is one half of the process- cryptography. You must be wondering what’s the other half; it’s called decryption. Let us understand the history of cryptography. The word can be traced back to the greek root Kryptos which means ‘hidden’. To understand this in simple terms, cryptography is a method of protecting information and communications using codes. Let us dive deeper into understanding the basics of cryptography.Â
## 1. What is Cryptography
We all have grown up watching movies for entertainment. In movies where two detectives or investigators are meeting outside and exchanging information, they always use certain code words. Only when both of them confirm the code word do they proceed with the transaction. Now if you are wondering what is cryptography- it’s exactly that.
By definition, cryptography is a way of conducting a protected communication between two parties using mathematical concepts and various rule-based calculations which in computer science terms are called algorithms. These algorithms transform ordinary plain text, information, messages in a series of text that is hard to decipher using a sequence of code so that it can only be accessed by those for whom it is intended to read and process.
## 2. Cryptography Techniques
So how does this work? Let’s understand the different cryptography algorithms. As we have understood so far, cryptography involves two parts namely, encryption and decryption using a cipher aka the key. A cipher suite uses a unique algorithm for encryption- encoding while sending and uses another algorithm for message authentication at the receiver’s end. The entire process, embedded in protocols and software codes that are written to run operating systems and electronic devices, needs the generation of public and private keys for encryption and decryption. There are three types of cryptographic algorithms. Let us understand how cryptography works
## 3. Cryptography Types
1. Symmetric key cryptography:  In this type of cryptography, both the sender and receiver use a single key during the communication or transfer of information. The sender uses this key for encrypting plain text and sends the ciphered text to the receiver through the communication channel. The receiver uses the same key for decrypting the ciphered text on the other end. Upon deciphering he recovers the plain text.Â
2. Asymmetric key cryptography: this type of cryptography is also referred to as public-key cryptography. Here we use two related keys; public and private. The public key is used for encryption and the private key is used for decryption. Both keys are different. Even if the public is known by everyone, only the intended individual will be able to decipher the text as only he has the unique private key. Asymmetric cryptography algorithm is used in WhatsApp communication.Â
3. Hash function: In this type of cryptography, no key is used during the transfer of information. A fixed length of the hash value is used depending on the plain text which makes it impossible for contents of plain text to be recovered.  Hash functions are commonly used by many operating systems to encrypt passwords. To put it in simple terms, the major difference between symmetric and asymmetric cryptography is in the key used by the sender and the receiver. If both of them use the same key for transmission and receiver, then it is Symmetric key cryptography and if they use different keys then it is called asymmetric key cryptography.Â
## 4. Application of Cryptography
1. To maintain confidentiality in the storage of information: It allows the users to share confidential information securely without having to worry about an interception by hackers.Â
2. Reliability in transmission: traditionally, a conventional approach used to be taken to carry out a checksum of the information to be transferred and communicate this respective checksum in an encrypted format. Upon receiving both the checksum and the encrypted data the data is again checksummed and compared to the original data after completing the deciphering at the receiver’s end.Â
3. Authentication of identity: It is highly useful as confirmation of individual identity. Â Innovative systems make use of strong cryptographic methods along with physical methods of individuals resulting in a highly secure and reliable way of verification of identity.Â
## 5. Examples of CryptographyÂ
One of the most commonly used and seen examples of cryptography is WhatsApp that promises end-to-end encryption. It involves asymmetry methods of cryptography. I.e. only the intended individual has the private key. Upon installation of Whatsapp on the devices, public keys are registered with the server and the messages are communicated securely using unique private keys.Â
With the increasing usage of digital transactions using UPI, digital wallets, internet banking, end-to-end secure payments is a need of the hour. The respective banks and the payment gateways use cryptography for these transactions to avoid the risk of malware and hackers. cryptography technology protects sensitive and confidential information of users such as account details, CVV, etc. Generation of one time passwords- OTP helps in confirming the identity of the user.Â
## 6. Uses of Cryptography
Imagine this as a lock and key mechanism, wherein you need a unique key to unlock a lock- likewise, a unique sequence of code is mixed in the input and output can only be deciphered using this key. Cryptography is used for creating a digital signature, cryptographic key generation, banking transaction cards, computer passwords, e-commerce transactions, web browsing, confidential communications, etc.
The new age is all about technology and you must get on this bandwagon to be a part of innovative solutions. Jigsaw Academy has a variety of courses that will help you learn cryptography, machine learning, artificial intelligence, data science, and much moreÂ
## Conclusion
This article gave you an in-depth understanding of the concept of cryptography, its origin, types of cryptography, techniques, and algorithms that are used in cryptography. The importance of cryptography in today’s world is beyond description with data breaches, cyber-attacks getting more and more frequent.  With an increase in digitization and influence of the internet, cryptography becomes of utmost importance as it provides a secure way of transactions.Â
So, have you made up your mind to make a career in Cyber Security? Visit our Master Certificate in Cyber Security (Red Team) for further help. It is the first program in offensive technologies in India and allows learners to practice in a real-time simulated ecosystem, that will give you an edge in this competitive world.
Elliptic Curve Cryptography: An Overview
You Might Also Like | 1,441 | 7,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-33 | latest | en | 0.902897 |
https://it.hinative.com/dictionaries/cube | 1,638,669,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00095.warc.gz | 381,125,440 | 14,694 | # Domande riguardo a frasi esempio con, spiegazione d'utilizzo di "Cube"s
## Il significato di "Cube" In varie frasi ed espressioni.
Q: Che cosa significa The cube root of...?
A: https://s3.amazonaws.com/static.graphemica.com/glyphs/i500s/000/009/178/original/221B-500x500.png?1275323386
Q: Che cosa significa Still, eating a cube of congealed blood isn’t in my book of cures when I’m trying my darndest not to hurl!?
A: To "hurl" is slang for "vomit" or "throw up" in this context.
He's trying not to throw up/vomit, and eating a cube of blood is making it very difficult for him to not vomit.
#3
http://learnersdictionary.com/definition/hurl
Q: Che cosa significa cube?
A: a shape
## Parole simili a "Cube" e le sue differenze
Q: Qual è la differenza tra He's heading to that cube. e He's heading for that cube. ?
would mean that it is in a more drastic measure as in he is in danger or something could happen
Q: Qual è la differenza tra A cube has six surfaces. e Cube has six surfaces. ?
A: They mean the same thing.
But, “A cube has six surfaces” is grammatically correct.
“Cube has six surfaces.” is incomplete, and not a grammatically correct sentence.
Q: Qual è la differenza tra cube e cubic ?
A: cube= a noun
Q: Come si dice in Inglese (Stati Uniti)? The cube has 6 faces and 12 sides. Am I right?
What about edge? How can I use it in describing shapes?
Thank you .
A: faces and sides are the same here. so there’s 6 faces/sides.
and there’s 12 edges.
Q: Come si dice in Inglese (Stati Uniti)? cube
A: Check the question to view the answer
Q: Come si dice in Inglese (Stati Uniti)? cube advertising
A: Can you describe what cube advertising is in more detail?
## Altre domande riguardo "Cube"
Q: I usually make curry with curry cube. Could you tell me the good way to say カレーのルー in English? sembra naturale?
A: そうだね！rouxはちゃんと「一から」という感じがあるので、分からない人があるかも、大抵の人は見当をつけると思う。You can also say "curry mix," which gives the right idea too :)
Q: "This is the cube. It was used part of the scene in the movie in which the characters killed."
"This is the cube. It was used part of the scene in the movie where the characters killed in." sembra naturale?
A: Something more natural sounding would be:
"This is the cube. It was used [in the] part [of] the movie in which the characters [were] killed."
"This is the cube. It was used [in the] part of the movie where the characters [were] killed."
Q: do we bring our cube van to your company(or your place)? sembra naturale?
A: both work
Q: Finally I got a first magnetic cube! This is pretty awesome and my new main cube. sembra naturale?
A: It's sound a little bit unnatural. You could remove this 'pretty'.
## cube
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Domande suggerite | 817 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.845963 |
https://www.dsprelated.com/showthread/comp.dsp/109531-1.php | 1,713,278,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00719.warc.gz | 664,911,784 | 20,011 | Questions regarding polyphase filter
Started by March 2, 2009
I am try to get a better understand of polyphae filter. There are several
questions I wonder if I can get some help:
1) Relationship of polyphase filter and Different signal
interpolation/decimation methods. One thing I am a bit confused about is
the relationship between different curve fitting type interpolation methods
(i.e., cubic bspline, cubic polynomial, linear etc.) and polyphase filter.
In some literation, there was some section detail of how to compute the
cubic polynomial coefficients, then another section describe the structure
of the polyphase filter used. Is it common to use cubic polynomial
interpolation, then followed by polyphase filter?
2) Variable interpolation/decimation factor for a given polyphase filter
structure. I understand this can be achieved by using Farrow type filter
structure. However, also use the conventional polyphase filter structure
and achieve multiple (not continuous) interpolation/decimation factors
without updating the polyphase filter coefficients?
For example, if I design the original filter with bandwidth <= 1/32; will
it be possible for me to get interpolation of 2,3,4,5..up to 32?
Thanks for the help.
On Mar 2, 3:22 pm, "bcarmaint" <bing_carma...@yahoo.com> wrote:
> I am try to get a better understand of polyphae filter. There are several > questions I wonder if I can get some help: > > 1) Relationship of polyphase filter and Different signal > interpolation/decimation methods. One thing I am a bit confused about is > the relationship between different curve fitting type interpolation methods > (i.e., cubic bspline, cubic polynomial, linear etc.) and polyphase filter. > In some literation, there was some section detail of how to compute the > cubic polynomial coefficients, then another section describe the structure > of the polyphase filter used. Is it common to use cubic polynomial > interpolation, then followed by polyphase filter? > > 2) Variable interpolation/decimation factor for a given polyphase filter > structure. I understand this can be achieved by using Farrow type filter > structure. However, also use the conventional polyphase filter structure > and achieve multiple (not continuous) interpolation/decimation factors > without updating the polyphase filter coefficients? > > For example, if I design the original filter with bandwidth <= 1/32; will > it be possible for me to get interpolation of 2,3,4,5..up to 32? > > Thanks for the help.
I think you are confusing a method for implementation versus filters that arise from different approximation methods. A polyphase filter is a way of implementing an FIR filter if you use it as a filter for upsampling and/or downsampling, so that you reduce the number of taps that you compute. A farrow filter is a filter that implements fractional delay via polynomial approximation. Hope this helps, Julius
>On Mar 2, 3:22 pm, "bcarmaint" <bing_carma...@yahoo.com> wrote: >> I am try to get a better understand of polyphae filter. There are
several
>> questions I wonder if I can get some help: >> >> 1) Relationship of polyphase filter and Different signal >> interpolation/decimation methods. One thing I am a bit confused about
is
>> the relationship between different curve fitting type interpolation
methods
>> (i.e., cubic bspline, cubic polynomial, linear etc.) and polyphase
filter.
>> In some literation, there was some section detail of how to compute
the
>> cubic polynomial coefficients, then another section describe the
structure
>> of the polyphase filter used. Is it common to use cubic polynomial >> interpolation, then followed by polyphase filter? >> >> 2) Variable interpolation/decimation factor for a given polyphase
filter
>> structure. I understand this can be achieved by using Farrow type
filter
>> structure. However, also use the conventional polyphase filter
structure
>> and achieve multiple (not continuous) interpolation/decimation factors >> without updating the polyphase filter coefficients? >> >> For example, if I design the original filter with bandwidth <= 1/32;
will
>> it be possible for me to get interpolation of 2,3,4,5..up to 32? >> >> Thanks for the help. > >I think you are confusing a method for implementation versus >filters that arise from different approximation methods. > >A polyphase filter is a way of implementing an FIR filter if you >use it as a filter for upsampling and/or downsampling, so that >you reduce the number of taps that you compute. > >A farrow filter is a filter that implements fractional delay via >polynomial approximation. > >Hope this helps, >Julius >
Julius, Thanks for the response. I understand the relationship between Farrow structure and polyphase filter. My questions was the relationship between different interpolation method, such as cubin bspline interpolation and polyphase filter. Thanks, Ben
bcarmaint wrote:
...
> My questions was the relationship between different interpolation method, > such as cubin bspline interpolation and polyphase filter.
A polyphase filter is not an interpolation technique. It is an efficient implementation method. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
On Mar 2, 7:11�pm, "bcarmaint" <bing_carma...@yahoo.com> wrote:
[snip]
> > Thanks for the response. I understand the relationship between Farrow > structure and polyphase filter. > > My questions was the relationship between different interpolation method, > such as cubin bspline interpolation and polyphase filter. > > Thanks, Ben
I'm with Jerry, I think you are confusing an efficient implementation method (polyphase filter) with a mathematical tool for approximation (cubic bspline interpolation). For example, you can implement a cubic bspline interpolator as a polyphase filter. So I don't understand what your question is, sorry. Reading your original question again, I think that where you are confused is between: * the formula for computing the coefficients of a filter to apply to a signal, for example for interpolation. * the efficient implementation of a filter, in fact any filter. The first one includes the cubic spline formula, etc etc etc. The second one includes polyphase structures, lattice filters, etc. As for your second question, yes, if your original filter has bandwidth <= 1/32 as a fraction of the digital bandwidth, then you can use it for variable rate interpolation, as long as that bandwidth is not so narrow that your signal is distorted. Hope this helps, Julius
>On Mar 2, 7:11=A0pm, "bcarmaint" <bing_carma...@yahoo.com> wrote: >[snip] >> >> Thanks for the response. I understand the relationship between Farrow >> structure and polyphase filter. >> >> My questions was the relationship between different interpolation
method,
>> such as cubin bspline interpolation and polyphase filter. >> >> Thanks, Ben > >I'm with Jerry, I think you are confusing an efficient implementation >method >(polyphase filter) with a mathematical tool for approximation (cubic >bspline >interpolation). For example, you can implement a cubic bspline >interpolator >as a polyphase filter. So I don't understand what your question is, >sorry. > >Reading your original question again, I think that where you are >confused >is between: >* the formula for computing the coefficients of a filter to apply to a >signal, > for example for interpolation. >* the efficient implementation of a filter, in fact any filter. > >The first one includes the cubic spline formula, etc etc etc. The >second >one includes polyphase structures, lattice filters, etc. > >As for your second question, yes, if your original filter has >bandwidth ><=3D 1/32 as a fraction of the digital bandwidth, then you can use it >for >variable rate interpolation, as long as that bandwidth is not so >narrow >that your signal is distorted. > >Hope this helps, >Julius >
Julius, Jerry Thanks for the clarifications about the second question. About my first question, I still have some confusion... I can't seem to relate a cubic spline coefficients with a 32 phase polyphase filter. Do you by any chance have some example in this regard? Ben
bcarmaint wrote:
>> On Mar 2, 7:11=A0pm, "bcarmaint" <bing_carma...@yahoo.com> wrote: >> [snip] >>> >>> Thanks for the response. I understand the relationship between >>> Farrow structure and polyphase filter. >>> >>> My questions was the relationship between different interpolation >>> method, such as cubin bspline interpolation and polyphase filter. >>> >>> Thanks, Ben >> >> I'm with Jerry, I think you are confusing an efficient implementation >> method >> (polyphase filter) with a mathematical tool for approximation (cubic >> bspline >> interpolation). For example, you can implement a cubic bspline >> interpolator >> as a polyphase filter. So I don't understand what your question is, >> sorry. >> >> Reading your original question again, I think that where you are >> confused >> is between: >> * the formula for computing the coefficients of a filter to apply to >> a signal, >> for example for interpolation. >> * the efficient implementation of a filter, in fact any filter. >> >> The first one includes the cubic spline formula, etc etc etc. The >> second >> one includes polyphase structures, lattice filters, etc. >> >> As for your second question, yes, if your original filter has >> bandwidth >> <=3D 1/32 as a fraction of the digital bandwidth, then you can use it >> for >> variable rate interpolation, as long as that bandwidth is not so >> narrow >> that your signal is distorted. >> >> Hope this helps, >> Julius >> > Julius, Jerry > > Thanks for the clarifications about the second question. About my > first question, I still have some confusion... I can't seem to relate > a cubic spline coefficients with a 32 phase polyphase filter. Do you > by any chance have some example in this regard? > > Ben
Ben, Here is a really simple way to get at such questions: 1) Figure out the filter you want / need. 2) Does the filter have a bunch of regular zeros in the unit sample response? (i.e. the coefficients?). Below, I will use the case where the coefficients alternate between non-zero and zero valued. 3) If 2 applies then you can think of doing something like this: - Get a sequence of samples for the input. Note that only half of them will be multiplied by nonzero coefficients for the next output. - Add an input sample. Note that only half of the "new" sample sequence will be multiplied by nonzero coefficients for the next output. AND note that these samples were the ones which were NOT operated on in the previous sample interval. So, half of the samples are operated on in the first sample instant and the other half of the samples are operated on in the next sample instant. So, if you effectively commutate the sample sequence so that there are two paths with half the sample rate and you apply these decimated sequences to two identical filters then the sum of the outputs of the two filters will have the original sample rate and will be the same as if you had not done this and just used a single filter at the original sample rate. Thus, the idea of a polyphase filter is born. It's all about the coefficient structure having lots of zeros in it....... Is the filter you're interested in like that? If so, then consider a polyphase implementation. If not, forgeddaboudit. As others have pointed out, polyphase is but an implementation thing. First you have to decide that it's useful in your situation. Fred
On Mar 3, 6:56�am, "bcarmaint" <bing_carma...@yahoo.com> wrote:
> >On Mar 2, 7:11=A0pm, "bcarmaint" <bing_carma...@yahoo.com> wrote: > >[snip] > > >> Thanks for the response. I understand the relationship between Farrow > >> structure and polyphase filter. > > >> My questions was the relationship between different interpolation > method, > >> such as cubin bspline interpolation and polyphase filter. > > >> Thanks, Ben > > >I'm with Jerry, I think you are confusing an efficient implementation > >method > >(polyphase filter) with a mathematical tool for approximation (cubic > >bspline > >interpolation). �For example, you can implement a cubic bspline > >interpolator > >as a polyphase filter. �So I don't understand what your question is, > >sorry. > > >Reading your original question again, I think that where you are > >confused > >is between: > >* the formula for computing the coefficients of a filter to apply to a > >signal, > > �for example for interpolation. > >* the efficient implementation of a filter, in fact any filter. > > >The first one includes the cubic spline formula, etc etc etc. �The > >second > >one includes polyphase structures, lattice filters, etc. > > >As for your second question, yes, if your original filter has > >bandwidth > ><=3D 1/32 as a fraction of the digital bandwidth, then you can use it > >for > >variable rate interpolation, as long as that bandwidth is not so > >narrow > >that your signal is distorted. > > >Hope this helps, > >Julius > > Julius, Jerry > > Thanks for the clarifications about the second question. About my first > question, I still have some confusion... I can't seem to relate a cubic > spline coefficients with a 32 phase polyphase filter. Do you by any chance > have some example in this regard? > > Ben
Ben, A generic explanation of the resampling process is as follows: 1. Upsample to a higher rate (i.e., insert zeros between existing samples) 2. Use an interpolator to figure out what the values of the samples should be at the new points (where you inserted zeros) 3. Possibly downsample to achieve the desired rate Now you can implement step (2) of the process in one of several ways. One way is to use a linear combination of samples around the point you want - eg. x(t) = 0.5(x(t0)+x(t1)). Basically, the linear combination of samples with different weights is what constitutes a filter - so what you have is an "interpolating filter". Now, you could also choose not to be restricted to a linear technique for interpolation, i.e., you could use square, cubic, and higher order terms of the samples to do the interpolation. For example, x(t) = a0.x(t0)^2 + b0.x(t0) + a1.x(t1)^2 + b1.x(t1) + c. What you have here is a polynomial interpolation technique. In a sense, the linear interpolating technique is a specific case of the polynomial technique, with the order of the polynomial restricted to one. So when do you choose one over the other? The linear filtering technique is a low computational complexity technique and you can use a number of filter design techniques to design your interpolating filter. On the other hand, in situations where the spacing between your samples is not constant, or, if the spacing between your interpolating instant and the sampling instant changes, then the filter needs to change. The filtering technique falls short because your filter was designed assuming a particular fixed spacing of samples on the time axis and does not hold if that changes. The greatest advantage of polynomial interpolation is that it can be implemented in a way that factors in the time interval between sampled and interpolated points - for example, x(t) = x(t0) + (x(t1)-x(t0)).(t-t0)/(t1-t0) This has the advantage that the interpolation holds even if the relative spacing between the interpolating instant, t, and sampling instants t0 and t1, changes. In the case of a communications receiver for example, the receiver clock can be at a slightly different frequency from the transmitter clock - so assuming that the two clocks are in sync at the first sample, the second sample is off by dT, the third by 2dT, etc. If you can estimate this difference, you could use a polynomial interpolator to interpolate existing received samples to the transmitter rate. And as others have said, a polyphase filter is simply a structure that allows for efficient implementation of the filter. - Ravi
>On Mar 3, 6:56=A0am, "bcarmaint" <bing_carma...@yahoo.com> wrote: >> >On Mar 2, 7:11=3DA0pm, "bcarmaint" <bing_carma...@yahoo.com> wrote: >> >[snip] >> >> >> Thanks for the response. I understand the relationship between
Farrow
>> >> structure and polyphase filter. >> >> >> My questions was the relationship between different interpolation >> method, >> >> such as cubin bspline interpolation and polyphase filter. >> >> >> Thanks, Ben >> >> >I'm with Jerry, I think you are confusing an efficient implementation >> >method >> >(polyphase filter) with a mathematical tool for approximation (cubic >> >bspline >> >interpolation). =A0For example, you can implement a cubic bspline >> >interpolator >> >as a polyphase filter. =A0So I don't understand what your question
is,
>> >sorry. >> >> >Reading your original question again, I think that where you are >> >confused >> >is between: >> >* the formula for computing the coefficients of a filter to apply to
a
>> >signal, >> > =A0for example for interpolation. >> >* the efficient implementation of a filter, in fact any filter. >> >> >The first one includes the cubic spline formula, etc etc etc. =A0The >> >second >> >one includes polyphase structures, lattice filters, etc. >> >> >As for your second question, yes, if your original filter has >> >bandwidth >> ><=3D3D 1/32 as a fraction of the digital bandwidth, then you can use
it
>> >for >> >variable rate interpolation, as long as that bandwidth is not so >> >narrow >> >that your signal is distorted. >> >> >Hope this helps, >> >Julius >> >> Julius, Jerry >> >> Thanks for the clarifications about the second question. About my
first
>> question, I still have some confusion... I can't seem to relate a
cubic
>> spline coefficients with a 32 phase polyphase filter. Do you by any
chanc=
>e >> have some example in this regard? >> >> Ben > >Ben, > >A generic explanation of the resampling process is as follows: >1. Upsample to a higher rate (i.e., insert zeros between existing >samples) >2. Use an interpolator to figure out what the values of the samples >should be at the new points (where you inserted zeros) >3. Possibly downsample to achieve the desired rate > >Now you can implement step (2) of the process in one of several ways. >One way is to use a linear combination of samples around the point you >want - eg. x(t) =3D 0.5(x(t0)+x(t1)). Basically, the linear combination >of samples with different weights is what constitutes a filter - so >what you have is an "interpolating filter". > >Now, you could also choose not to be restricted to a linear technique >for interpolation, i.e., you could use square, cubic, and higher order >terms of the samples to do the interpolation. For example, > >x(t) =3D a0.x(t0)^2 + b0.x(t0) + a1.x(t1)^2 + b1.x(t1) + c. > >What you have here is a polynomial interpolation technique. In a >sense, the linear interpolating technique is a specific case of the >polynomial technique, with the order of the polynomial restricted to >one. > >So when do you choose one over the other? The linear filtering >technique is a low computational complexity technique and you can use >a number of filter design techniques to design your interpolating >filter. On the other hand, in situations where the spacing between >your samples is not constant, or, if the spacing between your >interpolating instant and the sampling instant changes, then the >filter needs to change. The filtering technique falls short because >your filter was designed assuming a particular fixed spacing of >samples on the time axis and does not hold if that changes. > >The greatest advantage of polynomial interpolation is that it can be >implemented in a way that factors in the time interval between sampled >and interpolated points - for example, > >x(t) =3D x(t0) + (x(t1)-x(t0)).(t-t0)/(t1-t0) > >This has the advantage that the interpolation holds even if the >relative spacing between the interpolating instant, t, and sampling >instants t0 and t1, changes. In the case of a communications receiver >for example, the receiver clock can be at a slightly different >frequency from the transmitter clock - so assuming that the two clocks >are in sync at the first sample, the second sample is off by dT, the >third by 2dT, etc. If you can estimate this difference, you could use >a polynomial interpolator to interpolate existing received samples to >the transmitter rate. > >And as others have said, a polyphase filter is simply a structure that >allows for efficient implementation of the filter. > >- Ravi >
Ravi, Thanks a lot. I think I confused "interpolation filter" and the polyphae implementation of the anti-aliasing filter used after the interpolation. Here is my understanding of a overall interpolation procedure: 1) First I will need interpolate the signal (may need to use both upsampling and downsampling to achieve the required factor); in this process, I may choose different interpolation techniques you described above. 2) Implement anti-aliasing/anti-imaging lowpass filter to reduce artifects. In this step, I can choose the cutoff bandwidth to be the smallest possible bandwidth I need to support (determined by the highest interpolation factor, for example, if the factor is 32, then the cutoff bandwidth should be 1/32); A typical filter like this may have several hundre taps, and a good choise to implement this filter is polyphase filter. With this structure, I can support the filtering of interpolation of 2,3,4...32 discrete steps. Does this looks correct to you? Thanks for the help from all of you, Regards, Ben
On Mar 3, 7:56�pm, "bcarmaint" <bing_carma...@yahoo.com> wrote: | 5,314 | 22,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-18 | latest | en | 0.893305 |
https://rebab.net/how-many-grams-in-a-fluid-ounce/ | 1,656,264,253,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00730.warc.gz | 522,269,457 | 6,271 | Ounces come Grams conversion table Ounces (oz) Grams (g) Kilograms+Grams (kg+g)
20 oz 566.99 g 0 kg 566.99 g
30 oz 850.49 g 0 kg 850.49 g
40 oz 1133.98 g 1 kg 133.98 g
50 oz 1417.48 g 1 kg 417.48 g
Additionally, just how much fluid is 30 grams? Liquid procedures 1 cup 8 liquid ounces 1/2 pint
1 teaspoon 1/6 liquid ounce 5 grams
1 tablespoon 1/2 liquid ounce 15 grams
2 tablespoons 1 liquid ounce 30 grams
8 tablespoons 4 fluid ounces 1/4 pint
In this regard, how numerous grams are in a fl oz?
29.5735296875
How plenty of grams is 32 liquid ounces?
Ounces come Grams table Ounces Grams
29 oz 822.14
30 oz 850.49
31 oz 878.84
32 oz 907.18
Alexe NahrlichProfessional
## How plenty of G is a cup of water?
How countless grams the water the water measure room in 1 U.S. cup? The prize is: The adjust of 1 cup us ( U.S. cup ) unit in a water measure up measure equates to = into 236.59 g ( gram that water ) as per the identical measure and also for the same water measure type.
You are watching: How many grams in a fluid ounce
Alcina BoutinProfessional
## How carry out you measure up grams?
The most accurate way to measure grams is to usage a scale. Pick a digital or mechanical range that provides the metric system, then push the tare button to zero that out. Ar your article on the center of the scale, wait because that the scale"s digital display screen or needle to involved a stop, and record the grams.
## How do you transform liquids come Grams?
To convert from grams come milliliters, division the grams by the density instead the multiplying. The thickness of water is 1 g/mL. If a substance"s density is greater than 1 g/mL then that substance is more dense 보다 pure water, and also would sink in it.
Ivanete HardwickExplainer
## How plenty of cups is 100 grams?
While a gram is a metric weight measurement, a cup is a volume measurement. These varieties of measurement counter are useful for recipes and cooking. However, various ingredients have various conversion equivalents. For example, 100 grams of water is same to approximately 0.423 cup.
Zandra BasterrecheaExplainer
## What go 100 grams weigh?
100 grams = 0.22046 pounds.
Nube Fernandez De CordobaExplainer
## What is 50g ML?
g to ml switch table:
1 gram = 1 ml 21 grams = 21 ml 41 grams = 41 ml
7 grams = 7 ml 27 grams = 27 ml 47 grams = 47 ml
8 grams = 8 ml 28 grams = 28 ml 48 grams = 48 ml
9 grams = 9 ml 29 grams = 29 ml 49 grams = 49 ml
10 grams = 10 ml 30 grams = 30 ml 50 grams = 50 ml
Anam AbendañoPundit
## How countless cups is 60g?
Cream Cheese/Soft Cheese
us cups lot in Grams quantity in Ounces
1/4 cup 30g 1 oz
1/3 cup 40g 1 1/2 oz
1/2 cup 60g 2 oz
2/3 cup 80g 3 oz
Contemplacion HennzePundit
## How lot does 1 fluid oz weigh?
A U.S. fluid ounce of water will weight approx. 29.57 g, when 1 ounce the water will certainly weigh 28.35 g, and an royal fluid ounce will certainly weight approx. 28.4 g, i beg your pardon is only a difference from the avoirdupois ounce that water by 50 milligrams, which is an insignificant difference, specifically for baking.
## How many fl oz is 200g?
Basic ounces come grams weight conversions
1/2 oz 15g
6 oz 170g
7 oz 200g
8 oz 225g
9 oz 255g
Garofita EscobarPundit
## How lot does a fluid ounce sweet in grams?
29.57 grams
Lijiao JasminePundit
## How much does 1 liquid ounce sweet in grams?
How countless grams in 1 liquid ounce? The prize is 29.5735296875. Us assume you space converting between gram and ounce liquid>. You can view more details on each measurement unit: grams or fluid ounce The SI derived unit for volume is the cubic meter.
Chia DauderTeacher
## What is 50 grams in liquid ounces?
Grams to Ounces switch table
Grams (g) Ounces (oz)
30 g 1.0582 oz
40 g 1.4110 oz
50 g 1.7637 oz
60 g 2.1164 oz
Zenoviy GragerSupporter
## How much is 100 grams that liquid?
Volume of 100 Grams of Water
100 Grams that Water =
0.42 U.S. Cups
0.35 Imperial Cups
0.40 Metric Cups
100.00 Milliliters
Hamama MatthewsSupporter
## How many grams is a dry measuring cup?
Dry measures
3 teaspoons 1 tablespoon 14.3 grams
5 1/3 tablespoons 1/3 cup 75.6 grams
8 tablespoons 1/2 cup 113.4 grams
12 tablespoons 3/4 cup .375 pound
32 tablespoons 2 cups 1 pound
Iftikhar TukvachinskySupporter
## How many grams is a tespoon of powder?
One teaspoon that powdered sugar converted come gram equals to 2.60 g. How many grams of powdered sugar room in 1 teaspoon? The answer is: The change of 1 tsp ( teaspoon ) unit in a powdered street measure amounts to = into 2.60 g ( gram ) together per the identical measure and for the exact same powdered street type.
Savu BeñaranBeginner
## How many ml is one ounce?
1 US liquid ounce is equal to 29.5735296 milliliters and also 1 Imperial fluid ounce is equal to 28.4130625 milliliters. The symbol is "fl oz". To convert between US and UK liquid ounces, please inspect fluid ounces conversion calculator tool.
Narine RanjitsinhjiBeginner
## How execute you measure up an ounce of powder?
The Chemistry of absent Candy
The United claims customary system specifies that 1 tbsp. Is same to one-half that a fluid ounce. However, a powder, such as sugar or salt, is measured in ounces (oz.) together weight. Come make an exact conversion the ounces come tablespoons, you require to recognize the density of the powder.
See more: The Two Factors That Combine To Keep The Planets In Orbit Are
Mircea AramayoBeginner
## How many cups is 250 grams?
250 grams flour amounts to 2 cups. | 1,575 | 5,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-27 | latest | en | 0.844437 |
https://www.goconqr.com/quiz/3287601/wave | 1,638,097,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00385.warc.gz | 732,926,266 | 15,089 | # WAVE
Quiz by Izatul Husna, updated more than 1 year ago
Created by Izatul Husna over 6 years ago
8
0
GOOD LUCK
## Resource summary
### Question 1
Question
A pendulum is oscillating with amplitude 2cm and frequency 1.5Hz. Which of the following graphs correctly shows the displacement of the pendulum bob against time?
### Question 2
Question
Which of the following quantities will decrease when an oscillating system undergoes damped oscillation?
• Speed
• Amplitude
• Frequency
• Wave length
### Question 3
Question
Damping of an oscillating system is due to
• the system is losing energy to the surrounding.
• the system has been set up wrongly.
• the system is not oscillating in vacuum
### Question 4
Question
A force oscillation is
• A system oscillates with maximum amplitude.
• A system oscillates with decreasing amplitude.
• A system oscillates with the help of an external force.
• A system oscillates without any external force act on it.
### Question 5
Question
When an oscillating spring reaches resonance state, it
• Stop oscillates.
• Experiencing damped oscillation.
• Oscillates with maximum amplitude.
• Oscillates with higher frequency.
### Question 6
Question
The natural frequency of a spring is f Herz. Resonance happens if the frequency of the external force that applied on it is
• 0.5f Hz
• f Hz
• 2f Hz
• 4f Hz
### Question 7
Question
Five pendulums are hung at a wire PQ. The pendulum X is displaced and then released. Which of the other pendulum will oscillate with greatest amplitude?
• Pendulum A
• Pendulum B
• Pendulum C
• Pendulum D
### Question 8
Question
A thin piece of glass is broken when a soprano is singing with a high pitch. This probably is caused by a physics phenomena called
• Diffraction
• Interference
• Damping
• Resonance
### Question 9
Question
It is observed that a car will vibrate intensely at a particular speed. This is probably due to
• Inertia
• Resonance
• Damping
• Diffraction
### Question 10
Question
A water wave is produced from a source of frequency 25 Hz. If the wavelength is 0.8 cm, what is the velocity of propagation of the water wave?
• 0.2m
• 20m
• 31.25m
• 3.125m
### Question 11
Question
Which of the following quantities will cause the amplitude of a wave changes?
• Type of the wave
• Energy of the wave
• Wavelength of the wave
• Frequency of the wave
### Question 12
Question
The wavelength of a propagating wave is not affected by
• the frequency of the wave
• the velocity of the wave
• the energy of the wave
• the medium of wave that propagates
### Question 13
Question
The figure above shows a transverse wave. The wavelength of the wave is equal to
### Question 14
Question
The main difference between transverse wave and longitudinal wave is
• The wave pattern
• The direction of the vibration of the particles in the wave
• Transverse wave propagates faster than longitudinal wave.
• Transverse wave can travels in vacuum while longitudinal wave cannot
### Question 15
Question
Which of the following waves is longitudinal wave?
• Microwave
• Light wave
• Sound wave
### Question 16
Question
The figure above shows the simulation of longitudinal wave by using a slinky spring. What is the wavelength of the longitudinal wave?
• 15cm
• 10cm
• 5cm
• Can't be determined from the diagram
### Question 17
Question
The figure above shows the simulation of transverse wave by using a slinky spring. What is the wavelength of the transverse wave?
• 25cm
• 20cm
• 5cm
• 12.5cm
### Question 18
Question
Figure above shows the pattern of wavefronts produced by bob dipper at a frequency of 25Hz in a ripple tank. Find the speed of the wave.
• 5m
• 0.05m
• 1.25m
• 0.3m
### Question 19
Question
In a ripple tank experiment, waves are produced at the RATE of 24 waves per second and the distance between 5 successive ripples is 20cm. The experiment is repeated by changing the frequency of the vibrator, and the new distance between 5 successive ripples is 12cm. What is the new frequency of the vibrator?
• 100Hz
• 60Hz
• 20Hz
• 40Hz
### Question 20
Question
The end coil of a slinky spring is connected to a motor which is moving forward and backward at a frequency of 4Hz. Given that the distance between four successive rarefactions is 60cm. What is the speed of the speed of wave produced?
• 2.4 ms-1
• 0.8 ms-1
• 2.5 ms-1
• 1.0 ms-1
### Question 21
Question
A sound wave of wavelength 4mm travels in air at a frequency of 6.0 × 104 Hz. Find the time take by the wave to travel 60m
• 1 s
• 0.5 s
• 0.25 s
• 0.8 s
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A View from the Bridge Quotes | 1,303 | 4,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.865221 |
https://m.wikihow.com/Measure-Liquids-without-a-Measuring-Cup | 1,586,150,852,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00329.warc.gz | 541,877,758 | 56,926 | # How to Measure Liquids without a Measuring Cup
Co-authored by wikiHow Staff
Updated: July 9, 2019
Measuring cups are generally considered to be essential items in a pantry. In particular, they are useful for measuring the volume of fluids. Fortunately, if you find yourself in a situation without a measuring cup, there are other simple ways to determine the amount of liquid that you need.
### Method 1 of 3: Estimating Using Size Comparisons
1. 1
Use an object as a reference point. If you are stuck without a measuring device, it can be useful to have some visual aids in your head as a reference for the correct amount. Here are some good ones to remember:
• A teaspoon is about the size of the tip of your finger
• A tablespoon is about the size of an ice cube
• 1/4 cup is about the size of a large egg
• 1/2 cup is about the size of a tennis ball
• A full cup is about the size of a baseball, an apple or a fist[1]
2. 2
Choose an appropriate vessel to pour your liquid into. Ideally, you would use your hands as they can be cupped to create a rounded shape. However, this may not be appropriate for sticky liquids. [2]Try to select a transparent vessel that you can easily imagine your visual aid just fitting into.
• For example, if you are measuring a 1/4 cup, it may be useful to use a tall glass that an egg would just fit into. A wider glass, on the other hand, may be more appropriate for a 1/2 or full cup.
3. 3
Place your vessel on a flat surface and bring yourself down to eye level. This will help you to clearly see the amount being poured. Pour the fluid slowly into your vessel.
• When you think you may have the right amount, stop and compare it to the size of your visual aid.
• Make adjustments to the amount in the vessel if necessary.
4. 4
Take a look at the amount of fluid in the vessel and commit it to memory. This will make future estimations easier, as it provides you with a point of reference. [3] It is helpful to continue using the same vessels for the certain measurements (e.g. the tall glass again for a 1/4 cup).
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### Method 2 of 3: Using a Kitchen Scale
1. 1
Use a kitchen scale to measure the correct amount of liquid. In general, it is fine to weigh your liquid using an ordinary kitchen scale, using water as the assumed density.[4]
• Most liquids, such as milk and orange juice, will have a similar density to water. However, keep in mind that some liquids may be significantly denser (such as honey or syrup) so the reading may not be appropriate for these.
• To provide greater accuracy, some kitchen scales give you the option to select different liquids, such as milk. The scale then calculates the volume based on the density of the selected liquid. If you have a scale with this feature, ensure that it is set to measure the correct liquid. [5]
2. 2
Calculate the weight of your fluid. If you are using an ordinary scale, you will need to work out the correct weight for your liquid. It is worth remembering that one fluid ounce of water corresponds to exactly one ounce in water. This principle also applies to liters (1 milliliter of water is 1 gram in weight). [6]
• Use this as your key metric when measuring your liquid. For example, if you need half a cup of water, it should weigh 4 oz or 125g.
3. 3
Select a glass or container to use to measure your liquid in. Place your container on the scale, making sure it is in the center position.
• Do not add any liquid to your container yet. It is important to have your container empty at this stage, as you will need to set your scale up to exclude the container’s weight from measurements. [7]
4. 4
Calibrate your scale to exclude the container in measurements. Look for the button on your scale button labeled “tare” or “zero”.
• Once this is pressed, the weight of your container should display as zero on your scale. This will ensure that the measurement of your liquid is accurate.[8]
5. 5
Pour your liquid into your container. Do this slowly, pausing to take note of the weight. Stop pouring once your scale displays the weight or volume that you need. If you go over the correct amount, pour the excess into the sink.
6. 6
Measure any additional liquids you need for your recipe. If you are using an ordinary scale and plan on mixing liquids together, you can measure these in the same container. Keep the container on the scale and calculate the new amount that you need by adding together the amounts of both liquids. Pour the new liquid into the container until you reach the correct combined amount.
• Keep in mind that if you are using a kitchen scale that provides the option of measuring different liquids, you will need to change your settings and start a new measurement.
• If you are measuring water and want to measure milk, for example, set aside your container of water, select the milk option on your scale and begin a new measurement with another container.
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### Method 3 of 3: Using Tablespoons and Teaspoons
1. 1
Work out how many tablespoons you will need. An easy way to do this is to remember that one cup equates to 16 tablespoons. This can be used as a simple metric to calculate how many tablespoons you need.
• For example, if you need half a cup, you would require 8 tablespoons of fluid.[9]
2. 2
Use a tablespoon to measure out the liquid you need. Measure your liquid over a vessel to prevent a mess. Pouring slowly and steadily to avoid excess spillage into the vessel, fill your tablespoon with the liquid.
• Transfer to the vessel and repeat until you have measured the amount you need in tablespoons.
3. 3
Use a teaspoon to refine the volume. Some recipes may call for more precise measurements. In this case, you can use teaspoons to get the exact amount needed.
• One teaspoon is equivalent to 1/6 of a fluid ounce or 4.7ml.
4. 4
Commit the amount of fluid in the vessel to memory. This will help to develop your ability to estimate measurements.
• If you are using a glass or plastic vessel may alternatively wish to use a marker on the outside of the vessel to indicate measurements. [10]This will save you re-measuring the tablespoon amounts in the future. For example, if you measured out a quarter of a cup (4 tablespoons), you would write “1/4” on the vessel.
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## Tips
• If you are using an old recipe, keep in mind that it may be using an imperial cup as its reference. Imperial cups are larger than the standard US cups, equating to 9.6 ounces. This means you would measure out 19 tablespoons, rather than 16.[11]
Thanks!
• Recipes from other countries also may differ slightly in measurements. For example, the standard cup for the UK, New Zealand, Australia, Canada and South Africa is 250ml (8.4 fluid ounces). [12]
Thanks!
• If the recipe is all in cups, e.g. two cups of flour, half a cup of sugar, a cup of milk, you can use a cup! Any recipes with parts or various amounts of the same quantity you can use one container to measure all your ingredients. You risk only a larger or small end result.
Thanks!
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## About This Article
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 12 references.
34 votes - 40%
Co-authors: 6
Updated: July 9, 2019
Views: 129,123
Article SummaryX
To measure liquids without a measuring cup, imagine an object of the same volume to help you visualize the correct amount. For example, a tablespoon is about the size of an ice cube and a cup is about the size of a baseball or apple. For 1/2 cup, imagine a tennis ball and for a 1/4 cup, picture a large egg. Then, find a glass that you could easily fit your object into, like a highball glass or whisky tumbler. When you pour your liquid, imagine the object in your glass and fill it with the rough volume of your object. For more tips, including how to use a kitchen scale to calculate the right amount of liquid, read on!
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Thanks to all authors for creating a page that has been read 129,123 times.
## Did this article help you?
Advertisement | 1,923 | 8,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-16 | latest | en | 0.928728 |
https://cboard.cprogramming.com/cplusplus-programming/78129-averaging-algorithm.html | 1,512,974,401,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00266.warc.gz | 531,323,126 | 16,144 | 1. ## Averaging Algorithm
Hi,
This is my first time posting here. I'm taking a programming class, and one of our assignments is to write a program that calculates the average of a group of numbers without the use of division. The code itself isnt so much the problem, but I am having trouble coming up with an algorithm to do it. I know this isnt so much a programming question but a math question, but I'm hoping theres a math whiz here who can give me a hand. Even a hint would help really.
2. Which sort of average
- mean
- median
- mode
http://mathworld.wolfram.com
3. There's no way to compute a mean of an arbitrary number of elements without using division. If you end up thinking you've done so, then you've implemented a division algorithm. So let's pray your teacher meant median or mode. (But then why would the teacher bother specifying not to use division?)
If you know beforehand the size of the group of numbers, then you can just multiply. For example, if you know there will be ten numbers to average, you could just multiply the sum by 0.1.
4. You might want to consider the mathematical relation log(x/y) = log(x)-log(y) (if x and y are both positive).
5. You don't need division to perform division!
Using Newton-Rhapson with f(x) == 1/x - N, you can approximate 1/N like this:
X0 = first approximation, close to but smaller than 1/N
Xk+1 = Xk * (2 - Xk * N)
If Xk has z digits of precision, Xk+1 will have around 2*z digits of precision (ie it converges to 1/N quadratically), so complexity for z digits of 1/N is O(log z M(N)) where M(N) is multiplication cost.
Then take the sum, and multiply by this number to the required degree of precision!
I have no idea if this is cheating, though...
6. Gotta love these pointless homework questions which ask you to solve fairly trivial problems by restricting the allowed set of operators you can use.
Like the "how to do addition without using +" problem which happens every so often.
7. Could he not just find out how many numbers there are in the list that he needs to find the average of? And then use Rashakil Fol's idea?
8. yeah, just take the number of numbers you have and a decimal place to it to the left.
9. Originally Posted by c89c
yeah, just take the number of numbers you have and a decimal place to it to the left.
That only works if the number is 10.
10. Yeah I was trying to figure out ways around that after I posted.
Could you not just cheat and make sure that you can only have a number of values between a certain range.
11. This is what I made that does that:
Code:
```#include <iostream.h>
#include <windows.h>
#include <cstdlib>
using namespace std;
int main()
{
int numvalues;
int values [20];
int intermediary;
int x = 0;
float y;
cout << "Please enter the number of values you have (1-20): ";
cin >> numvalues;
cout << endl;
for (int i = 0; i < numvalues; ++i)
{
cout << "Enter Value number " << i+1 << ": ";
cin >> intermediary;
values [i] = intermediary;
}
for (int j = 0; j < numvalues; ++j)
{
x += values [j];
}
cout << endl << "The total of your values is: "<< x << endl;
cin.ignore();
cin.get();
switch (numvalues) {
case 1:
y = 1;
break;
case 2:
y = 0.5;
break;
case 3:
y = 0.3333333333;
break;
case 4:
y = 0.25;
break;
case 5:
y = 0.2;
break;
case 6:
y = 0.1666666666;
break;
case 7:
y = 0.1428571428;
break;
case 8:
y = 0.125;
break;
case 9:
y = 0.1111111111;
break;
case 10:
y = 0.1;
break;
case 11:
y = 0.0909090909;
break;
case 12:
y = 0.0833333333;
break;
case 13:
y = 0.0769230769;
break;
case 14:
y = 0.0714285714;
break;
case 15:
y = 0.0051282051;
break;
case 16:
y = 0.0051282051;
break;
case 17:
y = 0.0588235294;
break;
case 18:
y = 0.0555555555;
break;
case 19:
y = 0.0526315789;
break;
case 20:
y = 0.5;
break;
default:
cout << "You have entered too many values you will have to restart the program!"<<endl;
cout << endl;
cout << "Press [ENTER] to exit...";
cin.get();
return 1;
}
for (int k = 0; k < 4; ++k)
{
system("CLS");
Sleep (600);
cout << ".";
Sleep (600);
cout << ".";
Sleep (600);
}
cout << endl << endl << "The answer is: " << answer << endl;
cout << endl << "Press [ENTER] to close...";
cin.get();
return 0;
}```
12. I like the thing where you let the user wait 1.2 seconds for an operation performed in less than 5 milliseconds (probably even shorter time)
But I don't really see the point in finding the average without using division, on the other hand I don't see any point in writing the programs we gets as assignment in my school. Although we are allowed to use division in our average-algorithms
13. bumfluf your approach won't work for number of values > 20.
I hinted at one approach earlier; looks like I need to spell it out;
Code:
```#include <cmath>
#include <iostream>
int main()
{
double sum;
int n;
// somehow initialise previous two values ...
double average;
if (n > 0)
{
average = std::exp(std::log(std::fabs(sum)) - std::log((double)n)) * ((sum > 0) : 1 : -1);
std::cout << "Average is " << average << '\n';
}
else
{
std::cout << "Cannot compute average if number of values < 0\n";
}
}```
You might also wish to consider what the expression "sum*std::pow((double)n, -1.0)" yields.
14. as an approximation:
Code:
```#include<stdio.h>
main(){
int a=35;
int b=46;
while((b-a)>1){
a++;
b--;
}
printf("%d",a);
}```
Code:
```#include<stdio.h>
main(){
float a=35;
float b=46;
while((b-a)>0.001){
a+=0.001;
b-=0.001;
}
printf("%f",a);
}```
15. Originally Posted by grumpy
bumfluf your approach won't work for number of values > 20.
His program does work with values > 20! I compiled and ran it. It works perfect.
@ bumfluf: use <iostream> instead of <iostream.h>! | 1,621 | 5,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-51 | latest | en | 0.921766 |
http://popflock.com/learn?s=Altitude | 1,603,166,940,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107869785.9/warc/CC-MAIN-20201020021700-20201020051700-00034.warc.gz | 84,493,099 | 26,042 | Altitude
Get Altitude essential facts below. View Videos or join the Altitude discussion. Add Altitude to your PopFlock.com topic list for future reference or share this resource on social media.
Altitude
Altitude or height (also sometimes known as 'depth') is defined based on the context in which it is used (aviation, geometry, geographical survey, sport, atmospheric pressure, and many more). As a general definition, altitude is a distance measurement, usually in the vertical or "up" direction, between a reference datum and a point or object. The reference datum also often varies according to the context. Although the term altitude is commonly used to mean the height above sea level of a location, in geography the term elevation is often preferred for this usage.
Vertical distance measurements in the "down" direction are commonly referred to as depth.
## In aviation
An airplane in its cruising altitude.
Vertical distance comparison
In aviation, the term altitude can have several meanings, and is always qualified by explicitly adding a modifier (e.g. "true altitude"), or implicitly through the context of the communication. Parties exchanging altitude information must be clear which definition is being used.[1]
Aviation altitude is measured using either mean sea level (MSL) or local ground level (above ground level, or AGL) as the reference datum.
Pressure altitude divided by 100 feet (30 m) is the flight level, and is used above the transition altitude (18,000 feet (5,500 m) in the US, but may be as low as 3,000 feet (910 m) in other jurisdictions); so when the altimeter reads 18,000 ft on the standard pressure setting the aircraft is said to be at "Flight level 180". When flying at a flight level, the altimeter is always set to standard pressure (29.92 inHg or 1013.25 hPa).
On the flight deck, the definitive instrument for measuring altitude is the pressure altimeter, which is an aneroid barometer with a front face indicating distance (feet or metres) instead of atmospheric pressure.
There are several types of altitude in aviation:
• Indicated altitude is the reading on the altimeter when it is set to the local barometric pressure at mean sea level. In UK aviation radiotelephony usage, the vertical distance of a level, a point or an object considered as a point, measured from mean sea level; this is referred to over the radio as altitude.(see QNH)[2]
• Absolute altitude is the vertical distance of the aircraft above the terrain over which it is flying.[1]:ii It can be measured using a radar altimeter (or "absolute altimeter").[1] Also referred to as "radar height" or feet/metres above ground level (AGL).
• True altitude is the actual elevation above mean sea level.[1]:ii It is indicated altitude corrected for non-standard temperature and pressure.
• Height is the vertical distance above a reference point, commonly the terrain elevation.radiotelephony usage, the vertical distance of a level, a point or an object considered as a point, measured from a specified datum; this is referred to over the radio as height, where the specified datum is the airfield elevation (see QFE)[2]
• Pressure altitude is the elevation above a standard datum air-pressure plane (typically, 1013.25 millibars or 29.92" Hg). Pressure altitude is used to indicate "flight level" which is the standard for altitude reporting in the Class A airspace (above roughly 18,000 feet). Pressure altitude and indicated altitude are the same when the altimeter setting is 29.92" Hg or 1013.25 millibars.
• Density altitude is the altitude corrected for non-ISA International Standard Atmosphere atmospheric conditions. Aircraft performance depends on density altitude, which is affected by barometric pressure, humidity and temperature. On a very hot day, density altitude at an airport (especially one at a high elevation) may be so high as to preclude takeoff, particularly for helicopters or a heavily loaded aircraft.
These types of altitude can be explained more simply as various ways of measuring the altitude:
• Indicated altitude - the altitude shown on the altimeter.
• Absolute altitude - altitude in terms of the distance above the ground directly below
• True altitude - altitude in terms of elevation above sea level
• Height - vertical distance above a certain point
• Pressure altitude - the air pressure in terms of altitude in the International Standard Atmosphere
• Density altitude - the density of the air in terms of altitude in the International Standard Atmosphere in the air
## In atmospheric studies
### Atmospheric layers
The Earth's atmosphere is divided into several altitude regions. These regions start and finish at varying heights depending on season and distance from the poles. The altitudes stated below are averages:[3]
• Troposphere: surface to 8,000 metres (5.0 mi) at the poles, 18,000 metres (11 mi) at the Equator, ending at the Tropopause
• Stratosphere: Troposphere to 50 kilometres (31 mi)
• Mesosphere: Stratosphere to 85 kilometres (53 mi)
• Thermosphere: Mesosphere to 675 kilometres (419 mi)
• Exosphere: Thermosphere to 10,000 kilometres (6,200 mi)
The Kármán line, at an altitude of 100 kilometres (62 mi) above sea level, by convention defines represents the demarcation between the atmosphere and space.[4] The thermosphere and exosphere (along with the higher parts of the mesosphere) are regions of the atmosphere that are conventionally defined as space.
### High altitude and low pressure
Regions on the Earth's surface (or in its atmosphere) that are high above mean sea level are referred to as high altitude. High altitude is sometimes defined to begin at 2,400 meters (8,000 ft) above sea level.[5][6][7]
At high altitude, atmospheric pressure is lower than that at sea level. This is due to two competing physical effects: gravity, which causes the air to be as close as possible to the ground; and the heat content of the air, which causes the molecules to bounce off each other and expand.[8]
### Temperature profile
The temperature profile of the atmosphere is a result of an interaction between radiation and convection. Sunlight in the visible spectrum hits the ground and heats it. The ground then heats the air at the surface. If radiation were the only way to transfer heat from the ground to space, the greenhouse effect of gases in the atmosphere would keep the ground at roughly 333 K (60 °C; 140 °F), and the temperature would decay exponentially with height.[9]
However, when air is hot, it tends to expand, which lowers its density. Thus, hot air tends to rise and transfer heat upward. This is the process of convection. Convection comes to equilibrium when a parcel of air at a given altitude has the same density as its surroundings. Air is a poor conductor of heat, so a parcel of air will rise and fall without exchanging heat. This is known as an adiabatic process, which has a characteristic pressure-temperature curve. As the pressure gets lower, the temperature decreases. The rate of decrease of temperature with elevation is known as the adiabatic lapse rate, which is approximately 9.8 °C per kilometer (or 5.4 °F [3.0 °C] per 1000 feet) of altitude.[9]
Note that the presence of water in the atmosphere complicates the process of convection. Water vapor contains latent heat of vaporization. As air rises and cools, it eventually becomes saturated and cannot hold its quantity of water vapor. The water vapor condenses (forming clouds), and releases heat, which changes the lapse rate from the dry adiabatic lapse rate to the moist adiabatic lapse rate (5.5 °C per kilometer or 3 °F [1.7 °C] per 1000 feet).[10] As an average, the International Civil Aviation Organization (ICAO) defines an international standard atmosphere (ISA) with a temperature lapse rate of 6.49 °C per kilometer (3.56 °F per 1,000 feet).[11] The actual lapse rate can vary by altitude and by location.
Finally, note that only the troposphere (up to approximately 11 kilometres (36,000 ft) of altitude) in the Earth's atmosphere undergoes notable convection; in the stratosphere, there is little vertical convection.[12]
## Effects on organisms
### Humans
Medicine recognizes that altitudes above 1,500 metres (4,900 ft) start to affect humans,[13] and there is no record of humans living at extreme altitudes above 5,500-6,000 metres (18,000-19,700 ft) for more than two years.[14] As the altitude increases, atmospheric pressure decreases, which affects humans by reducing the partial pressure of oxygen.[15] The lack of oxygen above 2,400 metres (8,000 ft) can cause serious illnesses such as altitude sickness, high altitude pulmonary edema, and high altitude cerebral edema.[7] The higher the altitude, the more likely are serious effects.[7] The human body can adapt to high altitude by breathing faster, having a higher heart rate, and adjusting its blood chemistry.[16][17] It can take days or weeks to adapt to high altitude. However, above 8,000 metres (26,000 ft), (in the "death zone"), altitude acclimatization becomes impossible.[18]
There is a significantly lower overall mortality rate for permanent residents at higher altitudes.[19] Additionally, there is a dose response relationship between increasing elevation and decreasing obesity prevalence in the United States.[20] In addition, the recent hypothesis suggests that high altitude could be protective against Alzheimer's disease via action of erythropoietin, a hormone released by kidney in response to hypoxia.[21] However, people living at higher elevations have a statistically significant higher rate of suicide.[22] The cause for the increased suicide risk is unknown so far.[22]
#### Athletes
For athletes, high altitude produces two contradictory effects on performance. For explosive events (sprints up to 400 metres, long jump, triple jump) the reduction in atmospheric pressure signifies less atmospheric resistance, which generally results in improved athletic performance.[23] For endurance events (races of 5,000 metres or more) the predominant effect is the reduction in oxygen which generally reduces the athlete's performance at high altitude. Sports organizations acknowledge the effects of altitude on performance: the International Association of Athletic Federations (IAAF), for example, marks record performances achieved at an altitude greater than 1,000 metres (3,300 ft) with the letter "A".[24]
Athletes also can take advantage of altitude acclimatization to increase their performance. The same changes that help the body cope with high altitude increase performance back at sea level.[25][26] These changes are the basis of altitude training which forms an integral part of the training of athletes in a number of endurance sports including track and field, distance running, triathlon, cycling and swimming.
### Other organisms
Decreased oxygen availability and decreased temperature make life at high altitude challenging. Despite these environmental conditions, many species have been successfully adapted at high altitudes. Animals have developed physiological adaptations to enhance oxygen uptake and delivery to tissues which can be used to sustain metabolism. The strategies used by animals to adapt to high altitude depend on their morphology and phylogeny. For example, small mammals face the challenge of maintaining body heat in cold temperatures, due to their small volume to surface area ratio. As oxygen is used as a source of metabolic heat production, the hypobaric hypoxia at high altitudes is problematic.
There is also a general trend of smaller body sizes and lower species richness at high altitudes, likely due to lower oxygen partial pressures.[27] These factors may decrease productivity in high altitude habitats, meaning there will be less energy available for consumption, growth, and activity.[28]
However, some species, such as birds, thrive at high altitude.[29] Birds thrive because of physiological features that are advantageous for high-altitude flight.
## References
1. ^ a b c d Air Navigation. Department of the Air Force. 1 December 1989. AFM 51-40.
2. ^ a b Radiotelephony Manual. UK Civil Aviation Authority. 1 January 1995. ISBN 978-0-86039-601-7. CAP413.
3. ^ "Layers of the Atmosphere". JetStream, the National Weather Service Online Weather School. National Weather Service. Archived from the original on 19 December 2005. Retrieved 2005.
4. ^ Dr. S. Sanz Fernández de Córdoba (24 June 2004). "The 100 km Boundary for Astronautics". Fédération Aéronautique Internationale. Archived from the original on 9 August 2011.
5. ^ Webster's New World Medical Dictionary. Wiley. 2008. ISBN 978-0-470-18928-3.
6. ^ "An Altitude Tutorial". International Society for Mountain Medicine. Archived from the original on 19 July 2011. Retrieved 2011.
7. ^ a b c Cymerman, A; Rock, PB (1994). "Medical Problems in High Mountain Environments. A Handbook for Medical Officers". USARIEM-TN94-2. U.S. Army Research Inst. of Environmental Medicine Thermal and Mountain Medicine Division Technical Report. Cite journal requires `|journal=` (help)
8. ^ "Atmospheric pressure". NOVA Online Everest. Public Broadcasting Service. Archived from the original on 25 January 2009. Retrieved 2009.
9. ^ a b Goody, Richard M.; Walker, James C.G. (1972). "Atmospheric Temperatures" (PDF). Atmospheres. Prentice-Hall.
10. ^ "Dry Adiabatic Lapse Rate". tpub.com. Archived from the original on 3 June 2016. Retrieved 2016.
11. ^ Manual of the ICAO Standard Atmosphere (extended to 80 kilometres (262 500 feet)) (Third ed.). International Civil Aviation Organization. 1993. ISBN 978-92-9194-004-2. Doc 7488-CD.
12. ^ "The stratosphere: overview". UCAR. Retrieved 2016.
13. ^ "Non-Physician Altitude Tutorial". International Society for Mountain Medicine. Archived from the original on 23 December 2005. Retrieved 2005.
14. ^ West, JB (2002). "Highest permanent human habitation". High Altitude Medical Biology. 3 (4): 401-407. doi:10.1089/15270290260512882. PMID 12631426.
15. ^ Peacock, Andrew J (17 October 1998). "Oxygen at high altitude". British Medical Journal. 317 (7165): 1063-1066. doi:10.1136/bmj.317.7165.1063. PMC 1114067. PMID 9774298.
16. ^ Young, Andrew J.; Reeves, John T. (2002). "21". Human Adaptation to High Terrestrial Altitude. In: Medical Aspects of Harsh Environments. 2. Borden Institute, Washington, DC. Archived from the original on 11 January 2009.
17. ^ Muza, SR; Fulco, CS; Cymerman, A (2004). "Altitude Acclimatization Guide". U.S. Army Research Inst. Of Environmental Medicine Thermal and Mountain Medicine Division Technical Report (USARIEM-TN-04-05). Retrieved 2009.
18. ^ "Everest:The Death Zone". Nova. PBS. 24 February 1998.
19. ^ West, John B. (January 2011). "Exciting Times in the Study of Permanent Residents of High Altitude". High Altitude Medicine & Biology. 12 (1): 1. doi:10.1089/ham.2011.12101. PMID 21452955.
20. ^ Voss, JD; Masuoka, P; Webber, BJ; Scher, AI; Atkinson, RL (2013). "Association of Elevation, Urbanization and Ambient Temperature with Obesity Prevalence in the United States". International Journal of Obesity. 37 (10): 1407-1412. doi:10.1038/ijo.2013.5. PMID 23357956.
21. ^ Ismailov, RM (July-September 2013). "Erythropoietin and epidemiology of Alzheimer disease". Alzheimer Dis. Assoc. Disord. 27 (3): 204-6. doi:10.1097/WAD.0b013e31827b61b8. PMID 23314061.
22. ^ a b Brenner, Barry; Cheng, David; Clark, Sunday; Camargo, Carlos A., Jr (Spring 2011). "Positive Association between Altitude and Suicide in 2584 U.S. Counties". High Altitude Medicine & Biology. 12 (1): 31-5. doi:10.1089/ham.2010.1058. PMC 3114154. PMID 21214344.
23. ^ Ward-Smith, AJ (1983). "The influence of aerodynamic and biomechanical factors on long jump performance". Journal of Biomechanics. 16 (8): 655-658. doi:10.1016/0021-9290(83)90116-1. PMID 6643537.
24. ^ "IAAF World Indoor Lists 2012" (PDF). IAAF Statistics Office. 9 March 2012. Archived from the original (PDF) on 22 October 2013.
25. ^ Wehrlin, JP; Zuest, P; Hallén, J; Marti, B (June 2006). "Live high--train low for 24 days increases hemoglobin mass and red cell volume in elite endurance athletes". J. Appl. Physiol. 100 (6): 1938-45. doi:10.1152/japplphysiol.01284.2005. PMID 16497842.
26. ^ Gore, CJ; Clark, SA; Saunders, PU (September 2007). "Nonhematological mechanisms of improved sea-level performance after hypoxic exposure". Med Sci Sports Exerc. 39 (9): 1600-9. doi:10.1249/mss.0b013e3180de49d3. PMID 17805094.
27. ^ Jacobsen, Dean (24 September 2007). "Low oxygen pressure as a driving factor for the altitudinal decline in taxon richness of stream macroinvertebrates". Oecologia. 154 (4): 795-807. Bibcode:2008Oecol.154..795J. doi:10.1007/s00442-007-0877-x. PMID 17960424.
28. ^ Rasmussen, Joseph B.; Robinson, Michael D.; Hontela, Alice; Heath, Daniel D. (8 July 2011). "Metabolic traits of westslope cutthroat trout, introduced rainbow trout and their hybrids in an ecotonal hybrid zone along an elevation gradient". Biological Journal of the Linnean Society. 105: 56-72. doi:10.1111/j.1095-8312.2011.01768.x.
29. ^ McCracken, K. G.; Barger, CP; Bulgarella, M; Johnson, KP; et al. (October 2009). "Parallel evolution in the major haemoglobin genes of eight species of Andean waterfowl". Molecular Evolution. 18 (19): 3992-4005. doi:10.1111/j.1365-294X.2009.04352.x. PMID 19754505. | 4,256 | 17,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | longest | en | 0.905557 |
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• # Artykuł - szczegóły
## Open Mathematics
2012 | 10 | 6 | 1940-1943
## A new method of proof of Filippov’s theorem based on the viability theorem
EN
### Abstrakty
EN
Filippov’s theorem implies that, given an absolutely continuous function y: [t 0; T] → ℝd and a set-valued map F(t, x) measurable in t and l(t)-Lipschitz in x, for any initial condition x 0, there exists a solution x(·) to the differential inclusion x′(t) ∈ F(t, x(t)) starting from x 0 at the time t 0 and satisfying the estimation $$\left| {x(t) - y(t)} \right| \leqslant r(t) = \left| {x_0 - y(t_0 )} \right|e^{\int_{t_0 }^t {l(s)ds} } + \int_{t_0 }^t \gamma (s)e^{\int_s^t {l(\tau )d\tau } } ds,$$ where the function γ(·) is the estimation of dist(y′(t), F(t, y(t))) ≤ γ(t). Setting P(t) = {x ∈ ℝn: |x −y(t)| ≤ r(t)}, we may formulate the conclusion in Filippov’s theorem as x(t) ∈ P(t). We calculate the contingent derivative DP(t, x)(1) and verify the tangential condition F(t, x) ∩ DP(t, x)(1) ≠ ø. It allows to obtain Filippov’s theorem from a viability result for tubes.
EN
1940-1943
wydano
2012-12-01
online
2012-10-12
### Twórcy
autor
• Faculty of Mathematics and Computer Sciences, Nicholas Copernicus University, Chopina 12/18, 87-100, Toruń, Poland
autor
• Faculty of Mathematics and Computer Sciences, Nicholas Copernicus University, Chopina 12/18, 87-100, Toruń, Poland
### Bibliografia
• [1] Aubin J.-P., Cellina A., Differential Inclusions, Grundlehren Math. Wiss., 264, Springer, Berlin, 1984 http://dx.doi.org/10.1007/978-3-642-69512-4[Crossref]
• [2] Aubin J.-P., Frankowska H., Set-Valued Analysis, Systems Control Found. Appl., 2, Birkhäuser, Boston, 1990
• [3] Filippov A.F., Classical solutions of differential equations with multi-valued right-hand side, SIAM J. Control, 1967, 5(4), 609–621 http://dx.doi.org/10.1137/0305040[Crossref]
• [4] Frankowska H., Plaskacz S., Rzezuchowski T., Measurable viability theorems and the Hamilton-Jacobi-Bellman equation, J. Differential Equations, 1995, 116(2), 265–305 http://dx.doi.org/10.1006/jdeq.1995.1036[Crossref] | 806 | 2,274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-33 | latest | en | 0.433541 |
http://mathhelpboards.com/business-mathematics-18/maximizing-profit-utility-functions-23043-2.html | 1,516,305,769,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00631.warc.gz | 220,161,771 | 18,998 | # Thread: Maximizing profit and utility functions
1. Originally Posted by mrjericho1991
without constraint. Yes I know but tell me what values do I need to input for partial derivatives?
You need to equate the first partials to zero, and then solve the resulting system. We have:
$\displaystyle P(x,y)=100x+80y+2xy-x^2-2y^2-5000$
And so we compute the first partials, and equation them to zero to obtain the system:
$\displaystyle 100+2y-2x=0$
$\displaystyle 80+2x-4y=0$
So, what critical point do you get from solving the above linear system?
2. As a follow-up, I would begin by adding the two equations to eliminate $x$ and we obtain:
$\displaystyle 180-2y=0\implies y=90$
Now, substituting for $y$ into the first equation, there results:
$\displaystyle 100+2(90)-2x=0\implies x=140$
And so our critical point is:
$\displaystyle (x,y)=(140,90)$
Now, in order to use the second partials test for relative extrema, we need to compute:
$\displaystyle P_{xx}(x,y)=-2$
$\displaystyle P_{yy}(x,y)=-4$
$\displaystyle P_{xy}(x,y)=2$
And we find:
$\displaystyle D(x,y)=P_{xx}(x,y)P_{yy}(x,y)-\left(P_{xy}(x,y)\right)^2=8-4=4>0$
And so we conclude that the critical point is at the global maximum, which is:
$\displaystyle P(140,90)=5600$
For the second problem, we have the objective function:
$\displaystyle U(x,y)=xy$
Subject to the constraint:
$\displaystyle g(x,y)=5x+10y-100=0$
Using Lagrange multipliers, we obtain the system:
$\displaystyle y=\lambda(5)$
$\displaystyle x=\lambda(10)$
We find this implies:
$\displaystyle x=2y$
Substituting for $x$ into the constraint, we have:
$\displaystyle 5(2y)+10y-100=0\implies (x,y)=(10,5)$
We find:
$\displaystyle U(10,5)=50$
Picking another point on the constraint, such as $(12,4)$, we the find:
$\displaystyle U(12,4)=48<50$
And so we may conclude that:
$\displaystyle U_{\max}=50$
We also find that:
$\displaystyle \lambda=1$ | 588 | 1,910 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-05 | latest | en | 0.752315 |
https://www.coursehero.com/file/5948867/L05M/ | 1,516,104,058,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00642.warc.gz | 892,753,097 | 23,677 | # L05M - Determinants Consider the two linear equations for...
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Unformatted text preview: Determinants Consider the two linear equations for two unknowns x, y xa1 + ya1 = b1 1 2 xa2 + ya2 = b2 1 2 Do xa1a2 + ya1a2 = b1a2 12 22 2 xa2a1 + ya2a1 = b2a1 12 22 2 and subtract: x(a1a2 − a2a1) = b1a2 − b2a1 . 12 12 2 2 Do xa1a2 + ya1a2 = b1a2 11 21 1 xa2a1 + ya2a1 = b2a1 11 21 1 ( ∗x ) ( ∗) and subtract: (∗y ) y (a1a2 − a2a1) = b2a1 − b1a2 . 12 12 1 1 We see that (∗x), (∗y ) have a solution no matter what b1, b2 are if and only if the determinant a1 a1 def 1 2 1 2 = (a1a2 − a2a1) 12 a2 a2 1 2 of the Coefficient Matrix a1 a1 1 2 a2 a2 1 2 of (∗) is not zero. In that case we get b1a2−b2a1 x= 12 22 a1a2−a1a1 2 2 b2a1−b1a2 y= 11 21 a1a2−a1a1 2 2 2 ...
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L05M - Determinants Consider the two linear equations for...
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Ask a homework question - tutors are online | 497 | 1,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-05 | latest | en | 0.699182 |
http://au.metamath.org/mpegif/sspr.html | 1,534,558,344,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213264.47/warc/CC-MAIN-20180818021014-20180818041014-00653.warc.gz | 33,411,066 | 5,349 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sspr Structured version Unicode version
Theorem sspr 3964
Description: The subsets of a pair. (Contributed by NM, 16-Mar-2006.) (Proof shortened by Mario Carneiro, 2-Jul-2016.)
Assertion
Ref Expression
sspr
Proof of Theorem sspr
StepHypRef Expression
1 uncom 3493 . . . . 5
2 un0 3654 . . . . 5
31, 2eqtri 2458 . . . 4
43sseq2i 3375 . . 3
5 0ss 3658 . . . 4
65biantrur 494 . . 3
74, 6bitr3i 244 . 2
8 ssunpr 3963 . 2
9 uncom 3493 . . . . . 6
10 un0 3654 . . . . . 6
119, 10eqtri 2458 . . . . 5
1211eqeq2i 2448 . . . 4
1312orbi2i 507 . . 3
14 uncom 3493 . . . . . 6
15 un0 3654 . . . . . 6
1614, 15eqtri 2458 . . . . 5
1716eqeq2i 2448 . . . 4
183eqeq2i 2448 . . . 4
1917, 18orbi12i 509 . . 3
2013, 19orbi12i 509 . 2
217, 8, 203bitri 264 1
Colors of variables: wff set class Syntax hints: wb 178 wo 359 wa 360 wceq 1653 cun 3320 wss 3322 c0 3630 csn 3816 cpr 3817 This theorem is referenced by: sstp 3965 pwpr 4013 indistopon 17070 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1556 ax-5 1567 ax-17 1627 ax-9 1667 ax-8 1688 ax-6 1745 ax-7 1750 ax-11 1762 ax-12 1951 ax-ext 2419 This theorem depends on definitions: df-bi 179 df-or 361 df-an 362 df-tru 1329 df-ex 1552 df-nf 1555 df-sb 1660 df-clab 2425 df-cleq 2431 df-clel 2434 df-nfc 2563 df-ral 2712 df-v 2960 df-dif 3325 df-un 3327 df-in 3329 df-ss 3336 df-nul 3631 df-sn 3822 df-pr 3823
Copyright terms: Public domain W3C validator | 789 | 1,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-34 | latest | en | 0.124892 |
http://community.boredofstudies.org/12/mathematics/385075/halp.html | 1,542,581,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744750.80/warc/CC-MAIN-20181118221818-20181119003237-00027.warc.gz | 66,967,758 | 13,241 | 1. ## halp
(Cambridge Mathematics HSC 2U)
5C
(3) A particle is moving with acceleration function a = 8. Two seconds after time zero, it is stationary at the origin.
(a) Integrate to find the velocity function.
(b) Integrate again to find the displacement function.
so like I got (a), velocity is 8t-16, but idk how to get the constant for (b)
2. ## Re: halp
Originally Posted by sodiumziha
Two seconds after time zero, it is stationary at the origin.
This means that when $t=2;\,v=0,\,x=0$.
Your integral of velocity will be
$x =4t^2-16t+C$
Knowing that the particle is at the origin after two seconds, you can work out the constant.
3. ## Re: halp
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• | 242 | 880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-47 | latest | en | 0.858619 |
https://www.aft.com/support/product-tips/calculating-flow-through-an-orifice-using-asme-mfc-3m-1989-and-irrecoverable-pressure-drop-equations | 1,726,733,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651995.50/warc/CC-MAIN-20240919061514-20240919091514-00663.warc.gz | 572,220,655 | 28,040 | ## AFT Blog
Welcome to the Applied Flow Technology Blog where you will find the latest news and training on how to use AFT Fathom, AFT Arrow, AFT Impulse, AFT xStream and other AFT software products.
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# Calculating Flow Through an Orifice Using ASME MFC-3M-1989 and Irrecoverable Pressure Drop Equations
Recently, a customer calculated the mass flow rate of a fluid through an orifice using both the ASME standard MFC-3M-1989 and AFT Arrow. He was puzzled when this mass flow rate calculation differed by approximately 100 lbm/hr between the two methods, so he reached out to AFT for help in determining the reason for the discrepancy. The answer for the difference in flow rates on a basic level is that these two calculations are, fundamentally, not the same thing; ASME correlates differential pressure at pressure taps for purposes of flow measurement, while AFT Fathom and Arrow calculate the irrecoverable pressure drop across the orifice for purposes of system pressure loss.
ASME MFC-3M-1989 ORIFICE FLOWMETER
The ASME standard utilizes the differential pressure measured across pressure taps located very near the orifice in areas specified by the standard. In this region, the static pressure changes due to irrecoverable pressure losses from increased friction and turbulence in the flow, as well as pressure changes that are a result of the vena contracta and complex flow patterns as the fluid progresses through the orifice. It is crucial to note that this measured pressure differential is not equal to the irrecoverable pressure drop. The pressure profile particular to flow through a given orifice and its immediate upstream and downstream piping will vary based on several factors that, aside from computational fluid dynamics (CFD) modeling, can only be determined with experimental data. The ASME standard, then, is applicable only to systems that follow the pressure tap and orifice type, placement, and installation guidelines established by the standard. Figure 1 from the ASME standard (Reference 1) shows the configuration of an orifice with both D and D/2 taps and flange taps.
The ASME standard provides Equation 1 to calculate mass flow of a liquid through an ASME orifice flowmeter when the orifice flowmeter diameter is at the temperature of the flowing liquid:
Where:
EQUATION 1: Flow through ASME Orifice Flowmeter (Reference 1, page 10)
For D and D/2 taps, the equation for C (the discharge coefficient) to be used in Equation 1 is:
EQUATION 2: Discharge Coefficient for D and D/2 Pressure Taps (Reference 1, page 30)
To convert from hw to a true pressure drop, the following relationship between head and pressure is used:
IRRECOVERABLE PRESSURE DROP EQUATIONS FOR ORIFICE FLOW
AFT software calculates the irrecoverable pressure drop across an orifice using several methods. Perhaps the most popular method uses a K factor (Equations 3 and 4). The K factor is calculated using correlations from various hydraulic resistance sources. In the following example, the K factor was calculated to be 52.6 from I.E. Idelchik’s Handbook of Hydraulic Resistance (Reference 2, page 221). Crane Technical Paper 410 (Reference 3, page 4-5) was also used, and a K factor of 56.0 was calculated using this reference. After the K factor is calculated, it is used in the following equation to calculate head loss:
Where
EQUATION 3: Irrecoverable Pressure Drop in terms of Head Loss (Reference 3, page 2-7)
Converting head in Equation 3 into pressure using the following equation gives Equation 4.
EQUATION 4: Irrecoverable Pressure Drop in terms of Pressure
SAMPLE CALCULATIONS
To demonstrate the appropriate use of both of these calculation methods, an example calculation was created. The flow was specified at 100 gpm with a 1 inch orifice diameter and a 2.25 inch upstream pipe diameter. The differential pressure across pressure taps was calculated using the equation given in the ASME standard, and the irrecoverable pressure drop across the orifice was calculated using spreadsheet calculations. These calculations were confirmed in AFT Fathom. This principle applies to gas and steam flow, but an incompressible fluid (water at 50°F) was used here for simplicity as varying densities complicate these calculations.
*Note: the ASME standard provides correlations for three types of pressure taps (flange, D and D/2, and corner taps), but the difference in calculated differential pressure was within 0.2 psid for all three pressure tap types, so the correlation for D and D/2 taps was used.
The results of this analysis are shown below:
FIGURE 2: ASME Orifice Flowmeter Pressure Differential for 100 gpm Water Flow
(from Equation 1)
FIGURE 3: Irrecoverable Pressure Drop Across an Orifice with 100 gpm Water Flow
(from Equation 4)
As can be seen from the results shown in Figures 2 and 3, the pressure differential measured at the flowmeter pressure taps is different than the irrecoverable pressure drop.
The ASME standard does provide a correlation to relate the differential pressure at the orifice flowmeter to the irrecoverable pressure loss of the orifice flowmeter. This equation is shown in Equation 5.
EQUATION 5: Irrecoverable Pressure Drop Across an ASME Orifice Flowmeter
(Reference 1, page 32)
FIGURE 4: Irrecoverable Pressure Drop Across an Orifice using ASME Equation 5
This irrecoverable pressure drop predicted by the ASME standard for the orifice flowmeter (Figure 4) is very close to that calculated in Figure 3. This further illustrates that the ASME flowmeter differential pressure (Figure 2) is not the same thing as the irrecoverable pressure drop across an orifice (Figures 3 and 4). AFT software uses Equation 4 to calculate the irrecoverable pressure drop across an orifice.
CONCLUSION
When it comes to assessing flow rate through an orifice, ASME standard MFC-3M 1989 relates a pressure differential across pressure taps while AFT software calculates an irrecoverable pressure drop associated with the orifice. Therefore, while both of these methods ultimately calculate flow through an orifice, the equations associated with each were developed to be applied differently, and the user must be cautious to ensure that his/her input is applicable to the particular calculation method used.
REFERENCES
1. ASME MFC-3M-1989. “Measurement of Fluid Flow in Pipes Using Orifice, Nozzle, and Venturi”. Date of Issuance: January 31, 1990.
2. Idelchik, I.E. Handbook of Hydraulic Resistance. 3rd edition.
3. Crane Co. “Flow of Fluids through Valves, Fittings, and Pipe Technical Paper No. 410”. Reprinted 11/09.
#### Related Posts
Guest - Eric Elizondo on Thursday, 28 July 2016 07:43
So you are looking at a ~10 to 12% difference between a typical flow orifice calc. and arrow. Is there some factor on arrow that you can correct to get it closer to matching the typical orifice calculation.
So you are looking at a ~10 to 12% difference between a typical flow orifice calc. and arrow. Is there some factor on arrow that you can correct to get it closer to matching the typical orifice calculation.
on Thursday, 28 July 2016 09:39
Hi Eric and thank you for the question.
The main point here is that the dP you are measuring across the orifice IS NOT equal to the total, irrecoverable pressure loss. Therefore, there is no factor that can be applied to Arrow, because the dP you have measured is not relevant to a necessary input in Arrow.
To expand on this idea, think about the upstream and downstream pressures you measure in the field in accordance with the standard. This pressure is very near the orifice where there is a vena contracta and other physical phenomena occurring. If you measure a static pressure of 200 psia at this point and input this as the upstream pressure in Arrow, you are telling Arrow that the static pressure is 200 psia, but Arrow is associating this static pressure with the pipe diameter you have modeled, not the actual flow diameter (due to the vena contracta). This means that Arrow calculates a different velocity at this point than actually exists in your system, which, according to the fundamental equations, results in a different overall irrecoverable pressure drop.
Fortunately, you can easily correlate the pressure loss measured across the orifice according to the standard, to the standard's equation for overall irrecoverable pressure loss (this is Equation 5 in Figure 4 in the article).
This overall, irrecoverable pressure loss is what Arrow needs to accurately represent your system. Once you have calculated this overall irrecoverable pressure loss, you can model the orifice in Arrow to take this pressure drop, and a factor to get Arrow to match up with your field results will be unnecessary because Arrow will now match the hydraulics of your system (assuming your system is modeled accurately otherwise).
I hope this clears up your question, but please let me know if you have any follow-up questions. You can always call or email our support system, as well. Thanks Eric!
Hi Eric and thank you for the question. The main point here is that the dP you are measuring across the orifice IS NOT equal to the total, irrecoverable pressure loss. Therefore, there is no factor that can be applied to Arrow, because the dP you have measured is not relevant to a necessary input in Arrow. To expand on this idea, think about the upstream and downstream pressures you measure in the field in accordance with the standard. This pressure is very near the orifice where there is a vena contracta and other physical phenomena occurring. If you measure a static pressure of 200 psia at this point and input this as the upstream pressure in Arrow, you are telling Arrow that the static pressure is 200 psia, but Arrow is associating this static pressure with the pipe diameter you have modeled, not the actual flow diameter (due to the vena contracta). This means that Arrow calculates a different velocity at this point than actually exists in your system, which, according to the fundamental equations, results in a different overall irrecoverable pressure drop. Fortunately, you can easily correlate the pressure loss measured across the orifice according to the standard, to the standard's equation for overall irrecoverable pressure loss (this is Equation 5 in Figure 4 in the article). This overall, irrecoverable pressure loss is what Arrow needs to accurately represent your system. Once you have calculated this overall irrecoverable pressure loss, you can model the orifice in Arrow to take this pressure drop, and a factor to get Arrow to match up with your field results will be unnecessary because Arrow will now match the hydraulics of your system (assuming your system is modeled accurately otherwise). I hope this clears up your question, but please let me know if you have any follow-up questions. You can always call or email our support system, as well. Thanks Eric!
Guest - Jesus on Friday, 04 August 2017 18:59
Erin What standard(s) should I use to measure the uncertainty if i want to measure the flow with an orifice plate 2" in a 4" pipe line.
Erin What standard(s) should I use to measure the uncertainty if i want to measure the flow with an orifice plate 2" in a 4" pipe line.
on Monday, 07 August 2017 09:59
Hi Jesus,
A definitive answer to your question is outside the scope of this blog and the advice I can provide as AFT software is used in a vast range of industries that all may have different procedures/standards to determine the uncertainty in flow rate measurement. However, the more recent version of the standard referenced in this blog, ASME MFC-3M-2004, does provide some calculations for measuring the uncertainty of flow rate measurement on pages 9 and 10, including referring users to ASME MFC-2M and ISO/TR 5168 for more detailed information. There is a section entitled "Practical Computation of the Uncertainty" that will hopefully prove helpful to you.
Additionally, once you have determined the uncertainty, it is a good idea to model the system with the upper and lower possible flow rate values to observe how your system pressures and flows (elsewhere in the system, depending on how your system is configured) will be affected with these different flows. You may find that, even with the uncertainty in flow rate, the system requirements are still met. You may also find that more analysis and/or system modifications need to be done to ensure that your system will function as desired.
I hope this clears up your question, but please let me know if you have any follow-up questions. You can always call or email our support system, as well. Thanks Jesus!
Thursday, 19 September 2024
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Verify email * | 2,809 | 12,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.931252 |
https://socratic.org/questions/how-do-you-factor-the-expression-9x-2-65x-14 | 1,579,933,884,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251669967.70/warc/CC-MAIN-20200125041318-20200125070318-00549.warc.gz | 643,937,673 | 6,158 | # How do you factor the expression 9x^2 + 65x + 14?
Mar 26, 2016
(9x + 2)(x + 7)
#### Explanation:
Use the new systematic AC Method to factor trinomials (Google, Yahoo).
$y = 9 {x}^{2} + 65 x + 14 =$9(x + p)(x + q)
Converted trinomial: $y ' = {x}^{2} + 65 x + 126 =$ (x + p')(x + q').
p' and q' have same sign, because ac > 0.
Compose factor pairs of (ac = 126) --> (2, 63). This sum is 65 = b. Then p' = 2 and q' = 63.
Back to original trinomial y, $p = \frac{p '}{a} = \frac{2}{9}$ and $q = \frac{q '}{a} = \frac{63}{9} = 7$.
Factored form $y = 9 \left(x + \frac{2}{9}\right) \left(x + 7\right) = \left(9 x + 2\right) \left(x + 7\right)$ | 277 | 643 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-05 | latest | en | 0.597378 |
https://wiingy.com/resources/calculators/mile-to-kilometre/ | 1,680,040,511,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00774.warc.gz | 692,557,102 | 38,891 | # Convert Mile to kilometre
Conversion of Mile to Kilometre: Miles and kilometres are units used to measure the length and distance in the metric system. A stretch of 1.60934 kilometres makes a mile. This article will help you to convert miles to kilometres.
Miles
mi
Kilometers
km
## What is a Mile?
In the International System of Units (SI), a mile is a unit of linear measure equal to 1,760 yards, approximately 1.609 kilometres. A mile is a customary unit for the measurement of distance mainly in the United States. It’s denoted as ‘mi.’
Common Usage:
• Used to measure the distance travelled from one state to another.
• Used to measure the length of a road from one city to another.
• Used to measure the length of a river.
## What is Kilometre?
In the International System of Units (SI), kilometre is a unit of length that is equal to 1000 metres. It is denoted as ‘km.’
Common Usage:
• Used to measure the distance from one city to another city.
• Used to measure the distance of a runway.
• Used to measure the distance covered by a person in a marathon.
## How to convert Mile to Kilometre?
To convert mile to kilometre, multiply the given miles by 1.60934 to get the value in kilometre. As
1 mile = 1.60934 kilometre
The Formula to convert mile to kilometre,
X mile = X × 1.60934 kilometres
Where X is the value in miles.
## Examples of the conversion of Mile to Kilometre
Problem 1: Convert 6 miles to kilometres.
Solution 1:
Step 1: The given value of miles is 6.
Step 2: To convert miles to kilometre, substitute the given values at the required places in the conversion formula.
X miles = X × 1.60934 kilometre
Hence,
6 mi = 6 × 1.60934 km
= 9.65606 km
Therefore, 6 miles is 9.65606 kilometres.
Problem 2: Convert 360 miles to kilometres.
Solution 2:
Step 1: The given value of miles is 360.
Step 2: To convert miles to kilometre, substitute the given values at the required places in the conversion formula.
X mile = X× 1.60934 kilometre
Hence,
360 mi = 360 × 1.60934
= 579.364
Therefore, 360 miles is 579.364 kilometres.
Problem 3: Convert 450 miles to kilometres.
Solution 3:
Step 1: The given value of miles is 450.
Step 2: To convert miles to kilometres, substitute the given values at the required places in the conversion formula.
X mile = X × 1.60934 kilometre
Hence,
450 mi = 450 × 1.60934 km
= 724.205 km
Therefore, 450 miles is 724.205 kilometres.
Problem 4: Convert 80 miles to kilometres.
Solution 4:
Step 1: The given value of miles is 80.
Step 2: To convert miles to kilometres, substitute the given values at the required places in the conversion formula.
X mile = X × 1.60934 kilometre
Hence,
80 mi = 80 × 1.60934 km
= 128.748 km
Therefore, 80 miles is 128.748 kilometres.
Problem 5: Convert 77 miles to kilometres.
Solution 5:
Step 1: The given value of miles is 77.
Step 2: To convert miles to kilometres, substitute the given values at the required places in the conversion formula.
X mile = X × 1.60934 kilometre
Hence,
77 mi = 77 × 1.60934 km
= 123.919 km
Therefore, 77 miles is 123.919 kilometres.
## FAQs on the conversion ofMile to Kilometre
What is 1 km equal to in miles?
1 km is 0.621371 miles.
How many Kilometres Makes 1 mile?
1.60934 kilometres makes a mile.
What distance is 5 miles in km?
5 miles is 8.04672 kilometres.
Does 2 km equal 1 mile?
2 km is 1.24274 miles.
What is a mile?
In the International System of Units (SI), a mile is a unit of linear measure equal to 1,760 yards, approximately 1.609 kilometres. A mile is a customary unit for the measurement of distance mainly in the United States. It’s denoted as ‘mi.’
Convert 25 miles to km.
Step 1: The given value of miles is 25.
Step 2: To convert miles to kilometres, substitute the given values at the required places in the conversion formula.
X mile = X × 1.60934 kilometre
Hence,
25 mi = 25 × 1.60934 km
= 40.2336 km
Therefore, 25 miles is 40.2336 kilometres.
Convert 28 miles to km.
Step 1: The given value of miles is 28.
Step 2: To convert miles to kilometres, substitute the given values at the required places in the conversion formula.
X mile = X × 1.60934 kilometre
Hence,
28 mi = 28 × 1.60934 km
= 45.0616 km
Therefore, 28 miles is 45.0616 kilometres.
## Related conversion
We hope this article on the conversion of mile to kilometre is helpful to you. If you have any queries related to this post, ping us through the comment section below and we will get back to you as soon as possible.
References
The SI Base Units
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# Editorialist: Additional restrictions should be placed on drivers’ lic
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22 Dec 2016, 20:50
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Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others.
(C) Teenagers drive, on average, over twice as far each year as other drivers.
(D) Teenagers cause car accidents that are more serious than those caused by others.
(E) Teenagers are likely to drive with more passengers than the average driver.
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22 Dec 2016, 20:51
in other forums the OA is D
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23 Dec 2016, 23:08
YangYichen wrote:
Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others.
(C) Teenagers drive, on average, over twice as far each year as other drivers.
(D) Teenagers cause car accidents that are more serious than those caused by others.
(E) Teenagers are likely to drive with more passengers than the average driver.
only B and D comes close ...B is not a driving skill..i will also bet on D as well...
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25 Dec 2016, 01:24
Should differ weakening from not weakening
(A) Teenagers tend to drive older and less stable cars than other drivers (problem not in skills- Weakens)
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others (problem not in skills- Weakens)
(C) Teenagers drive, on average, over twice as far each year as other drivers (problem not in skills- Weakens)
(D) Teenagers cause car accidents that are more serious than those caused by others (can be skill problem - Not weakening)
(E) Teenagers are likely to drive with more passengers than the average driver (problem not in skills- Weakens)
D
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04 Jan 2017, 19:01
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Top Contributor
The OA is D. Let us discuss why -
Premise - teenagers make up 7% of registered drivers BUT are responsible for 14% of fatalities. (clearly the proportion of fatalities is twice that of number of drivers).
Conclusion - Teenagers lack basic driving skills.
Option A - Incorrect.
This provides an alternate explanation for 14% fatalities.
this suggests that the higher fatality rate is due to old and less stable cars, and probably not due to lack of basic driving skills.
Option B - Incorrect.
this option also provides an alternate explanation for the higher fatality rate.
this suggests that the higher fatality rate is due to the fact that they do not wear seat belts, and probably not due to lack of basic driving skills.
Option C - Incorrect.
this option also provides an alternate explanation for the higher fatality rate.
If they drive twice as far, the likelihood that teenagers and their co-passengers will be involved in a fatal accident also increases.
this answer option strengthens the argument that they lack basic driving skills.
Option E - Incorrect.
this option also provides an alternate explanation for the higher fatality rate.
'More passengers' suggests why the fatality rate is higher; If a teenager is involved in a fatal car accident, it is likely that more people would also be involved in that.
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04 Jan 2017, 19:47
OA is D. Can someone change the OA from B to D?
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17 Jan 2017, 17:26
"not wearing a seat belt" isn't that a basic driving skill?
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17 Jan 2017, 22:56
How is the answer here D?
As far as I understand, 4 answer choices "weaken" the idea that teens lack basic driving skills.
The only clear A/C that claims teens lack these basic skills is:
- If teens are less likely to use seat belts and shoulder straps
How about looking at other A/C
A) If teens drive older and less stable cars, how is this an indicator of their driving skills?
C) If teens drive further than older drivers, again, how does this indicate how good/bad their driving skills are?
E) If teens drive with more passengers, I don't see how this again could impact the driver's skills
The way I'm reading this, I think this is a very poorly worded question
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18 Jan 2017, 02:07
Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers. Weakens as fatalities could be due to older car and has nothing to do with driver's skill.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others. Weakens as this is not related to driving skills
(C) Teenagers drive, on average, over twice as far each year as other drivers. No impact as this is not relevant to driving skills
(D) Teenagers cause car accidents that are more serious than those caused by others. Strengthen. As teenage drivers lacks skills, their accidents are more severe in nature. so this is the answer
(E) Teenagers are likely to drive with more passengers than the average driver. No impact as this is not relevant to driving skills
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18 Jan 2017, 14:41
deenesh2309 wrote:
Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers. Weakens as fatalities could be due to older car and has nothing to do with driver's skill.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others. Weakens as this is not related to driving skills
(C) Teenagers drive, on average, over twice as far each year as other drivers. No impact as this is not relevant to driving skills
(D) Teenagers cause car accidents that are more serious than those caused by others. Strengthen. As teenage drivers lacks skills, their accidents are more severe in nature. so this is the answer
(E) Teenagers are likely to drive with more passengers than the average driver. No impact as this is not relevant to driving skills
Correct me if I'm wrong, but the way I'm reading this question is: "Find 4 A/C that Weaken the argument, and the A/C that does not do this is your answer". Technically, A/Cs that are irrelevant to driving skills (i.e. A, C, E) do not Weaken the idea that teens lack basic driving skills.
To clarify, the Q does not ask: tell me which A/C "Strengthens" the idea that teens lack basic driving skills. The Q asks "tells me which does not weaken". This can be done 2 ways: 1) by "strengthening" ; 2) and by not weakening
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18 Jan 2017, 16:56
(D) Teenagers cause car accidents that are more serious than those caused by others.
Since this answer choice does not have any relation witht the skills of the teeenager driving the car which was involved in the acccident ,This answer choice is he appropriate answer to the question asked.
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19 Jan 2017, 00:50
LakerFan24 wrote:
deenesh2309 wrote:
Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers. Weakens as fatalities could be due to older car and has nothing to do with driver's skill.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others. Weakens as this is not related to driving skills
(C) Teenagers drive, on average, over twice as far each year as other drivers. No impact as this is not relevant to driving skills
(D) Teenagers cause car accidents that are more serious than those caused by others. Strengthen. As teenage drivers lacks skills, their accidents are more severe in nature. so this is the answer
(E) Teenagers are likely to drive with more passengers than the average driver. No impact as this is not relevant to driving skills
Correct me if I'm wrong, but the way I'm reading this question is: "Find 4 A/C that Weaken the argument, and the A/C that does not do this is your answer". Technically, A/Cs that are irrelevant to driving skills (i.e. A, C, E) do not Weaken the idea that teens lack basic driving skills.
To clarify, the Q does not ask: tell me which A/C "Strengthens" the idea that teens lack basic driving skills. The Q asks "tells me which does not weaken". This can be done 2 ways: 1) by "strengthening" ; 2) and by not weakening
You are right. Other than D, the rest of the options does weakens the claim that teens lack driving skills. Only D does support/strengthen the argument and hence the correct answer.
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19 Jan 2017, 02:55
YangYichen wrote:
Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others.
(C) Teenagers drive, on average, over twice as far each year as other drivers.
(D) Teenagers cause car accidents that are more serious than those caused by others.
(E) Teenagers are likely to drive with more passengers than the average driver.
- Even though drivers of age nineteen and younger make up only 7 percent of registered drivers, they are responsible for over 14 percent of traffic fatalities.
We need to find that option which does not weaken "teenagers lack basic skills". Rest all options will weaken this. So they will give alternative explanation on why young drivers make up only 7% of registered drivers but are responsible for 14% fatalities.
(A) Teenagers tend to drive older and less stable cars than other drivers.
The car could be the reason. The teenagers' basic skills may not be a problem.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others.
Not taking safety precautions could be the reason. The teenagers' basic skills may not be a problem.
(C) Teenagers drive, on average, over twice as far each year as other drivers.
They drive more and hence are more prone to accidents. The teenagers' basic skills may not be a problem.
(D) Teenagers cause car accidents that are more serious than those caused by others.
Here the problem is teenagers' skills.
(E) Teenagers are likely to drive with more passengers than the average driver.
More people in the car is the reason for more fatalities. The teenagers basic skills may not be a problem.
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22 Jan 2017, 17:10
deenesh2309 wrote:
LakerFan24 wrote:
deenesh2309 wrote:
Editorialist: Additional restrictions should be placed on drivers’ licenses of teenagers because teenagers lack basic skills. Even though drivers of age nineteen and younger make up only 7 percent of registered driers, there are responsible for over 14 percent of traffic fatalities.
Each of the following, if true, weakens the argument that teenagers lack basic driving skill EXCEPT:
(A) Teenagers tend to drive older and less stable cars than other drivers. Weakens as fatalities could be due to older car and has nothing to do with driver's skill.
(B) Teenagers and their passengers are less likely to use seat belts and shoulder straps than others. Weakens as this is not related to driving skills
(C) Teenagers drive, on average, over twice as far each year as other drivers. No impact as this is not relevant to driving skills
(D) Teenagers cause car accidents that are more serious than those caused by others. Strengthen. As teenage drivers lacks skills, their accidents are more severe in nature. so this is the answer
(E) Teenagers are likely to drive with more passengers than the average driver. No impact as this is not relevant to driving skills
Correct me if I'm wrong, but the way I'm reading this question is: "Find 4 A/C that Weaken the argument, and the A/C that does not do this is your answer". Technically, A/Cs that are irrelevant to driving skills (i.e. A, C, E) do not Weaken the idea that teens lack basic driving skills.
To clarify, the Q does not ask: tell me which A/C "Strengthens" the idea that teens lack basic driving skills. The Q asks "tells me which does not weaken". This can be done 2 ways: 1) by "strengthening" ; 2) and by not weakening
You are right. Other than D, the rest of the options does weakens the claim that teens lack driving skills. Only D does support/strengthen the argument and hence the correct answer.
Maybe I'm reading this the wrong way, but I can't see how that makes sense.
Again, the Conclusion is that teens LACK driving skills. We do NOT want to weaken this (hence "Weaken EXCEPT". Therefore, there "should" be 4 A/C that "WEAKEN" the argument, and we want the A/C that does NOT do this. So, if I want to "Strengthen" this, I could say, "yeah, they don't put on their seatbelts (B), so yes they do lack basic driving skills...this is one of the first safety precautions/things you learn when you begin driving, or I could say "yeah, they lack driving skills b/c they cause really serious car accidents so they must not know what they're doing (D).
BUT the main problem I have w/ the A/C below is that I do not believe any of them WEAKEN the argument, which is where I'm getting confused
- (A) If you told me "Teens LACK driving skills" and then say "They drive older and less stable cars" I'd say "this sentence does not WEAKEN the idea that teens suck at driving b/c maybe they can't afford newer/more stable cars, but they may still be careful drivers".
- (C) If you told me "they drive a lot more than other drivers", I'd say "well yeah but again this doesn't weaken the idea that teens can't drive well b/c this tells me nothing about their driving skills sucking -- what if they're really careful on the road?",
- (E) If you told me "hey they have a lot of passengers", I'd say "again this doesn't WEAKEN the idea that teens cannot drive well -- maybe their passengers are not a distraction to the driver, maybe the driver has good driving skills still"
VeritasPrepKarishma, can you please explain where I'm going wrong here?
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22 Jan 2017, 21:26
1
LakerFan24 wrote:
Maybe I'm reading this the wrong way, but I can't see how that makes sense.
Again, the Conclusion is that teens LACK driving skills. We do NOT want to weaken this (hence "Weaken EXCEPT". Therefore, there "should" be 4 A/C that "WEAKEN" the argument, and we want the A/C that does NOT do this. So, if I want to "Strengthen" this, I could say, "yeah, they don't put on their seatbelts (B), so yes they do lack basic driving skills...this is one of the first safety precautions/things you learn when you begin driving, or I could say "yeah, they lack driving skills b/c they cause really serious car accidents so they must not know what they're doing (D).
BUT the main problem I have w/ the A/C below is that I do not believe any of them WEAKEN the argument, which is where I'm getting confused
- (A) If you told me "Teens LACK driving skills" and then say "They drive older and less stable cars" I'd say "this sentence does not WEAKEN the idea that teens suck at driving b/c maybe they can't afford newer/more stable cars, but they may still be careful drivers".
- (C) If you told me "they drive a lot more than other drivers", I'd say "well yeah but again this doesn't weaken the idea that teens can't drive well b/c this tells me nothing about their driving skills sucking -- what if they're really careful on the road?",
- (E) If you told me "hey they have a lot of passengers", I'd say "again this doesn't WEAKEN the idea that teens cannot drive well -- maybe their passengers are not a distraction to the driver, maybe the driver has good driving skills still"
VeritasPrepKarishma, can you please explain where I'm going wrong here?
The reason A and C weaken the argument is the data on which the conclusion is based.
Even though drivers of age nineteen and younger make up only 7 percent of registered drivers, they are responsible for over 14 percent of traffic fatalities.
This makes the author argue that teens lack basic skills.
But if he is told that teens drive old cars, could that explain that they are responsible for over 14% fatalities? Yes. Old and less stable cars are likely to lead to more fatalities if they are in an accident. So it weakens the author's claim that teens lack basic skills.
Same logic for (C).
(C) Teenagers drive, on average, over twice as far each year as other drivers.
If teenagers drive more, the chance of an accident increases. It is just about probabilities. If a person drives 20 km a day, the probability of an accident is higher than if the same person travels 10 km every day.
So it weakens the author's claim that teens lack basic skills. The extra drive could be the reason for the higher number of fatalities.
_________________
Karishma
Veritas Prep GMAT Instructor
Manager
Joined: 26 Dec 2015
Posts: 232
Location: United States (CA)
Concentration: Finance, Strategy
WE: Investment Banking (Venture Capital)
### Show Tags
22 Jan 2017, 22:18
VeritasPrepKarishma wrote:
LakerFan24 wrote:
Maybe I'm reading this the wrong way, but I can't see how that makes sense.
Again, the Conclusion is that teens LACK driving skills. We do NOT want to weaken this (hence "Weaken EXCEPT". Therefore, there "should" be 4 A/C that "WEAKEN" the argument, and we want the A/C that does NOT do this. So, if I want to "Strengthen" this, I could say, "yeah, they don't put on their seatbelts (B), so yes they do lack basic driving skills...this is one of the first safety precautions/things you learn when you begin driving, or I could say "yeah, they lack driving skills b/c they cause really serious car accidents so they must not know what they're doing (D).
BUT the main problem I have w/ the A/C below is that I do not believe any of them WEAKEN the argument, which is where I'm getting confused
- (A) If you told me "Teens LACK driving skills" and then say "They drive older and less stable cars" I'd say "this sentence does not WEAKEN the idea that teens suck at driving b/c maybe they can't afford newer/more stable cars, but they may still be careful drivers".
- (C) If you told me "they drive a lot more than other drivers", I'd say "well yeah but again this doesn't weaken the idea that teens can't drive well b/c this tells me nothing about their driving skills sucking -- what if they're really careful on the road?",
- (E) If you told me "hey they have a lot of passengers", I'd say "again this doesn't WEAKEN the idea that teens cannot drive well -- maybe their passengers are not a distraction to the driver, maybe the driver has good driving skills still"
VeritasPrepKarishma, can you please explain where I'm going wrong here?
The reason A and C weaken the argument is the data on which the conclusion is based.
Even though drivers of age nineteen and younger make up only 7 percent of registered drivers, they are responsible for over 14 percent of traffic fatalities.
This makes the author argue that teens lack basic skills.
But if he is told that teens drive old cars, could that explain that they are responsible for over 14% fatalities? Yes. Old and less stable cars are likely to lead to more fatalities if they are in an accident. So it weakens the author's claim that teens lack basic skills.
Same logic for (C).
(C) Teenagers drive, on average, over twice as far each year as other drivers.
If teenagers drive more, the chance of an accident increases. It is just about probabilities. If a person drives 20 km a day, the probability of an accident is higher than if the same person travels 10 km every day.
So it weakens the author's claim that teens lack basic skills. The extra drive could be the reason for the higher number of fatalities.
+1 Kudos. Thank you so much - makes complete sense now
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### Show Tags
19 Nov 2019, 23:32
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: Editorialist: Additional restrictions should be placed on drivers’ lic [#permalink] 19 Nov 2019, 23:32
Display posts from previous: Sort by | 5,894 | 24,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-51 | latest | en | 0.915732 |
https://music.stoyanstefanov.com/2017/01/16/table-of-intervals-part-1/ | 1,685,835,645,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00532.warc.gz | 433,815,638 | 13,212 | # Table of intervals: part 1
Hello and welcome to this 2-part blog post on intervals. I mean intervals such as “major third” and “augmented fifth”, things like this. Extremely important building block in Music Theory. You need intervals to transpose stuff for example. Or, much more importantly, to figure out how to make chords (triads or more complicated ones).
Eventually with practice you should have these intervals memorized but to get off the ground you need some help.
(psst, do you want to learn about the Circle of fifths too?)
## Tl;DR
Starting with the end in mind, here’s all you need in order to figure out any interval ever:
In this first post you’ll learn how to draw most of this table, in the second – how to finish and use it.
You should be able to draw this on your own so you don’t rely on referring to a piece of paper (that you lost last week, most likely) and also so that you memorize the intervals sooner. DIY FTW!
## Mother’s little helpers
You’ll need some aids. The first one is the C major scale. Pancakes all the way!
From C to shiny C:
Next, you need a keyboard. You know how to draw a keyboard, right? Rectangles for the white keys, then the black keys go in groups of 2, nothing, 3, nothing, 2, nothing, 3, nothing… Finally the white key next to the group of 2 (that look like Chopsticks) is C. The white key next to the cluster of 3 (that look like a Fork) is F. Chopsticks – C, Fork – F. Fill in the rest using your mad alphabet skills.
So there – these are the visual aids: a staff with C major and a keyboard:
## The table of intervals
Time for the table. It starts with two groups. The intervals belong to one of the two groups.
Each group gets two more columns: “#” for “number of lines and spaces on the staff” and “d” for distance in terms of semitones on the keyboard.
Now you need to put the numbers 1 through 8 in the # column. 1 goes first on top left.
2 is top right.
3 goes under 2.
4 goes to the left of 3.
5 is under 4.
6 is to the right of 5.
7 under 6.
And finally 8 is bottom left.
Did you see a pattern? It’s a snake!
Why 8 do you ask? 8 is the number of lines and spaces between C and C. More on that later.
Let’s add some names to those numbers, starting with the oddballs: 1 is called unison and 8 is called an octave.
Unison means “the same sound”. So the first C and the first C are the same, the interval between them is a unison. Why 1 and not 0? Because when you count the number of lines and spaces you start with the first note. You put your finger on the first note and say “one”. As you move up, you increment. Don’t worry, you’ll see a lot more of this later.
Next, let’s fill in the rest of the interval names in the left column (group 1). They are pretty boring: fourth and fifth.
Now the right column (group 2). It’s all-too-boring too. Second, third, sixth, seventh.
Back to the discussion above, let’s learn to count lines and spaces.
Put your finger on the first note, say “one” and as you move up you increment for each line or space. So C is 1 and it’s on a line. You move up – it’s a space. You increment 1 – it’s a 2. The note there is D.
Next one up is E. It’s 3. F – 4. And so on.
Now it’s time to populate the last column left – the one with distances in terms of semitones.
First is the unison. It’s the same C note. There are 0 semitones between the C and the same C. So put a 0.
Next comes the second interval called “second” (wow!).
The number of lines and spaces between C and D is 2, we already established this and put in the # column.
Now what about the distance in semitones? From C you go to C♯ which is one semitone, they you go to D which is another semitone. Two in total.
So since there are two semitones between C and D, you put 2 in the table.
Next, the third interval. Between C and E. (BTW, we use C major to figure out the table because it’s the easiest, but these intervals then apply to any scale or starting point.)
There are 3 lines/spaces when you start from C and count C = 1, D = 2, E = 3.
And there are 4 semitones between C and E (C♯, D, D♯, E).
The semitone distance from C to E is 4 and that’s the number that goes in the table.
Next number if 5. Why? Look at the keyboard above and count the semitones between C and F.
Next number is 7. Why? There are C♯, D, D♯, E, F, F♯, G (7 in total) semitones between C and G.
The distance from C to A is 9 semitones as you can see keyboard picture above.
To wrap it up – C to B is 11 semitones and C to C is 12.
“Group 1” and “group 2” aren’t particularly cool names. We need something better. Let’s put a P in the left group.
P stands for “perfect”. Let’s call the whole group the perfect group.
The second group is called the Major group.
Now let’s stop here for now and see how to use what we have so fat in the next post.
## Aside: Why the name Perfect? Cool story.
Imagine a string. Pluck it. Now another string that’s exactly half as long the first. Pluck. The twice-as-short string makes a very similar sound only higher (an octave above). If you play the two together, they sound good even to the pre-historic human. Perfect actually. (Today we call it Perfect 8th.)
Now the unison is two strings of the same length vibrating. They sound the same and they sound good together. Boo-boo the Conqueror of the Cave as well as Bach both will agree they sound perfect together. (Only Bach would say the two sounds are in a perfect unison.)
If you shorten the second string to be 1/3 of the length of the first, the two still sound good together. A perfect 5th. Ask all heavy metal and punk guitar players – this interval is just perfect for most songs. When you play C and G together with a distorted guitar it sounds awesome. C and G#? Not so much.
If you shorten the string by a 1/4 and you play both they still sound good, albeit borderline. Pushing the limit of what your ear would call perfect. This is the perfect forth. Anything else is not so perfect. Therefore it needs a different name – let’s go with Major.
## Guitar harmonics
Octave, fifth and fourth (the perfect) also happen to coincide (well, no coincidence really) with where the harmonics of a guitar string are easiest to produce. Harmonics is when you hold your finger but not press on the string above 12th, 7th and 5th fret. 12th fret (the octave) is the easiest and perfectiest, the 5th fret is borderline (the perfect fourth interval).
12th fret of the guitar is exactly the middle of the string, so no surprise.
## Pentatonic? ‘nother cool story.
Pentatonic scale is 5 (penta) notes before you reach an octave. For the longest time in human history it was the only scale before our ears became sophisticated enough to find a few more notes musical. All Chinese music, all Scottish, all blues, all everything for a long while just used a pentatonic.
Example, common in blues guitar – A C D E G. 5. You know what’s cool? You make up this scale only using perfect 5th intervals.
C to G. G to D, D to A, A to E. | 1,805 | 6,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-23 | longest | en | 0.930242 |
https://www.physicsforums.com/threads/fourier-series-of-a-function.531412/ | 1,623,865,097,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00144.warc.gz | 866,492,203 | 14,579 | # Fourier series of a function
I'm trying to find Fourier series for the following function:
$$f(x) = \begin{cases}1, & \mbox{if x \in (-\frac{\pi}{2}+2\pi n,\frac{\pi}{2}+2\pi n) } \\ -1, & \mbox{if x \in [\frac{\pi}{2}+2\pi n,\frac{3\pi}{2} + 2\pi n]} \end{cases}$$
This is how I calculated a_n and b_n:
So I got the following series: $$\sum_{n=1}^{\infty} \cos{nx}\frac{4(-1)^{n+1}}{\pi(2n-1)}$$
But when I checked if it converges to f(x) at point \pi I get that it diverges, however all requirements of Fourier theorem are met.
What am I doing wrong?
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LCKurtz
Homework Helper
Gold Member
I'm trying to find Fourier series for the following function:
$$f(x) = \begin{cases}1, & \mbox{if x \in (-\frac{\pi}{2}+2\pi n,\frac{\pi}{2}+2\pi n) } \\ -1, & \mbox{if x \in [\frac{\pi}{2}+2\pi n,\frac{3\pi}{2} + 2\pi n]} \end{cases}$$
This is how I calculated a_n and b_n:
So I got the following series: $$\sum_{n=1}^{\infty} \cos{nx}\frac{4(-1)^{n+1}}{\pi(2n-1)}$$
But when I checked if it converges to f(x) at point \pi I get that it diverges, however all requirements of Fourier theorem are met.
What am I doing wrong?
The cos(nx) term in the sum should be cos((2n-1)x) since you only pick up the odd subscripts. Otherwise it is correct. You could have saved half the work by noting that since the function is period 2pi, you could have used the formula on (-pi,pi) and observed the function is even. That's why the bn are 0. And using the half range integral for an would have made less work there too.
Thanks, indeed it saved me a lot of work! | 567 | 1,632 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-25 | latest | en | 0.844508 |
http://mathhelpforum.com/calculus/170930-calculating-definate-intregals.html | 1,480,779,693,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00224-ip-10-31-129-80.ec2.internal.warc.gz | 180,769,997 | 10,133 | 1. ## calculating definate intregals
Let .
Find
and Let .
Find
and
So I found the first one, which is just 0.
How do I find the seecond one? I don't get the logic of it...
2. $
\int_{-1}^{-3.5}(10f(x)-3)dx=-\int_{-3.5}^{-1}(10f(x)-3)dx=-10\int_{-3.5}^{-1}f(x)dx+3\int_{-3.5}^{-1}dx
$
3. wouldn't it just be -10 (0) + 3(0) which is just zero? but that answer is wrong...
4. $
\int_{-3.5}^{-1}dx=x|_{-3.5}^{-1}=(-1)-(-3.5)=2.5
$
5. Sorry I still don't get it. What is dx or x in this problem? If I wsa given that, I could figure it ou, I know how to be integration... | 233 | 573 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-50 | longest | en | 0.873665 |
https://searchoracle.techtarget.com/answer/Three-consecutive-months-in-fiscal-year | 1,619,144,852,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00408.warc.gz | 596,991,297 | 45,935 | Q
Problem solve Get help with specific problems with your technologies, process and projects.
# Three consecutive months in fiscal year
## I have a conveyance_claim table with claim date and type, one record per month. I have to check in financial year (April to March) if three continuous months have claims of the same type. How can I find it?
I have a conveyance_claim table with claim date and type (C=car, B=bike, etc.), one record per month. I have to check in financial year (April to March) if three continuous months have claims of the same type. How can I find it?
Fiscal years are traditionally named after the calendar year in which they end. So the year that runs from April 2005 to March 2006 is called Fiscal 2006 (even though most of it is in calendar 2005). However, all of this is a red herring.
We don't need to do anything special with fiscal months. We can refer to them by their ordinary calendar year and month numbers. All we need to do is find three consecutive ones in a row. This is accomplished with a three-table self-join.
```select t1.c_date as claimdate1
, t2.c_date as claimdate2
, t3.c_date as claimdate3
from conveyance_claim as t1
inner
join conveyance_claim as t2
on t1.c_type = t2.c_type
and year(t1.c_date)*12+month(t1.c_date)
= year(t2.c_date)*12+month(t2.c_date) - 1
inner
join conveyance_claim as t3
on t2.c_type = t3.c_type
and year(t2.c_date)*12+month(t2.c_date)
= year(t3.c_date)*12+month(t3.c_date) - 1```
The expressions which multiply the year by 12 and add the month are intended only to produce a sequential number, which we will not need to display, but which will allow us to subtract 1 from January 2005, for example, and be able to compare it successfully to December 2004.
To run this query so that it takes effect only within a certain fiscal year, use a condition like this:
``` where t1.c_date >= '2005-04-01'
and t1.c_date < '2006-02-01'```
The date range may be restricted to April to February, because after February a three-consecutive-month run cannot occur in that fiscal year.
Close | 537 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-17 | latest | en | 0.948614 |
https://studylib.net/doc/11619158/statistics-402---assignment-1-due-january-26--2011 | 1,620,682,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989749.3/warc/CC-MAIN-20210510204511-20210510234511-00263.warc.gz | 529,411,067 | 11,597 | # STATISTICS 402 - Assignment 1 Due January 26, 2011
advertisement
```STATISTICS 402 - Assignment 1
Due January 26, 2011
Farmer Jones wants to make an informed decision about which variety of corn to use. She decides
to run an experiment, a field trial, to compare two corn varieties, A and B. She has a field with 36
plots available for the experiment. On 18 plots she will plant variety A and on the other 18 plots
she will plant variety B. She wishes to see if the two varieties have equal yields, on average, or if
the two varieties differ significantly. How should she assign varieties to plots?
Convenience Assignment
It is easiest to plant one variety on 18 plots on one side of the field and the other variety
on the 18 plots on the other side. On the next page is a picture of such and assignment
and the yields, in bushels per acre, which would result for each plot. The summary of
those yields is given below
Variety
A
B
n
18
18
Sample mean
144.9
141.8
Sample std. dev.
8.29
7.65
Based on this assignment, by convenience, is there a statistically significant difference in
sample mean yields between the two corn varieties? Perform the appropriate test of
hypothesis.
Do you think that the result gives a fair comparison between the two varieties? Explain
briefly. Simply saying that it is not a random assignment is not a sufficient answer to this
question.
Systematic Assignment
Many people think that an alternating sequence is a random, or at least an unbiased,
sequence. On the next page is a picture of an alternating pattern, like a checkerboard, and
the yields, in bushels per acre, which would result for each plot. The summary of those
yields is given below.
Variety
A
B
n
18
18
Sample mean
142.3
144.5
Sample std. dev.
5.75
5.37
Based on this assignment, alternating, is there a statistically significant difference in
sample mean yields between the two corn varieties? Perform the appropriate test of
hypothesis.
Think about the results of the two tests of significance, the convenience assignment and
the alternating assignment. Some may find it a bit disturbing that B has a higher sample
mean in one study and A has a higher sample mean in the other. Of course, this could be
due to chance variation. It could also be due to a poor assignment of treatments to plots.
Random Assignment
How, physically, would you randomly assign varieties to plots? Come up with a
randomization scheme to assign variety A to 18 plots and variety B to the remaining 18
plots. Describe how to accomplish your randomization so that someone else can follow
your instructions and get a randomized assignment. Record your assignments in the grid.
1
Once you have completed your random assignment, consult “The Truth” – this sheet
gives the yield for each plot using either variety. “The Truth” was used to fill in the yields
for the plots in the convenience and alternating patterns you looked at earlier. In general,
“The Truth” is not available since it requires knowing what would happen to the same
plot of land using each of the treatments.
Write down the yields for your random assignment – if you have an A in the row 1,
column 1 plot then you would put down 130 whereas if you have a B in the row 1,
column 1 plot you would put down 118 for the yield. Repeat for all plots.
Based on this assignment, at random, is there a statistically significant difference in
sample mean yields between the two corn varieties? Perform a test of hypothesis.
Examine “The Truth” more closely. What is the relationship between the yields for the
two varieties?
A
130
A
149
A
141
A
150
A
139
A
155
Convenience Assignment
A
A
B
B
149 139 155 137
A
A
B
B
133 152 131 147
A
A
B
B
156 137 146 132
A
A
B
B
142 155 136 152
A
A
B
B
155 139 147 137
A
A
B
B
138 150 137 145
B
145
B
136
B
148
B
133
B
153
B
136
A
130
B
137
A
141
B
138
A
139
B
143
Systematic Assignment
B
A
B
A
137 139 155 149
A
B
A
B
133 140 143 147
B
A
B
A
144 137 146 144
A
B
A
B
142 143 148 152
B
A
B
A
143 139 147 149
A
B
A
B
138 138 149 145
B
145
A
148
B
148
A
145
B
153
A
148
Randomized Assignment
1
2
3
4
5
6
1
2
3
4
5
6
2
``` | 1,156 | 4,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-21 | latest | en | 0.936245 |
http://bonzoband.tk/tot-measure-ruler-sheets.html | 1,563,380,419,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183536-00071.warc.gz | 24,963,939 | 4,150 | # Tot measure ruler sheets
Ruler measure
## Tot measure ruler sheets
This tot is really handy when you need to measure something but don’ t have a ruler around. Sweet Pea is 6 years old. Measurement To The Nearest Inches. Turnitin provides instructors with the tools to prevent plagiarism engage students in the writing process, provide personalized feedback. Riddle Solutions Answer; Which letter of the alphabet has the most water? com is fun with amateur videos nipples, boobs videos, natural breasts, boobs bouncing, ruler boobs movies, hard nipples, amateur milf, tot female breasts nipples. Obituaries for the last 7 days on measure Your Life Moments. It is usually made from a solid piece of wood, but I.
Watchdog: What time of measure day is the same forwards, backwards , when written in a capital letters upside down? Δ v = V tot e * ln[ measure R]. pour découvrir une panoplie de contenus intéressants qui touchent la cuisine les voyages, la tot santé, les sorties, la mode la. The variables are the velocity change required by the tot mission ( Δ measure v the propulsion sheets system' s exhaust velocity ( V e), , delta- V) measure the spacecraft' s mass ratio measure ( R). The Child' s Estate.
C: What kind of dog keeps the best time? Math Worksheets provided by sheets Math- Aids. For More Dynamically Created Measurement Worksheets Go to Math- Aids. From: The addition working chart or finger chart is a commonly used Montessori tot tot material. Prerequisite: This purpose of this activity is to help with the memorization of addition measure facts.
This video playlist is measure over 40 minutes long and will show you some of the details that I think are important to getting this cabinet right. Remember the mass ratio is the spacecraft' s wet mass ( mass fully loaded with propellant) divided by the dry mass. Home Education, Volume 1 of the Charlotte Mason Series. Tot measure ruler sheets. History of the ruler Indians ( London: Edward & Charles Dilly, 1775). Please enjoy this nautical glossary of sailing terms. Some are ones that we sheets use in everyday language tot -. Search the history of over 349 billion web pages on the Internet. Visitez le ruler ruler nouveau site web de tot Salut Bonjour!
Showing top 8 worksheets sheets in the category - Measurement To tot The Nearest Inches. A Method Of Education II. Ready- to- print Measurement Standard Ruler Worksheet with answer sheets. ruler PLEASE be extremely careful with all power tools. com - Canada' s most comprehensive job search engine. Ulysses James Joyce- - I - - Stately bearing a bowl of lather on which a ruler mirror , measure plump Buck Mulligan came from sheets the measure stairhead a razor lay sheets crossed. Everything about fundamental spacecraft design revolves around the Tsiolkovsky rocket equation.
Find your dream job today! Preface Part 1 Some Preliminary measure Considerations I. Project Gutenberg' s The Essays of Montaigne Complete, by tot Michel de Montaigne This eBook is for the use of anyone anywhere at no cost ruler with almost ruler no restrictions whatsoever. Some of the worksheets displayed are Name measuring to the nearest half inch Measuring lengths tot to the nearest quarter inch, Measure each bar to the nearest quarter inch, Measure each bar to the nearest half sheets inch, Measuring to the nearest fourth inch Measuring with a ruler. Tot measure ruler sheets. Your child should already be very familiar with the concept of addition and be able ruler to write all of the numerals from zero to nine. There are some ruler templates ( in PDF PS) that can be printed out with paper can be disposable.
## Ruler sheets
Learning how to measure with a ruler is a skill kids will use throughout their whole lives! Build measurement skills with this practice page. Use the printable ruler ( not to scale) to help you measure each of these shoelaces. Search metadata Search text contents Search TV news captions Search archived web sites Advanced Search.
``tot measure ruler sheets``
This is a partial glossary of nautical terms; some remain current, while many date from the 17th to 19th centuries. See also Wiktionary' s nautical terms, Category: Nautical terms, and Nautical metaphors in English. See the Further reading section for additional words and references. | 908 | 4,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-30 | latest | en | 0.842275 |
http://perplexus.info/show.php?pid=3441&cid=26096 | 1,544,796,603,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825916.52/warc/CC-MAIN-20181214140721-20181214162221-00480.warc.gz | 219,852,039 | 4,389 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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Variation on a Classic (Posted on 2005-08-23)
In the classic problem you are given a triangle ABC with points D on AB, E on BC, and F on AC such that |AD|=2|DB|, |BE|=2|EC|, and |CF|=2|FA|. The lines AE, BF, and CD enclose a triangle inside triangle ABC. You are to find the area of this enclosed triangle relative to that of ABC. The answer is 1/7.
What if everything is the same except |BE|=|EC| and |CF|=3|FA|. What is the area of the enclosed triangle relative to that of ABC?
See The Solution Submitted by McWorter No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): According to Geometer's Sketchpad... (spoiler) | Comment 4 of 12 |
If D is on the extended line AB and F on the extended line CA, rather than on their respective segments (E has to be the midpoint of segment BC; the numbers don't work out otherwise), the new triangle FED has an area 25/12 that of triangle ABC.
However, if for example, D is on segment AB but F is again on the extended line CA, not on segment CA, the new triangle, DEF, is 16/15 the area of ABC.
Posted by Charlie on 2005-08-23 20:19:18
Search: Search body:
Forums (0) | 348 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-51 | latest | en | 0.88722 |
https://piersonforcongress.com/what-is-1-4-times-3/ | 1,638,843,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363332.1/warc/CC-MAIN-20211207014802-20211207044802-00137.warc.gz | 496,430,692 | 5,109 | ### Multiplying fractions
A portion is a part of a whole. In the critical lesson, friend learned just how to include and subtract fractions. Yet that’s not the only kind of mathematics you have the right to do with fractions. There space times when it will be helpful to main point fractions too.
You are watching: What is 1/4 times 3
Click through the slideshow come learn just how to write a multiplication difficulty with fractions.
Let's set up a multiplication instance with fractions. Suppose you drink 2/4 the a pot of coffee every morning.
But her doctor just told you that you need to cut down her coffee drinking by half.
Now you require to figure out how much 1/2 that 2/4 the a pot the coffee is.
This may not look prefer a multiplication problem. Yet when girlfriend see the word of with fractions, it method you must multiply.
To set up the example, we'll simply replace the word of v a multiplication sign.
Now our example is ready to be solved.
Unlike continual multiplication, which gives you a larger number...
Unlike regular multiplication, which gives you a larger number...multiplying fractions will usually offer you a smaller number.
So when we multiply 1/2 time 2/4...
So as soon as we main point 1/2 times 2/4...our answer will be smaller than 2/4.
Here's another example. Let's say you have actually 3/5 of a cup of chocolate filling.
You desire to put an equal amount of pour it until it is full in each of these 4 cupcakes.
You could say that you desire to put 1/4 that 3/5 of a cup of filling in each cupcake.
Just choose we walk before, we'll change the word of into a multiplication sign.
And currently our fountain are prepared to it is in multiplied.
Try This!
Try setup up the multiplication trouble below. Don't worry about solving that yet!
A cooking recipes calls because that 2/3 the a cup the milk. You desire to reduced the cooking recipes in half.
Note: back our example says the exactly answer is 2/3 x 1/2, remember, with multiplying bespeak does no matter. 1/2 x 2/3 would additionally be correct.
### Solving multiplication troubles with fractions
Now that we know exactly how to collection up multiplication difficulties with fractions, let's practice solving a few. If you feel comfortable multiplying totality numbers, you're prepared to main point fractions.
Click v slideshow come learn just how to multiply 2 fractions.
Let's main point to find 1/2 of 7/10.
Just prefer we walk earlier, we'll replace the word of through a multiplication sign. Now we're prepared to multiply.
First, we'll multiply the numerators: 1 and also 7.
1 time 7 amounts to 7, therefore we'll write 7 to the best of the numerators.
See more: What Do You Call A Pizza Maker ? What Do You Call A Pizza Maker
When we added fractions, the denominators continued to be the same. However when we multiply, the denominators gain multiplied too. | 658 | 2,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-49 | latest | en | 0.942221 |
http://www.smithsonianmag.com/smart-news/your-probability-of-dying-doubles-every-eight-years-180948228/ | 1,501,194,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549429548.55/warc/CC-MAIN-20170727222533-20170728002533-00177.warc.gz | 574,734,388 | 23,832 | # Your Probability of Dying Doubles Every Eight Years
## It’s a brand new year, and the chances that this one will be your last might be higher than you suspect
smithsonian.com
It's a brand new year, and the chances that this one will be your last might be higher than you suspect. Not surprisingly, the older we get, the greater our odds of dying in any given year. The math shows, in fact, that the likelihood that we kick the bucket before our next birthday doubles every eight years until that chance becomes a reality.
The blog Gravity and Levity explained this unsettling fact few years back:
This startling fact was first noticed by the British actuary Benjamin Gompertz in 1825 and is now called the “Gompertz Law of human mortality.” Your probability of dying during a given year doubles every 8 years. For me, a 25-year-old American, the probability of dying during the next year is a fairly minuscule 0.03% — about 1 in 3,000. When I’m 33 it will be about 1 in 1,500, when I’m 42 it will be about 1 in 750, and so on. By the time I reach age 100 (and I do plan on it) the probability of living to 101 will only be about 50%. This is seriously fast growth — my mortality rate is increasing exponentially with age.
Real-world data collected each year by the U.S. Centers for Disease Control follow the Gompertz Law almost perfectly, Gravity and Levity says. And it doesn't matter whether a country's average lifespan is on the longer or shorter side—the law still holds true. Assuming you live in the U.S. or a country with a similar average lifespan, you can use this simple calculator to estimate your chances of making it to your next birthday.
More from Smithsonian.com: | 398 | 1,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-30 | longest | en | 0.923883 |
https://www.quesba.com/questions/matthew-mason-took-three-year-12000-debenture-earning-7-per-annum-mason-s-m-1595535 | 1,702,224,463,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00805.warc.gz | 1,036,215,560 | 28,046 | # Matthew Mason took out a three-year \$12000 debenture earning 7% per annum. Mason’s marginal tax...
Matthew Mason took out a three-year \$12000 debenture earning 7% per annum. Mason’s marginal tax rate is 30%.
a How much interest will Mason receive: i per year? ii in total?
b How much of the interest earned will be payable in tax?
c Calculate Mason’s rate of return after tax.
d If inflation was measured at 4% p.a., what is the net gain achieved by Mason’s investment in real terms?
e State one advantage and one disadvantage of debentures as an investment for individuals
Jan 27 2022| 02:17 PM |
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• Need any last minute study tips? | 817 | 3,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.923029 |
https://www.gamedeveloper.com/design/randomness-serendipity-and-gamification | 1,723,204,672,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00331.warc.gz | 602,209,147 | 74,771 | # Randomness, Serendipity and GamificationRandomness, Serendipity and Gamification
Randomness can be fun and can lead to many serendipitous moments in games as well as the real world.
Andrzej Marczewski, Blogger
November 18, 2014
Recently I have been trying to write a few games, just for fun, with my daughter. There area couple of card games and I am trying to make a single player board game.
Making a game with a 7 year old is an interesting experience. I got asked to go into her room as she had a new game she wanted to play. She had made a game board with LEGO. You had to get from one end to the other, with certain bricks having modifier effects on the game (move faster, move slower etc). She said I could go first, so I asked “how do we decide how many spaces to move?”. I expected a dice or something, but no. Her answer was “just pick a number between one and ten”!
I was baffled, but did as I was told. I chose the biggest ten, assuming she would do the same, as this was the quickest way to win. However, she chose seven – she genuinely randomly chose a number in her head!
It was at this point I began to realise just how important randomness is to enjoying games. I couldn’t just choose random numbers, it made no sense to me as I could win by just choosing the biggest or the “right number”. Of course, I let her win, but still! If you have my sort of brain, where you like logic, you analyse the situation and work out the best solution – i.e. how to get to the win state fastest. Now, if there had been a dice added, that would change the dynamic. Rather than being able to choose the best route, you would have to work out based on the random number you had been assigned by the dice.
This got me thinking about randomness in general and how it is or could be applied to normally predictable things. So I built a random story builder. It has about 15 books in it from Project Gutenberg and randomly chooses paragraphs from the books and puts them on the screen. Of course most of the outcome is rubbish, but every now and then you get a great combination of a Grimm fairy tale, Dantes Divine Comedy and Frankenstein that makes you chuckle. It has no practical use (except as a more interesting version of Lorem Ipsum), but it is a bit of random fun.
Another example of this sort of fun is Rory’s Story Cubes. These are a series of dice with icons on them. You role the dice and have to build a story using the icons to inspire you. I play this with my daughter all the time. I also use them to try and inspire new ideas when my mind is drawing a blank. The random nature of the dice means that you have to remove all of your pre programmed ideas of structure and go with the flow.
Randomness, Serendipity and Gamification
Steve Jobs, it turns out, was a great believer in serendipity. This is the idea that random happenings can lead to great and unexpected outcomes. He built the offices at Pixar studios with central toilets, ensuring that people from all over the company would randomly run into each other. Marissa Mayer famously banned working from home, citing (among other things) that incidental meetings around the office lead to new insights that would not happen otherwise. On the site, I have a link called “Page Roulette” which will just take you to a random page. You may like it, you may not – but it is fun to find out just because it is a journey into the unknown. I love the “random article” button on Wikipedia for this exact reason!
In gamification, we are no strangers to randomness, I have spoken about random reward schedules (and their potential dangers) in the past. But this is generally relegated to just surprises than changing something that is usually totally predictable into something that isn’t.
When planning your user journey, think of ways to take people out of the predictable linear from time to time. Give Free Spirit Explorers something to look forward to! When building your systems, think how you can create moments of serendipity, where ideas and people may randomly collide to create wonderful new innovations.
Originially posted on my Gamified UK Blog - http://www.gamified.co.uk/2014/11/10/randomness-serendipity-gamification/
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You May Also Like | 958 | 4,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-33 | latest | en | 0.987582 |
https://numbermatics.com/n/21977305/ | 1,586,281,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00534.warc.gz | 622,820,456 | 6,262 | # 21977305
## 21,977,305 is an odd composite number composed of four prime numbers multiplied together.
What does the number 21977305 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 24 divisors.
21977305 is an odd composite number. It is composed of four distinct prime numbers multiplied together. It has a total of twenty-four divisors.
## Prime factorization of 21977305:
### 5 × 7 × 232 × 1187
(5 × 7 × 23 × 23 × 1187)
See below for interesting mathematical facts about the number 21977305 from the Numbermatics database.
### Names of 21977305
• Cardinal: 21977305 can be written as Twenty-one million, nine hundred seventy-seven thousand, three hundred five.
### Scientific notation
• Scientific notation: 2.1977305 × 107
### Factors of 21977305
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 1222
### Divisors of 21977305
• Number of divisors d(n): 24
• Complete list of divisors:
• Sum of all divisors σ(n): 31534272
• Sum of proper divisors (its aliquot sum) s(n): 9556967
• 21977305 is a deficient number, because the sum of its proper divisors (9556967) is less than itself. Its deficiency is 12420338
### Bases of 21977305
• Binary: 10100111101011000110110012
• Base-36: D31SP
### Squares and roots of 21977305
• 21977305 squared (219773052) is 483001935063025
• 21977305 cubed (219773053) is 10615080842470294647625
• The square root of 21977305 is 4687.9958404419
• The cube root of 21977305 is 280.1075479713
### Scales and comparisons
How big is 21977305?
• 21,977,305 seconds is equal to 36 weeks, 2 days, 8 hours, 48 minutes, 25 seconds.
• To count from 1 to 21,977,305 would take you about two weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 21977305 cubic inches would be around 23.3 feet tall.
### Recreational maths with 21977305
• 21977305 backwards is 50377912
• The number of decimal digits it has is: 8
• The sum of 21977305's digits is 34
• More coming soon!
HTML: To link to this page, just copy and paste the link below into your blog, web page or email.
BBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below: | 741 | 2,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-16 | latest | en | 0.841923 |
https://www.coursehero.com/file/6124071/Calc07-3day2/ | 1,498,270,447,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320209.66/warc/CC-MAIN-20170624013626-20170624033626-00306.warc.gz | 834,782,463 | 163,241 | Calc07_3day2
# Calc07_3day2 - 7.3 day 2 Disk and Washer Methods Limerick...
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Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2003 7.3 day 2 Disk and Washer Methods Limerick Nuclear Generating Station, Pottstown, Pennsylvania
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y x = Suppose I start with this curve. My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape. So I put a piece of wood in a lathe and turn it to a shape to match the curve.
y x = How could we find the volume of the cone? One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes. The volume of each flat cylinder (disk) is: 2 the thickness r π In this case: r= the y value of the function thickness = a small change in x = dx ( 29 2 x dx
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y x = The volume of each flat cylinder (disk) is: 2 the thickness r π If we add the volumes, we get: ( 29 2 4 0 x dx 4 0 x dx = 4 2 0 2 x = 8 = ( 29 2 x dx
This application of the method of slicing is called the disk method . The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk.
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## This note was uploaded on 02/12/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.
### Page1 / 10
Calc07_3day2 - 7.3 day 2 Disk and Washer Methods Limerick...
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https://simplywall.st/stocks/pl/capital-goods/wse-ptn/poltronic-shares/news/calculating-the-fair-value-of-poltronic-sa-wseptn | 1,718,507,682,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00294.warc.gz | 474,231,821 | 44,341 | Stock Analysis
# Calculating The Fair Value Of Poltronic S.A. (WSE:PTN)
Today we'll do a simple run through of a valuation method used to estimate the attractiveness of Poltronic S.A. (WSE:PTN) as an investment opportunity by taking the forecast future cash flows of the company and discounting them back to today's value. The Discounted Cash Flow (DCF) model is the tool we will apply to do this. Models like these may appear beyond the comprehension of a lay person, but they're fairly easy to follow.
Remember though, that there are many ways to estimate a company's value, and a DCF is just one method. Anyone interested in learning a bit more about intrinsic value should have a read of the Simply Wall St analysis model.
View our latest analysis for Poltronic
### The calculation
We have to calculate the value of Poltronic slightly differently to other stocks because it is a trade distributors company. Instead of using free cash flows, which are hard to estimate and often not reported by analysts in this industry, dividends per share (DPS) payments are used. Unless a company pays out the majority of its FCF as a dividend, this method will typically underestimate the value of the stock. The 'Gordon Growth Model' is used, which simply assumes that dividend payments will continue to increase at a sustainable growth rate forever. For a number of reasons a very conservative growth rate is used that cannot exceed that of a company's Gross Domestic Product (GDP). In this case we used the 5-year average of the 10-year government bond yield (2.5%). The expected dividend per share is then discounted to today's value at a cost of equity of 10%. Compared to the current share price of zł1.1, the company appears about fair value at a 13% discount to where the stock price trades currently. Remember though, that this is just an approximate valuation, and like any complex formula - garbage in, garbage out.
Value Per Share = Expected Dividend Per Share / (Discount Rate - Perpetual Growth Rate)
= zł0.1 / (10% – 2.5%)
= zł1.3
### Important assumptions
We would point out that the most important inputs to a discounted cash flow are the discount rate and of course the actual cash flows. You don't have to agree with these inputs, I recommend redoing the calculations yourself and playing with them. The DCF also does not consider the possible cyclicality of an industry, or a company's future capital requirements, so it does not give a full picture of a company's potential performance. Given that we are looking at Poltronic as potential shareholders, the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC) which accounts for debt. In this calculation we've used 10%, which is based on a levered beta of 1.521. Beta is a measure of a stock's volatility, compared to the market as a whole. We get our beta from the industry average beta of globally comparable companies, with an imposed limit between 0.8 and 2.0, which is a reasonable range for a stable business.
Valuation is only one side of the coin in terms of building your investment thesis, and it is only one of many factors that you need to assess for a company. The DCF model is not a perfect stock valuation tool. Rather it should be seen as a guide to "what assumptions need to be true for this stock to be under/overvalued?" For instance, if the terminal value growth rate is adjusted slightly, it can dramatically alter the overall result. For Poltronic, there are three further elements you should assess:
1. Risks: Case in point, we've spotted 5 warning signs for Poltronic you should be aware of, and 2 of them make us uncomfortable.
2. Other Solid Businesses: Low debt, high returns on equity and good past performance are fundamental to a strong business. Why not explore our interactive list of stocks with solid business fundamentals to see if there are other companies you may not have considered!
3. Other Top Analyst Picks: Interested to see what the analysts are thinking? Take a look at our interactive list of analysts' top stock picks to find out what they feel might have an attractive future outlook!
PS. Simply Wall St updates its DCF calculation for every Polish stock every day, so if you want to find the intrinsic value of any other stock just search here.
### Valuation is complex, but we're helping make it simple.
Find out whether Poltronic is potentially over or undervalued by checking out our comprehensive analysis, which includes fair value estimates, risks and warnings, dividends, insider transactions and financial health.
View the Free Analysis
Have feedback on this article? Concerned about the content? Get in touch with us directly. Alternatively, email editorial-team (at) simplywallst.com.
This article by Simply Wall St is general in nature. We provide commentary based on historical data and analyst forecasts only using an unbiased methodology and our articles are not intended to be financial advice. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. We aim to bring you long-term focused analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material. Simply Wall St has no position in any stocks mentioned. | 1,134 | 5,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.930706 |
https://myengineerings.com/sheet-pile-retaining-walls/ | 1,695,752,564,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510219.5/warc/CC-MAIN-20230926175325-20230926205325-00879.warc.gz | 454,546,703 | 24,407 | # sheet pile retaining walls
Sheet pile retaining walls are widely used for waterfront construction and consist of interlocking members that are driven into place. Individual sheet piles come in many different sizes and shapes. Sheet piles have an interlocking joint that enables the individual segments to be connected together to form a solid wall.
Many different types of design methods are used for sheet pile walls. below Figure shows the most common type of design method. In below Fig. , the term H represents the unsupported face of the sheet pile wall. this sheet pile wall is being used as a waterfront retaining structure and the level of water in front of the wall is at the same elevation as the groundwater table elevation behind the wall. For highly permeable soil, such as clean sand and gravel, this often occurs because the water can quickly flow underneath the wall in order to equalize the water levels.
, the termD represents that portion of the sheet pile wall that is anchored in soil. Also shown in above Fig. is a force designated as Ap. This represents a restraining force on the sheet pile wall due to the construction of a tieback, such as by using a rod that has a grouted end or is attached to an anchor block. Tieback anchors are often used in sheet pile wall construction in order to reduce the bending moments in the sheet pile. When tieback anchors are used, the sheet pile wall is typically referred to as an anchored bulkhead, while if no tiebacks are utilized, the wall is called a cantilevered sheet pile wall.
Sheet pile walls tend to be relatively flexible. Thus, , the design is based on active and passive earth pressures. For this analysis, a unit length (1 m or 1 ft) of sheet pile wall is assumed. The soil behind the wall is assumed to exert an active earth pressure on the sheet pile wall. At the groundwater table (Point A), the active earth pressure is equal to kA yd1, where kA=active earth pressure coefficient from Eq. (the friction between the sheet pile wall and the soil is usually neglected in the design analysis), yt= total unit weight of the soil above the groundwater table, and d1= depth from the ground surface to the groundwater table. At Point B in the figure Fig. 6.38, the active earth pressure equals kA ytd1 + kA yb d2, where yb= buoyant unit weight of the soil below the groundwater table and d2 depth from the groundwater table to the bottom of the sheet pile wall. For a sheet pile wall having assumed values of H and D , and using the calculated values of active earth pressure at Points A and B, the active earth pressure resultant force (PA), in kN per linear m of wall or lb per linear foot of wall, can be calculated
The soil in front of the wall is assumed to exert a passive earth pressure on the sheet pile wall. The passive earth pressure at Point C in the above Figure. is equal to kp bD, where the passive earth pressure coefficient (kp) can be calculated from Eq. . Similar to the analysis of cantilever retaining walls, if it is desirable to limit the amount of sheet pile wall translation, then a reduction factor can be applied to the passive pressure. Once the allowable passive pressure is known at Point C, the passive resultant force (Pp) can be readily calculated. As an alternative solution for
the passive pressure, Eq. can be used to calculate Pp with the buoyant unit
References:
• BUILDING DESIGN AND CONSTRUCTION HANDBOOK for Frederick S. Merritt & Jonathan T. Ricketts
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 768 | 3,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.934746 |
https://www.onlinemath4all.com/how-to-find-the-product-of-a-number.html | 1,716,063,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00196.warc.gz | 834,596,131 | 7,927 | # HOW TO FIND THE PRODUCT OF A NUMBER
In Math, you can find product of a number by multiplying it by another number.
For example, to find the product of a number 7, you have to multiply 7 by another number, say 4.
7 x 4 = 28
So, 28 is the product of 7 and 4.
The meaning of the word 'product' in Math is a number you get to by multiplying two or more numbers together.
To get a product of two or more numbers, we have to multiply them.
For example, to get a product of 9 and 4, multiply them.
9 x 4 = 36
So, the product of 9 and 4 is 36.
## Product of 1 and a Number
The product of 1 and any number is always equal to the number itself.
Examples :
1 x 12 = 12
25 x 1 = 25
1.5 x 1 = 1.5
1 x 0.87 = 0.87
## Product of Zero and a Number
The product of zero and any number is always equal to zero.
Examples :
0 x 14 = 0
5 x 0 = 0
2.6 x 0 = 0
0 x 0.98 = 0
## Product of 10 and a number
The product of 10 and any other whole number will always end in zero.
Examples :
10 x 4 = 40
13 x 10 = 130
152 x 10 = 1520
## Product of 5 and a Number
The product of 5 and any other whole number will always end in 5 or zero.
Examples :
5 x 4 = 20
7 x 5 = 35
5 x 12 = 60
## Product of a Decimal and 10
To get the product of a decimal and 10, move the decimal point one digit to the right.
Examples :
1.789 x 10 = 17.89
10 x 0.6 = 6
0.005 x 10 = 0.05
## Product of a Decimal and 100
To get the product of a decimal and 100, move the decimal point two digits to the right.
Examples :
1.789 x 100 = 178.9
100 x 0.75 = 75
0.0005 x 100 = 0.05
## Product of a Decimal and 1000
To get the product of a decimal and 1000, move the decimal point three digits to the right.
Examples :
1.7895 x 1000 = 1789.5
1000 x 0.075 = 75
0.0005 x 1000 = 0.5
## Product of a Number and Multiples of 10
Consider the following product of a number and a multiple of 10.
4.8 x 20
To get the product of 4.8 and 20, write 20 as a product of 2 and 10.
= 4.8 x 2 x 10
Multiply 4.8 and 2.
= 9.6 x 10
Since 9.6 is multiplied 10, the decimal point has to be moved 1 digit to the right.
= 96
So, the product of 4.8 and 20 is 96.
## Product of Any Number and an Even Number
The product of any whole number and an even number is always an even number.
Examples :
7 x 12 = 94
6 x 9 = 54
25 x 4 = 100
## Product of an Odd Number and an Odd Number
The product of an odd number and an odd number is always an odd number.
Examples :
7 x 5 = 35
11 x 3 = 33
7 x 13 = 91
## Solved Problems
Problem 1 :
Write three products of 7.
Solution :
7 x 2 = 14
7 x 3 = 21
7 x 4 = 28
Three products of 7 are 14, 21 and 28.
Problem 2 :
Find the product of 135 and 1.
Solution :
The product of any number and 1 is always equal to the number itself.
So, the product of 135 and 1 is 135.
Problem 3 :
Find the product of 0.785 and 0.
Solution :
The product of any number and zero is always equal to zero.
So, the product of 0.785 and 0 is 0.
Problem 4 :
Find the digit at ones place in the product of 225 and 17.
Solution :
The product of a number ends with 5 and any number will always end with 5 or zero.
Multiply the digit at ones place in both the numbers.
5 x 7 = 35
35 ends with 5.
Therefore, the digit at ones place in the product of 225 and 17 is 5.
Problem 5 :
Find the digit at ones place in the product of 174 and 30.
Solution :
The product of any number and a multiple of 10 will always end with zero.
Therefore, the digit at ones place in the product of 174 and 30 is 0.
Problem 6 :
Find the product of 0.2897 and 1000.
Solution :
Since 0.2897 is multiplied by 1000, the decimal point has to be shifted to the right by three places.
Therefore,
0.2897 x 1000 = 289.7
Problem 7 :
Will the digit at ones place in the product of 25 and 36 an even number or odd number?
Solution :
In the given two numbers 25 and 36, one of the numbers is even, that is 36.
The product of any number and an even number is always an even number.
Therefore, the digit at ones place in the product of 25 and 36 will be an even number.
Problem 8 :
Will the digit at ones place in the product of 17 and 53 an even number or odd number?
Solution :
Both 17 and 53 are odd numbers.
The product of an odd number and an odd number is always an odd number.
Therefore, the digit at ones place in the product of 17 and 53 will be an odd number.
Problem 9 :
Find the product of 0.2 and 400.
Solution :
To get the product of 0.2 and 400, write 400 as a product of 4 and 100.
= 0.2 x 4 x 100
Multiply 0.2 and 4.
= 0.8 x 100
Since 0.8 is multiplied 100, the decimal point has to be shifted to the right by two places.
= 80
So, the product of 0.2 and 400 is 80.
Problem 10 :
Find the product of 0.25 and 8000.
Solution :
To get the product of 0.25 and 8000, write 8000 as a product of 8 and 1000.
= 0.25 x 8 x 1000
Multiply 0.25 and 8.
= 2 x 1000
= 2000
So, the product of 0.25 and 8000 is 2000.
## Practice Questions
Kindly mail your feedback to v4formath@gmail.com
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The Mean Value Theorem Worksheet | 1,669 | 5,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-22 | latest | en | 0.852226 |
https://m.hanspub.org/journal/paper/23791 | 1,611,809,560,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00072.warc.gz | 434,247,813 | 15,874 | 灰色神经网络组合模型在能源需求中的应用
# 灰色神经网络组合模型在能源需求中的应用Application of the Gray Neural Network Combination Model in Forecasting Energy Demand of China
Abstract: Energy is the cornerstone of social development. With the development of economy, the demand for energy is increasing year by year. So it is very important to forecast the energy demand with high precision. In this paper, based on GM(1, 1) model and BP neural network model, a tandem type grey neural network model is established to forecast the national total energy demand using the resident consumption level index, the total energy production, the energy industry investment, the gross domestic product, the total energy available for consumption, and the total population at year-end as the influence indexes. The actual data of total energy consumption in 2000~2011 were selected as modeling data, and those in 2012~2016 were selected as testing data. Comparative analysis of GM(1, 1) and gray neural network prediction accuracy was made. The results show that the grey neural network model has the highest prediction accuracy and the mean variance of the prediction error is 0.5840, which is less than the mean variance of the single GM(1, 1) prediction error.
1. 引言
2. 研究方法
2.1. GM(1, 1)预测模型
1982年，邓聚龙教授首次提出了灰色理论(Grey system) [7] ，该理论对原始数作累加生成的的近似指数规律再进行建模的方法。灰色模型有严格的理论基础，最大的优点是所需数据少，一般只需要4个数据就能对未来数据进行预测，为此该理论被广泛应用于各个领域。能源需求量既有已知的信息，也有未确定的信息，可看做一个灰色系统。GM(1, 1)灰色模型 [8] 建立如下：
${x}^{\left(1\right)}\left(k\right)=\underset{i=1}{\overset{k}{\sum }}{x}^{\left(0\right)}\left(i\right)\left(k=1,2,\cdots ,n\right)$ . (1)
${Z}^{\left(1\right)}$${X}^{\left(1\right)}$ 的紧邻均值生成序列： ${Z}^{\left(1\right)}=\left({z}^{\left(1\right)}\left(2\right),{z}^{\left(1\right)}\left(3\right),\cdots ,{z}^{\left(1\right)}\left(n\right)\right)$ ，其中：
${z}^{\left(1\right)}\left(k\right)=0.5\left({x}^{\left(1\right)}\left(k\right)+0.5{x}^{\left(1\right)}\left(k-1\right)\right),\text{\hspace{0.17em}}k=2,3,\cdots ,n$ . (2)
${x}^{\left(0\right)}\left(k\right)+a{z}^{\left(1\right)}\left(k\right)=b,\text{\hspace{0.17em}}k=2,3,\cdots ,n$ . (3)
$u={\left[a,b\right]}^{\text{T}}$ 为参数列，且有
$B=\left[\begin{array}{cc}-{z}^{\left(1\right)}\left(2\right)& 1\\ -{z}^{\left(1\right)}\left(3\right)& 1\\ ⋮& ⋮\\ -{z}^{\left(1\right)}\left(n\right)& 1\end{array}\right]$ , $Y=\left[\begin{array}{l}{x}^{\left(0\right)}\left(2\right)\\ {x}^{\left(0\right)}\left(3\right)\\ ⋮\\ {x}^{\left(0\right)}\left(n\right)\end{array}\right]$(4)
$\stackrel{^}{u}={\left({B}^{\text{T}}B\right)}^{-1}{B}^{\text{T}}Y$ (5)
$B,Y,\stackrel{^}{u}$ 满足式(4)和式(5)，相应的白化微分方程为
$\frac{\text{d}{x}^{\left(1\right)}}{\text{d}t}+a{x}^{\left(1\right)}\left(t\right)=b$ . (6)
GM(1, 1)模型的时间序列为：
${\stackrel{^}{x}}^{\left(1\right)}\left(k+1\right)=\left({x}^{\left(0\right)}\left(1\right)-\frac{b}{a}\right){\text{e}}^{-ak}+\frac{b}{a},\text{\hspace{0.17em}}k=1,2,\cdots ,n$ (7)
${\stackrel{^}{x}}^{\left(1\right)}\left(k+1\right)={\stackrel{^}{x}}^{\left(1\right)}\left(k+1\right)-{\stackrel{^}{x}}^{\left(1\right)}\left(k\right)=\left(1-{\text{e}}^{a}\right)\left({\stackrel{^}{x}}^{\left(0\right)}\left(1\right)-\frac{b}{a}\right){\text{e}}^{-ak}$ (8)
$\epsilon \left(k\right)=\frac{{x}^{\left(0\right)}\left(k\right)-{\stackrel{^}{x}}^{\left(0\right)}\left(k\right)}{{x}^{\left(0\right)}\left(k\right)}$ (9)
2.2. BP神经网络模型
1985年，美国加州大学的鲁梅尔哈特和麦克莱兰等人提出BP神经网络 [9] ，神经网络已广泛应用于各个领域。从结构上看，该模型具有输入层，隐含层和输出层，是一种典型的多层前向型神经网络。该网络广泛应用于图像处理、模式识别、语言处理、预测分析等领域。
1) 神经元模型
$Y=f\left(wp+b\right)$ (10)
2) 激励函数
$f\left(x\right)=\frac{1}{1+{\text{e}}^{-cx}}$ (11)
3) BP神经网络的学习算法
BP神经网络的误差反向传播算法基本思想是对一定数量的样本对输入和期望输出进行学习，将输入数据传送至各个神经元，经隐含层和输出层计算后，输出层输出对应的预测值。如果预测值与实际值误差较大时，输出层进行反向传播该误差，进行权值和阈值的调整，不断修正误差，直至满足精度要求。其算法核心的思想是：权值和阈值的调整总是沿着误差函数最快的方向，即负梯度方向。
${W}_{1}={\left({w}_{ij}^{1}\right)}_{l×m},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\cdots ,l,\text{\hspace{0.17em}}j=1,2,\cdots ,m$ (12)
${W}_{2}={\left({w}_{ij}^{2}\right)}_{n×l},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\cdots ,n,\text{\hspace{0.17em}}j=1,2,\cdots ,l$ (13)
${O}_{j}=f\left(\underset{i=1}{\overset{m}{\sum }}{w}_{ji}^{2}{x}_{i}-{\theta }_{l}^{1}\right)=f\left(ne{t}_{j}\right),\text{\hspace{0.17em}}j=1,2,\cdots ,l$ (14)
${z}_{k}=g\left(\underset{j=1}{\overset{l}{\sum }}{w}_{kj}^{2}{O}_{j}-{\theta }_{k}^{2}\right)=g\left(ne{t}_{k}\right),\text{\hspace{0.17em}}k=1,2,\cdots ,n$ (15)
$E=\frac{1}{2}\underset{k=1}{\overset{n}{\sum }}{\left({y}_{k}-{z}_{k}\right)}^{2}$ (16)
$\Delta {w}_{ji}^{1}=-\eta \frac{\partial E}{\partial {w}_{ji}^{1}}$ , $\Delta {w}_{kj}^{2}=-\eta \frac{\partial E}{\partial {w}_{kj}^{2}}$ (17)
${R}^{2}=\frac{{\left(l\underset{i=0}{\overset{l}{\sum }}{\stackrel{^}{y}}_{i}{y}_{i}-\underset{i=0}{\overset{l}{\sum }}{\stackrel{^}{y}}_{i}\underset{i=0}{\overset{l}{\sum }}{y}_{i}\right)}^{2}}{\left(l\underset{i=0}{\overset{l}{\sum }}{\stackrel{^}{y}}_{i}^{2}-{\left(\underset{i=0}{\overset{l}{\sum }}{\stackrel{^}{y}}_{i}^{2}\right)}^{2}\right)\left(l\underset{i=0}{\overset{l}{\sum }}{y}_{i}^{2}-{\left(\underset{i=0}{\overset{l}{\sum }}{y}_{i}^{2}\right)}^{2}\right)}$ , (18)
2.3. 组合灰色神经网络模型
$\delta =\sqrt{\underset{k=1}{\overset{n}{\sum }}{\left({x}^{\left(0\right)}\left(k\right)-{\stackrel{^}{x}}^{\left(0\right)}\left(k\right)\right)}^{2}/n}$ (19)
3. 能源需求实例分析
3.1. 数据收集
Figure 1. Gray network structure diagram
3.2. 灰色GM(1, 1)预测模型
${\stackrel{^}{x}}^{\left(1\right)}\left(k+1\right)=2048563.152{\text{e}}^{0.0825k}-1901599.152,\text{\hspace{0.17em}}k=1,2,\cdots ,n$
Table 1. Energy demand indicators
Table 2. Comparison of forecast and actual value of total energy demand in 2012~2016 based on different models
3.3. 组合灰色神经网络预测模型
1) 模型的输入、输出参数的选择
2) 模型的结构设计
3) 模型网络参数的选取及参数设定
3.4. 模型精度比较
4. 结论
Figure 2. Gray neural network topology
Figure 3. Energy demand forecast based on different models versus real value chart
Figure 4. Comparison chart of relative error between predicted and actual value of energy demand based on different models
1) 由于全国能源需求总量具有波动性和非线性的特点，建立了灰色GM(1, 1)预测模型和灰色神经网络模型，并以2000~2011年的能源需求量基础数据作为训练样本，以2012~2016年的数据为检验样本，经计算预测值与真实值的误差均在13%以内，表明建立的模型是有效的。
2) 以全国能源需求为例分别利用GM(1, 1)和灰色神经网络对能源需求进行预测，其预测误差的均方差分别为：12.3462，0.5840，可以看出灰色神经网络对能源需求总量预测的精度最高。
3) 灰色神经网络模型具有更好的预测效果，能够弥补单一模型的不足。
[1] 周扬, 吴文祥, 胡莹, 等. 基于组合模型的能源需求预测[J]. 中国人口•资源与环境, 2010, 20(4): 63-68.
[2] 索瑞霞, 王福林. 组合预测模型在能源消费预测中的应用[J]. 数学的实践与认识, 2010, 40(18): 80-85.
[3] 张帆. 基于GM-GRNN的能源需求预测模型研究[J]. 湖北汽车工业学院学报, 2012, 26(3): 65-69.
[4] 崔吉峰, 乞建勋, 杨尚东. 基于粒子群改进BP神经网络的组合预测模型及其应用[J]. 中南大学学报(自然科学版), 2009, 40(1): 190-194.
[5] 赵荣. 云南省某州(市)能源预测模型的研究[D]: [硕士学位论文]. 昆明: 昆明理工大学, 2017.
[6] 虞亚平. 京津冀能源需求预测分析及发展对策研究[D]: [硕士学位论文]. 天津: 天津理工大学, 2017.
[7] 邓聚龙. 灰色系统基本方法[M]. 武汉: 华中工学院出版社, 1987: 96-108.
[8] 李剑波, 鲜学福. 基于灰色神经网络模型的重庆能源需求预测[J]. 西南大学学报(自然科学版), 2016, 38(6): 136-141.
[9] 李柏年, 吴礼斌. MATLAB数据分析方法[M]. 北京: 机械工业出版社, 2012: 170-172.
[10] 中国省统计局. 中国统计年鉴(2017) [M]. 北京: 中国统计出版社, 2017.
Top | 3,369 | 6,963 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 72, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-04 | latest | en | 0.637281 |
https://cpep.org/mathematics/2324937-a-rabbit-can-run-35-miles-per-hour-a-fox-can-run-21-miles-in-half-an-h.html | 1,631,887,497,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055645.75/warc/CC-MAIN-20210917120628-20210917150628-00211.warc.gz | 235,345,185 | 7,537 | 19 July, 08:53
# A rabbit can run 35 miles per hour. A fox can run 21 miles in half an hour. Which animal is faster and how much
0
1. 19 July, 10:44
0
Step-by-step explanation:
The fox is faster. Because he is a fox.
Also because he can (assuming he doesn't fatigue or get caught by hounds in the second half of the hour) run 42 miles in 1 hour.
The rabbit runs 35 miler per hour.
Velocity = Distance/time
Fox velocity = 21 miles/0.5 hour
Dividing by 0.5 is the same as multiplying by 2, so ...
Fox velocity = 42 miles/hour | 157 | 532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-39 | latest | en | 0.929118 |
https://blog.csdn.net/hetianbiao/article/details/52350566 | 1,532,121,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591831.57/warc/CC-MAIN-20180720193850-20180720213850-00283.warc.gz | 599,376,435 | 10,191 | # Problem 1 Times 2
function y = times2(x)
% Modify the line below so that the output y is twice the incoming value x
y = x*2;
% After you fix the code, press the "Submit" button, and you're on your way.
end
# Problem 2 Make the vector [1 2 3 4 5 6 7 8 9 10]
function x = oneToTen
x = [1:1:10];
end
# Problem 3. Find the sum of all the numbers of the input vector
function y = vecsum(x)
y = sum(x);
end
# Problem 4. Make a checkerboard matrix
function a = checkerboard(n)
a = ones(n);
a(1:2:n, 2:2:n) = 0;
a(2:2:n, 1:2:n) = 0;
end
# Problem 5. Triangle Numbers
function t = triangle(n)
t = sum([1:n]);
end
# Problem 6. Select every other element of a vector
function y = everyOther(x)
y = x(1:2:length(x));
end
# Problem 7. Column Removal
function B = column_removal(A,n)
A(:,n) = [] ;
B = A;
end
# Problem 8. Add two numbers
function c = add_two_numbers(a,b)
c = a+b;
end
# Problem 9. Who Has the Most Change?
function b = most_change(a)
price = [0.25; 0.05; 0.1; 0.01];
c = a*price;
b= find(c==max(c));
end
ps: 每次找index这种题都不是很顺利
# Problem 10. Determine whether a vector is monotonically increasing
function tf = mono_increase(x)
tf = true;
for i = 1:length(x)-1
result = x(i+1)>x(i);
if isequal(result, false)
tf = false;
end
end
end
08-31 219
08-24 1775
#### computer network 5th problem solutions
2018年06月08日 172KB 下载
#### CSAPP 答案-Instructor's Manual of CSAPP
2009年12月17日 231KB 下载
#### Cody's data cleaning techniques using SAS 2nd Edition
2009年08月19日 1.81MB 下载
#### COMPUTER NETWORKS
2009年09月13日 1.4MB 下载
#### computer network 5th andre problem solutions-CSDN下载
2018年03月14日 152KB 下载
#### Computer_Network 5th & Computer_Network 5th andre problem solutions
2017年11月14日 6.86MB 下载
#### computer network 5th andre problem solutions-CSDN下载.2018_03_16
2018年03月16日 152KB 下载
#### computer network 5th andre problem solutions
2012年11月22日 170KB 下载 | 661 | 1,882 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-30 | latest | en | 0.300723 |
http://modb.oce.ulg.ac.be/backup/toros/dea98/node23.html | 1,542,133,072,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00317.warc.gz | 231,649,583 | 2,743 | Next: Spatial smoothing Up: The nested models Previous: Open sea boundary conditions
# The interpolation scheme
In order to interpolate data from the coarse grid to the fine one in the best possible way, a "conservative" interpolation formula has been developped.
We chose a local second order scheme using three contiguous coarse data points to infer data in the fine cells corresponding to the center coarse cell. The parameters of the parabolic function have been chosen in order to insure that the mean value of the interpolated data on the fine cells corresponding to a coarse cell is equal to the data in this coarse cell. Imposing this condition on three successive coarse points yields the value of parameters for the parabolic interpolation formula.
If we denote the quadratic interpolation, we have :
And the conservation of mean value conditions on three successive cells give, according to the notations of figure 4.2 :
Figure 4.2: Notations for the interpolation formula
Introducing the expression of the interpolation formula in the conservation conditions yields the value of the interpolation parameters a, b and c :
We can see that those parameters are similar to those we would have obtained had we simply interpolated the values of the coarse data on three successive points except that the independant term c differs by a second derivative-like quantity. This means the "simple" quadratic interpolation curve is smoothed somewhat with the introduction of our mean conserving conditions.
Supposing we have 2N+1 fine cells corresponding to one coarse cell, the fine cell interpolated values are given by :
We then can check that the mean value on a coarse cell is conserved :
Figure 4.3 shows the result of this interpolation formula on four different Fourrier modes on the extent of one coarse cell. Frequencies of the presented modes are lower or higher than one coarse cell's size.
Figure 4.3: Results of the interpolation ; stars indicate the interpolated fine values and + signs mark the corresponding coarse cell value.
Additionally, an interpolation normally to the boundaries must be performed in the case of scalar variables. Effectively, due to the choice of grid type, the scalar points at the boundary of the nested model don't match with scalar points of the coarse model. We have than decided to perform a linear interpolation normally to the boundary when necessary. This interpolation scheme is represented at figure 4.4.
Figure 4.4: Linear interpolation normally to the boundaries
Next: Spatial smoothing Up: The nested models Previous: Open sea boundary conditions
Wed Jun 10 10:53:44 DFT 1998 | 514 | 2,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-47 | latest | en | 0.857602 |
https://topprowellness.com/louisxvii-terlampir-boleh-ngasal-tpi-jwbnny-diapus/ | 1,660,533,973,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00403.warc.gz | 519,079,132 | 15,715 | # @LouisXVII, terlampir .., boleh ngasal (tpi jwbnny diapus),
@LouisXVII
terlampir ..
boleh ngasal (tpi jwbnny diapus)
cos 50 × cos 543( 40 × 40 × 40 )
= ( 40 × 40 × 40 )cos 50 × cos 543
= ( 1.600 × 40 )cos 50 × cos 543
= 64.000 cos 50 × cos 543
PLEASE HELP TOMORROW KO NA PO ITO KAILANGAN :(( 1. What is the sum of the interior angles of a triangle? A. 180° B.... | 156 | 371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-33 | latest | en | 0.346114 |
https://yufeizhao.com/blog/ | 1,679,537,925,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00267.warc.gz | 1,173,587,192 | 43,691 | # Blog
## Schildkraut: Equiangular lines and large multiplicity of fixed second eigenvalue
March 6, 2023
I am excited to share and discuss the latest work by an MIT undergraduate senior, Carl Schildkraut, titled Equiangular lines and large multiplicity of fixed second eigenvalue. In his paper, Carl offers new constructions on two related problems:
• equiangular lines,
• graphs with high second eigenvalue multiplicity.
### Equiangular lines
In my previous work on equiangular lines (with Zilin Jiang, Jonathan Tidor, Yuan Yao, and Shengtong Zhang) (previously reported on this blog and MIT News), we solved the following problem:
Problem (Equiangular lines with a fixed angle). For a given fixed angle, in sufficiently high dimensions, determine the maximum possible number of lines that can be pairwise separated by the given angle.
The answer can be phrased in terms of the spectral radius order $k(\lambda)$, defined to be the smallest integer $k$ so that there exists a $k$-vertex graph $G$ whose adjacency matrix has top eigenvalue exactly $\lambda$. Furthermore, we define $k(\lambda) = \infty$ if no such graph $G$ exists.
Theorem (Jiang–Tidor–Yao–Zhang–Zhao). For $\alpha \in (0,1)$. Let $\lambda = (1-\alpha)/(2\alpha)$ and let $k(\lambda)$ be its spectral radius order. Then the maximum number $N_\alpha(d)$ of equiangular lines in $\mathbb{R}^d$ with common angle $\arccos \alpha$ satisfies:
(a) If $k(\lambda) < \infty$, then $N_\alpha(d) = \lfloor k(d-1) / (k-1) \rfloor$ for all sufficiently large $d$;
(b) If $k(\lambda) = \infty$, then $N_\alpha(d) = d + o(d)$ as $d \to \infty$.
In particular, this result determines the first order growth rate of $N_\alpha(d)$ for all $\alpha$. Even better, when the spectral radius order is finite, our result determines the answer exactly in all sufficiently large dimensions. (There is a tantalizing open question on how high the dimension needs to be.) However, when the spectral radius order is infinite, our result gives only an asymptotic $d + o(d)$. Can we hope for a more precise answer?
Jiang and Polyanskii conjectured that $N_\alpha(d) = d + O_\alpha(1)$ when $k(\lambda) = \infty$, i.e., the $o(d)$ error term should be at most constant depending on $\lambda$. They verified this conjecture when $\lambda$ is either (a) not a totally real algebraic integer or (b) a totally real algebraic integer that is not the largest among its Galois conjugates. We also reiterated this conjecture in our paper.
Carl Schildkraut’s new paper gives a counterexample to this conjecture.
Theorem (Schildkraut). There exist infinitely many $\alpha$ such that $N_{\alpha}(d) \ge d + \Omega_\alpha(\log \log d)$.
I found this result to be quite surprising. I had expected the opposite to be true.
### Second eigenvalue multiplicity
One of the key ingredients in our equiangular lines paper is a bound on the second eigenvalue multiplicity.
Theorem (Jiang-Tidor-Yao-Zhang-Zhao). The adjacency matrix of every connected $n$-vertex graph with maximum degree $\Delta$ has second eigenvalue multiplicity at most $O_\Delta(n/\log \log n)$.
The fact that the second eigenvalue multiplicity is sublinear in the number of vertices plays a crucial role in our solution to the equiangular lines with a fixed angle problem.
The second eigenvalue multiplicity bound leaves many open questions.
How tight is the bound? The quantitative bound is the bottleneck being able to improve the “sufficiently large dimension” part of the equiangular lines result. The $O_\Delta(n/\log\log n)$ bound is still best known (though, see the work of McKenzie–Rasmussen–Srivastava for an improvement to $O_\Delta(n/(\log n)^c)$ in the case of regular graphs). See my work with Milan Haiman, Carl Schildkraut, Shengtong Zhang (previously reported on this blog) for the current best lower bound constructions, giving $\Omega(\sqrt{n /\log n})$.
Here is a question relevant to the equiangular lines problem with a fixed angle (appearing as Question 6.4 in our equiangular lines paper):
Question. Fix $\Delta, \lambda >0$. What is the maximum multiplicity that $\lambda$ as the second eigenvalue of a connected $n$-vertex graph with maximum degree at most $\Delta$?
We suggested the possibility that perhaps the answer is always at most a constant depending on $\Delta$ and $\lambda$. I really thought that the answer should be bounded. Carl’s paper shows otherwise.
Theorem (Schildkraut). For infinitely many values of $\lambda$ and $d$, there exist sequences of connected $d$-regular $n$-vertex graphs whose second eigenvalue is exactly $\lambda$ and has multiplicity at least $\Omega_{d, \lambda}(\log \log n)$.
So now we know that the answer to the earlier question is between $\Omega(\log\log n)$ and $O(n/\log \log n)$. There is still a huge gap.
Carl managed to get his construction many families of $\lambda$, including some with $k(\lambda) = \infty$, which implies his equiangular lines result stated earlier.
Carl’s construction is quite elegant, incorporating several clever ideas in spectral graph theory. One of the ideas used is the 2-lift technique from the seminal Bilu-Linial paper. This is a way to transform a graph into another graph with twice as many vertices and in a way that allows us to control the spectrum.
For Carl’s construction, in order to perform the 2-lift and control the eigenvalues, given a $d$-regular bipartite graph, he needs to select a regular subgraph in a way that is somewhat random. There needs to be enough independent sources of randomness so that he can apply concentration bounds. He does this by, roughly speaking, partitioning the graph into short even cycles, and then for each even cycle, independently choosing one of the two alternating sets of edges.
There is another catch, which is that in order for his concentration bounds to go through, he needs to first make the graph large enough so that the relevant eigenvectors are all somewhat flat. He does this by repeatedly performing the Ramanujan lift from the seminal work of Marcus–Spielman–Srivastava. (Carl doesn’t need the full strength of the MSS result, but why resist applying such a beautiful theorem.)
This is really nice construction. These interesting ideas will surely lead to further developments in spectral graph theory.
## Nearly all k-SAT functions are unate
September 19, 2022
The Boolean satisfiability problem, specifically k-SAT, occupies a central place in theoretical computer science. Here is a question that had been open until just now: what does a typical k-SAT function look like? Bollobás, Brightwell, and Leader (2003) initiated the study of this question, and they conjectured that a typical k-SAT function is unate (meaning monotone up to first negating some variables; more on this below.)
In a new preprint with the above coauthors (Dingding and Nitya are PhD students), we completely settle the Bollobás–Brightwell–Leader conjecture.
Theorem. Fix k. A $1-o(1)$ proportion of all k-SAT functions on n Boolean variables are unate, as $n \to \infty$.
Some special cases were known earlier: $k=2$ by Allen (2007) and $k=3$ by Ilinca and Kahn (2012).
As a reminder, a k-SAT function is a function $f \colon \{ 0, 1 \}^n \to \{0,1\}$ that is expressible as a formula of the form
$f(x_1, \dots, x_n) = C_1 \land C_2 \land \cdots \land C_m$
(here $\land$ = AND) where each “clause” $C_i$ is an OR of at most $k$ literals (each “literal” is some $x_i$ or its negation $\overline{x_i}$). Such a function or formula is called monotone if the negatives literals $\overline{x_i}$ do not appear. Furthermore, the function/formula is called unate if each variable $x_i$ either appears only in its positive form or only in its negative form, but not both simultaneously.
Here is an example of a unate 3-SAT function:
$(x_1 \lor \overline{x_2} \lor x_3) \land (x_1 \lor \overline{x_2} \lor \overline{x_4}) \land \cdots \land (x_3 \lor \overline{x_4} \lor x_5)$
(the variables $x_1$, $x_3$, $x_5$ appear only in their positive form, and the variables $x_2$, $x_4$ only appear in their negative form).
Our new paper is the second half of a two-part work proving the Bollobás-Brightwell-Leader conjecture.
The first part, by Dingding Dong, Nitya Mani, and myself (reported earlier on this blog) reduces the Bollobás–Brightwell–Leader conjecture to a Turán problem on partially directed hypergraphs. The reduction applies the hypergraph container method, a major recent development in combinatorics.
I presented the first part in an Oberwolfach meeting earlier this year, which led to a successful collaboration with József Balogh and Bernard Lidický that resulted in a complete solution to the problem.
In this short new paper we solve the relevant Turán problem. Our solution is partly inspired by the method of flag algebras introduced by Razborov.
We actually prove a stronger statement. A k-SAT formula is said to be minimal if deleting any clause changes the function. We prove:
Theorem. Fix k. A $1−o(1)$ proportion of all minimal k-SAT formulae on n Boolean variables are unate, as $n \to \infty$.
As suggested by Bollobás, Brightwell, and Leader, these results open doors to a theory of random k-SAT functions. For example, we now know the following.
Corollary. A typical k-SAT function admits a unique minimal k-SAT formula, and furthermore the formula has $(1/2 + o(1)) \binom{n}{k}$ clauses.
This model is very different from that of random k-SAT formulae where clauses are added at random (e.g., the recent breakthrough of Ding–Sun–Sly on the satisfiability conjecture). Rather, our result concerns a random k-SAT formula conditioned on minimality. In this light, our results are analogous to the theory of Erdős–Rényi random graphs $G(n, p)$ with constant edge probability $p$. It would be interesting to further study the behavior of sparser random k-SAT formulae conditioned on minimality. This potentially leads to a rich source of problems on thresholds and typical structures.
## Kwan–Sah–Sawhney–Simkin: High-girth Steiner triple systems
January 13, 2022
There is an impressive new preprint titled High-Girth Steiner Triple Systems by Matthew Kwan, Ashwin Sah, Mehtaab Sawhney, and Michael Simkin. They prove a 1973 Erdős conjecture showing the existence of Steiner triple systems with arbitrarily high girth.
An order n triple system is defined to be a set of unordered triples of ${1, \dots, n}$. Furthermore, it is a Steiner triple system if every pair of vertices is contained in exactly one triple.
An example of a Steiner triple system is the Fano plane, illustrated below.
Here each line or circle represents a triple containing its elements. Note that every pair of vertices lies in exactly one line/circle (this is what makes it Steiner).
Steiner triple systems are old and fascinating objects that occupy a central place in combinatorics. For example, see this Quanta Magazine article for a popular account on Steiner triple systems and recent breakthroughs in design theory.
In the past decade, since Peter Keevash’s 2014 breakthrough proving the existence of designs, there has been a flurry of exciting developments by many researchers settling longstanding open problems related to designs and packings. Several related results were also featured in Quanta magazine: Ringel’s conjecture, Erdős–Faber–Lovász conjecture, n-queens problem.
The new work proves the existence of Steiner triple systems with a desirable additional property: high girth.
In a graph, the girth is defined to be the length of a shortest cycle. The girth of a triple system is defined as follows (see the first page of the paper for some motivation behind this definition).
Definition. The girth of a triple system is the smallest positive integer g so that there exist g triples occupying no more than $g+2$ vertices. (If no such g exists, then we say that the girth is infinite.)
Here is the main result of the new paper, which settles the Erdős conjecture.
Theorem (Kwan–Sah–Sawhney–Simkin). For all $n \equiv 1,3 \pmod{6}$, there exists a Steiner triple system whose girth tends to infinity as $n \to \infty$.
They build on previous works by Glock, Kühn, Lo, and Osthus and Bohman and Warnke proving the existence of “approximate” Steiner triple systems (meaning having $1-o(1)$ fraction of the maximum possible number of triples).
Before this work, it was not even known if there existed infinitely many Steiner triple systems with girth ≥ 10.
The proof uses a stunning array of techniques at the frontier of extremal and probabilistic combinatorics, including recently developed tools for random greedy algorithms, as well as the iterative absorption method.
## Graphs with high second eigenvalue multiplicity
September 28, 2021
With Milan Haiman, Carl Schildkraut, and Shengtong Zhang (all undergrads at MIT), we just posted our new paper Graphs with high second eigenvalue multiplicity.
Question. How high can the second eigenvalue multiplicity of a connected bounded degree graph get?
In my earlier paper solving the equiangular lines problem (with fixed angle and high dimensions), we proved:
Theorem. (Jiang, Tidor, Yao, Zhang, and Zhao) A connected bounded-degree $n$-vertex graph has second eigenvalue multiplicity $O(n/\log\log n)$.
The fact that the second eigenvalue has sublinear multiplicity played a central role in our equiangular lines proof (as well as in our follow-up paper on spherical two-distance sets).
While the spectral gap (the difference between the top two eigenvalues) has been intensely studied due to the connections to graph expansion, not much is known about second eigenvalue multiplicities. The above theorem is the first result of this form about general classes of graphs.
Questions. Can the bound in the theorem be improved? What about when restricted to regular graphs, or Cayley graphs?
In particular, is $\le n^{1-c}$ true? We don’t know.
Why should we care? This appears to be a fundamental question in spectral graph theory. We saw that it is important for the equiangular lines problem (and maybe more applications yet to be discovered). In particular, improving the bounds in our theorem will improve the effectiveness of our equiangular lines theorem, which holds in “sufficiently high” dimension (how high is enough).
Eigenvalue multiplicities have already played an important role in Riemannian geometry, e.g., via connections to Gromov’s theorem on groups of polynomial growth; see the discussions in this paper by James Lee and Yury Makarychev.
Previously, the best example we knew has second eigenvalue multiplicity $\ge c n^{1/3}$. Basically any connected Cayley graph on $\mathrm{PSL}(2, p)$ has this property, since a classic result of Frobenius tells us that every non-trivial irreducible representation of this group has dimension $\ge (p-1)/2$.
In our new paper, we construct several families of $n$-vertices, connected, and bounded degree graphs:
• with second eigenvalue multiplicity $\ge c \sqrt{n/\log n}$;
• Cayley graphs, with second eigenvalue multiplicity $\ge c n^{2/5}$;
• with approximate second eigenvalue multiplicity $\ge c n/\log\log n$, thereby demonstrating a barrier to the moment/trace methods used in our earlier paper.
It would be interesting to find other constructions with second eigenvalue multiplicity exceeding $\sqrt{n}$. This appears to be a barrier for group representation theoretic methods, since every representation of an order $n$ group has dimension $\le \sqrt{n}$.
Another interesting recent paper by Theo McKenzie, Peter Rasmussen, and Nikhil Srivastava showed that connected bounded-degree $n$-vertex regular graphs have second eigenvalue multiplicity $\le n/(\log n)^c$, improving our $O(n/\log\log n)$ bound for general graphs. For general non-regular graphs, their result works for the normalized (random walk) adjacency matrix (rather than the usual adjacency matrix). Their result introduced the following novel observation: a closed walk of length $2k$ starting at an arbitrary vertex in a bounded degree graph typically covers $\ge k^c$ vertices.
## Enumerating k-SAT functions
July 21, 2021
In a new paper coauthored with graduate students Dingding Dong and Nitya Mani titled Enumerating k-SAT functions, we study these fundamental questions about k-SAT functions:
• How many k-SAT functions on n boolean variables are there?
• What does a typical such function look like?
These questions about k-SAT functions were first studied by Bollobás, Brightwell, and Leader (2003), who made the following conjecture.
Conjecture. Fix $k \ge 2$. Almost all k-SAT functions are unate.
Here an unate k-SAT function is one that can be turned monotone by first negating some subset of variables.
More precisely, the conjecture says that, as $n \to \infty$, $1-o(1)$ fraction of all k-SAT functions on n boolean variables are unate.
An easy argument shows that there are $(1+o(1))2^{\binom{n}{k} + n}$ unate functions. So the above conjecture can be equivalently restated as:
Conjecture. Fix $k \ge 2$. The number of k-SAT functions on n boolean variables is $(1+o(1))2^{\binom{n}{k} + n}$.
The conjecture was previously proved for $k=2$ (Allen 2007) and $k=3$ (Ilinca and Kahn 2012) using the (hyper)graph regularity method.
In our new paper, using the hypergraph container method, we show that the k-SAT conjecture is essentially equivalent to a variant of an extremal hypergraph problem. Such extremal problems are related to hypergraph Turán density problems, which are notoriously difficult to solve in general, and for which there are lots of open problems and very few answers.
We solved our Turán problem for all $k \le 4$, thereby confirming the conjecture for a new value $k=4$.
Our solution to the Turán problem for $k=4$ uses a recent result in extremal graph theory by Füredi and Maleki (2017) determining the minimum density of triangular edges in a graph of given edge density.
Here is a Turán density conjecture whose resolution would imply the above conjecture for $k=5$, which is still open.
Definition. A 5-PDG (partially directed 5-graph) is obtained from a 5-uniform hypergraph by “directing” some subset of its edges. Here every directed edge is “directed towards” one of its vertices.
Conjecture. There exists a constant $\theta > \log_2 3$ such that for every n-vertex 5-PDG with A undirected edges and B directed edges, if there do not exist three edges of the form
12345
1234 6
123 56
with at least one of these three edges directed towards 4, 5, or 6, then
$A + \theta B \le (1+o(1))\binom{n}{5}.$
There are actually additional patterns that one could forbid as well for the purpose solving the k-SAT enumeration problem. In the paper we give an explicit description of all forbidden patterns. But the above single forbidden pattern might just be enough (it is enough for $k \le 4$).
## Mathematical tools for large graphs
July 12, 2021
Author’s note: This article was written intended for a general scientific audience.
Graphs and networks form the language and foundation of much of our scientific and mathematical studies, from biological networks, to algorithms, to machine learning applications, to pure mathematics. As the graphs and networks that arise get larger and larger, it is ever more important to develop and understand tools to analyze large graphs.
### Graph regularity lemma
My research on large graphs has its origins in the work of Endre Szemerédi in the 1970s, for which he won the 2012 Abel Prize, a lifetime achievement prize in Mathematics viewed as equivalent to the Nobel Prize. Szemerédi is a giant of combinatorics, and his ideas still reverberate throughout the field today.
Szemerédi was interested in an old conjecture of Paul Erdős and Pál Turán from the 1930s. The conjecture says that
if you take an infinite set of natural numbers, provided that your set is large enough, namely that it occupies a positive fraction of all natural numbers, then you can always find arbitrarily long arithmetic progressions.
For instance, if we take roughly one in every million natural numbers, then it is conjectured that one can find among them k numbers forming a sequence with equal gaps, regardless of how large k is.
This turns out to be a very hard problem. Before the work of Szemerédi, the only partial progress, due to Fields Medalist Klaus Roth in the 1950’s is that one can find three numbers forming a sequence with equal gaps. Even finding four numbers turned out to be an extremely hard challenge until Szemerédi cracked open the entire problem, leading to a resolution of the Erdős–Turán conjecture. His result is now known as Szemerédi’s theorem. Szemerédi’s proof was deep and complex. Through his work, Szemerédi introduced important ideas to mathematics, one of which is the graph regularity lemma.
Szemerédi’s graph regularity lemma is a powerful structural tool that allows us to analyze very large graphs. Roughly speaking, the regularity lemma says that
if we are given a large graph, no matter how the graph looks like, we can always divide the vertices of the graph into a small number of parts, so that the edges of the graph look like they are situated randomly between the parts.
Between different pairs of parts of vertices, the density of edges could be different. For example, perhaps there are five vertex parts A, B, C, D, E; between A and B the graph looks like a random graph with edge density 0.2; between A and C the graph looks like a random graph with edge density 0.5, and so on.
An important point to emphasize is that, with a fixed error tolerance, the number of vertex parts produced by the partition does not increase with the size of the graph. This property makes the graph regularity lemma particularly useful for very large graphs.
Szemerédi’s graph regularity lemma provides us a partition of the graph that turns out to be very useful for many applications. This tool is now a central technique in modern combinatorics research.
A useful analogy is the signal-versus-noise decomposition in signal processing. The “signal” of a graph is its vertex partition together with the edge density data. The “noise” is the residual random-like placement of the edges.
Szemerédi’s idea of viewing graph theory via the lens of this decomposition has had a profound impact in mathematics. This decomposition is nowadays called “structure versus pseudorandomness” (a phrase popularized by Fields Medalist Terry Tao). It has been extended far beyond graph theory. There are now deep extensions by Fields Medalist Timothy Gowers to what is called “higher order Fourier analysis” in number theory. The regularity method has also been extended to hypergraphs.
### Graph limits
László Lovász, who recently won the 2021 Abel Prize, has been developing a theory of graph limits with his collaborators over the past couple of decades. Lovász’s graph limits give us powerful tools to describe large graphs from the perspective of mathematical analysis, with applications ranging from combinatorics to probability to machine learning.
Why graph limits? Here is an analogy with our number system. Suppose our knowledge of the number line was limited to the rational numbers. We can already do a lot of mathematics with just the rational numbers. In fact, with just the language of rational numbers, we can talk about irrational numbers by expressing each irrational number as a limit of a sequence of rational numbers chosen to converge to the irrational number. This can be done but it would be cumbersome if we had to do this every time we wanted to use an irrational number. Luckily, the invention of real numbers solved this issue by filling the gaps on the real number line. In this way, the real numbers form the limits of rational numbers.
Likewise, graphs are discrete objects analogous to rational numbers. A sequence of graphs can also converge to a limiting object (what does it mean for a sequence of graphs to converge is a fascinating topic and beyond the scope of this article). These limiting objects are called “graph limits,” also known as “graphons.” Graph limits are actually simple analytic objects. They can be pictured as a grayscale image inside a square. They can also be represented as a function from the unit square to the unit interval (the word “graphon” is an amalgamation of the words “graph” and “function”). Given a sequence of graphs, we can express each graph as an adjacency matrix drawn as a black and white image of pixels. As the graphs get larger and larger, this sequence of pixel images looks closer and closer to a single image, which might be not just black and white but also could have various shades of gray. This final image is an example of a graph limit. (Actually, there are subtleties here that I am glossing over, such as permuting the ordering of the vertices before taking the limit.)
Conversely, given a graph limit, one can use it as a generative model for random graphs. Such random graphs converge to the given graph limit when the number of vertices increases. This random graph model generalizes the stochastic block model. An example of a problem in machine learning and statistics is how to recover the original graph limit given a sequence of samples from the model. There is active research work on this problem (although it is not the subject of my own research).
The mathematics underlying the theory of graph limits, notably the proof of their existence, hinges on Szemerédi’s graph regularity lemma. So these two topics are intertwined.
### Sparser graphs
Graph regularity lemma and graph limits both have an important limitation: they can only handle dense graphs, namely, graphs with a quadratic number of edges, i.e., edge density bounded from below by some positive constant. This limitation affects applications in pure and applied mathematics. Real life networks are generally far from dense. So there is great interest in developing graph theory tools to better understand sparser graphs. Sparse graphs have significantly more room for variability, leading to mathematical complications.
A core theme of my research is to tackle these problems by extending mathematical tools on large graphs from the dense setting to the sparse setting. My work extends Szemerédi’s graph regularity method from dense graphs to sparser graphs. I have also developed new theories of sparse graph limits. These results illuminate the world of sparse graphs along with many of their complexities.
I have worked extensively on extending graph regularity to sparser graphs. With David Conlon and Jacob Fox, we applied new graph theoretic insights to simplify the proof of Ben Green and Terry Tao’s celebrated theorem that the prime numbers contain arbitrarily long arithmetic progressions (see exposition). It turns out that, despite being a result in number theory, a core part of its proof concerns the combinatorics of sparse graphs and hypergraphs. Our new tools allow us to count patterns in sparse graphs in a setting that is simpler than the original Green–Tao work. In a different direction, we (together additionally with Benny Sudakov) have recently developed the sparse regularity method for graphs without 4-cycles, which are necessarily sparse.
My work on sparse graph limits, together with Christian Borgs, Henry Cohn, and Jennifer Chayes , developed a new $L^p$ theory of sparse graph limits. We were motivated by sparse graph models with power law degree distributions, which are popular in network theory due to their observed prevalence in the real world. Our work builds on the ideas of Bela Bollobás and Oliver Riordan , who undertook the first systematic study of sparse graph limits. Bollobás and Riordan also posed many conjectures in their paper, but these conjectures turned out to be all false due to a counterexample we found with my PhD students Ashwin Sah, Mehtaab Sawhney, and Jonathan Tidor. These results illustrate the richness of the world of sparse graph limits.
### Graph theory and additive combinatorics
A central theme of my research is that graph theory and additive combinatorics are connected via a bridge that allows many ideas and techniques to be transferred from one world to the other. Additive combinatorics is a subject that concerns problems at the intersection of combinatorics and number theory, such as Szemerédi’s theorem and the Green–Tao theorem, both of which are about finding patterns in sets of numbers.
When I started teaching as an MIT faculty, I developed a graduate-level math class titled Graph theory and additive combinatorics introducing the students to these two beautiful subjects and highlighting the themes that connect them. It gives me great pleasure to see many students in my class later producing excellent research on this topic.
In Fall 2019, I worked with MIT OpenCourseWare and filmed all my lecture videos for this class. They are now available for free on MIT OCW and YouTube. I have also made available my lecture notes, which I am currently in the process of editing into a textbook.
In my first lecture, I tell the class about the connections between graph theory and additive combinatorics as first observed by the work of Issai Schur from over a hundred years ago. Thanks to enormous research progress over the past century, we now understand a lot more, but there is still a ton of mystery that remains.
The world of large graphs is immense.
## How I manage my BibTeX references, and why I prefer not initializing first names
July 4, 2021
This is both a how-to guide, and a discussion of the rationale for some of my preferences.
I use BibTeX for compiling my bibliographic references when writing math papers in LaTeX. BibTeX, when used correctly, eases the burden of bibliographic management. For example, BibTeX automatically excludes uncited references and alphabetizes the entries.
Since a single paper can have a large number of references, it is useful to have a simple system to
• Retrieve complete and accurate bibliographic records, and
• Maintain consistency.
Here are some of my own processes and conventions, along with my rationale behind some of the choices. This blog post is not a beginner’s guide on BibTeX (here is a good guide). Rather, I focus on my own conventions and choices, specifically tailored for math research papers.
### Include author names as they appear in the original publication
It is common practice to shorten first names to initials in references. I do not agree with this practice. In many Asian cultures, a large fraction of the population use a fairly small number of surnames (e.g., Wang, Li, Kim). So first name initialization can lead to ambiguity. Readers are also less likely to distinguish common names and associate them with specific individuals, leading to less recognition and exposure, and this can have an adverse effect especially on early-career academics.
(Aside) As a grad student, every once in a while I was sent some combinatorics paper to referee, but these papers weren’t exactly close to anything I worked on. Looking back, I think perhaps the editors had sometimes intended to ask the other Y. Zhao in combinatorics.
(Another aside) While we are on the topic of names, it’s also worth pointing out that Chinese surnames tend to cluster late in the alphabet. This creates an interesting effect when you scroll to the bottom of a math department directory.
Economists, like mathematicians, also have co-authorship alphabetized by default. A 2006 research paper by economics professors from Stanford and Caltech showed that faculty in top 35 U.S. economics departments “with earlier surname initials are significantly more likely to receive tenure at top ten economics departments, are significantly more likely to become fellows of the Econometric Society, and, to a lesser extent, are more likely to receive the Clark Medal and the Nobel Prize,” even after controlling for “country of origin, ethnicity, religion or departmental fixed effects.” The same effects were not observed in the psychology profession, where authorship is not alphabetized.
To be clear, I am not at all advocating for abolishing the practice of ordering authors alphabetically. Quite the opposite. Ordering authors by contribution invites other conflicts and politics, which seems a lot worse. Ordering “randomly” doesn’t really make sense to me unless one can somehow enforce an independent random permutation of the authors every time the paper is read or mentioned. (End asides)
The practice of initializing first names was perhaps originally due to antiquated needs such as saving ink and space. Many journals still enforce it today in their house style.
(Example) First names are initialized not just in the references, but also sometimes in the author line, even in modern times. Up to the end of the 2006 volume, the highly regarded journal Geometric and Functional Analysis (GAFA) would typically initialize authors’ first names in the byline. There were occasional exceptions where the authors’ names were printed in full. For example, in the 2005 volume, exactly two out of 42 papers had the authors’ full names printed, and these were precisely the two papers (and only ones) having authors with Chinese names. It is likely that these authors requested to have their full names published.
When I first started writing math papers I followed what other papers did. But later on I realized that this practice does not make as much sense as I initially assumed it would. Nowadays, whenever I have control over my references (e.g., my arXiv uploads), I follow this rule:
Include author names as they appear in the original publication
Rationale and caveats:
• Why not just always use full names? Not all authors actually want their initializations expanded. Some authors may have chosen a specific stylization of their pen name (e.g., J. K. Rowling), or some specific capitalization. Also, sometimes, it is difficult to look up their full names. It is not my role to guess (just as it would never be proper for me to permute the order of authors in a paper I am citing). Following how they write their own name on their paper is the simplest and most consistent practice.
• Some authors may prefer to use different pseudonyms (or pen names) for different publications. I should not be “correcting” them. Some notable examples:
• Noga Alon published several papers under “A. Nilli.” Accompanying the discussion of one of such paper in Proofs from THE BOOK is a photo of a young girl with the caption “A. Nilli.”
• Victor Kac, after emmigrating from the Soviet Union, had to publish his work with his Soviet advisor Ernest Vinberg under the Italian pseudonyms Gatti and Viniberghi due to Soviet restrictions.
(I learned this fact from another fascinating economics paper on the effect of the collapse of the Soviet Union on the productivity of American mathematicians.)
Amazingly, the AMS Mathematical Reviews (MathSciNet) actually knows about author pseudonyms to high accuracy, including the examples above. There is a manual painstaking process by the Math Reviews team to assign each author to the correct person.
• In rare instances, the publisher makes a mistake in printing the authors’ names. For example, someone told me that a journal once erroneously expanded his name as the copy editor mistakenly thought that the author’s first name was provided in a shortened form, after the final galley proof was returned to the publisher (so that there was no further opportunity for the author to correct the last minute error introduced by the copy editor). This error arose from someone deciding how someone else’s name should be written.
When there is a publication error, which name should be used in bibliographic citations? Should it matter whether or not an erratum is published? In this specific case, I would use the correct name as intended by the author, but in general, I might not have access to the extra information. These situations are annoying but rare, and can be decided case by case.
So the simplest and most consistent solution is to just include author names as they appear in the original publication.
• It is okay if some entries use full names and some others use first initials.
• It is okay if the same person appears in the references multiple times differently.
• It is not okay to change how someone’s name is intended to written.
• Initializing first names leads to ambiguity.
I admit that I have not always followed this practice, but lately I have been making an effort to do so at least when I am in charge of managing the references. The BibTeX how-to guide here makes it easy to implement. The entries imported from MathSciNet always have author names as they were originally printed.
### BibTeX style file
I made a custom modification of the amsplain style file:
Compared to the standard amsplain style file, I made the following modifications:
• No longer displaying the MR number in the bibliographic entry;
• No longer replacing the authors by a long dash when the authors are identical to the previous entry.
To apply the new style file, I include the .bst file in the same folder as my LaTeX file, and insert the references in LaTeX using:
\bibliographystyle{amsplain0}
\bibliography{bib file name (without the .bib extension)}
Optionally, before uploading the LaTeX source to the arXiv or journal, I copy and paste the contents of the auxillary .bbl output to replace the above two lines. Then, I only have to upload a single .tex file rather than both .tex and .bbl files.
### Retrieving BibTeX entries from MathSciNet
For the most part, you do not need to enter BibTeX entries by hand. MathSciNet has already compiled all this information.
• Accessing MathSciNet requires a subscription, which many universities have. Common ways to access MathSciNet:
• From a device on campus network;
• Using university VPN;
• Using university library’s proxy url;
• Using AMS Remote access to first pair your browser with AMS when you’re on campus or on VPN, so that you can continue accessing MathSciNet and other AMS resources later even when off-campus and off-VPN.
• You can configure your browser to search for MathSciNet directly from the search bar. For example, in Chrome, in Settings / Search Engine you can add a new search engine with the following configurations
• Search engine: MR search by title
• Keyword: mr
• Query url: https://mathscinet.ams.org/mathscinet/search/publications.html?pg1=TI&s1=%s&extend=1
to search MathSciNet by title. For instance, typing mr large graph into the search bar yields a search for articles with both words “large” and “graph” in the title, whereas typing mr "large graph" into the search bar yields a search for papers with the phrase “large graph” in the title. You can also configure other custom searches by modifying the query url.
• A free lite version, MR Lookup, can be used without subscription, though it only displays the top three search results. It also gives the results in BibTeX format.
All articles listed on MathSciNet can be extracted as BibTeX entries, either individually or in bulk (this can be done on the MathSciNet website by selecting BibTeX from the drop down menu).
• I like using BibDesk (only on Mac; free and comes with TeXShop), which offers a nice interface for BibTeX management, including automatic import from MathSciNet (from the External–Web tab on the left). BibDesk also has other useful automated tools such as batch renaming entry keys according to authors-year.
I imagine there must be other similar BibTeX management tools with automatic MathSciNet import, but I have not personally used any of them. See this discussion.
Most entries from MathSciNet are accurate and consistently formatted, especially for published journal articles. MathSciNet has a dedicated staff for maintaining the quality of its entries. They follow the official AMS journal title abbreviations (best not to make up these abbreviations yourself). In contrast, other sources (e.g., Google Scholar search) often produce inconsistently formatted and inaccurate bibliographic entries.
Occasionally some entries need to be added or edited by hand. Here are some common scenarios in my experience.
• Pre-print and pre-published articles (including those appearing first on the publisher website but not yet assigned an issue) do not appear in MathSciNet. I add them manually as follows:
• For arXiv preprints:
@misc{Smi-ar,
AUTHOR = {John Smith},
TITLE = {I proved a theorem},
NOTE = {arXiv:2999.9999},
}
• For papers to appear (use official AMS journal title abbreviations)
@article{Smi-mj,
AUTHOR = {John Smith},
TITLE = {I proved an excellent theorem},
JOURNAL = {Math. J.},
NOTE = {to appear},
}
(It might annoy some people to have an @article entry without a year, but shrug, it compiles, and I haven’t found a more satisfactory solution.)
• Really old (like, pre-1940) and some foreign language articles/books are not listed on MathSciNet — I copy the citations from somewhere else and enter them as an @article/@book by hand.
• For non-standard entries like websites and blog posts, I add them as @misc similar to arXiv entries, using the note field as a catch-all. (Yes I know that there is specifically a @url entry type, but I find it easier and more predictable to use @misc. In any case, the choice of entry type does not show up in the output.)
• Capitalizations. By default, BibTeX changes the entire title to lower case except for the very first letter. To keep capitalizations of proper nouns, LaTeX commands, etc., surround those letters/words by {...}. BibTeX entries from MathSciNet usually gets this right automatically, but more care if needed when manually adding entries such as from the arXiv.
Modifications:
• MathSciNet usually gets it right but sometimes accented names need some manual editing to compile correctly. A notable exaple is that when “Erdős” appears in a title, one needs to surround the ő \H{o} by {...}, e.g., {E}rd{\H{o}}, or else BibTeX changes \H to \h thereby causing a compilation error. Similar issues also arise in the author field, e.g., with the Polish Ł \L. They can always be fixed by surrouding the problematic snippets by {...}.
• Some MathSciNet entries for books have extraneous data that should be deleted. It is usually clear what’s going on after seeing the output PDF.
• For some journals with reboots in its publication history, the official AMS journal title abbreviation sometimes includes a “series” number, e.g., Annals of Mathematics is actually Annals of Mathematics. Second Series abbreviated as Ann. of Math. (2), which is consistent with MathSciNet imports. Some people don’t like to keep the (2). I’m okay either way.
• Conference proceedings are the worst offenders on consistent formatting. They are not covered by AMS guidance and it is a free-for-all. Basically, I import from MathSciNet, compile and see what the PDF output looks like, and then delete extraneous data. If a conference proceeding article is superseded by a journal article, I use the journal article instead.
• MathSciNet BibTeX entries include a bunch of other extraneous information like the reviewer’s name. I could delete them but these do not show up in the output anyway, so I don’t bother. The only piece that does show up in the amsplain style is MR number, which does not show up in my modified amsplain0 style.
### Entry keys
The simplest cases:
• Single author: the first up to three letters of surname (preserving capitalization), followed by two-digit year. Example: Wil95 in
@article {Wil95,
AUTHOR = {Wiles, Andrew},
TITLE = {Modular elliptic curves and {F}ermat's last theorem},
JOURNAL = {Ann. of Math.},
VOLUME = {141},
YEAR = {1995},
NUMBER = {3},
PAGES = {443--551},
}
• Multiple authors: a string of initialized surnames followed by two-digit year. Example: ER60 in
@article {ER60,
AUTHOR = {Erd\H{o}s, P. and R\'{e}nyi, A.},
TITLE = {On the evolution of random graphs},
JOURNAL = {Magyar Tud. Akad. Mat. Kutat\'{o} Int. K\"{o}zl.},
VOLUME = {5},
YEAR = {1960},
PAGES = {17--61},
}
In BibDesk, I automatically assign keys via the following configuation: in BibDesk Preferences / Cite Key Format, select
Select one or more rows/entries, enter Cmd-K to automatically generate keys. For multi-author entries, this correctly generates the key, e.g., ER60 (unless there are more than nine authors). For single-author entries, a bit of manual work is needed to change from W95 to Wil95 (I prefer not leaving it like W95 since it leads to more conflicts and confusion).
Where it gets a bit more tricky:
• If the above rules produce conflicting keys:
• Add a suffix to the key with a few letters containing some informative data such as journal name or topic, e.g., [Wil95ann], [ER60rg], or
• Distinguish the keys by letters, e.g., [Wil95a] [Wil95b] (this is what BibDesk does automatically; but this system could potentially be more confusing later on).
• Pre-prints or pre-publications:
• Forget the year, e.g., [XYZ], and if disambiguation is needed, add a dash followed by some informative letters, e.g., [XYZ-gcd]
• Not sure which year to use? E.g., a proceeding published in 2000 for a conference held in 1999. Use the year field in BibTeX as imported by MathSciNet.
### What the BibTeX output looks like
Using the amsplain0 style file from the beginning, a citation \cite{Wil95} produces
References
[1] Andrew Wiles, Modular elliptic curves and Fermat’s last theorem, Ann. of Math. (2) 141 (1995), 443–551.
And that’s it! This is how I manage my BibTeX.
## The cylindrical width of transitive sets
January 28, 2021
The high dimensional sphere comes many surprising and counterintuitive features. Here is one of them.
We say that a subset $X$ of the unit sphere in $\mathbb{R}^d$ is transitive if, informally, one cannot distinguish one point of $X$ from another if we are allowed to rotate/reflect. Formally, we say that $X$ is transitive if the group of isometries of $X$ acts transitively on it. This is equivalent to saying that $X$ is the orbit of some point under the action of some subgroup of the orthogonal group.
Perhaps you, like me, can only visualize in three-dimensions. Here are some examples:
• Every regular polytope (platonic solid) is transitive.
• Start with an arbitrary vector, and consider all points obtained by permuting its coordinates, as well as switching the signs of some of the coordinates. The resulting set is transitive (this is a generalization of the permutahedron.) A specific example is given below.
• The whole sphere itself is transitive
The finite examples look like they can be fairly spread out on the sphere. However, this intuition turns out to be completely wrong in high dimensions. All finite transitive sets in high dimensions have small width, meaning that they lie close to some hyperplane (in sharp contrast to the whole unit sphere, which has width 2).
I had conjectured the following statement, which was proved by Ben Green.
Theorem (Green). Every finite transitive set of unit vectors in $\mathbb{R}^d$ lies within distance $O\bigl(\frac{1}{\sqrt{\log d}}\bigr)$ of some hyperplane.
This bound is actually tight for the example obtained by taking all signings and permutations of the unit vectors
$\frac{1}{\sqrt{H_d}}\Biggl(\pm 1, \frac{\pm 1}{\sqrt{2}}, \frac{\pm 1}{\sqrt{3}}, \ldots, \frac{\pm 1}{\sqrt{d}}\Biggr)$
where
$H_d = 1 + \frac{1}{2} + \cdots + \frac{1}{d} \sim \log d.$
Green’s proof, among other things, uses the classification of finite simple groups.
### Our new results: finite transitive sets have small cylindrical width
In our new paper, The cylindrical width of transitive sets, coauthored with Ashwin Sah and Mehtaab Sawhney, we generalize Green’s result to show that not only do transitive sets lie close to a hyperplane, but they also lie close to some codimension-$k$ subspace for every $k$ that is not too large.
Theorem (Sah–Sawhney–Zhao). There exists some constant $C$ so that for every $1 \le k \le d/(\log d)^C$, every finite transitive set of unit vectors in $\mathbb{R}^d$ lies within distance $O\bigl(\frac{1}{\sqrt{\log (d/k)}}\bigr)$ of codimension-$k$ subspace.
The bound $O\bigl(\frac{1}{\sqrt{\log (d/k)}}\bigr)$ is tight up to a constant factor. We conjecture that the hypothesis $1 \le k \le d/(\log d)^C$ can be removed:
Conjecture. For every $1 \le k \le d$, every finite transitive set of unit vectors in $\mathbb{R}^d$ lies within distance $O\bigl(\frac{1}{\sqrt{\log (2d/k)}}\bigr)$ of codimension-$k$ subspace.
Finally, we conjecture that not only do finite transitive sets lie close to subspaces, but they can be completely contained in a small cube. All examples known to us are consistent with this conjecture, even though I still find it very counterintuitive.
Conjecture (Finite transitive sets lie in a small cube). Every finite transitive set of unit vectors in $\mathbb{R}^d$ is contained in a cube of width $O\bigl(\frac{1}{\sqrt{\log d}}\bigr)$.
Even proving a bound of $o(1)$ (as $d\to \infty$) would be quite interesting.
Both conjectures are generalizations of Green’s result, but the relationship between the two conjectures is unclear.
## Ashwin Sah and Mehtaab Sawhney win the Morgan Prize
October 29, 2020
Congratulations to Ashwin Sah and Mehtaab Sawhney for winning the 2021 AMS-MAA-SIAM Frank and Brennie Morgan Prize for Outstanding Research in Mathematics by an Undergraduate Student! This is the most prestigious prize in the US for undergraduate mathematics research (though I must say that their work far surpasses that of “undergraduate research”). It is also the first time in the history of the Morgan Prize for it to be jointly awarded to more than one recipient.
From the prize citation:
The award recognizes the duo’s innovative results across a broad range of topics in combinatorics, discrete geometry, and probability.
The prize committee notes, “Working alongside one another, Sah and Sawhney settled longstanding conjectures and improved results by established mathematicians. The research of Sah and Sawhney is both deep and broad, tackling questions at the very forefront of current research, yet extending across the entire gamut of modern combinatorics, with significant contributions to extremal graph theory, graph limits, additive combinatorics, Ramsey theory, algebraic combinatorics, combinatorial geometry, random graphs and random matrix theory. They have demonstrated a significant amount of ingenuity, originality and technical ability, resulting in a research record, which is extremely rare for undergraduate students.”
Combined, Sah and Sawhney have authored 30 papers (11 together), and published in top journals, including Inventiones Mathematicae, Advances in Mathematics, Mathematical Proceedings of the Cambridge Philosophical Society, the Journal of Combinatorial Theory Series B, and Combinatorica.
(The paper count in the award citation was taken from earlier this year and is now already quite a bit out of date.)
I got to know both Ashwin and Mehtaab as students in my Putnam Seminar and graduate combinatorics class during their very first undergraduate semester at MIT. Since then, they have led an intensively productive and ongoing collaboration, producing an incredible amount of fantastic research results across a wide spectrum of topics in combinatorics, probability, number theory, and algorithms. It was a great pleasure for me to be working with Ashwin and Mehtaab on many of these projects. Some of their work were previously featured on this blog:
Ashwin and Mehtaab are now both first-year PhD students at MIT. I’m sure that we will see a lot more coming from them.
## Jain–Sah–Sawhney: Singularity of discrete random matrices
October 13, 2020
Vishesh Jain, Ashwin Sah, and Mehtaab Sawhney just posted two amazing papers: Singularity of discrete random matrices Part I and Part II.
The singularity problem for random matrices: what is the probability that a random $n\times n$ matrix with independent 0/1 entries is singular?
The first non-trivial bound on this problem was by Komlós in 1967. After a long series of works by many mathematicians including Kahn, Komlós, Szemerédi, Tao, Vu, Bourgain, Wood, Rudelson, and Vershynin, a breakthrough was recently achieved by Tikhomirov, who showed that, when the matrix entries are iid uniform from $\{0,1\}$, the probability of singularity if $(1/2 + o(1))^n$, which is tight since with probability $2^{-n}$ the first row is zero.
The new papers of Jain–Sah–Sawhney consider a natural extension of the singularity problem where each entry of the $n\times n$ matrix is iid: 1 with probability $p$ and 0 with probability $1-p$, for some value of $p$ other than 1/2.
A folklore conjecture in this area states that the main reason for the singularity of such random matrices is the presence of two equal rows/columns or some zero row/column:
Conjecture. For a fixed $0 < p < 1$, the probability that a random $n\times n$ matrix with iid Bernoulli(p) entries is singular is $1+o(1)$ times the probability that the matrix has one of the following: (a) two equal rows, (b) two equal columns, (c) a zero row, (d) a zero column.
Note that even Tikhomirov’s breakthrough result does not give precise enough results to answer this conjecture. However, we now have the following incredible result.
Theorem. (Jain–Sah–Sawhney) The above conjecture is true for all $p \ne 1/2$.
(The case $p = 1/2$ remains open.)
In fact, they prove something much more general: namely that the above result remains true if the random matrix entries are iid discrete random variables that are non-uniform on its support. Furthermore, they prove nearly tight bounds on the probability of these matrices having small least singular values. Very impressive!
## Gunby: Upper tails for random regular graphs
October 5, 2020
Ben Gunby, a final-year PhD student at Harvard whom I advise, just posted his paper Upper tails of subgraph counts in sparse regular graphs. This is a substantial paper that unveils surprising new phenomena for the problem of large deviations in random graphs.
Problem. What is the probability that a random graph has far more copies of some given subgraph than expected?
This is called the “upper tail” problem for random graphs. It is a problem of central interest in probabilistic combinatorics. There was a lot of work on this problem for the classic Erdős–Rényi random graph $G(n,p)$ (with edges appearing independently at random), with a lot of exciting recent developments. Ben considers the upper tail problem for random regular graphs, another natural random graph model. Random regular graphs are usually harder to analyze since the random edge appearance are no longer independent, and this increased difficulty also shows up for the upper tail problem.
Let me highlight one specific result in Ben’s paper (Theorem 2.9).
Theorem. (Gunby) Let $G_n^d$ be a random $n$-vertex $d$-regular graph, where $d = d(n)$ depends on $n$. Assume that $dn$ is even and and $n^{14/15}\log n \le d = o(n)$. Let $X$ denote the number of copies of in $G_n^d$. Then for every fixed $\delta > 0$ (here $\sim$ means up to $1+o(1)$ factor),
$-\log \mathrm{Pr}(X \ge (1+\delta)\mathbb{E} X) \sim \frac{(18\delta)^{1/3}}{2} \frac{d^3}{n} \left(\log \frac{n}{d}\right)^{2/3} \left(\log\log \frac{n}{d}\right)^{1/3}$
This result is surprising. The appearance of the fractional power of the log factor has not been observed in previous works on the upper tail problem. It is even more impressive that Ben was able to prove such a sharp estimate.
Understanding these large deviation probabilities roughly corresponds to understanding the “dominant reason” for rare events. Roughly speaking, the fractional power of log appears because the dominant reason is not the presence of some fixed subgraph (e.g., a large clique or hub), unlike all earlier results, which have asymptotics of the form $n^2 p^d \log(1/p)$. Something else more subtle is happening. While we still do not understand the complete picture, Ben’s result provides strong evidence why the upper tail problem for random regular graphs may be substantially more intricate than that of $G(n,p)$.
In this blog post, I will tell you some history and background of the upper tail problem in random graphs. I will try to explain why Ben’s result is surprising. Lastly, I will also mention a recent paper I have with Yang Liu on the upper tail problem in random hypergraphs, where a different new phenomenon is observed.
### Chasing the infamous upper tail
Roughly, and intuitively, to gain a precise understanding of the upper tail problem, we are really asking:
What are the dominant reasons for the rare event that a random graph $G(n,p)$ has way too many triangles?
We can now say confidently the following:
It is primarily because the such a rare random graph contains roughly a large clique or a large hub.
Here a “hub” is a set of vertices adjacent to all other vertices. It is not hard to see that planting a large clique or hub in a $G(n,p)$ would automatically generate many additional triangles (these triangles may use also the edges of the random graph).
For example, the following theorem is the culmination of a long line of work by many researchers.
(We use the following asymptotic notation, which is standard in this line of work: $f \ll g$ means $f = o(g)$.)
Theorem. Let $X$ denote the number of triangles in $G(n,p)$. Then, for $n^{-1/2} \ll p \ll 1$ and fixed $\delta > 0$, one has
$-\log \mathrm{Pr}(X \ge (1+\delta) \mathbb{E} X) \sim \min \left\{\frac{\delta}{3}, \frac{\delta^{2/3}}{2}\right\} n^2p^2\log\frac{1}{p}$
The problem of estimating the upper tail $\mathrm{Pr}(X \ge (1+\delta) \mathbb{E})$ has a long history. Janson and Rucinski called this problem the infamous upper tail, which had resisted a wide range of techniques that researchers have used to tackle this problem. This upper tail problem was an excellent litmus test in the development of concentration inequalities.
Several significant breakthroughs then occurred on this problem over the past decade.
DeMarco–Kahn and Chatterjee independently determined the order of magnitude of the log-probability in the above upper tail problem, for triangle counts in a random graph. Their work closed a log-factor gap left by previous methods developed by Kim–Vu.
The next goal was then to pinpoint the tail probability. This led to further developments of powerful tools.
The seminal work of Chatterjee and Varadhan developed a large deviation principle for the upper tail problem in random graphs, thereby reducing the problem to a variational problem in the space of graphons, i.e., graph limits. Their work paved the way for much further development, which has two complementary strands:
1. Extending the large deviation framework to sparser graphs. Chatterjee and Varadhan’s proof uses Szemerédi’s graph regularity lemma, which only allows us to handle dense graphs (here for random graphs $G(n,p)$ dense means constant $p$, whereas sparse means $p = o(1)$). Further work, starting with Chatterjee–Dembo’s “non-linear large deviations”, as well as further improvements by Eldan, Cook–Dembo, Augeri, Harel–Mousset–Samotij, and Basak–Basu, allow us to handle much sparser densities of graphs.
For certain subgraphs, including cliques, we now have a very precise understanding thanks to Harel–Mousset–Samotij, who gave a short and clever container-inspired combinatorial argument in the case of triangles.
2. Solving the variational problem. The above large deviation frameworks reduces the problem of estimating the tail probability to another problem—a variational problem in the space of graphons.
Variational problem for upper tails: Among edge-weighted graphs with some prescribed subgraph density, determine the minimum possible relative entropy (a certain non-linear function of the edge-weights)
These variational problems are more difficult than classical problems in extremal graph theory. In fact, despite some progress, we still do not understand how to solve such problems exactly in the “dense” regime.
For sparse graphs $G(n,p)$ with $p \to 0$, the variational problem was solved asymptotically in the case of clique subgraphs by Lubetzky–Zhao, and for more general subgraph counts by Bhattacharya–Ganguly–Lubetzky–Zhao.
The two complementary strands together pinpoint upper tail probabilities. Similar to the case of triangles, the main message is, roughly speaking,
The dominant reason for a random graph to have too many copies of some fixed $H$ is that it contains a large clique or a large hub.
Theorem. Let $H$ be a graph with maximum degree $\Delta$. Let $\delta > 0$. Let $X$ be the number of copies of $H$ in $G(n,p)$. Provided that $n^{-\alpha_H} \le p = o(1)$ (where $\alpha_H$ is some explicit constant that has improved over time), one has
$-\log \mathrm{Pr} (X \ge (1+\delta)\mathbb{E} X) \sim c(H, \delta) p^{\Delta} n^2 \log \frac{1}{p},$
where $c(H,\delta) > 0$ is an explicit constant determined in Bhattacharya–Ganguly–Lubetzky–Zhao.
While this is a pretty satisfactory formula, our understanding is still not complete.
1. Is the result valid for the “full” range of $p$?
2. What precise statements can you make about the conditioned random graph? (see Harel–Mousset–Samotij)
### Gunby’s work on random regular graphs
Ben Gunby’s work studies the upper tail problem for random regular graphs:
Problem. Let $X$ be the number of copies of some given subgraph $H$ in a random $d$-regular graph $G_n^d$. Estimate $-\log \mathrm{Pr}(X \ge (1+\delta)\mathbb{E} X)$, when $n^{1-c} \le d = o(n)$.
There is a similar dichotomy of steps as above:
1. Proving a large deviations framework. The upper bound to the probability can be deduced from the corresponding theorems for $G(n,p)$, while the lower bound requires new and lengthy arguments.
2. Solving the variational problem – this is where new difficulties and innovations lie.
A recent paper of Bhattacharya–Dembo (this is by Sohom Bhattacharya and not my coauthor Bhaswar Bhattacharya mentioned earlier) essentially solved the variational problem when $H$ is a regular graph (Ben had also independently worked out the case when $H$ is a cycle).
Ben Gunby’s new paper solves the variational problem for “most” graphs $H$ (those that satisfy some technical condition), which includes all graph $H$ with average degree greater than 4, as well as pretty much all small graphs.
For most graphs $H$, the upper tail has “expected” behavior, meaning that the dominant reason for seeing many copies of $H$ is, very roughly speaking, the presence of some large planted subgraph. The resulting tail probability formulas also resemble what we have seen earlier:
$-\log \mathrm{Pr}(X \ge (1+\delta) \mathbb{E} X) \sim \rho(H, \delta) n^2 (d/n)^{2+\gamma(H)} \log(n/d).$
for some graph parameter $\gamma(H)$, whose role here is far from obvious.
For every graph $H$, Ben manages to bound the log-probablity within a log-factor gap:
$n^2 (d/n)^{2+\gamma(H)} \lesssim -\log \mathrm{Pr}(X \ge (1+\delta) \mathbb{E} X) \lesssim n^2 (d/n)^{2+\gamma(H)} \log(n/d)$
The most surprising part of Ben’s paper is that for some $H$, the upper tail is not “expected” as above, i.e., the asymptotic formula two display equations earlier is incorrect for this $H$.
For the graph , Ben solves the variational problem, which led to the formula at the beginning of this post. The solution turns out to come not from “planting a subgraph” but rather from some other more delicate construction.
It remains to be seen what happens to other graphs $H$. Will we need even more intricate constructions? What does this mean for techniques towards solving the upper tail problem for the full range of degree parameters? There is still lot that we do not understand.
### Large deviations for random hypergraphs
In a recent paper with Yang Liu (to appear in Random Structures & Algorithms), we considered the extension of the upper tail problem to random hypergraphs. Yang Liu is a PhD student at Stanford. We started this work while Yang was an undergraduate student at MIT.
Consider the random $k$-uniform hypergraph $G^{(k)}(n,p)$, where every unordered $k$-tuple of vertices appears as an edge independently with probability $p$.
Problem. Given a $k$-uniform hypergraph $H$. Let $X$ denote the number of copies of $X$ in $G^{(k)}(n,p)$. Estimate $\mathrm{Pr}(X \ge (1+\delta)\mathbb{E} X)$.
As before, there are two complementary strands (1) developing large deviation framework and (2) solving the variational problem. Some of the work towards (1) for random graphs can be transferred over to the hypergraph setting. Towards (2), we solve the variational problem asymptotically for some $H$, and make a conjecture for what we believe should happen for general $H$.
We show that one cannot naively extend the upper tail results from random graphs to random hypergraphs. In our paper, we take the readers on a long discussion through several iterations of naive conjectures and counterexamples, eventually building up to our final conjecture, which takes some effort to state, but very roughly, says that
The dominant reason for upper tails in random hypergraphs is the presence of a collection of “stable mixed hubs.”
For both random regular graphs and random hypergraphs, these recent results highlight unexpected phenomena. They illustrate the richness of the problem and lead us to more questions than answers.
## Joints of varieties
September 12, 2020
The summer has come to an end. The temperature in Boston seems to have dropped all the sudden. Our fall semester has just started, though without all the new and returning students roaming around campus this time (most classes are online this term). I quite miss the view from my MIT office, and all the wonderful people that I would run into the department corridors.
I’ve been working from home for the past several months and connecting with my students via Zoom. In this and the upcoming several posts, I would like to tell you about some of the things that we have been working on recently.
With PhD student Jonathan Tidor and undergraduate student Hung-Hsun Hans Yu, we recently solved the joints problem for varieties by a developing a new extension of the polynomial method. (Also see my talk video and slides on this work.)
I discussed the joints problem in an earlier blog post. Recall that the joints problem asks:
Joints problem: At most how many joints can $N$ lines in $\mathbb{R}^3$ can make?
Here a joint is a point contained in three non-coplanar lines.
Here is an example of a configuration of lines making many joints. Take k generic planes in space, pairwise intersect them to form $N = \binom{k}{2}$ lines, and triplewise intersect to form $J = \binom{k}{3} \sim \frac{\sqrt{2}}{3} N^{3/2}$ joints.
The joints problem was raised in the early 1990’s. It is a problem in incidence geometry, which concerns possible incidence patterns of objects like points and lines. The joints problem has also received attention from harmonic analysts due to its connections to the Kakeya problem.
The big breakthrough on the joints problem was due to Guth and Katz, who showed that the above example is optimal up to a constant factor:
Joints theorem (Guth–Katz). $N$ lines in $\mathbb{R}^3$ form $O(N^{3/2})$ joints.
Guth and Katz solved the problem using a clever technique known as the polynomial method, which was further developed in their subsequent breakthrough on the Erdős distinct distances problem. The polynomial method has since then grown into a powerful and indispensible tool in incidence geometry, and it also led to important recent developments in harmonic analysis.
I learned much of what I know about the polynomial method from taking a course by Larry Guth while I was a graduate student. Guth later developed the course notes into a beautifully written textbook, which I highly recommend. A feature of the expository style of this book, which I wish more authors would adopt, is that it frequently motivates ideas by first explaining false leads and wrong paths. As I was reading this book, I felt like that I was being taken on a journey as a travel companion and given a lens towards the thinking process, rather than somehow being magically teleported to the destination.
In my previous paper on the joints problem with Hans Yu, which I discussed in earlier blog post, we sharpened the Guth–Katz theorem to the optimal constant.
Theorem (Yu–Zhao). $N$ lines in $\mathbb{R}^3$ have $\le \frac{\sqrt{2}}{3} N^{3/2}$ joints.
Everything I have said so far are also true in arbitrary dimensions and over arbitrary fields, but I am only stating them in $\mathbb{R}^3$ here to keep things simple and concrete.
### Joints of flats
Our new work solves the following extension of the joints problem.
Joints problem for planes. At most how many joints can $N$ planes in $\mathbb{R}^6$ make?
Here a joint is a point contained in a triple of 2-dimensional planes not all contained in some hyperplane.
The example given earlier for joints of lines extends easily to joints of planes.
Example. Take $k$ generic 4-flats in $\mathbb{R}^6$, pairwise intersect them to form $N = \binom{k}{2}$ planes, and triplewise intersect them to form $\binom{k}{3} = \Theta(N^{3/2})$ joints.
Joints of planes concerns incidences between high dimensional objects, which in general tend to be more intricate compared to incidence problems about points and lines.
I first learned about this problem from a paper of Ben Yang, who did this work as a PhD student of Larry Guth. Yang proved that $N$ planes in $\mathbb{R}^6$ form at most $N^{3/2+o(1)}$ joints. His proof uses several important recent developments in incidence geometry, and in particular uses the polynomial partitioning technique of Guth and Katz (as well as subsequent extensions due to Solymosi–Tao and Guth). This is a fantastic result, though with two important limitations due to the nature of the method:
1. There is an error term in the exponent $N^{3/2+o(1)}$ that is conjectured not to be there. One would like a bound of the form $O(N^{3/2})$
2. The result is restricted to planes in $\mathbb{R}^6$ and does not work if we replace $\mathbb{R}$ by an arbitrary field. This is because the polynomial partitioning method uses topological tools, in particular the ham sandwich theorem, which crucially requires the topology of the reals.
Our work overcomes both difficulties. As a special case, we prove (here $\mathbb{F}$ stands for an arbitrary field, and the hidden constants do not depend on $\mathbb{F}$):
Joints of planes (Tidor–Yu–Zhao). $N$ planes in $\mathbb{F}^6$ have $O(N^{3/2})$ joints.
And more generally, our results hold for varieties instead of flats. Here a joint is a point contained in a triple of varieties, such that it is a regular point on each of the three varieties and the three tangent plane do not lie inside a common hyperplane.
Joints of varieties (Tidor–Yu–Zhao). A set of 2-dimensional varieties in $\mathbb{F}^6$ of total degree $N$ has $O(N^{3/2})$ joints.
The above statements are special cases of our results. The joints problem for lines is known to hold under various extensions (arbitrary dimension and fields, several sets/multijoints, counting joints with multiplicities), and our theorem holds in these generalities as well.
### Proof sketch of the joints theorem for lines
Here is a sketch of the proof that $N$ lines in $\mathbb{R}^3$ make $O(N^{3/2})$ joints (following the simplifications due to Kaplan–Sharir–Shustin and Quilodrán; also see Guth’s notes for details of the proof).
The first two steps below are both hallmarks of the polynomial method, and they are present in nearly every application of the method. The third step is a more joints-specific argument.
1. Parameter counting. The dimension of the vector space of polynomials in $\mathbb{R}[x,y,z]$ up to degree $n$ is $\binom{n+3}{3}$. We deduce that if there are $J$ joints and $J < \binom{n+3}{3}$, then one can find a nonzero polynomial $g$ of degree $\le C J^{1/3}$ (for some constant $C$) that vanishes on all the joints (as there are more coefficients to choose from than there are linear constraints). Take such a polynomial $g$ with the minimum possible degree.
2. Vanishing lemma. A nonzero single variable polynomial cannot vanish more times than its degree.
3. If all lines have $> CJ^{1/3}$ joints, then the vanishing lemma implies that $g$ vanishes on all the lines. At each joint, $g$ vanishes identically along three independent lines, and hence its directional derivatives vanish in three independent directions, and thus the gradient $\nabla g$ vanishes at the joint. Therefore, the three partial derivatives $\partial_x g$, $\partial_y g$, $\partial_z g$ all vanish at every joint, and at least one of them is a nonzero polynomial and has smaller degree than $g$, thereby contradicting the minimality of the degree of $g$. Therefore, the initial assumption must be false, namely that some line has $\le C J^{1/3}$ joints. We can then remove this line, deleting $\le CJ^{1/3}$ joints, and proceed by induction to prove that $J = O(N^{3/2})$.
### Some wishful thinking
It is natural to try to extend the above proof for the joints problem for planes. One quickly runs into the following conundrum:
How does one generalize the vanishing lemma to 2-variable polynomials?
The vanishing lemma is such an elementary fact about how polynomial behaves on a line. Yet it plays a crucial role in the polynomial method. We would be able to adapt the above proof to joints of planes if only something like the following would be true (they are not):
Wishful thinking 1. Show that every nonzero polynomial $g(x,y)$ of degree $n$ has $O(n^2)$ zeros.
Well, this is obviously nonsense. The zero set of $g(x,y)$ is typically a curve with infinitely many points.
How else might we try to extend the vanishing lemma to two dimensions? Bézout’s theorem is one such extension. It tells us that two polynomials $g(x,y)$ and $h(x,y)$ without common factors cannot have more than $(\deg g)(\deg h)$. So, perhaps, instead of trying to find one polynomial vanishing at all joints, we try to find a pair of polynomials.
Wishing thinking 2. Show that one can find two polynomials $g, h \in \mathbb{R}[x_1, x_2, x_3, x_4, x_5, x_6]$ vanishing at all $J$ joints and $(\deg g)(\deg h) = O(J^{1/3})$, and such that $g$ and $h$ have no common factors when restricted to any of the given planes.
This approach is a natural one to try for the joints of planes problem, and is an approach that many people have thought about. It also seems extremely hard to get it to work (likely impossible). It seems closely related to the inverse Bézout problem, which already illustrates some of the difficulties encountered.
Instead of asking a polynomial to vanish at every joint, we can demand more. Namely that we can ask the polynomial to vanish at each joint to some high order (equivalently, asking some derivatives of the polynomial to vanish at the joints). This idea (sometimes called the “method of multiplicities”) has shown to be useful in a variety of contexts. Perhaps we can modify Wishing thinking 1 to account for these multiplicities?
Wishful thinking 3. Show that every nonzero polynomial $g(x,y)$ of degree $n$ can vanish to order $s$ at no more than $O(n^2/s^2)$ points.
The reason why one might hope for such a statement is that there are $\binom{n+2}{2} \approx n^2/2$ degrees of freedom for choosing $g$, and this number is roughly the same as the number of constraints associated to forcing the polynomial to vanish to order $s$ at $n^2/s^2$ points.
However, Wishful thinking 3 is still too good to be true. For example, the polynomial $g(x,y) = y^s$ is a low-degree polynomial that vanishes to order $s$ at every point on the $x$-axis.
So what was wrong with the parameter counting intuition? It must be the case that some of the linear constraints associated to these high order vanishing constraints are linearly dependent.
### Extending the polynomial method
So far all I have told you are what doesn’t work, and I haven’t said anything about what does work. Where we left off above is a good starting point of what we actually do, which I will not describe here in this blog post but instead refer you to either my talk or our paper for details. We have to account for the linear dependencies between these higher order vanishing conditions, and we come up with a process of assigning non-redundant vanishing conditions (each in the form of some higher order derivative operator evaluated at a joint). We prove a new vanishing lemma tailored for this joints problem. Rather vaguely, it has the following form:
A new vanishing lemma. Given a finite set of planes in $\mathbb{R}^6$ forming some joints, and a positive integer $n$, and an integer (“handicap”) attached to each joint, we run a certain process that assigns, on each plane, a total of $\binom{n+2}{2}$ derivative operators across the joints on this plane. Then for every nonzero polynomial $g(x_1, \dots, x_6)$ of degree at most $n$, there is some joint $p$, three planes $F_1, F_2, F_3$ containing $p$, and a derivative operator $D_i$ assigned to $p$ on $F_i$ through our process for each $i =1,2,3$, so that $D_1 D_2 D_3 g(p) \ne 0$.
This new vanishing lemma overcomes the earlier difficulty and allows us to extend the polynomial method to solve the joints problem for planes.
Let me finish with an open problem: while we can prove the optimal constant for the joints theorem for lines, we do not yet know how to prove the optimal constant for the joints theorem for planes.
Conjecture. $N$ planes in $\mathbb{F}^6$ have $\le (\frac{\sqrt{2}}{3} + o(1)) N^{3/2}$ joints.
## Ashwin Sah's new diagonal Ramsey number upper bound
May 20, 2020
Ashwin Sah just proved a new upper bound to diagonal Ramsey numbers. See his preprint on the arXiv. This is the first improvement since Conlon’s upper bound published in Annals of Math in 2009, which in turn built on earlier work of Thomason (1988).
Obtaining asymptotics of Ramsey numbers is perhaps the central open problem of Ramsey theory in combinatorics. Research on this problem had led us to a lot of new mathematics that have proven to be useful elsewhere. Most notably, Erdős’ proof of a lower bound to Ramsey numbers (1947) initiated the probabilistic method, by now an indispensable method in combinatorics.
Ashwin’s improvement relies on new insights into graph quasirandomness properties.
Congratulations Ashwin! This is an impressive accomplishment.
Ashwin had just finished his undergraduate studies at MIT. I’m happy that he will be staying at MIT for his PhD. Ashwin already has an impressive list of papers. I had previously mentioned Ashwin on this blog reporting on our joint work (together with Mehtaab Sawhney and David Stoner) on our work on independent sets and the reverse Sidorenko inequality.
## Joints tightened
November 21, 2019
I’m happy to announce a new paper titled Joints tightened coauthored with Hung-Hsun Hans Yu, an undergraduate student at MIT. In this paper, we determine the tight constant in the joints problem.
The joints problem is a classic problem in incidence geometry. A typical question in incidence geometry concerns what kinds of configurations can be built using just points and lines.
Suppose you are allowed to draw L lines in space. What is the maximum possible number of points that lie on three lines (let’s call these points “triple intersections” for now)?
It turns out this is not a great question, since you can “cheat” by drawing a 2-dimensional picture from a grid as below. This configuration of L lines has $\Theta(L^2)$ triple intersections, and you cannot do much better since L lines have at most $L^2$ intersections.
The issue with the above example is that it is really a 2-dimensional configuration, whereas we really want to ask a 3-dimensional question. One way to make the problem much more interesting is only count “truly 3-dimensional” intersection points.
Given a collection of lines in space, we define a joint to be a point that is contained in three of our lines and not all lying on the same plane.
Joints problem. What is the maximum number of joints that can be formed by L lines in 3-dimensional space?
The joints problem was raised in early 1990’s, and it turned out to be a very interesting problem! Besides its intrinsic interest as a combinatorial geometry problem, it also caught the attention of analysts as the joints problem is connected to the Kakeya problem, a central problem in analysis (this connection was popularized by Wolff).
Here are some examples of configurations of lines that have a lot of joints.
Example 1. The easiest example to visualize is by considering a $k \times k \times k$ grid of lines, which has $L = 3k^2$ lines and $J = k^3 = (L/3)^{3/2}$ joints:
Example 2. Actually, there is another example that has even more joints (by a constant factor). Think about a tetrahedron, formed by 4 faces (planes), and whose edges form 6 lines, and the vertices form 4 joints. We can generalize the tetrahedron example by taking k generic planes in space, and have the planes pairwise intersect to make $L = \binom{k}{2}$ lines, and triplewise intersect to make $J = \binom{k}{3}$ joints. (This example appeared in the original paper on the joints problem.)
In both examples, the number of joints is $\Theta(L^{3/2})$. A matching upper bound on the number of joints turned out to be much less obvious. A stunning breakthrough took place in 2008 when Guth and Katz proved an $O(L^{3/2})$ upper bound on the number of joints in using the polynomial method. Their solution was partly inspired by Dvir’s stunning solution to the finite field Kakeya problem, and it was also a precursor to Guth and Katz’s subsequent celebrated solution of the Erdős distinct distances problem.
Larry Guth conjectured in his book The Polynomial Methods in Combinatorics that the second example above is best possible. I first learned about this very nice conjecture as a graduate student in Guth’s class on the polynomial method.
The main result of our paper is a new improved upper bound on the joints problem, matching the constant in the second example above. More precisely, we prove that L lines in space make at most $\frac{\sqrt{2}}{3}L^{3/2}$ joints, thereby confirming Guth’s conjecture up to a $1+o(1)$ factor. Our results also hold in arbitrary dimensions and over arbitrary fields.
It was quite unexpected to us that such a tight bound can even be proved, especially given how long it took just to obtain an upper bound of the right order of magnitude on the joints problem. In pretty much all other classical problems in incidence geometry (i.e., the Szemerédi–Trotter theorem), the optimal constant factor is not known. In fact, our result might be the first one in incidence geometry that obtains a tight constant factor.
## Equiangular lines with a fixed angle
July 30, 2019
I’m happy to announce our new paper Equiangular lines with a fixed angle joint with four MIT coauthors: Zilin Jiang, Jonathan Tidor, Yuan Yao, and Shengtong Zhang.
Zilin is an Instructor (postdoc), Jonathan is my PhD student who just finished his second year, and Yuan and Shengtong are undergraduates who just finished their second and first respective years.
Our paper solves a longstanding problem in discrete geometry concerning equiangular lines, which are configurations of lines in space that are pairwise separated by the same angle. How many such lines can exist simultaneously in a given dimension?
We determine, for every fixed angle, the maximum number of equiangular lines with the given angle in high dimensions.
Our paper builds on a long sequence of earlier ideas. There were some important progress on this problem by Igor Balla, Felix Dräxler, Peter Keevash, and Benny Sudakov, as featured in this earlier Quanta Magazine article written for a lay audience.
In the end we finished off the problem in a clean and crisp manner, in a 10-page paper with a self-contained proof.
It has been known that the study of equiangular lines is closed linked to spectral graph theory, which seeks to understand graphs via their spectrum (i.e., eigenvalues). We prove our main result via a new insight in spectral graph theory, namely that a bounded degree graph must have sublinear second eigenvalue multiplicity.
## Impartial digraphs
June 27, 2019
I just finished a new paper, Impartial digraphs, coauthored with Yunkun Zhou, who just completed his undergraduate studies at MIT and will be moving to Stanford to start his PhD this fall.
The problem (along with the conjecture that we proved) was proposed by Jacob Fox, Hao Huang, and Choongbum Lee. Ron Graham had also told me about this problem with much enthusiasm during tea time one afternoon at the Berkeley Simons Institute a couple of years ago, after learning about it from Jacob.
I really enjoyed working on this paper with Yunkun. The problem is natural and simple to state, the answer appears somewhat unexpected (at least initially), and the solution ended up being quite tricky yet clean, employing a variety of tools in combinatorics, ranging from analytic methods (graph limits), algebraic considerations (polynomial identities and factorization), as well basic combinatorial techniques such as bijections.
We are going to talk about directed graphs (which graph theorists like to abbreviate as digraphs). These are graphs where every edge is given an orientation.
A tournament is a directed graph where there is a directed edge between every pair of vertices (think of it as recording the outcome of a round-robin tournament where every player plays against every other player).
Given a directed graph H (think of it as a pattern), we can ask: how many different copies of the pattern H appear in a given tournament?
Look at the following 4-vertex tournament H. It has the following curious property: all tournaments of a fixed size contains the same number of copies of H, no matter how the edges in the tournament are oriented.
Hmm, why is this true? And are there other directed graphs H with the same property, namely that all tournaments of a given size contain the same number of copies of H?
In the good old mathematical tradition, we came up with a name for directed graphs H with this curious property: we called them impartial (as in: impartial judges of a tournament).
Okay. So why is the above directed graph impartial? Are there any other impartial directed graphs? What are all the impartial directed graphs?
Here is a simple example of an impartial digraph:
Here is another one:
Indeed, given any 10-vertex tournament, we can count the number of copies of the first edge (it doesn’t depend on the tournament), and then count the number of copies of the second edge among the remaining 8 vertices (it still doesn’t depend on the tournament).
Once these two edges are placed in a tournament, there are two ways to join their tips, but they result in identical patterns (one flipped from the other):
It must then follow that this directed graph is impartial as well:
We can follow the same logic and continue this procedure, iterate, and build up even larger impartial directed graphs. Each step, take two copies of the previous tree (keeping the same edge orientations), and then add a new edge between a twin pair of vertices.
We call any directed graph that can building up as above a recursively bridge-mirrored tree.
The same argument as earlier shows that every recursively bridge-mirrored tree is impartial.
Are there other impartial digraphs? Did we miss any?
It turns out, there are actually no others. We have found them all, and that is the main result of our paper:
Theorem. A directed graph is impartial if and only if every component is recursively bridge-mirrored trees.
Just a word about how about this problem is related to a bigger picture. It is related to a number of other graph homomorphism inequalities that are central in extremal graph theory. It can be considered as the “equality case” for a directed analog of an open problem known as Sidorenko’s conjecture. See our paper for some further discussions and open problems.
## A reverse Sidorenko inequality
September 26, 2018
Together with undergraduates Ashwin Sah, Mehtaab Sawhney, and David Stoner (the same team that proved Kahn’s conjecture on independent sets that I blogged about earlier, we are excited to announce our new paper A reverse Sidorenko inequality. This paper solves several open problems concerning graph colorings and homomorphisms, including one of my favorite problems regarding maximizing the number of q-colorings in a d-regular graph. I had previously blogged about some of these problems.
It has been truly a pleasure working with Ashwin, Mehtaab, and David on this project. I have learned so much from them.
Here are some highlights of our new results.
### 1. Graph colorings.
Among d-regular graphs, which graph has the most number of proper q-colorings, exponentially normalized by the number of vertices of G?
We show that the extremal graph is the complete bipartite graph $K_{d,d}$ (or disjoint unions thereof, as taking disjoint copies does not change the quantity due to the normalization).
### 2. Graph homomorphisms.
Among d-regular graphs, which triangle-free graph G has the most number of homomorphisms into a fixed H, again exponentially normalized by the number of vertices of G?
We show that the extremal graph G is the complete bipartite graph $K_{d,d}$.
The triangle-free hypothesis on G is best possible in general. For certain specific H, such as $K_q$, corresponding to proper q-colorings as earlier, the triangle-free hypothesis can be dropped. It remains a wide open problem to determine for which H can one drop the triangle-free hypothesis on G.
### 3. Partition functions of Ferromagnetic spin models.
Among d-regular graphs, which graph maximizes the log-partition function per vertex of a given Ferromagnetic spin model (e.g., Ising model)?
We show that the extremal graph is the clique $K_{d+1}$.
For each setting, we establish our results more generally for irregular graphs as well, similar to our earlier work on independent sets.
Our results can be interpreted as a reverse form of the inequality in Sidorenko’s conjecture, an important open problem in graph theory stating a certain positive correlation on two-variable functions.
One can also view our results as a graphical analog of the Brascamp–Lieb inequalities, a central result in analysis.
This paper resolves one of my favorite open problems on this topic (the number of graph colorings). It also points us to many other open problems. Let me conclude by highlighting one of them (mentioned in #2 earlier). I’ll state it in a simpler form for d-regular graphs, but it can be stated more generally as well.
Open problem. Classify all H with the following property: among d-regular graphs G, the number of homomorphisms from G to H, exponentially normalized by the number of vertices of G, is maximized by $G = K_{d,d}$.
Our results show that $H = K_q$ works, even when some of its vertices are looped. Generalizing this case, we conjecture that all antiferromagnetic models H have the same property (though we know that this would not be the complete class, even if true).
## The number of independent sets in an irregular graph
May 12, 2018
I am excited to announce our new paper, The number of independent sets in an irregular graph, coauthored with these three amazing collaborators:
Ashwin is a freshman at MIT, Mehtaab is a sophomore at MIT, and David is a junior at Harvard.
The paper solves a conjecture made by Jeff Kahn in 2001 concerning the number of independent sets in a graph.
An independent set of a graph is a subset of vertices with no two adjacent.
Since many problems in combinatorics can be naturally phrased in terms of independent sets in a graph or hypergraph, getting good bounds on the number of independent sets is a problem of central importance.
Instead of giving the exact statement of the conjecture (now theorem), which you can find in the abstract of our paper, let me highlight a specific instance of a problem addressed by the theorem:
Question. Consider the family of graphs with no isolated vertices and
• exactly a third of the edges have degree 3 on both endpoints, and
• a third of the edges have degree 4 on both endpoints,
• the remaining third of the edges have degree 3 on one endpoint and degree 4 on the other endpoint.
What is the smallest constant c such that every n-vertex graph satisfying the above properties has at most $c^n$ independent sets?
In other words, letting $i(G)$ denote the number of independent sets and $v(G)$ the number of vertices of $G$, maximize the quantity $i(G)^{1/v(G)}$ over all graphs G in the above family.
(See the end of this post for the answer)
In summer 2009, as an undergraduate participating in the wonderful Research Experience for Undergraduates (REU) run by Joe Gallian in Duluth, MN, I learned about Kahn’s conjecture (somewhat accidentally actually, as I was working on a different problem that eventually needed some bounds on the number of independent sets, and so I looked up the literature and learned, slightly frustratingly at the time, that it was an open problem). It was on this problem that I had written one of my very first math research papers.
I had been thinking about this problem on and off since then. A couple years ago, Joe Gallian invited me to write an article for a special issue of the American Mathematical Monthly dedicated to undergraduate research (see my previous blog post on this article), where I described old and new developments on the problem and collected a long list of open problems and conjectures on the topic (one of them being Kahn’s conjecture).
This spring, I had the privilege with working with Ashwin, Mehtaab, and David, three energetic and fearless undergraduate students, finally turning Kahn’s conjecture into a theorem. Fearless indeed, as our proof ended up involving quite a formidable amount of computation (especially for such an innocent looking problem), and my three coauthors deserve credit for doing the heavy-lifting. I’ve been told that there had been many late night marathon sessions in an MIT Building 2 classroom where they tore apart one inequality after another.
This is certainly not the end of the story, as many more interesting problems on the subject remain unsolved—my favorite one being the analogous problem for colorings, e.g., instead of counting the number of independent sets, how about we count the number of ways to color the vertices of the graph with 3 colors so that no two adjacent vertices receive the same color? (See my previous blog post for some more discussion.)
Anyway, here is the the answer to the question above. The number of independent sets in an n-vertex graph satisfying the above properties is at most $c^n$, where the optimal constant for c is $15^{4/63} \times 23^{1/21} \times 31^{1/28} = 1.558\dots$, or, more helpfully, the maximum is attained by the following graph (in general, the theorem says that the maximizer is always a disjoint union of complete bipartite graphs): | 21,958 | 95,481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-14 | longest | en | 0.783014 |
http://www.jiskha.com/display.cgi?id=1329077503 | 1,493,577,862,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125841.92/warc/CC-MAIN-20170423031205-00499-ip-10-145-167-34.ec2.internal.warc.gz | 562,766,804 | 3,584 | posted by on .
determine the domain of the real valued function f(x) = sqrt(5 - (sqrtx))
sqrt(x) has domain x >= 0
sqrt(5-sqrt(x)) has domain 5 - sqrt(x) >= 0, or sqrt(x) <= 5
so, the domain of f(x) is 0 <= x <= 25 | 79 | 218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-17 | latest | en | 0.793145 |
https://www.manhattaneliteprep.com/blog/weakening-questions-vs-strengthening-questions/ | 1,726,386,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00719.warc.gz | 811,030,839 | 6,774 | # GMAT Verbal: Critical Reasoning - Weakening vs. Strengthening
## GMAT Critical Reasoning: Weakening Questions
Q27.
Save-a-Tot Corporation is a manufacturer of safety seats for bicycles and automobiles. Children often die unnecessarily in collisions because they were not properly fastened into seats that conform to their fragile bodies. Save-a-Tot has recently designed a new type of safety seat that is able to conform more closely than ever to cures in the spines and necks of small children. This new design also cushions the head in a better way. Save-a-Tot claims that the usage of these safety seats will decrease child mortality rates in serious collisions by up to 60 percent.
Which of the following, if true, represent the strongest challenge to Save-a-Tot’s claim?
(A) The child safety seats that Save-a-Tot has designed are made of a lighter plastic compound that turns brittle when it is exposed to cold temperatures over a relatively short period of time.
(B) The government demands that Save-a-Tot produce these child safety seats to very strict specifications.
(C) By providing Save-a-Tot seats free with new cars, automobile sales will increase.
(D) The proposed child safety seats will add too much weight to bicycles.
(E) As production costs increase, Save-a-Tot will have to raise the price of its safety seats.
*(A) is the correct answer. We’re looking for something that undermines the Save-a-Tot corporation claims. (D) is the most tempting wrong answer because it suggests that the seats won’t work on bicycles.
Q29.
Passengers must exit airplanes swiftly after accidents, since gases released following accidents are toxic to humans and often explode soon after being release. In order to prevent passenger deaths from gas inhalation,, safety officials recommend that passengers be provided with smoke hoods that prevent inhalation of the gases.
Which of the following, if true, constitutes the strongest reason not to require implementation of the safety officials’ recommendation?
(A) Test evacuations showed that putting on the smoke hoods added considerably to the overall time it takes passengers to leave the cabin.
(B) Some airlines are unwilling to buy the smoke hoods b/c they consider them to be prohibitively expensive.
(C) Although the smoke hoods protect passengers from the toxic gases, they can do nothing to prevent thee gases form igniting.
(D) Some experience flyers fail to pay attention to the safety instructions given on every commercial flight before takeoff.
(E) In many airplane accidents, passengers who were able to reach emergency exits were overcome by toxic gases before they could exit the airplane.
*This is a weakener question. It is a proposal as well. We are looking for something that contradicts the proposal. What is the desired effect: to prevent passenger death from gas inhalation. The correct answer is (A) because it weakens the proposal and states that the proposal will not work. (B) does not tell us that the plan will not work. (E) is a strengthener and supports the proposal. (D) is not in the scope of the argument.
## GMAT Critical Reasoning: Strengthening Questions
Q30.
Finance Minister: last year was disastrous for our manufacturing section, which has traditionally contributed about 75% of our national budget. It is therefore encouraging that there is evidence that the IT sector is growing stronger. Taxes from the IT sector accounted for 15% of our national budge-up from 8% last year.
On the basis of the statement above, what best supports the above conclusion?
(A) The increase in taxes from the IT sector could have merely been the result of new laws imposed on the IT sector.
(B) The profits of the IT sector remained at a steady level despite the fact that it paid more taxes to the national government.
(C) The rise in the percentage of taxes that the IT sector contributed to the national government was insignificant in actual dollar terms.
(D) It is difficult to determine whether the jump from 8-15% tax contribution by the IT sector will be ongoing.
(E) The information given above does not fairly compare the contribution of taxes paid by different industries to the nation government.
*The keywords here that indicate that it’s a strengthening question is “ best supports.” The key issue is to read the question stem carefully. We need to think about what the conclusion is of the argument: there is evidence that the IT sector is growing stronger; so we’re looking for something that supports this argument. Answer (E) is wrong because it talks about other industries. The correct answer is (B).
(D) is wrong because it is a weakener that contradicts the conclusion. (D) is a classic trap answer because it’s not a strengthener at all but a weakener. (C) and (A) are weakener and contradict the conclusion. | 987 | 4,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.949919 |
https://brilliant.org/discussions/thread/find-the-integral/ | 1,531,932,410,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590295.61/warc/CC-MAIN-20180718154631-20180718174631-00194.warc.gz | 614,224,428 | 14,480 | # Find the integral
$\large \int \dfrac{ (\sin x + \cos x)^4}{(\cos x - \sin x)^4} \, dx = \, ?$
Note by Alaa Yousof
2 years, 4 months ago
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2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
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For simplicity sake, let $$s$$ and $$c$$ denote the functions $$\sin x , \cos x$$ respectively, then we have $$s^2 + c^2 = 1 , 2cs = \sin(2x)$$.
Then, $\begin{eqnarray} (s+c)^4 &=& s^4 + c^4 + 4cs(s^2 + c^2) + 6(cs)^2 = (s^2 + c^2)^2- 2(cs)^2 + 4cs + 6(cs)^2 = 1 + 4cs + 4(cs)^2 \\ (s-c)^4 &= & s^4 + c^4 - 4cs(s^2 + c^2) + 6(cs)^2 = (s^2 + c^2)^2- 2(cs)^2 - 4cs + 6(cs)^2 = 1 - 4cs + 4(cs)^2\end{eqnarray}$
Taking their ratio gives: $\begin{eqnarray} \dfrac{1 + 4cs + 4(cs)^2}{1 - 4cs + 4(cs)^2 } &=& 1 + \dfrac{8sc}{1 - 4cs + 4(cs)^2} \\ &=&1 + \dfrac{4\sin(2x)}{1 - 2\sin(2x) + \sin^2(2x)} = 1 + 4 \cdot \dfrac{\sin (2x)}{(\sin(2x) - 1)^2} \\ &=& 1 + 4 \left [\dfrac1{\sin(2x) - 1}+ \dfrac1{(\sin(2x) - 1)^2} \right ] \end{eqnarray}$
So we're integrating the final expression in the equation above. Use half angle tangent substitution, we have $$t = \tan(x) \Rightarrow dx = \dfrac{dt}{1+t^2} , \sin(2x) = \dfrac{2t}{1+t^2}$$. Use partial fractions to finish it off, note that $$2t - 1 - t^2 = -(t-1)^2$$.
Can you finish it off from here?
- 2 years, 4 months ago
no what are the next steps please
- 2 years, 3 months ago
What do you mean?
- 2 years, 3 months ago
how will we use the partial fractions
- 2 years, 3 months ago
I don't know where exactly you're stuck on. Please show your steps.
- 2 years, 3 months ago | 968 | 2,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-30 | latest | en | 0.598363 |
https://mvtrinh.wordpress.com/2011/06/20/b-x-de-abc/ | 1,532,172,558,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00005.warc.gz | 716,955,445 | 16,048 | ## B x DE = ABC
$A,B, \textup{and } C$ are consecutive digits. $D \textup{ and }E$ are also consecutive digits. $A,B,C,D,\textup{and }E$ are all different digits. If $B\times DE=ABC$, find the sum of $A + B + C + D + E$. (Note: $C \textup { and }D$ are not necessarily consecutive digits.
Source: mathcontest.olemiss.edu 10/29/2007
SOLUTION
Let $A=a$. Then, $B=a+1 \textup { and } C=a+2$ . Let $D=d$, then $E=d+1$.
Since $D$ is in the tens place and $E$ in the ones place, the value of $DE$ is represented by
$DE=10d+1\left (d+1\right )$
$=11d+1$
Since $A$ is in the hundreds place, $B$ in the tens place, and $C$ in the ones place, we represent the value of $ABC$ by
$ABC=100a+10\left (a+1\right )+1\left (a+2\right )$
$=100a+10a+10+a+2$
$=111a+12$
If $B\times DE=ABC$, then we have the following equation:
$\left (a+1\right )\left (11d+1\right )=111a+12$
Let’s try a few values for $a$.
a=0
$\left (0+1\right )\left (11d+1\right )=111\times 0+12$
$11d+1=12$
$11d=11$
$d=1$
Not a solution because it is conflicting with $b=1$.
a=1
$\left (1+1\right )\left (11d+1\right )=111\times 1+12$
$2\left (11d+1\right )=123$
Not a solution because the left side is even but the right side is odd.
a=2
$\left (2+1\right )\left (11d+1\right )=111\times 2+12$
$3\left (11d+1\right )=234$
$33d+3=234$
$33d=231$
$d=7$
Check
$A=2,\;B=3,\;C=4,\;D=7,\;E=8$
$B\times DE=ABC$
$3\times 78=234$?
$234=234$
YES
Thus, $A+B+C+D+E=2+3+4+7+8=24$. | 617 | 1,430 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 42, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2018-30 | latest | en | 0.689625 |
https://www.pinterest.co.uk/alexlynn698/maths/ | 1,618,083,443,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00635.warc.gz | 1,071,002,856 | 60,709 | # maths
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Do you remember playing with magnetic or wooden shapes when you were little? I had a set of wooden shapes in different sizes and colours which could be used to make all sorts of pictures and patterns. This is my DIY version, which I made for my class recently as we've been learning about shapes. I started with seven or eight sheets of colourful paper and being very lazy, I cut the shapes out mainly freehand and holding the whole bundle together at the same time! The only shape I didn't cut…
These Counting Activities: Building Pizzas can be set up as cookie sheet activities, a magnet center, or math center. This set includes two pizzas, 12 build a pizza instruction cards, and lots of build a pizza building pieces (all in color). In this activity, students watch the pizza toppings pile up as they work on basic counting skills and following directions. Students count out the correct number of pizza toppings indicated on the build a pizza instruction cards to build their…
Jack and the Beanstalk Finger Gym. In this activity you have to put the right amount of beans onto the numbered leaves using the tweezers.
Early Years ideas from Tishylishy. 72,736 likes · 31,674 talking about this. Sharing photos, provision enhancements and outcomes from my EYFS class and...
Introducing Sorting: Teaching Young Toddlers - a first lesson in sorting with a 16 month old! Learn tips and tricks for introducing sorting to toddlers
2D shapes with masking tape #blacktray #tuffspot #lego
Take Aim! I would stick or write the numbers on a wall and then use a wet sponge to throw at the numbers.
Challenge Table. Ch match household objects to its mathematical shape
The current Tuff Spot Tuesday theme (15th September) is: Farm and Harvest Tuff Spot. Time to explore farms or harvest time within your Tuff Spot. You could
Early Years ideas from Tishylishy. 72,736 likes · 31,674 talking about this. Sharing photos, provision enhancements and outcomes from my EYFS class and...
Interactive maths display - autumn - addition
Learn with Play at Home. Play based learning ideas and activities for kids. | 599 | 2,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-17 | latest | en | 0.953155 |
http://wiki.erepublik.com/index.php?title=Health_building&direction=next&oldid=531628 | 1,580,284,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251789055.93/warc/CC-MAIN-20200129071944-20200129101944-00194.warc.gz | 181,475,569 | 9,991 | # Health building
Languages:
## What is the change?
From day 1576 at 00:00, the houses from your storage will be replaced by a temporary Health Building that will increase your recoverable health by 50 wellness. You will have this building for a maximum number of 500 days. The days will be calculated according to the number and quality of the houses you have in your storage. However, don't worry - it's just an extra building. If you have bought any wellness building, it will remain like it was before.
## For how long will I have this building?
To calculate the number of days your health building will have, you need to calculate the wellness given to you by all houses in total, sum up the days and divide it by 50.
• House q1 gives 10 wellness for 10 days = total 100 wellness
• House q2 gives 20 wellness for 20 days = total 400 wellness
• House q3 gives 30 wellness for 30 days = total 900 wellness
• House q4 gives 40 wellness for 40 days = total 1600 wellness
• House q5 gives 50 wellness for 50 days = total 2500 wellness
Examples:
• If you have a quality 5 house with 50 days left, that house can give you 2500 wellness in total and if you divide it by 50, you will get 50 days for the temporary Health Building.
• If you have ten quality 5 houses with 49 days left, these houses can give you 24500 wellness in total, and if you divide it by 50, you will get 490 days for the temporary Health Building.
• If you have one quality 1 house with only 1 day left, that house will give you 10 wellness but the temporary Health Building will be adjusted in your favor to 1 day.
• If you have two quality 1 houses that can give you 200 wellness, three quality 2 houses that can give you 1200 wellness and four quality 3 houses can give you 3600 wellness. This sums up to a total wellness of 5000 and if you divide by 50, you will get 100 days for Temporary Health Building. | 463 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-05 | latest | en | 0.953517 |
https://bestexmorvatt.netlify.app/kosuta43463ro/interest-rate-and-effective-interest-rate-144.html | 1,713,927,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00536.warc.gz | 107,943,506 | 11,290 | Interest rate and effective interest rate
2 Sep 2019 Effective interest rate is a crucial term in finance as it helps to compare varying financial products that calculate interest on a compounding basis. 3 Oct 2017 The effective interest rate on the other hand is the true interest rate you are paying based on the outstanding loan amount calculated after each They convert between nominal and annual effective interest rates. If the annual nominal interest rate is known, the corresponding annual effective rate can be
The effective annual interest rate is the interest rate that is actually earned or paid on an investment, loan or other financial product due to the result of 12 Dec 2013 EFFECTIVE INTEREST RATE “A measure of the time value of money that fully reflects the effects of compounding” is the definition of effective The effective interest rate on any form of debt is that discount rate at which the present value of the payments made by the borrower equals the amount actually. Interest rate risk (IRR) is defined as the potential for changing market interest rates to adversely affect a bank's earnings or capital protection. Two previous issues The effective interest rate is calculated by dividing average total debt outstanding by total interest expenses (including capitalised interest). W
over a given period. Simply put, the effective annual interest rate is the rate of interest
Calculating Effective Interest Rate To calculate the effective interest rate, use the following formula: (1 plus i/n) to the nth power minus 1 where n is the compounding periods. So, for a 25 percent interest rate, you would calculate (1 plus.25/12) to the 12th power minus 1, which equals 28.073 percent. If you have an investment earning a nominal interest rate of 7% per year and you will be getting interest compounded monthly and you want to know effective rate for one year, enter 7% and 12 and 1. If you are getting interest compounded quarterly on your investment, enter 7% and 4 and 1. The effective interest rate is the interest rate on a loan or financial product restated from the nominal interest rate as an interest rate with annual compound interest payable in arrears. It is used to compare the annual interest between loans with different compounding terms (daily, monthly, quarterly, semi-annually, annually, or other). The Effective Annual Rate (EAR) is the interest rate that is adjusted for compounding over a given period. Simply put, the effective annual interest rate is the rate of interest that an investor can earn (or pay) in a year after taking into consideration compounding.
The term “interest rate” is one of the most commonly used phrases in fixed-income investment lexicon. The different types of interest rates, including real, nominal, effective and annual, are
An interest rate takes two forms: nominal interest rate and effective interest rate. The nominal interest rate does not take into account the compounding period. If you are shopping around for a personal loan, you have no doubt seen banks advertise two different interest rates: Annual Flat Rate and Effective Interest Rate 5 Feb 2019 The effective interest rate is the usage rate that a borrower actually pays on a loan. It can also be considered the market rate of interest or the The effective interest rate is the true rate of interest earned. It can also mean the market interest rate, the yield to maturity, the discount rate, the internal rate of 17 Oct 2019 The effective rate is how much interest you will really owe or receive once compounding is considered. APR is the annual percentage rate: the The nominal rate is the interest rate as stated, usually compounded more than once per year. The effective rate (or effective annual rate) is a rate that, Effective rates are what borrowers or investors actually pay or receive, depending on whether or how frequently interest is compounded. When interest is
The effective interest rate is the true rate of interest earned. It can also mean the market interest rate, the yield to maturity, the discount rate, the internal rate of
in interest is due. 8.243% The effective rate is equal to the interest actually paid divided by the principal. If the interest is compounded quarterly, then interest is charged at the rate of 2% every 3 months. So, if a car loan carries a nominal rate of 5 percent and the rate of inflation is 6 percent, the real rate of interest will be -1%. Effective Interest Rates. Effective interest rates incorporate the concept of compounding interest. It’s the rate you’ll earn or pay on a loan or an investment over a certain period. Effective interest rates aim to give you a fuller picture by taking these things into consideration. And, in the same scenario above, you’ve actually been paying for a loan of \$100,000 when you only got an initial loan of \$98,000. This translates to an effective interest rate of over 17% per annum.
6 Jun 2019 The effective annual interest rate is the rate of interest an investor earns in a year after accounting for the effects of compounding. How Does the
12 Dec 2013 EFFECTIVE INTEREST RATE “A measure of the time value of money that fully reflects the effects of compounding” is the definition of effective
The term effective interest rate is used to describe the actual rate of interest received when compounding is applied to a nominal rate of interest. The effective Historical Listing of Maximum Effective Interest Rates (Home Loan Rate). As Deteremined Pursuant to T.C.A. 47-15-103. Month Effective Rate. April, 2020 5.77 Guide to Effective Interest Rate .Here we discuss annual equivalent rate (AER) in detail. Also we will discuss how AER works with examples. – How should you manage your debt? Page 2. Nominal vs. effective interest rates . Nominal interest rate: rate What is the effective interest rate? Definition of Effective Interest Rate. The effective interest rate is the true rate of interest earned. It can also mean the market interest rate, the yield to maturity, the discount rate, the internal rate of return, the annual percentage rate (APR), and the targeted or required interest rate. Effective Annual Interest Rate: The effective annual interest rate is the interest rate that is actually earned or paid on an investment, loan or other financial product due to the result of The term “interest rate” is one of the most commonly used phrases in fixed-income investment lexicon. The different types of interest rates, including real, nominal, effective and annual, are | 1,349 | 6,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-18 | latest | en | 0.949946 |
https://discourse.processing.org/t/generative-terrain-customization/39396 | 1,670,253,224,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00061.warc.gz | 238,055,962 | 13,101 | # Generative terrain customization
Hi, after following along some tutorials I created a terrain generator using perlin noise and triangle_stripes. It is working and it runs.
For further iterations I would like to slow down the speed of the generated terrain and make the triangles stripes less visible/ less bold at the horizon of the terrain (the end of it) and have the triangle_stripes more visible in the font of the screen (the newly generated terrain)! Any ideas/ recommendations on how to achieve that?
Like in this example:
Here is the code I extracted from Daniel Schiffermans tutorial code:
int cols, rows;
// variable scl for scale
int scl = 20;
int w = 2000;
int h = 1600;
float flying = 0;
//2dimensional array to store for every single point a z value generated by an algorythm
//in java thats how you define a 2dimensional array
float[][] terrain;
void setup() {
size(600,600, P3D);
// columnes is the width of the grid devided by that scale
cols = w / scl;
rows = h / scl;
terrain = new float[cols][rows];
float yoff = 0;
for (int y = 0; y < rows; y++) {
float xoff = 0;
for (int x = 0; x < cols; x++) {
terrain[x][y] = map(noise(xoff,yoff), 0, 1, -100, 100);
xoff += 0.2;
}
yoff += 0.2;
}
}
void draw() {
flying -= 0.1;
float yoff = flying;
for (int y = 0; y < rows; y++) {
float xoff = 0;
for (int x = 0; x < cols; x++) {
terrain[x][y] = map(noise(xoff,yoff), 0, 1, -100, 100);
xoff += 0.2;
}
yoff += 0.2;
}
background(0);
stroke(255);
noFill();
translate(width/2, height/2+50);
rotateX(PI/3);
translate(-w/2,-h/2);
//nested loop for every single point on the grid
for (int y = 0; y < rows-1; y++) {
beginShape(TRIANGLE_STRIP);
for (int x = 0; x < cols; x++) {
vertex(x*scl, y*scl, terrain[x][y]);
vertex(x*scl, (y+1)*scl, terrain[x][y+1]);
//rect(x*scl, y*scl,scl,scl);
}
endShape();
}
}
This will change the stroke color as step increments:
`stroke(step*255/78.0); // step 0 to 78 becomes 0 to 255 ( for color)’
There is also the map() function to re-map a number from one range to another:
https://processing.org/reference/map_.html
Scrutinize the above carefully and understand what it is doing and where it may be useful.
A search in the forum for terrain will yield many results.
I encourage you to do some exploration of the Processing resources and experimenting with the code.
Resources here:
https://processing.org/
Make use of println() statements to see what the variables are doing in your code and consider what to do to change them in a loop or over time.
:)
2 Likes
This is Daniel Shiffman’s code: p5.js Web Editor
If you really understood the code, slowing it down should not be an issue.
My bet is that this is an assignement
1 Like
Hello @fijthecreator
Once you figure out how this is working:
You can then start to play with multiple colors:
Via dovetailing with this tutorial on Happy Coding:
BTW, kindly tag this as homework.
4 Likes
I thought this looked familiar.
You should be aware that posting another person’s code and claiming it as your own is frowned upon.
This appears to be the exact same code in Dan Shiffman’s tutorial on 3D Terrain Generation as seen here:
Not cool.
As described in my post, this is code I extracted from a tutorial. I just changed the false label of my code and specified where I got this code from. In no way I intended to pretend this would be my self-written code. I will definitely pay more attention to that in further posts.
1 Like
A question that came up while working on this project: Is it possible to create a similar terrain generator in p5js?
I experimented a little and got to this far:
The issues I encountered are the following, the TRIANGLE_STRIPES aren’t all connected, as u can see, resulting in more kind of terrain stripes/ lines instead of a terrain mesh. The other issue I couldn’t find a workaround for is the flying effect from the processing sketch.
1 Like
Hi
It is possible!
There are already versions of this out there and I will leave that search to you.
If you are converting the code from Processing to p5.js be aware of this:
https://p5js.org/examples/arrays-array-2d.html
I remember the above because I did this a while back and learned this.
:)
Thanks a lot. Instead of just copying one of the examples of a good-looking terrain generator in p5js, I tried to apply my learnings from different tutorials, such as daniel Schiffman’s, to my own terrain generator.
My current results look like this:
And the code like this:
//global variables:
//random array to preload the noise levels which will serve as the z coordinates
var terrainValues = [];
//tells us how high the terrain will be
var multiP = 140;
var xoff = 0;
var yoff = 0;
//tells us how fast the illusion of movement will be & how smooth the terrain will be
var inc = 0.1;
var marchingStep = 0;
function setup() {
createCanvas(900, 600, WEBGL);
angleMode(DEGREES);
//nested for loop to create, get noise values and store them in the 2 dimensional array
for(var y = 0; y < 60; y++){
//create new empty array while we previously just declared a random on dimensional array and later referred to a 2 dimensional array
terrainValues.push([]);
xoff = 0;
for(var x = 0; x <60; x++){
terrainValues[y][x] = map(noise(xoff, yoff), 0, 1, -multiP, multiP);
xoff = xoff + inc;
}
yoff = yoff + inc;
}
}
function draw() {
background(0);
// set initial stroke to transparent
stroke(255,5);
noFill();
// reset the y offset
yoff = 0;
// nested for loop to replace x array value with added marchingStep -> illusion of endlessness
//compute new value for landscape and replace the values in our terrain array with our marchingStep added in
for(var y = 0; y < 60; y++){
xoff = 0;
for(var x = 0; x < 60; x++){
terrainValues[y][x] = map(noise(xoff + marchingStep, yoff), 0, 1, -multiP, multiP);
xoff = xoff + inc;
}
yoff = yoff + inc;
}
//rotate around x-axis by 70 degrees to get the horizontal view
rotateX(70);
//translate to poisition stripe mesh in center of the screen - negative because we want to transalte the mesh in the negativ direction
translate(-width/3, -height/3)
//outter loop needs to be y while I want the grid to start first with y, then do all the x's. Go to the next y and do all the x's again, and so on.
//nested for loop to create a 2 dimensional triangle stripe mesh
for(var y = 0; y < 60; y++) {
//for every single row begin Shape
beginShape(TRIANGLE_STRIP);
for(var x = 0; x < 60; x++){
vertex(x *10, y *10, terrainValues[x][y]);
vertex(x *10, (y +1) * 10, terrainValues[x][y+1]);
}
endShape();
//decreases the alpha value of the stroke each time we increment the y loop
stroke(255,y*2);
}
//increment the small offset we add to the perlin noise calculation for terrain values in each draw loop
marchingStep -= 0.1;
}
Things I would like to change, but fail to do so:
1. I would like the terrain mesh to fill the viewport width without leaving the “empty” spaces to the left and the right of the mesh. (I first tried to scale up the mesh but I couldn’t figure out how to eliminate the “empty” space on both sides.)
2. I would like to rotate the mesh even further to 80 degrees, but while I do so the first line of my mesh is visible and disturbs the look of the terrain generation like this:
Does anybody have a recommendation on how to conquer these challenges?
Move the stroke() call to before the beginShape() in your drawing loop. That will fix your bright leading edge.
If you want the terrain to fill the screen, then you either have to generate a wider grid or just zoom in more.
I increased the size of the mesh, but this unfortunately lowers the framerate and lets the terrain appear laggy (while I think all the points are drawn constantly in the draw function). Is there a way to optimize the performance (framerate) while increasing the size of the mesh?
An alternative is to spread out the back sample points so that your are sampling in a fan shape rather than a strict rectangular grid. Whether that would be acceptable or not depends on your actual goal for this project.
Another choice would be to store the noise values in an array and only generate new back rows as they are needed. Your current code calls the noise() function 3600 times per draw() (or more if you’re trying to cover more of the screen) which has a cost.
1 Like
The second option seems very interesting and probably best fitting to my project. While it would not only solve the width issue, it would also enhance the performance of my terrain generator (which I’d like to embed into a webpage).
Unfortunately, I’m still learning and don’t know how to go about storing the noise values in an array that only generates new back rows as needed. Any chance to show me/ give me a starting point?
It turns out that it doesn’t matter how fast or slow noise() is, it’s the rendering that slows things down. So the only way to get speed is to pack each row into a p5.Geometry mesh. I spent way too long on this, but here’s your terrain fly-over:
//global variables:
//random array to preload the noise levels which will serve as the z coordinates
let terrainValues = [];
//tells us how high the terrain will be
let vScale = 10;
let nCols = 90;
let nRows = 60;
let steps = 10; // how many steps (frames) per row, fewer is faster
let ns = 0.1; // horizontal noise() scale
function makeStrip( y, r1, r2 ) {
let s = new p5.Geometry( 1, 1 );
s.gid = "land"+str(y);
for( let x=0; x<nCols-1; x++ ) {
s.vertices.push( new p5.Vector( x, 0, r1[x] ), new p5.Vector( x, 1, r2[x] ) );
}
for( let x=0; x<nCols-2; x++ ) {
s.faces.push( [ 2*x, 2*x+1, 2*x+2 ] );
}
s.computeNormals();
return s;
}
let mesh = [];
function setup() {
createCanvas(900, 600, WEBGL);
angleMode(DEGREES);
//nested for loop to create, get noise values and store them in the 2 dimensional array
for(let y = 0; y < nRows; y++){
//create new empty array while we previously just declared a random on dimensional array and later referred to a 2 dimensional array
terrainValues.push([]);
for(let x = 0; x <nCols; x++){
terrainValues[y][x] = map(noise(x*ns, (-y-1)*ns), 0, 1, -vScale, vScale);
}
}
for( let y=0; y<nRows-2; y++ ) {
mesh.push( makeStrip( -y-1, terrainValues[y], terrainValues[y+1] ) );
}
}
function draw() {
background(0);
let yPos = frameCount/steps;
let yFrac = yPos % 1;
noFill();
if( (frameCount % steps) == 0 ) {
for( let y=nRows-1; y>0; y-- ) {
terrainValues[y] = terrainValues[y-1];
mesh[y] = mesh[y-1];
}
terrainValues[0] = [];
for(let x = 0; x < nCols; x++){
terrainValues[0][x] = map(noise(x*ns, yPos*ns), 0, 1, -vScale, vScale);
}
mesh[0] = makeStrip( yPos, terrainValues[0], terrainValues[1] );
}
//rotate around x-axis by 70 degrees to get the horizontal view
rotateX(70);
scale( width/nCols*1.8 );
translate( -(nCols-1)*0.5, -(nRows-1)*0.6 );
for( y=0; y<nRows-2; y++ ) {
push();
translate( 0, y+yFrac, 0 );
fill( 255, y*2 );
model( mesh[y] );
pop();
}
}
I just saw that I left it with filled triangles. Change the fill to stroke near the bottom to go back to full wireframe.
1 Like
Thank you so much for your effort! Amazing how smooth the fly-over is compared to my previous sketches. Unfortunately, there are a lot of new concepts that I’m not too familiar with yet. I will definitely take the time to understand the code, its methods and functions.
Once again, thanks.
After further experimentations, I got very close to my initial goal.
The challenges that I currently face, are shown here:
While I included a windowResize() function, I would like the terrain to always stick to the bottom of the windowHeight, so the first line of the mesh is not visible. Is there any way to accomplish that?
Furthermore, I do still struggle to resize the mesh without its performance being significantly decreased. I would like the mesh to appear endless, so it fills the whole windowWitdh.
Anyone able to help?
Here is my code:
//global variables:
//random array to preload the noise levels which will serve as the z coordinates
var terrainValues = [];
//tells us how high the mountains will be
var multiP = 150;
var xoff = 0;
var yoff = 0;
//tells us how smooth the terrain will be
var inc = 0.08;
var marchingStep = 0;
function setup() {
var canvas = createCanvas(windowWidth, windowHeight, WEBGL);
canvas.style('display', 'block');
angleMode(DEGREES);
//nested for loop to create, get noise values and store them in the 2 dimensional array
for(var y = 0; y < 60; y++){
//create new empty array while we previously just declared a random on dimensional array and later referred to a 2 dimensional array
terrainValues.push([]);
xoff = 0;
for(var x = 0; x < 60; x++){
terrainValues[y][x] = map(noise(xoff, yoff), 0, 1, -multiP, multiP);
xoff = xoff + inc;
}
yoff = yoff + inc;
}
}
function windowResized() {
resizeCanvas(windowWidth, windowHeight);
}
function draw() {
background(0);
// set initial stroke to transparent
stroke(255, 5);
noFill();
// reset the y offset
yoff = 0;
// nested for loop to replace x array value with added marchingStep -> illusion of endlessness
//compute new value for landscape and replace the values in our terrain array with our marchingStep added in
for(var y = 0; y < 60; y++){
xoff = 0;
for(var x = 0; x < 60; x++){
terrainValues[y][x] = map(noise(xoff + marchingStep, yoff), 0, 1, -multiP, multiP);
xoff = xoff + inc;
}
yoff = yoff + inc;
}
//rotate around x-axis by 80 degrees to get the horizontal view
rotateX(80);
//translate to poisition stripe mesh in center of the screen - negative because we want to transalte the mesh in the negativ direction
translate(-width/2, -height/2, -250)
scale(width/600,1)
//outter loop needs to be y while I want the grid to start first with y, then do all the x's. Go to the next y and do all the x's again, and so on.
//nested for loop to create a 2 dimensional triangle stripe mesh
for(var y = 0; y < 60; y++) {
//for every single row begin Shape
beginShape(TRIANGLE_STRIP);
for(var x = 0; x < 60; x++){
vertex(x *10, y *10, terrainValues[x][y]);
vertex(x *10, (y +1) * 10, terrainValues[x][y+1]);
}
endShape();
//decreases the alpha value of the stroke each time we increment the y loop
stroke(255,y*2);
}
//increment the small offset we add to the perlin noise calculation for terrain values in each draw loop
marchingStep -= 0.08;
} | 3,917 | 14,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.669693 |
https://www.pdf-archive.com/2016/02/20/week4/ | 1,571,235,534,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00243.warc.gz | 1,043,012,006 | 11,420 | # PDF Archive
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### Document preview
Assignment Week 4: An SDP based randomized algorithm for
the Correlation Clustering problem
The objective of this exercise is to design an algorithm for the correlation clustering problem. Given an
undirected graph G = (V, E) without loops, for each edge e = {i, j} ∈ E there are two non-negative
numbers we+ , we− ≥ 0 representing how similar and dissimilar the nodes i and j are, respectively. For
S ⊆ V , let E(S) be the set of edges with both endpoints in S, that is, E(S) = {{i, j} ∈ E : i, j ∈ S}.
The goal is to find a partition S of V that maximizes
X
X
f (S) =
we+ +
we− .
(1)
S∈S:e∈E(S)
e∈E\∪S∈S E(S)
In words, the objective is to find a partition that maximizes the total similarity inside each set of the
partition plus the dissimilarity between nodes in different sets of the partition.
Consider the following simple algorithm:
Algorithm 1
Let S1 = {{i} : i ∈ V } the partition that considers each vertex as a single cluster, and S2 = {V }, that
is every vertex in the same cluster. Compute the values f (S1 ) and f (S2 ) of these two partitions, and
output the best among this two.
• Question 1. Compute the values f (S1 ), f (S2 ) in terms of the weights w− and w+ .
For S1 there are no edges in any E(S) with S = {i} ∈ S1 , while for S2 all edges are in E(V ). So
we find
X
f (S1 ) =
we− ,
e∈E
f (S2 ) =
X
we+ .
(2)
e∈E
• Question 2. Conclude that previous algorithm is a 1/2-approximation.
We have the following bound for any S:
X
X
f (S) ≤
we+ +
we− = f (S1 ) + f (S2 ) ≤ 2 max(f (S1 ), f (S2 )) ,
e
(3)
e
and thus also for the optimal S ∗
max(f (S1 ), f (S2 )) ≥ 1/2f (S ∗ ) = 1/2 OPT ,
(4)
so we have 1/2-approximation.
Let B = {e` : ` ∈ {1, 2, . . . , n}} be the canonical basis in Rn , where n = |V |. For every vertex
i ∈ V there is a vector xi that is equal to ek if node i is assigned to cluster k. Consider the
following program:
X
+
max
w{i,j}
xi · xj + w{i,j}
(1 − xi · xj ) : xi ∈ B, ∀i ∈ V .
(5)
{i,j}∈E
• Question 3. Explain why this program is a formulation of the correlation clustering problem.
We have two cases. Either i and j belong to the same cluster, or either two different clusters. In
the first case, there is some cluster k so that both i and j belong to k. We have xi = xj = ek and
xi · xj = 1. Also e = {i, j} ∈ E(Sk ). The edge e then contributes we+ to the first term of both
objective (1) and objective (5). In the second case we have xi · xj = 0, since different basis vectors
1
are orthogonal, as well as e ∈ E \ ∪E(S) and the edge e contributes we− to the second term of
both objective (1) and objective (5). It follows that objective (1) and objective (5) are identical.
The formulation is relaxed to obtain the following vector program:
X
+
max
w{i,j}
vi · vj + w{i,j}
(1 − vi · vj )
.
(6)
{i,j}∈E
subject to
vi · vi = 1 ,
∀i ∈ V,
vi · vj ≥ 0 ,
∀i, j ∈ V,
n
vi ∈ R ,
(7)
∀i ∈ V .
Consider the following algorithm:
Algorithm SDP
Solve the the previous relaxation to obtain vectors {vi : i ∈ V }, with objective value equal to
Z. Draw independently two random hyperplanes with normals r1 and r2 . This determines four
regions,
R1 = {i ∈ V : r1 · vi ≥ 0 and r2 · vi ≥ 0},
R2 = {i ∈ V : r1 · vi ≥ 0 and r2 · vi < 0},
(8)
R3 = {i ∈ V : r1 · vi < 0 and r2 · vi ≥ 0},
R4 = {i ∈ V : r1 · vi < 0 and r2 · vi < 0},
and output the partition R = {R1 , R2 , R3 , R4 }.
In the following, the goal is to analyse this algorithm, and to prove that it is a 3/4-approximation.
• Question 4. Let X{i,j} be the random variable that is equal to 1 if the vectors vi and vj lie on the
same side of the two random hyperplanes, and zero otherwise. Using an argument similar to the
2
one used for Max-Cut, prove that Prob(X{i,j} = 1) = 1 − 1/πθ{i,j} , where θ{i,j} = arccos(vi ·vj )
is the angle between vectors vi and vj .
As in the lectures the change that one random hyperplane separates the two vectors is θ{i,j} /π,
so the change that one hyperplane does not separate them is 1 − θ{i,j} /π. Since the hyperplanes
are chosen independently the change that neither of them separates them is
(1 − θ{i,j} /π)2 .
• Question 5. Let f (R) =
P
{i,j}∈E
(9)
+
w{i,j}
X{i,j} + w{i,j}
(1 − X{i,j} ) the value of the partition R,
and denote g(θ) = (1 − θ/π)2 the probability function computed before. Prove that the expected
value of f (R), denoted by E(f (R)), is
X
+
w{i,j}
g(θ{i,j} ) + w{i,j}
(1 − g(θ{i,j} )).
(10)
{i,j}∈E
We have immediately
E(f (R)) =
X
+
w{i,j}
E(X{i,j} ) + w{i,j}
(1 − E(X{i,j} ))
{i,j}∈E
=
X
+
w{i,j}
g(θ{i,j} ) + w{i,j}
(1 − g(θ{i,j} )).
{i,j}∈E
The following lemma will be helpful to conclude the analysis (You don’t need to prove it.)
Lemma. For θ ∈ [0, π/2], g(θ) ≥ 3/4 cos(θ) and 1 − g(θ) ≥ 3/4(1 − cos(θ)).
2
(11)
• Question 6. Using the lemma conclude that E(f (R)) ≥ 3/4 Z, and that the algorithm is a 3/4approximation.
Using Q5 and the lemma we find:
X
+
E(f (R)) =
w{i,j}
g(θ{i,j} ) + w{i,j}
(1 − g(θ{i,j} ))
{i,j}∈E
≥ 3/4
X
+
w{i,j}
cos(θ{i,j} ) + w{i,j}
(1 − cos(θ{i,j} )) = 3/4 Z ≥ 3/4 OPT ,
(12)
{i,j}∈E
so that
E(f (R)) ≥ 3/4OPT .
3
(13) | 1,908 | 5,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-43 | longest | en | 0.89761 |
http://xrpp.iucr.org/Cb/ch6o1v0001/sec6o1o2/ | 1,553,201,554,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202572.29/warc/CC-MAIN-20190321193403-20190321215403-00479.warc.gz | 393,060,082 | 12,653 | International
Tables for
Crystallography
Volume C
Mathematical, physical and chemical tables
Edited by E. Prince
International Tables for Crystallography (2006). Vol. C, ch. 6.1, pp. 590-593
## Section 6.1.2. Magnetic scattering of neutrons
P. J. Browna
### 6.1.2. Magnetic scattering of neutrons
| top | pdf |
#### 6.1.2.1. Glossary of symbols
| top | pdf |
Neutron mass Electron mass γ Neutron magnetic moment in nuclear magnetons (−1.91) Bohr magneton Nuclear magneton Classical electron radius Electron momentum operator Electron spin operator Neutron spin operator Magnetization density operator k Scattering vector (H/2π) A unit vector parallel to k A lattice vector g A reciprocal-lattice vector (h/2π) Propagation vector for a magnetic structure A unit vector parallel to the neutron spin direction q, q′ Initial and final states of the scatterer σ, σ′ Initial and final states of the neutron Eq Energy of the state q
#### 6.1.2.2. General formulae for the magnetic cross section
| top | pdf |
The cross section for elastic magnetic scattering of neutrons is given in the Born approximation by V(R) is the potential of a neutron at R in the field of the scatterer. If the field is due to N electrons whose positions are given by , then V(R) is more simply written in terms of a magnetization density operator , which gives the magnetic moment per unit volume at r due to both the electron's spin and orbital motions. The potential of (6.1.2.2) can then be written (Trammell, 1953) giving for the cross section, from (6.1.2.1), The unit-cell magnetic structure factor M(k) is defined as For periodic magnetic structures, where P is a periodic function with a period of unity, which describes how the magnitude and direction of the magnetization density, defined within one chemical unit cell by , propagates through the lattice. The magnetic structure factor m(k) is then given by where is the jth term in the Fourier expansion of P defined by and the scattering cross section given in terms of the magnetic interaction vector , is Equation (6.1.2.9) leads to two independent scattering cross sections: one for scattering of the neutron with no change in spin state (σ′ = σ) proportional to , and the other to scattering with a change of neutron spin (spin flip scattering') proportional to . The sum over all final spin states gives
#### 6.1.2.3. Calculation of magnetic structure factors and cross sections
| top | pdf |
If the magnetization within the unit cell can be assigned to independent atoms so that each has a total moment aligned in the direction of the axial unit vector , then the unit-cell structure factor can be written and are the rotations and translations associated with the jth element of the space group and is an operator that reverses all the components of moment whenever the element j includes time reversal in the magnetic space group. is the magnetic form factor of the ith atom (see Subsection 6.1.2.3).
The vector part of the magnetic structure factor can be factored out so that where is now a scalar. For collinear structures, all the atomic moments are either parallel or antiparallel to , which in this case is independent of k. The intensity of a magnetic Bragg reflection is proportional to and where α is the angle between the moment direction and the scattering vector k. The factor , often referred to as , is the means by which the moment direction in a magnetic structure can be determined from intensity measurements. If the intensities are obtained from measurements on polycrystalline samples then the average of over all the different k contributing to the powder line must be taken. the sum being over all rotations of the point group. is given for different crystal symmetries by Shirane (1959). For uniaxial groups, the result is where ψ and are the angles between the unique axis and the scattering vector and moment direction, respectively. For cubic groups independent of the moment direction and of the direction of k.
#### 6.1.2.4. The magnetic form factor
| top | pdf |
The magnetic form factor introduced in (6.1.2.11) is determined by the distribution of magnetization within a single atom. It can be defined by where q now represents a state of an individual atom.
In the majority of cases, the magnetization of an atom or ion is due to a single open atomic shell: the d shell for transition metals, the 4f shell for rare earths, and the 5f shell for actinides. Magnetic form factors are calculated from the radial wavefunctions of the electrons in the open shells. The integrals from which the form factors are obtained are where U(r) is the radial wavefunction for the atom and is the lth-order spherical Bessel function. Within the dipole approximation (spherical symmetry), the magnetic form factor is given by where g is the Landé splitting factor (Lovesey, 1984). Higher approximations are needed if the orbital contribution is large and to describe departures from spherical symmetry. They involve terms in etc. Fig. 6.1.2.1 shows the integrals , and for Fe2+ and in Fig. 6.1.2.2 the spherical spin-only form factors for 3d, 4d, 4f, and 5f electrons are compared. Tables of magnetic form factors are given in Section 4.4.5 .
Figure 6.1.2.1 | top | pdf |The integrals , , and for the Fe2+ ion plotted against . The integrals have been calculated from wavefunctions given by Clementi & Roetti (1974).
Figure 6.1.2.2 | top | pdf |Comparison of 3d, 4d, 4f, and 5f form factors. The 3d form factor is for Co, and the 4d for Rh, both calculated from wavefunctions given by Clementi & Roetti (1974). The 4f form factor is for Gd3+ calculated by Freeman & Desclaux (1972) and the 5f is that for U3+ given by Desclaux & Freeman (1978).
#### 6.1.2.5. The scattering cross section for polarized neutrons
| top | pdf |
The cross section for scattering of neutrons with an arbitrary spin direction is obtained from (6.1.2.9) but adding also nuclear scattering given by the nuclear structure factor , which is assumed to be spin independent. In this case, the scattering without change of spin direction is and, for the spin flip scattering, with .
The cross section I++ implies interference between the nuclear and the magnetic scattering when both occur for the same k. This interference is exploited for the production of polarized neutrons, and for the determination of magnetic structure factors using polarized neutrons.
In the classical method for determining magnetic structure factors with polarized neutrons (Nathans, Shull, Shirane & Andresen, 1959), the flipping ratio' R, which is the ratio between the cross sections for oppositely polarized neutrons, is measured: In this equation, is a unit vector parallel to the polarization direction. P is the neutron polarization defined as where and are the expectation values of the neutron spin parallel and antiparallel to averaged over all the neutrons in the beam. e is the flipping efficiency' defined as e = (2f − 1), where f is the fraction of the neutron spins that are reversed by the flipping process. Equation (6.1.2.21) is considerably simplified when both and are real and the polarization direction is parallel to the magnetization direction, as in a sample magnetized by an external field. The flipping ratio' then becomes with , ρ being the angle between the magnetization direction and the scattering vector. The solution to this equation is the relative signs of and are determined by whether R is greater or less than unity. The uncertainty in the sign of the square root in (6.1.2.23) corresponds to not knowing whether or vice versa.
#### 6.1.2.6. Rotation of the polarization of the scattered neutrons
| top | pdf |
Whenever the neutron spin direction is not parallel to the magnetic interaction vector Q(k), the direction of polarization is changed in the scattering process. The general formulae for the scattered polarization are given by Blume (1963). The result for most cases of interest can be inferred by calculating the components of the scattered neutron's spin in the x, y, and z directions for a neutron whose spin is initially parallel to z. For simplicity, y is taken parallel to k; x and z define a plane that contains Q(k). From (6.1.2.18),
It is clear from this set of equations that and are zero if . Three simple cases may be taken as examples of the use of (6.1.2.24):
(a) A magnetic reflection from a simple antiferromagnet for which Q(k) is real, F(k) = 0; under these conditions, showing that the direction of polarization is turned through an angle in the xy plane where is the angle between Q(k) and the initial polarization direction. (b) A satellite reflection from a magnetic structure described by a circular helix for which = = 0; in this case, and the scattered polarization is parallel to the scattering vector independent of its initial direction. (c) A mixed magnetic and nuclear reflection from a Cr2O3-type antiferromagnet for which Q(k) is imaginary, , is real. Then, so that in this case the final polarization has components along all three directions.
### References
Blume, M. (1963). Polarization effects in the magnetic elastic scattering of slow neutrons. Phys. Rev. 130, 1670–1676.
Clementi, E. & Roetti, C. (1974). Roothaan–Hartree–Fock atomic wavefunctions. Basis functions and their coefficients for ground and certain excited states of neutral and ionized atoms. At. Data Nucl. Data Tables, 14, 177–478.
Desclaux, J. P. & Freeman, A. J. (1978). Dirac–Fock studies of some electronic properties of actinide ions. J. Magn. Magn. Mater. 8, 119–129.
Freeman, A. J. & Desclaux, J. P. (1972). Neutron magnetic form factor of gadolinium. Int. J. Magn. 3, 311–317.
Lovesey, S. W. (1984). Theory of neutron scattering from condensed matter. Vol. 2. Polarization effects and magnetic scattering. The International Series of Monographs on Physics No. 72. Oxford University Press.
Nathans, R., Shull, C. G., Shirane, G. & Andresen, A. (1959). The use of polarised neutrons in determining the magnetic scattering by iron and nickel. J. Phys. Chem. Solids, 10, 138–146.
Shirane, G. (1959). A note on the magnetic intensities of powder neutron diffraction. Acta Cryst. 12, 282–285.
Trammell, G. T. (1953). Magnetic scattering of neutrons from rare earth ions. Phys. Rev. 92, 1387–1393. | 2,461 | 10,290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-13 | latest | en | 0.825839 |
http://www.greenspun.com/bboard/q-and-a-fetch-msg.tcl?msg_id=000OJ9 | 1,534,676,752,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215077.71/warc/CC-MAIN-20180819110157-20180819130157-00178.warc.gz | 505,564,551 | 9,183 | ### Probability of Systemic Failure: Another Point of View
greenspun.com : LUSENET : TimeBomb 2000 (Y2000) : One Thread
There has been a number of posts that analyze the probability of systemic failure by concentrating on a few key industries and postulating that if any one of them were to suffer a critical failure, economic paralysis would result. While this thesis may indeed be true, the debate is diverted to the actual probabilities of success one ascribes to each of these key industries. The net result is that the debate quickly gets bogged down into a futile number crunching exercise because these probabilities are essentially unknown.
I think that the probability-of-systemic-failure calculation, may be of more practical value by searching for those independent economic (and even social) ACTIVITIES that form essential "spokes" that keep the economic wheel balanced and spinning without a wobble. So just like a wheel could keep on turning upon losing a few spokes, it would quickly become unbalanced and begin to wobble which would quickly lead to catastrophic physical (and economic) failure.
The benefit of approaching the calculation this way is that we put the greatest emphasis on the SYSTEMIC nature of the problem which, in fact, goes to the heart of y2k, and the least emphasis on the probabilistic aspects which, I think we would all agree, is the least understood. Then by assigning very optimistic success probabilities to each of these "spokes" it can be shown that we are facing a huge risk of severe economic deterioration and eventual collapse. There are only two selection criteria for picking an economic (or social) activity to become a "spoke"; it should be independent of all others, and should be considered a vital economic activity. Given below is my guess as to which activities would qualify together with my optimistic "Success Probabilities". Note that the chance of at least one critical failure is 62%. If I assign my "best guess" probabilities that number rises to over 99%.
CRITICAL FAILURE OPTIMISTIC SUCCESS PROBABILITY Financial 92.17 banking 98 Fed wire transfer 99 tax proceeds (IRS/Customs) 95
Power 96.05 generation 98 transmission 99 distribution 99
Telecom 98.5 Food Production 98 Water Delivery 98
Waste Disposal 97.02 sewage 98 solid waste 99
Transportation 88.52 rail 95 truck 98 surface docking facilities 98 air 98 pipelines 99
Government Oversight 91.23 Military/National guard /police/fire protection 99 FAA 97 All other support such as SS, Medicare, Medicaid, etc. 95
Extraction and Mining 96.06 coal 99 gas 99 oil (domestic) 99 metals 99
External Factors 66.34 oil importation 95 foreign banking 95 " manufacturing 95 " air traffic control 95 " docking facilities 95 " surface shipping 95 " ground shipping 95 other critical raw material imports 95
Public Support of Infrastructure 91.75 urban rioting 97 bank runs 97 job lockout (from bankruptcies or fear of leaving home) 98 war/terrorism/series of natural disasters 99.5
OVERALL PROBABILITY OF AT LEAST ONE CRITICAL FAILURE: 61.6%
-- Dr. Roger Altman (rogaltman@aol.com), January 15, 1999
Sorry, my table got all re-arranged upon transmission. To understand it now, read the first row entry as the general category. For instance, "External Factors 66.34" is the subheading together with the OVERALL "Success Probability" for all the items TAKEN TOGETHER UNDER THAT SUBHEADING. Following after that are each category for that subheading. So, continuing with this example, oil importation 95%, foreign banking 95%, foreign manufacturing 95%, etc.
Oh boy, computers, computers...
-- Dr. Roger Altman (rogaltman@aol.com), January 15, 1999.
All I see here is more BS from someone who likes to hear themselves talk. Is anyone else in favor of banning further posts on Time Bomb 2000 until September? I mean, there is only so much postulating we can take before we explode like a tick.
-- Mojo Rising (Yesca@HT.com), January 15, 1999.
Mojo, feel free to drop off the face of the earth. Everything is speculation which is one of the greatest benefits of this forum. Posts on either side of the y2k thin, red line are welcome and essential.
Your ability to lurk or post here is your decision and no one is twisting your arm or banging your head against the wall to force you to stay.
Dr. Altman, I appreciate your work and I thank you for your post.
Mike =================================================================
-- Michael Taylor (mtdesign3@aol.com), January 15, 1999.
oh heaven's no, Mike, I want to watch Mojo explode like a tick!
Arlin
"...the least emphasis [is] on the probabilistic aspects which, I think we would all agree, is the least understood."
Roger, I disagree. The nature of complex systems' failures is not well understood.
Consider the following from Margaret Wheatley's Leading Through the Unknowns of Y2K:
"Complex systems failures share a set of distinguishing features:
• the longer they unravel, the more extensive their effects
• costs always far exceed what has been budgeted for amelioration
• as effects materialize, unknown interdependencies become visible
• the more that problems come into focus, the fuzzier they appear
• past experiences with simple systems dont apply
• cause and effect are impossible to track
• consequently, there is no one to blame
• These features describe the most frightening realization about complex systemic problems: They are inherently uncontrollable. They cannot be understood sufficiently either before or even while they are occurring, therefore prediction and control are impossible. Traditional approaches to management simply dont work."
Also, the suggestion to search for those independent economic (and even social) ACTIVITIES that form essential "spokes" that keep the economic wheel balanced and spinning without a wobble, discounts the invisible interconnecting processes that structure an economy, an ecology.
So, I ask you, kind sir, "Might there be something missing from your algorithm?"
and
Is there an activity within the "ACTIVITIES" that allows order to unfold?
~C~
-- Critt Jarvis (Wilmington, NC) (critt@critt.com), January 16, 1999.
I am replying to Critt Jarvis' comments. First, thanks for pointing out Margaret Wheatley's article entitled, "Leading Through the Unknowns of Y2K". I think if you re-read her article you will find that in large part what she is referring are the unknown complexities and interrelationships related to SOLVING the Y2K problem. However, she does make some reference to economic activity as another example of such incomprehensibe complexity that no one expert can really understand the entire system.
So I cannot understand how you come away thinking that we know MORE about probabilities than I expressed in my original comments. Since her article barely mentions economics it is not particularly relevant to my point, and to the extent that she uses it to bolster her 'complexity' message, it boslsters mine (unknown probabilities) as well.
To get back to the point: probability is not the issue because, as I have demonstrated by setting obsurdly optimistic probability-of-success values, the SYSTEMIC nature of the y2k issue VIRTUALLY GUARANTEES that we will have critical failures. Now once a critical failure occurs then all those "hidden complexities and interrelationships" that Wheatly refers to will surely rear their ugly heads to exacerbate and compound existing problems which will RAPIDLY accelerate economic collapse.
To site just one example: if sewage is not removed from urban areas, how long would it take to precipitate a severe emergency through rioting, and disease? And because death will rapidly follow, how long will it take disease to spread to surrounding areas? Do we need to know what probability sewage plant have to succeed with their y2k remediation? No we don't because a calamity that I have just described can come from almost ANY SOURCE. The point is WE ARE EXTREMELY VULNERABLE TO A CALAMITY. Let that anthropologists of the twenty second century pick apart our remains to figure out how it all started.
-- Dr. Roger Altman (rogaltman@aol.com), January 16, 1999.
I have an idea... Why don't you all go and try to figure the probability of which one of you Fat Heads will hit the bottom of a canyon floor if you all jump off a thousand foot ledge at the same time? I guess the only real way to be sure is to have you do some "theoretical testing".
The Lemming bus is now leaving...
-- Mojo Rising (FatHeadsBegone@Y@K.com), January 16, 1999.
Mojo: thank you for your detailed analysis on why there won't be a problem. Now tuck your head back in your ass and go off to sleep.
-- a (a@a.a), January 16, 1999.
Mojo - they'd all hit the floor at the same time BOZOID (Newtonian Physics) - didn't you learn anything at school yesterday???
Mojo drooping more like...
Andy
Two digits. One mechanism. The smallest mistake.
"The conveniences and comforts of humanity in general will be linked up by one mechanism, which will produce comforts and conveniences beyond human imagination. But the smallest mistake will bring the whole mechanism to a certain collapse. In this way the end of the world will be brought about."
Pir-o-Murshid Inayat Khan, 1922 (Sufi Prophet)
-- Andy (2000EOD@prodigy.net), January 16, 1999.
Oh! Forgot to add, if you jumped off at the same time Mojo Drooping, you would hit the floor LAST because you are an AIR HEAD (Mojoian Physics.)
Andy the doombrooder :)
"We're doomed I tell ye, doomed!"
Private Frazer, Dad's Army, Walmington-On-Sea Home Guard, 1939 (Undertaker)
-- Andy (2000EOD@prodigy.net), January 16, 1999.
Double </b> to correct for earlier <b> where there should have been a </b>
-- No Spam Please (anon@ymous.com), January 16, 1999.
Whatever you said/did Thanks. Some people just won't pick up after themselves.
S.O.B.
-- sweetolebob (La) (buffgun@hotmail.com), January 16, 1999.
Thanks no spam, I'm new to this stuff!
-- Andy (2000EOD@prodigy.net), January 16, 1999.
No spam, can you help me out. You recently posted a very good example of a banking transaction, a very detailed one - I have been trying to find it, can you remember which thread it was on? Many thanks if you can.
Andy
-- Andy (2000EOD@prodigy.net), January 16, 1999.
S.O.B.,
Oh, Andy just made one teeny little typo.
- - -
Andy,
Two HTML-newbie suggestions from others in this forum:
(1) As soon as you post something with HTML, exit back to main page, then click on the thread again to see it with your new entry, and check whether looks all right. {I started doing this ... mmmmm ... sometime last month. :-}
(2) When you're going to post something with HTML, first write it in a file on your own disk with some editor, then open that file with your browser to see how it looks. If OK, copy to submission window.
- - - -
Andy (cont.),
Creating that example was fun.
Am now searching ... searching ... searching ...
-- No Spam Please (anon@ymous.com), January 17, 1999.
Andy,
>example of a banking transaction, a very detailed one - I have been trying to find it, can you remember which thread it was on?
Will noncompliant systems reinfect compliant systems at http://www.greenspun.com/bboard/q-and-a-fetch-msg.tcl?msg_id=000Llt
-- No Spam Please (anon@ymous.com), January 17, 1999.
Hey thanks No Spam - I'm resaerching an article I'm writing on the imported data problem and may use your example for discussion if that's ok with you.
Thanks, Andy
-- Andy (2000EOD@prodigy.net), January 18, 1999.
Andy -- Okay.
-- No Spam Please (anon@ymous.com), January 18, 1999.
There's another way of looking at this "Death of a Thousand Cuts" scenario. Notice that the analysis, as presented, makes this very unlikely (and hidden) assumption: none of the experienced failures will be repaired.
I don't think that's the case. Certainly some failures will be repaired. How many? How soon? Who knows?
Given this completely unquantifiable variable, the calculation provides no new information. The probability of a critical failure, somewhere, sometime, of unknown duration, remains indeterminate.
-- Tom Carey (tomcarey@mindspring.com), January 19, 1999.
Tom:
Thanks for your response. It's a good criticism, but don't be tempted to dismiss my attempt to quantify the "death-by-a-thousand-cuts" scenario too quickly. Remember, the model represents a "snapshot" of the "probability-dynamic" caused by the SYSTEMIC nature of y2k. Sure there will be repairs. After all, with every passing day the world is incrementally closer to fixing all y2k problems. But the key question you should think about is the following:
Given the overwhelming probability that there will be critical economic failures as we move into the next millennium, will the TREND of the probabilitiy-of-critical-failure FOR THE ECONOMY AS A WHOLE worsen or improve?
I submit to you that if I were to prepare new data similar to the type I prepared in my original comments, the average of the "Probability-of-Success" values in that table would be LOWER not higher (which is really what you are implying -- otherwise what you're saying makes no sense at all) for the following reasons:
1) Non-productive (meaning military, police, firefighting which are expense items) segments of our enonomy would be put into force at a much greater rate than normal for crowd control and security, which would create a much higher than average economic drain.
2) An initial economic dislocation of the magnitude I forsee would have a significant impact on bankruptcy, worker productivity, job absentia, all of which lower productivity.
3) To the extent that y2k hits computers directly, productivity will be directly affected ("just-in-time-inventory-control" is but one example).
4) Trade and commerce will slow down drastically, particularly those elements dependent upon overseas goods (and who isn't these days). Thus, vast sectors of our economy will be extremely vulnerable to failure.
5) Consumer confidence is likely to take a drastic hit (stock market plunge) which will restrict economic activity, raising unemployment and reducing corporate earnings.
6) Even a FEW key corporate/government agency failures are likely to have a domino effect on the rest of the economy ("division-of-labor" argument).
7) The ability to continue fixing y2k problems will be substantially more difficult because programmers will be forced to work under "fix-on-failue" conditions -- a repair mode with HUGE disadvantages.
8) Even successfully operating contingency plans will significantly reduce corporate/government agency productivity (by definition, at least) which will further hamper economic activity.
There are many more reasons I could list, but please don't miss the point here. While no one can predict the EXTENT to which the items I've enumerated will effect the probabilities in question, they WILL BE AFFECTED and to argue otherwise puts you in the wishful thinking camp. Now, if that's where you want to wait out the rest of 1999, well, that's your business, but an argument such as that serves no useful purpose on this thread. 5)
-- Dr. Roger Altman (rogaltman@aol.com), January 19, 1999.
Yes, but aside from that, Doctor...?
Actually I think you're right on this. I think stuff will happen, and in a serious way.
I just don't think it's possible to assign a numerical probability to any of these contingencies. And qualitative estimates, whether yours, or mine, or Milne's, or Bilbo Baggins', characterize the estimator more than they do the extrinsic situation. This is not a fault. And Pascal's Wager still holds.
-- Tom Carey (tomcarey@mindspring.com), January 21, 1999.
Tom:
I thought the uncertainty of the probabilities was the point. By assigning ridiculously high "Probability-of-Success" values to each economic sector, and still have the overall number come out indicating the likelihood for at least one of these sectors will fail IS BETTER than bouncing qualitative scenarios back and fourth.
-- Dr. Roger Altman (rogaltman@aol.com), January 21, 1999. | 3,664 | 16,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-34 | latest | en | 0.955618 |
https://academia.fandom.com/wiki/Optimal_Classification/Rypka/Equations/TruthTableSize | 1,721,055,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00209.warc.gz | 72,997,207 | 38,399 | ## Truth table size-related equations
These equations determine the truth table size from the highest value of logic and the number of characteristics in the set and in the group for the bounded class[1] of elements and denote the range of the truth table values for the set and the group.[2]
### Group size
${\displaystyle Q = V^C }$, where:
• Q is the group size[3]
• V is the highest value of logic in the group,
• C is the highest number of characteristics in the group.
### Set size
${\displaystyle R = V^K}$, where:
• R is the set size[4]
• V is the highest value of logic in the group,
• K is the highest number of characteristics in the set.
## notes
1. See page 158, Table II of the primary reference
2. See page 172 of the primary reference.
3. group - Columns C1 to C5 (0 to C, C=5)
4. Target set - Columns C1 to C3 (0 to K, K=3) | 226 | 847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.823773 |
https://id.scribd.com/document/20867078/CGS-09-5116-S4-prelim-P2 | 1,571,096,332,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655554.2/warc/CC-MAIN-20191014223147-20191015010647-00065.warc.gz | 536,592,173 | 72,182 | Anda di halaman 1dari 14
# Section A
## 1. (a) Explain what is meant by the centre of gravity of an object. [1]
……………………………………………………………………………………………..
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(b) The diagram shows a cross-sectional view of a traffic safety cone with the
centre of gravity of the cone indicated by the point C which is 0.10 m above
its base.
C
x
0.10 m
## (i) Assuming that the gravitational force on a mass of 1.0 kg is 10 N,
calculate the weight of the cone if its total mass is 4.0 kg. [2]
(ii) Explain TWO features in the design of the traffic cone that improve the
stability of the cone. [2]
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This paper consists of 12 printed pages (including the cover page), and 3 lined pages.
2
## 2. (a) State the principle of moments for a body in equilibrium. [1]
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## (b) Fig. 2.1 shows a person crossing a bridge. .
2.1
The bridge is supported at points A and B. The length of the bridge is 6.0 m and its
weight is 2000 N. The centre of gravity of the bridge is at its midpoint. The weight of the
person is 750 N. The person is 4.0 m from point B.
(i) Calculate the moment of the weight of the bridge about point B. [2]
(ii) Calculate the moment of the weight of the person about point B. [1]
(iii) Calculate the force FA exerted by the ground on the bridge at point A. [2]
(iv) Calculate the force FB exerted by the ground on the bridge at point B. [1]
09/5116/S4 Prelim/P2
3
3. A beaker containing hot liquid is placed on a table in a room. The temperature of the
beaker and its contents changes as shown in Fig. 3.
4 8 time/min
Fig. 3
(a) Explain why the temperature of the beaker and its contents remain constant
over the section labelled BC in Fig. 3 . [2]
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## (b) What is the room temperature ? [1]
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(c) If the average rate of heat released during section AB is 60 Joules in a second.
What is the amount of energy released during this section? [2]
09/5116/S4 Prelim/P2
4
4. People like barbecuing a lot during winter because they could get warm from the fire
at the same time. Meat is pierced on iron forks and grilled over the fire at a distance
20 cm. The meat will be cooked up in 10 minutes. Some people would also wrap the
sweet potatoes by using aluminium foil and place them alongside the fire.
(a) Explain why the meat has to be pierce on an iron fork during barbecuing?
………………………………………………………………………………………….
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……………………………………………………………………………………….. [2]
(b) Explain how heat is transferred from the fire to the meat.
………………………………………………………………………………………….
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……………………………………………………………………………………… [2]
## (c) Suggest one advantage of wrapping the sweet potatoes by using
aluminium foil.
……………………………………………………………………………………… [1]
09/5116/S4 Prelim/P2
5
5. (a) Fig. 5.1 shows a graph of the variation the vertical displacement s of the
ripple tank vibrator with distance d travelled by the wave.
Fig. 5.1
## (ii) What is the amplitude of the wave? [1]
(b) Fig. 5.2 below shows a graph of the variation the vertical displacement s
of the ripple tank vibrator with time t travelled by the wave.
Fig. 5.2
## (i) Determine the frequency of the vibrator. [2]
(ii) Calculate the speed of the ripples. State clearly the formula used [2]
09/5116/S4 Prelim/P2
6
6. The diagram below is drawn to half scale. It shows a ray of light leaving
the head of an object AB and striking a converging lens L. The lens forms
an image of the object AB on the screen S.
S
L
A
On the diagram,
## (a) locate the image of AB and label it I. [2]
(b) complete the path of the given ray after it passes through the lens. [1]
(c) draw a suitable light ray which enables you to find the position of the
focal point of the lens. Label this point F. [1]
## Focal length of the lens = …………………………
09/5116/S4 Prelim/P2
7
7.
Fig. 7.1
In Fig. 7.1 shown above, with the switch, S, opened, find the
(a) effective resistance between P and Q. [1]
## The switch now is closed. Find the
(d) effective resistance between the P and Q. [1]
## (e) the reading of the voltmeter. [1]
09/5116/S4 Prelim/P2
8
8. An unqualified electrical engineer connects three electric fans each rated “220 V, 600
W” at home. He makes a domestic wiring as shown in the diagram below. The mains
supply is at 220 V and L, N and E denote the live, neutral and earth wire respectively.
(a) The electrical engineer forgot to connect the Earth wire in this electrical wiring.
On the diagram above, show by drawing how the Earth wire should
be connected. [1]
(b)
What electric component is missing inside the fan and what is the use of it?
[2]
………………………………………………………………………………………………
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(c) After correcting the mistakes, calculate the total current drawn from the mains
supply when all the fans are switched on. [2]
(d) Calculate the cost of using the three electric fans for 2 hours given that 1 kWh
costs \$0.15. [2]
09/5116/S4 Prelim/P2
9
Section B
9. The diagram shows the variation of the speed of a lorry as it travels along a level road.
The lorry begins to brake at the time t = 20s, and comes to rest when t = 24s.
The lorry has a mass of 1000 kg and the forward driving force on the wheels is 1200 N.
## (iii) the power supplied by the driving force. [1]
(b) (i) Calculate the kinetic energy of the lorry while it is travelling at
the constant speed. [2]
## (ii) When the lorry is travelling at a constant speed there is no
change in its kinetic energy. Suggest what happens to the
energy that is provided by the engine. [1]
(c ) During braking the speed of the lorry decreases uniformly. The engine no longer
provides a driving force.
## (i) Calculate the deceleration of the lorry between t = 20 s and t= 24 s. [1]
(ii) Calculate the total braking force acting on the lorry during this period. [2]
(iii) Explain why the power dissipated in the brakes to slow the
lorry down is greater at the beginning of the braking period [1]
than in the end.
09/5116/S4 Prelim/P2
10
10. (a) With the aid of a diagram, explain how an electric current may be used to
magnetized a steel bar. [3]
(b) State two ways to increase the magnetism of a permanent magnet. [2]
(c) The figure shows two chains of clips made of two different metals (iron and
steel) that are attached to two ends of a bar magnet.
## (ii) With reference to forces acting on the clips,
th
explain why a 5 clip added to chain B will not stay on the chain. [2]
(iii) When both chains are slowly removed by pulling the top-most clip
away from the magnet, which chain does not collapse immediately? [2]
Why?
09/5116/S4 Prelim/P2
11
11. (a) Describe with the aid of a labelled diagram an experiment you could perform to
show that a wire carrying an electric current perpendicular to a magnetic field
experiences a force ; indicate on the diagram the direction of the force.
[5]
(b) A wire PQRS connected to a battery (not shown) is placed through a piece of
card as shown in Fig. 11. An electric current flows in the direction shown.
P S
card
Q R
Fig. 11
(i) Sketch on the card the magnetic field patterns around the wires. [2]
## (ii) Indicate the direction of the force acting on PQ. [1]
(iii) Suggest two factors which may affect the magnitude of the
force on PQ. [2]
End of Paper
09/5116/S4 Prelim/P2
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09/5116/S4 Prelim/P2 | 2,434 | 11,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-43 | latest | en | 0.830938 |
https://www.enotes.com/homework-help/coordinates-0-0-mark-take-off-point-rocket-473998 | 1,498,411,553,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320545.67/warc/CC-MAIN-20170625170634-20170625190634-00202.warc.gz | 876,225,111 | 13,030 | # The coordinates (0,0) mark the take-off point for a rocket constructed as part of a science class. The positive x direction from (0,0) is considered to be East. a) Find the equation of the...
The coordinates (0,0) mark the take-off point for a rocket constructed as part of a science class. The positive x direction from (0,0) is considered to be East.
a) Find the equation of the rocket's path if it rises at a rate of 5m vertically for every 1m in an easterly direction.
b) A second rocket is fired 2m vertically above from where the first rocket was launched. It rises at a rate of 13m for every 2m in an easterly direction. Find the equation describing its path.
### 1 Answer
txmedteach | High School Teacher | (Level 3) Associate Educator
Posted on
Here, it's best to understand the idea of slope. It is defined as the ratio of the change in the vertical dimension over the corresponding change in the horizontal dimension. As an equation, it looks like this if given two points on the line `(x_1,y_1)` and `(x_2,y_2)` :
`m = (Deltay)/(Deltax) = (y_2-y_1)/(x_2-x_1)`
However, slope is only one part of the picture. You can have many different lines of the same slope! So, you'll need another parameter. For us, the convenient choice will be the y-intercept, we'll call b, meaning the value of the y coordinate when x=0. Knowing the slope and y-intercept, we can produce the following equation for the line:
`y = mx + b`
Armed with this tool, we can now solve the problem.
For the first part, we are given that for any horizontal change of one unit, the vertical change will be 5 units. We are given in the problem that when x = 0, y = 0, as well, considering the starting point is (0,0). So, we have our slope (`m = 5/1`) and we have our y-intercept: 0.
Our equation is therefore the following:
`y = 5/1 x + 0`
`y = 5x`
For (b), the exact same concept applies. We know that for each horizontal change of 2 units, the vertical change is 13 units, giving us a slope of 13/2. At the starting point (x = 0), we are given that it is fired 2 units from the ground (giving a y-intercept of 2). Therefore, we get the following equation describing the motion of the rocket:
`y = 13/2 x + 2` | 600 | 2,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-26 | longest | en | 0.926069 |
https://www.varsitytutors.com/hspt_math-help/how-to-solve-two-step-equations?page=4 | 1,582,469,855,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00336.warc.gz | 935,467,402 | 47,032 | # HSPT Math : How to solve two-step equations
## Example Questions
### Example Question #41 : Use Parentheses, Brackets, Or Braces In Numerical Expressions And Evaluate These Expressions: Ccss.Math.Content.5.Oa.A.1
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the division.
### Example Question #44 : Use Parentheses, Brackets, Or Braces In Numerical Expressions And Evaluate These Expressions: Ccss.Math.Content.5.Oa.A.1
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first, followed by the multiplication, and then the addition.
### Example Question #45 : Use Parentheses, Brackets, Or Braces In Numerical Expressions And Evaluate These Expressions: Ccss.Math.Content.5.Oa.A.1
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the division.
### Example Question #46 : Use Parentheses, Brackets, Or Braces In Numerical Expressions And Evaluate These Expressions: Ccss.Math.Content.5.Oa.A.1
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first, followed by the multiplication, and then addition.
### Example Question #31 : How To Solve Two Step Equations
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication.
### Example Question #21 : Common Core Math: Grade 5
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication, and then the addition.
### Example Question #21 : Common Core Math: Grade 5
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the subtraction.
### Example Question #21 : Use Parentheses, Brackets, Or Braces In Numerical Expressions And Evaluate These Expressions: Ccss.Math.Content.5.Oa.A.1
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first followed by the multiplication.
### Example Question #21 : Use Parentheses, Brackets, Or Braces In Numerical Expressions And Evaluate These Expressions: Ccss.Math.Content.5.Oa.A.1
Solve:
Explanation:
When solving this problem, remember order of operations PEMDAS. The parentheses come first, followed by the multiplication, and then the division.
Solve: | 561 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-10 | latest | en | 0.727667 |
http://www.techiedelight.com/2-vertex-connectivity-graph/ | 1,510,940,526,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803848.60/warc/CC-MAIN-20171117170336-20171117190336-00106.warc.gz | 510,940,537 | 16,971 | # 2-Vertex Connectivity in the graph
Given a undirected connected graph, check if the graph is 2-vertex connected or not.
A connected graph G is said to be 2-vertex-connected (or 2-connected) if it has more than 2 vertices and remains connected on removal of any vertices. Any such vertex whose removal will disconnected the graph is called Articulation point.
For example, Consider below connected graph on the left, if we remove vertex 3 or vertex 4 from the graph, the graph will be disconnected in two connected component. So we can say that 3 and 4 are the Articulation points and graph is not 2-vertex connected. If we add edges (0 -> 1), (0 -> 5) and (2 -> 4) in the graph, it will become 2-vertex connected (check graph on the right).
We can find Articulation points in a graph using DFS. We can say that the graph is 2-vertex connected if and only if for every vertex u in the graph, there is at-least one back-edge that is going out of subtree rooted at u to some ancestor of u. When we say subtree rooted at u, we mean all u’s descendants (excluding vertex u). In other words, when we backtrack from a vertex u, we need to ensure that there is some back-edge from some descendant (children) of u to some ancestor (parent or above) of u. There is an exception to this rule for the root of the tree. If the root has more than one children, then it is an articulation point, otherwise not.
Please note that vertex u and v might be confusing to readers in this post. So please read the post carefully and remember that for an edge u -> v, to check weather or not u is articulation point or not we run DFS on v, not on u.
How should we modify DFS so that we can check if there is a back-edge going out of every sub-tree rooted at u?
We can modify DFS such that DFS(v) returns the smallest arrival time to which there is an back edge from the sub tree rooted at v (including v) to some ancestor of vertex u. For example, let arrival(v) be the arrival time of vertex v in the DFS. Then if there is a back out of the sub tree rooted at v, it’s to something visited before v & therefore with a smaller arrival value. Remember for a back edge u -> v in a graph,
`arrival[u] > arrival[v]`
Suppose there are four edges going out of sub-tree rooted at v to vertex a, b, c and d and with arrival time arrival(a), arrival(b), arrival(c) and arrival(d) respectively. We look at their four arrival times & consider the smallest among them keeping in mind that the back-edge goes to an ancestor of vertex u (and not to vertex u itself) and that will be the value returned by DFS(v). But before returning, check if min(arrival(a), arrival(b), arrival(c), arrival(d)) is more than the arrival(u). If yes, then that means that no back-edge is going out of the sub tree rooted at v and u is an articulation point.
Time complexity of above solution will be O(n + m) where n is number of vertices and m is number of edges in the graph.
Applications – To check if the graph is biconnected or not. A biconnected graph is a connected graph on two or more vertices having no articulation vertices.
Please use ideone or C++ Shell or any other online compiler link to post code in comments.
Like us? Please spread the word and help us grow. Happy coding 🙂 | 776 | 3,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-47 | longest | en | 0.923499 |
https://www.mathnasium.com/math-centers/fulshear/news/number-sense-is-the-key-to-success-in-math-674463944 | 1,726,452,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00063.warc.gz | 801,116,899 | 16,694 | # Number Sense Is the Key to Success in Math!
Jan 2, 2021 | Fulshear
Regardless of age or skill level, there are almost always gaps in a child’s foundational understanding that, if addressed, would boost their success in math.
When we give assessments to children in our math learning centers, we see that a high proportion of their knowledge gaps stems from a lack of number sense. Number sense is “the ability to appreciate the size and scale of numbers in the context of the question at hand.” In essence, it’s the understanding of what numbers mean, how they work together and how they connect to real life. It’s critical that children have a firm grasp of number sense, because it underlies all mathematical comprehension and ability. Typical math tutoring does not teach number sense, acting as a bandage rather than a cure for the problem.
# How do I know if my child has number sense?
Context provides a framework of ideas and information that can be used in a meaningful way. If a child doesn’t understand the context of a mathematical concept or principle when it’s taught to them, they’re likely to say they “don’t get it” or that it “doesn’t make sense.” Sound familiar?
Without number sense, a child feels increasingly lost as they move up through the school grades. Year after year, the disparity grows between what they are being taught and what they truly understand. Children who lack number sense work longer and harder than they should have to in order to solve math problems. Students in middle- and high-school-level math courses — even if they seem to be doing OK — can still lack number sense, because they never got a good grip on it when they were younger. This is part of why so many children need a math tutor.
Children who have good number sense are able to solve problems like these quickly in their heads:
What is 7% of 250?
99 + 99 + 99 = ?
Which is the largest?
Whether it’s in the early or later years, eventually, the child without number sense will “hit a wall.” This can manifest as the inability to understand new material, do certain types of problems, low grades on timed tests, homework frustration, or the mistaken belief that they “can’t do math.” In fact, it’s often problems with number sense that cause children to give up on math. It’s a pity, because it doesn’t have to be that way. Anyone, of any age, can be taught number sense! It’s never too late to learn.
# How do we learn number sense?
The important thing to realize is that number sense doesn’t just come naturally. It’s a learned skill. Teachers can instill number sense if they have the right materials (curriculum and tools) and use the right instruction. This is why Mathnasium invested 40+ years and millions of dollars refining the Mathnasium Method. By teaching number sense in a way that makes sense to children, Mathnasium supports children’s classroom education.
In addition to teachers, parents and other caregivers also play a role in children’s mathematical growth. It’s often as simple as spending a few minutes each day asking number sense-related questions. You can do this on the way to and from school, at the market, or even when enjoying a sporting event together. It’s a great way to instill in your child that math is an important part of everyday life. For great tips on how to do this, see this article.
# What are the components of number sense?
Generally, we break down number sense into three basics: counting, wholes and parts, and proportional thinking. But there are three other principles/skills that are important to fully possess number sense. We’ll break them all down here:
# Counting
When children fully understand counting, they can count from any number, to any number, by any number, both forward and backward. This may sound basic — like any child would eventually learn this ability, but truly, it’s not. That’s why so many children are still using their fingers to count long past when they should be doing the math in their head (mental math).
# Wholes and Parts
We all know the principle “The whole equals the sum of its parts.” A full comprehension of this principle leads to success with problem-solving and fractions — the nemesis of many children (and many adults, too).
# Proportional Thinking
Proportional thinking is thinking “according to amount.” It’s a sense of scale. It helps children to understand problems such as: Five is what part of 15? If you can buy 3 candies with \$2, how many candies can you buy with \$10? Children who have learned proportional thinking can answer these questions quickly and easily.
# Quantity and Denomination
The quantity is the amount of something — the number of things. The denomination is the name-value of that thing. In order to solve math problems, you need to know both.
# The “Law of SAMEness”
This leads to Mathnasium’s “Law of SAMEness”: We can add and subtract only things that have the same name (same denomination). Here’s the classic Mathnasium example: When you add two bananas and three apples, do you get five banapples? Of course not. You can, however, get five pieces of fruit. “Fruit” is the denomination (name) that both apples and bananas share.
# Numerical-Computational Fluency
Numerical fluency is the ability to “effortlessly recall — to know by heart.” While memorization is the way that many of us learned, it does notteach the mathematical thinking and problem-solving skills that are needed to build higher-order math skills. A student may memorize their 12 times tables, but a numerically fluent student uses computational strategies to quickly and efficiently answer questions beyond what is considered normal. For example, a student with numerical fluency can solve problems like 14 x 12 = ? without the need for pencil and paper.
We like to say: “It is fairly easy to forget that which you have memorized, and it’s nearly impossible to forget that which you have learned.”
Mathnasium places enormous emphasis on number sense because it truly is the key to mathematical understanding. In partnership with schoolteachers and parents, the expert math instructors in our centers grow and build upon the power of number sense to foster true conceptual understanding in children. That critical understanding leads to math mastery and to a love and appreciation of math that will positively impact their entire lives. | 1,337 | 6,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.956185 |
http://at.metamath.org/ilegif/mmtheorems52.html | 1,674,807,177,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00586.warc.gz | 3,055,457 | 15,255 | Home Intuitionistic Logic ExplorerTheorem List (p. 52 of 102) < Previous Next > Browser slow? Try the Unicode version. Mirrors > Metamath Home Page > ILE Home Page > Theorem List Contents > Recent Proofs This page: Page List
Theorem List for Intuitionistic Logic Explorer - 5101-5200 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremf1co 5101 Composition of one-to-one functions. Exercise 30 of [TakeutiZaring] p. 25. (Contributed by NM, 28-May-1998.)
Theoremfoeq1 5102 Equality theorem for onto functions. (Contributed by NM, 1-Aug-1994.)
Theoremfoeq2 5103 Equality theorem for onto functions. (Contributed by NM, 1-Aug-1994.)
Theoremfoeq3 5104 Equality theorem for onto functions. (Contributed by NM, 1-Aug-1994.)
Theoremnffo 5105 Bound-variable hypothesis builder for an onto function. (Contributed by NM, 16-May-2004.)
Theoremfof 5106 An onto mapping is a mapping. (Contributed by NM, 3-Aug-1994.)
Theoremfofun 5107 An onto mapping is a function. (Contributed by NM, 29-Mar-2008.)
Theoremfofn 5108 An onto mapping is a function on its domain. (Contributed by NM, 16-Dec-2008.)
Theoremforn 5109 The codomain of an onto function is its range. (Contributed by NM, 3-Aug-1994.)
Theoremdffo2 5110 Alternate definition of an onto function. (Contributed by NM, 22-Mar-2006.)
Theoremfoima 5111 The image of the domain of an onto function. (Contributed by NM, 29-Nov-2002.)
Theoremdffn4 5112 A function maps onto its range. (Contributed by NM, 10-May-1998.)
Theoremfunforn 5113 A function maps its domain onto its range. (Contributed by NM, 23-Jul-2004.)
Theoremfodmrnu 5114 An onto function has unique domain and range. (Contributed by NM, 5-Nov-2006.)
Theoremfores 5115 Restriction of a function. (Contributed by NM, 4-Mar-1997.)
Theoremfoco 5116 Composition of onto functions. (Contributed by NM, 22-Mar-2006.)
Theoremf1oeq1 5117 Equality theorem for one-to-one onto functions. (Contributed by NM, 10-Feb-1997.)
Theoremf1oeq2 5118 Equality theorem for one-to-one onto functions. (Contributed by NM, 10-Feb-1997.)
Theoremf1oeq3 5119 Equality theorem for one-to-one onto functions. (Contributed by NM, 10-Feb-1997.)
Theoremf1oeq23 5120 Equality theorem for one-to-one onto functions. (Contributed by FL, 14-Jul-2012.)
Theoremf1eq123d 5121 Equality deduction for one-to-one functions. (Contributed by Mario Carneiro, 27-Jan-2017.)
Theoremfoeq123d 5122 Equality deduction for onto functions. (Contributed by Mario Carneiro, 27-Jan-2017.)
Theoremf1oeq123d 5123 Equality deduction for one-to-one onto functions. (Contributed by Mario Carneiro, 27-Jan-2017.)
Theoremnff1o 5124 Bound-variable hypothesis builder for a one-to-one onto function. (Contributed by NM, 16-May-2004.)
Theoremf1of1 5125 A one-to-one onto mapping is a one-to-one mapping. (Contributed by NM, 12-Dec-2003.)
Theoremf1of 5126 A one-to-one onto mapping is a mapping. (Contributed by NM, 12-Dec-2003.)
Theoremf1ofn 5127 A one-to-one onto mapping is function on its domain. (Contributed by NM, 12-Dec-2003.)
Theoremf1ofun 5128 A one-to-one onto mapping is a function. (Contributed by NM, 12-Dec-2003.)
Theoremf1orel 5129 A one-to-one onto mapping is a relation. (Contributed by NM, 13-Dec-2003.)
Theoremf1odm 5130 The domain of a one-to-one onto mapping. (Contributed by NM, 8-Mar-2014.)
Theoremdff1o2 5131 Alternate definition of one-to-one onto function. (Contributed by NM, 10-Feb-1997.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremdff1o3 5132 Alternate definition of one-to-one onto function. (Contributed by NM, 25-Mar-1998.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremf1ofo 5133 A one-to-one onto function is an onto function. (Contributed by NM, 28-Apr-2004.)
Theoremdff1o4 5134 Alternate definition of one-to-one onto function. (Contributed by NM, 25-Mar-1998.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremdff1o5 5135 Alternate definition of one-to-one onto function. (Contributed by NM, 10-Dec-2003.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremf1orn 5136 A one-to-one function maps onto its range. (Contributed by NM, 13-Aug-2004.)
Theoremf1f1orn 5137 A one-to-one function maps one-to-one onto its range. (Contributed by NM, 4-Sep-2004.)
Theoremf1oabexg 5138* The class of all 1-1-onto functions mapping one set to another is a set. (Contributed by Paul Chapman, 25-Feb-2008.)
Theoremf1ocnv 5139 The converse of a one-to-one onto function is also one-to-one onto. (Contributed by NM, 11-Feb-1997.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremf1ocnvb 5140 A relation is a one-to-one onto function iff its converse is a one-to-one onto function with domain and range interchanged. (Contributed by NM, 8-Dec-2003.)
Theoremf1ores 5141 The restriction of a one-to-one function maps one-to-one onto the image. (Contributed by NM, 25-Mar-1998.)
Theoremf1orescnv 5142 The converse of a one-to-one-onto restricted function. (Contributed by Paul Chapman, 21-Apr-2008.)
Theoremf1imacnv 5143 Preimage of an image. (Contributed by NM, 30-Sep-2004.)
Theoremfoimacnv 5144 A reverse version of f1imacnv 5143. (Contributed by Jeff Hankins, 16-Jul-2009.)
Theoremfoun 5145 The union of two onto functions with disjoint domains is an onto function. (Contributed by Mario Carneiro, 22-Jun-2016.)
Theoremf1oun 5146 The union of two one-to-one onto functions with disjoint domains and ranges. (Contributed by NM, 26-Mar-1998.)
Theoremfun11iun 5147* The union of a chain (with respect to inclusion) of one-to-one functions is a one-to-one function. (Contributed by Mario Carneiro, 20-May-2013.) (Revised by Mario Carneiro, 24-Jun-2015.)
Theoremresdif 5148 The restriction of a one-to-one onto function to a difference maps onto the difference of the images. (Contributed by Paul Chapman, 11-Apr-2009.)
Theoremf1oco 5149 Composition of one-to-one onto functions. (Contributed by NM, 19-Mar-1998.)
Theoremf1cnv 5150 The converse of an injective function is bijective. (Contributed by FL, 11-Nov-2011.)
Theoremfuncocnv2 5151 Composition with the converse. (Contributed by Jeff Madsen, 2-Sep-2009.)
Theoremfococnv2 5152 The composition of an onto function and its converse. (Contributed by Stefan O'Rear, 12-Feb-2015.)
Theoremf1ococnv2 5153 The composition of a one-to-one onto function and its converse equals the identity relation restricted to the function's range. (Contributed by NM, 13-Dec-2003.) (Proof shortened by Stefan O'Rear, 12-Feb-2015.)
Theoremf1cocnv2 5154 Composition of an injective function with its converse. (Contributed by FL, 11-Nov-2011.)
Theoremf1ococnv1 5155 The composition of a one-to-one onto function's converse and itself equals the identity relation restricted to the function's domain. (Contributed by NM, 13-Dec-2003.)
Theoremf1cocnv1 5156 Composition of an injective function with its converse. (Contributed by FL, 11-Nov-2011.)
Theoremfuncoeqres 5157 Re-express a constraint on a composition as a constraint on the composand. (Contributed by Stefan O'Rear, 7-Mar-2015.)
Theoremffoss 5158* Relationship between a mapping and an onto mapping. Figure 38 of [Enderton] p. 145. (Contributed by NM, 10-May-1998.)
Theoremf11o 5159* Relationship between one-to-one and one-to-one onto function. (Contributed by NM, 4-Apr-1998.)
Theoremf10 5160 The empty set maps one-to-one into any class. (Contributed by NM, 7-Apr-1998.)
Theoremf1o00 5161 One-to-one onto mapping of the empty set. (Contributed by NM, 15-Apr-1998.)
Theoremfo00 5162 Onto mapping of the empty set. (Contributed by NM, 22-Mar-2006.)
Theoremf1o0 5163 One-to-one onto mapping of the empty set. (Contributed by NM, 10-Sep-2004.)
Theoremf1oi 5164 A restriction of the identity relation is a one-to-one onto function. (Contributed by NM, 30-Apr-1998.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremf1ovi 5165 The identity relation is a one-to-one onto function on the universe. (Contributed by NM, 16-May-2004.)
Theoremf1osn 5166 A singleton of an ordered pair is one-to-one onto function. (Contributed by NM, 18-May-1998.) (Proof shortened by Andrew Salmon, 22-Oct-2011.)
Theoremf1osng 5167 A singleton of an ordered pair is one-to-one onto function. (Contributed by Mario Carneiro, 12-Jan-2013.)
Theoremf1oprg 5168 An unordered pair of ordered pairs with different elements is a one-to-one onto function. (Contributed by Alexander van der Vekens, 14-Aug-2017.)
Theoremtz6.12-2 5169* Function value when is not a function. Theorem 6.12(2) of [TakeutiZaring] p. 27. (Contributed by NM, 30-Apr-2004.) (Proof shortened by Mario Carneiro, 31-Aug-2015.)
Theoremfveu 5170* The value of a function at a unique point. (Contributed by Scott Fenton, 6-Oct-2017.)
Theorembrprcneu 5171* If is a proper class, then there is no unique binary relationship with as the first element. (Contributed by Scott Fenton, 7-Oct-2017.)
Theoremfvprc 5172 A function's value at a proper class is the empty set. (Contributed by NM, 20-May-1998.)
Theoremfv2 5173* Alternate definition of function value. Definition 10.11 of [Quine] p. 68. (Contributed by NM, 30-Apr-2004.) (Proof shortened by Andrew Salmon, 17-Sep-2011.) (Revised by Mario Carneiro, 31-Aug-2015.)
Theoremdffv3g 5174* A definition of function value in terms of iota. (Contributed by Jim Kingdon, 29-Dec-2018.)
Theoremdffv4g 5175* The previous definition of function value, from before the operator was introduced. Although based on the idea embodied by Definition 10.2 of [Quine] p. 65 (see args 4694), this definition apparently does not appear in the literature. (Contributed by NM, 1-Aug-1994.)
Theoremelfv 5176* Membership in a function value. (Contributed by NM, 30-Apr-2004.)
Theoremfveq1 5177 Equality theorem for function value. (Contributed by NM, 29-Dec-1996.)
Theoremfveq2 5178 Equality theorem for function value. (Contributed by NM, 29-Dec-1996.)
Theoremfveq1i 5179 Equality inference for function value. (Contributed by NM, 2-Sep-2003.)
Theoremfveq1d 5180 Equality deduction for function value. (Contributed by NM, 2-Sep-2003.)
Theoremfveq2i 5181 Equality inference for function value. (Contributed by NM, 28-Jul-1999.)
Theoremfveq2d 5182 Equality deduction for function value. (Contributed by NM, 29-May-1999.)
Theoremfveq12i 5183 Equality deduction for function value. (Contributed by FL, 27-Jun-2014.)
Theoremfveq12d 5184 Equality deduction for function value. (Contributed by FL, 22-Dec-2008.)
Theoremnffv 5185 Bound-variable hypothesis builder for function value. (Contributed by NM, 14-Nov-1995.) (Revised by Mario Carneiro, 15-Oct-2016.)
Theoremnffvmpt1 5186* Bound-variable hypothesis builder for mapping, special case. (Contributed by Mario Carneiro, 25-Dec-2016.)
Theoremnffvd 5187 Deduction version of bound-variable hypothesis builder nffv 5185. (Contributed by NM, 10-Nov-2005.) (Revised by Mario Carneiro, 15-Oct-2016.)
Theoremfunfveu 5188* A function has one value given an argument in its domain. (Contributed by Jim Kingdon, 29-Dec-2018.)
Theoremfvss 5189* The value of a function is a subset of if every element that could be a candidate for the value is a subset of . (Contributed by Mario Carneiro, 24-May-2019.)
Theoremfvssunirng 5190 The result of a function value is always a subset of the union of the range, if the input is a set. (Contributed by Stefan O'Rear, 2-Nov-2014.) (Revised by Mario Carneiro, 24-May-2019.)
Theoremrelfvssunirn 5191 The result of a function value is always a subset of the union of the range, even if it is invalid and thus empty. (Contributed by Stefan O'Rear, 2-Nov-2014.) (Revised by Mario Carneiro, 24-May-2019.)
Theoremfunfvex 5192 The value of a function exists. A special case of Corollary 6.13 of [TakeutiZaring] p. 27. (Contributed by Jim Kingdon, 29-Dec-2018.)
Theoremrelrnfvex 5193 If a function has a set range, then the function value exists unconditional on the domain. (Contributed by Mario Carneiro, 24-May-2019.)
Theoremfvexg 5194 Evaluating a set function at a set exists. (Contributed by Mario Carneiro and Jim Kingdon, 28-May-2019.)
Theoremfvex 5195 Evaluating a set function at a set exists. (Contributed by Mario Carneiro and Jim Kingdon, 28-May-2019.)
Theoremsefvex 5196 If a function is set-like, then the function value exists if the input does. (Contributed by Mario Carneiro, 24-May-2019.)
Se
Theoremfv3 5197* Alternate definition of the value of a function. Definition 6.11 of [TakeutiZaring] p. 26. (Contributed by NM, 30-Apr-2004.) (Revised by Mario Carneiro, 31-Aug-2015.)
Theoremfvres 5198 The value of a restricted function. (Contributed by NM, 2-Aug-1994.)
Theoremfunssfv 5199 The value of a member of the domain of a subclass of a function. (Contributed by NM, 15-Aug-1994.)
Theoremtz6.12-1 5200* Function value. Theorem 6.12(1) of [TakeutiZaring] p. 27. (Contributed by NM, 30-Apr-2004.)
Page List
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Copyright terms: Public domain < Previous Next > | 4,731 | 14,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-06 | latest | en | 0.763996 |
https://cxc.cfa.harvard.edu/ciao4.2/ahelp/grpgetgrpnum.html | 1,669,901,336,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00206.warc.gz | 212,649,560 | 15,012 | AHELP for CIAO 4.2
# grpGetGrpNum
Context: group
## Synopsis
Calculate the group number for each element in the array.
## Syntax
`Integer_Type grpnum = grpGetGrpNum( Array_Type grouping )`
## Description
This function calculates which group each element in the input array belongs to, where the groups are numbered from 1. The return value is an array whose length equals that of the input data (the grouping argument) and each element within a group contains the same value. As an example,
`grpGetGrpNum( [1,-1,1,-1,1,1] )`
will return the array
`[1,1,2,2,3,4]`
since the groups consist of the first two elements, then the next two elements, with the last two elements being un-grouped.
The group module is not available by default; to use it in a S-Lang program, it must be loaded using the S-Lang require() function:
` require("group");`
## Example 1
`slsh> igrp = grpGetGrpNum( grp );`
The igrp array will contain the group number for each element of the input array, for the grouping scheme stored in the grp array, and has the same size as the grp array.
## Example 2
```slsh> x = [0.5:6.0:0.05];
slsh> y = 3 + 30 * exp( - (x-2.0)^2 / 0.1 );
slsh> ( grp, qual ) = grpNumCounts( y, 15 );
slsh> yavg = grpGetGroupSum( y, grp ) / grpGetChansPerGroup( grp );
slsh> igrp = grpGetGrpNum( grp );
slsh> i = where( grp == 1 and qual == 0 );```
Here we take the function
`y = 3 + 30 * exp( -(x-2)^2 / 0.1 )`
and group it by 15 counts per group.
## Bugs
See the bugs page for the group library on the CIAO website for an up-to-date listing of known bugs. | 478 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.690929 |
https://www.convert-measurement-units.com/convert+Nanosievert+per+year+to+Sievert+per+second.php | 1,642,513,804,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00109.warc.gz | 782,055,521 | 12,779 | Convert nSv/y to Sv/s (Nanosievert per year to Sievert per second)
## Nanosievert per year into Sievert per second
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Nanosievert+per+year+to+Sievert+per+second.php
## How many Sievert per second make 1 Nanosievert per year?
1 Nanosievert per year [nSv/y] = 0.000 000 000 000 000 031 688 087 814 029 Sievert per second [Sv/s] - Measurement calculator that can be used to convert Nanosievert per year to Sievert per second, among others.
# Convert Nanosievert per year to Sievert per second (nSv/y to Sv/s):
1. Choose the right category from the selection list, in this case 'Radiation dose'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Nanosievert per year [nSv/y]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Sievert per second [Sv/s]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '679 Nanosievert per year'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Nanosievert per year' or 'nSv/y'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Radiation dose'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '16 nSv/y to Sv/s' or '11 nSv/y into Sv/s' or '10 Nanosievert per year -> Sievert per second' or '16 nSv/y = Sv/s' or '94 Nanosievert per year to Sv/s' or '17 nSv/y to Sievert per second' or '74 Nanosievert per year into Sievert per second'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(42 * 83) nSv/y'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '679 Nanosievert per year + 2037 Sievert per second' or '39mm x 32cm x 89dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 3.160 493 798 4×1021. For this form of presentation, the number will be segmented into an exponent, here 21, and the actual number, here 3.160 493 798 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 3.160 493 798 4E+21. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 3 160 493 798 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 1,003 | 4,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-05 | latest | en | 0.782767 |
https://physics.stackexchange.com/questions/197507/casimir-effect-for-spinning-casimir-plates | 1,701,381,818,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00494.warc.gz | 513,089,248 | 43,079 | # Casimir effect for spinning Casimir plates
I recently thought of the following experiment. Let's say I have two plates in vacuum facing each other. Now, due to the Casimir effect, there will be some internal attraction between the plates. Now let's say we spin the plates while facing each other about their combined centre of mass.
The shortest path between the plates is no longer a straight line as the geometry changes when undergoing acceleration. What is the new shortest path between the plates and force between them due to the change in geometry?
• Why is spacetime curved? As far as I know this can be handled with special relativity. Aug 3, 2015 at 19:27
• @Javier I massively edited the thought experiment hopefully now it does have elements of general relativity :) Aug 6, 2015 at 13:25
• I don't understand what the Casimir energy has to do with this. The problem can be formulated independently of the Casimir effect by demanding a spinning ring of radius R. Are the two points under consideration on the ring or are they exterior to the ring but in close proximity to the spinning ring?
– user58089
Aug 6, 2015 at 13:45
• So my understanding of the Casimir effect is as follows: If I have two plates with a gap .. They are attracted to each other ... Now similarly the ring should also experience this force which is dependent on the distance between two points (on the ring) ... I hope this explains what I'm thinking (please help me rephrase the experiment if you can) Aug 6, 2015 at 13:51
• I tried to solve it in Born coordinates, but I am afraid it is a bit too involved. I am getting like $r^2 (1 - 2\omega^2 r^2) R'' + r (1 - 4\omega^2 r^2) R' - k_z^2 + [r^2( \alpha_t^2 - \omega^2 \alpha_\phi^2- \omega \alpha_{\phi t}) + \alpha_\phi^2]R = 0$ just for one part of the solution. Maybe you could try to apply a coordinate transform to the stress energy tensor using Born coordinates and see what happens? I suspect it may not be enough, since the transformation applies to $\langle 0 \vert \hat{T}_{ab} \vert 0 \rangle$ as a whole, but it might be worth a try, perhaps Aug 10, 2015 at 11:18
In layman's terms, the Casimir effect is an outside pressure pushing the plates together. It comes from modes of quantum fields that have longer wavelengths than the separation of the plates. Therefore, these modes can no longer be excited by vacuum fluctuations.
Since special relativity is a basic ingredient in QFT, these fields are homogeneous and isotropic. Thefore a constant Lorentz transformation will not alter the measurement of the Casimir effect.
In general relativity, we allow for non-constant Lorentz transformations and even for fully general diffeomorphisms. Then, there are new effects to consider, most prominently the Uhruh effect.
The derivations of the Unruh effect assumes constant acceleration for simplicity. A rotating frame implies non-constant acceleration for parts of your plates. A naive consideration in your thought experiment could go like
• Let us consider only the outermost edge of each plate
• These edges are accelerated towards the center of mass
• If we are allowed to consider infinitesimal time slices, the acceleration is approximately constant for any given time slice
• The acceleration towards the c.o.m. implies (in the rest frame of the edge) Unruh radiation, reducing the magnitude of the acceleration
• Therefore the measured attraction between the plates should be reduced
The above argument assumes that we are allowed to approximate the acceleration to be constant at each instant, which might not hold. Also, the Unruh radiation only exists in the rest frame of the edges, which are different frames for each edge, so concluding that there is less attraction from just considering one edge is quite bold. Finally, I did not carefully consider the transition from the rest frames to the lab frame.
All these caveats could be solved by careful consideration, but that is beyond the scope of my interest for this thought experiment :-P
• I suppose 50 bounty points aren't enough to tempt you ... :( Aug 7, 2015 at 17:43
• @AnantSaxena No, I'm afraid they are not. Looking into these questions in detail easily involves hours of work and from my experience QFT in GR backgrounds can be so confusing that even then I'm not sure I'd get the answer right... Aug 10, 2015 at 12:58
• I'm glad that you atleast told me the considerations ... Cheers and thank you! Aug 10, 2015 at 14:29 | 1,035 | 4,456 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-50 | longest | en | 0.938467 |
http://www.chegg.com/homework-help/stats-engineers-and-scientists-2nd-edition-chapter-7.3-solutions-9780073127798 | 1,455,316,481,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701165378.58/warc/CC-MAIN-20160205193925-00157-ip-10-236-182-209.ec2.internal.warc.gz | 332,923,006 | 16,090 | View more editions
# TEXTBOOK SOLUTIONS FOR STATS Engineers and Scientists 2nd Edition
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 8
1. (a) .
.
• Step 2 of 8
(b) .
• Step 3 of 8
(c) .
.
There are n − 2 = 23 degrees of freedom. t23,.025 = 2.069.
Therefore a 95% confidence interval for β0 is 4.782923 ± 2.069(0.18085), or (4.409, 5.157).
The 95% confidence interval for β1 is 3.082115 ± 2.069(0.064590), or (2.948, 3.216).
• Step 4 of 8
(d) . There are 25 − 2 = 23 degrees of freedom.
The null and alternate hypotheses are H01 ≤ 3 versus H1: β1 > 3.
t = (3.082115 − 3)/0.064590 = 1.271.
Since the alternate hypothesis is of the form β1 > b, the P-value is the area to the right of t = 1.271.
From the t table, 0.10 < P < 0.25. A computer package gives P = 0.108.
We cannot conclude that the claim is false.
• Step 5 of 8
(e) . There are 25 − 2 = 23 degrees of freedom.
The null and alternate hypotheses are H0: β0 ≥ 5.5 versus H1: β0 < 5.5.
t = (4.782923 − 5.5)/0.18085 = −3.965.
Since the alternate hypothesis is of the form β0 < b, the P-value is the area to the left of t = −3.965.
From the t table, P < 0.0005. A computer package gives P = 0.00031.
We can conclude that the claim is false.
• Step 6 of 8
(f) .
. There are 23 degrees of freedom. t23 = 2.807.
Therefore a 99% confidence interval for the mean response is 9.4060955 ± 2.807(0.109803), or (9.098,9.714).
• Step 7 of 8
(g) .
. There are 23 degrees of freedom. t23,.005 = 2.807.
Therefore a 99% prediction interval is 9.4060955 ± 2.807(0.478535), or (8.063,10.749).
• Step 8 of 8
(h) The confidence interval is more useful, since it is concerned with the true length of the spring, while the prediction interval is concerned with the next measurement of the length.
Corresponding Textbook
STATS Engineers and Scientists | 2nd Edition
9780073127798ISBN-13: 0073127795ISBN: Authors: | 704 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-07 | latest | en | 0.786551 |
https://www.coursehero.com/file/6030793/Quiz-7-solutions/ | 1,513,481,549,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948592846.98/warc/CC-MAIN-20171217015850-20171217041850-00347.warc.gz | 693,780,319 | 45,262 | Quiz 7 solutions
# Quiz 7 solutions - Name Quiz 7(30 pts 1 Sketch a sinusoid...
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Name______________________ Quiz 7 1. Sketch a sinusoid with amplitude A = 3.0 and frequency ω = 1.0, as well as the steady state time response of a first order system with τ = 3.0 sec and K = 1.0 to this sinusoidal input; plot the curves versus ω t. Make sure your sketch shows the amplitudes reasonably correctly. Recall that at steady state: () 1 2 1 KA y( t ) sin t tan ω ωτ =− + . We need to find the amplitude and phase of the response. ωτ = 3 1. estimation method 2 1 1 M( ) = + 2 11 1 1 1 90 3 3 3 19 1 0 9 113 M . * == = = = + + The amplitude is reduced by a factor of 3. At ωτ = 1, φ = -45°. At ωτ = 10, φ = -90°. The delay is between 45° and 90°. On a log scale, one can make a straight line approximation (see below) for the phase shift
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Ask a homework question - tutors are online | 396 | 1,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-51 | latest | en | 0.837947 |
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