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https://community.glideapps.com/t/weeknum-and-year/69262 | 1,726,723,502,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00267.warc.gz | 155,798,574 | 10,179 | Weeknum and Year
Hello,
Can someone tell me if it is possible to use the weeknum function and also use a formula to add the year of that week?
Try this in a math column.
`YEAR(Date)*10^2+WEEKNUM(date)`
2 Likes
Or you can use a Format Date column
Use `yyyy` to return the year
`W` to return the number of the week
and `WW` to return the padded number
I personally prefer the math column approach Jeff suggested, but sometimes you need more than just a number
Thank you!
But I want to show me somenthing like “2024-1”, is that possible?
You can create two Math columns, one for Year YEAR(Date) and one for Week WEEKNUM(date), then create a Template column with this:
X-Y where X is the Year column and Y the Week column.
1 Like
It works. Thank Youuu!
1 Like
Or one JavaScript column:
``````var ddd = new Date();
var onejan = new Date(ddd.getFullYear(), 0, 1);
var week = Math.ceil((((ddd.getTime() - onejan.getTime()) / 86400000) + onejan.getDay() + 1) / 7);
return ddd.getFullYear()+'-'+week
``````
1 Like
If you need to get a date from an existing column, use this:
``````
var ddd = new Date(p1);
var yyy = ddd.getFullYear();
var onejan = new Date(yyy, 0, 1);
var week = Math.ceil((((ddd.getTime() - onejan.getTime()) / 86400000) + onejan.getDay() + 1) / 7);
return yyy+'-'+week
``````
Using my solution will speed up your App two times… which will be helpful if you have lots of rows to compute.
1 Like
Thank youu!!!
And for month? Can you show me the code, please?
You need only one column to make your request… not 3
Oh, do you mean month instead of week?
Yesss
``````var ddd = new Date(p1);
var yyy = ddd.getFullYear();
var mmm = ddd.getMonth()+1;
var onejan = new Date(yyy, 0, 1);
var week = Math.ceil((((ddd.getTime() - onejan.getTime()) / 86400000) + onejan.getDay() + 1) / 7);
return yyy+'-'+mmm+'-'+week
``````
1 Like
Thank youuu!!!
1 Like
Please test the JavaScript method above in iOS to see if it works correctly. It’s a nice solution, but sometimes Safari doesn’t like the Date functions for Glide.
1 Like
I have seen JS date functions fail on IOS in the past, so I try to avoid them.
The only method I feel 100% confident about is the Glide Date Math method.
3 Likes
Nope… Glide dates failed many times for me… javascript… never… but the secret is, that you need to stick to java from the beginning… meaning get all your dates from the JavaScript column… do all the computing in JavaScript columns… and convert to localeString when you need to display it…
This is absolutely necessary if you work with users from different time zones and importing/exporting data to external sources.
@Darren_Murphy, how do you think the math column dates work??? I think it is a JavaScript code…
Javascript is the slowest column ever, it’s best to avoid it if possible.
As for 2024-1, the fastest and the most reliable is math+template
Format Date is slower, but that’s one column
And JavaScript is simply the slowest, so not recommended
1 Like
@tuzin, I don’t know where you are getting your information… LOL
JavaScript always beats the Math column… did you even do testing before writing your statement???
Here is my test:
Math column 114 milliseconds, JavaScript 112 milliseconds.
And if you wanna count on the reliability of the math column… LOL, only if the source is Glide… otherwise, it is the same as JavaScript… that’s why I always use JavaScript as my calculation and format base… Good luck trying to export Math column results outside and relying on the calculations based on that. | 958 | 3,580 | {"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} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.642872 |
https://es.mathworks.com/matlabcentral/answers/503487-for-loop-array-question?s_tid=prof_contriblnk | 1,713,382,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00893.warc.gz | 221,535,755 | 27,002 | # For Loop /Array question
1 visualización (últimos 30 días)
Tyler Young el 3 de Feb. de 2020
Comentada: Tyler Young el 4 de Feb. de 2020
How would i go about creating a for loop to create an array where each value in X is equal to its associated row value * 2 plus its associated column value * 3 +1. Size of array is 50x50 ?
##### 2 comentariosMostrar NingunoOcultar Ninguno
James Tursa el 3 de Feb. de 2020
What have you done so far? What specific problems are you having with your code? Do you know how to write a for-loop? Do you know how to pre-allocate a 50x50 matrix?
Tyler Young el 4 de Feb. de 2020
I know how to write simpler for loops where I can identify a "start" point for my indx and run through a simple array or vector to "end" but Im struggling to comprehend how to build the value of this equation.
Iniciar sesión para comentar.
Hiro Yoshino el 4 de Feb. de 2020
X = zeros(50, 50);
[m, n] = size(X); % m x n = 50, 50
for i = 1:m % Row
for j = 1:n % Column
X(i, j) = i * 2 + j * 3 + 1;
end
end
(m, n) could be anytihng.
The point here is size function. Good luck!
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Tyler Young el 4 de Feb. de 2020
Thank you, that does work, and helps to see it done that way too. I ended up going about it differently, but I was orginally thinking of using "zeros"
for row = 1:1:50
for col = 1:1:50
X(row,col) = [row * 2 + col * 3 + 1]
end
end
Iniciar sesión para comentar.
### Más respuestas (1)
KSSV el 4 de Feb. de 2020
Check the below deom code where each element of a matrix is sum of its row and column position. Extend this to your case.
m = 5 ; % number of rows
n = 5 ; % number of columns
iwant = zeros(m,n) ; % initialize the required matrix
% loop
for i = 1:m % loop for row
for j = 1:n % loop for column
iwant(i,j) = i+j ; %(i,j)th element is sum of i and j
end
end
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Tyler Young el 4 de Feb. de 2020
Thank you, appreciate the multiple ways of seeing this done.
Iniciar sesión para comentar.
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Translated by | 699 | 2,335 | {"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} | 3.09375 | 3 | CC-MAIN-2024-18 | latest | en | 0.518443 |
https://www.physicsforums.com/threads/determining-appropriate-focal-length.912848/ | 1,713,642,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00498.warc.gz | 851,236,088 | 17,527 | # Determining Appropriate Focal Length
• I
• MxwllsPersuasns
In summary: If the diode laser you have doesn't have a correction lens, you can make one by soldering a small lens to the front of the diode. Assuming the object is infinitely far away, the equation 1/ƒ = 1/μ or focal length would be the same as the distance from the lens to the image.In summary, the focal length should be the same as the distance from the lens to the image.
MxwllsPersuasns
So I'm working on a project where we're deciding a new lens to use for our laser diode. I need to determine the appropriate focal length based on the equation: 1/ƒ = 1/μ + 1/σ where ƒ is our focal length, μ is the distance from the lens to image and σ is distance from lens to object. For the purposes of our project we've determined we can assume that our object is infinitely far away and thus we have 1/ƒ = 1/μ or our focal length should be the same as the distance from our lens to our image.
Now I just need some help finding that distance. I was looking on the schematic diagram for our laser diode and wanted to determine that distance. I imagined the distance from the "image" to the lens for a laser (as opposed to a camera) would be from the point of emission of the laser to the lens. I've attached a picture of this schematic, if anyone can help me find the distance σ I would truly appreciate it.
** The little distance indicator that I drew in pencil on the diagram is where I believed the distance σ would be. But there's no way to determine this distance from the info given.
Sorry but I coulnd't understand the phisical situation.
You have a laser a lens and a plane where the image will be formed, right? Where should the object be?
Yes we have laser light coming out of a laser diode (what I interpret to be the "image" -- please correct me if I am wrong) which then passes through and gets refracted by a lens and which then travels through a crystal and through a beamsplitter into photo-detectors. Now we can assume the object that the laser is hitting is far enough away such that we can neglect the term associated with it. This then means the equation 1/ƒ = 1/σ tells us the appropriate focal length will be the length from the "image" (again, where the laser is emitted, I believe) to the lens.
I need to find that distance.
From the LASER spec sheet, look up the beam diameter and the beam divergence angle. From these, calculate where a point would have to be to match the listed values. Note that the beam from a bare diode laser is elliptical, not circular; i.e. there is a lot of astigmatism. Some, not all, prepackaged diode lasers have a built-in correction lens to yield a somewhat circular beam.
## 1. What is focal length?
Focal length is the distance between the lens of a camera and the image sensor when the subject is in focus. It is typically measured in millimeters (mm) and determines the magnification and angle of view of the image.
## 2. How do you determine the appropriate focal length for a photo?
The appropriate focal length for a photo depends on the subject, the desired composition, and the type of camera being used. Generally, a shorter focal length (e.g. 35mm) is better for capturing a wide angle view, while a longer focal length (e.g. 200mm) is better for close-up shots. It is also important to consider the crop factor of the camera, as this will affect the effective focal length.
## 3. What is the difference between a fixed and zoom focal length?
A fixed focal length, also known as a prime lens, has a fixed focal length and cannot be adjusted. A zoom lens, on the other hand, has a range of focal lengths that can be adjusted by the photographer. Prime lenses are generally sharper and faster (allowing for a larger aperture) than zoom lenses, but zoom lenses offer more versatility in terms of composition.
## 4. How does focal length affect depth of field?
Focal length is one of the factors that affects depth of field, or the area of the photo that is in focus. A longer focal length will result in a shallower depth of field, while a shorter focal length will result in a deeper depth of field. This is why portrait photographers often use longer focal lengths to achieve a shallow depth of field and blur the background.
## 5. Can I change the focal length of my smartphone camera?
Some smartphone cameras have a fixed focal length, meaning it cannot be changed. However, some newer models have multiple lenses with different focal lengths that can be switched between, allowing for more versatility in composition. It is important to check the specifications of your specific smartphone camera to see if it has this feature.
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2K | 1,196 | 5,039 | {"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} | 3.4375 | 3 | CC-MAIN-2024-18 | latest | en | 0.949192 |
http://mathhelpforum.com/differential-equations/160709-method-characteristics-prove-u-v-solution-print.html | 1,503,157,237,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105455.37/warc/CC-MAIN-20170819143637-20170819163637-00434.warc.gz | 279,039,284 | 5,129 | # Method of Characteristics: Prove the u and v is a solution
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 23rd 2010, 09:41 AM
bugatti79
Method of Characteristics: Prove the u and v is a solution
Dear Folks,
Can anyone get me started on how to approach the following question:
Consider the following 1st order linear PDE $a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} + c(x,y)u=0$
If u and v are 2 solutions shown that a) u + v and b) ku for any constant k are also solutions?
Just a tip would be good, I want to tackle it myself.
Thanks
• Oct 23rd 2010, 10:35 AM
Ackbeet
Just plug u + v into the pde and use the linearity of everything. Do the same for ku.
• Oct 23rd 2010, 10:51 AM
bugatti79
but we dont know what the unknown functions u and v are so that we can substitute back into PDE.... Thats my understanding of things...
• Oct 23rd 2010, 11:00 AM
Ackbeet
That's true, we don't know what u and v are. At least, we don't have a closed-form expression for them. However, we do know they are solutions of the pde, and that's all we need. Try this:
$\displaystyle a(x,y)\frac{\partial}{\partial x}(u+v)+b(x,y)\frac{\partial}{\partial y}(u+v)+c(x,y)(u+v)=?$
• Oct 23rd 2010, 12:03 PM
bugatti79
Quote:
Originally Posted by Ackbeet
That's true, we don't know what u and v are. At least, we don't have a closed-form expression for them. However, we do know they are solutions of the pde, and that's all we need. Try this:
$\displaystyle a(x,y)\frac{\partial}{\partial x}(u+v)+b(x,y)\frac{\partial}{\partial y}(u+v)+c(x,y)(u+v)=?$
Well I could multiply out the brackets like
$a(x,y)\frac{\partial u(x,y)}{\partial x} +a (x,y)\frac{\partial v(x,y)}{\partial x}.....$
Put all v terms on one side and u terms the other side....
I have difficulty seeing the 'link'..... (Shake)
• Oct 23rd 2010, 12:07 PM
Ackbeet
No need to put them on different sides of the equation. Just group them differently. You can separate out all the u terms, and shove them to the left. Group the v terms, and shove them to the right, but all terms are still on the LHS of the equation. Write this out. Then what can you say?
• Oct 23rd 2010, 12:23 PM
bugatti79
Quote:
Originally Posted by Ackbeet
No need to put them on different sides of the equation. Just group them differently. You can separate out all the u terms, and shove them to the left. Group the v terms, and shove them to the right, but all terms are still on the LHS of the equation. Write this out. Then what can you say?
$a(x,y)\frac{\partial u(x,y)}{\partial x} +b (x,y)\frac{\partial u(x,y)}{\partial x} +c(x,y)u+a(x,y)\frac{\partial v(x,y)}{\partial x} +b (x,y)\frac{\partial v(x,y)}{\partial x}+c(x,y)v=0$
Ok, just multiplying out the brackets and grouping like terms etc and all this is equal to 0 gives me the impression that u + v is also a solution.....but its still not a 'proof'.
I merely substituted and rearranged.... (Doh)
• Oct 23rd 2010, 12:26 PM
Ackbeet
But $u$ is a solution, and hence
$a(x,y)\dfrac{\partial u(x,y)}{\partial x} +b (x,y)\dfrac{\partial u(x,y)}{\partial x} +c(x,y)u=0.$
Also, $v$ is a solution, and hence
$a(x,y)\dfrac{\partial v(x,y)}{\partial x} +b (x,y)\dfrac{\partial v(x,y)}{\partial x}+c(x,y)v=0.$
So you see why you're done now?
• Oct 23rd 2010, 12:35 PM
bugatti79
Quote:
Originally Posted by Ackbeet
But $u$ is a solution, and hence
$a(x,y)\dfrac{\partial u(x,y)}{\partial x} +b (x,y)\dfrac{\partial u(x,y)}{\partial x} +c(x,y)u=0.$
Also, $v$ is a solution, and hence
$a(x,y)\dfrac{\partial v(x,y)}{\partial x} +b (x,y)\dfrac{\partial v(x,y)}{\partial x}+c(x,y)v=0.$
So you see why you're done now?
Yes sure!!! From above 0+0 = 0. So obvious. I was just fixated on some elaborate proof. Its a pleasure learning from you, thanks (Happy)
• Oct 23rd 2010, 12:53 PM
Ackbeet
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?
• Oct 23rd 2010, 12:56 PM
bugatti79
Quote:
Originally Posted by Ackbeet
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?
I will have a look at it tomorrow...Im going for a pint of guinneas now :-)
Thanks
• Oct 23rd 2010, 01:05 PM
Ackbeet
Sounds good! I always take Sundays off from the MHF, but I'll be back tomorrow.
• Oct 24th 2010, 06:10 AM
bugatti79
Quote:
Originally Posted by Ackbeet
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?
Well, because of the unknown function u is squared, the 1st order PDE becomes nonlinear. But not sure how to prove it mathematically...
I could multiply out the terms and get something like the 'partial terms' + c(x,y)u^2+c(x,y)v^2=0
I dont see how it would fail....
Also, I have attempted the ku problem...see attached jpg (part 2).
Thanks
• Oct 25th 2010, 02:10 AM
Ackbeet
No, no. Your solution for ku is not correct. You DON'T know that all those partial derivatives are zero. In fact, they are probably nonzero. However, you do know that k, a constant, behaves in a certain fashion under derivative signs, right? I mean, how could you simplify the expression
$\dfrac{\partial(ku)}{\partial x}?$
• Oct 25th 2010, 03:13 AM
bugatti79
Quote:
Originally Posted by Ackbeet
No, no. Your solution for ku is not correct. You DON'T know that all those partial derivatives are zero. In fact, they are probably nonzero. However, you do know that k, a constant, behaves in a certain fashion under derivative signs, right? I mean, how could you simplify the expression
$\dfrac{\partial(ku)}{\partial x}?$
Your right, they cant be 0, poor judgement on my part. I guess you can pull out the k's and divide across by k. And you could do this for any constant k. See attached... Hope im right!
I also replied to your query about what if u was squared?... Thanks
Attachment 19464
Show 40 post(s) from this thread on one page
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https://brainly.com/question/171248 | 1,485,192,051,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282935.68/warc/CC-MAIN-20170116095122-00085-ip-10-171-10-70.ec2.internal.warc.gz | 780,800,098 | 9,445 | 2014-11-02T16:17:43-05:00
You can show your work like when you did in elementary school taking out the decimal point. Then put the decimal point when you get the answer.
2014-11-02T16:39:11-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
That's going to be very, very difficult, because 2 x 43.50 is NOT 435.
Now, if you want to solve 2 x 43.5 and show your work, that's easy.
Just take a blank sheet of paper and a pencil, write down 43.50, then
write down 2 under it,and do the multiplication. All of your work is
right there on the paper. | 212 | 809 | {"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 | 3 | CC-MAIN-2017-04 | latest | en | 0.936681 |
http://www.jiskha.com/display.cgi?id=1253476922 | 1,369,258,700,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702448584/warc/CC-MAIN-20130516110728-00009-ip-10-60-113-184.ec2.internal.warc.gz | 547,293,572 | 3,057 | Wednesday
May 22, 2013
# Homework Help: Math
Posted by B.B. on Sunday, September 20, 2009 at 4:02pm.
Solve the linear equation: -8.3=x+6.1
Is this correct?
Thanks.
• Math - annaiiz, Sunday, September 20, 2009 at 4:15pm
yes it is
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https://www.speedysoft.com/uri/blog/the-largest-known-prime-number/ | 1,721,120,463,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00648.warc.gz | 860,598,010 | 20,407 | # The largest known prime number
The Largest Known Primes website claims that the number 232582657-1 is the largest known prime number. How can this number be the largest known prime number? Is the next prime number after 232582657-1 not a prime? Is it not a number? Is it not known? Can’t it be calculated? I can write a simple algorithm that will print the first n prime numbers for every n. Are there not enough prime numbers? Are they not infinite? Is the term “the largest known prime number” well defined? Is it a number? Is it real? Does it have a factorial? Can’t its factorial be substracted by one? Can’t the result be factorized into prime factors? And if it can, is the largest prime factor not larger than “the largest known prime number”? Is it not known?
The answer to the question “is 232582657-1 the largest known prime number?” depends on who’s asking and who’s answering the question. If you’re asking me, 232582657-1 is definitely not the largest known prime. If it is a prime, and I’m not sure it is, then a larger prime can also be calculated. Therefore, it is already known. Or at least, it is not unknown. Maybe it’s just another example of an unknown unknown. In any case, the answer to this question is not deterministic.
It appears to me that any language, whether human language, mathematical language or computer language, contains ambiguities, paradoxes and vague definitions. No language is both deterministic and fully consistent. If a language is inconsistent, then in what sense can it be deterministic? Which leads me to the conclusion that determinism doesn’t exist even in theory. With any given number, there is some uncertainty whether it is or is not a prime. | 380 | 1,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.937307 |
https://whatisconvert.com/474-knots-in-miles-hour | 1,623,572,030,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00406.warc.gz | 557,068,816 | 7,923 | # What is 474 Knots in Miles/Hour?
## Convert 474 Knots to Miles/Hour
To calculate 474 Knots to the corresponding value in Miles/Hour, multiply the quantity in Knots by 1.1507794480225 (conversion factor). In this case we should multiply 474 Knots by 1.1507794480225 to get the equivalent result in Miles/Hour:
474 Knots x 1.1507794480225 = 545.46945836269 Miles/Hour
474 Knots is equivalent to 545.46945836269 Miles/Hour.
## How to convert from Knots to Miles/Hour
The conversion factor from Knots to Miles/Hour is 1.1507794480225. To find out how many Knots in Miles/Hour, multiply by the conversion factor or use the Velocity converter above. Four hundred seventy-four Knots is equivalent to five hundred forty-five point four six nine Miles/Hour.
## Definition of Knot
The knot is a unit of speed equal to one nautical mile (1.852 km) per hour, approximately 1.151 mph. The ISO Standard symbol for the knot is kn. The same symbol is preferred by the IEEE; kt is also common. The knot is a non-SI unit that is "accepted for use with the SI". Worldwide, the knot is used in meteorology, and in maritime and air navigation—for example, a vessel travelling at 1 knot along a meridian travels approximately one minute of geographic latitude in one hour. Etymologically, the term derives from counting the number of knots in the line that unspooled from the reel of a chip log in a specific time.
## Definition of Mile/Hour
Miles per hour (abbreviated mph, MPH or mi/h) is an imperial and United States customary unit of speed expressing the number of statute miles covered in one hour. Although kilometres per hour is now the most widely used measure of speed, miles per hour remains the standard unit for speed limits in the United States, the United Kingdom, Antigua & Barbuda and Puerto Rico, although the latter two use kilometres for long distances.
## Using the Knots to Miles/Hour converter you can get answers to questions like the following:
• How many Miles/Hour are in 474 Knots?
• 474 Knots is equal to how many Miles/Hour?
• How to convert 474 Knots to Miles/Hour?
• How many is 474 Knots in Miles/Hour?
• What is 474 Knots in Miles/Hour?
• How much is 474 Knots in Miles/Hour?
• How many mph are in 474 kt?
• 474 kt is equal to how many mph?
• How to convert 474 kt to mph?
• How many is 474 kt in mph?
• What is 474 kt in mph?
• How much is 474 kt in mph? | 606 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-25 | latest | en | 0.887097 |
https://physics.stackexchange.com/questions/458329/do-strings-propagate-through-space-time-or-do-they-make-space-time | 1,656,741,557,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103984681.57/warc/CC-MAIN-20220702040603-20220702070603-00232.warc.gz | 512,764,285 | 67,593 | # Do strings propagate through space time or do they make space time?
In the beginning of string theory textbooks, strings are said to live in a background "target" space time. They then propagate through this space time. Strings also have a spin 2 ("graviton") mode, and can scatter off of each other. So in one sense, it seems like strings live "in" space time. Gravitational waves can then be thought of as coherent states of "string plane waves," constructed from asymptotic states.
But somehow, don't these strings also affect the space time they are in? It is often said that string theory has "no free parameters," but that different laws of physics (low energy effect QFTs) are given by different compactifications of extra dimensions. In some fuzzy sense, somehow these different compacifictions are minima of some potential energy (although I don't see how). What do these compactifications have to do the strings?
In other words, it seems as though a space time manifold is an input to your theory, and strings propagate in it. However it also seems like somehow the strings can select which space time they are in, because there are different possible compacifications of extra dimensions. So do strings live in space time, or do they "make" space time, somehow?
(I am a person who knows QFT but not much about string theory. I welcome any math used in an answer.)
• This sounds like a question about background independence, which AFAICT is controversial.
– user4552
Feb 1, 2019 at 22:53
• @BenCrowell You are correct in so far that if we had an explicit formula for how string vacua are selected, then that would have to be a background-independent formula by necessity. Nevertheless, even in the absence of a universal bg-independent formulation of string theory, we can still talk about how spacetime appears in string theory and what mechanisms exist to change from one spacetime to another, which is what I've attempted to explain in my answer. Feb 1, 2019 at 22:58
There are at least two different ways to think about the ontology of spacetime in string theory, with neither being any more correct than the other:
1. The non-linear $$\sigma$$ model view: One may view string theory as being a theory essentially defined by the stringy scattering amplitude as the sum over compact 2d manifolds on which certain conformal field theories live. All you need is a conformal field theory with non-anomalous Weyl symmetry on the manifold and the definition of the stringy amplitude, and you've got a string theory. Turns out that at least a certain subset of these theories - superconformal, $$\mathcal{N}=2$$ theories - are often naturally expressed in terms of an action of fields whose target space is a ten-dimensional manifold.
In the general spirit of non-linear $$\sigma$$-models, we identify this target space with spacetime, and when you think about 2d manifolds, you'll realize you could claim that they are the worldsheets of 1d objects. But the formalism a priori doesn't require you to make this interpretation - we did emphatically not start from the assumption that we would be making a theory about strings floating through space to reach this point. Adherents of this ontology view both "strings" and "spacetime" as emergent aspects of "string theory", with the "fundamental" objects of the theory being abstract (S)CFTs living on donuts.
Let me note that there is not a unique spacetime even for a fixed choice of CFTs, since there are mirror models, which are two different choices for the target space that nevertheless produce exactly the same string theory.
2. The "strings floating through space" view: This approach is how string theory is usually taught: We start with the idea that we want to quantize the theory of a classical string floating through a spacetime, and jump through all sorts of ad hoc quantization procedures until we finally figure out that the only consistently quantizable theories are those with a vanishing Weyl anomaly, which in this approach is usually expressed in terms of an ordering constant arising due to operator ordering ambiguities during quantization. Here "spacetime" and "string" are fundamental objects, and the CFT aspect is something we learn along the way.
If we take the CFT viewpoint, then different compactifications are different choices for the CFT living on the worldsheet. If we take the other viewpoint, then different compactifications are different choices for the spacetime the string propagates through. No matter what, we will realize that string theory is intimately related to conformal field theory - and conformal field theories have a concept that's called deformation by marginal operators associated with their renormalization group flow. In an effective field theory POV (i.e. from the POV of the effective QFT on spacetime associated to a QFT), such a deformation corresponds to the variation of some vacuum expectation value of some field (or combination of fields).
But, more importantly, it also corresponds to changing the shape of the target space of the CFT! That is, it directly corresponds to the so-called moduli of the target space. So a smooth variation in the CFT is associated to a variation in spacetime structure (but not a smooth one in the ordinary sense of classical geometry - the variation in moduli can cause the topology of the target space to change, and topology changes can't be smooth). If we understood the dynamics that governed the deformations of the CFT, then we would also understand how string compactifications are "selected", but AFAIK no universally accepted idea of what these dynamics should be exists among string theorists. Nevertheless, this seems strongly suggestive of a renormalization group flow-like mechanic ultimately determining the structure of spacetime.
For a lucid exposition of the "string theory as CFT on donuts" approach, I highly recommend Greene's "String Theory on Calabi-Yau Manifolds". | 1,256 | 5,960 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-27 | latest | en | 0.96211 |
https://math.stackexchange.com/questions/1510769/difference-between-epimorphism-isomorphism-endomorphism-and-automorphism-with | 1,643,054,266,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00218.warc.gz | 432,412,866 | 35,777 | # Difference between epimorphism, isomorphism, endomorphism and automorphism (with examples)
Can somebody please explain me the difference between linear transformations such as epimorphism, isomorphism, endomorphism or automorphism?
I would appreciate if somebody can explain the idea with examples or guide to some good source to clear the concept.
For any algebraic structure, a homomorphism preserves the structure, and some types of homomorphisms are:
• Epimorphism: a homomorphism that is surjective (AKA onto)
• Monomorphism: a homomorphism that is injective (AKA one-to-one, 1-1, or univalent)
• Isomorphism: a homomorphism that is bijective (AKA 1-1 and onto); isomorphic objects are equivalent, but perhaps defined in different ways
• Endomorphism: a homomorphism from an object to itself
• Automorphism: a bijective endomorphism (an isomorphism from an object onto itself, essentially just a re-labeling of elements)
Note that these are common definitions in abstract algebra; in category theory, morphisms have generalized definitions which can in some cases be distinct from these (but are identical in the category of vector spaces).
So a linear transformation $$A\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$$ is a homomorphism since it preserves the vector space structure (vector addition, scalar addition and multiplication, scalar multiplication of vectors), e.g. $$A(av+w)=aA(v)+Aw$$. It is an epimorphism if its image is $$\mathbb{R}^{m}$$, a monomorphism if it has zero kernel, an endomorphism if $$n=m$$, and an automorphism (as well as an isomorphism) if all of these are true.
The below figure might be helpful. More details here.
• I'm new to math.stackexchange, can someone explain the downvote? I found that I could not embed the image without more reputation, but if I could it would have been self-explanatory I think. Feb 1 '18 at 3:47
• Added more details to perhaps address the downvote. Also, I cannot comment on the above answer, but as @al-jebr mentions, an endomorphism need not be injective. Feb 1 '18 at 19:47
• “For any algebraic structure... epimorphism” careful. Epimorphism means right-cancellable, and while every surjective homomorphism is an epimorphism, there are nonsurjective epimorphisms, e.g., the embedding $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is an epimorphism of rings (and of multiplicative semigroups). Monomorphism, left cancellable is almost always equivalent to injective, but not always. These are true for vector spaces and linear algebras, but saying “for any algebraic structure” is too general. Nov 10 '20 at 1:07
• Thanks Arturo, added a note that in categories besides vector spaces the category theoretic definitions may be distinct. Nov 11 '20 at 20:28
In linear algebra, an epimorphism between vector spaces is a surjective linear application $A:V_{1}\to V_{2}$, that is $\text{Im}(A)=V_{2}$. For example, $B:\mathbb{R}^{2}\to\mathbb{R}:(x,y)\mapsto x+y$.
In linear algebra, an endomorphism from a vector space to itself. For example, $B:\mathbb{R}\to\mathbb{R}:x\mapsto 2x$. Edited (30/08/2018): removed 'injective' and corrected the definition.
In linear algebra, an isomorphism between vector spaces is a both surjective and injective linear application $A:V_{1}\to V_{2}$, that is $\text{Ker}(A)=\{0_{V_{1}}\}$ and $\text{Im}(A)=V_{2}$. An automorphism is an isomorphism between a vector space and itself. For example, $B:\mathbb{R}\to\mathbb{R}:x\mapsto x$.
In a more general setting, a morphism $\phi$ between two groups $(G,\cdot)$ and $(H,\star)$ is an application $G\to H:g\mapsto \phi(g)$ such that, for all $g,g'\in G$, we have $\phi(g\cdot g')=\phi(g)\star\phi(g')$ and such that $\phi(e_{G})=e_{H}$ where $e_{I}$ is the identity element of $I=G,H$.
An epimorphism between such two groups is a surjective morphism $\phi:G\to H$, i.e. for any $h\in H$, there exists $g\in G$ such that $h=\phi(g)$.
An endomorphism between such two groups is an injective morphism $\phi:G\to H$, i.e. for all $g,g'\in G$ with $\phi(g)=\phi(g')$, it implies $g=g'$.
An isomorphism between two such groups is a both injective and surjective morphism. An automorphism is an isomorphism with $(H,\star)=(G,\cdot)$.
• Dear Corzer thanks for the relpy, can you explain what does {0_v1} means Nov 3 '15 at 19:13
• I use the notation $0_{V_{1}}$ to denote the null vector from the vector space $V_{1}$. For example, if $V_{1}=\mathbb{R}^{2}$, $0_{V_{1}}=(0,0)$. If $V_{1}=\mathbb{R}^{5}$, $0_{V_{1}}=(0,0,0,0,0)$. Nov 3 '15 at 23:21
• Endomorphisms are not necessarily injective. That is a monomorphism. Endomorphisms are maps from a mathematical object to itself. Nov 16 '17 at 2:04 | 1,344 | 4,647 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-05 | latest | en | 0.926357 |
http://ajediam.com/diamond_carat_weight.html | 1,500,964,316,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425082.56/warc/CC-MAIN-20170725062346-20170725082346-00318.warc.gz | 13,021,946 | 12,141 | 20.00 carat = 4,000 milligram = 4.00 gram
15.00 carat = 3,000 milligram = 3.00 gram
10.00 carat = 2,000 milligram = 2.00 gram
8.00 carat = 1,600 milligram = 1.60 gram
7.00 carat = 1,400 milligram = 1.40 gram
6.00 carat = 1,200 milligram = 1.20 gram
5.00 carat = 1,000 milligram = 1.00 gram
4.00 carat = 800 milligram = 0.80 gram
3.00 carat= 600 milligram = 0.60 gram
2.50 carat= 500 milligram = 0.50 gram
2.25 carat= 450 milligram = 0.45 gram
2.00 carat= 400 milligram = 0.40 gram
1.75 carat= 350 milligram = 0.35 gram
1.50 carat= 300 milligram = 0.30 gram
1.25 carat= 250 milligram = 0.25 gram
1.00 carat = 200 milligram = 0.20 gram
0.85 carat= 170 milligram = 0.17 gram
0.75 carat= 150 milligram = 0.15 gram
0.65 carat= 130 milligram = 0.13 gram
0.50 carat= 100 milligram = 0.10 gram
0.40 carat= 80 milligram = 0.08 gram
0.33 carat= 66 milligram = 0.06 gram
0.25 carat= 50 milligram = 0.05 gram
0.20 carat= 40 milligram = 0.04 gram
0.15 carat= 30 milligram = 0.03 gram
0.10 carat= 20 milligram = 0.02 gram
0.07 carat = 14 milligram = 0.014 gram
0.05 carat = 10 milligram = 0.010 gram
0.03 carat = 6 milligram = 0.006 gram
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here: | 918 | 2,700 | {"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-2017-30 | latest | en | 0.438908 |
http://link-springer-com-443.webvpn.fjmu.edu.cn/chapter/10.1007/978-0-387-68445-1_2 | 1,597,180,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738855.80/warc/CC-MAIN-20200811205740-20200811235740-00003.warc.gz | 67,235,785 | 14,329 | # Algebra
• Răzvan Gelca
• Titu Andreescu
Chapter
## Abstract
It is now time to split mathematics into branches. First, algebra. A section on algebraic identities hones computational skills. It is followed naturally by inequalities. In general, any inequality can be reduced to the problem of finding the minimum of a function. But this is a highly nontrivial matter, and that is what makes the subject exciting. We discuss the fact that squares are nonnegative, the Cauchy—Schwarz inequality, the triangle inequality, the arithmetic mean-geometric mean inequality, and also Sturm’s method for proving inequalities.
Our treatment of algebra continues with polynomials. We focus on the relations between zeros and coefficients, the properties of the derivative of a polynomial, problems about the location of the zeros in the complex plane or on the real axis, and methods for proving irreducibility of polynomials (such as the Eisenstein criterion). From all special polynomials we present the most important, the Chebyshev polynomials.
Linear algebra comes next. The first three sections, about operations with matrices, determinants, and the inverse of a matrix, insist on both the array structure of a matrix and the ring structure of the set of matrices. They are more elementary, as is the section on linear systems. The last three sections, about vector spaces and linear transformations, are more advanced, covering among other things the Cayley—Hamilton Theorem and the Perron—Frobenius Theorem.
The chapter concludes with a brief incursion into abstract algebra: binary operations, groups, and rings, really no further than the definition of a group or a ring.
## Keywords
Triangle Inequality Positive Real Number Binary Operation Identity Element Chebyshev Polynomial
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
## Authors and Affiliations
• Răzvan Gelca
• 1
• Titu Andreescu
• 2
1. 1.Department of Mathematics and StatisticsTexas Tech UniversityLubbockUSA
2. 2.School of Natural Sciences and MathematicsUniversity of Texas at DallasRichardsonUSA | 468 | 2,180 | {"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-2020-34 | latest | en | 0.909466 |
https://newpathworksheets.com/math/grade-8/equations-and-inequalities-2/mississippi-standards | 1,638,956,206,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00247.warc.gz | 479,944,158 | 9,149 | ◂Math Worksheets and Study Guides Eighth Grade. Equations and inequalities
The resources above correspond to the standards listed below:
Mississippi College & Career Readiness Standards
8.EE. Expressions and Equations (EE)
Analyze and solve linear equations and pairs of simultaneous linear equations
8.EE.7. Solve linear equations in one variable.
8.EE.7.a. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
8.EE.7.b. Solve linear equations and inequalities with rational number coefficients, including those whose solutions require expanding expressions using the distributive property and collecting like terms.
MS.CM8AI. Compacted Mathematics Grade 8 (with Algebra I)
CM8AI.A-CED. Algebra: Creating Equations (A-CED)
Create equations that describe numbers or relationships
A-CED.1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
A-CED.3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.
CM8AI.A-REI. Algebra: Reasoning with Equations and Inequalities (A-REI)
Understand solving equations as a process of reasoning and explain the reasoning
A-REI.1. Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
Solve equations and inequalities in one variable
A-REI.3. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
MS.CM8IM. Compacted Mathematics Grade 8 (with Integrated Math I)
CM8IM.A.CED. Algebra: Creating Equations (A-CED)
Create equations that describe numbers or relationships
A-CED.1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
A-CED.3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.
CM8IM.A-REI. Algebra: Reasoning with Equations and Inequalities (A-REI)
Solve equations and inequalities in one variable
A-REI.3. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. | 625 | 3,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-49 | latest | en | 0.895549 |
https://web2.0calc.com/questions/help-me-plz-i-need-help | 1,508,352,119,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823067.51/warc/CC-MAIN-20171018180631-20171018200631-00379.warc.gz | 874,753,974 | 5,854 | +0
# Help Me Plz!! I need help!! :(
0
91
1
I have this math problem bugging me. :( i am only in 5th grade plz help
Which value is the best estimate of 26% of 259?
A- 50
B- 52
C- 65
D- 88
Guest Aug 3, 2017
Sort:
#1
+76901
+2
26% of 259 would be very similar to 25% of 260 = (1/4) of 260 = 260 / 4 = 130 / 2 = 65
CPhill Aug 3, 2017
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https://www.clutchprep.com/physics/practice-problems/142381/when-we-say-that-momentum-is-conserved-we-mean-a-the-momentum-of-an-object-is-th | 1,632,148,115,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00225.warc.gz | 740,522,137 | 28,389 | Intro to Conservation of Momentum Video Lessons
Concept
# Problem: When we say that momentum is "conserved," we mean A) The momentum of an object is the same before and after a collision. B) The momentum of the universe increases; it never decreases. C) The momentum of an isolated system is constant. D) If there are no forces acting on an object, its momentum is zero.
###### FREE Expert Solution
Impulse:
The change in the momentum of a system is called impulse.
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###### Problem Details
When we say that momentum is "conserved," we mean
A) The momentum of an object is the same before and after a collision.
B) The momentum of the universe increases; it never decreases.
C) The momentum of an isolated system is constant.
D) If there are no forces acting on an object, its momentum is zero. | 183 | 821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | latest | en | 0.864551 |
http://professorwhiteford.blogspot.com/2012/04/ | 1,503,017,849,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104204.40/warc/CC-MAIN-20170818005345-20170818025345-00347.warc.gz | 334,644,458 | 28,863 | Friday, April 27, 2012
The Reasons for the Seasons
This coming Monday I have about an hour in my Teaching Math/Science class to teach the Earth Science part of the K - 6 science curriculum; clearly not enough time to "cover" or explore everything K - 6 students need to know and understand. So, like each area of the science curriculum for which I have about an hour, we explore a sample topic. (This is pretty much standard procedure for most teacher education programs). I always try to select something from which I think the students would benefit most in terms of the development of both their own understanding and the work they will be doing with K-6 students.
For Earth Science I do an activity with them that helps them develop a better understanding of why it is generally hotter near the Equator and colder near the poles. An activity appropriate for 5/6th grade students. We also explore the reasons for the seasons as well as day and night; all in ths pace of about an hour. The diagram above is typical of what you see in text books and can give rise to all kinds of misconceptions so we do something similar but make it into an activity.
The students sit in a circle and pass a globe around while I shine a flashlight on it from the center of the room. By carefully establishing some activity rules like; the globe is always tilted toward the same wall as it orbits the "Sun", and, the rays from the Sun must be kept horizontal, and,the Earth orbits the Sun and rotates on its axis counter clockwise, we can pretty much develop an understanding of the concepts of insolation, the seasons, and day and night.
We finish off the activity by exploring websites like this one, 24Time Zones.com that gives the time anywhere in the world as well as showing how darkness is coverning the Earth. You can see if it is tomorrow or today somewhere, whether it is day or night, and how long the days are at any location. Websites like these really allow us to help students develop a much more robust understanding of what are pretty difficult concepts.
Hopefully, my students will never teach their students that it is hotter at the equator because it is closer to the Sun; one of the most common misconceptions in the whole universe of science..
Thursday, April 26, 2012
Zero, Nought, or Nothing
Noughts and crosses is a popular game usually played with a pencil and paper especially during a long and boring lecture by two overly competitive students. I still call it this even though it is inexplicably called tic tac toe in the US (If you Google the origin of the name you'll find it has to do with the sound Roman pencil's made when they hot the ground - still inexplicable).
The word 'nought' is the British equivalent of zero and has given rise to some interesting phenomena such as the 'early noughties', the first ten years of the current millenium as in '04 and '08. Shakespeare loved 'nought', of course, as in " That it yields nought but shame and bitterness" from Henry V.
Today, the concept of zero is something we must focus on developong with young children as they learn to count and understand the intricacies of our base ten system. The idea of the concept of zero was first introduced as an "empty set" through the disastrous New Math of the 1960s. Math educators are still frequently erroneously tarred with the same brush when any mathematical ideas are suggested that differ from traditional mathematics; but that's another story.
When we teach children how to number name from 0 - 100 we must start with zero so that they recognize it as a number with a place on the number line. Later when children start to develop their sense of cardinality, the ability to recognize and name a quantity, they can start with 0 before they have any objects to count. The importance of developing this idea only really becomes apparent when they encounter the base ten system.
Traditionally, children have been taught to think of 0 as a "place hloder'. The problem with this idea is that it doesn't help in the development of the child's conceptual understanding of place value. For example, in the number 306, the zero stands for no tens. The idea of the "place holder" simply means that it keeps the 3 and 6 apart which does nothing to develop an understanding of what each digit in a number means. It's another example of the useless metaphors that are sprinkled throughout traditional math.
Monday, April 23, 2012
Farthings and Cents
Every so often a blog topic just lands on my desk. Today, Jonathan Silverman, my art education colleague extraordinaire, presented me with a 1939 farthing. It is identical to the one in the picture and brings back the most incredible childhood memories. We used to be able to by two "chews" for a farthing. A "Chew" was a candy I am certain was invented by an enterprising dentist.
Farthings were in use on the UK up until 1960 when they became no longer legal tender. A farthing was a quarter of a penny or 1/960th of a pound. This would be about 1/600th of a US cent at today's rate of exchange, roughly speaking. Eleven pence and three farthings was a popular price for an item when a shilling (12 pennies) sounded too much; just like \$9.99 in today's currency. The farthing was also used as a description of a high wheeler bicycle which, in the UK, is still called a Penny Farthing because of the relative sizes of the wheels.
Two farthings made a halfpenny, or ha'penny, which was removed from legal tender in the UK in 1969. Both the farthing and ha'penny had been around for 700 or so years before they met their demise. I can well remember the discussions we would have about how silly and annoying such pittances of coins were especially with the imminent demise of the Canadian cent. I wonder how much longer the US cent has to go? I wonder how many there are in "penny jars".
You could almost study the entire history of Britain by exploring changes in the simple farthing
Saturday, April 21, 2012
Probability and Confidence
Every time I teach I learn something new and never has this been truer than with my graduate class this semester. Last Thursday we were exploring issue of probability. We started out developing the concept linguistically trying to identify those things in our lives that were likely, probably or certain. It always amazes me how so many people don't like to say something is certain; even the sun rising tomorrow morning.
We talked about the difference between theoretical and experiential probability and how the latter is so much more interesting. We tossed coins and used an automatic on-line coin tosser that tosses a coin 1000 times in twenty seconds. I can't think of a better way of showing the law of large numbers.
We concluded class with a really interesting activity in which you place five cubes in a bag; the cubes can either be, say, red or blue. The task is then to work out how many of each by pulling one out, recording the color, putting it back in, shaking the bag and pulling another. This continues, not for a predetermined amount of time, but until the person pulling the cubes feels confident enough to infer the number of each color. I've done the activity many times before but have never had such an interesting discussion about it. We decided that not only did it demonstrate the relationship between probability and confidence, something that is taught in statistics classes, it also demonstrated something about one's personality. Some students must have pulled 30 or 40 cubes before they wanted to say how many of each color there were while others wanted to say after just 10 or 15. Isn't that neat? It's almost like a personality test. Some wanted certainty while others settled for a risk.
The sad thing is that we will not perhaps be able to do this in the future until 6th grade as probability is not included in the elementary school Common Core math curriculum. This is sad in a way because without the topic of probability math takes on an air of certainty. Perhaps this is not a bad thing for young children and something had to go?
On the other hand, when you stop to think about it, the concept of probability is involved in just about every aspect of human life. Every time we wonder if it will rain, or whether our favorite team will win, or whether we will get an A on a paper. In fact any time there is uncertainty.
Wednesday, April 18, 2012
A Custom Built Text
I did something this week that I've never done before; I designed my own custom textbook for my Teaching Elementary School Math and Science course. I have used a wonderful textbook by John Van de Walle in the course for many years but I wasn't using all of it. The book costs an amazing \$158 and so didn't want my students to spend all that money on a book I only partially used. The part I didn't use was pedagogy or how-to of teaching math. For this part of the course I use the Bridges program which is used extensively in school throughout Vermont and the US.
What I did want to use was the pedagogical content knowledge or 'what to teach' section of the book as I feel very strongly that my students need to understand the math they are teaching even if it is something as basic as subitizing.
So after several emails with the very helpful Pearson rep I now have a custom made book that includes only the pieces I need and with a really neat Matisse painting on the cover. You can even see some numbers in the picture if you look carefully; math and art are so interrelated. The book will only cost my students \$123 and will be completely up to date with Common Core info, unlike the used copies of previous editions that are available on-line. They can also get it conveniently at the college book store.
Subitizing, by the way is one of the "basics" of math. It is the ability to look at a number of objects and quantify them without having to count them. Many animals can do this, with crows being quite spectacularly good at it. Most people can subitize up to 5 to 7 objects randomly placed although it takes significantly longer to subitize quantities after 3. We arrange dots on dice and dominoes into patterns so that they are easier to recognize (or, perhaps, subitize). Double 6 dominoes are familiar but try looking at a set of double 12s; the 7 and 8 and 11 arrangements are quite a challenge.
Monday, April 16, 2012
Science or Engineering?
Last week I posted a blog entry about Dr. Ioannis Miauilis, the Director of the Boston Museum of Science and Engineering. Like Dr. Miauilis I have always believed that the study of the natural world is not enough to fulfill the "science" component of the elementary school curriculum. To ignore design technology (engineering) would be like teaching children to read without talking about literature, or teaching children to compute without teaching them how to problem solve.
To help my students differentiate between science and engineering
(design technology) we talk about the origin of the questions we ask, the sources of the inquiry. In science the questions arise out of the natural world; what is magnetism? how does it work? what are the five senses? what is the water cycle? In design technology, or engineering, the questions arise out of how we use our scientific knowledge to solve human problems. How do we use magnetism to enhance our lives? How do we use our knowledge of the water cycle to grow better crops.
In the science part of class today my students explored water through the concept of drops of water. In the engineering or design technology part of the class they used a copy of the local newspaper to support a small washer as high as they could above the tables.
An application that brings the two together in a unified way is to explore the motion of a pendulum in terms of what causes it to swing faster or slower and then design a pendulum clock that swings once every second.
The design technology part of the science of magnetism can be exemplified with Mag Lev trains or cow magnets. (My students would not believe there was such a thing, even though I was holding one, until I showed them one on a website).
Sunday, April 15, 2012
Starting the Journey to Becoming a Teacher
Yesterday, I took part in the Spring accepted students day at St. Michael's college. As an Education professor my job was to answer questions from students and their parents about our elementary school teacher education program.
I met some wonderful families, the sons and daughters full of excitement and anticipation, the mums and dads nervous and apprehensive; I remember having those same feelings so well ten years ago when my own daughter was searching for a college. For a couple of hours I answered questions about State teacher licensure reciprocity, about what our program was like in terms of field experiences, whether courses could be transferred from one college to another and how our program compared with others. Two prospective students stood out for me.
The first wanted to be a music teacher and wondered if we had a program leading to music licensure. I told her that we didn't and that not many colleges, especially in Vermont, offered that somewhat restricted licensing program. I suggested that she might want to think about getting her elementary education licensure through a double major in Education and Music. She clearly loved music and told me enthusiastically about all the instruments she played. I could just see her as a second grade teacher in 6 years time turning every student in her class, and probably the school, onto music. I have nothing against dedicated music teachers, of course, but when the classroom teacher is a musician it makes such a difference.
The second student seemed to want to be an elementary school teacher but his parents were terribly concerned about their son's writing skills. I told them how St. Michael's had the Writing Center to help students improve their writing but they seemed more interested in wanting to know how much their son would have to write; would the required papers be any more than five pages long? I tried to let them know gently that elementary teachers must be able to write since it is one of the main things they do. I suggested that his passion for becoming an elementary school teacher would really have to drive him to improve his writing skills if he wanted to succeed.
I wonder if both students will choose to attend St. Michael's. If they do, I wonder which one will choose to begin the journey of becoming an elementary school teacher. Perhaps they both will!
Friday, April 13, 2012
What Are Schools For?
I was listening to the local NPR radio station on the way to work this morning and the recent teacher strike in central Vermont was being discussed. The interviewer asked if parents were relieved that the strike was over suggesting that it must have been difficult for parents finding someone to watch their children for six days. The interviewee agreed saying it had put quite a strain on local families.
The exchange made me wonder, not for the first time, why we have schools. What is the purpose of the school, of education? Imagine being from another planet and listening to this exchange on NPR; what would you think? Perhaps you would think that schools are places where parents keep their children during the day so they can go to work. You might conclude that, as communities, they employ people to look after them and keep them gainfully occupied during the day. They employ people to take them to and from school and other people to feed them at lunch time.
All joking aside, the purpose of schooling has changed significantly over the years as times have changed, to use a line from Bob Dylan. Schools change in response to all kinds of local and global forces. The one I remember most was the changes in the science curriculum in the western world when the Russians launched Sputnik in 1957. Changes have occurred in what we expect students to learn, how we expect them to act and what we expect them to be able to do as a result of their education.
It used to be fairly easy to predict the future and have a sense of what students would need to know and be able to do but with the current rate of change and uncertainty about the future things are becoming more difficult. Gone are the days when a job out of college or high school was for life. One wonders if the Common Core, due to hit schools in 2014, might be out of date before it even arrives. I often wonder if those creating the Common Core have examined what comprises the curriculum carefully enough given the ever increasing rate of change we are experiencing.
In my own field of math education, arithmetic still takes up a significant part of the K - 6 math curriculum in the Common Core. Do we really still need to spend so much time teaching children how to subtract 38 from 92 with a pencil or pen? Shouldn't we be acknowledging that basic numeracy and problem solving skills are so much more important things to learn, and that cellphones, computers and calculators can now easily perform such menial calculations. Wouldn't it be better to spend the time helping students become quantitatively literate and be able to work out which one of the four operations is required to solve a particular problem?
Perhaps it's time that thinking and understanding replaced knowing as the primary goals of education?
Thursday, April 12, 2012
The Value of a Place
I can remember being told, as a student in primary school, to put a comma every three numbers when writing large numbers. I was told it would separate the hundreds, thousands, millions etc. I remember dutifully doing this with absolutely no sense of why or feeling any nearer to being able to read very large numbers.
Today, we still use the commas when writing large numbers but
unlike the instruction of the last century, we now teach why we put a comma every three numbers. It certainly does separate things but, more importantly, it shows the repetition of the ones, tens and hundreds every three numbers. First we have ones, tens and hundreds of ones, then we have ones, tens and hundreds of thousands and then we have ones, tens, and hundreds of millions, and so on. If we learn this repetition of the place value referents it makes reading and understanding large numbers much easier.
It also helps students develop a far more realistic idea of what zero, or nought, is. I remember learning 0 as a "place holder", a bit like a pot holder, perhaps. What did that really mean? The 0 holds a place when there is no number there, perhaps. What nonsense! The 0 means there are none of that particular referent in this number as in 23,405, where the 0 means there are 0 tens. In this number, 305, 877 the 0 means there are no ten-thousands; there are 3 hundred-thousands and 5 thousands but 0 ten-thousands.
'Nought', by the way is the British word for zero and became quite popular during the first decade of the current century with years like '03 and '06 which was referred to as the 'noughties'.
Tuesday, April 10, 2012
Three Friday 13ths This year
As you probably already know there are three Friday 13ths this year and they are exactly 13 weeks apart. The second is this coming Friday. The last time this happened was in 1984, the year made famous by George Orwell's novel of that name. I remember reading it in the mid '60s and wondering where I would be in 1984. Little did I know I would be in my third year in Vermont. The next time there will be 3 Friday 13ths in one year will be 2040. What will we all be doing then? Here are some interesting facts about 13. The Costa Concordia started sinking on the last Friday 13!
13th is an ordinal number, like first, second, third, fourth, fifth etc. We don't often talk about a thirteenth of something in a fractional sense but we do use third, fourth, fifth etc as fraction words. Yet another quirk of our strange English language. Fraction words are the same as ordinal words after you get beyond second/half. I remember working with a 2nd grade teacher once to teach fractions. We taught the students that a third as a fraction was one of three equal parts among other things. When the children lined up for recess one of the students looked up at me and said, "I'm third in line but I'm one of 16 equal parts!", referring to the 16 students in her class. We discussed the two meanings of the word 'third' but I was never sure she understood.
Numbers occur in our lives in lots of ways; cardinals (counting numbers), ordinals (sequence numbers) and nominals (naming numbers). Room numbers, house numbers, telephone numbers and just about any number used to identify something is a nominal number. If you can replace the number with a name then it is a nominal. There are also classifications such as shoe sizes, the Beaufort wind scale, the Richter scale and Mohs rock hardness scale as well as rates such as prices, and measures such as temperatures. Each numeral (the written grapheme) has a different meaning when used as a different type of number.
Next time you fill your car up with gas, or petrol, look at the pump to see how many different types of numbers there are. It will take your mind off the price of the gas or petrol.
Monday, April 9, 2012
Engineering for Kindergartners
This morning I gave a presentation at the STEM conference in Stowe, Vermont sponsored by the Vermont Science Teachers Association and the Vermont Council of Teachers of Mathematics. My presentation on teaching math to children with Down Syndrome went well but what really caught my attention was the keynote address by Dr. Ioannis Miauilis, President and Director of the Boston Museum of Science. He had the group of science and math teachers riveted from the moment he said that engineering should be a part of the K - 12 curriculum and it should begin in kindergarten.
The most interesting part of what he advocates for is the idea that we spend so much of our time in science education classes focusing on the natural world and so little time on the man-made world. We spend so much time teaching things that are so remote from students' lives, such as lessons about volcanoes and coral reefs to students in Vermont, and so little time on things that are part of children's lives such as how drinking a milk shake through a straw really works.
When I teach science education to my students I always try to make the distinction between science and design technology, which I think, is close to Dr. Miauilis' concept of engineering. In science the questions arise from the natural world whereas in design technology the questions arise from how we use science to deal with the natural world. In science we can study the properties of magnetism but in design technology we can use our knowledge of magnetism to solve problems. Even though there are standards for design technology in the Vermont Grade Level expectations I fear they are seldom addressed.
Perhaps it's time to implement, as Dr. Miauilis did when he was a professor at Tufts University, courses such as those on the science of cooking or the science of fishing. In other words, perhaps science needs to be more relevant to students' lives in a way that will attract more students into the formal study of the sciences and engineering. Such courses would have the same academic rigor as the pure science courses but would be much more relevant to students' lives and interests.
By the way, Dr. Miauilis is far more entertaining, engaging and humorous as a keynote speaker than he is in the YouTube video.
Friday, April 6, 2012
\$3.99.9 a Gallon
I've always wanted to ask for exactly one gallon of gas to see what would happen. How much change would I get? The only place in the US, in fact in the world, where the smallest legal tender denomination can be split into tenths is with gas or petrol prices. I suppose it is because the prices are very visible on large signs and gas stations tend to often be next to each other and, as we all know, a penny difference really matters!!!! I wonder what will happen in Canada now that they're eliminating the cent as legal tender?
The .9 is, of course, pretty meaningless since every oil company does it and it's usually so small that it's virtually impossible to see from a distance. Anyway, it's always there so you always ignore it. This is a wonderful example of just how quantified our lives are and how certain numbers are significant and other are not.
Perhaps, tomorrow, the price will have risen to 3.99.99!
The gas prices in the picture above are actually petrol/diesel prices in the UK. They are in pence and are for 1 liter of petrol. A liter is equal to about .26 US gallons so there are around 4 liters in a gallon. The rate of exchange between dollars and pounds sterling is around 1.6 dollars per pound. So the cost of petrol in gas equivalent gallons is roughly 4 x 1.48 x 1.6 which works out to around \$9.47 a gallon.
Somehow, \$9.46.9 doesn't seem to make a whole lot of difference.
What does make a difference is that 70% of cars in the UK are diesel and get about 70 mpg and places are much closer together so you don't have to drive so far.
Wednesday, April 4, 2012
Do the Math - Do The Art
One of the new Open Education sites that is devoted to education is the Kahn Academy. There are hundreds of short videos available to those wishing to brush up on some aspect of their math or many other subject areas. Generally, the ideas in the math videos seem to be explained well but it is certainly a contrast with the Vi Hart math videos which, I must admit, I much prefer.
One of the phrases that has always sort of annoyed me related to math is "do the math" which is used any time someone wants you to compute or calculate in some way. It's as if that's all math is.
I was watching one of the Kahn Academy math videos, the one titled "The Beauty of Algebra", when I could barely believe my ears. After a brief explanation of how to use algebra to find a percentage of a price of something he says, quite remarkably, "now do the math" as he calculates the exact reduction in price. So what has he been doing up to this point and what will he be doing after this point? Clearly he has not been doing math for the first several minutes of the video because he now decides to do the math.
Now I must "do the writing" to continue this entry so that you can "do the reading". We must also not forget to "do the driving" to get home after work, "do the cooking" when we get home and prepare the evening meal, and "do the sleeping" so that we are well rested for tomorrow.
And, of course, don't forget to "do the art" when you feel the need to doodle in a long meeting.
Monday, April 2, 2012
I have the most amazing graduate math education class this semester. The students just will not let anything go that they don't understand. I have always believed that we should not teach anything we don't understand so this group of students is really keeping me on my toes.
Two weeks ago we were exploring fractions. We first looked at fraction concepts such as the idea that fractions only have value when we know the size of the whole or referent to which they refer. (Would you rather have half the money in my right hand or a quarter of the money in my left hand?). We then went on to look at how to develop the fraction algorithms for the four operations. Addition and subtraction were fairly straight forward but, of course, multiplication and division are a completely different kettle of fish.
We also played around with fraction word problems which, again, is OK for addition and subtraction, but quite confusing for the other two operations. The other problem with fraction problems is that it's hard to get away from food; "if you ate 3/8 of your pizza.......?". And, of course, pizzas are only ever divided into eighths!
Then someone said "I was always told that "of" means multiply as in what is 1/2 of 1/3". Yes, it always has, I thought, but why? What is the conceptual connection between the word 'of ' and the operation of multiply? I asked several mathematicians and could only come up with nothing more than my original thought which was that it is just a linguistic connection as in "automobile means car" and "house means home".
My daughter, Marie, even suggested using the word "times" as in "a half of a quarter is a quarter, half a time" (4 groups of 3 is 3 four times). Then I started thinking about the area concept of multiplication as in the image above. in a 1 x 1 square an area which is 1/3 (blue side) by 3/4 (red side) provides a conceptual connection between of and multiplication. Voila................................. maybe. | 6,264 | 28,745 | {"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} | 2.578125 | 3 | CC-MAIN-2017-34 | longest | en | 0.954497 |
https://open.kattis.com/contests/nadc21practice2/problems/europeantrip | 1,624,567,597,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00478.warc.gz | 385,107,388 | 6,958 | NADC 2021 - Practice session 2
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Problem EEuropean Trip
After winning Vietlott, a woman decided to go on a trip to Europe! Let’s refer to her as Ms. Mask.
Like other women, Ms. Mask really loves shopping and guess where her first stop is? Of course, it is London, a dream land for shopaholics. She has already discovered three greatest shopping centers in London: Westfield Stratford City, Piccadily Arcade and Fortnum & Mason. On the Cartesian plane, these three shopping centers can be depicted by three points.
Ms. Mask wants to rent a house to stay during the whole trip, so that the total distance from her house to those shopping centers are as small as possible. Help her find an optimal position for her house, assuming that she can put her house everywhere, even in Green Park or on Thames River!
Input
The input consists of three lines, each line contains two integers $x$ and $y$ (between $0$ and $10^3$, inclusive) representing the coordinates of three shopping centers.
It is guaranteed that those three points are not collinear.
Output
Write in one line two real numbers $x$ and $y$ representing the place where Ms. Mask should hire a house and stay.
Let $P$ be the total distance from your point to three points given in the input, and $J$ be the total distance from jury’s point. Your answer is considered correct iff $P$ differs from $J$ at most $10^{-4}$ in term of either absolute or relative value.
Sample Input 1 Sample Output 1
0 0
1 0
0 1
0.211324865 0.211324865
Sample Input 2 Sample Output 2
174 711
980 989
976 384
803.563974893 697.742533711 | 474 | 1,780 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-25 | longest | en | 0.919332 |
https://brainly.in/question/85990 | 1,485,293,865,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285289.45/warc/CC-MAIN-20170116095125-00577-ip-10-171-10-70.ec2.internal.warc.gz | 793,181,738 | 10,071 | If points P(1,2) ,Q(0,0) and R(a,b) ,then (a) a = b (b) a = 2b (c) 2a = b (d) a = -b
1
by aromalsuper
Question incomplete. Are they collinear?
2015-03-13T15:12:24+05:30
As the question incomplete,
Assuming that the points P, Q and R are collinear.
So, the area of ΔPQR is 0
Area of triangle =
0 =
0 =
2a = b | 132 | 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} | 3.15625 | 3 | CC-MAIN-2017-04 | latest | en | 0.825731 |
http://www.11plusforparents.co.uk/Maths/decimals1.html | 1,542,519,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743968.63/warc/CC-MAIN-20181118052443-20181118074443-00225.warc.gz | 383,102,521 | 8,259 | # Decimals - The Decimal System
The word 'Decimal' is derived from the Latin word 'decem', meaning ten. The system has its roots in India at around 300BC although this system was also evident in China at around the same time. The system was further developed in Persia and Egypt with the familiar Hindu-Arabic numerals similar to what we use today being introduced to Europe around the 12th century.
The Decimal system is based on ten symbols 0,1,2,3,4,5,6,7,8, and 9. The position of these symbols, ie their place, in a number determines the value of the number. This is why Place Value is such an important concept to understand.
I would recommend reviewing Place Value before moving on.
Each place is ten times the value of the place to it's right, and consequently each place is ten times smaller than the place to it's left.
TTh
Th
H
T
U
1/10
1/100
## 0
The number 1 placed in the Hundreds column is worth 100, its value is 100, which is ten times greater than the 1 in the Tens column whose value is 10, which is in turn ten times greater than the 1 in the units column whose value is 1.
So far we have only looked at whole numbers, but the decimal system can also be used to represent numbers that are less than a whole number, less than one. The whole numbers and parts of whole numbers are separated by the decimal point. Any number placed to the left of the decimal point is a whole number and the value of any number placed to the right of the decimal point is less than a whole and each column to the right of the decimal point decreases by ten or is a tenth of the column to its left. These numbers to the left of the decimal point can be thought of as Decimal Fractions.
TTh
Th
H
T
U
1/10
1/100
## 1
The number 1 placed in the tenths column to the right of the decimal point has a value ten time smaller than the 1 placed in the Units column - one tenth. The number 1 placed in the Hundredths column to the right of the decimal point has a value ten time smaller than the 1 placed in the Tenths column - one hundredth.
It is important that children understand that the numbers placed to right of the decimal point are parts of a whole number - ensure that for 1.11 they say "One point one one" and not "One point Eleven" because the .11 does not mean eleven, This is a common misconception and you need to make sure that they fully understand this.
Write out some decimal numbers and have your child say them out loud to you.
### 2.63 is 'two point six three' not 'two point sixty three'
Decimal numbers can be thought of in terms of fractions - any number to the left of the decimal point is the whole number and anything to the right of the decimal point is the fraction - But the fraction of what?
Lets look at the first column to the right of the decimal point, any number placed here has a value of that number of tenths - so 0.1 is 1/10, 0.2 is 2/10, 0.5 is 5/10 and so on.
Any number placed in the second column to the right of the decimal point has a value of that number of hundredths - so 0.01 is 1/100, 0.02 is 2/100, 0.5 is 5/100 and so on. | 767 | 3,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-47 | longest | en | 0.942368 |
https://scoop.eduncle.com/q10-an-astronaut-in-a-spaceship-has-a-mass-75kg-in-the-rest-frame-of-the-spaceship-the-spaceship-s-length | 1,686,346,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00544.warc.gz | 580,572,800 | 39,408 | • Want FREE Counselling for Exam Preparation?
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# Q10. an astronaut in a spaceship has a mass 75kg in the rest frame of the spaceship. the spaceship's length in its rest frame is 1150 m while its length as obse
Q10. An astronaut in a spaceship has a mass 75kg in the rest frame of the spaceship. The spaceship's length in its rest frame is 1150 m while its length as observed from the earth is 325 m The total energy of the astronaut in the earth's frame will be approximately: (a) 75 (c) 2.4x10 J (b) 750 J (d) 6.8x108 J
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• ##### Somnath
See the attachment.
• ##### Ruby negi Best Answer
i did both methods for funding total energy... do the method in exam that need less time or that require less calculation.... best regards.
finding**
### Do You Want Better RANK in Your Exam?
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• Last 5 Years Question Papers & Answers | 320 | 1,200 | {"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-2023-23 | latest | en | 0.83419 |
https://www.tutorialspoint.com/what-is-a-multidimensional-array-in-c-language | 1,709,017,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00221.warc.gz | 1,061,850,154 | 23,533 | What is a multidimensional array in C language?
C language allows arrays of three (or) more dimensions. This is a multidimensional array.
The exact limit is determined by the compiler.
The syntax is as follows −
datatype arrayname [size1] [size2] ----- [sizen];
For example, for three – dimensional array −
int a[3] [3] [3];
Number of elements = 3*3*3 = 27 elements
Example
Following is the C program to compute the row sum and column sum of a 5 x 5 array by using run time compilation −
Live Demo
void main(){
//Declaring array and variables//
int A[5][5],i,j,row=0,column=0;
//Reading elements into the array//
printf("Enter elements into the array : ");
for(i=0;i<5;i++){
for(j=0;j<5;j++){
printf("A[%d][%d] : ",i,j);
scanf("%d",&A[i][j]);
}
}
//Computing sum of elements in all rows//
for(i=0;i<5;i++){
for(j=0;j<5;j++){
row=row+A[i][j];
}
printf("The sum of elements in row number %d is : %d",i,row);
row=0;
}
//Computing sum of elements in all columns//
for(j=0;j<5;j++){
for(i=0;i<5;i++){
column=column+A[i][j];
}
printf("The sum of elements in column number %d is : %d",i,column);
column=0;
}
}
Output
When the above program is executed, it produces the following result −
Enter elements into the array:
A[0][0] : 1
A[0][1] : 2
A[0][2] : 4
A[0][3] : 3
A[0][4] : 5
A[1][0] : 2
A[1][1] : 5
A[1][2] : 6
A[1][3] : 7
A[1][4] : 2
A[2][0] : 3
A[2][1] : 6
A[2][2] : 2
A[2][3] : 6
A[2][4] : 7
A[3][0] : 2
A[3][1] : 7
A[3][2] : 4
A[3][3] : 3
A[3][4] : 1
A[4][0] : 4
A[4][1] : 5
A[4][2] : 6
A[4][3] : 7
A[4][4] : 8
The sum of elements in row number 0 is: 15
The sum of elements in row number 1 is: 22
The sum of elements in row number 2 is: 24
The sum of elements in row number 3 is: 17
The sum of elements in row number 4 is: 30
The sum of elements in column number 5 is: 12
The sum of elements in column number 5 is: 25
The sum of elements in column number 5 is: 22
The sum of elements in column number 5 is: 26
The sum of elements in column number 5 is: 23
Updated on: 08-Mar-2021
208 Views
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https://homeworkmules.com/a-person-invests-5000-in-treasury-notes-and-bonds-the-notes-pay-8-annual-interest-and-the-bonds-pay-10-annual-interest-if-the-annual-in-come-is-480-how-much-is-invested-in-treasury-notes/ | 1,718,337,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00182.warc.gz | 271,004,176 | 17,393 | # A person invests \$5000 in treasury notes and bonds. The notes pay 8% annual interest and the bonds pay 10% annual interest. If the annual in- come is \$480, how much is invested in treasury notes?
Investment in treasury notes is ##\$1000##.
Let the money invested in treasury notes be ##\$x##. Annual income on it at ##8%## per annum for a year will be ##8/100 xx x##.
Then money invested in bonds at ##10%## would be ##\$(5000-x)## and annual income on it for a year will be ##10/100 xx (5000-x)##.
As total yield is ##480##, we have
##(8x)/100+(10(5000-x))/100=480## – multiplying both sides by ##100##
or ##8x+10(5000-x)=48000##
or ##8x+50000-10x=48000##
or ##-2x=48000-50000=-2000##
i.e. ##x=-2000/-2=1000##
Hence, investment in treasury notes is ##\$1000##.
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Customers referred by a friend | 924 | 3,617 | {"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-2024-26 | latest | en | 0.904145 |
https://abidakon.com/sumifs-google-sheets/ | 1,726,790,063,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00350.warc.gz | 60,465,026 | 19,883 | # SUMIFS Google Sheets – A Complete And Depth Guide In 2022
Written by Abid Akon
#### Here's what we'll cover:
When working on a dataset with multiple values and conditions, you may want to add up the values that apply to a particular condition on your worksheet.
To do this, you would use the SUMIF and SUMIFS functions. If you aren’t familiar with this function, it is quite easy to learn and understand.
Basically, the SUMIF function allows you to add up values that are based on a single condition. But when you need to add up values with multiple conditions, the SUMIFS function is used.
This article is a complete guide on how the SUMIFS function is used and the logic and explanation behind every application of the SUMIFS function. To get the most out of this guide, be sure to stick around to the end!
## What Does the SUMIFS Google Sheets Function Do?
As said earlier, the SUMIFS function sums up values that apply to multiple conditions in your worksheet.
The function operates by analyzing the dataset within a specific cell range, then it extracts the values that satisfy the given condition, and then it adds these values together.
The final result is returned to the selected cell.
## How to Use SUMIFS in Google Sheets
Before we start, we have to understand the syntax of this function and the parameters needed for its formula to work accurately.
The SUMIFS function takes the syntax stated below.
=SUMIFS(sum_range, criteria_range1, criteria1,[ criteria_range2, criteria2, … criteria_range_n, criteria_])
Where
• The sum range is the cell range that contains the needed values.
• The criteria range1 is the cell range where you want to search for the first condition.
• Criteria1 is the first condition that criteria range1 searches for.
• The other included parameters (criteria_range2, criteria2) are not compulsory. They can be added when you have to search for other conditions.
• Criteria_range_n implies that you can add as many criteria ranges and their criteria as you want.
## How To Use the SUMIFS Function in Google Sheets
The SUMIFS functions can be used in different ways. It can be used with texts, numbers, dates, wildcards, and operators as criteria.
Note that an array formula cannot be used with the SUMIFS function.
Now that we understand the dos and don’ts when using the SUMIFS function, we’ll practicalize it on a sample worksheet.
In the sample worksheet below, I have a dataset of purchase dates, sales representatives on duty, customer names, items purchased and the sales made in May.
I would show you to know the total of these transactions based on some conditions here.
I’ll show you some ways to use the SUMIFS function on your worksheet.
## Using Text Conditions
In the first case, we want to know the total sales made on Dresses that were sold by Joan. That means we have two criteria. The Item Purchased is “Dress” while the Sales Representative is “Joan”.
Therefore, the required parameters would take the following values.
• Sum_range is the range of the sales made, that is, E2:E17.
• Criteria_range1 is the range of the sales representatives, that is, B2:B17.
• Criteria1 will be “Joan”.
• Criteria_range2 is the range of the items purchased, that is, D2:D17.
• Criteria2 is “Dress”.
The formula would look like this.
=SUMIFS(E2:E17, B2:B17, “Joan”, D2:D17, “Dress”)
1. Enter the above formula into an empty cell. This is where the result would be returned. In this case, I selected cell F3.
2. Hit the Enter key and the total sales made by Joan on dresses are returned into cell F3.
## Explanation of the SUMIFS and Logic in This Formula
In case you don’t understand how the SUMIF formula returned that value, I’ll explain it properly here.
The SUMIFS function scanned the ranges B2:B17 and D2:D17, for cells that match the given criteria (Joan and Dress).
Then, when it finds a row that satisfies the conditions, the function moves to range E2:E17 to find the sales made on the item.
After that, it adds up the sales and returns the result in cell F3.
## Using SUMIFS with Date Condition.
Another way to use the SUMIF function is to apply a date as its condition. Using the same worksheet, I want to use the dates as the criteria now.
Let’s assume that I want to know the total sales made by Malik before May 15th 2022.
Now the conditions are: Date is “<15/05/2022” while the Sales Representative is “Malik”.
In this case, the parameters are
• Sum_range is the range of the sales made, that is, E2:E17.
• Criteria_range1 is the range of the date, that is, A2:A17.
• Criteria1 will be “<15/05/2022”.
• Criteria_range2 is the range of the sales representatives, that is, B2:B17.
• Criteria2 is “Malik”.
The formula would look like this.
=SUMIFS(E2:E17, A2:A17, “<15/05/2022”, B2:B17, “Malik”)
1. Enter the above formula into an empty cell. This is where the result would be returned. In this case, I selected cell F3.
2. Hit the Enter key and the total sales made by Malik before the 15th of May 2022, is returned to cell F3.
## Explanation of the Formula.
The SUMIFS function scanned the ranges A2:A17 and B2:B17, for cells that match the given criteria (<15/05/2022 and Malik).
Then, when it finds a row that satisfies the conditions, the function moves to range E2:E17 to find the sales made before the specified date by Malik.
After that, it adds up the sales and returns the result in cell F3.
## Using SUMIFS with Number Condition.
Lastly, you can use a number as a condition for the SUMIFS function. Let’s say we want to know the total sales made on some units sold by Amy that are more than or equal to 5 units.
That means we have two conditions, which are: units sold is “>=5”, and the Sales Representative is “Amy”.
In this case, the parameters are
• Sum_range is the range of the sales made, that is, E2:E17.
• Criteria_range1 is the range of the sales representative, that is, B2:B17.
• Criteria1 will be “Amy”.
• Criteria_range2 is the range of the units sold, that is, C2:C17.
• Criteria2 is “>=5”.
The formula used would look like this.
=SUMIFS(E2:E17, B2:B17, “Amy”, C2:C17, “>=5”)
1. Enter the above formula into an empty cell. This is where the result would be returned. In this case, I selected cell F3.
2. Hit the Enter key and the total sales made on units above 5 units by Amy, is returned into cell F3.
## Explanation of the Formula
The SUMIFS function scanned the ranges B2:B17 and C2:C17, for cells that match the given criteria (Amy and >=5, respectively).
Then, when it finds a row that satisfies the conditions, the function moves to range E2:E17 to find the sales made on the units.
After that, it adds up the sales and returns the result in cell F3.
## Multiple Conditions in One Column (SUMIFS OR Google Sheets Logic).
Google Sheets can’t use SUMIF to add multiple conditions that are in a single column. The easiest way to do this is by adding two SUMIFS calculations together. But each calculation should have different conditions but the same sum range.
Only then would you be able to sum multiple criteria in a single column in Google Sheets.
## Frequently Asked Questions on Google Sheets SUMIFS (FAQs).
### 1. Can You Add Two SUMIFS Together? / Can You Use or With SUMIFS?
Yes, you can. It can be done by using a direct formula into two SUMIFS statements and adding them together. For example:
=SUMIFS(D2:D17, C2:C17, “True”, E2:E17″True”) + SUMIFS(D2:D17, C2:C17, “False”, E2:E17, “True”)
### 2. How Do You Do Sumifs With Multiple Criteria in Google Sheets?
Yes, you can. If there are multiple criteria in different columns, you can include these criteria in a SUMIF formula. Like in the calculations treated above.
=SUMIFS(E2:E17, B2:B17, “Joan”, D2:D17, “Dress”)
There are multiple criteria and ranges in one SUMIF formula.
## Final Thoughts
The SUMIFS function is very dynamic due to its capability to accommodate multiple criteria and ranges in a single SUMIF statement.
This function would come in handy for technical sum calculations in Google Sheets. If you have any questions or suggestions, let us know in the comments section below and you’ll get a response soon.
Now you know about SUMIFS Function In Google Sheets. I hope you found this guide useful. Thanks for reading.
#### Abid Akon
I am a tech content writer. I really love to talk about different Modern Technology. It is very important that people know how to fix their tech related problem. That’s why I am writing articles to share my tech knowledge with you.
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### Abid Akon
Hi, I am Abid. I really love to talk about different Modern Technology. It is very important that people know how to fix their tech related problem. That’s why I’ve created this website to share my technology keeping knowledge with you. Welcome to ”abidakon.com”
### Abid Akon
My Personal Favorites | 2,204 | 9,244 | {"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} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.857565 |
www1.bmdgc.com | 1,679,421,683,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00286.warc.gz | 100,257,105 | 7,361 | # Most popular based on COSMOSMOTION and excel
2022-10-01
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Dynamic analysis of shaper based on COSMOSMOTION and excel
preface
Dynamic Analysis of mechanical system is a very old research topic, and there are many solutions now. With the continuous development of larger computer software, such as tuna and pet food, a simple, practical, intuitive and accurate research method is needed. The author found that the modeling of the three-dimensional design software SolidWorks is relatively simple and easy for beginners. In addition, the full-function motion simulation software COSMOSMOTION is seamlessly integrated in the later version, which can carry out complete kinematic simulation and dynamic static analysis of complex mechanical systems. If a large number of mechanical system motion and dynamic parameters (such as kinetic energy curve of each component, system balance torque curve, etc.) obtained from the simulation are processed by Excel spreadsheet, the dynamic model of the mechanical system can be established, and then the real motion law of the mechanical system in the stable operation stage can be easily solved by using the numerical solution of differential equations -- difference method
based on this idea, the author took the shaper as an example, and finally obtained a relatively simple mechanical system power credit after repeated experiments. Plastic has become the core material composition analysis method of most electronic and electrical products. That is, first use SolidWorks software to establish the three-dimensional model of mechanical system, then use COSMOSMOTION software to carry out motion simulation, and output the simulation results to excel spreadsheet. Then, these simulation results are analyzed and processed in Excel spreadsheet, so as to obtain the equivalent parameters of the mechanical system (such as equivalent moment of inertia, equivalent torque, etc.), establish the dynamic equation of the mechanical system, and finally use the formula calculation function of Excel to solve the numerical solution of the dynamic equation - the real motion law of the mechanical system
the biggest advantage of this mechanical system dynamics analysis method is that SolidWorks operation technology is easy to master; COSMOSMOTION simulation has powerful functions, and the definitions of constraints, forces, moments, motions and other concepts are consistent with those in mechanical principles, which is easy to understand; Excel software is commonly used; The whole solution process does not need programming
the application of this method in dynamic analysis of mechanical system of shaper is introduced below
1 establish a virtual prototype
b650 shaper is mainly composed of ram, rocker arm, large helical gear, gear speed change/reduction device, belt drive, motor, workbench, feeding device and body. According to the measured data, all parts are manufactured and assembled in SolidWorks, and finally the virtual prototype shown in Figure 1 is obtained (the body and other parts in the figure have been hidden). The main transmission route can be reflected in Figure 1: motor belt transmission (small belt pulley and large belt pulley) - spline shaft - sliding gear set - gear shaft - large helical gear - slider 1 (not visible in the figure) - rocker arm and slider 2 - connector, pins 1 and 2 - ram and cutter bar
Figure 1 b650 shaper main drive
for the convenience of analysis, it is assumed that the gear speed change device of the shaper is in the meshing state shown in the figure. The transmission parameters are as follows: the reference diameter of the small pulley is 75mm, the reference diameter of the large pulley is 355mm, the number of gears engaged between the spline shaft and the gear shaft is 55 and 36 respectively, and the number of gears engaged between the gear shaft and the large helical gear is 21 and 84 respectively
2 motion simulation
cosm0smotion software has strong simulation function, supports a variety of constraints and virtual constraints, and can define various motions according to displacement, speed or acceleration, including fixed value, step, harmonic, spline curve and function. COSMOSMOTION can be used to simulate the precise motion of various complex mechanical systems and carry out dynamic static analysis
the key to the success of simulation lies in the setting of simulation parameters. COSMOSMOTION simulation settings include dividing moving and stationary components, adding kinematic pair constraints, defining prime mover motion, adding working resistance, and so on
2.1 parts grouping
in COSMOSMOTION, parts need to be divided into two categories: moving parts and static parts. The author puts the parts related to motion analysis into the moving parts group, and the parts irrelevant to motion analysis or fixed during motion analysis are set as static parts (the V-belt can be set as static parts). For convenience of observation, static parts that affect observation are usually compressed or hidden
when entering the COSMOSMOTION interface, the software will automatically add constraints for components according to the matching relationship between components in the assembly drawing. However, some constraints should be added, deleted and adjusted according to the specific analysis object and content
c0smosmosmotion's "fixed pair" constraint is used to lock two rigid members so that they cannot make relative motion, which is equivalent to welding two members together in the real world. In the shaper studied by the author, there is no relative movement between the cutter bar, the connecting piece and the ram, between the pin 1, pin 2 and the connecting piece, and between the large pulley, the sliding gear set and the spline shaft, so the constraint between them adopts "fixed pair"
"rotating pair" constraint only allows the relative rotation of 1 degree of freedom between 2 rigid members. In the shaper, there is only one relative rotation between the motor and the small pulley (including the motor rotor), between the spline shaft and the bearing (equivalent to the frame), between the gear shaft and the bearing (equivalent to the frame), between the large helical gear (crank) and the frame (body), between the large helical gear (equivalent to the crank) and the slider L, between the slider 2 and the frame (body), and between the rocker arm and the connecting piece (or pin 1), so the constraint between them adopts "rotating pair"
"moving pair" constraint only allows the relative movement of 1 degree of freedom between 2 rigid members. In the shaper, there is only one relative movement between the slider L, slider 2 and the rocker arm, and between the ram and the frame (body), so the constraint between them adopts the "moving pair". In COSMOSMOTION software, the dynamic simulation of gear transmission and belt transmission depends on "coupling". The belt drive is used between the small belt pulley and the large belt pulley in the shaper, and the gear drive is used between the sliding gear set (or spline shaft) and the gear shaft, and between the gear shaft and the large helical gear. Therefore, the "coupling" method should be used to define the lifting angle between them α Sports relationship. The transmission ratios of the three "coupling" are 355/75, 36/55 and 84/21 respectively
2.3 input motion
for the convenience of analysis, the large helical gear (equivalent to the crank of the guide rod mechanism) is selected as the prime mover in the system motion analysis, and its speed can be set according to the speed N4 in the real system
therefore, in the COSMOSMOTION interface, set the speed of the large helical gear to "constant value" 720 (°)/s
the speed of the prime mover can also be selected. Because when establishing the equivalent dynamic model, the ratio of the speed of each component to the prime mover is used in the calculation of the equivalent moment of inertia and equivalent moment of the system, which is independent of the real speed
because the tool bar only has working resistance when cutting the workpiece, it is necessary to carry out motion simulation first to obtain the displacement curve of the tool bar (Fig. 2), so as to determine the functional relationship between the working resistance P and the simulation time t:
in this way, the working resistance can be added to the "tool bar" parts in the form of function in COSMOSMOTION, The expression is: if (time-0.0472:0, 7000, if (time-0.256 25:7000, 0, if (time-0.5:0, 0, 0)))
2.5 simulation and result output
at this point, the simulation can be run and the total kinetic energy curve (including moving kinetic energy and rotating kinetic energy) of each part of the shaper can be output to excel. In addition, it is also necessary to simulate the movement of the mechanical system under the action of "no working resistance and gravity" and "with working resistance and gravity" respectively, and output the balance couple moment curve acting on the large helical gear under these two conditions, as shown in Figure 3, so that the pressure film can be used to convert the oil pressure into the change of the resistance of the strain gauge
Figure 3 balance couple moment and equivalent moment
3 system dynamics analysis
any mechanical system can be simplified into the following equivalent dynamics model:
3.1 calculation of equivalent moment
in dynamic statics analysis, if the system is not affected by working resistance and gravity, the system balance moment is caused by the inertia force and inertia couple moment of each component; Besides, the balance moment of the system under the action of working resistance and gravity also includes the influence of gravity and working resistance. Therefore, the equivalent combined torque of working resistance and gravity is obtained by subtracting the balance torque under the former working condition from the balance torque under the latter working condition (the author has verified the correctness of this conclusion). According to this equivalent relationship, the equivalent torque (resistance torque) curve of working resistance and gravity in the shaper shown in Figure 3 can be obtained through Excel analysis and calculation
assuming that the equivalent driving torque is a constant torque, its value can be calculated by the formula
3.2 system equivalent moment of inertia
according to the principle that the kinetic energy of the system is equal before and after equivalence, the equivalent moment of inertia of the system can be calculated
according to formula (4), the system equivalent moment of inertia curve shown in figure 4:
system equivalent moment of inertia
can be obtained in Excel spreadsheet
3.3 solve the real speed of the equivalent member
from the above, the equivalent dynamic model of the shaper is formula (3), in which the system equivalent moment and equivalent moment of inertia have been calculated in the form of curves (Fig. 3 and Fig. 4). Now the angular velocity of the equivalent member can be calculated by the difference method ω， That is, the actual speed of the large helical gear in the real system. The solution formula is as follows:
(5) the actual speed curve on the large helical gear of the shaper system can be obtained in the Excel spreadsheet, as shown in Figure 5. It is worth noting that if ω If the initial value of is too small, negative rotation speed will appear in the solution process (unreasonable)
4 Conclusion
this paper makes full use of the modeling function of SolidWorks and the motion simulation function of cosmosm0tion, and combines the analysis and calculation and chart display functions of Excel to realize the dynamic analysis of the shaper system. The software used in this research method is simple and easy to learn. The method used is practical and reliable, and the results obtained are accurate and true. It is suitable for the kinematics and dynamics of any mechanical system. (end)
Related Topics | 2,416 | 12,025 | {"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-2023-14 | latest | en | 0.925064 |
https://answers-ph.com/math/question2386477 | 1,656,471,947,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00164.warc.gz | 152,732,744 | 41,188 | • Accueil
• Math
• Find the sum of the first 4 multiples of 3....
Find the sum of the first 4 multiples of 3.
• Réponse publiée par: cyrishlayno
Hello !
The first 4 four multiples of 3 is...
M = {0 , 3 , 6 , 9}
Now , you find the sum of the first 4 multiples of 3.
Sn = (n/2) × (A1 + An)
S4 = (4/2) × (0 + 9)
S4 = 2 × 9
S4 = 18
Final result : the sum of the first 4 multiples of 3 is 18.
I hope I have collaborated !
• Réponse publiée par: nelspas422
14 kilogram of water she contain
• Réponse publiée par: nelspas422
Wawakasan ko ang kwento sa pamamagitan ng pagpapangiti sa mga mambabasa o taga panood
Connaissez-vous la bonne réponse?
Find the sum of the first 4 multiples of 3.... | 247 | 702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-27 | latest | en | 0.393349 |
http://www.thescienceforum.com/physics/15773-relativity-twins-paradox-print.html | 1,601,121,529,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00298.warc.gz | 222,642,802 | 2,410 | # relativity and the twins paradox
• September 1st, 2009, 01:13 PM
rgba
Hi there!
I've been reading Stephen Hawking's book, a brief history of time.
If one of two twins goes to live on top of a mountain, and the second twin goes to live at sea level, the first twin will age faster than the second. As far as I can understand this, the first twin will age faster because light reaches him first (the difference is so minuscle because the height of a mountain is so tiny relative to the speed of light). Is that correct?
I also read another version of the paradox, which states that if a twin (called A) gets into a spaceship and travels into space at close to the speed of light, while his twin (called B) stays on earth, that on the twin A's return to earth we will see that the twin B has aged considerably.
I'm confused, because I thought the longer it takes light to reach you, the faster time?
Or maybe I'm not getting the whole concept wrong? The speed of light is constant no matter how fast you are moving.
Hope someone can explain this a bit better!
Thanks, looking forward to a response!
• September 1st, 2009, 02:03 PM
Chronman
• September 1st, 2009, 02:17 PM
rgba
Wow thanks Chronman, that looks like a really neat article! :) | 306 | 1,246 | {"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-2020-40 | latest | en | 0.965159 |
https://community.letsthink.org.uk/thinkingmathslessons/chapter/episode-1-33/ | 1,652,951,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00118.warc.gz | 230,527,489 | 21,533 | Lesson 11 Setters and solvers
# Episode 1
Reasoning Resources: Worksheet 1, Calculators
Whole class preparation
Hiding a digit here makes routine number work impossible and focuses the attention on the mathematical ideas involved. This is even more so if two or more digits are hidden.
The pupil must combine the value of the individual digit with the value of its position to keep a sense of overall magnitude. A basic step in numeracy is ease in comparing two numbers so that 4899 is quickly seen as smaller than 6110, even though most of the digits are larger.
Rehearse with the class: what a digit is; the alternative words for ‘digit’ such as ‘figure’ and ‘numeral’; How is a digit different from a number? and How many digits are there? Tell a story of historians finding sheets of calculations in a castle with some digits damaged or erased, and having to construct them, and making mistakes.
Start with very simple addition calculations with one or two missing digits, such as 32+ ?=35 and 72 + ?? = 97. Let pupils talk about their methods, which could simply be from ‘adding on’ to'l just know it’ referring to knowledge of number bonds.
Introduce the word “inverse” as a mathematical word for the natural language of doing the opposite or reverse operation. Clarify where they have only used the units position. Pupils should also notice that they can find individual missing digits separately in the units and tens columns.Then work with the whole class on the first three questions A,B and C on Worksheet 1. The questions ‘across; A, B and C, are of different types. So are the questions ‘down; A, D and G. One question can be solved through steps of reasoning, one is impossible to solve, and the other has more than one possible correct answer.
Pair work
Coordinating place value while carrying out an operation or its inverse may involve breaking up or combining numbers either side of tens, variously labelled bridging, trading, or decomposition. Give out Worksheet 1 to pairs to work on the remaining questions, They should see which can be solved and then sort them into the three types - one solution, many solutions, cannot be solved. Pairs who finish early could check with other pairs and then give names or labels to each of the three types of problem.
Allow pupils to use calculators but suggest their best use is for checking answers occasionally. Calculators will be more useful for Episode 3, which deals with multiplication and division. During the lesson highlight instances where thinking through questions is much easier and faster than trial and error on the calculator.
To include impossible sums and sums that allow more than one answer fosters a critical outlook, based on belief in consistency in mathematics itself. Whole class sharing and discussion
Subtraction examples: 3?-?2 =45 and ?7 - 83 =2? are impossible. 28 — 3?=09 and 6? —?8 = 23 have unique answers 12? -56 = 6? and ?1 —?2 = 69 have more than one answer each. Agree as a class which sums have unique, multiple or impossible solutions. They share alternative methods for finding solutions and how these combine using place value with inverse operations.
Encourage pupils to explain in words the conditions on the ‘units position’ and ‘tens position; and the least and most that the number can be (the maximum and minimum possible total of addition of 2 two-digit numbers). The agenda here is for them to realise they are using a chain of reasoning, each step of which is fully proved to be true. That is a form of proof.
Elaboration and extension
• Write some subtraction questions on the board in the same vertical format, leading to the sum recognition that the top number must be the bigger one.
• Write three-digit addition and subtraction questions in a horizontal format, allowing the pupils either to organise them themselves or to solve them without such organisation. | 827 | 3,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-21 | latest | en | 0.931253 |
http://aseba.wikidot.com/en:thymiotictactoe | 1,723,168,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00415.warc.gz | 1,918,950 | 7,548 | Thymio plays Tic-tac-toe
# Introduction :
This project was completed as part of the optional computer course 2013-2014 at the Beaulieu High School of Lousanne. The purpose of this rated work was to implement an application allowing two people to play "Tic-tac-toe" (Noughts and Crosses) with Thymios robots.
## Material :
• 2 Thymios
• A3/A2 paper sheet
• Ballpen
# Approach
## General concept :
This project uses two Thymios. The first is used to draw the shapes on the 3 x 3 Tic-tac-toe area, it is called "designer". The second deals with the management of user turns and boxes (wins/draw), the user actions, verification of their validity as well as the sending of the selected coordinates to the designer robot, this robot is called "master".
## Operation of the "designer" :
The designer code is separated into 3 blocks, the first represents the basic motor functions setting the engine speed for forward, backward, rotate, etc. These are used in the second block, represented by loops in an //onevent timer // allowing the Thymio to create such complex figures as a cross, a square, rotations and linear movement. All figures are based on incrementing a counter that indicates to the Thymio which motor function to use and when (based on the Thymio drawing model).
The third block is capable of giving a set of instructions for the second block to perform according to its destination square. It could be summarised thus: reception of the destination square by the prox.comm box → moving on the x-axis → movement on the y-axis → drawing of the appropriate shape depending on the player's turn. The designer uses the following features of the Thymio:
• Motors
• Inter-robot communication
• Ballpen for drawing
• Leds for visual information
## Operation of the "master" :
The behaviour of the master can be described thus: start of the game → perform the moves → game ends with a win/draw. A move is composed of different parts: beginning of the move → column/x-coordinate selection → confirmation → line/y-coordinate selection→ confirmation → validation/refusal of the coordinate entry → checking the status of the boxes (win/ or not) → If not: sending the "designer" the destination coordinate, waiting for the response of the designer and end of the round, if no win: end.
The choice of the column and row is made with the side buttons of the Thymio, confirmation with the centre button. Information is currently passed via light and sound, the leds and the speaker indicates: which player must play (bottom right: green/red), is the player selecting the column/row (bottom left: green/violet), what is the current value of x/y (/ / leds.circle//)
The game is managed by a list (array) of 9 integers, allowing a real 3 x 3 table to be simulated. Each of these integers can take 3 values:-1 if it was played by the player 2, 0 if it is empty, 1 if it was played by player 1. For a coordinate to be accepted, the destination box must be empty, otherwise an exception is raised. This array also serves to manage the victory of one of the players, it is enough when verifying to total each row/column/diagonal. If one of these sums gives 3 or - 3, we know that one of the two players has won. If the coordinate is validated, it is sent to the designer in the form "xy", and is there extracted.
The master therefore uses the following features of the Thymio:
• Inter-robot communication
• LEDs and speaker to transmit the information concerning the progress of the game
• Buttons to manage the game
# Implementation
Here is a typical example of a game:
## Discussion and possible improvements :
### Concerning the "designer" :
Motor rotation speed varies depending on the state of charge of the robot. This implies that when the robot should turn through a right angle, the angle turned is not always correct. It is similar for the rest of the drawings even if the effect is less noticeable. In the instructions the motor speed passes through zero between each change of speed in order to keep the shapes as similar as possible over time.
### Concerning the "master" :
The onevent button seems to trigger much too often, certainly at a frequency of 50 Hz. This implies that one simple touch of a button executes the associated loop about 3 to 5 times. To circumvent this problem, I use a timer with a flag allowing to have direct control over the duration of a button touch and the time between 2 touches. The whens seem to want to trigger randomly once the condition is met. I therefore did not rely on this possibility during the creation of the master.:
To make less random and more accurate drawings, it would have been possible to change the way the designer moves on the game area. For example, one could represent the boxes by black lines and make sure that the designer follows them. However, such a change would require a rethink of the whole operation of the designer. For the master, one might wonder about the idea of a feature that allows to play new games maintaining an evolving tally, because it is rare in real life to play only one game of tic tac toe. A way should however be found so as not to confuse the games, for example by changing the pen.
Adding a playing heuristic to the master, allowing to chose if to play against thymio or against another user. | 1,171 | 5,303 | {"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} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.9202 |
https://www.physicsforums.com/threads/energy-conservation-law-question-with-capacitor.1050177/ | 1,713,481,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00010.warc.gz | 847,312,197 | 19,355 | # Energy conservation law question with capacitor
• annamal
In summary: !) as the first term in the equation, and the work done by the non-conservative forces can appear (again with a sign!) as the second term.
annamal
Homework Statement
We wish to use the energy in a capacitor to accelerate a car of mass 1000 kg from 0-15 km/hr; if the voltage is initially 500 V; what is the area if the dielectric constant = 3.4 and d = 0.0005 inches between the capacitor plates?
Relevant Equations
Energy of capacitor = change in kinetic energy
I was wondering why energy of capacitor does not equal change in kinetic energy PLUS change in potential energy where potential energy is the change in the potential energy of the charges. I believe that should be so because energy conservation = change in kinetic energy plus change in potential energy. So why in this problem would we leave potential energy out?
annamal said:
energy conservation = change in kinetic energy plus change in potential energy.
Energy conservation: Total energy remains constant.
So, final kinetic energy + final potential energy= initial kinetic energy plus initial potential energy
or, change in kinetic energy plus change in potential energy = 0.
So why in this problem would we leave potential energy out?
The energy of the charges in the capacitor is the potential energy of the system.
annamal
The energy stored in a capacitor IS electrical potential energy!
When a charged capacitor is fully discharged, all of the capacitor's electrical potential energy is converted to some other form(s) of energy.
If we assume all the capacitor's potential energy is converted to kinetic energy, then the law of conservation of energy tells us:
(change in capacitor's energy) + (change in kinetic energy) = 0
Remember, a change is positive if a quantity increases; and a change is negative if a quantity decreases.
Edit: Ahah! @TSny just beat me to it.
annamal
Ok, I had a follow up question. How come in the work energy theorem, work = change in kinetic energy; but in the energy conservation law work = change in kinetic energy + change in potential energy?
The work done by a conservative force is the negative of the change in potential energy. If an object is displaced under the influence of a conservative force from A to B, ##W_{AB}=\int_A^B \mathbf{F}\cdot d\mathbf{s}.## The change in the potential energy from which this force is derived is ##\Delta U=U_B-U_A##. Since ##\mathbf{F}=-\mathbf{\nabla}U##, $$W_{AB}=\int_A^B \mathbf{F}\cdot d\mathbf{s}=\int_A^B(- \mathbf{\nabla}U)\cdot d\mathbf{s}=-(U_B-U_A)=-\Delta U.$$So starting with the work-energy theorem, you get energy conservation: \begin{align} & \Delta K=W_{AB}\nonumber \\ & \Delta K=-\Delta U\nonumber \\ & \Delta K+\Delta U=0. \nonumber \\ \end{align}
ChiralSuperfields
work = change in kinetic energy + change in potential energy?
Your equation isn’t always correct (but conservation of energy is!). You have to be very careful about what you mean by ‘work’.
What's always correct is the 'work-energy' theorem: work done by the resultant force = Δ(KE).
E.g. you push a block uphill (no friction) from rest, giving the block a kinetic energy (KE) of 1000J and increasing its gravitational potential energy (GPE) by 2000J. You have done 3000J of work.
Here are 2 ways to write the energy-conservation equation:
1) Work done by you = Δ(KE) + Δ(GPE)
3000J = 1000J + 2000J
2) (Work done by you) + (work done by gravity) = Δ(KE)
3000 + (-2000) = 1000J
(Because work done by gravity = -Δ(GPE) = -2000J.)
For a block sliding (no friction) downhill - with no force applied by you - you could write
1) Δ(KE) + Δ(GPE) = 0 or
2) Work done by gravity = Δ(KE)
They mean the same because work done by gravity = -Δ(GPE).
Does the work energy theorem: delta W = delta KE apply with external and nonconservative forces as well or does that formula only work if there are only conservative forces and external forces = 0?
First, there is no "delta W". The work (not change in work) is equal to the change in kinetic energy. This is because the work is a quantity that describes what happens in a process, in the transition from initital to final state. The kinetic energy, on the other hand, is a quantity that describes the state and has different values in the initial and final states. So the change in the state (the "delta") depends on what happens in the transition from initial to final (the work done by the forces).
And second, the change in kinetic energy of a system is equal to the work done by all forces, internal and external to the system. It does not matter if they are or not conservative.
The conservative character is important only for defining a potential energy. It is not relevant to the work-energy theorem.
annamal said:
Does the work energy theorem: delta W = delta KE apply with external and nonconservative forces as well or does that formula only work if there are only conservative forces and external forces = 0?
It works with non-conservative forces as well. The general form of the W-E theorem is
$$\Delta K = W_{Net}$$ where ## W_{Net}=\sum_i\int\mathbf{F}_i\cdot d\mathbf{s}## is the sum of all the forces, conservative and non-conservative. Textbooks often split ##W_{Net}## in two parts, one that includes the work ##W_C## done on the object by all the conservative forces and one that includes the work ##W_{NC}## done by all the non-conservative forces. Then the W-E theorem becomes, $$\Delta K = W_{C}+W_{NC}.$$ The work done by the conservative forces can then appear (with a sign change) on the other side of the equation as the change in potential energy $$\Delta K +\Delta U= W_{NC}.$$It seems to me that you are using the term "external force" incorrectly. As the name implies, external forces are outside the system and cross the system boundary to do work on it. You cannot say that a force is external if you have not defined the system and its boundaries.
For example, you drop a book of mass ##m## on the floor from height ##h##. If you choose the book only as your system, the force of gravity ##mg## is external to the book because it exerted by the Earth and that is not part of the system. The force of gravity does work ##W= mgh## and the W-E theorem says $$\Delta K = mgh.$$ Note that a single-component system does not have potential energy. Potential energy requires at least two components and depends on the relative configuration of these, i.e. their relative position in space.
If you choose the Earth and the book as your system, there are no external forces acting on the system to do external work. (We ignore the gravitational attraction of the Moon, the Sun, Jupiter, etc.) Therefore, $$W_{Net} = 0$$ However, the potential energy of the system changes because the book moves closer to the center of the Earth. What is work done by gravity in the one-component system is change in potential energy on the other side of the equation in the two-component system $$0=\Delta K +\Delta U=\Delta K - mgh.$$Whatever system you use, the W-E theorem gives the same result.
annamal
## 1. What is the energy conservation law in regards to a capacitor?
The energy conservation law states that energy cannot be created or destroyed, but it can be transformed from one form to another. In the case of a capacitor, the energy is stored in the electric field between the two plates.
## 2. How does a capacitor conserve energy?
A capacitor conserves energy by storing it in the form of an electric field. When a capacitor is charged, work is done to separate the charges on the two plates, creating an electric potential difference. This potential difference stores the energy in the capacitor until it is discharged.
## 3. Can energy be lost in a capacitor?
No, energy cannot be lost in a capacitor. As per the energy conservation law, the energy stored in a capacitor cannot be destroyed. However, some energy may be lost due to factors such as resistance in the circuit or leakage from the capacitor itself.
## 4. How does the energy conservation law apply to the charging and discharging of a capacitor?
When a capacitor is charged, the energy is stored in the electric field between the plates. As the capacitor discharges, the energy is released and the electric field decreases. However, the total amount of energy remains the same, as per the energy conservation law.
## 5. How does the energy conservation law affect the behavior of a capacitor in a circuit?
The energy conservation law dictates that the total energy in a closed system must remain constant. This means that the energy stored in a capacitor must be equal to the energy supplied to it by the circuit. As a result, the capacitor will charge and discharge in a predictable manner in order to conserve energy.
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# How do gearmotors impact reflected mass inertia from the load?
Sunday - 05/05/2019 22:55
In any drive system (that is, a motor driving a load), there is going to be inertia. Specifically, the motor will have inertia as will the load.
For a direct drive system (a direct motor-to-load connection) the individual inertias can just be added together in a straightforward way. However, for any other type of setup involving some additional mechanical drive components such as couplings or gears, the relationship changes.
Specifically, when looking at how gears impact a drive system’s inertia, there are some basic ways to look at it. The basics; any gear will reduce the load inertia reflected to the motor by a factor of the square of the gear ratio. When looking at a gearmotor, this is basically just a motor with an embedded gearset in it, so the calculations and formulas for calculating inertia are the same.
In high performance motion control applications, it’s ideal for the reflected load inertia to equal the motor inertia. If inertia matching is the only concern, then the gear ratio can be calculated as:
N = √ (JLoad/ J Motor)
where N is the gear ratio, JLoad is the inertia of the driven load, and JMotor is the inertia of the motor.
Note: Another way to calculate gear ratio is by reference to the individual gears. For instance, if NIn is the number of gear teeth on the input gear and NOut is the number of gear teeth on the output gear, the gear ratio is then:
N = NOut / NIn
As for reflected inertia itself, this is best understood as the inertia of the load that is translated back to the motor through various drive components, in this case the gearing of the gearmotor.
So in a simple direct-drive system, the total inertia would be calculated like this:
JTotal = JLoad + JMotor
However, for a geared drive system such as a gearmotor, the total inertia is calculated using the equation:
JTotal = ( JLoad / N2 ) + JMotor
This is because, as stated earlier, the reflected load inertia is equal to the load inertia divided by the square of the gear ratio.
Total notes of this article: 0 in 0 rating
Click on stars to rate this article | 566 | 2,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-21 | latest | en | 0.889871 |
https://cqlife.net/teoria-de-la-relatividad | 1,675,054,468,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00464.warc.gz | 211,199,710 | 7,460 | theory of relativity
2022
We explain what the theory of relativity is and what are the two theories that compose it. Also, who was Albert Einstein.
The theory of relativity was developed by the physicist Albert Einstein.
What is the Theory of Relativity?
It is known as Theory of Relativity or even Einstein's Theory, the set of scientific formulations developed at the beginning of the 20th century by the physicist Albert Einstein (1879-1955). Its objective was to resolve the theoretical incompatibility that exists between the two main fields of physics: mechanics. Newtonian and the electromagnetism.
These formulations were published as two different scientific theories:
• Theory of special relativity. A treatise on physical of movement of bodies in the absence of gravitational forces or gravity, where Maxwell's equations referring to electromagnetism were made compatible with those of Newton referring to motion.
• Theory of general relativity. A theoretical approach to the gravity and of both inertial and non-inertial reference frames. Generalizes the theory of special relativity and replaces theNewtonian gravity in cases where the gravitational fields are very strong.
The basic foundations of the Theory of Relativity can be summarized in that the location in thespace Yweather (referred to as space-time, a kind of four-dimensional matrix proposed by Hermann Minkowski) of a given phenomenon, will always depend on the speed at which the observer moves.
Or put in simpler terms: that things can be perceived in a very different way depending on the observer's point of view, even with regard to dimensions that until now were thought absolute, such as time or space.
This, which apparently seems like something simple, allowed to rethink the way in which contemporary physics understood time and space. In addition, it opened the door to a whole series of new equations around phenomena that, initially, seem to go against common sense.
These phenomena include spatial contraction, time dilation, the universal speed limit (equivalent to thespeed of light). On the other hand, Einstein discovered the equivalence between mass and energy, which he expressed in the famous formula E = m.c2 (Energy equals mass times velocity squared).
Importance of the Theory of Relativity
Thanks to the theory of relativity, it was discovered how to harness nuclear energy.
Einstein's Theories refounded modern physics. They were quickly adopted by all the major centers of thought and study of physics in the world.
They also had an important impact on the philosophy, since among other things they denied the existence of an absolute time and allowed to think, in serious terms, matters that were once exclusive to fantasy and reverie, such as time manipulation or high-speed space travel.
Although its specific explanation may be tedious or complicated, it should be noted that the Theory of Relativity has been proven. In fact, it has been put into practice in matters as complex as the atomic Energy (the atomic bombs, for example).
In addition, there is evidence of the slight but undeniable differences in aging and the passage of time that occur between astronauts and inhabitants of the Earth, since the latter, being more subject to the planet's gravity, live the time faster.
Einstein's theories allowed the emergence of cosmology, which is a branch of physics dedicated to determining the conditions oforigin of the universe. His observations on the curvature of light were publicly verified in 1919, in the framework of asolar eclipse.
Albert Einstein's biography
Einstein fled to the United States shortly before Nazism came to power in Germany.
Albert Einstein was born in Ulm, Germany, in 1879. Son of Hermann Einstein and Pauline Koch, family Of Jewish descent, his intellectual development is said to have been delayed and as a child he was notoriously slow to express himself, leading his parents to think that he had some kind of mental retardation.
As a child he was shy, patient and methodical. Later he demonstrated a notorious talent for natural Sciences, despite never having been a brilliant student. The rigidity of the German educational system in imperial times always worked against him and it is said that he had some altercations with the authorities.
He graduated from professor of math and physicist, and his first wife was the radical feminist Mileva Marić. His genius went unnoticed and wasted until the beginning of the 20th century when he published his first essays about physics.
The brilliance of his contributions to the field did not prevent him from being afraid of the anti-Semitic policies proclaimed by the German Nazi regime (which later came to power in 1933). For that reason Einstein fled to the United States in 1932, along with his second wife, his cousin Eva Loewenthal.
In the United States he obtained nationalization and continued his studies, focused on a theory that would unify the four fundamental interactions of the nature. But this work was left unfinished.
At age 76, Einstein suffered an abdominal aortic aneurysm and died on April 18, 1955, at Princeton Hospital. His body was cremated that same day. Previously, the hospital pathologist removed his brain, without permission from the family, in order to preserve it for future studies regarding his incredible intelligence.
!-- GDPR --> | 1,059 | 5,384 | {"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} | 2.796875 | 3 | CC-MAIN-2023-06 | longest | en | 0.961091 |
https://math.stackexchange.com/questions/2876979/check-the-statements-about-irreducibility | 1,669,991,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00418.warc.gz | 427,831,891 | 38,919 | # Check the statements about irreducibility
We have the ring $$R$$ and the polynomial $$a=x^3+x+1$$ in $$R[x]$$. I want to check the following statements:
1. If $$R=\mathbb{R}$$ then $$a$$ is irreducible in $$\mathbb{R}[x]$$.
This statement is false, since according to Wolfram the polynomial has a real solution, right? But how can we calculate this root without Wolfram, by hand?
2. If $$R=\mathbb{Z}_5$$ then $$a$$ is irreducible in $$R[x]$$.
The possible roots are $$0,1,2,3,4$$. We substitute these in $$a$$ and we get the following: \begin{align*}&0^3+0+1=1\neq 0\pmod 5 \\ &1^3+1+1=3\neq 0\pmod 5 \\ &2^3+2+1=11\equiv 1\pmod 5\neq 0\pmod 5 \\ &3^3+3+1=31\equiv 1\pmod 5\neq 0\pmod 5 \\ &4^3+4+1=69\equiv 4\pmod 5\neq 0\pmod 5\end{align*} So since none of these elements is a root of $$a$$, the statement is correct, right?
3. If $$R=\mathbb{C}$$ then $$a$$ is irreducible in $$R[x]$$.
This statement is wrong, because from the first statement we have that $$a$$ is reducible in $$\mathbb{R}[x]$$, and so it is also in $$\mathbb{C}[x]$$. Is this correct?
4. If $$R=\mathbb{Q}$$ then $$a$$ is not irreducible in $$R[x]$$.
Since this is a cubic polynomial, it is reducible if and only if it has roots. By the rational root test, the only possible rational roots are $$\pm 1$$. Since neither of these is a root, it follows that $$a$$ is irreducible over $$\mathbb{Q}$$, right?
5. If $$R=\mathbb{C}$$ then $$a$$ has no root in $$R$$.
This statement is wrong, since from statemenet 3 we have that $$a$$ is reducible in $$\mathbb{C}[x]$$ and so it has roots in $$\mathbb{C}$$, or not?
Let's let aside the five element field, for the moment.
Cardan's formula for the roots of the polynomial $x^3+px+q$ require to compute $$\Delta=\frac{p^3}{27}+\frac{q^2}{4}$$ In your case $$\Delta=\frac{1}{27}+\frac{1}{4}>0$$ so the polynomial has a single real root, precisely $$r=\sqrt[3]{-\frac{1}{2}+\sqrt{\Delta}}+\sqrt[3]{-\frac{1}{2}-\sqrt{\Delta}}$$ This shows that $a$ is reducible over $\mathbb{R}$. Of course it is reducible over $\mathbb{C}$ and has three complex roots that you can, in principle, compute by factoring out $x-r$.
The polynomial is, however, irreducible over $\mathbb{Q}$, because the only possible rational roots are $1$ and $-1$, which aren't roots by direct substitution.
Let $F$ be a field.
Theorem. A polynomial $f(x)\in F[x]$ of degree $2$ or $3$ is irreducible if and only if it has no roots in $F$.
Proof. If $f(x)$ has a root $r$, then it is divisible by $x-r$, so it is reducible. If $f(x)$ is reducible, then an irreducible factor must have degree $1$ (just count the degrees). QED
This can be applied to the case $\mathbb{Z}_5$: no element is a root, so the polynomial is irreducible.
Important note. The above criterion does not extend to polynomials of degree $>3$.
Over the reals there is a simpler criterion, instead of considering Cardan's formula.
Theorem. A polynomial of odd degree in $\mathbb{R}[x]$ has at least a real root.
This follows from continuity of polynomials as functions and the fact that the limit of a monic polynomial of odd degree at $-\infty$ is $-\infty$ and the limit at $\infty$ is $\infty$. The intermediate value theorem allows us to conclude.
If you know that $\mathbb{C}$ is algebraically closed, you can also classify the irreducible polynomials over $\mathbb{R}$: a polynomial in $\mathbb{R}[x]$ is irreducible if and only if it has degree $1$ or has degree $2$ and negative discriminant.
1. Notice that the polynomial is of odd degree, thus it must have a zero by the intermediate value theorem.
2. Yes, you are correct.
3. Follows from 1.
4. Indeed you are right.
5. Follows from 1.
• 1. Why does it hold that a polynomial of odd degree must have a zero? Could you explain it further to me? Aug 9, 2018 at 8:47
• Sure: 1. Any polynomial function is smooth 2. If the degree is odd, then $\lim_{x\to+\infty}p(x)=+\infty$ and $\lim_{x\to-\infty}p(x)=-\infty$. 3. Thus by the intermediate value theorem any value between $-\infty$ and $+\infty$ must be attained (at least once). In particular the value zero. Aug 9, 2018 at 8:55
• I understand it now! Thanks for explaining!! Does the statements 3 and 5 follow from 1 as I wrote above? Aug 9, 2018 at 9:00
• Yes and yes. Once you have reducibility over a smaller field, reducibility over a larger field is immediate. Same goes for roots. Aug 9, 2018 at 9:33
• Ah ok!! Thank you!! So, the correct statement is the second one, right? Aug 9, 2018 at 9:40 | 1,427 | 4,483 | {"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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-49 | latest | en | 0.77541 |
http://openstudy.com/updates/505374e4e4b0a91cdf441087 | 1,449,000,819,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398471436.90/warc/CC-MAIN-20151124205431-00212-ip-10-71-132-137.ec2.internal.warc.gz | 167,220,849 | 9,623 | ## Darlene23_ 3 years ago A gym membership costs \$25 to join and \$15 each month. Write and use an algebraic expression to find the cost of the gym membership for 6 months.
1. onix12
\[15x+25=y, 15(6)+25=cost of gym membership\]
2. Darlene23_
thankyou (: | 79 | 259 | {"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} | 2.796875 | 3 | CC-MAIN-2015-48 | longest | en | 0.850686 |
https://busfoundation.org/answers-on-questions/how-heavy-is-a-bus.html | 1,642,778,432,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303385.49/warc/CC-MAIN-20220121131830-20220121161830-00339.warc.gz | 187,566,672 | 11,223 | ## How heavy is a bus in tons?
The big yellow bus pictured below is a type ”C” school bus. This equates to up to 14 tons.
## How much does a bus weight kg?
The larger (Type-D) school buses are around 33,000 lbs (15,000 kg ).
## How heavy is a bus UK?
United Kingdom
Dimension Value
Length 18.75 metres (61 ft 6 in)
Width 2.55 metres (8 ft 4 in)
Height 4.95 metres (16 ft 3 in)
Mass 12,000 kilograms (26,455 lb)
## How heavy is a school bus in KG?
You can expect an average school bus weight to be about 24,387 pounds (11,062 kg ).
## What weighs a ton example?
Show an object that weighs about a pound, such as a loaf of bread or a small melon. There are 2,000 pounds in a ton. A small car weighs about 1 ton. The largest mammal is the blue whale, which can weigh about 150 tons, or about 300,000 pounds.
## What weighs about 3 tons?
How heavy is 3 tons? The tongue of a Blue Whale can weigh up to 3 tons.
## What things weigh half a ton?
A Grand Piano, of the kind used for concert performances in large music halls, weighs 0.50 tons. Well-made pianos such as this may last 75 to 100 years. In other words, 0.50 tons is 1.1 times the weight of a Horse, and the weight of a Horse is 0.91 times that amount.
You might be interested: Readers ask: Fortnite How To Thank Bus Driver?
## What is weight of a school bus?
There are four types of buses — Type A, Type B, Type C and Type D — ranging from 6,000 pounds to more than 33,000 pounds. A conventional school bus is a Type C bus, which can weigh from 19,501 to 26,000 pounds, according to School Transportation News.
## How much does a house weigh?
(That would be one awfully big scale, but house movers tell me that most houses weigh in at between 80,000 and 160,000 pounds. Of course, this is sans foundation and concrete floor slabs, and house movers typically are dealing with older, smaller structures that have been stripped of many components (but not always).
## How much does a bus cost to buy UK?
Transport for London has released the costs for buying the New Bus for London fleet, and despite years of soothing reassurances from the Mayor that they’ll cost less than normal hybrids, they’ll actually cost a bit more. If you look at the current cost of a bus, £250,000, roughly speaking, buys you a new bendy bus.
## Why are London buses red?
In 1907 one company, the powers that be at London General Omnibus Company had a genius idea. They decided to paint the entire fleet red, making their buses stand out from their rivals, and place numbers on the front of the bus to tell people the route it would be taking.
## What are London buses called?
The name London General was replaced by London Transport, which became synonymous with the red London bus.
## What is type1 bus?
Basically, the Type –A school bus is constructed using a vehicle with a cutaway front section. The driver’s door is on the left side. Type A- 1 has a weight rating of less than 10000 lb and Type A-2 has a gross vehicle weight rating of 10000 lb or more. The bus is designed to carry more than 10 people on board.
You might be interested: How Many Seats On A City Bus?
## What is a Type 3 School Bus?
TYPE III: Type III school buses and type III Head Start buses are restricted to passenger cars, station wagons, vans, and buses having a maximum manufacturer’s rated seating capacity of ten or fewer people, including the driver, and a gross vehicle weight rating of 10,000 pounds or less.
## What is class A bus?
A ” Class A” bus is an articulated bus. They are considered combination vehicles towing greater than 10,000lbs so they need a ” Class A” license. | 899 | 3,628 | {"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-2022-05 | longest | en | 0.953041 |
https://linguistics.stackexchange.com/questions/4143/when-are-numbers-nouns | 1,718,406,874,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00414.warc.gz | 345,591,679 | 39,936 | # When are numbers nouns?
In my native language, Portuguese, numbers have officially been in various classes, from adjectives and nouns to "quantifiers" and determiners.
I'm thinking that perhaps we can't group them all, simply because they behave differently. For example, "ten" works differently from "million". Multiplying by 2, you get "twenty" and "two millions". A million is a noun, like a potato or a third; while "ten" is different. "Twelve" and "a dozen" are the same in meaning, but grammatically behave in different ways.
Why? In other languages, this happens with other words. Some may agree in gender/number with the object they quantify, like nouns do. Can we grammatically divide numbers in different classes? How would this work? Would another class be required? How would this system vary from language to language (I'd appreciate if you'd make a comparison between English and some other language of your choice)?
• As a monolingual English speaker, and writing off the cuff, I would say that English numeral terms are nouns when they stand for specific abstract quantities or for numerals. In "One and one make two," all the numeral terms are nouns that stand for abstract quantities. In "Write a three in the third box," the numeral term "three" stands for a numeral, i.e. a symbol that usually stands for a specific abstract quantity. Commented Aug 9, 2013 at 23:32
• I'm going to try to summarise the different questions i think you're asking here, just to make sure i'm following: (1) Why don't number words behave in a syntactically uniform manner? (2) If number words do cross-cut syntactic categories, which categories to they belong to (this raises interesting questions about the syntax/semantics interface, e.g. if number words are semantically uniform, why don't they behave the same syntactically?) (3) Do we need a distinct syntactic category 'numeral'? (4) To what extent to numeral systems differ cross-linguistically? Commented Aug 9, 2013 at 23:59
• @James, in "Write a three in the third box.", wouldn't the "three" be a noun, as it is substitutable be "word"? Commented Aug 10, 2013 at 14:28
• @PElliot, precisely my questions; only yours are written better. :) Should I edit the post and replace the last paragraph with those? Commented Aug 10, 2013 at 15:04
• If you like, but it's probably just me. I just wanted to check that i understood. Greville Corbett (1978: academia.edu/746864/…) argues all numerals are somewhere between nouns and adjectives - he tentatively claims that, universally, the numerals which are more noun-like are numerically higher than the numerals which are more adjective-like. For example, the numerals in English which take an article are numerically higher than those which don't. It's an interesting claim and it's totally mysterious to me why it would hold. Commented Aug 10, 2013 at 16:28
Each word is classified by what is permissible in the instances in which it appears. In English, "three" can be a noun, an adjective, a pronoun, or a numeral. I will use French as the second language in my examples. If a word can be replaced by another word without changing the grammar of the sentence, the two words have the same grammatical class in that instance.
### Noun
• I have a three.
• I have a (playing) card (with the value three).
• J'ai un trois.
• J'ai une carte (à jouer) (avec la valeur trois).
• I have three cats.
• I have small cats.
### Determiner (French)
• J'ai trois chats. (I have three cats.)
• J'ai des chats. (I have some cats.)
• J'ai les trois chats. (I have the three cats.)
• J'ai les petits chats. (I have the small cats.)
### Pronoun
• Three are eating.
• They are eating.
• Trois mangent.
• Ils mangent.
### Numeral
• One, two, three.
• One, two, four.
• Un, deux, trois.
• Un, deux, quatre.
Note: "numeral" can be hard to differentiate from the other classes. It doesn't work if you replace "three" with a word from a different class:
• One, two, three
• One, two, *card
• One, two, *some
• One, two, *small
• One, two, *they
So, "three" isn't a noun, an adjective, a pronoun, and a numeral at the same time, it is only one of those at a time. You should now be able to apply the same process in Portuguese to work out when numbers are nouns.
In regards to your question about the declination of numbers (i.e. whether they are invariable, can be singular or plural, etc.), note that some nouns are invariable (in certain contexts), and some have to be declined.
• I am drinking milk.
Not:
• I am drinking *a milk. (allowable if "a milk" is the ellipsis of "a serving of milk")
• I am drinking *milks.
It seems from your question that in Portuguese, "ten" is invariable, while "million" can be singular or plural. This doesn't change the grammatical class. | 1,186 | 4,778 | {"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-2024-26 | latest | en | 0.951703 |
https://foreach.id/EN/common/power/Btuth%7Chour-to-erg%7Csecond.html | 1,618,083,987,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00515.warc.gz | 377,900,937 | 10,906 | # Convert Btu (th)/hour to erg/second (Btu (th)/h to erg/s)
Batch Convert
• Btu (th)/hour [Btu (th)/h]
• erg/second [erg/s]
Copy
_
Copy
• Btu (th)/hour [Btu (th)/h]
• erg/second [erg/s]
## Btu (th)/hour to Erg/second (Btu (th)/h to erg/s)
### Btu (th)/hour (Symbol or Abbreviation: Btu (th)/h)
Btu (th)/hour is one of power units. Btu (th)/hour abbreviated or symbolized by Btu (th)/h. The value of 1 Btu (th)/hour is equal to 0.29287 watt. In its relation with erg/second, 1 Btu (th)/hour is equal to 2928700 erg/second.
#### Relation with other units
1 Btu (th)/hour equals to 0.29287 watt
1 Btu (th)/hour equals to 0.00029287 kilowatt
1 Btu (th)/hour equals to 2.9287e-7 megawatt
1 Btu (th)/hour equals to 292,870,000,000 picowatt
1 Btu (th)/hour equals to 292,870,000 nanowatt
1 Btu (th)/hour equals to 292,870 microwatt
1 Btu (th)/hour equals to 292.87 milliwatt
1 Btu (th)/hour equals to 29.287 centiwatt
1 Btu (th)/hour equals to 2.9287 deciwatt
1 Btu (th)/hour equals to 0.029287 dekawatt
1 Btu (th)/hour equals to 0.0029287 hectowatt
1 Btu (th)/hour equals to 2.9287e-10 gigawatt
1 Btu (th)/hour equals to 0.00039275 horsepower
1 Btu (th)/hour equals to 0.0003982 horsepower (metric)
1 Btu (th)/hour equals to 0.000029856 horsepower (boiler)
1 Btu (th)/hour equals to 0.00039259 horsepower (electric)
1 Btu (th)/hour equals to 0.00039257 horsepower (water)
1 Btu (th)/hour equals to 0.0003982 pferdestarke
1 Btu (th)/hour equals to 0.99933 Btu (IT)/hour
1 Btu (th)/hour equals to 0.016656 Btu (IT)/minute
1 Btu (th)/hour equals to 0.00027759 Btu (IT)/second
1 Btu (th)/hour equals to 0.016667 Btu (th)/minute
1 Btu (th)/hour equals to 0.00027778 Btu (th)/second
1 Btu (th)/hour equals to 9.9932e-7 MBtu (IT)/hour
1 Btu (th)/hour equals to 0.00099933 MBH
1 Btu (th)/hour equals to 0.000083278 ton (refrigeration)
1 Btu (th)/hour equals to 0.25183 kilocalorie (IT)/hour
1 Btu (th)/hour equals to 0.0041971 kilocalorie (IT)/minute
1 Btu (th)/hour equals to 0.000069952 kilocalorie (IT)/second
1 Btu (th)/hour equals to 0.252 kilocalorie (th)/hour
1 Btu (th)/hour equals to 0.0041999 kilocalorie (th)/minute
1 Btu (th)/hour equals to 0.000069999 kilocalorie (th)/second
1 Btu (th)/hour equals to 251.83 calorie (IT)/hour
1 Btu (th)/hour equals to 4.1971 calorie (IT)/minute
1 Btu (th)/hour equals to 0.069952 calorie (IT)/second
1 Btu (th)/hour equals to 252 calorie (th)/hour
1 Btu (th)/hour equals to 4.1999 calorie (th)/minute
1 Btu (th)/hour equals to 0.069999 calorie (th)/second
1 Btu (th)/hour equals to 777.65 foot pound-force/hour
1 Btu (th)/hour equals to 12.961 foot pound-force/minute
1 Btu (th)/hour equals to 0.21601 foot pound-force/second
1 Btu (th)/hour equals to 2,928,700 erg/second
1 Btu (th)/hour equals to 0.00029287 kilovolt ampere
1 Btu (th)/hour equals to 0.29287 volt ampere
1 Btu (th)/hour equals to 0.29287 newton meter/second
1 Btu (th)/hour equals to 0.29287 joule/second
1 Btu (th)/hour equals to 2.9287e-10 gigajoule/second
1 Btu (th)/hour equals to 2.9287e-7 megajoule/second
1 Btu (th)/hour equals to 0.00029287 kilojoule/second
1 Btu (th)/hour equals to 0.0029287 hectojoule/second
1 Btu (th)/hour equals to 0.029287 dekajoule/second
1 Btu (th)/hour equals to 2.9287 decijoule/second
1 Btu (th)/hour equals to 29.287 centijoule/second
1 Btu (th)/hour equals to 292.87 millijoule/second
1 Btu (th)/hour equals to 292,870 microjoule/second
1 Btu (th)/hour equals to 292,870,000 nanojoule/second
1 Btu (th)/hour equals to 292,870,000,000 picojoule/second
1 Btu (th)/hour equals to 1,054.3 joule/hour
1 Btu (th)/hour equals to 17.572 joule/minute
1 Btu (th)/hour equals to 1.0543 kilojoule/hour
1 Btu (th)/hour equals to 0.017572 kilojoule/minute
### Erg/second (Symbol or Abbreviation: erg/s)
Erg/second is one of power units. Erg/second abbreviated or symbolized by erg/s. The value of 1 erg/second is equal to 1e-7 watt. In its relation with Btu (th)/hour, 1 erg/second is equal to 3.4144e-7 Btu (th)/hour.
#### Relation with other units
1 erg/second equals to 1e-7 watt
1 erg/second equals to 1e-10 kilowatt
1 erg/second equals to 1e-13 megawatt
1 erg/second equals to 100,000 picowatt
1 erg/second equals to 100 nanowatt
1 erg/second equals to 0.1 microwatt
1 erg/second equals to 0.0001 milliwatt
1 erg/second equals to 0.00001 centiwatt
1 erg/second equals to 0.000001 deciwatt
1 erg/second equals to 1e-8 dekawatt
1 erg/second equals to 1e-9 hectowatt
1 erg/second equals to 1e-16 gigawatt
1 erg/second equals to 1.341e-10 horsepower
1 erg/second equals to 1.3596e-10 horsepower (metric)
1 erg/second equals to 1.0194e-11 horsepower (boiler)
1 erg/second equals to 1.3405e-10 horsepower (electric)
1 erg/second equals to 1.3404e-10 horsepower (water)
1 erg/second equals to 1.3596e-10 pferdestarke
1 erg/second equals to 3.4121e-7 Btu (IT)/hour
1 erg/second equals to 5.6869e-9 Btu (IT)/minute
1 erg/second equals to 9.4782e-11 Btu (IT)/second
1 erg/second equals to 3.4144e-7 Btu (th)/hour
1 erg/second equals to 5.6907e-9 Btu (th)/minute
1 erg/second equals to 9.4845e-11 Btu (th)/second
1 erg/second equals to 3.4121e-13 MBtu (IT)/hour
1 erg/second equals to 3.4121e-10 MBH
1 erg/second equals to 2.8435e-11 ton (refrigeration)
1 erg/second equals to 8.5985e-8 kilocalorie (IT)/hour
1 erg/second equals to 1.4331e-9 kilocalorie (IT)/minute
1 erg/second equals to 2.3885e-11 kilocalorie (IT)/second
1 erg/second equals to 8.6042e-8 kilocalorie (th)/hour
1 erg/second equals to 1.434e-9 kilocalorie (th)/minute
1 erg/second equals to 2.3901e-11 kilocalorie (th)/second
1 erg/second equals to 0.000085985 calorie (IT)/hour
1 erg/second equals to 0.0000014331 calorie (IT)/minute
1 erg/second equals to 2.3885e-8 calorie (IT)/second
1 erg/second equals to 0.000086042 calorie (th)/hour
1 erg/second equals to 0.000001434 calorie (th)/minute
1 erg/second equals to 2.3901e-8 calorie (th)/second
1 erg/second equals to 0.00026552 foot pound-force/hour
1 erg/second equals to 0.0000044254 foot pound-force/minute
1 erg/second equals to 7.3756e-8 foot pound-force/second
1 erg/second equals to 1e-10 kilovolt ampere
1 erg/second equals to 1e-7 volt ampere
1 erg/second equals to 1e-7 newton meter/second
1 erg/second equals to 1e-7 joule/second
1 erg/second equals to 1e-16 gigajoule/second
1 erg/second equals to 1e-13 megajoule/second
1 erg/second equals to 1e-10 kilojoule/second
1 erg/second equals to 1e-9 hectojoule/second
1 erg/second equals to 1e-8 dekajoule/second
1 erg/second equals to 0.000001 decijoule/second
1 erg/second equals to 0.00001 centijoule/second
1 erg/second equals to 0.0001 millijoule/second
1 erg/second equals to 0.1 microjoule/second
1 erg/second equals to 100 nanojoule/second
1 erg/second equals to 100,000 picojoule/second
1 erg/second equals to 0.00036 joule/hour
1 erg/second equals to 0.000006 joule/minute
1 erg/second equals to 3.6e-7 kilojoule/hour
1 erg/second equals to 6e-9 kilojoule/minute
### How to convert Btu (th)/hour to Erg/second (Btu (th)/h to erg/s):
#### Conversion Table for Btu (th)/hour to Erg/second (Btu (th)/h to erg/s)
Btu (th)/hour (Btu (th)/h) erg/second (erg/s)
0.01 Btu (th)/h 29,287 erg/s
0.1 Btu (th)/h 292,870 erg/s
1 Btu (th)/h 2,928,700 erg/s
2 Btu (th)/h 5,857,500 erg/s
3 Btu (th)/h 8,786,200 erg/s
4 Btu (th)/h 11,715,000 erg/s
5 Btu (th)/h 14,644,000 erg/s
6 Btu (th)/h 17,572,000 erg/s
7 Btu (th)/h 20,501,000 erg/s
8 Btu (th)/h 23,430,000 erg/s
9 Btu (th)/h 26,359,000 erg/s
10 Btu (th)/h 29,287,000 erg/s
20 Btu (th)/h 58,575,000 erg/s
25 Btu (th)/h 73,219,000 erg/s
50 Btu (th)/h 146,440,000 erg/s
75 Btu (th)/h 219,660,000 erg/s
100 Btu (th)/h 292,870,000 erg/s
250 Btu (th)/h 732,190,000 erg/s
500 Btu (th)/h 1,464,400,000 erg/s
750 Btu (th)/h 2,196,600,000 erg/s
1,000 Btu (th)/h 2,928,700,000 erg/s
100,000 Btu (th)/h 292,870,000,000 erg/s
1,000,000,000 Btu (th)/h 2,928,700,000,000,000 erg/s
1,000,000,000,000 Btu (th)/h 2,928,700,000,000,000,000 erg/s
#### Conversion Table for Erg/second to Btu (th)/hour (erg/s to Btu (th)/h)
erg/second (erg/s) Btu (th)/hour (Btu (th)/h)
0.01 erg/s 3.4144e-9 Btu (th)/h
0.1 erg/s 3.4144e-8 Btu (th)/h
1 erg/s 3.4144e-7 Btu (th)/h
2 erg/s 6.8289e-7 Btu (th)/h
3 erg/s 0.0000010243 Btu (th)/h
4 erg/s 0.0000013658 Btu (th)/h
5 erg/s 0.0000017072 Btu (th)/h
6 erg/s 0.0000020487 Btu (th)/h
7 erg/s 0.0000023901 Btu (th)/h
8 erg/s 0.0000027315 Btu (th)/h
9 erg/s 0.000003073 Btu (th)/h
10 erg/s 0.0000034144 Btu (th)/h
20 erg/s 0.0000068289 Btu (th)/h
25 erg/s 0.0000085361 Btu (th)/h
50 erg/s 0.000017072 Btu (th)/h
75 erg/s 0.000025608 Btu (th)/h
100 erg/s 0.000034144 Btu (th)/h
250 erg/s 0.000085361 Btu (th)/h
500 erg/s 0.00017072 Btu (th)/h
750 erg/s 0.00025608 Btu (th)/h
1,000 erg/s 0.00034144 Btu (th)/h
100,000 erg/s 0.034144 Btu (th)/h
1,000,000,000 erg/s 341.44 Btu (th)/h
1,000,000,000,000 erg/s 341,440 Btu (th)/h
#### Steps to Convert Btu (th)/hour to Erg/second (Btu (th)/h to erg/s)
1. Example: Convert 37 Btu (th)/hour to erg/second (37 Btu (th)/h to erg/s).
2. 1 Btu (th)/hour is equivalent to 2928700 erg/second (1 Btu (th)/h is equivalent to 2928700 erg/s).
3. 37 Btu (th)/hour (Btu (th)/h) is equivalent to 37 times 2928700 erg/second (erg/s).
4. Retrieved 37 Btu (th)/hour is equivalent to 108360000 erg/second (37 Btu (th)/h is equivalent to 108360000 erg/s).
▸▸
▸▸ | 3,762 | 9,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-17 | latest | en | 0.797168 |
http://rhinoceros.helpmax.net/en/solids/cylindertubepipe/ | 1,660,011,218,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00460.warc.gz | 46,436,117 | 6,658 | Cylinder
Steps
• Follow the prompts for the selected option. If no option is specified, the default option is used.
Options
Note: The options for the base circle are the same as for the command.
DirectionConstraint
None
After creating the base circle
• Pick
the end of the cylinder.
Solid > Cylinder Solid > Cylinder Related topics…
Tube
Creates a closed cylinder with a concentric cylindrical hole.
Steps:
• Follow the prompts for the selected option. If no option is specified, the default option is used.
After drawing the base circle from one of the options
1. Pick the radius for the second tube wall.
2. Pick the end of the tube.
Options for the base circle
Note: The options for the base circle are the same as for the command.
DirectionConstraint
None
Solid > Tube Solid > Tube Related topics…
Pipe
Creates a surface with a circular profile around a curve.
Steps:
1. Select
a curve.
2. Pick
the start radius at the beginning of the pipe.
3. If the curve is closed, pick the radius for the pipe.
4. Pick the radius at the end of the pipe.
5. Pick a point for the next radius or press Enter
to end the command.
Options
ChainEdges
Cap
Specifies how to cap the ends.
None
No cap.
Flat
Cap with planar surface.
Round
Cap with hemispherical surface.
Thick
No
The pipe has only one wall.
Yes
The pipe has two walls.
1. Pick the first start radius.
2. Pick the second start radius.
3. Pick the first end radius.
4. Pick the second end radius.
ShapeBlending
Local
The pipe radius stays constant at the ends and changes more rapidly in the middle.
If the curve is a polycurve of lines and arcs, local blending fits the rail and makes a single surface.
Global
The radius is linearly blended from one end to the other, creating pipes that taper from one radius to the other.
If the curve is a polycurve of lines and arcs, the result is a polysurface with joined surfaces created from the polycurve segments.
Solid > Pipe (Right click) Solid > Pipe Related topics…
MeshCylinder
Creates a polygon mesh cylinder from a base circle and a length.
Options
Note: The mesh version has additional options for the number of mesh faces.
VerticalFaces
The number of faces from the base to the top.
AroundFaces
The number of faces around the circumference.
Mesh > Mesh Cylinder Mesh > Polygon Mesh Primitives > Cylinder Related topics… | 532 | 2,391 | {"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} | 2.578125 | 3 | CC-MAIN-2022-33 | latest | en | 0.84248 |
https://archi-monarch.com/geometry-in-architecture/ | 1,723,135,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00291.warc.gz | 81,784,717 | 42,525 | ## GEOMETRY IN ARCHITECTURE
If you want to know about the scale in architecture or form in architecture or proportion in architecture, please click the link.
Geometry is a fundamental aspect of architecture and is used to create aesthetically pleasing designs, plan functional spaces, and ensure structural stability. It involves the use of mathematical concepts like points, lines, angles, and shapes to create architectural forms.
Geometric shapes, such as circles, squares, triangles, and polyghedrons, are often used as building blocks for architectural design and construction. In architecture, geometry is used to determine the proportions of a building, to create symmetrical and asymmetrical forms, and to plan spatial arrangements and circulation patterns.
• Geometry is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. Geometry can be seen also as a structural science.
• Architectural design is based on geometric structures developed out of the idea of transformations. These transformations are visible as design concepts through history of architecture.
“For without symmetry and proportion no temple can have a regular plan”.
#### Vitruvius (De Architectura, or Ten Books on Architecture)
In “De Architectura,” Vitruvius discussed the use of geometry in architecture, describing how the principles of mathematics were used to create well-proportioned and aesthetically pleasing buildings. He also described the use of modular systems, symmetry, and other geometric concepts in architectural design.
Today, “De Architectura” is still widely read and respected as a source of historical and architectural information. It continues to influence architects and designers, who continue to draw inspiration from its timeless insights and principles.
### 1) Important concepts in geometry
The following are some of the important concepts in geometry relevant to architecture:
1. Points: the most basic element in geometry, representing a location in space.
2. Lines: a series of connected points, forming a path between two points.
3. Angles: the measure of the separation between two lines that share a common endpoint.
4. Shapes: two-dimensional forms created by the combination of lines, including squares, rectangles, triangles, circles, and polyggons.
5. Solids: three-dimensional objects created by the combination of shapes, including cubes, cylinders, cones, and spheres.
6. Proportions: the relationship between the size of one part of an object and the size of another part or the whole object.
7. Symmetry: the balance and consistency of design elements on either side of an axis.
8. Tessellation: the arrangement of shapes in a repeated pattern to cover a surface.
9. Perspective: the use of optical illusion to create the illusion of depth and space on a two-dimensional surface.
10. Modular systems: a method of creating proportion and rhythm in architectural design by dividing a space or form into repeating units.
These concepts are used by architects and engineers to create aesthetically pleasing, functional, and safe buildings.
### 2) Geometry uses in architecture
• Another way architects use geometry is by certain shapes of buildings. Architects must know the perimeter and area of the shapes to create the building.
• Another way architects use geometry is by using the Pythagorean Theorem for the design and measurements of building structures.
### 3) Why geometry in architecture?
• Architecture begins with geometry. Since earliest times, architects have relied on mathematical principles.
#### Greek theatre
##### Rome
• The Colosseum, the most important roman amphitheater, was used for gladiatorial contests and public spectacles such as mock sea battles.
Taj Mahal, mausoleum, India, reflection symmetry
Versailles, symbol of the system of absolute monarchy of the Ancien Régime
#### ii) Some modern building….
Here are a few examples of modern buildings that use geometry in their design:
1. Burj Khalifa, Dubai: The world’s tallest building features a sleek, triangular shape that narrows as it rises, creating a distinct geometric form that is visible from a distance.
2. Guggenheim Museum, Bilbao: The building’s distinctive, flowing form was created using a complex system of geometric shapes, including spirals and curves, to create a fluid, organic feel.
3. CCTV Headquarters, Beijing: This iconic building features a looping, geometric form that resembles a twisted, rectangular tube, creating a highly recognizable and distinctive silhouette.
4. Sydney Opera House, Sydney: The building’s iconic sails are created using a series of interconnected shells that are precisely modeled using geometric principles, creating a harmonious and recognizable form.
5. Petronas Towers, Kuala Lumpur: The twin towers feature a distinctive geometric form that resembles two interlocking squares, creating a recognizable and iconic silhouette.
These are just a few examples of how geometry is used in modern architecture to create aesthetically pleasing and functional buildings.
In summary, geometry is an essential tool in architecture, providing a way to create aesthetically pleasing, functional, and safe buildings that reflect the goals and values of their designers.
If you want to know also about the elements of design or objective of design or origin of architecture or principles of design, please click the link.
## Related video
error: Content is protected !! | 1,062 | 5,509 | {"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} | 3.859375 | 4 | CC-MAIN-2024-33 | latest | en | 0.928154 |
http://www.physicsforums.com/showthread.php?t=230620 | 1,369,412,781,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704752145/warc/CC-MAIN-20130516114552-00065-ip-10-60-113-184.ec2.internal.warc.gz | 655,636,183 | 9,467 | ## Vertical spring - elevator question
A mass is attached to a spring supported from the ceiling of an elevator. We pull down on the mass and let it to vibrate. If the elevator starts to accelerate(fixed accelerate) upward,
1) How the maximum velocity changes?
2) How the amplitude changes?
3) How the total energy changes?
I think the amplitude and maximum velocity does not change. Because the acceleration doesn't change the net force but only slide down the equilibrium point. Am i right?
PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
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Quote by Volcano A mass is attached to a spring supported from the ceiling of an elevator. We pull down on the mass and let it to vibrate. If the elevator starts to accelerate(fixed accelerate) upward, 1) How the maximum velocity changes? 2) How the amplitude changes? 3) How the total energy changes? I think the amplitude and maximum velocity does not change. Because the acceleration doesn't change the net force but only slide down the equilibrium point. Am i right?
I would agree with your choice with respect to the amplitude. However, in terms of the maximum velocity, it depends on your frame of reference, what are you measuring the velocity relative to.
P.S. We have Homework Forums for all your textbook questions.
___ |M| === \_\ /_/ \_\ _|_| |....| /\ |....| .| |....| .| moving upward |__.|
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## Vertical spring - elevator question
Quote by Crazy Tosser ___ |M| === \_\ /_/ \_\ _|_| |....| /\ |....| .| |....| .| moving upward |__.|
Your diagram is wrong, the mass is hanging down from the ceiling, inside the elevator, but thanks for your contribution anyway...
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Quote by Hootenanny Your diagram is wrong, the mass is hanging down from the ceiling, inside the elevator, but thanks for your contribution anyway...
Edit: It wouldn't actually make any difference to the answer, but it's best not to confuse the matter
Quote by Hootenanny Your diagram is wrong, the mass is hanging down from the ceiling, inside the elevator, but thanks for your contribution anyway...
oops D=
__________________
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$$| \amalg| \cdot \cdot \cdot \cdot |M| \cdot \cdot \cdot | \amalg |$$
$$| \amalg| \cdot \cdot \cdot \cdot \bigsqcup \cdot \cdot \cdot \cdot | \amalg |$$
$$| \amalg| \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot | \amalg | \Uparrow Moving Up$$
$$| \amalg| \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot | \amalg |$$
$$| \amalg| \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot | \amalg |$$
$$| \amalg| \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot | \amalg |$$
$$| \amalg| \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot | \amalg |$$
__________________
what are you measuring the velocity relative to
it is releative to elevator. But honestly i can not explain with equations. My choice is completely instinctive. By te way, it would be nice to see the pictures with post. Latex is hard for figures.
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Quote by Volcano it is releative to elevator. But honestly i can not explain with equations. My choice is completely instinctive. By te way, it would be nice to see the pictures with post. Latex is hard for figures.
Then you are correct, if your measuring the velocity of the mass with respect to the elevator. Obviously, if the velocity is measured relative to some other 'fixed' point outside the elevator then this will not be the case.
I want to understand the effects of adding force and adding mass while it is vibrate. As you are approved, additional force on motion is not change the amplitude and maximum velocity. Now I wonder, how the mass change the amplitude and max velocity? Now there is not an elevator. The same spring and mass attached to the ceiling of a door instead of an elevator and vibrating. While the mass in bottom position, an additional mass attached to other one suddenly. What happens now? I think, as previous problem, the equilibrium point slides down. The amplitude will not change because net force was not change. But maximum velocity will reduce because period will increase and distance was not change. Am I right now?
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Gold Member | 1,150 | 4,668 | {"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} | 3.578125 | 4 | CC-MAIN-2013-20 | longest | en | 0.853012 |
https://scholar.archive.org/work/g4jvv343jnhp3jfrgbcvp5bety | 1,652,867,608,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00430.warc.gz | 578,854,629 | 5,743 | ### Higher Order Unification via Explicit Substitutions
Gilles Dowek, Thérèse Hardin, Claude Kirchner
2000 Information and Computation
Higher order uni cation is equational uni cation for -conversion. But it is not rst order equational uni cation, as substitution has to avoid capture. Thus the methods for equational uni cation (such as narrowing) built upon grafting (i.e. substitution without renaming), cannot be used for higher order uni cation, which needs speci c algorithms. Our goal in this paper is to reduce higher order uni cation to rst order equational uni cation in a suitable theory. This is achieved by replacing
more » ... itutionby grafting, but this replacementis not straightforward as it raises two major problems. First, some uni cation problems have solutions with grafting but no solution with substitution. Then equational uni cation algorithms rest upon the fact that grafting and reduction commute. But grafting and -reduction do not commute in -calculus and reducing an equation may change the set of its solutions. This di culty comes from the interaction between the substitutions initiated by -reduction and the ones initiated by the uni cation process. Two kinds of variables are involved: those of -conversion and those of uni cation. So, we need to set up a calculus which distinguishes these two kinds of variables and such that reduction and grafting commute. For this purpose, the application of a substitution of a reduction variable to a uni cation one must be delayed until this variable is instantiated. Such a separation and delay are provided by a calculus of explicit substitutions. Uni cation in such a calculus can be performed by well-known algorithms such as narrowing, but we present a specialised algorithm for greater e ciency. At last we show how to relate uni cation in -calculus and in a calculus with explicit substitutions. Thus we come up with a new higher order uni cation algorithm which eliminates some burdens of the previous algorithms, in particular the functional handling of scopes. Huet's algorithm can be seen as a speci c strategy for our algorithm, since each of its steps can be decomposed into elementary ones, leading to a more atomic description of the uni cation process. Also, solved forms in -calculus can easily be computed from solved forms in -calculus. | 483 | 2,334 | {"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-2022-21 | latest | en | 0.918987 |
https://direct.physicsclassroom.com/mop/Electric-Circuits/Voltage/QG2help | 1,713,427,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817200.22/warc/CC-MAIN-20240418061950-20240418091950-00727.warc.gz | 182,643,857 | 26,932 | # Electric Circuits - Mission EC2 Detailed Help
If the electrical circuit in your iPod player were analogous to a water circuit at a water park, then the voltage (or the battery) would be comparable to _____.
Electric Potential and Charge Flow: Electric potential is a location dependent quantity that describes the amount of electric potential energy per charge possessed by a charged object at a given location. In order for charge to flow from one location to another location, there must be a difference in electric potential between the two locations. If there is a difference in electric potential between the two locations, then charge will spontaneously move from the location of high potential to low potential.
An analogy comparing the flow of water in a water park to the flow of charge in a circuit is often used in a physics class. Both situations involve the flow of a fluid through pipes or a conduit. In the case of water flow, such flow would not be possible unless two ends of the pipe were held at different water pressure or gravitational potential. That is, the potential energy possessed per kg of water at the top of a slide must be greater than the the potential energy possessed per kg of water at the bottom of a slide. With this difference in pressure or gravitational potential established between the two locations, water naturally flows through its circuit from high potential location to low potential location. In the same manner, there must be a difference in electric pressure or electric potential energy between two ends of an external circuit in order for charge to flow through the pipes or wires. Of course to maintain such a difference in water pressure or potential at a water park, a water pump must do work upon the water to move it from the low energy to the high energy location. And in the same way, a charge pump (a.k.a. a battery) must do work upon a charge in a circuit to move it from the low energy to the high energy location. | 390 | 1,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.934776 |
https://www.teacherspayteachers.com/Product/Geometry-Secant-Tangent-Theorem-Doodle-Graphic-Organizer-3920509 | 1,553,395,461,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00387.warc.gz | 881,804,592 | 23,120 | # Geometry: Secant & Tangent Theorem Doodle Graphic Organizer
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Doodle graphic organizer used to develop an understanding of the theorems associated with circle secants and tangents. Allow your students the ability to use visual cues to aid in memory and recall. The document prints on standard paper but can be trimmed to fit a composition notebook (that's how I use it). A key is included.
I have used math journals in my classroom for almost 20 years. I implemented interactive journals and struggled with the cut and paste and time involved in their creation. While I know there is a benefit, it did not work for me. These notes are the answer for my teaching style. While students are required to fill in vital information, they are not required to color. My students love getting an interesting handout instead of a generic math worksheet.
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Systems of Equations in 2 Variables
Solving Systems of Equations in 3 Variables
Cramer’s Rule
Sets of Real Numbers
Evaluating Determinants
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Polynomial Division: Long & Synthetic
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Logarithms and their Properties
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Referene & Coterminal Angles
Rational Expressions & Equations
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Square Root Functions
Standard Form of a Linear Equation
Graphing Piecewise Functions
Graphing Absolute Value Functions
Graphing Circles using Standard Form
Ellipses: Graphing & Equations
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Points, Lines, & Planes Packet
Angles & Parallel Lines and Transversals
Triangle Circumcenter, Incenter, Centroid Theorems
Triangle Angle-Sum & Exterior Angle Theorems
Isosceles & Equilateral Triangles
Triangle Congruence: CPCTC, SSS, SAS, AAS, ASA
Perpendicular Bisector & Angle Bisector Theorems
Triangle Inequality Theorem
Polygon Interior Angle Sum & Exterior Angle Sum Theorems
Parallelogram Theorems & Tests for Parallelograms
Rectangle Theorems
Rhombi & Squares
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Circle Terminology
Circle Central Angles & Arc Measure
Circle Arc Length
Circle Arcs & Chord Theorems
Circle Inscribed Angle Theorems
Tangets to a Circle
Secant & Tangent Theorems
Special Segments in a Circle Theorems
TRIGONOMETRY
Right Triangle Trig SOH CAH TOA
Finding Exact Trig Values
Trig Functions DEGREES Packet: Graphing & Equations
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 752 | 3,168 | {"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-2019-13 | latest | en | 0.813627 |
https://www.emathhelp.net/en/calculators/probability-statistics/ | 1,725,790,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00629.warc.gz | 733,162,460 | 6,372 | # Calculators - Probability/Statistics
## Average Calculator
For the given group of numbers, the calculator will find their average (arithmetic mean), with steps shown.
## Geometric Mean Calculator
For the given group of values, the calculator will find their geometric mean, with steps shown.
## Harmonic Mean Calculator
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For the given set of observations, the calculator will find their standard deviation (either sample or population), with steps shown.
## Sample/Population Coefficient of Variation Calculator
For the given data set, the calculator will find the sample or the population coefficient of variation (CV), with steps shown.
## Sample/Population Covariance Calculator
For the given two sets of values, the calculator will find the covariance between them (either sample or population), with steps shown.
## Correlation Coefficient Calculator
For the given two sets of values, the calculator will find the Pearson correlation coefficient between them (either sample or population), with steps shown.
## Percentile Calculator
For the given set of data, the calculator will find percentile no. $p$, with steps shown.
## Median Calculator
The calculator will find the median ($50$th percentile, second quartile) for the given sample data, with steps shown.
## Lower Quartile Calculator
For the given set of data, the calculator will find the lower (first) quartile ($25$th percentile), with steps shown.
## Upper Quartile Calculator
For the given set of data, the calculator will find the upper (third) quartile ($75$th percentile), with steps shown.
## Interquartile Range Calculator
For the given set of data, the calculator will find the interquartile range (IQR), with steps shown.
## Mode Calculator
For the given set of data, the calculator will find the mode, with steps shown.
## Range of a Data Set Calculator
For the given set of observations, the calculator will find their range, with steps shown.
## Five Number Summary Calculator
The calculator will find the five number summary for the given set of data (minimum value, first quartile, median, third quartile, maximum value), with steps shown.
## Box and Whisker Plot Calculator
The calculator will create the box and whisker plot for the given set of data, with steps shown. Two sets of data are supported.
## Percentile Rank, Class Rank Calculator
For the given set of data and score, the calculator will find the percentile class rank, with steps shown.
## Binomial Distribution Calculator
The calculator will find the simple and cumulative probabilities, as well as the mean, variance, and standard deviation of the binomial distribution.
## Poisson Distribution Calculator
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For the given mean and standard deviation, the calculator will find various probabilities for the random variable, and vice versa: for the specified probability, it will find the values of the random variable (inverse operation).
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The calculator will find the z-score (standardized score) given the unstandardized value, mean, and standard deviation of the population, with steps shown.
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The calculator will find the margin of error from the given sample size and distribution, with steps shown.
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The calculator will find the p-value for two-tailed, right-tailed and left-tailed tests from normal, Student's (T-distribution), chi-squared, and Fisher (F-distribution) distributions.
## Linear Regression Calculator
The calculator will find the line of best fit for the given set of paired data using the least squares method, with steps shown. | 957 | 4,888 | {"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} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.82695 |
https://www.coursehero.com/file/5567597/Graded-Homework-17/ | 1,498,379,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320443.31/warc/CC-MAIN-20170625064745-20170625084745-00620.warc.gz | 841,214,135 | 24,821 | Graded Homework 17
# Graded Homework 17 - Graded Homework 17 You have submitted...
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You have submitted this Homework 1 time (including this time). You may submit this Homework a total of 40 times and receive full credit. This homework will be due October 28 at 3am. This homework contains prediction/confidence interval questions and material pertaining to (violations of) assumptions about the error term, material that can be found on pages 75-79 in your course packet. Note: there is an equivalent version of the prediction and confidence interval formulas on your formula sheet compared to the reader page 78 and 79 you may find more helpful on a few of the problems. Question #1 It has been computed that the 95% confidence interval is [144.4, 154.2] for the average exam score when a student spent 10 hours on average per week studying for the class. The 90% prediction interval for a student who spent 10 hours on average per week studying for the class will be wider cannot be determined based on the provided information. narrower
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## This note was uploaded on 08/25/2009 for the course ECON 203 taught by Professor Petry during the Fall '08 term at University of Illinois at Urbana–Champaign.
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Ask a homework question - tutors are online | 375 | 1,667 | {"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-2017-26 | longest | en | 0.937106 |
https://jacobsphysics.blogspot.fi/2009_11_01_archive.html | 1,490,913,459,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218203536.73/warc/CC-MAIN-20170322213003-00495-ip-10-233-31-227.ec2.internal.warc.gz | 813,752,650 | 17,539 | Buy that special someone an AP Physics prep book: 5 Steps to a 5: AP Physics 1
Visit Burrito Girl's handmade ceramics shop, The Muddy Rabbit: Yarn bowls, tea sets, dinner ware...
## 30 November 2009
### Live PV diagrams!
It’s time to teach the ideal gas law, heat engines, and PV diagrams in AP physics. A lot of AP teachers are a bit intimidated by these topics. They’re more abstract than mechanics, and are farther divorced from our experience than, say, electricity and magnetism or waves and optics. I hope this next series of posts can help out.
Certainly your students have studied the ideal gas law in chemistry. But chances are, all they did (in their minds) was plug numbers into the equation PV=nRT. A major first step in teaching this unit is to give your class a firm understanding of the physical meaning of each of these variables.
You may or may not have seen Pasco’s heat engine / gas law apparatus. It consists of a low-friction piston that attaches to a metal cylinder – see the pasco.com picture above. I use it to demonstrate the ideal gas law and PV diagrams with live data collection.
The three variables to measure are pressure, volume, and temperature of the gas in the cylinder. I attach a Vernier pressure probe to one of the ports on the front to measure, um, pressure. Temperature can be taken care of with a Vernier temperature probe inserted into the hole in the stopper on top of the metal cylinder. It’s only volume measurement that’s truly tricky.
If you can measure the height of the piston, then the volume of the gas under the piston and in the cylinder can be calculated. Pasco provides an instruction packet that suggests the use of a rotary motion sensor or smart pulley to get the piston’s position. I don’t do that, though it should work fine.
Instead, I mount a motion detector above the piston. By measuring the height of the detector above the piston’s lowest point, I can set the Logger Pro software to calculate the volume of the gas automatically from the motion detector reading. Thus, I’m collecting volume, temperature, and pressure data as many as 20 times per second.
In the picture to the right, you can see me using this apparatus to demonstrate PV diagrams at last summer’s AP Summer Institute at the University of Georgia. (Thanks to Laura Englebert, a physics teacher from the Atlanta area, for sending me the pictures. Woo-hoo!) I told Logger Pro to graph pressure on the y-axis and volume on the x-axis. Then, I slowly raised the piston, taking care not to let my hand get in the way of the motion detector. The graph showed a nicely hyperbolic curve – an isothermal process. But then I let the piston compress the gas rapidly. When gas compresses (or expands) quickly enough that there’s not enough time for heat to flow into or out of the gas, the process is adiabatic. Adiabatic compression on a PV diagram should jump to a higher isotherm, because the temperature goes up. Sure enough, while the process happens too fast to define the adiabatic curve, you can see that the graph ends up at a higher product of PV.
If you’re a bit lost in that last paragraph, don’t worry, it will make more sense once you get a chance to study the four major types of thermodynamic process that are tested on the AP exam – isothermal, adiabatic, isobaric, and isovolumetric. The 5 Steps book (now in a new and much-edited edition!) gives a good, short, readable treatment of these processes.
But note anyway that ANY portion of the ideal gas law can be tested experimentally! The linear relationship between pressure and temperature at constant volume? Plunge the metal gas cylinder into boiling water while keeping the piston from expanding. The linear relationship between volume and temperature at constant pressure? Do the same thing, but instead allow the piston to rise. And the experiment I previously described shows the inverse relationship between pressure and volume at constant temperature! Cool, eh?
## 18 November 2009
### Mailbag -- thermodynamics sign convention?
From Jonathan Kirby, an Atlantan:
"I have a quick question for you. We are just getting into thermodynamics, and I was wondering if I should teach "ΔU = Q-W" where W is the work done BY the system, or if I should teach "ΔU = Q+W" where W is the work done ON the system. Which way would be better (if either) for the AP Test?"
The AP test changed to the ΔU = Q+W route in about 2002. Don't even mention the other way, unless your textbook does, in which case, good luck. :-)
(When I've used such a textbook, I've just repeated the correct definition over and over, and prayed.)
GCJ
## 13 November 2009
### Follow-up to multiple choice test corrections
Those of you who have attended my workshops know that, in Jacobs Physics, test corrections are one of the two most important components of the course. Sometimes, though, even the test corrections need correction.
Instead of assigning another round of “correction corrections,” I tend to just give the whole class a quiz when I find consistent misunderstandings. For example, consider the two multiple choice questions below. These were originally AAPT Physics Bowl questions, I believe…
1. A 2 kg object initially moving with a constant velocity is subjected to a force of magnitude F in the direction of motion. A graph of F as a function of time t is shown. What is the increase, if any, in the velocity of the object during the time the force is applied?
(A) 0 m/s
(B) 2.0 m/s
(C) 3.0 m/s
(D) 4.0 m/s
(E) 6.0 m/s
2. A deliveryman moves 10 cartons from the sidewalk, along a 10-meter ramp to a loading dock, which is 1.5 meters above the sidewalk. If each carton has a mass of 25 kg, what is the total work done by the deliveryman on the cartons to move them to the loading dock?
(A) 2500 J
(B) 3750 J
(C) 10 000 J
(D) 25 000 J
(E) 37 500 J
Many students showed an iffy grasp of these two questions on their test corrections. So, I posted to our class folder early last night. I noted that we would take a follow-up quiz today on these problems. I wrote the quiz to address specifically the mistakes that I had repeatedly seen on the first attempt at corrections. Here’s the quiz:
1. (a) What’s wrong with the statement “Work is done both up and to the right in order to move the boxes up the incline?”
(b) What is the direction of the force necessary to carry one box up the incline at constant speed? Justify your answer. Your justification should include a free body diagram.
2. (a) Explain why the average force during the time interval t = 1 s to t = 5 s is NOT 1.0 N.
(b) How do you get impulse from this graph WITHOUT trying to find an average force?
## 09 November 2009
### Going over a test
I know it's happened to you and it's frustrated you. You give back a test, you discuss one of the more frequently missed questions, hoping for a teachable moment. But half the class is rooting through the rest of the test, sitting back with a vacant expression, or simply absent mentally. What to do?
One option, which I've discussed before, is to allow corrections for half credit. Then there's no need for you to take too much class time to go over the test -- it's the students' job to figure out what they missed, and to convince you they understand now. A related idea is to announce a "fundamentals quiz" over commonly missed concepts from the test. Either way, the students are forced to think about the test beyond just "what did I get?"
Of course, test corrections are time- and manpower-intensive. You have to give time in or out of class to get the corrections done, you have to grade them as thoroughly as you would a test. I only do corrections in my AP class -- I find the general class moves slowly enough that those who missed important points will pick them up soon.
So how do I go over a test in general physics? Well, keeping my comments brief and to the point helps. But the key little trick is to HOLD THE TESTS IN MY HAND while I go over them.
Here I'm playing with the students' minds. They desperately want their tests back, but only so they can see the grade. Once they see that grade, their mind is done for a while, and they don't want to think about physics. So I use the grade as a carrot. I dangle the papers with the grades on them right in front of the class. Not obviously or obnoxiously, of course, but they are never sure when I'm going to shut up and hand out the tests. And, they're nervous about what they did right or wrong.
So they listen. And ask questions. They want to hear what I say, so they can figure out whether they were right or wrong. The same discussion AFTER I give the tests back would be fruitless.
How do I know this technique works? Well, I don't for sure. But I do note that folks occasionally note to their friends whether they did or didn't make the mistakes that I discussed... so they must have paid some attention.
GCJ | 2,054 | 8,921 | {"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-2017-13 | longest | en | 0.918867 |
https://canadam.math.ca/2021/program/abs/gt | 1,632,630,381,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057796.87/warc/CC-MAIN-20210926022920-20210926052920-00461.warc.gz | 208,330,790 | 4,761 | Game Theory
[PDF]
ALEXANDER CLOW, St. Francis Xavier University
From Poset Games to Partially Ordered Games [PDF]
Combinatorial games are 2 player games of no chance and perfect information. Under the normal play condition the player to have no move on their turn loses and a game is impartial if for every position both players have the same set of moves. In their paper Advances in Finding Ideal Play on Poset Games Clow and Finbow showed that poset games (a class of impartial combinatorial games) under normal play can be reduced to much simpler poset games in such a way that the score of the game (nimber) and optimal moves are preserved. This talk generalizes these results to all normal play games. In doing so both the notions of ordinal sums (whose partial order is a total order of cardinality $2$) and disjoint sums of games are generalized. Furthermore the Colon Principle, which deals with ordinal sums of games is generalized.
MOZHGAN FARAHANI, Memorial University of Newfoundland
The deduction game to capture robbers [PDF]
We study a variation of the game of cops and robbers on graphs in which the cops must capture an invisible robber in one move. Cops know each others' initial locations, but they can only communicate if they are on the same vertex. Thus, the challenge for the cops is to deduce the other cops' movement and move accordingly in order to capture the robbers and guarantee a win. We call this game the deduction game to capture robbers. In this talk, we introduce the deduction number as the minimum number of cops needed to capture all robbers and discuss the deduction number for some classes of graphs.
RYAN HAYWARD, University of Alberta
Let's Play Hex: Some Open Problems [PDF]
For the regular nxn Hex board, what are some winning first moves? What about for the misere version of this game, Reverse Hex? Is the world's strongest Hexbot stronger than the world's strongest Hex human? What happens when a deterministic Hex player plays a random Hex player on a board whose shape (more rows than columns) favours the random player? What's a best strategy for Dark Hex (also known as Kriegspiel Hex, where you don't see your opponent's moves) on the 4x4 board? I will discuss these open problems.
MASOOD MASJOODY, Simon Fraser University
Confining the Robber on Cographs [PDF]
In this talk, the notions of trapping and confining the robber on a graph, and the corresponding cop numbers are introduced. The latter two are easily seen to be lower bounds for the (regular) cop number of a graph. We present some structurally necessary conditions for graphs $G$ not containing the path on $k$ vertices ($P_k$-free graphs) so that $k-3$ cops do not have a strategy to capture or confine the robber on $G$. We show that for planar cographs and planar $P_5$-free graphs the confining cop number is at most one and two, respectively, and that the number of vertices of connected cographs having a confining cop number $\ge2$ has a tight lower-bound of eight. We conclude by posing two conjectures concerning the confining cop number of $P_5$-free graphs and the smallest planar graph of confining cop number of three.
JÉRÉMIE TURCOTTE, McGill University
Finding the smallest 4-cop-win graph(s) [PDF]
We study extremal graphs for the game of Cops and Robbers. It is well known that the smallest connected graph on which 3 cops are needed to capture the robber is the Petersen graph. Using both formal and computational methods, we determine the minimum order of connected 4-cop-win graphs, which confirms a conjecture of Andreae (1986), and later of Baird et al. (2014), and work towards the uniqueness of such graphs. Joint work with Samuel Yvon. | 846 | 3,691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-39 | latest | en | 0.930627 |
https://mathexamination.com/lab/unknotting-number.php | 1,623,730,509,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00556.warc.gz | 365,008,998 | 8,501 | ## Take My Unknotting Number Lab
As mentioned over, I used to compose a basic as well as uncomplicated math lab with only Unknotting Number However, the simpler you make your lab, the easier it ends up being to get stuck at the end of it, after that at the start. This can be really aggravating, and all this can take place to you since you are utilizing Unknotting Number and/or Modular Equations improperly.
With Modular Equations, you are currently utilizing the wrong formula when you get stuck at the beginning, otherwise, after that you are likely in a stumbling block, and there is no feasible escape. This will only worsen as the issue comes to be extra intricate, however then there is the inquiry of how to wage the issue. There is no other way to correctly deal with solving this type of mathematics trouble without having the ability to right away see what is going on.
It is clear that Unknotting Number as well as Modular Formulas are hard to find out, as well as it does take practice to create your very own sense of intuition. However when you wish to fix a mathematics problem, you have to make use of a device, as well as the devices for learning are utilized when you are stuck, as well as they are not used when you make the incorrect relocation. This is where lab Help Service is available in.
As an example, what is wrong with the inquiry is incorrect ideas, such as obtaining a partial value when you do not have enough working parts to complete the whole task. There is a very good reason that this was wrong, as well as it refers reasoning, not instinct. Reasoning enables you to comply with a detailed procedure that makes good sense, and when you make an incorrect step, you are usually forced to either attempt to go forward as well as fix the error, or attempt to go backward and do an in reverse step.
One more circumstances is when the pupil does not comprehend a step of a procedure. These are both sensible failures, as well as there is no chance around them. Even when you are stuck in an area that does not allow you to make any kind of type of move, such as a triangle, it is still vital to understand why you are stuck, so that you can make a much better step as well as go from the action you are stuck at to the following area.
With this in mind, the very best way to fix a stuck scenario is to just take the advance, as opposed to trying to go backward. Both procedures are various in their strategy, but they have some basic similarities. Nonetheless, when they are attempted together, you can rapidly tell which one is better at fixing the problem, and you can additionally tell which one is extra powerful.
Let's speak about the first instance, which relates to the Unknotting Number math lab. This is not also challenging, so allow's initial go over exactly how to begin. Take the adhering to procedure of connecting a part to a panel to be used as a body. This would call for 3 dimensions, and would be something you would need to attach as part of the panel.
Currently, you would certainly have an extra dimension, however that doesn't mean that you can simply maintain that measurement as well as go from there. When you made your very first step, you can quickly forget about the dimension, and afterwards you would need to go back as well as backtrack your actions.
Nevertheless, instead of remembering the additional measurement, you can utilize what is called a "psychological shortcut" to aid you remember that additional dimension. As you make your first step, imagine yourself taking the dimension as well as connecting it to the part you want to connect to, and after that see exactly how that makes you really feel when you duplicate the procedure.
Visualisation is a very effective method, and is something that you should not miss over. Picture what it would feel like to actually affix the component and have the ability to go from there, without the dimension.
Currently, let's consider the second instance. Let's take the same procedure as in the past, now the pupil needs to remember that they are going to return one step. If you tell them that they need to return one action, however then you get rid of the idea of needing to return one action, then they won't know exactly how to proceed with the issue, they won't understand where to look for that step, and the procedure will certainly be a mess.
Instead, use a mental shortcut like the mental diagram to psychologically show them that they are going to move back one step. and also put them in a setting where they can move forward from there. without having to think about the missing a step.
## Hire Someone To Do Your Unknotting Number Lab
" Unknotting Number - Need Aid With a Mathematics lab?" However, numerous students have actually had an issue realizing the ideas of straight Unknotting Number. The good news is, there is a new format for straight Unknotting Number that can be utilized to show straight Unknotting Number to trainees that deal with this concept. Students can make use of the lab Assist Solution to help them find out brand-new techniques in direct Unknotting Number without facing a mountain of problems and without having to take an examination on their ideas.
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## Pay Someone To Take My Unknotting Number Lab
Unknotting Number is used in a great deal of colleges. Some teachers, however, do not utilize it extremely efficiently or utilize it improperly. This can have an adverse effect on the pupil's discovering.
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An assistance service can make it easier for you to handle your Unknotting Number class. Additionally, you can discover more about yourself and assist you prosper. Locate the very best tutoring service as well as you will have the ability to take your research abilities to the following level. | 2,500 | 12,454 | {"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-2021-25 | latest | en | 0.977893 |
http://www.netlib.org/lapack/explore-3.2-html/zggrqf.f.html | 1,544,841,444,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00312.warc.gz | 429,256,725 | 4,376 | ```001: SUBROUTINE ZGGRQF( M, P, N, A, LDA, TAUA, B, LDB, TAUB, WORK,
002: \$ LWORK, INFO )
003: *
004: * -- LAPACK routine (version 3.2) --
005: * Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
006: * November 2006
007: *
008: * .. Scalar Arguments ..
009: INTEGER INFO, LDA, LDB, LWORK, M, N, P
010: * ..
011: * .. Array Arguments ..
012: COMPLEX*16 A( LDA, * ), B( LDB, * ), TAUA( * ), TAUB( * ),
013: \$ WORK( * )
014: * ..
015: *
016: * Purpose
017: * =======
018: *
019: * ZGGRQF computes a generalized RQ factorization of an M-by-N matrix A
020: * and a P-by-N matrix B:
021: *
022: * A = R*Q, B = Z*T*Q,
023: *
024: * where Q is an N-by-N unitary matrix, Z is a P-by-P unitary
025: * matrix, and R and T assume one of the forms:
026: *
027: * if M <= N, R = ( 0 R12 ) M, or if M > N, R = ( R11 ) M-N,
028: * N-M M ( R21 ) N
029: * N
030: *
031: * where R12 or R21 is upper triangular, and
032: *
033: * if P >= N, T = ( T11 ) N , or if P < N, T = ( T11 T12 ) P,
034: * ( 0 ) P-N P N-P
035: * N
036: *
037: * where T11 is upper triangular.
038: *
039: * In particular, if B is square and nonsingular, the GRQ factorization
040: * of A and B implicitly gives the RQ factorization of A*inv(B):
041: *
042: * A*inv(B) = (R*inv(T))*Z'
043: *
044: * where inv(B) denotes the inverse of the matrix B, and Z' denotes the
045: * conjugate transpose of the matrix Z.
046: *
047: * Arguments
048: * =========
049: *
050: * M (input) INTEGER
051: * The number of rows of the matrix A. M >= 0.
052: *
053: * P (input) INTEGER
054: * The number of rows of the matrix B. P >= 0.
055: *
056: * N (input) INTEGER
057: * The number of columns of the matrices A and B. N >= 0.
058: *
059: * A (input/output) COMPLEX*16 array, dimension (LDA,N)
060: * On entry, the M-by-N matrix A.
061: * On exit, if M <= N, the upper triangle of the subarray
062: * A(1:M,N-M+1:N) contains the M-by-M upper triangular matrix R;
063: * if M > N, the elements on and above the (M-N)-th subdiagonal
064: * contain the M-by-N upper trapezoidal matrix R; the remaining
065: * elements, with the array TAUA, represent the unitary
066: * matrix Q as a product of elementary reflectors (see Further
067: * Details).
068: *
069: * LDA (input) INTEGER
070: * The leading dimension of the array A. LDA >= max(1,M).
071: *
072: * TAUA (output) COMPLEX*16 array, dimension (min(M,N))
073: * The scalar factors of the elementary reflectors which
074: * represent the unitary matrix Q (see Further Details).
075: *
076: * B (input/output) COMPLEX*16 array, dimension (LDB,N)
077: * On entry, the P-by-N matrix B.
078: * On exit, the elements on and above the diagonal of the array
079: * contain the min(P,N)-by-N upper trapezoidal matrix T (T is
080: * upper triangular if P >= N); the elements below the diagonal,
081: * with the array TAUB, represent the unitary matrix Z as a
082: * product of elementary reflectors (see Further Details).
083: *
084: * LDB (input) INTEGER
085: * The leading dimension of the array B. LDB >= max(1,P).
086: *
087: * TAUB (output) COMPLEX*16 array, dimension (min(P,N))
088: * The scalar factors of the elementary reflectors which
089: * represent the unitary matrix Z (see Further Details).
090: *
091: * WORK (workspace/output) COMPLEX*16 array, dimension (MAX(1,LWORK))
092: * On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
093: *
094: * LWORK (input) INTEGER
095: * The dimension of the array WORK. LWORK >= max(1,N,M,P).
096: * For optimum performance LWORK >= max(N,M,P)*max(NB1,NB2,NB3),
097: * where NB1 is the optimal blocksize for the RQ factorization
098: * of an M-by-N matrix, NB2 is the optimal blocksize for the
099: * QR factorization of a P-by-N matrix, and NB3 is the optimal
100: * blocksize for a call of ZUNMRQ.
101: *
102: * If LWORK = -1, then a workspace query is assumed; the routine
103: * only calculates the optimal size of the WORK array, returns
104: * this value as the first entry of the WORK array, and no error
105: * message related to LWORK is issued by XERBLA.
106: *
107: * INFO (output) INTEGER
108: * = 0: successful exit
109: * < 0: if INFO=-i, the i-th argument had an illegal value.
110: *
111: * Further Details
112: * ===============
113: *
114: * The matrix Q is represented as a product of elementary reflectors
115: *
116: * Q = H(1) H(2) . . . H(k), where k = min(m,n).
117: *
118: * Each H(i) has the form
119: *
120: * H(i) = I - taua * v * v'
121: *
122: * where taua is a complex scalar, and v is a complex vector with
123: * v(n-k+i+1:n) = 0 and v(n-k+i) = 1; v(1:n-k+i-1) is stored on exit in
124: * A(m-k+i,1:n-k+i-1), and taua in TAUA(i).
125: * To form Q explicitly, use LAPACK subroutine ZUNGRQ.
126: * To use Q to update another matrix, use LAPACK subroutine ZUNMRQ.
127: *
128: * The matrix Z is represented as a product of elementary reflectors
129: *
130: * Z = H(1) H(2) . . . H(k), where k = min(p,n).
131: *
132: * Each H(i) has the form
133: *
134: * H(i) = I - taub * v * v'
135: *
136: * where taub is a complex scalar, and v is a complex vector with
137: * v(1:i-1) = 0 and v(i) = 1; v(i+1:p) is stored on exit in B(i+1:p,i),
138: * and taub in TAUB(i).
139: * To form Z explicitly, use LAPACK subroutine ZUNGQR.
140: * To use Z to update another matrix, use LAPACK subroutine ZUNMQR.
141: *
142: * =====================================================================
143: *
144: * .. Local Scalars ..
145: LOGICAL LQUERY
146: INTEGER LOPT, LWKOPT, NB, NB1, NB2, NB3
147: * ..
148: * .. External Subroutines ..
149: EXTERNAL XERBLA, ZGEQRF, ZGERQF, ZUNMRQ
150: * ..
151: * .. External Functions ..
152: INTEGER ILAENV
153: EXTERNAL ILAENV
154: * ..
155: * .. Intrinsic Functions ..
156: INTRINSIC INT, MAX, MIN
157: * ..
158: * .. Executable Statements ..
159: *
160: * Test the input parameters
161: *
162: INFO = 0
163: NB1 = ILAENV( 1, 'ZGERQF', ' ', M, N, -1, -1 )
164: NB2 = ILAENV( 1, 'ZGEQRF', ' ', P, N, -1, -1 )
165: NB3 = ILAENV( 1, 'ZUNMRQ', ' ', M, N, P, -1 )
166: NB = MAX( NB1, NB2, NB3 )
167: LWKOPT = MAX( N, M, P )*NB
168: WORK( 1 ) = LWKOPT
169: LQUERY = ( LWORK.EQ.-1 )
170: IF( M.LT.0 ) THEN
171: INFO = -1
172: ELSE IF( P.LT.0 ) THEN
173: INFO = -2
174: ELSE IF( N.LT.0 ) THEN
175: INFO = -3
176: ELSE IF( LDA.LT.MAX( 1, M ) ) THEN
177: INFO = -5
178: ELSE IF( LDB.LT.MAX( 1, P ) ) THEN
179: INFO = -8
180: ELSE IF( LWORK.LT.MAX( 1, M, P, N ) .AND. .NOT.LQUERY ) THEN
181: INFO = -11
182: END IF
183: IF( INFO.NE.0 ) THEN
184: CALL XERBLA( 'ZGGRQF', -INFO )
185: RETURN
186: ELSE IF( LQUERY ) THEN
187: RETURN
188: END IF
189: *
190: * RQ factorization of M-by-N matrix A: A = R*Q
191: *
192: CALL ZGERQF( M, N, A, LDA, TAUA, WORK, LWORK, INFO )
193: LOPT = WORK( 1 )
194: *
195: * Update B := B*Q'
196: *
197: CALL ZUNMRQ( 'Right', 'Conjugate Transpose', P, N, MIN( M, N ),
198: \$ A( MAX( 1, M-N+1 ), 1 ), LDA, TAUA, B, LDB, WORK,
199: \$ LWORK, INFO )
200: LOPT = MAX( LOPT, INT( WORK( 1 ) ) )
201: *
202: * QR factorization of P-by-N matrix B: B = Z*T
203: *
204: CALL ZGEQRF( P, N, B, LDB, TAUB, WORK, LWORK, INFO )
205: WORK( 1 ) = MAX( LOPT, INT( WORK( 1 ) ) )
206: *
207: RETURN
208: *
209: * End of ZGGRQF
210: *
211: END
212: ``` | 2,748 | 8,335 | {"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-2018-51 | latest | en | 0.465362 |
http://www.jiskha.com/search/index.cgi?query=Probability&page=5 | 1,477,126,872,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718866.34/warc/CC-MAIN-20161020183838-00016-ip-10-171-6-4.ec2.internal.warc.gz | 540,932,928 | 14,777 | Saturday
October 22, 2016
# Search: Probability
Number of results: 9,636
Math
If I have a random stack of 5 quarters, the 1st 3 quarters are heads. What is the probability the the 4th quarter would be another head What is the probability the the 4th quarter would be a tail Is there a way to show the probability or odds that you would continue to keep ...
September 23, 2015 by Jeff
maths!
6) There are three research and development projects focusing on the development of three new products A, B and C. The markets into which the new products are to be launched are identical. However, consumer preference is at present, unknown. A probability assessment has been ...
December 9, 2009 by mozza
probability
a teacher on her way to work passes through 3 stoplights each morning. the distances between the stop lights are great and the lights operate independently of each other. if the probability of the red lights are .4,.8 and .6 for each light, what is the probability to the ...
February 25, 2014 by Lindsey
statistics
The number of independent errors in a transmitted message of n symbols follows a Binomial distribution. The probability of an error in transmission is .001. If 2,000 symbols were sent, a) Find the expected number of errors. b) Write out the formula for the distribution of ...
September 4, 2011 by Layla
Algebra
Can anybody help me on this problem for my Algebra 2 class, I am having a hard time tyring to solve it step by step. The probability that an archer hits the target is p = 0.8, so the probability that he misses the target is q = 0.2. It is known that in this situation the ...
August 11, 2015 by Selena
Algebra
Can anybody help me on this problem for my Algebra 2 class, I am having a hard time tyring to solve it step by step. The probability that an archer hits the target is p = 0.8, so the probability that he misses the target is q = 0.2. It is known that in this situation the ...
August 12, 2015 by Selena
Math - Probability
You invent a new system to make excellent racing tires for the Indianapolis 500/ Your tires are good for an average of 105 miles, with a standard deviation of 7.5 miles. What is the probability that one of your tires can go for at least 44 laps? What is the probability that ...
July 9, 2012 by Laura
help... probability
The Smith family has 4 children that are all girls. What is the probability that the next one will be a girl? You plan on having four children and you want all boys. If you have children, what is the probability you will get four boys?
October 16, 2007 by maggie r
Statistics
(a) With n=12 and p =0.4 find the binomial probability that p(9) by using a binomial probability table. (b) np ¡Ý5, nq¡Ü5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial, is np < 5 or nq < 5 than state the ...
June 1, 2013 by Jenn
stastistics
With n=13 and p=0.7, find the binomial probability P(9)by using a binomial probability table. If np> and nq>5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial,if np<5 of nq<5 then state that the normal ...
November 22, 2013 by angela
statistics
With n=13 and p= 0.7, find the binomial probability p(9) by using a binomial probability table. If np> and nq> 5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial, if np<5 or nq<5 then state that the ...
November 25, 2013 by angela
Probability
Suppose that the probability is 1 in 3,900,000 that a single auto trip in the United States will result in the death of the driver. Over a lifetime, an average U.S. driver takes 50,000 trips. Assume that events are independent: (a) What is the probability of such a driver ...
October 25, 2010 by Laura
probability
Suppose that the probability is 1 in 3,900,000 that a single auto trip in the United States will result in the death of the driver. Over a lifetime, an average U.S. driver takes 50,000 trips. Assume that events are independent: (a) What is the probability of such a driver ...
October 26, 2010 by Laura
Probability
The probability that Kate goes to the cinema is 0.2.If Kate does not go Claire goes. If Kate goes to the cinema the probability that she is late home is 0.3. If Clair goes to the cinema the probability that she is late home is 0.6. Calculate the probability that..... The ...
February 22, 2014 by Anonymous
Probability
The probability that Kate goes to the cinema is 0.2.If Kate does not go Claire goes. If Kate goes to the cinema the probability that she is late home is 0.3. If Clair goes to the cinema the probability that she is late home is 0.6. Calculate the probability that..... The ...
February 22, 2014 by Anonymous
Probability
Two group of male students cost their past on a particular proposal.the result are as follows: Group infavour against A 128 32 B 96 48 <a> if a student in favour of proposal is selected for a post. What is the probability that he is from group A ? if a student is chosen ...
February 4, 2012 by Johnson
bio
The probability that their baby will be a carrier of the gene for PKU is (A) ___3/4_______. The probability that their baby will be affected by PKU is (B) __________. If both genotypes were Pp, the probability that their baby would be affected by PKU would be (C) ___1/4______...
May 30, 2014 by carrie
MATH (Probability)
Jackson and Alice are the boss and secretary respectively working in an office for a firm. At any particular time during office hours, the probability that Jackson is in the office is 0.82 and probability that Alice is in the office is 0.98. The probability that Jackson is in ...
August 31, 2011 by FAKAAPO
Statistics
Bart is taking two classes, backpacking and biology. His probability of passing backpacking is 0.54, that of failing biology is 0.83, and that of passing at least one of the two courses is 0.65. Find the probability of the following. A. Joe will pass both courses. B. Joe will ...
April 2, 2007 by Mary
probability help!!
buses arrive at a bus stop at 15minute interval,starting from 7am if a passenger arrived at a times that is his uniformly distributed (a):find the probability that the passenger wait less than 5minute for a bus? (b):find the probability that the passenger wait more than ...
May 7, 2016 by dam
Mathematics
Please help... In a 200 m swimming race between 8 boys, the probability of Omar winning the race if he is in one of the two outside lanes is 1/4. If he is in any of the other lanes, the probability of Omar winning is 1/3. If the lanes are drawn at random, what is the overall ...
May 29, 2007 by Elizabeth
statistics
S) Suppose we want to determine the (binomial) probability (p) of getting 5 heads in 15 flips of a 2-sided coin. Using the Binomial table in the appendix of the text, what values of n, x, and p would we use to look up this probability, and what would be the probability? n=...
February 17, 2015 by slomomo
Probability
Find the probability of obtaining between 50 and 55 tails inclusive when a fair coin is flipped 80 times. (answer accurate to four decimal places.)
November 22, 2009 by Anna
College Algebra
The letters in the word probability are written on separate slips of paper. Find the probability of drawing two i's in a row without replacement.
December 11, 2013 by Sarah
Choose the term to complete this sentence?
1 Describes how likely it is that an event will happen based on all the possible outcomes. Theoretical probability or Experimental probability
June 3, 2014 by nonononono
Math
Determine the probability of drawing two cards from a deck( without replacement) and getting the same suit using 1) conditional probability 2) combinations
July 14, 2015 by Helen
Math (Probability)
A spinner is divided into x equal sectors, numbered 1 through x . The probability of spinning a number less than four is 1/10. Determine the value of x
August 18, 2015 by Zena
probability
The probability of mary,esther and john coming to school late on monday are 1/4,2/5 and1/3 respectively.draw a tree diagram to represent the information .
April 21, 2016 by Starn
probability theory
3 students; A, B and C, are in a swimming race. A and B have the same probability of winning and each is twice as likely to win as C. Find the probability that B or C wins the swimming race. 3) Given three boxes as follows: Box A contains 3 red, 2 white and 2 blue balls. Box B...
November 20, 2015 by charles
I am having trouble with this homework question ,
Here is the probability model for the blood type of a randomly chosen person in the United States. Blood type O A B AB Probability 0.53 0.21 0.03 0.23 What is the probability that a randomly chosen American does not have type O blood? % Round to the nearest 0.01%
October 1, 2016 by Amanda
statistics
When someone buys a ticket for an airline flight there is a 0.0995 probability that the person will not show up for the flight (based on data from the IBM research paper by Lawrence, Hong, and Cherrier). An agent for Air America want to book 24 persons on an airplane that can ...
March 1, 2012 by meagan
probability
What is the probability that an 8 digit numbers contains "97"?
June 20, 2013 by steph
probability
Find the probability of P(E or F) if E and F are mutually exclusive P(E)=0.41 and P(F)=0.47
June 17, 2014 by carol
Probability
What is the probability of choosing an odd # from A that is AnB? A{1,2,3,4,7,9} B{3,7,9,11,12,15}
July 30, 2015 by Seany
Probability
What is the probability of choosing an odd # from A that is AnB? A{1,2,3,4,7,9} B{3,7,9,11,12,15}
August 6, 2015 by Seany
math (ms sue)
Suppose that one voter in the 2010 UK General Election was selected at random. The probability that the person voted for the Conservative Party was 0.36, the probability that the person voted for the Labour Party was 0.29, and the probability that the person voted for the ...
March 30, 2016 by Sam
Math - Probability
Annabelle wants to move to a new city. If her chance of getting a job in Montreal is 10%, and her chance of getting a job in Vancouver is 35%: 3a) What is the probability that Annabelle will be offered both jobs? 3b) What is the probability that she will only get one of the jobs?
May 19, 2013 by Jess
Counting and Probability
One member of the debate team is going to be chosen President. Each member is equally likely to be chosen. The probability that a boy is chosen is 2/3 the probability that a girl is chosen. Girls make up what fraction of the debate team?
November 16, 2015 by Mr. Alexander
math
The company will use machine A,B,C,D with probabilities 0.23,0.45,0.19 and 0.13 respectively.From past experience it is known that the probability of defective tablet produced by the machine are 0.11,0.2,0.07 and 0.05 respectively.Suppose a defective tablet is produced：...
May 15, 2016 by koro
probability
The probability that Alysha will miss school today is 0.15. The probability that Carla will miss school today is 0.07. The probability that both of these students will NOT miss school today is what?
November 14, 2009 by Anna
statitics
you are going to play two games the probability you win the first game is 0.60 if you win the first game the probability you will win the second game is 0.75 if you lose the first game the probability you win the second game is 0.55. what is the probability you win exactly one...
July 29, 2016 by bernie
statistic
You draw one card from a shuffled standard deck or cards (cards are in four suits-red diamonds, red heart, black clubs and black spades-and are numbered 2,3,4,5,6,7,8,9,10, J,O,K,A). A.What is the probability of drawing a red card B.What is the probability of drawing a queen c...
March 14, 2014 by Helen
statistic
You draw one card from a shuffled standard deck or cards (cards are in four suits-red diamonds, red heart, black clubs and black spades-and are numbered 2,3,4,5,6,7,8,9,10, J,O,K,A). A.What is the probability of drawing a red card B.What is the probability of drawing a queen c...
March 16, 2014 by HELEN
probability
Before leaving for work, Victor checks the weather report in order to decide whether to carry an umbrella. The forecast is “rain" with probability 20% and “no rain" with probability 80%. If the forecast is “rain", the probability of actually having rain on that day is 80%. On ...
February 12, 2014 by abc
Telecommunications
We send a frame of 256 bits. If the probability that a bit changes in transmission is 0.001 and each bit is independent. a. What is the probability that exactly 6 bits change? b. What is the probability that exactly 0 bits change ?
October 18, 2013 by liam
math
An ancient Korean drinking game involves a 14-sided die. The players roll the die in turn and must submit to whatever humiliation is written on the up-face: something like "Keep still when tickled on face." Six of the 14 faces are squares. Let's call them A, B, C, D, E, and F ...
May 15, 2016 by AnonYmous
stat
In a certain high school, the probability that a student drops out is 0.08, and the probability that a dropout gets a high-school equivalency diploma (GED) is 0.33. What is the probability that a randomly selected student gets a GED?
February 17, 2012 by shan
Maths
1)A two-figure number is written down at random. Find the probability that a)the number is greater than 44 b)the number is less than 100 2)A letter is picked at random from the english alphabet. Find the probability that a)the letter is a vowel (my answer=5/26) b)the letter ...
January 28, 2008 by Anonymous
Economics
The company estimates the probability of no damage to be 0.60, the probability of damage between \$0 and \$10,000 to be 0.25, and the probability of damage between \$10,000 and \$25,000 to be 0.12. If the company wants to make a profit of \$200 above the expected cost, what should...
July 1, 2012 by Sonya
stats again
ok I get it, so once I get the prob figred for one person, how do I figure it if additional attempts are made and the question is about the prob of at least one of the additional attempts say n = 6 getting the same results as the first Once you have the probability of passing ...
July 5, 2007 by holly
Maths
The probability of passing geography in the examination is 7/10 and the probability of passing agriculture after one has passed geography is 5/7. The probability of passing agriculture after one failed geography is 1/6. Calculate the probability of a student passing agriculture.
February 12, 2016 by Christopher
statistics
Use of normal distribution to approximate the desired probability. Find the probability of getting at least 30 fives in 200 tosses of a fair 6 sided die.
October 8, 2010 by Lori
Probability and statistics
4. The probability that a prime number occurs on at least 2 tosses when a balanced dice is tossed 5 times independently is equal to. Choose one answer a. 26/32 b. 30/32 c. 6/32 d. 16/32
March 2, 2012 by Arun
maths(Probability)
Two fairly six~sided dice are rolled,find the probability that their total score is (a)2 (b)7 (c)a prime number (d)a perfect square. Solutions. (a)2/6=1/3 (b)7/6
February 11, 2015 by eliphasjnr
math
At a quality control checkpoint on a manufacturing assembly line, 8% of the items failed check A, 10% failed check B, and 2% failed both checks A and B. a. If a product failed check A, what is the probability that it also failed check B? b. If a product failed check B, what is...
January 9, 2010 by crystal
Math
At a quality control checkpoint on a manafacturing assembly line, 10% of the items failed check A, 12% failed check B, and 3% failed both checks A and B. a. If a product failed check A, what is the probability that it also failed check B? b. If a product failed check B, what ...
December 8, 2010 by Rena
Math
At a quality control checkpoint on a manafacturing assembly line, 10% of the items failed check A, 12% failed check B, and 3% failed both checks A and B. a. If a product failed check A, what is the probability that it also failed check B? b. If a product failed check B, what ...
December 8, 2010 by Rena
Statistics (41)
A market research firm conducts telephone surveys with a 42% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions? In other words, what is the probability that the ...
May 6, 2014 by Mia
Economics (40)
A market research firm conducts telephone surveys with a 42% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions? In other words, what is the probability that the ...
May 9, 2014 by Mia
math probability
Assume that every time you buy an item of the Hong Kong Disney series, you receive one of the four types of cards, each with a cartoon character Mickey, Minnie, Donald and Daisy with an equal probability. Over a period of time, you buy 6 items of the series. What is the ...
September 27, 2016 by matthew
math (stats)
According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. [Round your answers to three decimal places, for example: 0.123] Compute the probability that a randomly ...
October 2, 2016 by b
Alg2/trig
Hi, I hope you guys don't mind if I post a couple probability questions. A child returns a five-volume set of books to a bookshelf. The child is not able to read, and so cannot distinguish one volume from another. What is the probability that the books are shelved in the ...
March 24, 2009 by Chris
Stats
The probability that a person has blue eyes is 16%. Three unrelated people are selected at random. a. Find the probability that all three have blue eyes: b. Find the probability that none of the three have blue eyes c. Find the probability that one of the three has blue eyes
April 30, 2013 by Dave
math
Marley drives to work every day and passes two independently operated traffic lights. The probability that both lights are red is 0.35. The probability that the first light is red is 0.48. What is the probability that the second light is red, given that the first light is red?
May 9, 2014 by nyshaw
Statistics
10 A color candy was chosen randomly out of a bag. Below are the results: Color Probability Blue 0.30 Red 0.10 Green 0.15 Yellow 0.20 Orange ??? a. What is the probability of choosing a yellow candy? b. What is the probability that the candy is blue, red, or green? c. What is ...
November 29, 2014 by Den
statistics
I am having trouble figuring out how to do this type of problem. Here is the current problem I am working on: the probability that a tomato seed will germinate is 60%. A gardener plants in batches of 12. A.) what is the probaility that exactly 10 seeds will germinate? B.) what...
January 2, 2010 by Pam
probability
Make a histogram showing the binomial probability distribution for the following: A waiter at the Green Spot resturant has learned from long experience that the probability that a lone diner will leave a tip is only 0.7. During one lunch hour he serves six people who are ...
March 4, 2016 by unclesam
Discrete Math
12 A computer science department has a probability of 0.35 that a senior receives a job offer in IT before graduation. Random select 8 senior students • What is the probability that 5 students received offers before graduation? • What is the probability only 1 student received...
April 30, 2015 by Peace
algebra
you are going to the Maryland state fair and would like to play some carnival games with the hopes of winning a prize .your mom said you are allowed to play one game only .your task is to analyze two carnival games and predict the probability of winning each game. you will ...
June 9, 2016 by ester
Quantitative Analysis for Business
The ages of a group of 50 women are approximately normally distributed with a mean of 48 years and a standard deviation of 5 years. One women is randomly selected from the group, and her age is observed. A) Find the probability that her age will fall between 56 and 60 years. B...
May 14, 2013 by Silvia
Probability - CHECK ANSWERS
I have an eight pack of crayons. Colors red, green, blue, orange, black, gray, pink, and purple. What is the probability of choosing the green crayon? 12.5 percent What is the probability of choosing the red, blue, or gray crayon? 37.5 percent What is the probability of not ...
October 15, 2010 by Jessica
Probability
Start with a standard, uniformly shuffled 52-card deck, and draw the cards one at a time. What is the probability of drawing a King strictly before getting a Spade?
July 10, 2011 by Anonymous
Stats
Use the normal distribution to approximate the following Binomial probability. A product is manufactured in batches of 100 and a defect rate of 8%. Find the probability of more then 7 defects.
June 11, 2015 by cindy
statistics
34. Two students (say A, and B) are challenged to solve a problem in statistics. Suppose that the probability that A will solve the problem is 40%, the probability that B will solve the problem is 55%, and the probability that A or B will solve the problem is 25%. The ...
June 30, 2015 by meri
statistics
The Denver Post stated that 80% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery chain introduces 66 new products, what is the probability that within 2 years , 7 or more fail? What is the probability that within 2 ...
March 23, 2012 by lola
Math
Suppose there is a bag containing 25 envelopes and 10 of the envelopes contain prizes. You reach in, without looking, and select 2 envelopes, without replacement. a) What is the probability that both of your picks are winner? b) What is the probability that none of your picks ...
March 22, 2013 by Anonymous
Algebra
Brandon flips two fair coins multiple times and records the frequency of the results in the following table. 50 trials|125trials|150trials HH 10 33 37 TH 9 31 38 HT 13 28 38 TT 18 33 37 DETERMINE THE EXPERIMENTAL PROBABILITY AND THEORETICAL PROBABILITY FOR EACH OUTCOME IN EACH...
May 22, 2016 by Blake
Statistics
Service times for customers at a post office follow some right-skewed distribution with mean 2.91 minutes and standard deviation 1.74 minutes. (a) Can you calculate the probability that the average service time for the next two customers is less than 2.64 minutes? If you can...
November 4, 2014 by Mel
Math
There is a 60% that your math teacher will give a pop quiz tonight and, due to the fact that you have a soccer game, there is an 85% chance you will not be able to do your math homework. You decide that you will do the homework if the probability of there being a pop quiz and ...
October 8, 2015 by Hannah
Probability
Before leaving for work, Victor checks the weather report in order to decide whether to carry an umbrella. On any given day, with probability 0.2 the forecast is “rain" and with probability 0.8 the forecast is “no rain". If the forecast is “rain", the probability of actually ...
February 13, 2015 by qwerty
There is a 60% that your math teacher will give a pop quiz tonight and, due to the fact that you have a soccer game, there is an 85% chance you will not be able to do your math homework. You decide that you will do the homework if the probability of there being a pop quiz and ...
October 8, 2015 by Hannah
math
What is the probability of getting an even number when rolling a six-sided number cube? 0.5 5% 2/6 30% You place the letters for the word smart in a bag. What is the probability of choosing a letter that is NOT a vowel? (remember vowels A, E, I, O, and U.) 0.2 1/5 80% 0.75 You...
April 27, 2016 by FluttershyK22
Math
Carol makes a spinner for a game. The spinner is divided into 10 equal sections. Each section is shaded 1 of these colors: green, red, yellow, or blue. On this spinner, the probability of spinning green is 3 times as great as the probability of spinning red. The probability of...
March 20, 2016 by Mason
Statistics
Suppose you conduct a study and find that the probability of having a baby boy is 60%. Now suppose three of your relatives are going to have babies. a) Build a tree diagram showing all the conditional probabilities and joint probabilities associated with the sex of the three ...
May 10, 2008 by Danny
Statistics
Suppose you conduct a study and find that the probability of having a baby boy is 60%. Now suppose three of your relatives are going to have babies. a) Build a tree diagram showing all the conditional probabilities and joint probabilities associated with the sex of the three ...
May 11, 2008 by Danny
Maths
The probability that Don's car passes the MOT test next time is 4/5, that Mag's car passes is 5/7 and that Joe's car passes is 1/2. Find the probability that a)all three will pass b)just two out of the three will pass c)no car passes? (might need a probability tree) PLEASE HELP!
February 4, 2008 by Anonymous
probabilities
A radioactive source of material emits a radioactive particle with probability 1/100 in each second. Let X be the number of particles emitted in one minute. 1. Determine the probability distribution, the expected value and the variance of X. 2. Calculate the probability that ...
April 26, 2016 by hassan
statistics
Find the indicated binomial probabilities. Round to the nearest 3 decimal places. In a local college, 20% of the math majors are women. Ten math majors are chosen at random. 1) What is the probability that exactly 2 are women? 2) What is the probability that 2 or less women ...
What is the probability of getting an even number when rolling a six-sided number cube? 0.5** 5% 2/6 30% You place the letters for the word smart in a bag. What is the probability of choosing a letter that is NOT a vowel? (remember vowels A, E, I, O, and U.) 0.2 1/5** 80% 0.75...
May 10, 2016 by #Blessed
math
of a set of 10 cards, 3 have red circles, 4 have yellow triangles, and 3 have green diamonds. you choose one card at random and then replace it. then you choose a second card. what is the probability that both cards have yellow triangles? Find the probability of obtaining the ...
April 16, 2007 by dillon
Math- probability
In a city, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is 3/4. If it's cloudy on any given day, then the probability that the next day will be cloudy is 2/3. a) In the long run, what ...
May 15, 2015 by Anonymous
statistics
Consider a binomial experiment with 20 trials and probability 0.45 on a single trial. Use the normal distribution to find the probability of exactly 10 successes. Round your answer to the thousandths place.
May 17, 2011 by coya
statistics
Consider a binomial experiment with 20 trials and probability 0.45 on a single trial. Use the normal distribution to find the probability of exactly 10 successes. Round your answer to the thousandths place.
August 2, 2011 by lee
PROBABILITY MATH
Find the probability of correctly answering the first 4 questions on a multiple choice test if random guesses are made and each question has 5 possible answers. Explain to me the answer?
November 9, 2011 by HM
Math Probability
Ben is greeting customers at a music store. Of the first 20 people he sees enter the store, 9 are wearing jackets and 11 are not. What is the experimental probability that the next person to enter the store will be wearing a jacket? Enter your answer as a simplified fraction. ...
Algebra
PLEASE HELP ME :( Brandon flips two fair coins multiple times and records the frequency of the results in the following table. 50 trials|125trials|150trials HH 10 33 37 TH 9 31 38 HT 13 28 38 TT 18 33 37 DETERMINE THE EXPERIMENTAL PROBABILITY AND THEORETICAL PROBABILITY FOR ...
May 23, 2016 by Blake
Brandon flips two fair coins multiple times and records the frequency of the results in the following table. 50 trials|125trials|150trials HH 10 33 37 TH 9 31 38 HT 13 28 38 TT 18 33 37 DETERMINE THE EXPERIMENTAL PROBABILITY AND THEORETICAL PROBABILITY FOR EACH OUTCOME IN EACH... | 7,190 | 28,169 | {"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-2016-44 | longest | en | 0.965667 |
http://www.nmr-relax.com/api/4.1/lib.diffusion.weights-module.html | 1,606,811,042,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00350.warc.gz | 143,352,737 | 5,364 | [frames] | no frames]
# Module weights
source code
Functions
calc_sphere_ci(data, diff_data) Weight for spherical diffusion. source code
calc_spheroid_ci(data, diff_data) Weights for spheroidal diffusion. source code
calc_spheroid_dci(data, diff_data) Weight gradient for spheroidal diffusion. source code
calc_spheroid_d2ci(data, diff_data) Weight Hessian for spheroidal diffusion. source code
calc_ellipsoid_ci(data, diff_data) Weight equations for ellipsoidal diffusion. source code
calc_ellipsoid_dci(data, diff_data) Weight gradient for ellipsoidal diffusion. source code
calc_ellipsoid_d2ci(data, diff_data) Weight Hessian for ellipsoidal diffusion. source code
Variables
__package__ = `'lib.diffusion'`
Imports: sqrt, outer
Function Details
### calc_sphere_ci(data, diff_data)
source code
Weight for spherical diffusion.
The equation is:
``` c0 = 1.
```
### calc_spheroid_ci(data, diff_data)
source code
Weights for spheroidal diffusion.
The equations are:
``` c-1 = 1/4 (3dz**2 - 1)**2,
c0 = 3dz**2 (1 - dz**2),
c1 = 3/4 (dz**2 - 1)**2,
```
where dz is the direction cosine of the unit bond vector along the z-axis of the diffusion tensor which is calculated as the dot product of the unit bond vector with a unit vector along Dpar.
### calc_spheroid_dci(data, diff_data)
source code
The equations are:
``` dc-1 ddz
---- = 3dz (3dz**2 - 1) --- ,
dOi dOi
dc0 ddz
--- = 6dz (1 - 2dz**2) --- ,
dOi dOi
dc1 ddz
--- = 3dz (dz**2 - 1) --- ,
dOi dOi
```
where the orientation parameter set O is {theta, phi}.
### calc_spheroid_d2ci(data, diff_data)
source code
Weight Hessian for spheroidal diffusion.
The equations are:
``` d2c-1 / ddz ddz d2dz \
------- = 3 |(9dz**2 - 1) --- . --- + dz (3dz**2 - 1) ------- | ,
dOi.dOj \ dOi dOj dOi.dOj /
d2c0 / ddz ddz d2dz \
------- = 6 |(1 - 6dz**2) --- . --- + dz (1 - 2dz**2) ------- | ,
dOi.dOj \ dOi dOj dOi.dOj /
d2c1 / ddz ddz d2dz \
------- = 3 |(3dz**2 - 1) --- . --- + dz (dz**2 - 1) ------- | ,
dOi.dOj \ dOi dOj dOi.dOj /
```
where the orientation parameter set O is {theta, phi}.
### calc_ellipsoid_ci(data, diff_data)
source code
Weight equations for ellipsoidal diffusion.
The equations are:
``` c-2 = 1/4 (d - e),
c-1 = 3dy**2.dz**2,
c0 = 3dx**2.dz**2,
c1 = 3dx**2.dy**2,
c2 = 1/4 (d + e),
```
where:
``` d = 3(dx**4 + dy**4 + dz**4) - 1,
e = 1/R [(1 + 3Dr)(dx**4 + 2dy**2.dz**2) + (1 - 3Dr)(dy**4 + 2dx**2.dz**2)
- 2(dz**4 + 2dx**2.dy**2)],
```
and where the factor R is defined as:
``` ___________
R = V 1 + 3Dr**2.
```
dx, dy, and dz are the direction cosines of the XH bond vector along the x, y, and z-axes of the diffusion tensor, calculated as the dot product of the unit bond vector and the unit vectors along Dx, Dy, and Dz respectively.
source code
# Oi partial derivatives
The equations are:
``` dc-2 / ddx ddy ddz \ de
---- = 3 | dx**3 --- + dy**3 --- + dz**3 --- | - --- ,
dOi \ dOi dOi dOi / dOi
dc-1 / ddz ddy \
---- = 6dy.dz | dy --- + dz --- | ,
dOi \ dOi dOi /
dc0 / ddz ddx \
--- = 6dx.dz | dx --- + dz --- | ,
dOi \ dOi dOi /
dc1 / ddy ddx \
--- = 6dx.dy | dx --- + dy --- | ,
dOi \ dOi dOi /
dc2 / ddx ddy ddz \ de
--- = 3 | dx**3 --- + dy**3 --- + dz**3 --- | + --- ,
dOi \ dOi dOi dOi / dOi
```
where:
``` de 1 / / ddx / ddz ddy \ \
--- = - | (1 + 3Dr) |dx**3 --- + dy.dz | dy --- + dz --- | |
dOi R \ \ dOi \ dOi dOi / /
/ ddy / ddz ddx \ \
+ (1 - 3Dr) | dy**3 --- + dx.dz | dx --- + dz --- | |
\ dOi \ dOi dOi / /
/ ddz / ddy ddx \ \ \
- 2 | dz**3 --- + dx.dy | dx --- + dy --- | | | ,
\ dOi \ dOi dOi / / /
```
and where the orietation parameter set O is:
``` O = {alpha, beta, gamma}.
```
# tm partial derivatives
The equations are:
``` dc-2
---- = 0,
dtm
dc-1
---- = 0,
dtm
dc0
--- = 0,
dtm
dc1
--- = 0,
dtm
dc2
--- = 0.
dtm
```
# Da partial derivatives
The equations are:
``` dc-2
---- = 0,
dDa
dc-1
---- = 0,
dDa
dc0
--- = 0,
dDa
dc1
--- = 0,
dDa
dc2
--- = 0.
dDa
```
# Dr partial derivatives
The equations are:
``` dc-2 3 de
---- = - - ---,
dDr 4 dDr
dc-1
---- = 0,
dDr
dc0
--- = 0,
dDr
dc1
--- = 0,
dDr
dc2 3 de
--- = - ---,
dDr 4 dDr
```
where:
``` de 1 / \
--- = ---- | (1 - Dr) (dx**4 + 2dy**2.dz**2) - (1 + Dr) (dy**4 + 2dx**2.dz**2) + 2Dr (dz**4 + 2dx**2.dy**2) | .
dDr R**3 \ /
```
### calc_ellipsoid_d2ci(data, diff_data)
source code
Weight Hessian for ellipsoidal diffusion.
# Oi-Oj partial derivatives
The equations are:
``` d2c-2 / / d2dx ddx ddx \ / d2dy ddy ddy \ / d2dz ddz ddz \ \ d2e
------- = 3 | dx**2 | dx ------- + 3 --- . --- | + dy**2 | dy ------- + 3 --- . --- | + dz**2 | dz ------- + 3 --- . --- | | - ------- ,
dOi.dOj \ \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj / / dOi.dOj
d2c-1 / d2dz ddz ddz \ / ddy ddz ddz ddy \ / d2dy ddy ddy \
------- = 6 dy**2 | dz ------- + --- . --- | + 12 dy.dz | --- . --- + --- . --- | + 6 dz**2 | dy ------- + --- . --- | ,
dOi.dOj \ dOi.dOj dOi dOj / \ dOi dOj dOi dOj / \ dOi.dOj dOi dOj /
d2c0 / d2dz ddz ddz \ / ddx ddz ddz ddx \ / d2dx ddx ddx \
------- = 6 dx**2 | dz ------- + --- . --- | + 12 dx.dz | --- . --- + --- . --- | + 6 dz**2 | dx ------- + --- . --- | ,
dOi.dOj \ dOi.dOj dOi dOj / \ dOi dOj dOi dOj / \ dOi.dOj dOi dOj /
d2c1 / d2dy ddy ddy \ / ddx ddy ddy ddx \ / d2dx ddx ddx \
------- = 6 dx**2 | dy ------- + --- . --- | + 12 dx.dy | --- . --- + --- . --- | + 6 dy**2 | dx ------- + --- . --- | ,
dOi.dOj \ dOi.dOj dOi dOj / \ dOi dOj dOi dOj / \ dOi.dOj dOi dOj /
d2c2 / / d2dx ddx ddx \ / d2dy ddy ddy \ / d2dz ddz ddz \ \ d2e
------- = 3 | dx**2 | dx ------- + 3 --- . --- | + dy**2 | dy ------- + 3 --- . --- | + dz**2 | dz ------- + 3 --- . --- | | + ------- ,
dOi.dOj \ \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj / / dOi.dOj
```
where:
``` d2e 1 / / / d2dx ddx ddx \ / d2dz ddz ddz \
------- = - | (1 + 3Dr) | dx**2 | dx ------- + 3 --- . --- | + dy**2 | dz ------- + --- . --- |
dOi.dOj R \ \ \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj /
/ d2dy ddy ddy \ / ddy ddz ddz ddy \ \
+ dz**2 | dy ------- + --- . --- | + 2dy.dz | --- . --- + --- . --- | |
\ dOi.dOj dOi dOj / \ dOi dOj dOi dOj / /
/ / d2dy ddy ddy \ / d2dz ddz ddz \
+ (1 - 3Dr) | dy**2 | dy ------- + 3 --- . --- | + dx**2 | dz ------- + --- . --- |
\ \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj /
/ d2dx ddx ddx \ / ddx ddz ddz ddx \ \
+ dz**2 | dx ------- + --- . --- | + 2dx.dz | --- . --- + --- . --- | |
\ dOi.dOj dOi dOj / \ dOi dOj dOi dOj / /
/ / d2dz ddz ddz \ / d2dy ddy ddy \
- 2 | dz**2 | dz ------- + 3 --- . --- | + dx**2 | dy ------- + --- . --- |
\ \ dOi.dOj dOi dOj / \ dOi.dOj dOi dOj /
/ d2dx ddx ddx \ / ddx ddy ddy ddx \ \ \
+ dy**2 | dx ------- + --- . --- | + 2dx.dy | --- . --- + --- . --- | | |
\ dOi.dOj dOi dOj / \ dOi dOj dOi dOj / / /
```
# Oi-tm partial derivatives
The equations are:
``` d2c-2
------- = 0,
dOi.dtm
d2c-1
------- = 0,
dOi.dtm
d2c0
------- = 0,
dOi.dtm
d2c1
------- = 0,
dOi.dtm
d2c2
------- = 0.
dOi.dtm
```
# Oi-Da partial derivatives
The equations are:
``` d2c-2
------- = 0,
dOi.dDa
d2c-1
------- = 0,
dOi.dDa
d2c0
------- = 0,
dOi.dDa
d2c1
------- = 0,
dOi.dDa
d2c2
------- = 0.
dOi.dDa
```
# Oi-Dr partial derivatives
The equations are:
``` d2c-2 d2e
------- = - 3 -------,
dOi.dDr dOi.dDr
d2c-1
------- = 0,
dOi.dDr
d2c0
------- = 0,
dOi.dDr
d2c1
------- = 0,
dOi.dDr
d2c2 d2e
------- = 3 -------,
dOi.dDr dOi.dDr
```
where:
``` d2e 1 / / ddx / ddz ddy \ \
------- = ---- | (1 - Dr) | dx**3 --- + dy.dz | dy --- + dz --- | |
dOi.dDr R**3 \ \ dOi \ dOi dOi / /
/ ddy / ddz ddx \ \
- (1 + Dr) | dy**3 --- + dx.dz | dx --- + dz --- | |
\ dOi \ dOi dOi / /
/ ddz / ddy ddx \ \ \
+ 2Dr | dz**3 --- + dx.dy | dx --- + dy --- | | |
\ dOi \ dOi dOi / / /
```
# tm-tm partial derivatives
The equations are:
``` d2c-2
----- = 0,
dtm2
d2c-1
----- = 0,
dtm2
d2c0
---- = 0,
dtm2
d2c1
---- = 0,
dtm2
d2c2
---- = 0.
dtm2
```
# tm-Da partial derivatives
The equations are:
``` d2c-2
------- = 0,
dtm.dDa
d2c-1
------- = 0,
dtm.dDa
d2c0
------- = 0,
dtm.dDa
d2c1
------- = 0,
dtm.dDa
d2c2
------- = 0.
dtm.dDa
```
# tm-Dr partial derivatives
The equations are:
``` d2c-2
------- = 0,
dtm.dDr
d2c-1
------- = 0,
dtm.dDr
d2c0
------- = 0,
dtm.dDr
d2c1
------- = 0,
dtm.dDr
d2c2
------- = 0.
dtm.dDr
```
# Da-Da partial derivatives
The equations are:
``` d2c-2
------ = 0,
dDa**2
d2c-1
------ = 0,
dDa**2
d2c0
------ = 0,
dDa**2
d2c1
------ = 0,
dDa**2
d2c2
------ = 0.
dDa**2
```
# Da-Dr partial derivatives
The equations are:
``` d2c-2
------- = 0,
dDa.dDr
d2c-1
------- = 0,
dDa.dDr
d2c0
------- = 0,
dDa.dDr
d2c1
------- = 0,
dDa.dDr
d2c2
------- = 0.
dDa.dDr
```
# Dr-Dr partial derivatives
The equations are:
``` d2c-2 3 d2e
------ = - - ------,
dDr**2 4 dDr**2
d2c-1
------ = 0,
dDr**2
d2c0
------ = 0,
dDr**2
d2c1
------ = 0,
dDr**2
d2c2 3 d2e
------ = - ------,
dDr**2 4 dDr**2
```
where:
``` d2e 1 / \
------ = ---- | (6Dr**2 - 9Dr - 1)(dx**4 + 2dy**2.dz**2) + (6Dr**2 + 9Dr - 1)(dy**4 + 2dx**2.dz**2) - 2(6Dr**2 - 1)(ddz*4 + 2dx**2.dy**2) |
dDr**2 R**5 \ /
```
Generated by Epydoc 3.0.1 on Fri Jun 14 11:29:05 2019 http://epydoc.sourceforge.net | 4,617 | 12,172 | {"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-2020-50 | latest | en | 0.61494 |
https://www.aqua-calc.com/convert/density/kilogram-per-cubic-meter-to-pound-per-cubic-decimeter | 1,579,812,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00072.warc.gz | 786,854,573 | 8,029 | # Convert kilograms per (cubic meter) to pounds per (cubic decimeter)
## kg/m³ to lb/dm³ (kg:kilogram, m:meter, lb:pound, dm:decimeter)
### Convert kg/m³ to lb/dm³
#### a density conversion table
to 100 with
kilo-
gram per cubic meter
pound per cubic deci-
meter
kilo-
gram per cubic meter
pound per cubic deci-
meter
kilo-
gram per cubic meter
pound per cubic deci-
meter
kilo-
gram per cubic meter
pound per cubic deci-
meter
kilo-
gram per cubic meter
pound per cubic deci-
meter
10.0210.0410.1610.1810.2
20.0220.0420.1620.1820.2
30.0230.1430.1630.1830.2
40.0240.1440.1640.1840.2
50.0250.1450.1650.1850.2
60.0260.1460.1660.1860.2
70.0270.1470.1670.1870.2
80.0280.1480.1680.1880.2
90.0290.1490.1690.2890.2
100.0300.1500.1700.2900.2
110.0310.1510.1710.2910.2
120.0320.1520.1720.2920.2
130.0330.1530.1730.2930.2
140.0340.1540.1740.2940.2
150.0350.1550.1750.2950.2
160.0360.1560.1760.2960.2
170.0370.1570.1770.2970.2
180.0380.1580.1780.2980.2
190.0390.1590.1790.2990.2
200.0400.1600.1800.21000.2
### kilograms per cubic meter to pounds per cubic decimeter conversion cards
• 1
through
20
kilograms per cubic meter
• 1 kg/m³ to lb/dm³ = 0 lb/dm³
• 2 kg/m³ to lb/dm³ = 0 lb/dm³
• 3 kg/m³ to lb/dm³ = 0 lb/dm³
• 4 kg/m³ to lb/dm³ = 0 lb/dm³
• 5 kg/m³ to lb/dm³ = 0 lb/dm³
• 6 kg/m³ to lb/dm³ = 0 lb/dm³
• 7 kg/m³ to lb/dm³ = 0 lb/dm³
• 8 kg/m³ to lb/dm³ = 0 lb/dm³
• 9 kg/m³ to lb/dm³ = 0 lb/dm³
• 10 kg/m³ to lb/dm³ = 0 lb/dm³
• 11 kg/m³ to lb/dm³ = 0 lb/dm³
• 12 kg/m³ to lb/dm³ = 0 lb/dm³
• 13 kg/m³ to lb/dm³ = 0 lb/dm³
• 14 kg/m³ to lb/dm³ = 0 lb/dm³
• 15 kg/m³ to lb/dm³ = 0 lb/dm³
• 16 kg/m³ to lb/dm³ = 0 lb/dm³
• 17 kg/m³ to lb/dm³ = 0 lb/dm³
• 18 kg/m³ to lb/dm³ = 0 lb/dm³
• 19 kg/m³ to lb/dm³ = 0 lb/dm³
• 20 kg/m³ to lb/dm³ = 0 lb/dm³
• 21
through
40
kilograms per cubic meter
• 21 kg/m³ to lb/dm³ = 0 lb/dm³
• 22 kg/m³ to lb/dm³ = 0 lb/dm³
• 23 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 24 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 25 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 26 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 27 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 28 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 29 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 30 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 31 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 32 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 33 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 34 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 35 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 36 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 37 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 38 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 39 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 40 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 41
through
60
kilograms per cubic meter
• 41 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 42 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 43 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 44 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 45 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 46 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 47 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 48 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 49 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 50 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 51 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 52 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 53 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 54 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 55 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 56 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 57 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 58 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 59 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 60 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 61
through
80
kilograms per cubic meter
• 61 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 62 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 63 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 64 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 65 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 66 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 67 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 68 kg/m³ to lb/dm³ = 0.1 lb/dm³
• 69 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 70 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 71 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 72 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 73 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 74 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 75 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 76 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 77 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 78 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 79 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 80 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 81
through
100
kilograms per cubic meter
• 81 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 82 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 83 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 84 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 85 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 86 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 87 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 88 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 89 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 90 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 91 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 92 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 93 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 94 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 95 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 96 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 97 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 98 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 99 kg/m³ to lb/dm³ = 0.2 lb/dm³
• 100 kg/m³ to lb/dm³ = 0.2 lb/dm³
#### Foods, Nutrients and Calories
FOOD LION, CANDIQUIK, CANDY COATING, VANILLA, UPC: 035826067760 weigh(s) 253.61 gram per (metric cup) or 8.47 ounce per (US cup), and contain(s) 533 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
PLUMP and TENDER MEDIUM GRAIN, UPC: 035200065207 contain(s) 378 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Copper, Cu, foods low in Copper, Cu, and Recommended Dietary Allowances (RDAs) for Copper
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Special Coarse Aragonite weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Ammonium dichromate [(NH4)2Cr2O7] weighs 2 115 kg/m³ (132.03514 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-123, liquid (R123) with temperature in the range of -28.89°C (-20.002°F) to 93.34°C (200.012°F)
#### Weights and Measurements
The grain per cubic centimeter density measurement unit is used to measure volume in cubic centimeters in order to estimate weight or mass in grains
The units of data measurement were introduced to manage and operate digital information.
µin/s² to fur/h² conversion table, µin/s² to fur/h² unit converter or convert between all units of acceleration measurement.
#### Calculators
Return on investment appreciation or rate of loss calculator | 3,266 | 6,537 | {"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} | 2.75 | 3 | CC-MAIN-2020-05 | latest | en | 0.106784 |
https://www.coursehero.com/file/6246110/3-volatility/ | 1,516,689,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891750.87/warc/CC-MAIN-20180123052242-20180123072242-00000.warc.gz | 831,852,713 | 92,327 | 3_volatility
# 3_volatility - VOLATILITY definition and other basics...
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VOLATILITY J. Wei, Department of Management, U of T 1 MGTD78 definition and other basics weighted moving average model GARCH(1, 1) maximum likelihood estimation of GARCH volatility forecasting and volatility term structure
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VOLATILITY J. Wei, Department of Management, U of T 2 MGTD78 Definition of volatility : standard deviation of the variable Variance rate (or variance) : square of volatility Calendar days or trading days (see Business Snapshot 9.1)? volatility seems to be generated by trading; we therefore count only trading days – 252 days in a year. Example: annual variance is 0.0625. Then annual volatility is 0.25; daily variance is 0.0625 / 252 = 0.000248; daily volatility is 0.25/sqrt(252) = 0.01575. Implied volatility: volatility implied from option prices (C71) VIX (Fig 9.1)
VOLATILITY IN FINANCIAL VARIABLES J. Wei, Department of Management, U of T 3 MGTD78 Black-Scholes model assumes normal distribution of returns Financial returns are usually not normal. Example in Table 9.2 (10-year period, 12 exchange rates): % days on which realized FX is n standard deviations away from mean: real world(%) Normal distribution(%) > 1 SD 25.04 31.73 > 2 SD 5.27 4.55 > 3 SD 1.34 0.27 > 4 SD 0.29 0.01 > 5 SD 0.08 0.00 > 6 SD 0.03 0.00 large / extreme changes are more likely than predicted by normal: heavy-tail or fat-tail distributions (see Figure 9.1)
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POWER LAW J. Wei, Department of Management, U of T 4 MGTD78 Alternative to normal distribution when extreme values are considered It is an approximation, applicable to many situations The Power Law: when x is large, the value of variable v approximately satisfies Prob(v > x) = Kx α , where K and α are constants Application: use not-so-extreme values of x to estimate K and α , then calculate probability for extreme realizations Example: the probability that v > 6 is 0.004 when K = 10.368 and = 4. Then we can calculate the α following:
STANDARD METHOD OF ESTIMATION J. Wei, Department of Management, U of T 5 MGTD78 Review of C71 (on day n, going back m days): where In risk management, we make three changes use percentage change (reflecting reality): assuming to be zero (for simplicity) replace m -1 by m (conforming to ML estimate) ( 29 u u 1 - m 1 σ m 1 i 2 i - n 2 n = - = u m 1 u , S S ln u m 1 i i n 1 - i i i = - = = and 1 - i 1 - i i i S S S u - = u u m 1 σ m 1 i 2 i - n 2 n = =
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WEIGHTING SCHEME J. Wei, Department of Management, U of T 6 MGTD78 Standard method assigns equal weight (1/m) to all observations Make sense to assign more weights to recent observations: The weighting scheme ( α i ) varies. The most common is to
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3_volatility - VOLATILITY definition and other basics...
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Ask a homework question - tutors are online | 900 | 3,373 | {"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-2018-05 | latest | en | 0.825903 |
https://git.skewed.de/count0/graph-tool/-/raw/1845dd071be88602eb3a122a5c88a8bcf5d79211/src/graph_tool/stats/__init__.py | 1,653,122,839,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00148.warc.gz | 338,735,640 | 4,015 | #! /usr/bin/env python # -*- coding: utf-8 -*- # # graph_tool -- a general graph manipulation python module # # Copyright (C) 2007-2011 Tiago de Paula Peixoto # # This program is free software: you can redistribute it and/or modify # it under the terms of the GNU General Public License as published by # the Free Software Foundation, either version 3 of the License, or # (at your option) any later version. # # This program is distributed in the hope that it will be useful, # but WITHOUT ANY WARRANTY; without even the implied warranty of # MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the # GNU General Public License for more details. # # You should have received a copy of the GNU General Public License # along with this program. If not, see . """ graph_tool.stats - Miscellaneous statistics ----------------------------------------------- Summary +++++++ .. autosummary:: :nosignatures: vertex_hist edge_hist vertex_average edge_average label_parallel_edges remove_parallel_edges label_self_loops remove_self_loops remove_labeled_edges distance_histogram Contents ++++++++ """ from .. dl_import import dl_import dl_import("import libgraph_tool_stats") from .. import _degree, _prop from numpy import * import numpy import sys __all__ = ["vertex_hist", "edge_hist", "vertex_average", "edge_average", "label_parallel_edges", "remove_parallel_edges", "label_self_loops", "remove_self_loops", "remove_labeled_edges", "distance_histogram"] def vertex_hist(g, deg, bins=[0, 1], float_count=True): """ Return the vertex histogram of the given degree type or property. Parameters ---------- g : :class:~graph_tool.Graph Graph to be used. deg : string or :class:~graph_tool.PropertyMap Degree or property to be used for the histogram. It can be either "in", "out" or "total", for in-, out-, or total degree of the vertices. It can also be a vertex property map. bins : list of bins (optional, default: [0, 1]) List of bins to be used for the histogram. The values given represent the edges of the bins (i.e. lower and upper bounds). If the list contains two values, this will be used to automatically create an appropriate bin range, with a constant width given by the second value, and starting from the first value. float_count : bool (optional, default: True) If True, the counts in each histogram bin will be returned as floats. If False, they will be returned as integers. Returns ------- counts : :class:~numpy.ndarray The bin counts. bins : :class:~numpy.ndarray The bin edges. See Also -------- edge_hist: Edge histograms. vertex_average: Average of vertex properties, degrees. edge_average: Average of edge properties. distance_histogram : Shortest-distance histogram. Notes ----- The algorithm runs in :math:O(|V|) time. If enabled during compilation, this algorithm runs in parallel. Examples -------- >>> from numpy.random import poisson, seed >>> seed(42) >>> g = gt.random_graph(1000, lambda: (poisson(5), poisson(5))) >>> print gt.vertex_hist(g, "out") [array([ 10., 30., 86., 138., 166., 154., 146., 129., 68., 36., 23., 8., 3., 2., 0., 1.]), array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], dtype=uint64)] """ ret = libgraph_tool_stats.\ get_vertex_histogram(g._Graph__graph, _degree(g, deg), [float(x) for x in bins]) return [array(ret[0], dtype="float64") if float_count else ret[0], ret[1]] def edge_hist(g, eprop, bins=[0, 1], float_count=True): """ Return the edge histogram of the given property. Parameters ---------- g : :class:~graph_tool.Graph Graph to be used. eprop : :class:~graph_tool.PropertyMap Edge property to be used for the histogram. bins : list of bins (optional, default: [0, 1]) List of bins to be used for the histogram. The values given represent the edges of the bins (i.e. lower and upper bounds). If the list contains two values, this will be used to automatically create an appropriate bin range, with a constant width given by the second value, and starting from the first value. float_count : bool (optional, default: True) If True, the counts in each histogram bin will be returned as floats. If False, they will be returned as integers. Returns ------- counts : :class:~numpy.ndarray The bin counts. bins : :class:~numpy.ndarray The bin edges. See Also -------- vertex_hist : Vertex histograms. vertex_average : Average of vertex properties, degrees. edge_average : Average of edge properties. distance_histogram : Shortest-distance histogram. Notes ----- The algorithm runs in :math:O(|E|) time. If enabled during compilation, this algorithm runs in parallel. Examples -------- >>> from numpy import arange >>> from numpy.random import random, seed >>> seed(42) >>> g = gt.random_graph(1000, lambda: (5, 5)) >>> eprop = g.new_edge_property("double") >>> eprop.get_array()[:] = random(g.num_edges()) >>> print gt.edge_hist(g, eprop, linspace(0, 1, 11)) [array([ 525., 504., 502., 502., 468., 499., 531., 471., 520., 478.]), array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ])] """ ret = libgraph_tool_stats.\ get_edge_histogram(g._Graph__graph, _prop("e", g, eprop), [float(x) for x in bins]) return [array(ret[0], dtype="float64") if float_count else ret[0], ret[1]] def vertex_average(g, deg): """ Return the average of the given degree or vertex property. Parameters ---------- g : :class:~graph_tool.Graph Graph to be used. deg : string or :class:~graph_tool.PropertyMap Degree or property to be used for the histogram. It can be either "in", "out" or "total", for in-, out-, or total degree of the vertices. It can also be a vertex property map. Returns ------- average : float The average of the given degree or property. std : float The standard deviation of the average. See Also -------- vertex_hist : Vertex histograms. edge_hist : Edge histograms. edge_average : Average of edge properties. distance_histogram : Shortest-distance histogram. Notes ----- The algorithm runs in :math:O(|V|) time. If enabled during compilation, this algorithm runs in parallel. Examples -------- >>> from numpy.random import poisson, seed >>> seed(42) >>> g = gt.random_graph(1000, lambda: (poisson(5), poisson(5))) >>> print gt.vertex_average(g, "in") (5.092, 0.07188557574367754) """ ret = libgraph_tool_stats.\ get_vertex_average(g._Graph__graph, _degree(g, deg)) return ret def edge_average(g, eprop): """ Return the average of the given degree or vertex property. Parameters ---------- g : :class:~graph_tool.Graph Graph to be used. eprop : :class:~graph_tool.PropertyMap Edge property to be used for the histogram. Returns ------- average : float The average of the given property. std : float The standard deviation of the average. See Also -------- vertex_hist : Vertex histograms. edge_hist : Edge histograms. vertex_average : Average of vertex degree, properties. distance_histogram : Shortest-distance histogram. Notes ----- The algorithm runs in :math:O(|E|) time. If enabled during compilation, this algorithm runs in parallel. Examples -------- >>> from numpy import arange >>> from numpy.random import random, seed >>> seed(42) >>> g = gt.random_graph(1000, lambda: (5, 5)) >>> eprop = g.new_edge_property("double") >>> eprop.get_array()[:] = random(g.num_edges()) >>> print gt.edge_average(g, eprop) (0.49674035434130187, 0.004094604069093868) """ ret = libgraph_tool_stats.\ get_edge_average(g._Graph__graph, _prop("e", g, eprop)) return ret def remove_labeled_edges(g, label): """Remove every edge e such that label[e] != 0.""" g.stash_filter(all=False, directed=True, reversed=True) libgraph_tool_stats.\ remove_labeled_edges(g._Graph__graph, _prop("e", g, label)) g.pop_filter(all=False, directed=True, reversed=True) def label_parallel_edges(g, mark_only=False, count_all=False, eprop=None): r"""Label edges which are parallel, i.e, have the same source and target vertices. For each parallel edge set :math:PE, the labelling starts from 0 to :math:|PE|-1. (If count_all==True, the range is 0 to :math:|PE| instead). If mark_only==True, all parallel edges are simply marked with the value 1. If the eprop parameter is given (a :class:~graph_tool.PropertyMap), the labelling is stored there.""" if eprop is None: if mark_only: eprop = g.new_edge_property("bool") else: eprop = g.new_edge_property("int32_t") libgraph_tool_stats.\ label_parallel_edges(g._Graph__graph, _prop("e", g, eprop), mark_only, count_all) return eprop def remove_parallel_edges(g): """Remove all parallel edges from the graph. Only one edge from each parallel edge set is left.""" eprop = label_parallel_edges(g) remove_labeled_edges(g, eprop) def label_self_loops(g, mark_only=False, eprop=None): """Label edges which are self-loops, i.e, the source and target vertices are the same. For each self-loop edge set :math:SL, the labelling starts from 0 to :math:|SL|-1. If mark_only == True, self-loops are labeled with 1 and others with 0. If the eprop parameter is given (a :class:~graph_tool.PropertyMap), the labelling is stored there.""" if eprop is None: if mark_only: eprop = g.new_edge_property("bool") else: eprop = g.new_edge_property("int32_t") libgraph_tool_stats.\ label_self_loops(g._Graph__graph, _prop("e", g, eprop), mark_only) return eprop def remove_self_loops(g): """Remove all self-loops edges from the graph.""" eprop = label_self_loops(g) remove_labeled_edges(g, eprop) def distance_histogram(g, weight=None, bins=[0, 1], samples=None, float_count=True): r""" Return the shortest-distance histogram for each vertex pair in the graph. Parameters ---------- g : :class:Graph Graph to be used. weight : :class:~graph_tool.PropertyMap (optional, default: None) Edge weights. bins : list of bins (optional, default: [0, 1]) List of bins to be used for the histogram. The values given represent the edges of the bins (i.e. lower and upper bounds). If the list contains two values, this will be used to automatically create an appropriate bin range, with a constant width given by the second value, and starting from the first value. samples : int (optional, default: None) If supplied, the distances will be randomly sampled from a number of source vertices given by this parameter. It samples == None (default), all pairs are used. float_count : bool (optional, default: True) If True, the counts in each histogram bin will be returned as floats. If False, they will be returned as integers. Returns ------- counts : :class:~numpy.ndarray The bin counts. bins : :class:~numpy.ndarray The bin edges. See Also -------- vertex_hist : Vertex histograms. edge_hist : Edge histograms. vertex_average : Average of vertex degree, properties. distance_histogram : Shortest-distance histogram. Notes ----- The algorithm runs in :math:O(V^2) time, or :math:O(V^2\log V) if weight != None. If samples is supplied, the complexities are :math:O(\text{samples}\times V) and :math:O(\text{samples}\times V\log V), respectively. If enabled during compilation, this algorithm runs in parallel. Examples -------- >>> from numpy.random import random, seed >>> seed(42) >>> g = gt.random_graph(100, lambda: (3, 3)) >>> hist = gt.distance_histogram(g) >>> print hist [array([ 0., 300., 861., 2165., 3801., 2576., 197.]), array([0, 1, 2, 3, 4, 5, 6, 7], dtype=uint64)] >>> hist = gt.distance_histogram(g, samples=10) >>> print hist [array([ 0., 30., 87., 221., 395., 234., 23.]), array([0, 1, 2, 3, 4, 5, 6, 7], dtype=uint64)] """ if samples != None: seed = numpy.random.randint(0, sys.maxint) ret = libgraph_tool_stats.\ sampled_distance_histogram(g._Graph__graph, _prop("e", g, weight), [float(x) for x in bins], samples, seed) else: ret = libgraph_tool_stats.\ distance_histogram(g._Graph__graph, _prop("e", g, weight), bins) return [array(ret[0], dtype="float64") if float_count else ret[0], ret[1]] | 3,022 | 11,672 | {"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} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.674971 |
https://community.freefem.org/t/pde-of-schrodinger-type/512 | 1,653,011,873,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00609.warc.gz | 223,364,639 | 4,187 | # PDE of Schrodinger type
Hello,
I wish to solve a linear PDE of the form:
-\Delta u + V u = f
in some 2d domain, where \Delta is the Laplacian. In addition to f(x,y), there is now another function V(x,y), which is specified analytically. The weak formulation is straightforward mathematically, but how do I include V in the stiffness matrix? In other words, I expect the syntax to be something like:
varf a(u,v) = int2d(Th) ( dx(u)dx(v) +dy(u)dy(v) +uVv )
But how do I actually specify V in the code, given its analytical form V(x,y)?
Best regards,
PhyzWiz
`func V = cos(x) * sin(y) + exp(x+y)`?
I see, that simple! Great, thank you very much.
PhyzWiz | 198 | 662 | {"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} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.853255 |
http://www.accountingtutorials.net/2013/04/bowman-company-produces-arthritis.html | 1,511,081,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805466.25/warc/CC-MAIN-20171119080836-20171119100836-00520.warc.gz | 336,967,784 | 12,601 | ## Pages
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### Bowman Company produces an arthritis
Price: \$3.50
Problem 6-46 Basic Flows, Equivalent Units
Bowman Company produces an arthritis medication that passes through two departments: Mixing and Tableting. Bowman uses the weighted average method. Data for February for Mixing is as follows: BWIP was zero; EWIP had 7,200 units, 50 percent complete; and 84,000 units were started. Tableting's data for February is as follows: BWIP was 4,800 units, 20 percent complete; and 2,400 units were in EWIP, 40 percent complete.
Instructions
1. For Mixing, calculate the (a) number of units transferred to Tableting, and (b) equivalent units of production.
2. For Tableting, calculate the number of units transferred out to Finished Goods.
3. Suppose that the units in the mixing department are measured in ounces,
while the units in Tableting are measured in bottles of 100 tablets, with a total weight
of eight ounces (excluding the bottle) and then repeat requirement 2 using this approach. | 248 | 1,044 | {"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} | 2.734375 | 3 | CC-MAIN-2017-47 | longest | en | 0.957001 |
https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1186/s13662-018-1933-z | 1,716,345,958,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00854.warc.gz | 65,731,695 | 98,770 | Theory and Modern Applications
# Control optimization and homoclinic bifurcation of a prey–predator model with ratio-dependent
## Abstract
In this paper, a predator–prey model with ratio-dependent and impulsive state feedback control is constructed, where the pest growth rate is related to an Allee effect. Firstly, the existence condition of the homoclinic cycle is obtained by analyzing the control parameter q. The existence, uniqueness and asymptotic stability of the periodic orbit are discussed by using the geometric theory of the differential equations, the method of successor functions and analog of the Poincaré criterion. Secondly, we formulate a control optimization with a minimal total cost in pest management, and we obtain an optimal economic threshold. Finally, we verify the main results by numerical simulation.
## 1 Introduction
Differential equations can be applied extensively in many fields, including economic development, environmental protection, population ecology, infectious diseases and pharmacokinetics, and the cultivation of micro-organisms [1,2,3,4,5,6]. In [3], Cui et al. proposed the fractional differential equations and analyzed the uniqueness of solution for boundary value problems. Many researchers have obtained very good achievements in the field of stochastic differential equations [7,8,9,10,11,12].
In recent decades, many researchers have found that the occurrence of some biological phenomena and the optimal control of some life phenomena were not continuous processes, but the transient behavior of an impulse, so we should use impulsive differential equations to describe these phenomena. The theory of impulsive differential equations is proposed on the basis of continuous differential equation theory, which is difficult but valuable. It is widely applied in the harvest description [13,14,15,16], ecological resources [17,18,19,20,21,22,23,24,25,26], pest control [27,28,29,30,31] and epidemiologic control [32,33,34,35,36]. Many scholars investigated state-dependent pulse differential equations in predator–prey model to simulate the pest management including the periodic release of natural enemies [37,38,39] and the periodic release of natural enemies combined with periodic spraying of pesticides [40,41,42,43,44,45,46]. On the other hand, the bifurcation theory has been widely applied in the continuous dynamic system [47,48,49]. However, its application in impulsive dynamical system was little.
In [50], Zhang et al. proposed a predator–prey model with multi-state pulse feedback control as follows:
$$\textstyle\begin{cases} \left . \textstyle\begin{array}{l} {\frac{dx(t)}{dt}=x(t) [r\ln \frac{K}{x(t)}-\rho y(t) ],} \\ {\frac{dy(t)}{dt}= [\mu \rho x(t)-d ]y(t),} \end{array}\displaystyle \right \} \quad x\neq h_{1},h_{2},\text{or }x=h_{1},y>y_{ML}, \\ \left . \textstyle\begin{array}{l} {\triangle x=0,} \\ {\triangle y=\zeta ,} \end{array}\displaystyle \right \} \quad x=h_{1},y\leq y_{ML}, \\ \left . \textstyle\begin{array}{l} {\triangle x=-px(t),} \\ {\triangle y=-qy(t)+\tau ,} \end{array}\displaystyle \right \} \quad x=h_{2}, \end{cases}$$
(1)
where $$h_{1}$$ and $$h_{2}$$ are the biological control level (i.e. the threshold with slight damage to the crops) and the chemical control level (i.e. pest economic injury threshold), respectively. $$y_{\mathrm{ML}}$$ represents the predator maintainable level at $$x=h_{1}$$. It is interesting and practically significant to involve chemical and biological controls at different economic thresholds, but there is a key problem to be identified in the model. Biological control can be adopted at $$x=h_{1}$$ and $$y< y_{\mathrm{ML}}$$, but for a higher pest density $$x=h$$, where $$h_{1}< h< h_{2}$$, there no control strategy is adopted, therefore the model is not flawless.
Based on above the factors, we propose a ratio-dependent predator–prey system with an integrated control strategy as follows:
$$\textstyle\begin{cases} \left . \textstyle\begin{array}{l} {\frac{dx_{1}(t_{1})}{dt_{1}}=rx_{1}(t_{1}) [x_{1}(t_{1})-\alpha _{1} ] [1-\frac{x_{1}(t_{1})}{K} ] -\frac{mx_{1}(t _{1})y_{1}(t_{1})}{x_{1}(t_{1})+mby_{1}(t_{1})},} \\ {\frac{dy_{1}(t_{1})}{dt_{1}}= [-\beta _{1}+\frac{cmx_{1}(t_{1})}{x _{1}(t_{1})+mby_{1}(t_{1})} ]y_{1}(t_{1}),} \end{array}\displaystyle \right \} \quad x_{1}< h', \\ \left . \textstyle\begin{array}{l} {\triangle x_{1}(t_{1})=-p_{1}(x_{1})x_{1}(t_{1}),} \\ {\triangle y_{1}(t_{1})=-q_{1}(x_{1})y_{1}(t_{1})+\tau _{1}(x_{1}),} \end{array}\displaystyle \right \} \quad x_{1}=h',y_{1}\leq h_{h}', \end{cases}$$
(2)
where $$\Delta x_{1}(t_{1})=x_{1}(t_{1}^{+})-x_{1}(t_{1})$$, $$\Delta y _{1}(t_{1})=y_{1}(t_{1}^{+})-y_{1}(t_{1})$$, $$x_{1}$$ and $$y_{1}$$ represent the density of the prey and predator at time $$t_{1}$$, respectively. r denotes the intrinsic rate growth of pests. $$h'\in [h_{1}',h_{2}']$$ represents the threshold of pests, where $$h_{1}'$$ and $$h_{2}'$$ are biological control level and chemical control level, respectively. $$K>0$$ represents the environment carrying capacity of pests. m denotes the attack rate of predator. b denotes the handling time of a single pest by natural enemies. c is the conversion ratio of consumed pests into viable natural enemy offspring. $$\beta _{1}$$ is the death rate of predator. The parameter $$\alpha _{1}$$ represents the survival threshold of the pest population for the Allee effect. $$0<\alpha _{1}<K$$ is the survival threshold of the pest population for the strong Allee effect, $$-K<\alpha _{1}<0$$ is the survival threshold of the pest population for the weak Allee effect. In this paper, only $$0<\alpha _{1}<K$$ is taken into consideration. For reducing the parameters, we nondimensionalize system (2) with the following scaling:
$$\frac{x_{1}}{K}=x, \quad\quad \frac{mby_{1}}{K}=y, \quad\quad rKt_{1}=t,$$
then the following form can be obtained:
$$\textstyle\begin{cases} \left . \textstyle\begin{array}{l} {\frac{dx(t)}{dt}=x(t) [x(t)-\alpha ] [1-x(t) ]-\frac{ \lambda x(t)y(t)}{x(t)+y(t)}}, \\ {\frac{dy(t)}{dt}=\frac{\lambda _{1}x(t)y(t)}{x(t)+y(t)}-\beta y(t),} \end{array}\displaystyle \right \}\quad x< h, \\ \left . \textstyle\begin{array}{l} {\triangle x(t)=-p(x)x(t),} \\ {\triangle y(t)=-q(x)y(t)+\tau (x),} \end{array}\displaystyle \right \} \quad x=h,y\leq y_{h}, \end{cases}$$
(3)
where $$\alpha =\frac{\alpha _{1}}{K}$$, $$\lambda =\frac{1}{rbK}$$, $$\lambda _{1}=\frac{cm}{rK}$$, $$\beta =\frac{\beta _{1}}{rK}$$, $$\tau =\frac{mb\tau _{1}}{K}$$, $$h=\frac{1}{K}h'$$, $$h_{1}=\frac{1}{K}h _{1}'$$, $$h_{2}=\frac{1}{K}h_{2}'$$, and they are all positive constants. We can refer to [51, 52] for the analysis of the term $$\frac{xy}{x+y}$$ at $$O(0,0)$$. The parameters $$p(x)$$, $$q(x)$$, $$\tau (x)$$ are continuous functions defined on $$[h_{1},h_{2}]$$, where $$\tau (h_{1})=\tau _{\max }$$, $$\tau (h_{2})=\tau _{\min }$$, $$p(h_{1})=0$$, $$p(h_{2})=p_{\max }$$, $$q(h_{1})=0$$, $$q(h_{2})=q_{\max }$$. In this paper, the functions p, q, τ are defined in a linear form [53], i.e.
$$\textstyle\begin{cases} p(x)=p_{\max }\frac{x-h_{1}}{h_{2}-h_{1}},\quad p_{\max }\in [0,1), \\ q(x)=q_{\max }\frac{x-h_{1}}{h_{2}-h_{1}},\quad q_{\max }\in [0,1), \\ \tau (x)=\tau _{\max }-(\tau _{\max }-\tau _{\min })\frac{x-h_{1}}{h_{2}-h _{1}}. \end{cases}$$
(4)
The organizational structure of this article is as follows. In Sect. 2, the existence of homocilinic cycle and existence, uniqueness and asymptotic stability of the periodic orbit of system (3) are proved. Furthermore, the optimization problem is formulated to reduce the total cost of pest control (i.e. the predator and the spraying agent). In Sect. 3, the numerical simulation of the concrete model is gradually carried out to verify the theoretical results. The final conclusion is drawn in Sect. 4.
## 2 Dynamical analysis of system (3)
### 2.1 Equilibria
Without impulse effect, system (3) becomes the following:
$$\textstyle\begin{cases} {\frac{dx}{dt}=x(x-\alpha )(1-x)-\frac{\lambda xy}{x+y},} \\ {\frac{dy}{dt}=\frac{\lambda _{1}xy}{x+y}-\beta y.} \end{cases}$$
(5)
### Theorem 2.1
([51])
System (5) has five equilibria: origin $$O(0,0)$$, two boundary equilibria: $$N_{1}(\alpha ,0)$$ and $$N_{2}(1,0)$$ and two positive equilibria: $$E_{1}(x_{1},y_{1})$$ and $$E_{2}(x_{2},y _{2})$$, where $$x_{1}=\frac{1+\alpha -\sqrt{(1-\alpha )^{2}-4\lambda (1-\frac{\beta }{\lambda _{1}})}}{2}$$, $$y_{1}=\frac{(\lambda _{1}- \beta )x_{1}}{\beta }$$, $$x_{2}=\frac{1+\alpha +\sqrt{(1-\alpha )^{2}-4 \lambda (1-\frac{\beta }{\lambda _{1}})}}{2}$$, $$y_{2}=\frac{(\lambda _{1}-\beta )x_{2}}{\beta }$$.
If $$(H_{1})$$ $$x(t_{0})=x_{0}<\alpha$$, $$y(t_{0})=y_{0}>0$$ holds, then the equilibrium $$O(0,0)$$ of system (5) is stable.
If $$(H_{2})$$ $$\lambda _{1}>\beta$$ holds, then the equilibrium $$N_{1}$$ is an unstable node and $$N_{2}$$ is a saddle point.
If $$(H_{3})$$ $$\lambda _{1}<\beta$$ holds, then the equilibrium $$N_{1}$$ is a saddle point and $$N_{2}$$ is a stable node.
If $$(H_{4})$$ $$\lambda _{1}>\beta >\frac{\alpha _{1}}{4\lambda }[4\lambda -(1-\alpha )^{2}]>0$$ holds, then equilibria $$E_{1}$$ and $$E_{2}$$ are positive equilibria.
If $$(H_{5})$$ $$\operatorname{Tr}(J(E_{2}))=x_{2}(1+\alpha -2x_{2})+\frac{\beta ( \lambda -\lambda _{1})(\lambda _{1}-\beta )}{\lambda _{1}^{2}}<0$$ holds, then point $$E_{1}$$ is a saddle point and point $$E_{2}$$ is a locally asymptotically stable node or focus (see Fig1).
### 2.2 Existence of the order one periodic orbit and homoclinic cycle of system (3)
By using the geometric theory of differential equations and the method of successor function, the existence of the order one periodic orbit and homoclinic cycle of system (3) is investigated in this section. For the actual biological significance, we always suppose that $$\max \{\frac{1+\alpha -\sqrt{(1-\alpha )^{2}-4\lambda (1-\frac{ \beta }{\lambda _{1}})}}{2},h_{1}\}<(1-p)h<h<\min \{\frac{1+\alpha +\sqrt{(1- \alpha )^{2}-4\lambda (1-\frac{\beta }{\lambda _{1}})}}{2},h_{2}\}$$ and conditions $$(H_{4})$$, $$(H_{5})$$ hold in the paper.
Set $$M=\{(x,y)|x=h,0\leq y\leq y_{h}\}$$ is called an impulsive set and set $$N=\{(x,y)|x=(1-p)h,\tau \leq y\leq (1-q)y_{h}+\tau \}$$ is called a phase set. For convenience, for any point L, let $$x_{L}$$ and $$y_{L}$$ denote its abscissa and ordinate, respectively. I is a continuous mapping which satisfies $$I(M)=N$$ and which is called an impulsive function. If $$L(h,y_{L})\in M$$, then the impulse function transfers the point L into $$L^{+}$$. We denote function $$F(J)=J^{-}$$ as the trajectory starting from the point $$J\in N$$ hits the impulse set M at the point $$J^{-}$$. In this paper, the tendency of trajectory is assumed to start from phase set. The unstable manifold and stable manifold of $$E_{1}$$ are denoted as $$T_{1}^{-}(t,E_{1})$$ and $$T_{1}^{+}(t,E_{1})$$, respectively. At time t, $$T_{1}^{-}(t,E_{1})$$ intersects phase set N at point $$A_{1}$$ and impulsive set M at point $$B_{1}$$. At time t, $$T_{1}^{+}(t,E_{1})$$ intersects phase set N at point $$A_{2}$$ and impulsive set M at point $$B_{2}$$. Isocline $$\varGamma _{1}$$: $$\frac{dy}{dt}=0$$ intersects phase set N at point A and impulsive set M at point B, where $$y_{A}=\frac{h(1-p)( \lambda _{1}-\beta )}{\beta }$$ and $$y_{B}=\frac{(\lambda _{1}-\beta )h}{ \beta }$$. The function $$\varphi (y,q)=(1-q)y+\tau$$ is monotonically decreasing about q, and monotonically increasing about y, thus there must exist a value $$q_{*}\in (0,1)$$ such that point B jumps to A after the impulse effect, i.e. $$\varphi (y_{B},q_{*})=(1-q_{*})y _{B}+\tau =y_{A}$$, we have $$q_{*}=p+\frac{\beta \tau }{(\lambda _{1}- \beta )h}$$, for convenience, we set $$\delta =p+\frac{\beta \tau }{( \lambda _{1}-\beta )h}$$. For any $$q\in (0,1)$$, the existence of order one periodic orbit of system (3) is to be proved in the cases of $$0< q\leq \delta$$ and $$\delta < q<1$$, respectively (see Fig. 2).
Case I $$0< q\leq \delta$$.
### Theorem 2.2
1. (a)
If conditions $$(H_{4})(H_{5})$$ and $$q=\delta$$ hold, then system (3) admits a unique order one periodic orbit.
2. (b)
If conditions $$(H_{4})(H_{5})$$ and $$0< q^{*}< q< q^{0}<\delta <1$$ hold, then system (3) admits a unique order one periodic orbit.
### Proof
The trajectory starting from the point $$A\in N$$ hits the point $$B\in M$$, then point $$B\in M$$ jumps to point $$B^{+}$$ after impulsive effect.
(a) Firstly, we prove the existence of the order one periodic orbit of system (3) in the case of $$q=\delta$$. In this case, $$B^{+}$$ coincides with point A, that is, $$\widehat{AB}$$ and $$\overline{BA}$$ constitute the order one periodic orbit.
(b) In this case, $$B^{+}$$ is above A. According to the impulsive equations of system (3), there surely has a value $$q^{*} \in (0,\delta )$$ satisfying $$\varphi (y_{B_{1}},q^{*})=(1-q^{*})y_{B _{1}}+\tau =y_{A_{2}}$$, that is, point $$B_{1}$$ jumps to point $$A_{2}$$ after impulse effect, also, there surely has a value $$q^{0}\in (q^{*},\delta )$$ satisfying $$\varphi (y_{B_{1}},q^{0})=(1-q ^{0})y_{B_{1}}+\tau =y_{A}$$. For any $$q\in (q^{*},q^{0})$$, point $$B_{1}$$ jumps to point $$B_{1}^{+}$$ after impulsive effect, we have $$(1-q^{0})y_{B_{1}}+\tau =y_{A}<(1-q)y_{B_{1}}+\tau =y_{B_{1}^{+}}<(1-q ^{*})y_{B_{1}}+\tau =y_{A_{2}}$$, that is, $$y_{A}< y_{B_{1}^{+}}< y_{A _{2}}$$. There must exist a trajectory going through point $$B_{1}^{+}$$ and intersecting the line $$x=h$$ at point $$C_{1}$$, and $$I(C_{1})=C_{1} ^{+}$$. In view of the disjointness of any two trajectories and the vector field of system (3), we have $$y_{B_{1}}< y_{C_{1}}< y _{B}$$, $$(1-q)y_{B_{1}}+\tau =y_{B_{1}^{+}}<(1-q)y_{C_{1}}+\tau =y_{C _{1}^{+}}$$, that is, $$y_{C_{1}^{+}}>y_{B_{1}^{+}}$$. Thus the successor function [54, Definition 3.3] of $$B_{1}^{+}$$: $$g(B_{1} ^{+})=y_{C_{1}^{+}}-y_{B_{1}^{+}}>0$$.
A point $$C((1-p)h, y_{A_{2}}-\varepsilon )$$ is selected in set N, where $$\varepsilon >0$$ is small enough, then point C is fully close to point $$A_{2}$$. Set $$F(C)=C_{2}\in M$$, we have $$y_{C_{2}}>y_{B_{1}}$$ due to continuous dependence of the solution on initial value and time, and $$C_{2}$$ is close enough to point $$B_{1}$$. Thus we have $$y_{C_{2} ^{+}}>y_{B_{1}^{+}}$$, and point $$C_{2}^{+}$$ is close enough to point $$B_{1}^{+}$$, then we have $$y_{C_{2}^{+}}< y_{C}$$, that is, $$g(C)=y_{C _{2}^{+}}-y_{C}<0$$. According to [54, Lemma 3.2, 3.3], there has a point $$P\in (B_{1}^{+},C)$$ such that $$g(P)=0$$, i.e. system (3) admits the order one periodic orbit, whose initial point is between point $$B_{1}^{+}$$ and point C in the set N (see Fig. 3(a)).
Next, we prove the uniqueness of the order one periodic orbit of system (3). We arbitrarily select two points K and Q in the $$\overline{AA_{2}}$$, where $$y_{A}< y_{K}< y_{Q}< y_{A_{2}}$$. Set $$F(K)=K_{1}\in M$$, $$F(Q)=Q_{1}\in M$$, then we have $$y_{Q_{1}}< y_{K_{1}}$$, and $$K_{1}$$ and $$Q_{1}$$ are, respectively, mapped to $$K_{1}^{+}\in N$$ and $$Q_{1}^{+}\in N$$ after the impulse effect, then the successor functions of K, Q satisfy
\begin{aligned} g(K)-g(Q) =&(y_{K_{1}^{+}}-y_{K})-(y_{Q_{1}^{+}}-y_{Q}) \\ =&(y_{Q}-y_{K})+(y_{K_{1}^{+}}-y_{Q_{1}^{+}}) \\ =&(y_{Q}-y_{K})+(1-q)y_{K_{1}}+\tau -(1-q)y_{Q_{1}}-\tau \\ =&(y_{Q}-y_{K})+(1-q) (y_{K_{1}}-y_{Q_{1}})>0. \end{aligned}
Thus the successor function is monotonically increasing in the $$\overline{AA_{2}}$$, then there is only one point $$P\in (A,A_{2})$$ such that $$g(P)=0$$, i.e. system (3) has only an order one periodic orbit when q satisfies $$0< q<\delta$$ (see Fig. 3(b)). This completes the proof. □
### Theorem 2.3
If conditions $$(H_{4})$$ and $$(H_{5})$$ and $$q=q^{*}$$ hold, then system (3) admits an order one homoclinic cycle; if $$q\in (0,q^{*})$$, system (3) does not admit an order one periodic orbit and the pests and natural enemy population will become extinct.
### Proof
When $$q=q^{*}$$, we have $$\varphi (y_{B_{1}}, q_{*})=(1-q^{*})y_{B_{1}}+ \tau =y_{A_{2}}$$, then the closed curve $$\overline{B_{1}A_{2}}\cup \widehat{A_{2}E_{1}}\cup \widehat{E_{1}B_{1}}$$ forms a cycle which passes through the saddle $$E_{1}$$. Thus system (3) admits an order one homoclinic cycle (see Fig. 4).
When $$q\in (0,q^{*})$$, we have $$(1-q)y_{B_{1}}+\tau =y_{B_{1}^{+}}>(1-q ^{*})y_{B_{1}}+\tau =y_{A_{2}}$$ and the trajectory of system (3) starting from $$B_{1}^{+}$$ will tend to origin when $$t\rightarrow +\infty$$, and it has no impulse effect. Thus system (3) has no order one periodic orbit and the pest and natural enemy population will become extinct. This completes the proof. □
Case II $$\delta < q<1$$.
### Theorem 2.4
If the conditions $$(H_{4})(H_{5})$$ and $$\delta < q<1$$ hold, then the system (3) admits the order one periodic orbit.
### Proof
In this case, we know that $$B^{+}$$ is below A. Set $$F(B^{+})=S_{1} \in M$$, according to the property of orbit of system (3), we know $$y_{B}>y_{S_{1}}$$, thus we have $$y_{B^{+}}>y_{S_{1}^{+}}$$, then the successor function of point $$B^{+}$$ is $$g(B^{+})=y_{S_{1}^{+}}-y_{B ^{+}}<0$$. A point $$H((1-p)h,\varepsilon )\in N$$ is selected, where $$\varepsilon <\tau$$, and set $$F(H)=H_{1}\in M$$, after impulsive effect, point $$H_{1}$$ jumps to $$H_{1}^{+}$$, then $$y_{H_{1}^{+}}=(1-q)y_{H_{1}}+ \tau >\varepsilon$$, that is, $$y_{H_{1}^{+}}>y_{H}$$, thus we have $$g(H)=y_{H_{1}^{+}}-y_{H}>0$$. By [54, Lemma 3.2, 3.3], there exists a point $$P\in (H,B^{+})$$ satisfying $$g(P)=0$$, that is, system (3) admits the order one periodic orbit, whose initial point is between H and $$B^{+}$$ in the set N. This completes the proof (see Fig. 5).
□
### 2.3 Stability of the order one periodic orbit of system (3)
Next, we study the stability of the order one periodic orbit of system (3).
### Theorem 2.5
If conditions $$(H_{4})$$, $$(H_{5})$$, and $$(1-\alpha )+\frac{\lambda h}{ \alpha }-\frac{\lambda _{1}\alpha y_{1}}{(1+h)^{2}}<0$$ and $$\vert \chi \vert <1$$ hold, then the order one periodic orbit of system (3) is orbitally asymptotically stable, where
$$\chi =\frac{ [(1-p)h-\alpha ] [1-(1-p)h ](h+ \eta _{1})}{(h-\alpha )(1-h)(h+\eta _{1})-\lambda \eta _{1}}.$$
### Proof
Let $$x=\xi (t)$$, $$y=\eta (t)$$ be a T-periodic orbit of system (3) and $$\xi _{1}=\xi (T)=h$$, $$\eta _{1}=\xi (T)$$, $$\xi _{0}= \xi (0)$$, $$\eta _{0}=\eta (0)$$, $$\xi _{1}^{+}=\xi (T^{+})$$, $$\eta _{1}^{+}= \eta (T^{+})$$, then we have
$$\xi _{1}^{+}=\xi _{0}=(1-p)h, \quad\quad \eta _{1}^{+}=\eta _{0}=(1-q)\eta _{1}+\tau .$$
Let $$P(x,y)=x(x-\alpha )(1-x)-\frac{\lambda xy}{x+y}$$, $$Q(x,y)=\frac{ \lambda _{1}xy}{x+y}-\beta y$$, $$\varPhi (x,y)=-px$$, $$\varPsi (x,y)=-qy+ \tau$$, $$\varUpsilon (x,y)=x-h$$.
Then
\begin{aligned}& \frac{\partial \varPhi }{\partial x}=-p, \quad\quad \frac{\partial \varPsi }{\partial x}=0, \quad\quad \frac{\partial \varPhi }{\partial y}=0,\quad\quad \frac{\partial \varPsi }{ \partial y}=-q, \quad\quad \frac{\partial \varUpsilon }{\partial x}=1, \frac{\partial \varUpsilon }{\partial y}=0, \\& \begin{aligned} \Delta _{1} &=\frac{P_{+}(\frac{\partial \varPsi }{\partial y}\frac{ \partial \varUpsilon }{\partial x}-\frac{\partial \varPsi }{\partial x}\frac{ \partial \varUpsilon }{\partial y}+\frac{\partial \varUpsilon }{\partial x})+Q _{+}(\frac{\partial \varPhi }{\partial x}\frac{\partial \varUpsilon }{ \partial y}-\frac{\partial \varPhi }{\partial y}\frac{\partial \varUpsilon }{\partial x}+\frac{\partial \varUpsilon }{\partial y})}{P\frac{\partial \varUpsilon }{\partial x}+Q\frac{\partial \varUpsilon }{\partial y}} \\ &=\frac{P(\xi _{1}^{+},\eta _{1}^{+})(-q\times 1-0\times 0+1)+Q(\xi _{1}^{+},\eta _{1}^{+})(-p\times 0-0\times 1+0)}{P(\xi _{1},\eta _{1}) \times 1+Q(\xi _{1},\eta _{1})\times 0} \\ &=\frac{(1-q)\xi _{0} [(\xi _{0}-\alpha )(1-\xi _{0})-\frac{\lambda \eta _{0}}{\xi _{0}+\eta _{0}} ]}{\xi _{1} [(\xi _{1}-\alpha )(1- \xi _{1})-\frac{\lambda \eta _{1}}{\xi _{1}+\eta _{1}} ]}, \end{aligned} \end{aligned}
and
\begin{aligned} \int _{0}^{T}\biggl(\frac{\partial P}{\partial x}+ \frac{\partial Q}{\partial y}\biggr) =& \int _{0}^{T} \biggl[(x-\alpha ) (1-x)- \frac{\lambda y}{x+y}+\frac{ \lambda _{1}x}{x+y}-\beta \biggr]\,dt \\ & {} + \int _{0}^{T} \biggl[x(1+\alpha -2x)+ \frac{\lambda xy}{(x+y)^{2}}-\frac{ \lambda _{1}xy}{(x+y)^{2}} \biggr]\,dt \\ =& \int _{0}^{T} \biggl[\frac{\dot{x}(t)}{x(t)}+ \frac{\dot{y}(t)}{y(t)} \biggr]\,dt+ \int _{0}^{T} \biggl[x(1+\alpha -2x)+ \frac{\lambda xy}{(x+y)^{2}}-\frac{ \lambda _{1}xy}{(x+y)^{2}} \biggr]\,dt \\ =&\ln \frac{x(T)y(T)}{x(0)y(0)}+ \int _{0}^{T} \biggl[x(1+\alpha -2x)+ \frac{ \lambda xy}{(x+y)^{2}}-\frac{\lambda _{1}xy}{(x+y)^{2}} \biggr]\,dt \\ =&\ln \frac{\xi _{1}\eta _{1}}{\xi _{0}\eta _{0}}+ \int _{0}^{T} \biggl[x(1+ \alpha -2x)+ \frac{\lambda xy}{(x+y)^{2}}- \frac{\lambda _{1} xy}{(x+y)^{2}} \biggr]\,dt. \end{aligned}
Furthermore,
\begin{aligned} \mu _{2} =&\Delta _{1}\exp \int _{0}^{T} \biggl[\frac{\partial P}{\partial x}\bigl(\xi (t), \eta (t)\bigr)+\frac{\partial Q}{\partial y}\bigl(\xi (t),\eta (t)\bigr) \biggr]\,dt \\ =&\frac{(1-q)\xi _{0} [(\xi _{0}-\alpha )(1-\xi _{0})-\frac{\lambda \eta _{0}}{\xi _{0}+\eta _{0}} ]}{\xi _{1} [(\xi _{1}-\alpha )(1- \xi _{1})-\frac{\lambda \eta _{1}}{\xi _{1}+\eta _{1}} ]} \\ & {} \cdot \exp \biggl[\ln \frac{\xi _{1}\eta _{1}}{\xi _{0}\eta _{0}}+ \int _{0}^{T} \biggl(x(1+\alpha -2x)+ \frac{\lambda xy}{(x+y)^{2}}-\frac{ \lambda _{1} xy}{(x+y)^{2}} \biggr)\,dt \biggr] \\ =&\frac{(1-q)\eta _{1} [(\xi _{0}-\alpha )(1-\xi _{0})-\frac{ \lambda \eta _{0}}{\xi _{0}+\eta _{0}} ]}{\eta _{0} [(\xi _{1}- \alpha )(1-\xi _{1})-\frac{\lambda \eta _{1}}{\xi _{1}+\eta _{1}} ]} \\ & {} \cdot \exp \biggl[ \int _{0}^{T} \biggl(x(1+\alpha -2x)+ \frac{\lambda xy}{(x+y)^{2}}-\frac{\lambda _{1} xy}{(x+y)^{2}} \biggr)\,dt \biggr]. \end{aligned}
For any point $$(x,y)\in \varTheta$$, $$\alpha < x<1$$ and $$y_{1}< y< h$$, we have $$(1-\alpha )+\frac{\lambda h}{\alpha }- \frac{\lambda _{1}\alpha y_{1}}{(1+h)^{2}}<0$$, so we can get
$$x(1+\alpha -2x)+\frac{\lambda xy}{(x+y)^{2}}- \frac{\lambda _{1}xy}{(x+y)^{2}}< (1-\alpha )+ \frac{\lambda h}{\alpha }-\frac{ \lambda _{1}\alpha y_{1}}{(1+h)^{2}}< 0,$$
and due to $$\vert \chi \vert <1$$, we have
$$\biggl\vert \frac{(1-q)\eta _{1} [(\xi _{0}-\alpha )(1-\xi _{0})-\frac{ \lambda \eta _{0}}{\xi _{0}+\eta _{0}} ]}{\eta _{0} [(\xi _{1}- \alpha )(1-\xi _{1})-\frac{\lambda \eta _{1}}{\xi _{1}+\eta _{1}} ]} \biggr\vert < 1.$$
Therefore $$\vert \mu _{2} \vert <1$$. According to the analog of the Poincaré criterion [55, Theorem 2.3], we know that the order one periodic orbit of system (3) is orbitally asymptotically stable. This completes the proof. □
### 2.4 Determination of optimal pest economic threshold
In order to determine the optimum release amount of natural enemies and the optimal frequency of spraying chemical pesticide, we formulate the following optimization problem and find the optimal economic threshold.
Let $$l_{1}$$ represent the unit cost of the biological control, $$l_{2}$$ the unit cost of the chemical control. Our final purpose is to minimize the expenses of per unit period. We denote by F the total expenses in a period of system (3), which is a function of the chemical control strength $$p(h)$$ and release amount of natural enemies $$\tau (h)$$. Then we have $$F(h)=l_{1}\tau (h)+l_{2}p(h)$$. Thus we formulate the following optimization model:
\begin{aligned}& \min \frac{F(h)}{T(h)}\\& \text{s.t. }h_{1}< h< h_{2}. \end{aligned}
Solving the objective function yields the optimum economic threshold $$h^{*}$$, which results in the optimum release amount of the predator $$\tau ^{*}=\tau (h^{*})$$, the optimal chemical control strength $$p^{*}=p(h^{*})$$ and the optimal control period of chemical control $$T^{*}=T(\tau ^{*},p^{*})$$. However, it is important to be noted that the optimal economic threshold $$h^{*}$$ is dependent on the ratio of $$\omega =\frac{l_{2}}{l_{1}}$$.
## 3 Simulations and optimization
In order to verify the theoretical results obtained in this paper, a specific example is presented in this section. Let $$\alpha =0.1$$, $$\lambda =0.5$$, $$\lambda _{1}=0.8$$, $$\beta =0.55$$, $$h_{1}=0.35$$, $$h_{2}=0.76$$, $$p _{\max }=0.3$$, $$\tau _{\max }=0.1$$, $$\tau _{\min }=0.01$$, $$q_{\max }=0.8$$. By a simple calculation, the saddle point and locally asymptotically stable point are $$E_{1}(0.3349,0.1522)$$ and $$E_{0}(0.7651,0.3478)$$, respectively. We carry out simulations by changing the main parameter h and fixing all other parameters. The control parameters p, q, τ are calculated by (4).
### 3.1 Numerical simulations
Taking parameters α, λ, β, $$\lambda _{1}$$ into system (5), we can obtain
$$\textstyle\begin{cases} {\frac{dx}{dt}=x(t) [x(t)-0.1 ] [1-x(t) ]-\frac{0.5 x(t)y(t)}{x(t)+y(t)},} \\ {\frac{dy}{dt}=\frac{0.8x(t)y(t)}{x(t)+y(t)}-0.55 y(t).} \end{cases}$$
(6)
Furthermore, Fig. 6(a) shows the phase portrait of $$x(t)$$ and $$y(t)$$, Fig. 6(b) shows time series of $$x(t)$$, Fig. 6(c) shows time series of $$y(t)$$. Let $$p=0.27$$, $$h=0.61$$, $$\tau =0.18$$, we shall get $$\delta =p+\frac{\beta \tau }{(\lambda _{1}-\beta )h}=0.27+\frac{0.55 \times 0.18}{(0.8-0.55)\times 0.61}=0.92$$. Let $$q=0.01$$ we shall get Fig. 7. Figures 7(a), 7(b) and 7(c) show that system (3) does not have order one periodic orbit and the pest and natural enemy population become extinct when $$0< q< q^{*}$$. Let $$q=0.6$$, $$q=0.7$$ and $$q=0.8$$ and we shall get a unique order one periodic solution and it is asymptotically stable (see Fig. 8). Furthermore, according to Fig. 8, we see that, as the parameter q increases, the period T becomes smaller. Figure 9 shows that system (3) admits the order one periodic orbit when $$q>\delta$$. According to Fig. 9, we see that, as the parameter q increases, the period T becomes smaller.
### 3.2 Optimum pest control level
The impulse period T of the order one periodic orbit varies with the economic threshold h, as is shown in Fig. 10. And Fig. 11 shows the variation of cost per unit time $$F/T$$ and the period T with the pest control level h. Assume $$l_{1}=10{,}000$$, $$l_{2}=100$$, we shall get $$\omega =\frac{l_{2}}{l_{1}}=1/100$$.
The optimal pest threshold is $$h^{*}=0.46$$, the optimal chemical control strength is $$p^{*}=0.0878$$, the optimal release amount of the natural enemies is $$\tau ^{*}=0.0737$$. It is important to note that the optimum economic threshold h is dependent on ω, that is, the larger ω is, the lower h is, as is illustrated in Fig. 12.
## 4 Conclusion
In this paper, a predator–prey model with ratio-dependent and impulsive state feedback control is proposed. We take control measures that combine biological control and chemical control. The aim of this study is to get optimum pest control level and minimize the cost of pest control.
Based on the above analysis, both the predator and the prey are extinct when $$0< q< q^{*}$$, that is, the order one periodic orbit does not exist when $$0< q< q^{*}$$, system (3) admits an order one homoclinic cycle for $$q=q^{*}$$; we obtain a unique and stable order one periodic orbit according to the geometric theory of the differential equations and the method of subsequence functions in the case of $$q^{*}< q<\delta$$; and system (3) admits the order one periodic orbit in the case of $$\delta < q<1$$.
In order to reduce the total cost of pest management, an optimization problem is formulated, and the optimal economic threshold and the minimum control cost are obtained. Finally, numerical simulation is carried out to verify the theoretical results.
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## Funding
The paper was supported by the National Natural Science Foundation of China (No. 11371230), Shandong Provincial Natural Science Foundation, China (No. S2015SF002), SDUST Research Fund (2014TDJH102), and Joint Innovative Center for Safe and Effective Mining Technology and Equipment of Coal Resources, Shandong Province of China.
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Shi, Z., Wang, J., Li, Q. et al. Control optimization and homoclinic bifurcation of a prey–predator model with ratio-dependent. Adv Differ Equ 2019, 2 (2019). https://doi.org/10.1186/s13662-018-1933-z
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https://ftp.aimsciences.org/article/doi/10.3934/dcds.2011.31.275 | 1,656,593,861,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00244.warc.gz | 315,219,582 | 16,147 | # American Institute of Mathematical Sciences
March 2011, 31(1): 275-299. doi: 10.3934/dcds.2011.31.275
## Preservation of homoclinic orbits under discretization of delay differential equations
1 School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China 2 School of Mathematics, Jilin University, Changchun 130012, China
Received February 2010 Revised November 2010 Published June 2011
In this paper, we propose a nondegenerate condition for a homoclinic orbit with respect to a parameter in delay differential equations. Based on this nondegeneracy we describe and investigate the regularity of the homoclinic orbit together with parameter. Then we show that a forward Euler method, when applied to a one-parameteric system of delay differential equations with a homoclinic orbit, also exhibits a closed loop of discrete homoclinic orbits. These discrete homoclinic orbits tend to the continuous one by the rate of $O(\varepsilon)$ as the step-size $\varepsilon$ goes to $0$. And the corresponding parameter varies periodically with respect to a phase parameter with period $\varepsilon$ while the orbit shifts its index after one revolution. We also show that at least two homoclinic tangencies occur on this loop. By numerical simulations, the theoretical results are illustrated, and the possibility of extending theoretical results to the implicit and higher order numerical schemes is discussed.
Citation: Yingxiang Xu, Yongkui Zou. Preservation of homoclinic orbits under discretization of delay differential equations. Discrete and Continuous Dynamical Systems, 2011, 31 (1) : 275-299. doi: 10.3934/dcds.2011.31.275
##### References:
[1] W.-J. Beyn, The effect of discretization on homoclinic orbits, in "Bifurcation: Analysis, Algorithms, Applications," Internat. Ser. Numer. Math., 79, Birkhäuser, Basel, (1987), 1-8. [2] W.-J. Beyn, The numerical computation of connecting orbits in dynamical systems, IMA J. Numer. Anal., 10 (1990), 379-405. doi: 10.1093/imanum/10.3.379. [3] W.-J. Beyn and J.-M. Kleinkauf, The numerical computation of homoclinic orbits for maps, SIAM J. Numer. Anal., 34 (1997), 1207-1236. doi: 10.1137/S0036142995281693. [4] W. A. Coppel, "Dichotomies in Stability Theory," Lecture Notes in Mathematics, 629, Springer-Verlag, Berlin-New York, 1978. [5] K. Engelborghs and E. J. Doedel, Stability of piecewise polynomial collocation for computing periodic solutions of delay differential equations, Numer. Math., 91 (2002), 627-648. doi: 10.1007/s002110100313. [6] K. Engelborghs, T. Luzyanina and D. Roose, Numerical bifurcation analysis of delay differential equations using DDE-BIFTOOL, ACM Trans. Math. Software, 28 (2002), 1-21. doi: 10.1145/513001.513002. [7] G. Farkas, Unstable manifolds for RFDEs under discretization: The Euler method, Comput. Math. Appl., 42 (2001), 1069-1081. doi: 10.1016/S0898-1221(01)00222-X. [8] G. Farkas, A numerical $C^1$-shadowing result for retarded functional differential equations, J. Compt. Appl. Math., 145 (2002), 269-289. doi: 10.1016/S0377-0427(01)00581-7. [9] G. Farkas, Nonexistence of uniform exponential dichotomies for delay equations, J. Differential Equations, 182 (2002), 266-268. [10] B. Fiedler and J. Scheurle, Discretization of homoclinic orbits, rapid forcing and 'invisible' chaos, Mem. Amer. Math. Soc., 119 (1996). [11] J. K. Hale and S. M. Lunel, "Introduction to Functional-Differential Equations," Applied Mathematical Sciences, 99, Springer-Verlag, New York, 1993. [12] J. K. Hale and W. Zhang, On uniformity of exponential dichotomies for delay equations, J. Differential Equations, 204 (2004), 1-4. [13] K. In't Hout and C. Lubich, Periodic orbits of delay differential equations under discretization, BIT, 38 (1998), 72-91. doi: 10.1007/BF02510918. [14] U. Kirehgraber, F. Lasagni, K. Nipp and D. Stoffer, On the application of invariant manifold theory, in particular to numerical analysis, in "Bifurcation and Chaos: Analysis, Algorithms, Applications," Internat. Ser. Numer. Math. (eds. R. Seydel et al.), 97, Birkhäuser, Basel, (1991), 189-197. [15] X.-B. Lin, Exponential dichotomies and homoclinic orbits in functional-differential equations, J. Differential Equations, 63 (1986), 227-254. [16] K. J. Palmer, "Shadowing in Dynamical Systems. Theory and Applications," Mathematics and its Applications, 501, Kluwer Academic Publishers, Dordrecht, 2000. [17] M. L. Peña, Exponential dichotomy for singularly perturbed linear functional-differential equations with small delays, Appl. Anal., 47 (1992), 213-225. doi: 10.1080/00036819208840141. [18] G. Samaey, K. Engelborghs and D. Roose, Numerical computation of connecting orbits in delay differential equations, Numer. Algorithms, 30 (2002), 335-352. doi: 10.1023/A:1020102317544. [19] Y.-K. Zou and W.-J. Beyn, On manifolds of connecting orbits in discretizations of dynamical systems, Nonlinear Anal., 52 (2003), 1499-1520. doi: 10.1016/S0362-546X(02)00269-9. [20] Y.-K. Zou and W.-J. Beyn, On the existence of transversal heteroclinic orbits in discretized dynamical systems, Nonlinearity, 17 (2004), 2275-2292. doi: 10.1088/0951-7715/17/6/014.
show all references
##### References:
[1] W.-J. Beyn, The effect of discretization on homoclinic orbits, in "Bifurcation: Analysis, Algorithms, Applications," Internat. Ser. Numer. Math., 79, Birkhäuser, Basel, (1987), 1-8. [2] W.-J. Beyn, The numerical computation of connecting orbits in dynamical systems, IMA J. Numer. Anal., 10 (1990), 379-405. doi: 10.1093/imanum/10.3.379. [3] W.-J. Beyn and J.-M. Kleinkauf, The numerical computation of homoclinic orbits for maps, SIAM J. Numer. Anal., 34 (1997), 1207-1236. doi: 10.1137/S0036142995281693. [4] W. A. Coppel, "Dichotomies in Stability Theory," Lecture Notes in Mathematics, 629, Springer-Verlag, Berlin-New York, 1978. [5] K. Engelborghs and E. J. Doedel, Stability of piecewise polynomial collocation for computing periodic solutions of delay differential equations, Numer. Math., 91 (2002), 627-648. doi: 10.1007/s002110100313. [6] K. Engelborghs, T. Luzyanina and D. Roose, Numerical bifurcation analysis of delay differential equations using DDE-BIFTOOL, ACM Trans. Math. Software, 28 (2002), 1-21. doi: 10.1145/513001.513002. [7] G. Farkas, Unstable manifolds for RFDEs under discretization: The Euler method, Comput. Math. Appl., 42 (2001), 1069-1081. doi: 10.1016/S0898-1221(01)00222-X. [8] G. Farkas, A numerical $C^1$-shadowing result for retarded functional differential equations, J. Compt. Appl. Math., 145 (2002), 269-289. doi: 10.1016/S0377-0427(01)00581-7. [9] G. Farkas, Nonexistence of uniform exponential dichotomies for delay equations, J. Differential Equations, 182 (2002), 266-268. [10] B. Fiedler and J. Scheurle, Discretization of homoclinic orbits, rapid forcing and 'invisible' chaos, Mem. Amer. Math. Soc., 119 (1996). [11] J. K. Hale and S. M. Lunel, "Introduction to Functional-Differential Equations," Applied Mathematical Sciences, 99, Springer-Verlag, New York, 1993. [12] J. K. Hale and W. Zhang, On uniformity of exponential dichotomies for delay equations, J. Differential Equations, 204 (2004), 1-4. [13] K. In't Hout and C. Lubich, Periodic orbits of delay differential equations under discretization, BIT, 38 (1998), 72-91. doi: 10.1007/BF02510918. [14] U. Kirehgraber, F. Lasagni, K. Nipp and D. Stoffer, On the application of invariant manifold theory, in particular to numerical analysis, in "Bifurcation and Chaos: Analysis, Algorithms, Applications," Internat. Ser. Numer. Math. (eds. R. Seydel et al.), 97, Birkhäuser, Basel, (1991), 189-197. [15] X.-B. Lin, Exponential dichotomies and homoclinic orbits in functional-differential equations, J. Differential Equations, 63 (1986), 227-254. [16] K. J. Palmer, "Shadowing in Dynamical Systems. Theory and Applications," Mathematics and its Applications, 501, Kluwer Academic Publishers, Dordrecht, 2000. [17] M. L. Peña, Exponential dichotomy for singularly perturbed linear functional-differential equations with small delays, Appl. Anal., 47 (1992), 213-225. doi: 10.1080/00036819208840141. [18] G. Samaey, K. Engelborghs and D. Roose, Numerical computation of connecting orbits in delay differential equations, Numer. Algorithms, 30 (2002), 335-352. doi: 10.1023/A:1020102317544. [19] Y.-K. Zou and W.-J. Beyn, On manifolds of connecting orbits in discretizations of dynamical systems, Nonlinear Anal., 52 (2003), 1499-1520. doi: 10.1016/S0362-546X(02)00269-9. [20] Y.-K. Zou and W.-J. Beyn, On the existence of transversal heteroclinic orbits in discretized dynamical systems, Nonlinearity, 17 (2004), 2275-2292. doi: 10.1088/0951-7715/17/6/014.
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Mathematics Asked on January 5, 2022
Let $$f,g$$ be holomorphic function in $$mathbb{D}$$ that are continuous in $$overline{mathbb{D}}$$. Show that if $$f=g$$ on $$|z|=1$$, then $$f=g$$
It seems like identity theorem. But they have to be equal on an open connected set. $$overline{mathbb{D}}$$ is not open so I cannot use the identity theorem. At least directly.
Let $$varphi=f-g$$, then $$varphi$$ is holomorphic on $$mathbb{D}$$ and thus $$forall zinmathbb{D},|varphi(z)|leqslantmax_{|u|=1}|varphi(u)|=0$$ and thus $$varphi=0$$ on $$mathbb{D}$$ and $$f=g$$ on $$overline{mathbb{D}}$$.
Answered by Tuvasbien on January 5, 2022
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### Convergence of $sqrt{1-sqrt{{1}-sqrt{{1}-sqrt{{1-sqrt{{1}}}…}}}}$
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### Is this series $sum_{n=0}^{infty} 2^{(-1)^n – n} = 3$ or $approx 3$
2 Asked on November 24, 2021 | 983 | 2,861 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-50 | latest | en | 0.803644 |
http://www.quickermaths.com/interesting-puzzle/ | 1,558,749,273,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257845.26/warc/CC-MAIN-20190525004721-20190525030721-00377.warc.gz | 335,897,658 | 13,629 | # Interesting Puzzle
Time to solve a very interesting and logical puzzle
I have picked up this puzzle from IBM monthly challenge. I will give the exact link of the site later (for those who are interested)
Background: It’s Thanksgiving, and time to stuff the turkey. But we’ve got a problem. The stuffing came in a cube measuring 12cm on a side. The hole in the turkey, however, will accommodate a rectangular block measuring 8cm by 8cm by 27cm. Due to time constraints (we can’t miss the parade on TV) we only want to cut the stuffing into four pieces.
The question:
How can we cut the 12cm cube into four pieces which will reassemble into an 8cm by 8cm by 27cm block?
If you have any puzzle, riddle, brainteaser, etc. you think we might enjoy, please send them in. All mails should be sent to: vineetpatawari@gmail.com
#### Vineet Patawari
Hi, I'm Vineet Patawari. I fell in love with numbers after being scared of them for quite some time. Now, I'm here to make you feel comfortable with numbers and help you get rid of Math Phobia!
## 4 thoughts to “Interesting Puzzle”
1. ANS IS 240 .BECAUSE AS PER THE QUES U CANNOT HAVE THE HT AS 27.SINCE HT OF THE CUBE IS ONLY 12.SO IT WOULD BE[( 12*12*12)-(8*8*12)]/4=240…………..
IF NOT I WOULD BE WAITING TO KNOW THE CORRECT ANS…..
2. arun chakravarthi says:
dandanaka ai danukunakka
3. Step 1: Facing the section 12cm wide by 12cm high, make a ‘staircase cut’ to get one ‘L’ shaped piece at the bottom and one inverted ‘L’ shaped piece at the top each with width 8cm and height 12 cm.
Step 2: Slide the upper inverted ‘L’ piece over the bottom ‘L’ piece to get a cuboid 8cm wide by 18cm high by 12 cm thick.
Step 3: Turn the block and looking at the face 12cm wide by 18cm high, again make a ‘staircase cut’ to get two ‘L’ shaped pieces as in step 1 each with width 8cm and height 18 cm.
Step 4: Slide the upper inverted ‘L’ piece over the bottom ‘L’ piece to get a cuboid 8 wide by 27cm high by 8 cm thick. | 567 | 1,966 | {"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-2019-22 | latest | en | 0.91706 |
https://math.stackexchange.com/questions/1791598/linear-transformation-of-n-dimensional-space-has-n1-eigenvectors | 1,561,064,755,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999273.79/warc/CC-MAIN-20190620210153-20190620232153-00385.warc.gz | 507,063,908 | 36,156 | # Linear transformation of n-dimensional space has n+1 eigenvectors
Linear transformation of $n$-dimensional space has $n+1$ eigenvectors, such as each $n$ of them are linear independent. Find all possible matrices, that can represent such transformation.
The bold part of question made me question everything I know about eigenvectors. Assume, we have linear transformation $A$:
1. What does it mean for A to have $n$ eigenvectors?
Eigenvector is vector $x$, that satisfies $Ax=\lambda x$. But any collinear vector $cx, c \ne 0$ is also eigenvector, so it seems to me we have infinite number of vectors.
2. How can $n$-dimensional A have $n+1$ vectors?
Okay, we can make some constrain on eigenvector, like $x$ is eigenvector if $|x| = 1$, and has 'positive' direction. Not very smart, because now eigenvectors depend on norm and direction selection. Now we either have $n$ distinct eigenvalues and $n$ eigenvectors, or some repeated eigenvalue, in which case eigenvector has some free constant in it. So it also becomes many vectors. So it should be either $n$ or infinite vectors.
3. Maybe after understanding 1 and 2 I will be able to solve original problem, but any hints are welcome.
Thanks to @Catalin Zara, I can prove that this linear transformation is diagonizable!
Let $v_1 ... v_{n+1}$ be our system of vectors. $n+1$ vectors in $n$ dimensional space are linear dependent: $$v_{n+1}=a_1v_1 + ... + a_nv_n$$
Apply linear transformation A to both sides of equation, assuming that $\lambda_n$ corresponds to $v_n$ and $v_{n+1}$: $$\lambda_n v_{n+1} = \lambda_1 a_1v_1 + ... + \lambda_n a_nv_n$$
Multipling first equation by $\lambda_n$ and subtructing from second: $$0=a_1(\lambda_1-\lambda_n)v_1 + ... + a_{n-1}(\lambda_{n-1}-\lambda_n)v_{n-1}$$
This must mean one of two things: all $a_1 ... a_{n-1}$ equals zero, or $\lambda_n$ equals zero. First cannot be true, because if it is: $v_{n+1}=a_nv_n$. Thus, we have one zero eigenvalue and it means matrix A is diagonalizable.
• The problem doesn't say that there are exactly $n+1$ eigenvectors. Can you show that the transformation is diagonalizable and has exactly one eigenspace? (By the way, that was Putnam 1988 - A6). – Catalin Zara May 19 '16 at 12:58
Your reasoning at the end is not correct. First of all, your last equation should read $$0 = a_1 (\lambda_1 - \lambda_n)v_1 + \ldots + a_{n-1}(\lambda_{n-1}-\lambda_n)v_{n-1}$$ From this we conclude that $a_i (\lambda_i - \lambda_n) = 0$ for all $i$, since the $v_i$ are linearly independent. If $a_i = 0$, then we have $$0 = a_1 v_1 + \ldots + a_{i-1}v_{i-1} + a_{i+1}v_{i+1} + \ldots - v_{n+1}$$ which is impossible because $v_1 , \ldots, v_{i-1}, v_{i+1}, \ldots v_{n+1}$ are linearly independent. It follows that $\lambda_i = \lambda_n$ for all $i$, so $A$ has only one eigenvalue, hence it is a multiple of the identity matrix.
• Oh, I missed it, thank you. But statement that it holds if $\lambda_n = 0$ is valid? So it is either diagonalizable with eigenvalue 0, or multiple of identity? – DoctorMoisha May 19 '16 at 15:10 | 932 | 3,057 | {"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} | 4.34375 | 4 | CC-MAIN-2019-26 | latest | en | 0.894206 |
http://brainly.com/question/280813 | 1,480,855,300,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541321.31/warc/CC-MAIN-20161202170901-00047-ip-10-31-129-80.ec2.internal.warc.gz | 36,770,041 | 10,035 | # I need help identifying the point and slope of y-2=1/4(x-3)
2
by dillalisha
the slope is rise over run in this case the slope is 1/4 and the staring point or (y intercept) is -3
Thank you | 61 | 190 | {"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-2016-50 | latest | en | 0.919103 |
https://docs.qgis.org/2.18/ko/docs/training_manual/basic_map/vector_data.html | 1,642,941,316,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00026.warc.gz | 271,639,294 | 9,029 | 3.1. Lesson: 벡터 데이터 작업¶
Vector data is arguably the most common kind of data you will find in the daily use of GIS. It describes geographic data in terms of points, that may be connected into lines and polygons. Every object in a vector dataset is called a feature, and is associated with data that describes that feature.
이 강의의 목표: 벡터 데이터의 구조와 맵에 벡터 데이터셋을 로드하는 방법을 배우기.
3.1.1. Follow Along: 레이어 속성 뷰¶
여러분이 작업할 데이터가 오브젝트가 어디에 위치하는지 나타낼 뿐만 아니라 어떤 오브젝트인지도 나타낸다는 사실을 알아야 합니다.
From the previous exercise, you should have the roads layer loaded in your map. What you can see right now is merely the position of the roads.
To see all the data available to you, with the roads layer selected in the Layers panel:
• Click on this button:
It will show you a table with more data about the roads layer. This extra data is called attribute data. The lines that you can see on your map represent where the roads go; this is the spatial data.
GIS에서 이 정의는 흔히 사용되므로, 잘 기억해두는 편이 좋습니다!
• 이제 속성 테이블을 닫으십시오.
Vector data represents features in terms of points, lines and polygons on a coordinate plane. It is usually used to store discrete features, like roads and city blocks.
The Shapefile is a specific file format that allows you to store GIS data in an associated group of files. Each layer consists of several files with the same name, but different file types. Shapefiles are easy to send back and forth, and most GIS software can read them.
Refer back to the introductory exercise in the previous section for instructions on how to add vector layers.
Load the data sets from the epsg4326 folder into your map following the same method:
• “places”
• “water”
• “rivers”
• “buildings”
Databases allow you to store a large volume of associated data in one file. You may already be familiar with a database management system (DBMS) such as Microsoft Access. GIS applications can also make use of databases. GIS-specific DBMSes (such as PostGIS) have extra functions, because they need to handle spatial data.
• Click on this icon:
(If you’re sure you can’t see it at all, check that the Manage Layers toolbar is enabled.)
It will give you a new dialog. In this dialog:
• Click the New button.
• In the same epsg4326 folder, you should find the file landuse.sqlite. Select it and click Open.
You will now see the first dialog again. Notice that the dropdown select above the three buttons now reads “landuse.sqlite@...”, followed by the path of the database file on your computer.
• Click the Connect button. You should see this in the previously empty box:
• Click on the landuse layer to select it, then click Add
주석
Remember to save the map often! The map file doesn’t contain any of the data directly, but it remembers which layers you loaded into your map.
사용자 레이어 목록에 있는 레이어들은 맵 상에 특정 순서대로 그려집니다. 목록 맨 아래에 있는 레이어를 첫 번째로 그리고, 맨 위에 있는 레이어를 마지막에 그립니다. 목록에서 레이어 순서를 바꾸면 맵 상에 그려지는 순서도 바꿀 수 있습니다.
주석
Depending on the version of QGIS that you are using, you may have a checkbox beneath your Layers list reading Control rendering order. This must be checked (switched on) so that moving the layers up and down in the Layers list will bring them to the front or send them to the back in the map. If your version of QGIS doesn’t have this option, then it is switched on by default and you don’t need to worry about it.
현재 맵 상에 레이어가 로드되는 순서가 전혀 논리적이지 않을 수도 있습니다. 다른 레이어들이 road 레이어를 덮어 완전히 가려져 있을 수도 있습니다.
예를 들어 레이어 순서를 이렇게 하면...
... “roads” 및 “places” 레이어가 “urban areas” 레이어 아래로 가기 때문에 보이지 않게 됩니다.
이 문제를 해결하려면,
• 레이어 목록에서 레이어를 클릭 & 드래그하십시오.
• 다음처럼 순서를 재배열하십시오.
이제 도로와 건물이 토지이용구역 위에 나타나는, 시각적으로 더 논리적인 맵을 볼 수 있습니다.
3.1.5. In Conclusion¶
이제 몇 가지 다른 소스들로부터 필요한 모든 레이어를 추가했습니다.
3.1.6. What’s Next?¶
레이어를 로드할 때 랜덤한 색상이 자동적으로 적용되므로, 현재 사용자 맵을 읽기 힘들 수도 있습니다. 사용자가 직접 색상 및 심볼을 적용하는 편이 좋습니다. 다음 강의에서 그 방법을 배울 것입니다. | 1,105 | 3,820 | {"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-2022-05 | latest | en | 0.821054 |
https://petrowiki.spe.org/index.php?title=Power_factor_and_capacitors&diff=46832&oldid=46827 | 1,643,192,251,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00185.warc.gz | 505,068,939 | 11,316 | You must log in to edit PetroWiki. Help with editing
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# Difference between revisions of "Power factor and capacitors"
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The electrical power required to drive a motor has three components: reactive power (Pr, kVAR), active power (Pa, kW), and apparent power (Pap, kVA). The active power is the actual amount of work done by the motor and measured for billing purposes. The reactive power is the power required to magnetize the motor winding or to create magnetic flux, and is not recordable. The apparent power is the vector sum of kilowatts and kilovars and is the total amount of energy furnished by the utility company.
## Power relationships
The power triangles shown in Fig. 1 illustrate the relationships between these terms.
## Power factor
The power factor (Fp) is the ratio of active power to apparent power:
(Eq. 1)
The power factor is "leading" in loads that are more capacitive and "lagging" in loads that are more inductive (e.g., motor or transformer windings). In a purely resistive load, Fops = 1 (unity), such that Pa = Pap (kW = kVA) and no reactive power is present. When Fp < unity, reactive power is present and more power is required to produce work, as seen in the following equation:
(Eq. 2)
## Reactive power
The reactive power of a motor is approximately the same from no load to full load. When a motor is operating at full load, the active/reactive power ratio is high, and thus the power factor of the motor is high. A lightly loaded motor has a low active/reactive power ratio, which causes the power factor to be low. At low power factors, more power will be required from the utility company than actually is needed by the load. This translates into higher energy cost and the need for larger generation units and transformers. Some utility companies charge a substantial penalty to their customers for low power factors (generally < 0.95). Also, low power factors might cause more voltage drop in the system, which causes the motors to operate sluggishly and the lights to dim.
It is essential that the power factor of the system be maintained as high as possible (close to unity). Removing the reactive power from the system can make this possible. Power-factor-correction capacitors are used for this purpose. A motor requires inductive or lagging reactive power for magnetizing. Capacitors provide capacitive or leading reactive power that cancels out the lagging reactive power when used for power-factor improvement. The power triangles in Fig. 2 show how capacitors can improve the power factor for a motor. The improved power factor changes the current required from the utility company, but not the one required by the motor.
## Capacitors
Capacitors should not be selected as a means of correcting poor power factors that are the result of oversized motors or unbalanced pumping units. Choosing a capacitor for this purpose might cause overcorrection, which can result in a leading power factor. A leading power factor, in turn, might cause overvoltages that would cause control-component failure or power-cable failure. This potential problem generally is avoided by connecting the capacitors downstream of the motor contactors and switching them on and off, along with the motor contactors.
Power factor correction capacitors could be applied to each individual motor to correct the power factor of that motor, or could be a single unit connected to the main bus of the switchgear. In the latter case, the unit should have power-factor-sensing circuits that automatically determine the amount of capacitance required for maintaining a preset power factor. The required amount of capacitors are automatically added to or removed from the switchgear bus to maintain the required power factor.
The cyclic kW load on a pumping-unit motor can cause the power factor to vary from 1.0 to near zero if excessive adverse pumping conditions exist.
## Nomenclature
Fp = power factor, cos θ Pa = active power, kW Pap = apparent power, kVA
## Noteworthy papers in OnePetro
Use this section to list papers in OnePetro that a reader who wants to learn more should definitely read | 904 | 4,354 | {"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-2022-05 | latest | en | 0.937147 |
https://www.exactlywhatistime.com/days-from-date/january-7/111-days | 1,718,324,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00814.warc.gz | 692,484,755 | 6,326 | # Monday April 28, 2025
Adding 111 days from Tuesday January 07, 2025 is Monday April 28, 2025 which is day number 118 of 2024. This page is designed to help you the steps to count 111, but understand how to convert and add time correctly.
• Days to add: 111
• Specific Date: Tuesday January 07, 2025
• Days from Tuesday January 07, 2025: Monday April 28, 2025
• Day of the year: 118
• Day of the week: Monday
• Month: April
• Year: 2024
## Calculating 111 days from Tuesday January 07, 2025 by hand
Attempting to add 111 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.
## Monday April 28, 2025 Stats
• Day of the week: Monday
• Month: April
• Day of the year: 118
## Counting 111 days forward from Tuesday January 07, 2025
Counting forward from today, Monday April 28, 2025 is 111 from now using our current calendar. 111 days is equivalent to:
111 days is also 2664 hours. Monday April 28, 2025 is 32% of the year completed.
## Within 111 days there are 2664 hours, 159840 minutes, or 9590400 seconds
Monday Monday April 28, 2025 is the 118 day of the year. At that time, we will be 32% through 2025.
## In 111 days, the Average Person Spent...
• 23842.8 hours Sleeping
• 3170.16 hours Eating and drinking
• 5194.8 hours Household activities
• 1545.12 hours Housework
• 1704.96 hours Food preparation and cleanup
• 532.8 hours Lawn and garden care
• 9324.0 hours Working and work-related activities
• 8578.08 hours Working
• 14039.28 hours Leisure and sports
• 7619.04 hours Watching television
## Famous Sporting and Music Events on April 28
• 1903 Newspaper publisher William Randolph Hearst (40) weds Millicent Veronica Willson (21) in New York City
• 2003 Andre Agassi recaptures the world no. 1 ranking to become the oldest top-ranked male in the history of the ATP rankings (33 years, 13 days) | 670 | 2,480 | {"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} | 3.671875 | 4 | CC-MAIN-2024-26 | latest | en | 0.926778 |
https://codereview.stackexchange.com/questions/113703/squaring-a-tree-in-clojure/113782 | 1,726,887,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00229.warc.gz | 149,404,432 | 43,349 | # Squaring a tree in Clojure
I am working on Problem 2.30 from Structure and Interpretation of Computer Programs. I book is in scheme, but I am doing the exercises in Clojure.
The problem is to write code that takes a tree of numbers and return a new tree with the numbers squared. I have done it in two ways:
1. Without higher-order functions:
(defn square-tree1 [tree]
(lazy-seq
(when-let [s (seq tree)]
(if (coll? (first s))
(cons (square-tree1 (first s))
(square-tree1 (rest s)))
(cons (square (first s))
(square-tree1 (rest s)))))))
2. Using map
(defn square-tree2 [tree]
(if (coll? tree)
(map square-tree2 tree)
(square tree)))
Where:
(defn square [x]
(* x x))
My tests for this are:
(deftest e2.30a
(testing "Ex 2.30a: square-tree"
(is (=
(square-tree1
(list 1
(list 2 (list 3 4) 5)
(list 6 7)))
'(1 (4 (9 16) 25) (36 49))))
(is (= (square-tree1 '()) '()))
(is (= (square-tree1 [1 2 3 4 5]) [1 4 9 16 25]))
(is (= (square-tree1 [1 [2 [3]]]) [1 [4 [9]]]))))
• I want the functions to work with lists and vector inputs.
• I am unsure if 1 is the proper usage of lazy-seq and whether there is a neater way to express this.
• Is (col? (first s)) the correct predicate here? I find it tricky in Clojure to test if the node in the tree is another sub-tree or if it is number. In Scheme it is easier because you have the pair? predicate. I would like to see if there is a better way.
Edit -- After a useful discussion with Timothy Pratley about clojure.walk
After looking at the source code of clojure.walk I have come up with a third square-tree function that extends the square-tree2, but now preserves the input collection types of the input tree like clojure.walk does:
(defn square-tree3
[tree]
(cond
(list? tree) (apply list (map square-tree3 tree))
(seq? tree) (doall (map square-tree3 tree))
(coll? tree) (into (empty tree) (map square-tree3 tree))
:else (square tree)))
So it extends the answer to the coll? predicate with a (into (empty tree)...) which puts the result of map into whatever type of collection tree is. It also adds cases for list and seq types, handling each of them in their own special way so that they replicate into the same collection type.
You can use postwalk:
(defn maybe-square [x]
(if (number? x)
(* x x)
x))
(clojure.walk/postwalk maybe-square [1 [2 [3]]])
=> [1 [4 [9]]]
Which is perhaps against the spirit of learning how to do it, so it is worth looking at the implementation code:
Which shows the correct way to test for various things you may expect in a tree (There are quite a few different tests you can take advantage of).
lazy-seq and map are cool, but it is probably more common to think of trees as data structures rather than lazy sequences; you will note that postwalk returns datastructures, not lazy sequences. So I recommend you also return datastructures (in your case vectors and lists, rather than a sequence).
• Thank you. In the wild I'd use zippers/tree-seq/walk for handling trees. Here I'm trying to do things explicitly for my own education. However the datastructure vs seq thing has confused me about Clojure. Are there some best practices for when a function should return a vector and when to return a seq? It feels with Clojure like everything starts with a vector, then I do a map or a filter and then everything becomes a seq and I've lost my vector. There is no map that retains data type? Commented Dec 13, 2015 at 18:16
• Seqs are used extensively in Clojure programs to great effect, but clearly there are situations where a vector is better than a seq, because of constant time lookup (calling nth on a seq is O(n), calling nth on a vector is O(1)). So the answer to your question "when should a function return a vector instead of a seq?" is when it is convenient to use that result in a way where O(1) access is important. As for "map that retains data type" Take a look at vec, mapv, into and empty. github.com/nathanmarz/specter is a library that has an explicit goal of preserving the input type. Commented Dec 13, 2015 at 18:53
• I recommend reading the source code of clojure.walk because it addresses these concerns that you raise about 'preserving types' in an idiomatic fashion. The code presents several design decisions relevant to the problem you are studying. So there is much to learn from it. Seeing you already wrote a solution on your own, you will be able to appreciate the tradeoffs, and identify why they might be good or bad. Commented Dec 13, 2015 at 19:16
• That was a useful exercise, thanks. I've added a new version of square-tree inspired by clojure.walk. Commented Dec 15, 2015 at 3:06
Starting with your square-tree2, we can distinguish vectors from other sequences as follows:
(defn square-tree2 [tree]
(if (coll? tree)
((if (vector? tree) vec identity) (map square-tree2 tree))
(square tree)))
• The map makes the horizontal explorations lazy.
• An enclosing lazy-seq would not make vertical exploration lazy - each level still involves a recursive call.
If you want to roll your own map, as square-tree1 does, you can simplify things by pulling it out into a distinct function:
(defn map-square [s]
(lazy-seq
(when (seq s)
(cons (square-tree (first s)) (map-square (rest s))))))
then
(defn square-tree [tree]
(if (coll? tree)
(map-square tree)
(square tree)))
These are mutually recursive, so we need to (declare square-tree) up front.
Edited to correct error in map-square. | 1,383 | 5,423 | {"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} | 3.3125 | 3 | CC-MAIN-2024-38 | latest | en | 0.841782 |
https://lcmgcf.com/factors-of-48/ | 1,618,626,004,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038098638.52/warc/CC-MAIN-20210417011815-20210417041815-00214.warc.gz | 429,916,471 | 5,508 | # Factors of 48 | Find the Factors of 48 by Factoring Calculator
Factoring Calculator calculates the factors and factor pairs of positive integers.
Factors of 48 can be calculated quickly with the help of Factoring Calculator i.e. 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 positive integers that divide 48 without a remainder.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. There are 10 integers that are factors of 48. The biggest factor of 48 is 48.
Factors of:
### Factor Tree of 48 to Calculate the Factors
48 2 24 2 12 2 6 2 3
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. There are 10 integers that are factors of 48. The biggest factor of 48 is 48.
Positive integers that divides 48 without a remainder are listed below.
• 1
• 2
• 3
• 4
• 6
• 8
• 12
• 16
• 24
• 48
• 1 × 48 = 48
• 2 × 24 = 48
• 3 × 16 = 48
• 4 × 12 = 48
• 6 × 8 = 48
• 8 × 6 = 48
• 12 × 4 = 48
• 16 × 3 = 48
• 24 × 2 = 48
• 48 × 1 = 48
### Factors of 48 Table
FactorFactor Number
1one
2two
3three
4four
6six
8eight
12twelve
16sixteen
24twenty four
48forty eight
### How to find Factors of 48?
As we know factors of 48 are all the numbers that can exactly divide the number 48 simply divide 48 by all the numbers up to 48 to see the ones that result in zero remainders. Numbers that divide without remainder are factors and in this case below are the factors
1, 2, 3, 4, 6, 8, 12, 16, 24, 48 are the factors and all of them can exactly divide number 48.
### Frequently Asked Questions on Factors of 48
1. What are the factors of 48?
Answer: Factors of 48 are the numbers that leave a remainder zero. The ones that can divide 48 exactly i.e. factors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
2.What are Factor Pairs of 48?
Answer:Factor Pairs of 48 are
• 1 × 48 = 48
• 2 × 24 = 48
• 3 × 16 = 48
• 4 × 12 = 48
• 6 × 8 = 48
• 8 × 6 = 48
• 12 × 4 = 48
• 16 × 3 = 48
• 24 × 2 = 48
• 48 × 1 = 48
3. What is meant by Factor Pairs?
Answer:Factor Pairs are numbers that when multiplied together will result in a given product. | 769 | 2,012 | {"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-2021-17 | longest | en | 0.875546 |
http://www.algebra.com/algebra/homework/NumericFractions/Numeric_Fractions.faq.question.457061.html | 1,369,312,211,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703306113/warc/CC-MAIN-20130516112146-00062-ip-10-60-113-184.ec2.internal.warc.gz | 315,577,473 | 5,413 | # SOLUTION: If 2(x -1) = 14, then x = If 5x + x² > 100, then x is not If 5x - 17 = -x + 7, then x =
Algebra -> Algebra -> Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: If 2(x -1) = 14, then x = If 5x + x² > 100, then x is not If 5x - 17 = -x + 7, then x = Log On
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Algebra: Numeric Fractions Solvers Lessons Answers archive Quiz In Depth
Question 457061: If 2(x -1) = 14, then x = If 5x + x² > 100, then x is not If 5x - 17 = -x + 7, then x = Found 2 solutions by Edwin McCravy, MathLover1:Answer by Edwin McCravy(8909) (Show Source): You can put this solution on YOUR website!```If 2(x -1) = 14, then 2x - 2 = 14 2x = 16 x = 8 ------------------------- If 5x + x² > 100, then x² + 5x - 100 > 0 which has critical values when x² + 5x - 100 = 0 approximately -12.8 or 7.8 Intervals (-oo,-12.8) (-12.8,7.8) (7.8,oo) Test values -13 0 8 expression at test value 4 -100 4 expression > 0? yes no yes x is not between and , inclusive --------------------------- If 5x - 17 = -x + 7, then 6x = 24 x = 4 --------------------------- Edwin``` Answer by MathLover1(6632) (Show Source): You can put this solution on YOUR website! If , then If , then is not .......use quadratic formula and solve solutions: or so, is not or If , then = | 490 | 1,439 | {"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-2013-20 | latest | en | 0.68855 |
https://fr.slideserve.com/maryrice/fluids-powerpoint-ppt-presentation | 1,632,841,474,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00496.warc.gz | 312,433,629 | 20,456 | # Fluids
Télécharger la présentation
## Fluids
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Fluids BELLWORK : What does it mean to be fluid?
2. Bellwork – 01/24/17 • You are on an ice-covered pond and the ice starts to crack – what do you do?
3. Bellwork – 02/01/16 • On ice and it starts to crack – what do you do? • Lay down flat • Distributes weight over more surface area
4. BELLWORK - Pressure • What is pressure?
5. TN Standards CLE.3202.1.8 – Investigate relationships among the pressure, temperature, and volume of gases / liquids CLE.3202.Inq.3 – Use appropriate tools and technology to collect precise and accurate data CLE.3202.Inq.6 – Communicate and defend scientific findings
6. Pressure [ Chpt 3 – section 3 ] • What is pressure? • Feeling forced to do something • A force needed to squash an object • Force exerted per unit area of contact
7. PhET simulation • Heat water to steam ( turn heat up at bottom ) • What happened to pressure in a closed container? • So why does the container ‘splode??
8. Balloons • While it is running, we’ll use balloons as an example of pressure • How does amount of gas in balloon relate to its size?
9. Balloons • What direction is the gas inside the balloon pointing to?
10. Balloons • What direction is the gas inside the balloon pointing to if really hot?
11. Balloons • What can we say about the pressure found in balloons of different sizes? ( start out the same size ) • Small one • Big one
12. Balloons • What can we say about the pressure found in balloons of different sizes? ( start out the same size ) • Small one • Big one PRESSURE IS LARGER IN THE BIGGER ONE
13. Challenge • What is a possible mathematical relationship for pressure? • Force and surface area are involved • What are independent and dependent variables?
14. Pressure • Force exerted per unit area of a surface • Pressure = F / A ( force divided by area )
15. Pressure - Example • Write down or Estimate your weight in lbs : • Estimate the area of your shoe-sole ( in2 ): • x “ wide & y “ long • Shoe area = • Pressure when standing on floor equals: • Weight / total shoe-area
16. Pressure - Example • Write down or Estimate your weight in lbs : • 190 lbs • Estimate the area of your shoe-sole ( in2 ): • Mine – 4.25 “ wide & 12 “ long • Shoe area = 51 x 2 = 102 in2 • Pressure when standing on floor equals: • Weight / total shoe-area • 1.9 psi exerted on floor
17. Pressure – Math Relationships • Force is directly proportional and surface area of ( contact ) is inversely proportional to pressure • P = F / A • Increase force increase pressure • Increase area DECREASE pressure
18. Buoyant Force • A buoyant force pushes the boat up • All fluids exert an upward buoyant force on matter
19. Buoyant Force • Archimedes’ principle states that an object floats if the buoyant force on the object is equal to the object’s weight • An object’s ability to float is based on the buoyant force, density of the fluid, and the density of the object
20. Hydrostatic Pressure • Pressure felt by a column of fluid ( so, just being under a certain amount of fluid ) • Does not matter how much volume of fluid there is – ONLY the depth • Pressure = fluid density * gravity * depth
21. Pascal’s Principle • An enclosed fluid exerts pressure equally in all directions • A change in the external pressure exerted on an enclosed fluid is transmitted equally and unchanged in all directions ( every point ) within that fluid
22. Work with Partners • Study Guides • Pg 12 • #2, 3 ( put in N / cm2)
23. Challenge / Exit Pass • Consider a submarine at the bottom of the ocean. • What is the water doing to the submarine • Why does the submarine not get crushed?
24. PLC Day / Upcoming in Week • Gonna watch stuff ‘splode ?? • Open-Notes Quiz Thursday/Friday
25. Exit Pass | 982 | 3,911 | {"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-2021-39 | latest | en | 0.817096 |
http://mathoverflow.net/questions/168398/time-inhomogeneous-markov-chains | 1,469,801,578,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830091.67/warc/CC-MAIN-20160723071030-00221-ip-10-185-27-174.ec2.internal.warc.gz | 153,433,844 | 17,554 | # Time-inhomogeneous Markov Chains
I'm trying to find out what is known about time-inhomogeneous ergodic Markov Chains where the transition matrix can vary over time. All textbooks and lecture notes I could find initially introduce Markov chains this way but then quickly restrict themselves to the time-homogeneous case where you have one transition matrix.
Obviously, in general such Markov chains might not converge to a unique stationary distribution, but I would be surprised if there isn't a large (sub)class of these chains where convergence is guaranteed. I'm particularly interested in theorems on the mixing time and convergence theorems that state when there exists a stationary distribution.
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Cross-posted from cstheory.stackexchange.com/q/22698/23060 – Daniel Moskovich May 28 '14 at 8:34
Note that you can homogenise the chain. If $X=(X_t:t \geq 0)$ is an inhomogeneous Markov chain on $E$ then $(X_t,t)$ is a homogeneous Markov chain on $E \times \mathbb Z^+$ (see Revuz and Yor Chapter III Excercise 1.10). – Bati May 28 '14 at 10:39
@Bati: Except for the fact that space-time chains have no stationary distributions – R W May 28 '14 at 11:50
@R W: You are right. – Bati May 28 '14 at 15:08
Stroock's Markov processes book is, as far as I know, the most readily accessible treatment of inhomogeneous Markov processes: he does all the basics in the context of simulated annealing, which is neat. Kleinrock's volume 1 is also of interest, though "buggy" IIRC.
In my experience the key object is the propagator $U(t) := \mathcal{TO}^* \int_0^t Q(s) \ ds$, where $Q$ is the time-dependent generator and $\mathcal{TO}^*$ is the formal adjoint or reverse time-ordering operator (see here and here for forward time-ordering). With this in hand, the transition kernel can be expressed as $P(s,t) = U^{-1}(s)U(t)$. Thus, e.g., an initial distribution $p(0)$ is propagated as $p(t) = p(0)U(t)$, which gives you the essential stuff to get a handle on (e.g.) mixing times.
One way to guarantee convergence is to have $U(t)$ varying within the group fixing a distribution, cf. Has the Lie group preserving a probability distribution been used in Bayesian statistics?
As another push towards your goal, the Dynkin martingale formula becomes (under suitable conditions, e.g., $t \mapsto f_t$ is $C^1$ and $f_t(X_t)$ is bounded) $$\mathbb{E}(f_t(X_t)-f_0(X_0)) = \mathbb{E}\int_0^t(\partial_s+Q(s))f_s \circ X_s \ ds$$ Here Rogers and Williams IV.20-21 is of interest, since the extension to inhomogeneous processes is trivial.
Finally, I will note that the cutest time-inhomogenous Markov process is the Poisson process, cf. http://blog.eqnets.com/2009/07/28/why-poissonian-traffic-models-matter-more-now-than-ever-part-4/
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I would like to add that in the field of differential equations on Banach spaces (which contain time continuous Markov chains as special cases) transition matrices that can vary over time become time-dependent operators. There is a well-developed (if not very elementary) theory for this kind of problems, which starts by replacing $C_0$-semigroups by objects called "evolution families". And yes, much is known the long time behavior of this kind of problems: see e.g. www.math.kit.edu/iana3/~schnaubelt/media/survey1.pdf – Delio Mugnolo May 29 '14 at 21:33
You don't find much about time-inhomogeneous Markov chains because it's extremely difficult to prove anything about them without strong additional assumptions, and it's not clear what additional assumptions make sense. The only literature I'm aware of is some quite recent papers by Jessica Zúñiga and Laurent Saloff-Coste; see this page for links. (If there is any prior literature on the problems you're interested in, I'm sure those papers have references.)
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Interestingly enough, one can prove Azume-Hoeffding and Talagrand-type concentration inequalities for inhomogeneous Markov chains using contraction-type conditions:
Concentration of measure inequalities for Markov chains and \Phi-mixing processes http://projecteuclid.org/euclid.aop/1019160125
Concentration inequalities for dependent random variables via the martingale method http://projecteuclid.org/euclid.aop/1229696598
Obtaining Measure Concentration from Markov Contraction http://dl.dropboxusercontent.com/u/3198145/mark-conc2.pdf
- | 1,092 | 4,305 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-30 | latest | en | 0.910849 |
https://doc.arcgis.com/en/cityengine/2020.0/cga/cga-array-type-operators.htm | 1,660,559,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00707.warc.gz | 218,060,497 | 9,404 | # Array type operators
### Index operator
Access array elements by indices or logical values.
See index operator.
[ ] 1d indexing ``````array1d = [1,2,3] array1d[1] // 2`````` 2d Indexing ``````array2d = [1,2;3,4] array2d[1,0] // 3``````
### Colon operator
Create a sequence of float values.
See colon operator.
: Binary ``[1:5] // [1,2,3,4,5]`` Ternary ``[1:2:5] // [1,3,5]``
### Equality operators
Check float[], bool[] or string[] for equality.
Two arrays are equal if they have equal dimensions and equal elements.
Return type: bool
== Equality ``````[1,2] == [1,2] // true ["a","b"] == ["a";"b"] // false [true,false] == [true,true] // false`````` != Inequality ``````[1,2] != [1;2] // true ["a","b"] != ["a";"b"] // true [true,true] != [true,true] // false``````
### Relational operators
Compare float[] or string[].
Two arrays are element-wise compared, row by row. If an element is less, the respective array is less. If the first rows are equal, the array with less columns is less. Otherwise the array with less rows is less.
Return type: bool
< Less ``````[1,2] < [1,3] // true ["a","b"] < ["a";"b"] // false`````` <= Less or equal ``````[1,2] <= [1,2] // true ["a";"b"] <= ["a"] // false`````` > Greater ``````[1,3] > [1,2] // true ["a";"b"] > ["a"] // true`````` >= Greater or equal ``````[1,2,3] >= [1,2] // true ["a";"b"] >= ["a";"b"] // true``````
## Element-wise array operators
Element-wise operators have a preceeding dot. They perform the operator on each element and return the results in an array with the same dimensions as the input array(s). The dimensions of two arrays must be equal.
### Logical operators
Element-wise logical operations on bool[].
Return type: bool[]
.! Negation ``.![true,false] // [false,true]`` .|| Logical Or ``````[true,false] .|| true // [true,true] false .|| [false;true] // [false;true] [false,true] .|| [false,false] // [false,true]`````` .&& Logical And ``````[false;true] .&& true // [false;true] false .&& [true;true;true] // [false;false;false] [true,false] .&& [false,true] // [false,false]``````
### Arithmetic operators
Element-wise arithmetic operations on float[].
Return type: float[]
.+ Plus ``````[1,2] .+ 1 // [2,3] 1 .+ [1;2] // [2;3] [1,2] .+ [2,1] // [3,3]`````` Unary plus ``.+[1,2] // [1,2]`` .- Minus ``````[1,2;3,4] .- 1 // [0,1;2,3] 3 .- [1,2] // [2,1] [3;2] .- [2;1] // [1;1]`````` Unary minus ``.-[1;2] // [-1;-2]`` .* Multiplication ``````[4,2] .* 2.5 // [10,5] -4 .* [0.25;-0.75;-1] // [-1;3;4] [1,2;3,4] .* [4,3;2,1] // [4,6;6,4]`````` ./ Division ``````[4;3] ./ 2 // [2;1.5] 12 ./ [2,3,4] // [6,4,3] [1.6,3.8] ./ [0.8,1.9] // [2,2]`````` .% Modulus (remainder) ``````[8;-5;0] .% 3 // [2;-2;0] 7 .% [2.7,-3] // [1.6,1] [-5,3.8] .% [-3,0.7] // [-2,0.3]``````
### Equality operators
Element-wise check float[], bool[] or string[] for equality.
Return type: bool[]
.== Equality ``````1 .== [1,2] // [true,false] ["a","b"] .== "b" // [false,true] [true;false] .== [true;false] // [true;true]`````` .!= Inequality ``````[1,1;2,2] .!= [1,2;1,2] // [false,true;true,false] ["a"] .!= "a" // [false] false .!= [true] // [true]``````
### Relational operators
Element-wise compare float[] or string[].
Return type: bool[]
.< Less ``````1 .< [1,2] // [false,true] ["a","b"] .< "b" // [true,false] [1,2;2,1] .< [2,1;1,2] // [true,false;false,true]`````` .<= Less or equal ``````1 .<= [1,2] // [true,true] ["a";"b";"c"] .<= "b" // [true;true;false] ["a","b"] .<= ["b","a"] // [true,false]`````` .> Greater ``````[1,2;3,4] .> [1,1;3,3] // [false,true;false,true] "a" .> ["a";"a"] // [false;false] ["a";"b"] .> "a" // [false;true]`````` .>= Greater or equal ``````[1;2;3] .>= [2;2;2] // [false;true;true] 2 .>= [1,2,3] // [true,true,false] ["a"] .>= "a" // [true]``````
### String concatenation operators
Element-wise concatenate string, string[] with string[], float[], bool[] or string[] with string, float, bool.
Return type: string[]
.+ String-string concatenation ``````["a","b"] .+ "c" // ["ac","bc"] "a" .+ ["b";"c"] // ["ab";"ac"] ["a","b"] .+ ["c","d"] // ["ac","bd"]`````` String-float concatenation ``````["a","b"] .+ 1 // ["a1","b1"] "a" .+ [1,2;3,4] // ["a1","a2";"a3","a4"] [1;2] .+ ["a";"b"] // ["1a";"2b"]`````` String-boolean concatenation ``````[true;false] .+ "a" // ["truea";"falsea"] "a" .+ [true] // ["atrue"] ["a";"b"] .+ [false;true] // ["afalse";"btrue"]``````
### Related
##### In this topic
1. Element-wise array operators | 1,730 | 4,464 | {"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-2022-33 | latest | en | 0.519614 |
https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-the-empty-product/ | 1,603,541,847,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00669.warc.gz | 713,491,227 | 14,602 | # Ask Uncle Colin: The Empty Product
Dear Uncle Colin,
Why does $0! = 1$ and not 0?
- Nothing Is Logical
Hi, NIL, and thanks for your message!
My best explanation for this - by which I mean, the one I can get some people to accept, goes like this:
• $4! = 4 \times 3 \times 2 \times 1 = 24$.
• To get to $3!$, you divide $4!$ by 4 and get $3! = 6$.
• To get to $2!$, you divide $3!$ by 3 and get $2! = 2$.
• To get to $1!$, you divide $2!$ by 2 and get $1! = 1$.
• And to get to $0!$, you divide $1!$ by 1 and get $0! = 1$.
(You can’t go any further, because you’d have to divide by zero - but you can extend the factorial function into non-integers using the gamma function.)
A similar argument works for powers:
• $3^2 = 3 \times 3 = 9$.
• To get to $3^1$, you divide $3^2$ by 3 and get $3^1 = 3$.
• To get to $3^0$, you divide $3^1$ by 3 and get $3^0 = 1$.
(You can continue this pattern into the negative numbers!)
In both cases, when you multiply nothing together, you get 1. (We say “1 is the multiplicative identity”).
It feels strange that the value of “no things” is 1 if you’re multiplying all of the no things together, and 0 if you’re adding them up, but it has to be that way for maths to work.
Hope that helps!
- Uncle Colin
## Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
#### Share
• ##### Barney Maunder-Taylor
A similar argument applies to the sequence 3^3^3^3, 3^3^3, 3^3, 3, then what? And after that??? In each case, do log base 3 to get from one term to the next. The next term is therefore … ?
• ##### Colin
Nice — although I suspect someone asking this question may not be entirely comfortable with logs 😉
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 573 | 1,887 | {"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} | 4.125 | 4 | CC-MAIN-2020-45 | longest | en | 0.891721 |
https://community.splunk.com:443/t5/Splunk-Search/Charting-Graphing-Multiple-data-sources-one-chart/m-p/26424 | 1,656,404,739,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00528.warc.gz | 231,820,740 | 36,584 | Splunk Search
## Charting / Graphing Multiple data sources one chart
Explorer
has anyone experimented with showing statistics for the same time slot over multiple time periods ?
e.g. imagine a chart that shows the number of transaction over a 24 hour period as a line graph. now imagine a second line (of a different color) that has the 24 hour transaction from yesterday, and another that has a weekly average for 24 hour transaction.
i appreciate any insights.
Tags (3)
1 Solution
Splunk Employee
Comparing week-over-week results used to a pain in Splunk, with complex date calculations. No more. Now there is a better way.
I wrote a convenient search command called "timewrap" that does it all, for arbitrary time periods.
``````... | timechart count span=1d | timewrap w
``````
That's it!
http://apps.splunk.com/app/1645/
Splunk Employee
Comparing week-over-week results used to a pain in Splunk, with complex date calculations. No more. Now there is a better way.
I wrote a convenient search command called "timewrap" that does it all, for arbitrary time periods.
``````... | timechart count span=1d | timewrap w
``````
That's it!
http://apps.splunk.com/app/1645/
Explorer
this is very insightful. there are lots of neat 'splunk moves' here that i didn't know existed. i appreciate you taking the time to write this up.
Splunk Employee
For the simple day over day case, the recipe for this solution is pretty well covered in:
To add a series for the weekly average, for each of the previous seven days at each time bucket, the search is significantly more involved. To do this, the first task is to `| append []` some results that are the computation of the average. The details of the inner search depend on what type of aggregate function that you're using, in this case, you're looking at count, so it's not too bad:
``````... | append [search earliest=-7d@d latest=@d ... | eval _time = ((_time - relative_time(now(), "-7d@d")) % 86400) + relative_time(now(), "@d") | bin _time span=15m | chart count by _time | eval metric = count/7 | eval marker = "weekly average"]
``````
Now you just have to glue this right before the final `| chart` in the day_over_day search in the referenced post, and you should have your answer.
Putting it all together:
``````<data> earliest=-1d@d latest=@h
| timechart span=15m count as metric | 579 | 2,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-27 | latest | en | 0.929374 |
https://ai-at-wvu.blogspot.com/2009/09/keys-and-keys2.html | 1,624,467,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00625.warc.gz | 94,698,310 | 16,094 | ## Monday, September 7, 2009
### KEYS and KEYS2
Within a model, there are chains of reasons that link inputs to the desired goals. As one might imagine, some of the links in the chain clash with others. Some of those clashes are most upstream; they are not dependent on other clashes. In the following chains of reasoning, the clashes are {e, -e}, {g, -g}, {j, -j}; the most upstream clashes are {e, -e}, {g, -g}
In order to optimize decision making about this model, we must first decide about these most upstream clashing reasons (the "keys"). Most of this model is completely irrelevant. In the context of reaching the goal, the only important discussions are the clashes {g,-g,j, -j}. Further, since {j, -j} are dependent on {g, -g}, then the core decision must be about variable g with two disputed values: true and false.
Setting the keys reduces the number of reachable states within the model. Formally, the reachable states reduce to the cross-product of all of the ranges of the collars. We call this the clumping effect. Only a small fraction of the possible states are actually reachable. This notion of $keys$ has been discovered and rediscovered
many times by many researchers. Historically, finding the keys has seen to be a very hard task. For example, finding the keys is analogous to finding the minimal environments of DeKleer's ATMS algorithm. Formally, this is logical abduction, which is an NP-hard task.
Our method for finding the keys uses a Bayesian sampling method. If a model contains keys then, by definition, those variables must appear in all solutions to that model. If model outputs are scored by some oracle, then the key variables are those with ranges that occur with very different frequencies in high/low scored model outputs. Therefore, we need not search for the keys- rather, we just need to keep frequency counts on how often ranges appear in best or rest outputs.
KEYS implements this Bayesian sampling methods. It has two main components - a greedy search and the BORE ranking heuristic. The greedy search explores a space of M mitigations over the course of M "eras". Initially, the entire set of mitigations is set randomly. During each era, one more mitigation is set to M_i=X_j, X_j being either true or false. In the original version of KEYS, the greedy search fixes one variable per era. Recent experiments use a newer version, called KEYS2, that fixes an increasing number of variables as the search progresses
(see below for details).
KEYS (and KEYS2), each era e generates a set as follows:
• [1:] MaxTries times repeat:
• Selected[1{\ldots}(e-1)] are settings from previous eras.
• Guessed are randomly selected values for unfixed mitigations.
• Input = selected from guessed.
• Call model to compute score=ddp(input);
• [2:] The MaxTries scores are divided into B "best" and the remainder become the "rest".
• [3:] The input mitigation values are then scored using BORE (described below).
• [4:] The top ranked mitigations (the default is one, but the user may fix multiple mitigations at once) are fixed and stored in selected[e].
The search moves to era e+1 and repeats steps 1,2,3,4. This process stops when every mitigation has a setting. The exact settings for MaxTries and B must be set via engineering judgment. After some experimentation, we used MaxTries=100 and B=10.
Pseudocode for the KEYS algorithm:
1. Procedure KEYS2. while FIXED_MITIGATIONS != TOTAL_MITIGATIONS3. for I:=1 to 1004. SELECTED[1...(I-1)] = best decisions up to this step5. GUESSED = random settings to the remaining mitigations6. INPUT = SELECTED + GUESSED7. SCORES= SCORE(INPUT)8. end for 9. for J:=1 to NUM_MITIGATIONS_TO_SET 10. TOP_MITIGATION = BORE(SCORES) 11. SELECTED[FIXED_MITIGATIONS++] = TOP_MITIGATION 12. end for 13. end while 14. return SELECTED \end{verbatim} }
KEYS ranks mitigations by combining a novel support-based Bayesian ranking measure. BORE (short for "best or rest") divides numeric scores seen over K runs and stores the top 10% in best and the remaining 90% scores in the set rest (the bes\$ set is computed by studying the delta of each score to the best score seen in any era). It then computes the probability that a value is found in best using Bayes theorem. The theorem uses evidence E and a prior probability P(H) for hypothesis H in{best, rest}, to calculate a posterior probability
When applying the theorem, likelihoods are computed from observed frequencies. These likelihoods (called "like" below) are then normalized to create probabilities. This normalization cancels out P(E) in Bayes theorem. For example, after K=10,000 runs are divided into 1,000 best solutions and 9,000 rest, the value mitigation31=false might appear 10 times in the best solutions, but only 5 times in the rest. Hence:
Previously, we have found that Bayes theorem is a poor ranking heuristic since it is easily distracted by low frequency evidence. For example, note how the probability of E belonging to the best class is moderately high even though its support is very low; i.e. P(best|E)=0.66 but freq(E|best) = 0.01.
To avoid the problem of unreliable low frequency evidence, we augment the equation with a support term. Support should increase as the frequency of a value increases, i.e. like(best|E) is a valid support measure. Hence, step 3 of our greedy search ranks values via
For each era, KEYS samples the model and fixes the top N=1 settings. Observations have suggested that N=1 is, perhaps, an overly conservative search policy. At least for the DDP models, we have observed that the improvement in costs and attainments generally increase for each era of KEYS. This lead to the speculation that we could jump further and faster into the solution space by fixing N=1 settings per era. Such a jump policy can be implemented as a small change to KEYS, which we call KEYS2.
• Standard KEYS assigns the value of one to NUM_MITIGATIONS_TO_SET (see the pseudocode above);
• Other variants of KEYS assigns larger values.
In era 1, KEYS2 behaves exactly the same as KEYS. In (say) era 3, KEYS2 will fix the top 3 ranked ranges. Since it sets more variables at each era, KEYS2 terminates earlier than KEYS. | 1,483 | 6,240 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.940456 |
https://statisticsglobe.com/create-vector-of-zeros-in-r | 1,717,009,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00527.warc.gz | 466,823,090 | 47,524 | # How to Create a Vector of Zeros in R (5 Examples)
In this R tutorial you’ll learn how to declare a vector or array containing only zeros.
The article will contain these topics:
Let’s get started.
## Example 1: Creating Vector of Zeros Using rep() Function
Example 1 illustrates how to create a vector consisting of zeros using the rep function.
```x1 <- rep(0, 5) # Applying rep() function
x1 # Printing vector to RStudio console
# 0 0 0 0 0```
The RStudio console returns the result of the previous R code: A vector of length five consisting only of zeros.
## Example 2: Creating Vector of Zeros Using rep() Function & 0L
Alternatively to the R syntax shown in Example 1, we can also specify 0L instead of 0 within the rep function:
```x2 <- rep(0L, 5) # Applying rep() function
x2 # Printing vector to RStudio console
# 0 0 0 0 0```
## Example 3: Creating Vector of Zeros Using numeric() Function
We can also use the numeric function to construct a vector of zeros. For this, we simply have to specify the length of the vector within the numeric function:
```x3 <- numeric(5) # Applying numeric() function
x3 # Printing vector to RStudio console
# 0 0 0 0 0```
## Example 4: Creating Vector of Zeros Using integer() Function
Similar to Example 3, we can use the integer function to create a vector of zeros.
```x4 <- integer(5) # Applying integer() function
x4 # Printing vector to RStudio console
# 0 0 0 0 0```
## Example 5: Creating Vector of Zeros Using c() Function
Last but not least, we can also use the c function to construct an array or vector containing zeros (even though the c function might be inconvenient for longer vector objects):
```x5 <- c(0, 0, 0, 0, 0) # Applying c() function
x5 # Printing vector to RStudio console
# 0 0 0 0 0```
## Video & Further Resources
Have a look at the following video of my YouTube channel. I’m explaining the R programming syntax of this article in the video:
Besides the video, you may want to have a look at the other articles of my website. Some articles about creating data objects are listed below.
To summarize: You learned in this tutorial how to create a vector of zeros in the R programming language. Please let me know in the comments, if you have further questions.
Subscribe to the Statistics Globe Newsletter | 599 | 2,501 | {"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} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.686033 |
https://www.jiskha.com/display.cgi?id=1360122847 | 1,503,298,186,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107720.63/warc/CC-MAIN-20170821060924-20170821080924-00285.warc.gz | 933,298,278 | 4,044 | # Calculus
posted by .
Explain, using the theorems, why the function is continuous at every number in its domain.
F(x)= 2x^2-x-3 / x^2 +9
A) F(x) is a polynomial, so it is continuous at every number in its domain.
B) F(x) is a rational function, so it is continuous at every number in its domain.
C) F(x) is a composition of functions that are continuous for all real numbers, so it is continuous at every number in its domain.
D) F(x) is not continuous at every number in its domain.
E) none of these
• Calculus -
The numerator is defined for every value of x you choose,
the denominator is always positive, so there is no danger of us dividing by zero, which would have caused a discontinuity.
so continuous for all values of x
B)
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6. ### algebra
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Find all values of a, B so that the function f(x)=1 if x>3, f(x)=ax-10 if x<3, and f(x)=B otherwise is continuous everywhere. I think a is continuous everywhere since ax-10 is a polynomial, and b is continuous at x=3.
More Similar Questions | 728 | 2,568 | {"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-2017-34 | longest | en | 0.852193 |
https://www.theregister.co.uk/AMP/2017/09/25/rubiks_cube_neural_network/ | 1,558,596,542,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257156.50/warc/CC-MAIN-20190523063645-20190523085645-00486.warc.gz | 956,870,112 | 10,527 | # Wanna get started with practical AI? Check out this chap's Rubik's Cube solving neural-net code
## Written in Python, it's not perfect – but it's pretty cool
28 SHARE
The Rubik’s Cube is one of those toys that just won't go away. Solving it is either something you can do in minutes to impress, or find so hard you end up using it as a paperweight.
There are several algorithms for solving the classic cube, which has a whopping 43,252,003,274,489,856,000 – about 43 quintillion – possible combinations. One state-of-the-art approach has shown the puzzle can be defeated in 20 or fewer moves, no matter the combination of the colored squares. This figure is dubbed God's Number.
These algorithms were crafted by mathematicians and expert human players. The computers just obey them as they would any other instructions. The machines would be lost without their programmers.
Can software learn how to solve the cube all on its own? It’s an interesting problem that Jeremy Pinto, a systems design engineering master’s student who recently graduated from the University of Waterloo, Canada, decided to crack. He wanted to see if a neural network could figure out how to solve a Rubik’s Cube.
### A simple AI for a simple game
The project is “just a proof of concept,” Pinto told The Register. The convolutional neural network (CNN) – a type of neural network normally used in computer vision – he coded for the job is very simple.
Pinto used Keras, a Python deep-learning library that interfaces with Google’s popular TensorFlow, to build his CNN.
It’s made of two layers: a convolutional layer and a feedforward layer. He imported open-source code, including MagicCube which simulates and visualizes Rubik’s Cubes, so he could generate and solve the puzzle virtually in software. The position of each colored square on the cube are encoded as an array of integers.
The cube is randomly shuffled from a solved state to a scrambled configuration in ten steps, ie: it's twisted and rotated randomly ten times. These ten steps are then fed into the CNN backwards, Pinto explained, effectively teaching it step by step how to go from a particular scrambled state to the solved state. At each step, the neural network guesses what the next move to solve the cube should be, then is told the next move, and it gradually learns from that.
It's a neat way of teaching the network how to go from a scrambled state to a solved cube without having to solve the puzzle yourself over and over, either by hand or by an algorithm, to train the CNN.
“It would be given the tenth move as input, and be trained to produce the ninth move, then the ninth move is given and it has to get to the eighth move. This carries on until it’s finally down to the last move and then it’s solved,” Pinto said.
After 12 hours of training using an Nvidia Geforce GTX 1050 graphics card with a Pascal GPU to accelerate the task, and around 50,000 games played, he realized the performance of the CNN plateaued. “That’s when you know to stop training, because it won’t really lead to anymore improvements,” Pinto said.
### Room for improvement
It was now ready for testing. Pinto gave the CNN new random combinations to crack, and to his surprise it managed to figure out, within seconds, how to solve the Rubik’s Cube when six or seven moves away, for about 75 per cent of the time. If you give it more moves than that, that is, you shuffle it more than seven times, it struggles. That's not a surprise seeing as it was trained from ten steps. Here's a short video of it in action:
“Surprisingly, the network works decently well for any position less than six moves away from solved. I found that by shuffling up to six moves, sometimes more, it is able to correctly solve it. It doesn't always work and can definitely use improvement, but it’s a good place to start,” Pinto said on his Github page.
The crude CNN is no match for the God’s number algorithm – it can’t find the most efficient solution for any arbitrary number of shuffles – but it’s a solid start for a small neural network. More training with many more steps will improve the system in its puzzle cracking.
It's also a fun practical starting point for anyone who wants to get into AI but needs some working code to get to grips with it. Folks with a more engineering bent, like those here at El Reg, may prefer real-world code to pages and pages of mathematics seen in most neural network papers and text books.
“Humans can generally do it in about five or six shuffles but at ten or more it gets too complex and it becomes a job for those specialized algorithms,” he said. “The entropy for the number of possible moves gets too big.”
At the moment it’s just a side project born from curiosity that took under three days to complete, but Pinto told us it’s “definitely worth exploring.”
“There are many ways to improve upon it," he said. "Using something like a reinforcement learning approach would be a great place to start. Also finding a way of exploiting the symmetries of the cube and invariance to colors would be good. Training larger networks would be also likely necessary. Finally, it would be interesting to inject known algorithms for specific cube solving to see if the network could pick up on it. I'm sure there are a million better ways to improve it.”
The source code can be found here if you want to find out how it works and how to write something similar. ® | 1,172 | 5,433 | {"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} | 3.015625 | 3 | CC-MAIN-2019-22 | longest | en | 0.954238 |
https://www.jiskha.com/display.cgi?id=1299030241 | 1,503,188,853,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00362.warc.gz | 946,724,650 | 3,742 | # high school chem 1
posted by .
Equation: 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)
question: What volume of NH3 is needed to react with 71.6 liters of oxygen?
• high school chem 1 -
Use the coefficients to convert L oxygen to L NH3.
71.6 L O2 x (4 moles NH3/5 moles O2) = 71.6 x (4/5) = ??
• high school chem 1 -
thanks Dr. Bob
## Similar Questions
1. ### chem
2) How many moles of nitrogen monoxide will be produced when 80.0 grams of oxygen are used?
2. ### Chemistry
Consider the following equation: 4NH3(g) + 5O2(g)----> 4NO(g)+6H2O(g) a. How many liters of oxygen are required to react with 2.5 L NH3?
3. ### Chemistry
given the balanced equation representing a reaction: 4NH3+5O2-->4NO+6H2O what is the minimum number of moles of O2 that are needed to completely react with 16 moles of NH3?
4. ### stoichiometry
Ammonia gas reacts with oxygen gas according to the following equation: 4NH3 + 5O2----4NO + 6H2O a. How many moles of oxygen gas are needed to react with 23 moles of ammonia?
5. ### High school chem 1
4NH3(g) + 5O2(g) -> 4NO(g) + 6H20(g) what mass of H2O can be produced from 350 liters of NH3?
6. ### chem 1 high school
equation: 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g) What mass of O2 is needed to produce 0.563g of NO?
7. ### Chemstry
4NH3+5O2--->4NO+6H2O determine the amount of oxygen needed to produce 1.2x10^4mol of nitrogen monoxide gas
8. ### Chemistry
How many liters of nitrogen oxide at STP are produced from the reaction of 59.g of NH3?
9. ### chemistry
4NH3 + 5O2 --> 4NO + 6H2O 170g of ammonia (NH3) are allowed to react with 384g of O2. how many moles of NO is produced?
10. ### college chemistry
how many L @STP if oxygen are needed to produce 45.0 grams of nitric oxide according to the following balanced chemical equation?
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$\int x^{n} dx$
$\int \frac{dx}{x}$
$\int (ax + b)^{n} dx$
$\int {dx \over ax + b}$
$\int {dx \over (ax + b)^2}$
$\int {xdx \over ax + b}$
$\int {x^2 dx \over ax + b}$
$\int {x^n dx \over ax + b}$
$\int {xdx \over (ax + b)^2}$
$\int {x^2dx \over (ax + b)^2}$
$\int {x^ndx \over (ax + b)^2}$
$\int {x^3dx \over (ax + b)^2}$
$\int {x^ndx \over (ax + b)^m}$
$\int {dx \over x(ax + b)}$
$\int {dx \over x(ax + b)^2}$
$\int {dx \over x(ax+b)^3}$
$\int {dx \over x(ax+b)^n}$
$\int {dx \over x^2(ax+b)^2}$
$\int {dx \over x^2(ax+b)}$
$\int {dx \over x^2(ax+b)^3}$
$\int {dx \over x^3(ax+b)}$
$\int {dx \over x^n(ax+b)^m}$
$\int {ax+b \over \alpha x+\beta}dx$
$\int {(ax+b)^2 \over \alpha x+\beta}dx$
$\int {ax+b \over (\alpha x+\beta)^2}dx$
$\int {(ax+b)^2 \over (\alpha x+\beta)^2}dx$
$\int {(ax+b)^n \over (\alpha x+\beta)^m}dx$
$\int {dx \over (ax+b)(\alpha x+\beta)}$
$\int {dx \over (ax+b)(\alpha x+\beta)^2}$
$\int {xdx \over (ax+b)(\alpha x+\beta)}$
$\int {xdx \over (x+a)(x+b)}$
$\int {xdx \over (x+a)(x+b)^2}$
$\int {dx \over (x+a)^2(x+b)^2}$
$\int {xdx \over (x+a)^2(x+b)^2}$
$\int {x^2dx \over (x+a)^2(x+b)^2}$
$\int {dx \over x(x+a)(x+b)}$
$\int {dx \over x^2+a^2}$
$\int {dx \over b^2x^2+a^2}$
$\int {xdx \over x^2+a^2}$
$\int {xdx \over (x^2+a^2)^2}$
$\int {dx \over (x^2+a^2)^2}$
$\int {dx \over (x^2+a^2)^3}$
$\int {dx \over (x^2+a^2)^n}$
$\int {xdx \over (x^2+a^2)^n}$
$\int {x^2dx \over x^2+a^2}$
$\int {x^2dx \over (x^2+a^2)^2}$
$\int {x^3dx \over (x^2+a^2)^2}$
$\int {x^2dx \over (x^2+a^2)^n}$
$\int {dx \over x(x^2+a^2)}$
$\int {dx \over x^2(x^2+a^2)}$
$\int {dx \over x(x^2+a^2)^2}$
$\int {dx \over a^2-x^2}$
$\int {dx \over a^2-b^2x^2}$
$\int {dx \over (a^2-x^2)^2}$
$\int {dx \over (a^2-x^2)^n}$
$\int {xdx \over a^2-x^2}$
$\int {x^2dx \over a^2-x^2}$
$\int {xdx \over (a^2-x^2)^2}$
$\int {x^2dx \over (a^2-x^2)^2}$
$\int {x^3dx \over a^2-x^2}$
$\int {x^3dx \over (a^2-x^2)^2}$
$\int {xdx \over (a^2-x^2)^n}$
$\int {dx \over x(a^2-x^2)}$
$\int {dx \over x(a^2-x^2)^2}$
$\int {dx \over x^2(a^2-x^2)}$
$\int {dx \over x^2(a^2-x^2)^2}$
$\int {dx \over x^3+a^3}$
$\int {dx \over (x^3+a^3)^2}$
$\int {xdx \over x^3+a^3}$
$\int {x^2dx \over x^3+a^3}$
$\int {dx \over x(x^3+a^3)}$
$\int {dx \over x(x^3+a^3)^2}$
$\int {dx \over x^4+a^4}$
$\int {dx \over x^4-a^4}$
$\int {xdx \over x^4+a^4}$
$\int {xdx \over x^4-a^4}$
$\int {dx \over x(x^4 \pm a^4)}$
$\int {dx \over x^2(x^4 \pm a^4)}$
$\int {dx \over x^2 \pm x + 1}$
$\int {dx \over (x^2+px+q)^2}$
$\int {dx \over x^2+px+q}$
$\int {xdx \over x^2+px+q}$
$\int {dx \over x(x^2+px+q)}$
$\int x^{n} dx = {x^{n+1} \over n + 1}, \quad n \in Z, \quad n \not = -1$
$\int \frac{dx}{x} = \ln|x|, \quad x \not = 0$
$\int (ax + b)^{n} dx = {(ax + b)^{n+1} \over a(n + 1)}, \quad n \in Z, \quad n \not = -1, \quad a \not = 0$
$\int {dx \over ax + b} = \frac{1}{a} \ln|ax + b|, \quad a \not = 0$
$\int {dx \over (ax + b)^2} = -\frac{1}{a} \frac{1}{ax + b}, \quad a \not = 0$
$\int {xdx \over ax + b} = \frac{x}{a} - \frac{b}{a^2}\ln|ax + b|, \quad a \not = 0$
$\int {x^2 dx \over ax + b} = \frac{x^2}{2a} - \frac{bx}{a^2} + \frac{b^2}{a^3}\ln|ax + b|, \quad a \not = 0$
$\int {x^n dx \over ax + b} = \sum\limits_{k=0}^{n-1} {(-1)^k \, b^k \, x^{n-k} \over (n-k) \, a^{k+1} } + {(-1)^n \, b^n \over a^{n+1}} \ln|ax + b|, \quad a \not = 0$
$\int {xdx \over (ax + b)^2} = \frac{b}{a^2(ax+b)} + \frac{1}{a^2}\ln|ax + b|, \quad a \not = 0$
$\int {x^2dx \over (ax + b)^2} = \frac{x}{a^2} - \frac{b^2}{a^3(ax+b)} - \frac{2b}{a^3}\ln|ax + b|, \quad a \not = 0$
$\int {x^ndx \over (ax + b)^2} = -\frac{x^n}{a(ax+b)} + \frac{n}{a} \int {x^{n-1} dx \over ax+b}, \quad a \not = 0$
$\int {x^3dx \over (ax + b)^2} = \frac{x^3}{3a} - \frac{bx^2}{2a^2} + \frac{b^2x}{a^3} - \frac{b^3}{a^4}\ln|ax + b|, \quad a \not = 0$
$\int {x^ndx \over (ax + b)^m} = -\frac{x^n}{(m-1)a(ax+b)^{m-1}} + \frac{n}{(m-1)a} \int {x^{n-1} dx \over (ax+b)^{m-1}}, \quad a, b \not = 0, \quad m \gt 1$
$\int {dx \over x(ax + b)} = \frac{1}{b} \ln\left|{x \over ax + b}\right|, \quad b \not = 0$
$\int {dx \over x(ax + b)^2} = \frac{1}{b(ax + b)} + \frac{1}{b^2} \ln\left|{x \over ax + b}\right|, \quad b \not = 0$
$\int {dx \over x(ax+b)^3} = {1 \over 2b(ax+b)^2} + {1 \over b^2(ax+b)} + \frac{1}{b^3} \ln\left|{x \over ax+b}\right|, \quad b \not = 0$
$\int {dx \over x(ax+b)^n} = \sum\limits_{k=1}^{n-1} {1 \over k\,b^{n-k}(ax+b)^k} + \frac{1}{b^n}\ln\left|{x \over ax+b}\right|, \quad b \not = 0$
$\int {dx \over x^2(ax+b)^2} = {a \over b^2(ax+b)} - {1 \over b^2x} - \frac{2a}{b^3} \ln\left|{x \over ax+b}\right|, \quad b \not = 0$
$\int {dx \over x^2(ax+b)} = - {1 \over bx} - \frac{a}{b^2} \ln\left|{x \over ax+b}\right|, \quad b \not = 0$
$\int {dx \over x^2(ax+b)^3} = - {a \over 2b^2(ax+b)^2} - {2a \over b^3(ax+b)} - \frac{1}{b^3x} - \frac{3a}{b^4} \ln\left|{x \over ax+b}\right|, \quad b \not = 0$
$\int {dx \over x^3(ax+b)} = \frac{a}{b^2x} - {1 \over 2bx^2} + \frac{a^2}{b^3} \ln\left|{x \over ax+b}\right|, \quad b \not = 0$
$\int {dx \over x^n(ax+b)^m} = - {1 \over (n-1)b\,x^{n-1}(ax+b)^{m-1}} - {(n+m-2)a \over b(n-1)} \int {dx \over x^{n-1}(ax+b)^m}, \quad b \not = 0, n \gt 0$
$\int {ax+b \over \alpha x+\beta}dx = \frac{ax}{\alpha} - {a\beta - \alpha b \over \alpha^2} \ln|\alpha x + \beta|, \quad \alpha \not = 0$
$\int {(ax+b)^2 \over \alpha x+\beta}dx = \frac{a^2 x^2}{2\alpha} + \frac{ax}{\alpha}\left(b - {a\beta - \alpha b \over \alpha}\right) + {(a\beta - \alpha b)^2 \over \alpha^3} \ln|\alpha x + \beta|, \quad \alpha \not = 0$
$\int {ax+b \over (\alpha x+\beta)^2}dx = {a\beta - \alpha b \over \alpha^2(\alpha x+\beta)} + \frac{a}{\alpha^2} \ln|\alpha x + \beta|, \quad \alpha \not = 0$
$\int {(ax+b)^2 \over (\alpha x+\beta)^2}dx = \frac{a^2x}{\alpha^2} - {(a\beta - \alpha b)^2 \over \alpha^3(\alpha x+\beta)} - \frac{2a(a\beta - \alpha b)}{\alpha^3 } \ln|\alpha x + \beta|, \quad \alpha \not = 0$
$\int {(ax+b)^n \over (\alpha x+\beta)^m}dx = - {(ax+b)^n \over (m-1)\alpha(\alpha x+\beta)^{m-1}} + {na \over (m-1)\alpha} \int {(ax+b)^{n-1} \over (\alpha x+\beta)^{m-1}} dx, \quad \alpha \not = 0, m \gt 1$
$\int {dx \over (ax+b)(\alpha x+\beta)} = {-1 \over a\beta - \alpha b} \ln\left|{\alpha x+\beta \over ax+b}\right|, \quad a\beta - \alpha b \not = 0$
$\int {dx \over (ax+b)(\alpha x+\beta)^2} = {1 \over (a\beta - \alpha b)(\alpha x + \beta)} + {a \over (a\beta - \alpha b)^2} \ln\left|{\alpha x+\beta \over ax+b}\right|, \quad a\beta - \alpha b \not = 0$
$\int {xdx \over (ax+b)(\alpha x+\beta)} = {-b \over a(a\beta - \alpha b)} \ln|ax+b| + {\beta \over \alpha(a\beta - \alpha b)} \ln|\alpha x + \beta|, \quad a\beta - \alpha b \not = 0, a \not = 0,\alpha \not = 0$
$\int {xdx \over (x+a)(x+b)} = \frac{a}{a-b}\ln|x+a| - \frac{b}{a-b}\ln|x+b|, \quad a \neq b$
$\int {xdx \over (x+a)(x+b)^2} = {b \over (a-b)(x+b)} - {a \over (a-b)^2} \ln\left|{a+x \over b+x}\right|, \quad a \neq b$
$\int {dx \over (x+a)^2(x+b)^2} = - \frac{1}{(a-b)^2}\left({1 \over x+a} + {1 \over x+b}\right) + \frac{2}{(a-b)^3}\ln\left|{a+x \over b+x}\right|, \quad a \neq b$
$\int {xdx \over (x+a)^2(x+b)^2} = \frac{1}{(a-b)^2}\left({a \over x+a} + {b \over x+b}\right) - \frac{a+b}{(a-b)^3}\ln\left|{a+x \over b+x}\right|, \quad a \neq b$
$\int {x^2dx \over (x+a)^2(x+b)^2} = - \frac{1}{(a-b)^2}\left({a^2 \over x+a} + {b^2 \over x+b}\right) + \frac{2ab}{(a-b)^3}\ln\left|{a+x \over b+x}\right|, \quad a \neq b$
$\int {dx \over x(x+a)(x+b)} = \frac{1}{ab}\ln|x| + {\ln|x+a| \over a(b-a)} - {\ln|x+b| \over b(b-a)}, \quad a \neq b, a \neq 0, b \neq 0$
$\int {dx \over x^2+a^2} = \frac1a arctg\frac{x}{a}, \quad a \neq 0$
$\int {dx \over b^2x^2+a^2} = \frac{1}{ab} arctg\frac{bx}{a}, \quad a \neq 0, b \neq 0$
$\int {xdx \over x^2+a^2} = \frac12 \ln(x^2+a^2)$
$\int {xdx \over (x^2+a^2)^2} = - {1 \over 2(x^2+a^2)}$
$\int {dx \over (x^2+a^2)^2} = {x \over 2a^2(x^2+a^2)} + \frac{1}{2a^3} arctg\frac{x}{a}, \quad a \neq 0$
$\int {dx \over (x^2+a^2)^3} = {x \over 4a^2(x^2+a^2)^2} + {3x \over 8a^4(x^2+a^2)} + \frac{3}{8a^5} arctg\frac{x}{a}, \quad a \neq 0$
$\int {dx \over (x^2+a^2)^n} = {x \over 2(n-1)a^2(x^2+a^2)^{n-1}} + {2n-3 \over (2n-2)a^2} \int {dx \over (x^2+a^2)^{n-1}}, \quad a \neq 0, n \gt 1$
$\int {xdx \over (x^2+a^2)^n} = - {1 \over 2(n-1)(x^2+a^2)^{n-1}}, \quad n \gt 1$
$\int {x^2dx \over x^2+a^2} = x - a\,\mathtt{arctg}\frac{x}{a}, \quad a \neq 0$
$\int {x^2dx \over (x^2+a^2)^2} = - {x \over 2(x^2+a^2)} + \frac{1}{2a}\,\mathtt{arctg}\frac{x}{a}, \quad a \neq 0$
$\int {x^3dx \over (x^2+a^2)^2} = {a^2 \over 2(x^2+a^2)} + \frac12 \ln(x^2+a^2)$
$\int {x^2dx \over (x^2+a^2)^n} = - {x \over 2(n-1)(x^2+a^2)^{n-1}} + \frac{1}{2(n-1)} \int {dx \over (x^2+a^2)^{n-1}}, \quad a \neq 0, n \gt 1$
$\int {dx \over x(x^2+a^2)} = \frac{1}{2a^2} \ln{x^2 \over x^2+a^2}, \quad a \neq 0$
$\int {dx \over x^2(x^2+a^2)} = - {1 \over a^2x} - \frac{1}{a^3}\,\mathtt{arctg}\frac{x}{a}, \quad a \neq 0$
$\int {dx \over x(x^2+a^2)^2} = {1 \over 2a^2(x^2+a^2)} + \frac{1}{2a^4} \ln{x^2 \over x^2+a^2}, \quad a \neq 0$
$\int {dx \over a^2-x^2} = \frac{1}{2a} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0$
$\int {dx \over a^2-b^2x^2} = \frac{1}{2ab} \ln\left|{a+bx \over a-bx}\right|, \quad a \neq 0, b \neq 0$
$\int {dx \over (a^2-x^2)^2} = {x \over 2a^2(a^2-x^2)} + \frac{1}{4a^3} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0$
$\int {dx \over (a^2-x^2)^n} = {x \over 2(n-1)a^2(a^2-x^2)^{n-1}} + {2n-3 \over (2n-2)a^2} \int {dx \over (a^2-x^2)^{n-1}}, \quad a \neq 0, n \gt 1$
$\int {xdx \over a^2-x^2} = - \frac12 \ln|a^2-x^2|$
$\int {x^2dx \over a^2-x^2} = - x + \frac{a}{2} \ln\left|{a+x \over a-x}\right|$
$\int {xdx \over (a^2-x^2)^2} = {1 \over 2(a^2-x^2)}$
$\int {x^2dx \over (a^2-x^2)^2} = {x \over 2(a^2-x^2)} - \frac{1}{4a} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0$
$\int {x^3dx \over a^2-x^2} = - {x^2 \over 2} - \frac{a^2}{2} \ln|a^2-x^2|$
$\int {x^3dx \over (a^2-x^2)^2} = {a^2 \over 2(a^2-x^2)} + \frac{1}{2} \ln|a^2-x^2|$
$\int {xdx \over (a^2-x^2)^n} = {1 \over (2n-2)(a^2-x^2)^{n-1}}, \quad n \gt 1$
$\int {dx \over x(a^2-x^2)} = \frac{1}{2a^2} \ln\left|{x^2 \over a^2-x^2}\right|, \quad a \neq 0$
$\int {dx \over x(a^2-x^2)^2} = {1 \over 2a^2(a^2-x^2)} + \frac{1}{2a^4} \ln\left|{x^2 \over a^2-x^2}\right|, \quad a \neq 0$
$\int {dx \over x^2(a^2-x^2)} = - {1 \over a^2 x} + \frac{1}{2a^3} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0$
$\int {dx \over x^2(a^2-x^2)^2} = - {1 \over a^4 x} + {x \over 2a^4(a^2-x^2)} + \frac{3}{4a^5} \ln\left|{a+x \over a-x}\right|, \quad a \neq 0$
$\int {dx \over x^3+a^3} = \frac{1}{6a^2}\ln{(x+a)^2 \over x^2-ax+a^2} + {1 \over a^2 \sqrt 3} \mathtt{arctg}\frac{2x-a}{a \sqrt 3}, \quad a \neq 0$
$\int {dx \over (x^3+a^3)^2} = {x \over 3a^3(x^3+a^3)} + {2 \over 3a^3} \int {dx \over x^3+a^3}$
$\int {xdx \over x^3+a^3} = - \frac{1}{6a}\ln{(x+a)^2 \over x^2-ax+a^2} + {1 \over a \sqrt 3} \mathtt{arctg}\frac{2x-a}{a \sqrt 3}, \quad a \neq 0$
$\int {x^2dx \over x^3+a^3} = \frac13 \ln|x^3+a^3|$
$\int {dx \over x(x^3+a^3)} = \frac{1}{3a^3} \ln\left|{x^3 \over x^3+a^3}\right|, \quad a \neq 0$
$\int {dx \over x(x^3+a^3)^2} = {1 \over 3a^3(x^3+a^3)} + \frac{1}{3a^6} \ln\left|{x^3 \over x^3+a^3}\right|, \quad a \neq 0$
$\int {dx \over x^4+a^4} = \frac{1}{4\sqrt{2}a^3} \ln\left|{x^2+ax\sqrt{2}+a^2 \over x^2-ax\sqrt{2}+a^2}\right| + \frac{1}{2\sqrt{2}a^3} \mathtt{arctg}\frac{x\sqrt{2}+a}{a} + \frac{1}{2\sqrt{2}a^3} \mathtt{arctg}\frac{x\sqrt{2}-a}{a}, \quad a \neq 0$
$\int {dx \over x^4-a^4} = \frac{-1}{2a^3} \mathtt{arctg}\frac{x}{a} - \frac{1}{4a^3} \ln\left|{x+a \over x-a}\right|, \quad a \neq 0$
$\int {xdx \over x^4+a^4} = \frac{1}{2a^2} \mathtt{arctg}\frac{x^2}{a^2}, \quad a \neq 0$
$\int {xdx \over x^4-a^4} = - \frac{1}{4a^2} \ln\left|{x^2+a^2 \over x^2-a^2}\right|, \quad a \neq 0$
$\int {dx \over x(x^4 \pm a^4)} = \pm \frac{1}{4a^4} \ln\left|{x^4 \over x^4 \pm a^4}\right|, \quad a \neq 0$
$\int {dx \over x^2(x^4 \pm a^4)} = \mp \frac{1}{a^4x} \mp \frac{1}{a^4} \int {x^2 dx \over x^4 \pm a^4}, \quad a \neq 0$
$\int {dx \over x^2 \pm x + 1} = \frac{2}{\sqrt3} \mathtt{arctg}\frac{2x \pm 1}{\sqrt3}$
$\int {dx \over (x^2+px+q)^2} = {2x+p \over (4q-p^2)(x^2+px+q)} + {2 \over 4q-p^2} \int {dx \over x^2+px+q}$
$\int {dx \over x^2+px+q} = \begin{cases} {2 \over \sqrt{4q-p^2}} \mathtt{arctg}\frac{2x+p}{\sqrt{4q-p^2}}, \quad p^2-4q \lt 0 \\ {1 \over \sqrt{p^2-4q}} \ln\left|{2x+p-\sqrt{p^2-4q} \over 2x+p+\sqrt{p^2-4q}}\right|, \quad p^2-4q \gt 0 \end{cases}$
$\int {xdx \over x^2+px+q} = \begin{cases} \frac12 \ln|x^2+px+q| - {p \over \sqrt{4q-p^2}} \mathtt{arctg}\frac{2x+p}{\sqrt{4q-p^2}}, \quad p^2-4q \lt 0 \\ \frac12 \ln|x^2+px+q| - {p \over 2\sqrt{p^2-4q}} \ln\left|{2x+p-\sqrt{p^2-4q} \over 2x+p+\sqrt{p^2-4q}}\right|, \quad p^2-4q \gt 0 \end{cases}$
$\int {dx \over x(x^2+px+q)} = \begin{cases} \frac{1}{2q} \ln\left|{x^2 \over x^2+px+q}\right| - {p \over q\sqrt{4q-p^2}} \mathtt{arctg}\frac{2x+p}{\sqrt{4q-p^2}}, \quad p^2-4q \lt 0 \\ \frac{1}{2q} \ln\left|{x^2 \over x^2+px+q}\right| - {p \over 2q\sqrt{p^2-4q}} \ln\left|{2x+p-\sqrt{p^2-4q} \over 2x+p+\sqrt{p^2-4q}}\right|, \quad p^2-4q \gt 0 \end{cases}$ | 7,013 | 13,166 | {"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} | 3.890625 | 4 | CC-MAIN-2021-21 | latest | en | 0.132189 |
http://ramparida.com/home/courses?category=maths&parent=19 | 1,618,667,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00014.warc.gz | 82,523,969 | 14,021 | ###### Ratings
• Knowing Our Numbers Ramparida Mumbai -
A number is a mathematical value used to numerate and measure various objects. Using the numbers we all can add, subtract, divide and multiply. Here we'll learn how to compare numbers, expand the number, and also learn about the smallest and the largest numbers.
16 Lessons 01:12:57 Hours English Beginner
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• Whole Numbers Ramparida Mumbai -
In mathematics, the whole number set is the most basic number set. This set contains the counting numbers 1, 2, 3,… and the number 0. This chapter begins with the discussion of the definition of whole numbers and natural numbers. You will be learning to differentiate between the whole numbers and the natural numbers. It teaches you that all whole numbers cannot be natural numbers, but all natural numbers are whole numbers.
19 Lessons 01:36:35 Hours English Beginner
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• Playing with Numbers Ramparida Mumbai -
Playing with Numbers comprises different fundamental concepts for building a solid conceptual foundation. In this chapter, you will be learning various concepts like factors and multiples, composite and prime numbers, common factors and common multiples, prime factorization, highest common factor (HCF) and lowest common factor (LCM).
24 Lessons 01:57:05 Hours English Beginner
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• Basic Geometrical Ideas Ramparida Mumbai -
Several important geometric terms that hold extreme importance in later grades are included in this chapter. The topics introduced will not only help build a foundation in geometry but will also help to easily grasp the higher-level concepts in the later grades. The main themes covered here are - basic geometry, polygons, triangles and quadrilateral.
17 Lessons 00:49:55 Hours English Beginner
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• Understanding Elementary Shapes Ramparida Mumbai -
This chapter deals mostly with the measurement of line segments, angles and their types, triangles and their classifications, polygons, quadrilaterals, and solid forms.
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• Integers Ramparida Mumbai -
A fraction represents a part of a whole or generally, any number of equal parts. Some of the important concepts that you will be learning in this chapter includes Types of Fractions, Addition of Fractions, Subtraction of Fractions, Multiplication of Fractions, Comparing Fractions, Simplest Form of Fractions and Representation of Fraction on a Number Line.
13 Lessons 00:54:37 Hours English Beginner
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0 Ratings | 535 | 2,526 | {"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} | 3.984375 | 4 | CC-MAIN-2021-17 | latest | en | 0.883359 |
https://perl6advent.wordpress.com/2012/12/21/ | 1,660,623,443,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00624.warc.gz | 415,806,682 | 16,132 | # Day 21 – Collatz Variations
The Collatz sequence is one of those interesting “simple” math problems that I’ve run into a number of times. Most recently a blog post on programming it in Racket showed up on Hacker News. As happens so often, I instantly wanted to implement it in Perl 6.
```sub collatz-sequence(Int \$start) {
\$start, { when * %% 2 { \$_ / 2 }; when * !%% 2 { 3 * \$_ + 1 }; } ... 1;
}
sub MAIN(Int \$min, Int \$max) {
say [max] (\$min..\$max).map({ +collatz-sequence(\$_) });
}
```
This is a very straightforward implementation of the Racket post’s `max-cycle-length-range` as a stand-alone p6 script. `collatz-sequence` generates the sequence using the p6 sequence operator. Start with the given number. If it is divisible by two, do so: `when * %% 2 { \$_ / 2 }`. If it is not, multiply by three and add 1: `when * !%% 2 { 3 * \$_ + 1 }`. Repeat this until the sequence reaches 1.
`MAIN(Int \$min, Int \$max)` sets up our main function to take two integers. Many times I don’t bother with argument types in p6, but this provides a nice feedback for users:
```> perl6 collatz.pl blue red
Usage:
collatz.pl <min> <max>
```
The core of it just maps the numbers from `\$min` to `\$max` (inclusive) to the length of the sequence (`+collatz-sequence`) and then says the max of the resulting list (`[max]`).
Personally I’m a big fan of using the sequence operator for tasks like this; it directly represents the algorithm constructing the Collatz sequence in a simple and elegant fashion. On the other hand, you should be able to memoize the recursive version for a speed increase. Maybe that would give it an edge over the sequence operator version?
Well, I was wildly wrong about that.
```sub collatz-length(\$start) {
given \$start {
when 1 { 1 }
when * !%% 2 { 1 + collatz-length(3 * \$_ + 1) }
when * %% 2 { 1 + collatz-length(\$_ / 2) }
}
}
sub MAIN(\$min, \$max) {
say [max] (\$min..\$max).map({ collatz-length(\$_) });
}
```
This recursive version, which makes no attempt whatsoever to be efficient, is actually better than twice as fast as the sequence operator version. In retrospect, this makes perfect sense: I was worried about the recursive version making a function call for every iteration, but the sequence version has to make two, one to calculate the next iteration and the other to check and see if the ending condition has been reached.
Well, once I’d gotten this far, I thought I’d better do things correctly. I wrote two framing scripts, one for timing all the available scripts, the other for testing them to make sure they work!
```my @numbers = 1..200, 10000..10200;
sub MAIN(Str \$perl6, *@scripts) {
my %results;
for @scripts -> \$script {
my \$start = now;
qqx/\$perl6 \$script { @numbers }/;
my \$end = now;
%results{\$script} = \$end - \$start;
}
for %results.pairs.sort(*.value) -> (:key(\$script), :value(\$time)) {
say "\$script: \$time seconds";
}
}
```
This script takes as an argument a string that can be used to call a Perl 6 executable and a list of scripts to run. It runs the scripts using the specified executable, and times them using p6’s `now` function. It then sorts the results into order and prints them. (A similar script I won’t post here tests each of them to make sure they are returning correct results.)
In the new framework, the Collatz script has changed a bit. Instead of taking a min and a max value and finding the longest Collatz sequence generated by a number in that range, it takes a series of numbers and generates and reports the length of the sequence for each of them. Here’s the sequence operator script in its full new version:
```sub collatz-length(Int \$start) {
+(\$start, { when * %% 2 { \$_ / 2 }; when * !%% 2 { 3 * \$_ + 1 }; } ... 1);
}
sub MAIN(*@numbers) {
for @numbers -> \$n {
say "\$n: " ~ collatz-length(\$n.Int);
}
}
```
For the rest of the scripts I will skip the `MAIN` sub, which is exactly the same in each of them.
Framework established, I redid the recursive version starting from the new sequence operator code.
```sub collatz-length(Int \$n) {
given \$n {
when 1 { 1 }
when * %% 2 { 1 + collatz-length(\$_ div 2) }
when * !%% 2 { 1 + collatz-length(3 * \$_ + 1) }
}
}
```
The sharp-eyed will notice this version is different from the first recursive version above in two significant ways. This time I made the argument `Int \$n`, which instantly turned up a bit of a bug in all implementations thus far: because I used `\$_ / 2`, most of the numbers in the sequence were actually rationals, not integers! This shouldn’t change the results, but is probably less efficient than using `Int`s. Thus the second difference about, it now uses `\$_ div 2` to divide by 2. This version remains a great improvement over the sequence operator version, running in 4.7 seconds instead of 13.3. Changing ` when * !%% 2` to a simple `default` shaves another .3 seconds off the running time.
Once I started wondering how much time was getting eaten up by the `when` statements, rewriting that bit using the ternary operator was an obvious choice.
```sub collatz-length(Int \$start) {
+(\$start, { \$_ %% 2 ?? \$_ div 2 !! 3 * \$_ + 1 } ... 1);
}
```
Timing results: Basic sequence 13.4 seconds. Sequence with `div` 11.5 seconds. Sequence with `div` and ternary 9.7 seconds.
That made me wonder what kind of performance I could get from a handcoded loop.
```sub collatz-length(Int \$n is copy) {
my \$length = 1;
while \$n != 1 {
\$n = \$n %% 2 ?? \$n div 2 !! 3 * \$n + 1;
\$length++;
}
\$length;
}
```
That’s by far the least elegant of these, I think, but it gets great performance: 3 seconds.
Switching back to the recursive approach, how about using the ternary operator there?
```sub collatz-length(Int \$n) {
return 1 if \$n == 1;
1 + (\$n %% 2 ?? collatz-length(\$n div 2) !! collatz-length(3 * \$n + 1));
}
```
This one just edges out the handcoded loop, 2.9 seconds.
Can we do better than that? How about memoization? `is cached` is supposed to be part of Perl 6; neither implementation has it yet, but last year’s Advent calendar has a Rakudo implementation that still works. Using the last version changed to `sub collatz-length(Int \$n) is cached {` works nicely, but takes 3.4 seconds to execute. Apparently the overhead of caching slows it down a bit. Interestingly, the non-ternary recursive version does speed up with `is cached`, from 4.4 seconds to 3.6 seconds.
```sub collatz-length(Int \$n) {
return 1 if \$n == 1;
state %lengths;
return %lengths{\$n} if %lengths{\$n}:exists;
%lengths{\$n} = 1 + (\$n %% 2 ?? collatz-length(\$n div 2) !! collatz-length(3 * \$n + 1));
}
```
Bingo! 2.7 seconds.
I’m sure there are lots of other interesting approaches for solving this problem, and encourage people to send them in. In the meantime, here’s my summary of results so far:
Script Rakudo Niecza bin/collatz-recursive-ternary-hand-cached.pl 2.5 1.7 bin/collatz-recursive-ternary.pl 3 1.7 bin/collatz-loop.pl 3.1 1.7 bin/collatz-recursive-ternary-cached.pl 3.2 N/A bin/collatz-recursive-default-cached.pl 3.5 N/A bin/collatz-recursive-default.pl 4.4 1.8 bin/collatz-recursive.pl 4.9 1.9 bin/collatz-sequence-ternary.pl 9.9 3.3 bin/collatz-sequence-div.pl 11.6 3.5 bin/collatz-sequence.pl 13.5 3.8
The table was generated from timing-table-generator.pl. | 2,022 | 7,310 | {"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-2022-33 | latest | en | 0.751944 |
http://math-blog.com/review-of-math-for-moms-and-dads/ | 1,475,248,135,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662251.92/warc/CC-MAIN-20160924173742-00002-ip-10-143-35-109.ec2.internal.warc.gz | 172,910,240 | 17,424 | # Review of Math for Moms and Dads
Last weekend I had a chance to read Math for Moms and Dads, which I received from Kaplan as a review copy. This book aims to providing a friendly guide for parents of children ages ten and up, who are struggling with mathematics.
Many parents face the challenge of helping their children with math homework, which for some stems in part to having developed a strong phobia or dislike of the subject themselves. Along with a psychological component, in many cases the challenge is augmented by a lack of basic skills (when it comes to knowing how to approach math problems and work their way through mathematical nomenclature). For some it’s like trying to help their child with French homework, when they don’t speak the language. Otherwise perfectly intelligent adults end up finding themselves worrying over problems that most math-savvy people would consider straightforward.
Math for Moms and Dads tries to solve this predicament by providing a vocabulary of essential terms, a very gentle introduction to problem solving and mathematical reasoning, fundamental concepts of elementary (primary) and middle school mathematics, and step-by-step solutions to basic exercises. It also stresses the importance of the parent-child and parent-teacher relationships when it comes to teaching and assisting with the learning of math. This book is very basic and relatively short, which means that it’s something most parents would be able to squeeze time into their schedule to read (which I feel is a positive element of this book). As someone with a passion for math, I’m biased and admit that I do not find this type of book terribly exciting myself, but I fully realize its usefulness for people who need a “less than scary” introduction (or refresher) to the subject.
The first chapter introduces the book and provides parents with a few pointers on how to use a calculator and when its usage is appropriate. The content on these pages will appear pretty obvious to a large number of readers, but this book tries not to make any assumptions, and as such it aims to cover concepts that many people might take for granted.
Chapter two details the mathematical vocabulary mentioned earlier in this article, and within this chapter parents will learn about fundamental math terminology, including terms such as absolute value, congruent, coordinate plane, diagonal, fraction, permutations and so on. The second part of the chapter provides the reader with more descriptive information about common, basic concepts like commutative and associative property, prime and composite numbers, rational and irrational numbers, union and intersection, linear and quadratic equations, etc.
Chapter three covers the basic rules necessary for resolving a variety of problems, including order of operations, exponents and their rules, properties of numbers, fraction and integer based arithmetic, expressions and equations, and so on.
Chapters four and five tackle the issue of solving homework exercises and preparing for math tests. Together these chapters help clarify how to approach mathematical problems, with examples that are solved in a step-by-step manner.
Chapter six is a pedagogical chapter about how to approach study, which covers topics such as how to create the right study conditions and find the ideal place in your house to turn into a homework area, as well as how to develop note taking and test preparation skills.
Chapter seven is entitled “When will I use this, anyway?”, and it attempts to convince both parents and their children that learning mathematics is an important and useful real world skill. I felt that this chapter (which is about a subject – the importance of math beyond the classroom – I believe strongly in) was on the weaker side, but it may still be useful to some.
Lastly chapter 8 deals with parent-teacher communication, a topic that I felt was important for this kind of book.
Should you feel that your own math skills are not your strongest suit or if you need a concise and easy to follow along with refresher course on numerous basic math topics, so that you can better assist your child with their studies, you will likely find this book right up your alley.
If you are a publisher and would like to have your books reviewed, please contact me at antonio@math-blog.com. As a policy, we will only publish reviews for book worth recommending, informing the publisher if a book doesn’t meet (in our opinion) the standard. | 859 | 4,500 | {"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-2016-40 | longest | en | 0.966217 |
http://www.cram.com/flashcards/trimester-1-science-final-study-guide-315256 | 1,527,364,631,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00204.warc.gz | 350,177,174 | 17,607 | • Shuffle
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### 8 Cards in this Set
• Front
• Back
what are the 4 characteristics of mass and weight? compare and contrast them MASS IS... •a measure of the amount of matter in an object •always constant for an object no matter where the object is in the universe •measured with a balance •expressed in kilograms, greams, and milligrams WEIGHT IS... •a measure of the gravitational force on an object •varied depending on where the object is in relation to the Earth (or any other large body in the universe) •measured with a spring scale •expressed in newtons COMPARE & CONTRAST •weight does not have to do with matter •mass is constant no matter where it is in the universe but weight is different depending on where you are •weight is measured with a spring scale while mass is measured with a balance •weight is expressed in Newtons while mass is expressed in grams, kilograms, and milligrams what is the formula for volume? length x width x height how would you find the volume of an irregular object? volume by displacement what is the formula for density, mass, or volume? D = m/V identify chemical and physical properties chemical properties are properties of matter that describe a substance based on its ability to change into a new substance with different properties physical properties are properties of matter that can be observed or measured without changing the identity of the matter identify chemical and physical changes. CHEMICAL CHANGES: the process of changing one or more substances into new substances with different properties PHYSICAL CHANGES:a change that affects one or more physical properties Name the factors that affect density temperature and pressure explain how temperature and pressure affect density. TEMPERATURE: •when the temperature increases, the colume increases and the density will decrease •when temperature decreases, the colume will decrease and the density will increase PRESSURE: •when pressure increases, the volume decreases then density will increase •when pressure decreases, the colume increases and the density will decrease | 535 | 2,551 | {"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} | 2.953125 | 3 | CC-MAIN-2018-22 | latest | en | 0.892996 |
https://help.geogebra.org/topic/how-can-i-convert-an-area-into-a-variable | 1,675,785,211,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00227.warc.gz | 321,419,539 | 13,853 | # How can I convert an area into a variable?
Cecilia Lucía Gigena shared this question 2 years ago
I have 2 triangles and I have used the Area function.
A_1=t1
A_1=t2
When I change the position of point B, triangle 1 disappears and triangle 2 shows up.
I want a text that says: area triangle = ...
And it has to tell t1 or t2 depending on which of those is shown.
So I want a number with if condition, that equals t1 or t2 depending on conditions to show triangle - and then use that number in the area text.
I tried e = t1 and nothing happens - it does not recognize I am refering to that area.
1
Maybe it's better to define just one triangle and just one rectange both conditionally according to y(B). You could do:
rect=Si(y(B) < y(A), Polígono(E, G, B, C), Polígono(E, A, D, C))
trian=Si(y(B) > y(A), Polígono(A, B, D), Polígono(A, B, G))
Doing so the rectangle and triangle are always defined, so you don't have any problem in your text referring to them.
chris
Files: area.ggb | 278 | 998 | {"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-2023-06 | latest | en | 0.919031 |
https://jodidurgin.com/category/teaching/math/ | 1,643,176,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00383.warc.gz | 387,935,987 | 20,827 | WOULD YOU LIKE ACCESS TO ALL THE FREEBIES FOR ELEMENTARY TEACHERS? ➔
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### How to Teach Rounding (3.NBT.A.1 & 4.NBT.A.3) in October 2022
If you are looking for ideas for how to teach rounding, then you found the right place! Rounding numbers is a math skill that can feel like it is easy to teach. Simply train you students how to circle a number, underline another, apply a rule, etc. Heck, I’ve even | 743 | 3,503 | {"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-2022-05 | latest | en | 0.940112 |
https://www.teacherspayteachers.com/Product/Cartesian-Coordinates-Map-Distances-and-Coordinates-Project-2220117 | 1,495,513,749,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607325.40/warc/CC-MAIN-20170523025728-20170523045728-00225.warc.gz | 945,244,524 | 26,970 | Total:
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# Cartesian Coordinates Map Distances and Coordinates Project
Subjects
Resource Types
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3.71 MB | 15 pages
### PRODUCT DESCRIPTION
Cartesian Coordinates: Map Distances and Coordinates Mini-Project is a mini-adventure that applies Cartesian map distances. Students first “travel” with their pirate companion. Students determine the path they will take, record distances and “sights” as they travel building a travel log. Students then create their own adventure and map. A big hit with students!
Multiple ways to use:
- Set up at a math-center with other cartesian coordinate and map distance activities
- Differentiate by providing to individual students
- Use as an assessment. How well do students understand how to identify coordinates on a Cartesian system and use these coordinates to determine distances?
For extra practice and assessment, you can have students use task cards for a scavenger hunt such as:
- Cartesian Coordinates: Practice and Review and
- Cartesian Coordinates: Find the Distance
A great addition to your middle school math curriculum and for differentiation.
This guide includes:
- Teacher guide
- Student handouts for engaging in activity
- Student handouts for creating their own adventure
Also available in a \$\$ saving Cartesian Coordinate Bundle of Notes, Practice and Project
This purchase is for one teacher only. This resource is not to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are interested in a site license, please contact me for a quote at docrunning@kulikuli.net. This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives.
You may also like:
- Rational vs. Irrational Numbers: 3 kinesthetic activities
- Rational and Irrational number activity bundle
- Secondary Math and Art Projects: Show students the "fun" of math with this 245 page, 9 project bundle with fun math such as tessellations, fibonacci spirals, statistics with Mondrian, and more (CCSS)
- Problem Solving: secondary math puzzlers - great for bell-ringers, time fillers, and challenges of the week
Coming soon…Algebra Puzzlers, Applying Statistics: A statistics research project, Real-world math projects, and more.
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Total Pages
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Included
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3 Days
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\$4.95 | 780 | 3,462 | {"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} | 3.421875 | 3 | CC-MAIN-2017-22 | longest | en | 0.888006 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-2-linear-equations-and-functions-2-3-graph-equations-of-lines-2-3-exercises-skill-practice-page-93/27 | 1,695,302,377,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00615.warc.gz | 889,316,407 | 15,895 | ## Algebra 2 (1st Edition)
We first let y equal zero to find the x intercept: $$2x-0=10 \\ x=-5$$ We now let x equal zero to find the y intercept: $$2(0)-y=10 \\ y =-10$$ | 60 | 171 | {"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} | 3.640625 | 4 | CC-MAIN-2023-40 | latest | en | 0.612837 |
http://courses.cs.vt.edu/~cs1104/Applications/app11.200.htm | 1,513,101,351,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517845.16/warc/CC-MAIN-20171212173259-20171212193259-00603.warc.gz | 71,752,688 | 1,685 | An Example of a BIG Numerical Problem
A very simple computational model of pollution in the air over the U.S. might try to compute the concentration of three pollutants at each point in a 3000 x 1500 x 25 grid.
From physics and chemistry a mathematical model of the interaction of these three pollutants can be derived. This mathematical model is a system of 3 time-dependent, nonlinear, partial differential equations.
A numerical method for this problem would compute values of each of the three unknowns so that the mathematical equations are approximately satisfied at each of the grid points.
So the total number of unknowns is
3 x 3000 x 1500 x 25 ~= 3 x 10 8
The classic algorithm requires O (n3) flops, where n is the number of unknowns. Hence, we need more than 1025 flops to solve the problem. On a teraflop machine, this would take at least
1025/1012 = 1013 seconds (nearly) = 317,000 years
Unfortunately, since the problem is time-dependent, we actually need to solve one of the above problems for each of hundreds or thousands of time steps!
flops: floating point operations
CS1104 Main Page
Last Updated 04/16/2000 | 270 | 1,138 | {"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-2017-51 | latest | en | 0.912053 |
https://www.hackmath.net/en/math-problem/50661 | 1,627,386,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153391.5/warc/CC-MAIN-20210727103626-20210727133626-00551.warc.gz | 830,470,165 | 12,017 | # Louis
Louis wants to carpet the rectangular floor of his basement. The basement has an area of 5,120 square feet. The width of the basement is 4/5 its length. What is the length of Louis's basement?
x = 80 ft
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
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#### You need to know the following knowledge to solve this word math problem:
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## Related math problems and questions:
• Rectangle
The length of the rectangle is 12 cm greater than 3 times its width. What dimensions and area this rectangle has if ts circumference is 104 cm.
• A map
A map with a scale of 1: 5,000 shows a rectangular field with an area of 18 ha. The length of the field is three times its width. The area of the field on the map is 72 cm square. What is the actual length and width of the field?
• Rectangle field
The field has a shape of a rectangle having a length of 119 m and a width of 19 m. , How many meters have to shorten its length and increase its width to maintain its area and circumference increased by 24 m?
• The hall
The hall had a rectangular ground plan one dimension 20 m longer than the other. After rebuilding the length of the hall declined by 5 m and the width has increased by 10 m. Floor area increased by 300 m2. What were the original dimensions of the hall?
• Square vs rectangle
Square and rectangle have the same area contents. The length of the rectangle is 9 greater and width 6 less than side of the square. Calculate the side of a square.
• The circumference
The circumference and width of the rectangle are in a ratio of 5: 1. its area is 216cm2. What is its length?
• The carpet
How many meters of carpet 90 cm wide need to cover floor room which has a rectangular shape with a lengths 4.8 m and 2.4 m if the number of pieces on the carpet is needed to be lowest?
• Rectangle
Calculate area of the rectangle if its length is 12 cm longer than its width and length is equal to the square of its width.
• Area of garden
If the width of the rectangular garden is decreased by 2 meters and its length is increased by 5 meters, the area of the rectangle will be 0.2 ares larger. If the width and the length of the garden will increase by 3 meters, its original size will increas
• Square to rectangle
What is the ratio of the area of a square of side x to the area of a rectangle of a rectangle of width 2 x and length 3
• A rectangular patio
A rectangular patio measures 20 ft by 30 ft. By adding x feet to the width and x feet to the length, the area is doubled. Find the new dimensions of the patio.
• The length 2
The length of a rectangle is 12y2 – 15y + 8 and the width is 7y – 11. Find the area of the rectangle
• Carpet
The room is 10 x 5 meters. You have the role of carpet width of 1 meter. Make rectangular cut of roll that piece of carpet will be longest possible and it fit into the room. How long is a piece of carpet? Note .: carpet will not be parallel with the diago
• Floor
Rectangular floor of living room has a length 5.4 meters and a circumference 17.2 meters. What is its width?
• Rectangular garden
The perimeter of Peter's rectangular garden is 98 meters. The width of the garden is 60% shorter than its length. Find the dimensions of the rectangular garden in meters. Find the garden area in square meters.
• Diagonal 20
Diagonal pathway for the rectangular town plaza whose length is 20 m longer than the width. if the pathway is 20 m shorter than twice the width. How long should the pathway be?
• The ABCD
The ABCD trapezoid has a base length of a = 120mm, c = 86mm and an area of S = 2575 mm2. Calculate the height of the trapezoid. | 945 | 3,855 | {"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-2021-31 | longest | en | 0.94832 |
https://essayzoo.org/coursework/apa/technology/python-program-coursework.php | 1,610,910,114,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00065.warc.gz | 337,652,642 | 7,080 | Not register? Register Now!
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# Python Program (Coursework Sample)
Instructions:
Module 3 - Case
CONDITIONAL, ITERATIONS, AND LOOPS
Case Assignment
Case 3 assignments review program flow control with conditionals, loops, and iterations.
Review the videos and read chapter 9, and 10 (sections 10.1 – 10.4 and 10.7 only) in the online book of "Python 2: For Beginners Only” and complete the following exercises from “Python 2: For Beginners Only.”
Exercise 9.2, Exercise 9.3
Exercise10.2, Exercise 10.3
Create a Word file named as “ITM205-Case 3-Exercises-YourFirstNameLastName”containing a copy of each of the exercises’ IDLE source codes and running results with clear exercise numbers marked on the page.
You can use the Snipping tools or screen print (ctrl + Print Screen) to show the Pythons editor’s (IDLE) code and results and demonstrate that your program executed correctly.
Write a summary document in Microsoft Word format named as “ITM205-Case3-Summary-YourFirstNameLastName” to show what you have accomplished.
source..
Content:
Students Name:
Topic:
Institution:
Date:
Exercise 9.2
Write a Python program that collects from the user a single integer. Have the program print "EVEN" if the integer is divisible by 2, and "ODD" if the integer is not divisible by two
num = 0
print("This program checks for EVEN or ODD numbers")
while (num != -1): #Continue looping unless user inputs -1
num = int(input("Enter a number: ")) #Get user input
if (num % 2 == 0): #Check if the number is divisible by 2 i.e. is EVEN
print("EVEN")
else...
Get the Whole Paper!
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Do you need a custom essay? Order right now:
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Need a Plagiarism Free Essay? | 644 | 2,609 | {"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-2021-04 | latest | en | 0.840902 |
https://www.philosophy-science-humanities-controversies.com/listview-details.php?id=746750&a=t&first_name=J.&author=Bigelow&concept=Second%20Order%20Logic,%20HOL | 1,516,380,818,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00247.warc.gz | 996,445,330 | 5,839 | Philosophy Lexicon of Arguments
2nd order Logic: Predicate logic of the 2nd order goes beyond predicate logic of the 1st level allowing quantification over properties and relations, and not just objects. Thus comparisons of the powerfulness of sets become possible. Problems which are expressed in everyday terms with terms such as "greater", "between", etc., and e.g. the specification of all the properties of an object require predicate logic of the 2nd order. Since the 2nd level logic is not complete (because there are, for example, an infinite number of properties of properties), one often tries to get on with the logic of the 1st order.
_____________
Annotation: The above characterizations of concepts are neither definitions nor exhausting presentations of problems related to them. Instead, they are intended to give a short introduction to the contributions below. – Lexicon of Arguments.
Author Item Summary Meta data
Books on Amazon
Logic 2nd Level/Bigelow/Pargetter: has many fascinating but also astonishing properties: we mention only one: for example, we can define the identity predicate with it without introducing another basic concept with axioms, simply by an abbreviation, for this we need Leibniz' law:
(x)(y) ((x = y) ⇔ (ψ)(ψx equi ψy)).
Two things are identical, iff everything that is one thing applies also to the other.
Logic of the 2nd level: the problem is of semantic, not of syntactic nature.
Question: Which language should we choose for our semantic theory?
---
I 161
Standard solution: frames a semantic theory for 2nd level logic within a 1st level logic. I.e. the meta language contains quantifiers of the 1st level and a predicate of being contained in a set (with axioms of the Zermelo-Fraenkelian set theory).
N.B.: then the 2nd level logic becomes equivalent to a fragment of set theory. For example, by a sentence like
(Eψ) (ψa)
Everyday language translation: "There is something that is a" - "It is somehow like a is" (There is somehow that a is).
That actually says that
(Ey)(a ε y)
Everyday language translation: "There is a set to which a belongs".
That is also logic of the 2nd level.
The semantics for this is as follows: for each predicate we attribute a set to it. So is a simple sentence
Fa
True, if the referent of a is an element of the set corresponding to F.
If F is replaced by a variable , this variable will go over all sets that a predicate such as F could refer to. Therefore,
(Eψ)(ψ a)
becomes true iff there is a set that a belongs to.
Leibniz' Law/Identity/Bigelow/Pargetter: can be rephrased like this:
Two things are identical iff they belong to exactly the same set.
Bigelow/Pargetter: Problem: this should not be taken as a definition of identity, because when defining sets the concept of identity was already used (circularly).
Solution: one could dispel the concerns of circularity by pointing out that identity has been defined in a language while set theory is used in a meta-language.
---
I 162
Circle/Bigelow/Pargetter: nevertheless the circularity is not harmless.
Logic of the 2nd level: Level/Bigelow/Pargetter: should not be treated as a notational variant of set theory. For example, the assertion that x is a set:
Set (x)
Then we can use logic of the 2nd Level to claim that all sets have something in common, that something exists, something that all sets are:
(Eψ)(x)(Set (x)ψx)
This is also true insofar as "being a set" is actually "something" that they all are.
Problem: this cannot be interpreted set theoretically, because otherwise we get a paradox:
There is a set that contains all the sets.
Cantor: showed that this is wrong. There is no set of all sets, no universal set.
Logic 2nd level/Bigelow/Pargetter: this is the reason why it must not be combined with set theory. ((s) Although all sets have something in common, this must not be a set, although otherwise all properties can be considered sets).
Solution/Bigelow/Pargetter: our semantics for a 2nd level logic must be a semantics in a 2nd level language, not a semantics in a 1st level language.
Universals/Bigelow/Pargetter: we are only concerned here, however, with the connection between universals and their instances.
Universal/Bigelow/Pargetter: can be denoted with a name (refers) and quantified via it with quantifiers of the 1st level.
Open sentence/predicate/Bigelow/Pargetter: not every open sentence corresponds to a universal, e.g. one cannot conclude from
There's something that these things are
(Eψ)(ψ x1 uψx2 u...
without additional premises on the fact that...
---
I 163
...there is something that these things have in common
(Ez)(x1 instantiates z u x2 instantiates z u.....)
Universals/Bigelow/Pargetter: this identifies universals as "recurrences" (see above recurrence). ((s) as "recurring" in various things).
Existence of 2nd level/Bigelow/Pargetter: of "something" that things "are" (that things are/(s) like things are) does not secure the existence of the 1st level of something that somehow stands by all these things.
Semantics/Bigelow/Pargetter: this shows that the primary arguments for universals are not semantic. ((s) Because they are not figurative truthmakers).
_____________
Explanation of symbols: Roman numerals indicate the source, arabic numerals indicate the page number. The corresponding books are indicated on the right hand side. ((s)…): Comment by the sender of the contribution.
Big I
J. Bigelow, R. Pargetter
Science and Necessity Cambridge 1990
> Counter arguments against Bigelow
> Suggest your own contribution | > Suggest a correction | > Export as BibTeX Datei
Ed. Martin Schulz, access date 2018-01-19 | 1,347 | 5,617 | {"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-2018-05 | longest | en | 0.929404 |
http://caruncho.eu/fundamentals-of-machine-component-design-juvinall-pdf/ | 1,529,466,715,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00054.warc.gz | 56,205,113 | 7,311 | Syllabus: UNIT I FUNDAMENTALS OF DESIGN Introduction to the design process - Phases of Design - Factors influencing machine design - Selection fundam
Syllabus: UNIT I FUNDAMENTALS OF DESIGN Introduction to the design process – Phases of Design – Factors influencing machine design – Selection fundamentals of machine component design juvinall pdf materials based on mechanical properties. Selection of materials based on mechanical properties. Design of Machine Elements by G. PSG College of Technology, Coimbatore, 2006.
Fundamentals of machine component design. This page was last edited on 19 August 2015, at 11:54. Machine Design An Integrated Approach 3rd Edition , Robert L. Machine_Design_An_Integrated_Approach 3rd Edition, Robert L. Machine Design An Integrated Approach 3rd Edition, Robert L. The Ebook starts from the next page : Enjoy ! Fundamentals of Machine Component Design, 4th Edition, Robert C.
Fundamentals of Machine Component Design 3rd 3dition, Robert C, Juvinall, Kurt M. Selection of Transmission chains and Sprockets. Design of pulleys and sprockets. Pressure angle in the normal and transverse plane- Equivalent number of teeth-forces and stresses.
Estimating the size of the helical gears. Straight bevel gear: Tooth terminology, tooth forces and stresses, equivalent number of teeth. Estimating the dimensions of pair of straight bevel gears. Worm Gear: Merits and demerits- terminology.
Introduction To Graph Theory – selling Today Creating Customer Value and ACT 10e L. Introductory Mathematical Analysis for Business — financial Accounting a Business Process Approach 2e by Jane L. Business Data Communications 4th by Gary B. Global Business Today, 5th edition 2002, step Field Engineering Methods Manual 3rd Ed. Personal Finance 9e Jack Kapoor — management Information Systems: Managing the Digital Firm, child Nursing Care 2e Marcia L.
Thermal capacity, materials-forces and stresses, efficiency, estimating the size of the worm gear pair. Cross helical: Terminology-helix angles-Estimating the size of the pair of cross helical gears. Ray diagram, kinematics layout -Design of sliding mesh gear box -Constant mesh gear box. Design of multi speed gear box. Cam Design: Types-pressure angle and under cutting base circle determination-forces and surface stresses. Note: If you have downloaded all the books in Design of Machine Elements, then no need to download this.
Introduction to Chemical Engineering Thermodynamics 7th Edition J. Modern Control Systems 12th Edition, Richard C. Introduction to Control Systems . Mathematical Models of Systems . Feedback Control System Characteristics . The Performance of Feedback Control Systems . The Stability of Linear Feedback Systems .
Worm Gear: Merits and demerits, contemporary Financial Management, fundamentals of Organizational Communication 7e Pamela S. CMOS Analog Circuit Design, payroll accounting 2009, statistics for Business and Economics 11e David R. Project Management for Information, book keeping and Accounting 3e Joel J. 3rd Ed by W. 12 By Douglas C. Principles of Accounting 1st by Meg Pollard, 13e by Stanley B. | 670 | 3,124 | {"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-2018-26 | latest | en | 0.75652 |
http://sourceforge.net/p/octave/data-smoothing/ci/77d7c26e16ceeb9b1083eca62cb3859f7c54d1e3/tree/inst/ddmat.m | 1,430,971,408,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430460196625.80/warc/CC-MAIN-20150501060316-00015-ip-10-235-10-82.ec2.internal.warc.gz | 185,061,322 | 7,969 | Learn how easy it is to sync an existing GitHub or Google Code repo to a SourceForge project!
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``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38``` ```function D = ddmat(x, d) %%function D = ddmat(x, d) %% Compute divided differencing matrix of order d %% %% Input %% x: vector of sampling positions %% d: order of diffferences %% Output %% D: the matrix; D * Y gives divided differences of order d %% %%References: Anal. Chem. (2003) 75, 3631. %% %% corrected the recursion multiplier; JJS 2/25/08 %% Copyright (C) 2003 Paul Eilers %% This program is free software; you can redistribute it and/or modify %% it under the terms of the GNU General Public License as published by %% the Free Software Foundation; either version 2 of the License, or %% (at your option) any later version. %% %% This program is distributed in the hope that it will be useful, %% but WITHOUT ANY WARRANTY; without even the implied warranty of %% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the %% GNU General Public License for more details. %% %% You should have received a copy of the GNU General Public License %% along with this program; If not, see . m = length(x); if d == 0 D = speye(m); else dx = x((d + 1):m) - x(1:(m - d)); V = spdiags(1 ./ dx, 0, m - d, m - d); D = d * V * diff(ddmat(x, d - 1)); end ``` | 450 | 1,488 | {"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-2015-18 | latest | en | 0.678882 |
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# Statement-1: A solid sphere and a hollow sphere of same material and same radius are floating in a liquid. Same percentage of volume of both the spheres is immersed in the liquid. <br> Statement-2: Up-thrust force acts on volume of liquid displaced. It has nothing to do whether the body is solid or hollow.
Updated On: 27-06-2022
Text Solution
Statement-1, is true, Statement-2 is true, Statement-2 is correct explanation for statement-1Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explantion for statement-1statement-1 is true, statement-2 is falsestatement-1 is false statement-2 is true
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
hello everyone a solid sphere and a hollow sphere of the same material and same radius are floating in a liquid of volume of the spheres is immersed in the liquid any statement to say is that of his force as on volume of liquid displaced it has nothing to do whether the body is solid or hollow to the questions given statement statement and statement to and is asking if both of the statement are independently true and if the statement to is the correct explanation of statement 1 the optimist force acting on a body that displaced some amount of water or liquid is given by and it will be equal to the density of liquid which is RO
X the volume displaced by the object which can be denoted by v x the gravitational acceleration which is ji as it is given in the statement one that the same percentage of volume for both the years is immersed in liquid therefore that the volume of the solid share which can be denoted by SS immersed in the liquid is equals to the volume of the holidays fair which can be denoted by HS emerged in the liquid as it is given in the question that they are immersed in the same liquid therefore RO and is constant G is also constant therefore we can write that the upward force or a thrust force acting on the solid sphere is equals to royel
X VSS in 2G and this is equals to zero and X VHS into G and this is equals to the upward force acting on the hollow sphere therefore we can write that fss is equals to feso4 the force is acting of words on both the square are equal therefore statement one is correct and statement to is also correct as we know that the first food at on volume of the liquid displaced as shown in equation number 154 both statements are independently to and statement to is the correct statement 1 so the correct option A thank you | 578 | 2,671 | {"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} | 3.109375 | 3 | CC-MAIN-2022-49 | latest | en | 0.935438 |
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# Excellent Exponents
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## Veronica Chavez
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#### Transcript of Excellent Exponents
THANK YOU Veronica Chavez Exponent Real World Application EXCELLENT
EXPONENTS Evaluate using the order of operations (PEMDAS)
1. 12 + 4 (2)
2. (25 + 5) ÷ 3(2)-5+10
Evaluate
3. (3) (3)
4. (-5) (-5)(-5)
5. (-3/4) (-3/4) Bell Work
Use your Math Journal to: = n x n x n = n³ Cube n n n = 3 x 3 = 3² = 9 3 Columns 3 Rows Exponents represented in models Write each number as a power of the given base.
I DO:
8; base 2 = _____
WE DO:
49; base 7 = _____
YOU DO:
-125: base -5 = _____ Writing Powers Exponential Form Evaluating Powers
Simplifying Expressions WE DO:
x – y(z·y²) for x=2, y= 4, z=3
YOU DO:
s + (t ͧ-1) for s=10, t=2, u=5 Work as a group to evaluate and/or simplify the exponent problems.
Once your group has an answer, raise the flag and wait to be called on.
Each student in the group must present a correct solution to obtain the total amount of points for each answer. EXPONENT JEOPARDY GAME RULES JEOPARDY GAME LINK
http://www.math-play.com/Exponents-Jeopardy/Exponents-Jeopardy.html LET’S PLAY JEOPARDY! Vocabulary:
Power: expression written with an exponent
Base: number that is used as a factor
Exponent: number that indicates how many times the base is used as a factor
Expanded Form: repeated multiplication of writing an exponent
Exponential Form: when a number or variable is written with a base and an exponent.
One example of how exponents can be applied to the real world is ... Exponent Song Please complete problems from textbook page 164 problems 1-10 even and #31, #42, #45.
Tomorrow quiz on exponents!! Individual Practice Homework Base Exponent 2 Vocabulary Lookout:
Power
Exponent
Base
Expanded Form
Exponential Form Exponents A certain bacterium splits into 2 bacteria every hour. There is 1 bacterium on a slide. How many bacteria will be on the slide after 5 hours? Application of Exponents 3 base = 3 exponent = 2 base = n exponent = 3 What Did We Learn? youtube.com/watch?v=QIZTruxt2rQ&playnext=1&list=PLB6BCB2537F70533E
Full transcript | 695 | 2,489 | {"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} | 3.875 | 4 | CC-MAIN-2018-30 | longest | en | 0.840798 |
https://www.scribd.com/doc/50656255/392 | 1,498,233,782,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320070.48/warc/CC-MAIN-20170623151757-20170623171757-00422.warc.gz | 899,243,809 | 54,085 | Introduction ..............................................................................2 AC Motors ................................................................................4 Force and Motion .....................................................................6 Energy ................................................................................... 11 Electrical Energy ................................................................... 13 AC Motor Construction......................................................... 17 Magnetism............................................................................. 23 Electromagnetism ................................................................. 25 Developing a Rotating Magnetic Field.................................. 31 Rotor Rotation ....................................................................... 37 Motor Specifications ............................................................. 42 NEMA Motor Characteristics ................................................ 46 Derating Factors .................................................................... 54 AC Motors and AC Drives .................................................... 56 Matching AC Motors to the Load ......................................... 61 Enclosures ............................................................................. 67 Mounting ............................................................................... 71 Siemens Motors ................................................................... 78 Above NEMA Motors ........................................................... 83 Review Answers ................................................................... 88 Final Exam ............................................................................. 89
1
Introduction
Welcome to another course in the STEP 2000 series, Siemens Technical Education Program, designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers AC Motors and related products. Upon completion of AC Motors you should be able to:
Explain the concepts of force, inertia, speed, and torque Explain the difference between work and power Describe the construction of a squirrel cage AC motor Describe the operation of a rotating magnetic field Calculate synchronous speed, slip, and rotor speed Plot starting torque, accelerating torque, breakdown torque, and full-load torque on a NEMA torque curve Apply derating factors as required by an application Describe the relationship between V/Hz, torque, and horsepower Match an AC motor to an application and its load Identify NEMA enclosures and mounting configurations Describe Siemens Medallion, PE-21 Plus, vertical pump, and IEC motors Describe torque characteristics and enclosures of Siemens above NEMA motors
2
This knowledge will help you better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. You should complete Basics of Electricity before attempting AC Motors. An understanding of many of the concepts covered in Basics of Electricity is required for AC Motors. You may also want to complete Basics of Control Components which discusses the application of control devices for start, stop, and thermal protection of AC motors. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts. Medallion and PE-21 Plus are trademarks of Siemens Energy & Automation, Inc. National Electrical Code® and NEC® are registered trademarks of the National Fire Protection Association, Quincy, MA 02269. Portions of the National Electrical Code are reprinted with permission from NFPA 70-1996, National Electrical Code Copyright, 1995, National Fire Protection Association, Quincy MA 02269. This reprinted material is not the complete and official position of the National Fire Protection Association on the referenced subject which is represented by the standard in its entirety. Underwriters Laboratories Inc. is a registered trademark of Underwriters Laboratories Inc., Northbrook, IL 60062. The abbreviation UL shall be understood to mean Underwriters Laboratories Inc. National Electrical Manufacturers Association is located at 2101 L. Street, N.W., Washington, D.C. 20037. The abbreviation NEMA is understood to mean National Electrical Manufacturers Association.
3
AC Motors
AC motors are used worldwide in many residential, commercial, industrial, and utility applications. Motors transform electrical energy into mechanical energy. An AC motor may be part of a pump or fan, or connected to some other form of mechanical equipment such as a winder, conveyor, or mixer. AC motors are found on a variety of applications from those that require a single motor to applications requiring several motors. Siemens manufactures a wide variety of motors for various applications. The material presented in this course will help in selection of a motor for a specific application.
Winder Pump
Conveyor
4
NEMA
Throughout this course reference is made to the National Electrical Manufacturers Association (NEMA). NEMA sets standards for a wide range of electrical products, including motors. NEMA is primarily associated with motors used in North America. The standards developed represent general industry practices and are supported by manufacturers of electrical equipment. These standards can be found in NEMA Standard Publication No. MG 1. Some large AC motors may not fall under NEMA standards. These motors are built to meet the requirements of a specific application. These are referred to as above NEMA motors.
IEC
The International Electrotechnical Commission (IEC) is another organization responsible for motor standards. IEC are a group of recommended electrical practices developed by committees from participating IEC countries. These standards are different than NEMA standards. IEC standards are associated with motors used in many countries, including motors used in North America. These standards can be found in IEC 34-1-16. Motors which meet or exceed these standards are referred to as IEC motors.
5
Force and Motion
Before discussing AC motors it is necessary to understand some of the basic terminology associated with motor operation. Many of these terms are familiar to us in some other context. Later in the course we will see how these terms apply to AC motors. Force In simple terms, a force is a push or a pull. Force may be caused by electromagnetism, gravity, or a combination of physical means. Net force is the vector sum of all forces that act on an object, including friction and gravity. When forces are applied in the same direction they are added. For example, if two 10 pound forces were applied in the same direction the net force would be 20 pounds.
10 LB Object 10 LB Net Force = 20 LB
Net force
If 10 pounds of force were applied in one direction and 20 pounds of force applied in the opposite direction, the net force would be 10 pounds and the object would move in the direction of the greater force.
10 LB Object 20 LB
Net Force = 10 LB
6
If 10 pounds of force were applied equally in both directions, the net force would be zero and the object would not move.
10 LB Object 10 LB
Net Force = 0 LB
Torque
Torque is a twisting or turning force that causes an object to rotate. For example, a force applied to the end of a lever causes a turning effect or torque at the pivot point.
Radius (Lever Distance) Pivot Point Force
Torque
Torque (T) is the product of force and radius (lever distance) and is typically measured in lb-ft (pound-feet). Torque = Force x Radius It can be seen that increasing force or increasing the radius increases torque. For example, if 10 pounds of force were applied to a lever 1 foot long there would be 10 lb-ft of torque. Increasing the force to 20 pounds, or the lever to two feet would increase the torque to 20 lb-ft. Inversely, a torque of 10 lb-ft with a two feet radius would yield 5 pounds of force.
Force 10 Lb
Radius 1 Foot Torque = 10 lb-ft
7
Inertia
Mechanical systems are subject to the law of inertia. The law of inertia states that an object will tend to remain in its current state of rest or motion unless acted upon by an external force. A soccer ball, for example, remains at rest until a player applies a force by kicking the ball. The ball will remain in motion until another force, such as friction or the goal net, stops it.
Friction
Because friction removes energy from a mechanical system, a continual force must be applied to keep an object in motion. The law of inertia is still valid, however, since the force applied is needed only to compensate for the energy lost. In the following illustration, a motor runs a conveyor. A large amount of force is applied to overcome inertia and start the system. Once the system is in motion only the energy required to compensate for various losses need be applied to keep the conveyor in motion. These losses include:
Friction within motor and conveyor bearings Wind losses in the motor and conveyor Friction between conveyor belt and rollers
Belt Package
Bearings Rollers
Motor
8
Speed
An object in motion travels a distance in a given time. Speed is the ratio of the distance traveled and the time it takes to travel the distance.
Speed =
Distance Time
A car, for example, may travel 60 miles in one hour. The speed of the car is 60 miles per hour (MPH).
60 MPH =
60 Miles 1 Hour
Speed of a rotating object
Speed also applies to a rotating object, such as the tire of a car or the shaft of a motor. The speed of a rotating object is a measurement of how long it takes a given point on the rotating object to make one complete revolution from its starting point. Speed of a rotating object is generally given in revolutions per minute (RPM). An object that makes ten complete revolutions in one minute has a speed of 10 RPM.
Starting Point
Pivot Point
Acceleration
An object can change speed. This change in speed is called acceleration. Acceleration only occurs when there is a change in the net force acting upon the object, which causes a change in velocity. A car increases speed from 30 MPH to 60 MPH. There has been a change in speed of 30 MPH. An object can also change from a higher to a lower speed. This is known as deceleration (negative acceleration).
30 MPH 60 MPH
9
Acceleration and deceleration also apply to rotating objects. A rotating object, for example, can accelerate from 10 RPM to 20 RPM, or decelerate from 20 RPM to 10 RPM.
10 RPM 20 RPM 20 RPM 10 RPM
Acceleration
Deceleration
Review 1
1. 2.
A ____________ is a push or a pull. An object has 20 pounds of force applied in one direction and 5 pounds of force applied in the opposite direction. The net force is ____________ pounds. A twisting or turning force that causes an object to rotate is known as ____________ . If 40 pounds of force were applied to a lever 2 feet long, the torque would be ____________ lb-ft. The law of ____________ states that an object will tend to remain in its current state of rest or motion unless acted upon by an external force. ____________ is the ratio of distance traveled and time. The speed of a rotating object is generally given in ____________ per ____________ . ____________ only occurs when there is a change in an objects state of motion.
3. 4. 5.
6. 7. 8.
10
Energy
Work
Whenever a force of any kind causes motion, work is accomplished. Work is generally expressed in foot-pounds and is defined by the product of the net force (F) applied and the distance (d) moved. If twice the force is applied, twice the work is done. If an object moves twice the distance, twice the work is done.
W=Fxd
Power Power is the rate of doing work, or work divided by time.
Power =
Force x Distance Time Work Time
Power =
11
Horsepower
Power can be expressed in foot-pounds per second, but is often expressed in horsepower (HP). This unit was defined in the 18th century by James Watt. Watt sold steam engines and was asked how many horses one steam engine would replace. He had horses walk around a wheel that would lift a weight. He found that each horse would average about 550 foot-pounds of work per second. One horsepower is equivalent to 550 foot-pounds per second or 33,000 foot-pounds per minute.
The following formula can be used to calculate horsepower when torque (in lb-feet) and speed are known. An increase of torque, speed, or both will cause an increase in horsepower.
HP =
T x RPM 5250
12
Electrical Energy
In an electrical circuit, voltage applied to a conductor will cause electrons to flow. Voltage is the force and electron flow is the motion.
Electron Flow = Motion
+ _
Battery
Voltage = Force
The rate at which work is done is called power and is represented by the symbol P. Power is measured in watts represented by the symbol W. The watt is defined as the rate work is done in a circuit when 1 amp flows with 1 volt applied. Power consumed in a resistor Power consumed in a resistor depends on the amount of current that passes through the resistor for a given voltage. This is expressed as voltage (E) times current (I).
P=ExI or P = EI
13
In the following simple circuit power consumed in the resistor can be calculated.
I=2 amps
+ _
12 volts
R=6 Ω
P = EI P = 12 volts x 2 amps P = 24 watts
Power in an AC circuit Resistance is not the only circuit property that affects power in an AC circuit. Capacitance and inductance also affect power. Power consumed by a resistor is dissipated in heat and not returned to the source. This power is used to do useful work and is called true power. True power is the rate at which energy is used and is measured in watts (W). Current in an AC circuit rises to peak values and diminishes to zero many times a second. The energy stored in the magnetic field of an inductor or plates of a capacitor is returned to the source when current changes direction. This power is not consumed and is called reactive power. Reactive power is measured in volt-amps reactive (VAR). Power in an AC circuit is the vector sum of true power and reactive power. This is called apparent power. Apparent power is measured in volt-amps (VA).
14
The formula for apparent power is:
P = EI
True power is calculated from a trigonometric function, the cosine of the phase angle (cos q as shown on the previous page). The formula for true power is:
P = EI cos θ
Power factor Power factor is the ratio of true power to apparent power in an AC circuit. Power factor is equal to the cosine q. PF = cos q Horsepower and kilowatts AC motors manufactured in the United States are generally rated in horsepower (HP). Equipment manufactured in Europe is generally rated in kilowatts (KW). Horsepower can be converted to kilowatts with the following formula: KW = .746 x HP For example, a 25 HP motor is equivalent to 18.65 KW. 18.65 KW = .746 x 25 HP Kilowatts can be converted to horsepower with the following formula: HP = 1.341 x KW The power formula for a single-phase system is:
KW = V x I x PF 1000
The power formula for three-phase power is:
KW = V x I x PF x 1.732 1000
Note that voltage, current, and power factor are provided by the motor manufacturer.
15
Three-phase power
Power is considered single-phase when it is operated by one voltage source. Single-phase power is used for small electrical demands such as found in the home. Three-phase power is produced by an alternating current power supply system equivalent to three voltage sources. Three-phase power is a continuous series of three overlapping AC voltages. Each voltage wave represents a phase and is offset by 120 electrical degrees. Three-phase power is used where a large quantity of electrical power is required, such as commercial and industrial applications.
+
Phase A Phase B Phase C
0
_ 120° 240° 360°
Review 2
1. 2. 3. 4. 5. 6.
____________ is accomplished whenever force causes motion. ____________ is the rate of doing work. In an electrical circuit the force required to cause electron flow is called ____________ . Power used to do work is called ____________ power. True power is measured in ____________ and apparent power is measured in ____________ . 18 KW is equivalent to ____________ HP .
16
AC Motor Construction
AC induction motors are commonly used in industrial applications. The following motor discussion will center around three-phase, 460 VAC, asynchronous, induction motors. An asynchronous motor is a type of motor where the speed of the rotor is other than the speed of the rotating magnetic field. This type of motor is illustrated below. The three basic parts of an AC motor are the rotor, stator, and enclosure.
Enclosure
Stator
Rotor
17
Stator construction
The stator and the rotor are electrical circuits that perform as electromagnets. The stator is the stationary electrical part of the motor. The stator core of a NEMA motor is made up of several hundred thin laminations.
Stator windings
Stator laminations are stacked together forming a hollow cylinder. Coils of insulated wire are inserted into slots of the stator core.
18
Each grouping of coils, together with the steel core it surrounds, form an electromagnet. Electromagnetism is the principle behind motor operation. The stator windings are connected directly to the power source.
Rotor construction
The rotor is the rotating part of the electromagnetic circuit. The most common type of rotor is the squirrel cage rotor. Other types of rotor construction will be mentioned later in the course. The construction of the squirrel cage rotor is reminiscent of rotating exercise wheels found in cages of pet rodents.
19
The rotor consists of a stack of steel laminations with evenly spaced conductor bars around the circumference.
The laminations are stacked together to form a rotor core. Aluminum is die cast in the slots of the rotor core to form a series of conductors around the perimeter of the rotor. Current flow through the conductors form the electromagnet. The conductor bars are mechanically and electrically connected with end rings. The rotor core mounts on a steel shaft to form a rotor assembly.
Steel Laminations
Shaft
Conductor Bars End Ring
20
Enclosure
The enclosure consists of a frame (or yoke) and two end brackets (or bearing housings). The stator is mounted inside the frame. The rotor fits inside the stator with a slight air gap separating it from the stator. There is no direct physical connection between the rotor and the stator.
The enclosure also protects the electrical and operating parts of the motor from harmful effects of the environment in which the motor operates. Bearings, mounted on the shaft, support the rotor and allow it to turn. A fan, also mounted on the shaft, is used on the motor shown below for cooling.
21
Review 3
1.
Identify the following components from the illustration: A. ____________ B. ____________ C._____________
A B
C
2. 3. 4. 5.
The ____________ and the ____________ are two parts of an electrical circuit that form an electromagnet. The ____________ is the stationary electrical part of an AC motor. The ____________ is the rotating electrical part of an AC motor. The ____________ ____________ rotor is the most common type of rotor used in AC motors.
22
Magnetism
The principles of magnetism play an important role in the operation of an AC motor. All magnets have two characteristics. They attract and hold metal objects like steel and iron. If free to move, like the compass needle, the magnet will assume roughly a north-south position.
S N
Magnetic lines of flux
We know that a magnet attracts an iron or steel object by an invisible force. The magnets invisible force is called lines of flux. These lines of flux make up an invisible magnetic field. Every magnet has two poles, one north pole and one south pole. Invisible magnetic lines of flux leave the north pole and enter the south pole. While the lines of flux are invisible, the effects of magnetic fields can be made visible. When a sheet of paper is placed on a magnet and iron filings loosely scattered over it, the filings will arrange themselves along the invisible lines of flux.
N
S
23
By drawing lines the way the iron filings have arranged themselves, the following illustration is obtained. Broken lines indicate the paths of magnetic flux lines. Field lines exist outside and inside the magnet. The magnetic lines of flux always form closed loops, leaving the north pole and entering the south pole. They return to the north pole through the magnet.
N
S
Unlike poles attract
The polarity of the magnetic field affects the interaction between separate magnets. For example, when the opposite poles of two magnets are brought within range of each other the lines of flux combine and tend to pull or attract the magnets.
N
S N
S
Like poles repel
When poles of like polarity of two magnets are brought within range of each other the lines of flux produce a force that tends to push or repel the magnets. For this reason it is said that unlike poles attract and like poles repel. The attracting and repelling action of the magnetic fields is important in the operation of AC motors.
N
S
S
N
24
Electromagnetism
When current flows through a conductor a magnetic field is produced around the conductor. The magnetic field is made up of lines of flux, just like a natural magnet. The size and strength of the magnetic field will increase and decrease as the current flow strength increases and decreases.
Current Flow
Increased Current Flow
25
Left-hand rule for conductors
A definite relationship exists between the direction of current flow and the direction of the magnetic field. The left-hand rule for conductors demonstrates this relationship. If a currentcarrying conductor is grasped with the left hand with the thumb pointing in the direction of electron flow, the fingers will point in the direction of the magnetic lines of flux.
In the following illustration it can be seen that when the electron flow is away from the viewer (indicated by the plus sign) the lines of flux flow in a counterclockwise direction around the conductor. When the electron flow reverses and current flow is towards the viewer (indicated by the dot) the lines of flux reverse direction and flow in a clockwise direction.
+
Current Flow Away From View Current Flow Toward View
26
Electromagnet
An electromagnet can be made by winding the conductor into a coil and applying a DC voltage. The lines of flux, formed by current flow through the conductor, combine to produce a larger and stronger magnetic field. The center of the coil is known as the core. In this simple electromagnet the core is air.
Air Core DC Voltage
Iron is a better conductor of flux than air. The air core of an electromagnet can be replaced by a piece of soft iron. When a piece of iron is placed in the center of the coil more lines of flux can flow and the magnetic field is strengthened.
Iron Core
DC Voltage
Number of turns
The strength of the magnetic field in the DC electromagnet can be increased by increasing the number of turns in the coil. The greater the number of turns the stronger the magnetic field will be.
DC Voltage
DC Voltage
5 Turns
10 Turns
27
Changing polarity
The magnetic field of an electromagnet has the same characteristics as a natural magnet, including a north and south pole. However, when the direction of current flow through the electromagnet changes, the polarity of the electromagnet changes. The polarity of an electromagnet connected to an AC source will change at the same frequency as the frequency of the AC source. This can be demonstrated in the following illustration. At Time 1 current flow is at zero. There is no magnetic field produced around the electromagnet. At Time 2 current is flowing in a positive direction. A magnetic field builds up around the electromagnet. The electromagnet assumes a polarity with the south pole on the top and the north pole on the bottom. At Time 3 current flow is at its peak positive value. The strength of the electromagnetic field is at its greatest value. At Time 4 current flow decreases and the magnetic field begins to collapse, until Time 5 when current flow and magnetic field are at zero. Current immediately begins to increase in the opposite direction. At Time 6 current is increasing in a negative direction. The polarity of the electromagnetic field has changed. The north pole is now on top and the south pole is on the bottom. The negative half of the cycle continues through Times 7 and 8, returning to zero at Time 9. This process will repeat 60 times a second with a 60 Hz AC power supply.
S
S
S
N 3 N 2 4 N 5
1 9
N
6 7 N
8
N
S
S
S
28
Induced voltage
A conductor moving through a magnetic field will have a voltage induced into it. This electrical principle is used in the operation of AC induction motors. In the following illustration an electromagnet is connected to an AC power source. Another electromagnet is placed above it. The second electromagnet is in a separate circuit. There is no physical connection between the two circuits. Voltage and current are zero in both circuits at Time 1. At Time 2 voltage and current are increasing in the bottom circuit. A magnetic field builds up in the bottom electromagnet. Lines of flux from the magnetic field building up in the bottom electromagnet cut across the top electromagnet. A voltage is induced in the top electromagnet and current flows through it. At Time 3 current flow has reached its peak. Maximum current is flowing in both circuits. The magnetic field around the coil continues to build up and collapse as the alternating current continues to increase and decrease. As the magnetic field moves through space, moving out from the coil as it builds up and back towards the coil as it collapses, lines of flux cut across the top coil. As current flows in the top electromagnet it creates its own magnetic field.
Ammeter 0 Ammeter 0 Ammeter 0
S
N
N
S
N
S
Time 1
Time 2
Time 3
29
Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field in the bottom electromagnet. Since opposite poles attract, the top electromagnet will follow the bottom electromagnet when it is moved.
Ammeter 0 Ammeter 0
S
N
S
N
N
S
N
S
Old Position
New Position
Review 4
1. 2.
Magnetic lines of flux leave the ____________ pole of a magnet and enter the ____________ pole. Identify which magnets will attract each other and which magnets will repel each other in the following illustration.
A. B. C. D.
3.
N N S N
S S N S
N S N N
S N S S
____________ ____________ ____________ ____________
When current flows through a conductor a ____________ ____________ is produced around the conductor. Which of the following will not increase the strength of the magnetic field? A. B. C. D. Increase the current flow Increase the number of turns in a coil Add an iron core to a coil Increase the frequency of the AC power supply
4.
30
Developing a Rotating Magnetic Field
The principles of electromagnetism explain the shaft rotation of an AC motor. Recall that the stator of an AC motor is a hollow cylinder in which coils of insulated wire are inserted.
Stator coil arrangement
The following schematic illustrates the relationship of the coils. In this example six coils are used, two coils for each of the three phases. The coils operate in pairs. The coils are wrapped around the soft iron core material of the stator. These coils are referred to as motor windings. Each motor winding becomes a separate electromagnet. The coils are wound in such a way that when current flows in them one coil is a north pole and its pair is a south pole. For example, if A1 were a north pole then A2 would be a south pole. When current reverses direction the polarity of the poles would also reverse.
Motor Windings A1 (N)
B2
C2 Iron Core
C1
(S) A2
B1
31
Power supply
The stator is connected to a 3-phase AC power supply. In the following illustration phase A is connected to phase A of the power supply. Phase B and C would also be connected to phases B and C of the power supply respectively.
A1 (N)
To Phase A C2
+
A
B
C
B2
0 _
C1 (S) A2 B1
To Phase A
Phase windings (A, B, and C) are placed 120° apart. In this example, a second set of three-phase windings is installed. The number of poles is determined by how many times a phase winding appears. In this example, each phase winding appears two times. This is a two-pole stator. If each phase winding appeared four times it would be a four-pole stator.
A1 B2 C2
C1
B1
A2
2-Pole Stator Winding
32
When AC voltage is applied to the stator, current flows through the windings. The magnetic field developed in a phase winding depends on the direction of current flow through that winding. The following chart is used here for explanation only. It will be used in the next few illustrations to demonstrate how a rotating magnetic field is developed. It assumes that a positive current flow in the A1, B1 and C1 windings result in a north pole.
Winding Current Flow Direction Positive Negative North South A1 South North A2 North South B1 South North B2 North South C1 South North C2
Start It is easier to visualize a magnetic field if a start time is picked when no current is flowing through one phase. In the following illustration, for example, a start time has been selected during which phase A has no current flow, phase B has current flow in a negative direction and phase C has current flow in a positive direction. Based on the above chart, B1 and C2 are south poles and B2 and C1 are north poles. Magnetic lines of flux leave the B2 north pole and enter the nearest south pole, C2. Magnetic lines of flux also leave the C1 north pole and enter the nearest south pole, B1. A magnetic field results, as indicated by the arrow.
A1 B2 C2
Resultant Magnetic Field Magnetic Lines Of Flux Current Flow In A Positive Direction
C1
B1
A2
C
A Current Flow At Zero B Start Current Flow In A Negative Direction
33
Time 1
If the field is evaluated at 60° intervals from the starting point, at Time 1, it can be seen that the field will rotate 60°. At Time 1 phase C has no current flow, phase A has current flow in a positive direction and phase B has current flow in a negative direction. Following the same logic as used for the starting point, windings A1 and B2 are north poles and windings A2 and B1 are south poles.
A1 B2 C2
C1
B1
A2
A1 B2 C2
C1
B1
A2
60°
C
Current Flow In A Positive Direction
A 60° B Current Flow In A Negative Direction Start 1 Current Flow At Zero
34
Time 2
At Time 2 the magnetic field has rotated 60°. Phase B has no current flow. Although current is decreasing in phase A it is still flowing in a positive direction. Phase C is now flowing in a negative direction. At start it was flowing in a positive direction. Current flow has changed directions in the phase C windings and the magnetic poles have reversed polarity.
A1 B2 C2 B2 A1 C2
C1
B1
C1 A1
B1
A2
A2
B2
60°
B1
C1
A2
C
Current Flow In A Positive Direction
A 60° B 120° Start 1 2 Current Flow At Zero Current Flow In A Negative Direction
360° rotation
At the end of six such time intervals the magnetic field will have rotated one full revolution or 360°. This process will repeat 60 times a second on a 60 Hz power supply.
A1 B2 C2 B2 A1 C2 B2 A1 C2 B2 A1 C2
C1
B1
C1 A1
B1
C1 A1
B1
C1 A1
B1
A2
A2
C2 B2
A2
C2 B2
A2
C2
B2
C1
B1
C1
B1
C1
B1
A2
A2
A2
C
A 60° B 360° Start 1 2 3 4 5 6
35
Synchronous speed
The speed of the rotating magnetic field is referred to as synchronous speed (NS). Synchronous speed is equal to 120 times the frequency (F), divided by the number of poles (P).
NS =
120F P
If the frequency of the applied power supply for the two-pole stator used in the previous example is 60 Hz, synchronous speed is 3600 RPM.
NS =
120 x 60 2
NS = 3600 RPM
The synchronous speed decreases as the number of poles increase. The following table shows the synchronous speed at 60 Hz for the corresponding number of poles.
No. of Poles Synchronous Speed 2 4 6 8 10 3600 1800 1200 900 720
36
Rotor Rotation
Permanent magnet
To see how a rotor works, a magnet mounted on a shaft can be substituted for the squirrel cage rotor. When the stator windings are energized a rotating magnetic field is established. The magnet has its own magnetic field that interacts with the rotating magnetic field of the stator. The north pole of the rotating magnetic field attracts the south pole of the magnet, and the south pole of the rotating magnetic field attracts the north pole of the magnet. As the rotating magnetic field rotates, it pulls the magnet along causing it to rotate. This design, used on some motors, is referred to as a permanent magnet synchronous motor.
Magnet
A1 B2
S
C2
Rotating Magnetic Field
C1
A2
B1
Induced voltage electromagnet
The squirrel cage rotor acts essentially the same as the magnet. When power is applied to the stator, current flows through the winding, causing an expanding electromagnetic field which cuts across the rotor bars.
Magnetic Field Of Coil A1 A1
B2
C2 Rotor Conductor Bars
C1
B1 Stator
Rotor
A2
37
When a conductor, such as a rotor bar, passes through a magnetic field a voltage (emf) is induced in the conductor. The induced voltage causes a current flow in the conductor. Current flows through the rotor bars and around the end ring. The current flow in the conductor bars produces magnetic fields around each rotor bar. Recall that in an AC circuit current continuously changes direction and amplitude. The resultant magnetic field of the stator and rotor continuously change. The squirrel cage rotor becomes an electromagnet with alternating north and south poles.
The following drawing illustrates one instant in time during which current flow through winding A1 produces a north pole. The expanding field cuts across an adjacent rotor bar, inducing a voltage. The resultant magnetic field in the rotor tooth produces a south pole. As the stator magnetic field rotates the rotor follows.
A1 N B2
S
C2
C1
B1
A2
38
Slip
There must be a relative difference in speed between the rotor and the rotating magnetic field. If the rotor and the rotating magnetic field were turning at the same speed no relative motion would exist between the two, therefore no lines of flux would be cut, and no voltage would be induced in the rotor. The difference in speed is called slip. Slip is necessary to produce torque. Slip is dependent on load. An increase in load will cause the rotor to slow down or increase slip. A decrease in load will cause the rotor to speed up or decrease slip. Slip is expressed as a percentage and can be determined with the following formula.
% Slip =
NS - NR NS
x 100
For example, a four-pole motor operated at 60 Hz has a synchronous speed (NS) of 1800 RPM. If the rotor speed at full load were 1750 RPM (NR), the slip is 2.8%.
% Slip =
1800 - 1765 x 100 1800
% Slip = 1.9%
Wound rotor motor
The discussion to this point has been centered on the more common squirrel cage rotor. Another type is the wound rotor. A major difference between the wound rotor motor and the squirrel cage rotor is the conductors of the wound rotor consist of wound coils instead of bars. These coils are connected through slip rings and brushes to external variable resistors. The rotating magnetic field induces a voltage in the rotor windings. Increasing the resistance of the rotor windings causes less current flow in the rotor windings, decreasing speed. Decreasing the resistance allows more current flow, speeding the motor up.
Wound Rotor Brush
Slip Ring
External Rotor Resistance
39
Synchronous motor
Another type of AC motor is the synchronous motor. The synchronous motor is not an induction motor. One type of synchronous motor is constructed somewhat like a squirrel cage rotor. In addition to rotor bars coil windings are added. The coil windings are connected to an external DC power supply by slip rings and brushes. On start AC is applied to the stator and the synchronous motor starts like a squirrel cage rotor. DC is applied to the rotor coils after the motor reaches maximum speed. This produces a strong constant magnetic field in the rotor which locks in step with the rotating magnetic field. The rotor turns at the same speed as synchronous speed (speed of the rotating magnetic field). There is no slip. Variations of synchronous motors include a permanent magnet rotor. The rotor is a permanent magnet and an external DC source is not required. These are found on small horsepower synchronous motors.
Rotor Bar Brush
Coil
External DC Power Supply Slip Ring
40
Review 5
1.
The following illustration represents a ____________ pole motor. If winding A1 is a south pole then winding A2 is a ____________ pole.
A1
B2
C2
C1
B1
A2
2. 3. 4. 5.
The speed of the rotating magnetic field is referred to as ____________ speed. The synchronous speed of a 60 Hz 4-pole motor is ____________ RPM. The difference in speed between the rotor and synchronous speed is ____________ . A 2-pole motor is operating on a 60 Hz power supply. The rotor is turning at 3450 RPM. Slip is _________ %.
41
Motor Specifications
Nameplate
The nameplate of a motor provides important information necessary for selection and application. The following drawing illustrates the nameplate of a sample 30 horsepower AC motor. Specifications are given for the load and operating conditions as well as motor protection and efficiency.
Voltage and amps
AC motors are designed to operate at standard voltages and frequencies. This motor is designed for use on 460 VAC systems. Full-load current for this motor is 34.9 amps.
42
RPM
Base speed is the nameplate speed, given in RPM, where the motor develops rated horsepower at rated voltage and frequency. It is an indication of how fast the output shaft will turn the connected equipment when fully loaded with proper voltage and frequency applied.
The base speed of this motor is 1765 RPM at 60 Hz. It is known that the synchronous speed of a 4-pole motor is 1800 RPM. When fully loaded there will be 1.9% slip. If the connected equipment is operating at less than full load, the output speed (RPM) will be slightly greater than nameplate.
% Slip =
1800 - 1765 x 100 1800
% Slip = 1.9%
Service factor
A motor designed to operate at its nameplate horsepower rating has a service factor of 1.0. This means the motor can operate at 100% of its rated horsepower. Some applications may require a motor to exceed the rated horsepower. In these cases a motor with a service factor of 1.15 can be specified. The service factor is a multiplier that may be applied to the rated power. A 1.15 service factor motor can be operated 15% higher than the motors nameplate horsepower. The 30 HP motor with a 1.15 service factor, for example can be operated at 34.5 HP It should be noted that . any motor operating continuously at a service factor greater than 1 will have a reduced life expectancy compared to operating it at its rated horsepower. In addition, performance characteristics, such as full load RPM and full load current, will be affected.
43
Class insulation
The National Electrical Manufacturers Association (NEMA) has established insulation classes to meet motor temperature requirements found in different operating environments. The four insulation classes are A, B, F, and H. Class F is commonly used. Class A is seldom used. Before a motor is started, its windings are at the temperature of the surrounding air. This is known as ambient temperature. NEMA has standardized on an ambient temperature of 40° C, or 104° F within a defined altitude range for all motor classes.
Temperature will rise in the motor as soon as it is started. Each insulation class has a specified allowable temperature rise. The combination of ambient temperature and allowed temperature rise equals the maximum winding temperature in a motor. A motor with Class F insulation, for example, has a maximum temperature rise of 105° C when operated at a 1.0 service factor. The maximum winding temperature is 145° C (40° ambient plus 105° rise). A margin is allowed to provide for a point at the center of the motors windings where the temperature is higher. This is referred to as the motors hot spot.
180° 160° 140° 120° 100° 80° 60° 40° 20° 0° 180° 160° 140° 120° 100° 80° 60° 40° 20° 0° 180° 160° 140° 120° 100° 80° 60° 40° 20° 0° 180° 160° 140° 120° 100° 80° 60° 40° 20° 0°
15°
10°
10° 80°
5° 60°
105°
125°
Class A 60° C Rise 5° C Hot Spot
Class B 80° C Rise 10° C Hot Spot
Class F 105° C Rise 10° C Hot Spot
Class H 125° C Rise 15° C Hot Spot
The operating temperature of a motor is important to efficient operation and long life. Operating a motor above the limits of the insulation class reduces the motors life expectancy. A 10° C increase in the operating temperature can decrease the motors insulation life expectancy as much as 50%.
44
Motor design
The National Electrical Manufacturers Association (NEMA) has established standards for motor construction and performance. NEMA design B motors are most commonly used.
Efficiency
AC motor efficiency is expressed as a percentage. It is an indication of how much input electrical energy is converted to output mechanical energy. The nominal efficiency of this motor is 93.6%. The higher the percentage the more efficiently the motor converts the incoming electrical power to mechanical horsepower. A 30 HP motor with a 93.6% efficiency would consume less energy than a 30 HP motor with an efficiency rating of 83%. This can mean a significant savings in energy cost. Lower operating temperature, longer life, and lower noise levels are typical benefits of high efficiency motors.
45
NEMA Motor Characteristics
Standard motor designs
Motors are designed with certain speed-torque characteristics to match speed-torque requirements of various loads. The four standard NEMA designs are NEMA A, NEMA B, NEMA C, and NEMA D. NEMA A is not used very often. NEMA B is most commonly used. NEMA C and NEMA D are used for specialized applications. A motor must be able to develop enough torque to start, accelerate and operate a load at rated speed. Using the sample 30 HP 1765 RPM motor discussed , previously, torque can be calculated by transposing the formula for horsepower.
HP = T=
T x RPM 5250 T= 30 x 5250 T = 89.2 Lb-Ft 1765
HP x 5250 RPM
Speed-torque curve for NEMA B motor
A graph, like the one shown below, shows the relationship between speed and torque the motor produces from the moment of start until the motor reaches full-load torque at rated speed. This graph represents a NEMA B motor.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
46
Starting torque
Starting torque (point A on the graph) is also referred to as locked rotor torque. This torque is developed when the rotor is held at rest with rated voltage and frequency applied. This condition occurs each time a motor is started. When rated voltage and frequency are applied to the stator there is a brief amount of time before the rotor turns. At this instant a NEMA B motor develops approximately 150% of its full-load torque. A 30 HP 1765 RPM motor, for example, will develop , approximately 133.8 Lb-Ft of torque.
300 275 250 225 200 175 150 A 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed Full-Load Torque 89.2 Lb-Ft Starting Torque 133.8 Lb-Ft
47
Accelerating and breakdown torque
The magnetic attraction of the rotating magnetic field will cause the rotor to accelerate. As the motor picks up speed torque decreases slightly until it reaches point B on the graph. As speed continues to increase from point B to point C torque increases until it reaches its maximum at approximately 200%. This torque is referred to as accelerating or pull up torque. Point C is the maximum torque a motor can produce. At this point a 30 HP motor will develop approximately 178.4 Lb-Ft of torque. If the motor were overloaded beyond the motors torque capability, it would stall or abruptly slow down at this point. This is referred to as breakdown or pullout torque.
300 275 250 225 200 175 150 A 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed Accelerating Torque Full-Load Torque 89.2 Lb-Ft Starting Torque
B
Breakdown (Pullout) Torque
C
178.4 Lb-Ft 133.8 Lb-Ft
48
Torque decreases rapidly as speed increases beyond breakdown torque (point C), until it reaches full-load torque at a speed slightly less than 100% synchronous speed. Full-load torque is the torque developed when the motor is operating with rated voltage, frequency and load. The speed at which full-load torque is produced is the slip speed or rated speed of the motor. Recall that slip is required to produce torque. If the synchronous speed of the motor is 1800 RPM and the amount of slip is 1.9%, the full-load rated speed of the motor is 1765 RPM. The full-load torque of the 1765 RPM 30 HP motor is 89.2 Lb-Ft. NEMA design B motors are general purpose single speed motors suited for applications that require normal starting and running torque such as conveyors, fans, centrifugal pumps, and machine tools.
300 275 250 225 200 175 150 A 125 100 75 50 25 0 0 Accelerating Torque Full-Load Torque Slip 1.9%
D
Breakdown (Pullout) Torque
C
Starting Torque
B
178.4 Lb-Ft 133.8 Lb-Ft
89.2 Lb-Ft
10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
49
Starting current is also referred to as locked rotor current, and is measured from the supply line at rated voltage and frequency with the rotor at rest. Full-load current is the current measured from the supply line at rated voltage, frequency and load with the rotor up to speed. Starting current is typically 600-650% of full-load current on a NEMA B motor. Starting current decreases to rated full-load current as the rotor comes up to speed.
700 600 500 400 300 200 100
Starting Current
NEMA A motor
NEMA sets limits of starting (locked rotor) current for NEMA design B motors. When special load torque or load inertia requirements result in special electrical designs that will yield higher locked rotor current (LRA), NEMA design A may result. This designation also cautions the selection of motor control components to avoid tripping protective devices during longer acceleration times or higher than normal starting current.
50
NEMA C motor
Starting torque of a NEMA design C motor is approximately 225%. A NEMA C, 1765 RPM, 30 HP motor will develop approximately 202.5 Lb-Ft of starting torque. Hard to start applications such as plunger pumps, heavily loaded conveyors, and compressors require this higher starting torque. Slip and full-load torque are about the same as a NEMA B motor. NEMA C applies to single speed motors from approximately 5 HP to 200 HP .
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
A
Breakdown Torque
B
202.5 Lb-Ft
51
NEMA D motor
The starting torque of a NEMA design D motor is approximately 280% of the motors full-load torque. A NEMA D, with a full-load rated speed of 1765 RPM, 30 HP motor will develop approximately 252 Lb-Ft of starting torque. Very hard to start applications, such as punch presses, cranes, hoists, and oil well pumps require this high starting torque. NEMA D motors have no true breakdown torque. After initial starting torque is reached torque decreases until full-load torque is reached. NEMA D motors typically are designed with 5 to 8% slip or 8 to 13% slip.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed Slip 8%
A
252 Lb-Ft
Multispeed and ASD (adjustable speed drive) Soft starts
These specialized motor designs are uniquely designed or selected to specific load requirements. NEMA design classifications are not applicable to these specialized motors. Various special configurations of motor controls are selected when starting/accelerating torques must be more accurately controlled, or when starting current must be limited. In the cases of part winding start or wye-delta start, the motor windings must be designed with unique connections for the special controls. In cases such as reduced voltage autotransformer or electronic soft starts, relatively standard motors may be approved for these special applications.
52
Review 6
1. 2. 3. 4.
A 30 HP motor with a 1.15 service factor can be operated at ____________ HP . A motor with Class F insulation has a maximum ____________ temperature rise. The starting torque of a NEMA B motor is approximately ____________ % of full-load torque. ____________ torque refers to point on a torque curve where a motor is overloaded beyond the motors torque capability, causing the motor to stall or abruptly slow down.
53
Derating Factors
Several factors can effect the operation and performance of an AC motor. These need to be considered when applying a motor. Voltage variation AC motors are designed to operate on standardized voltages and frequencies. The following table reflects NEMA standards.
60 Hz 115 VAC 200 VAC 230 VAC 460 VAC 575 VAC 50 Hz 380 VAC 400 VAC 415 VAC 220/380 VAC
A small variation in supply voltage can have a dramatic affect on motor performance. In the following chart, for example, when voltage is 10% below the rated voltage of the motor, the motor has 20% less starting torque. This reduced voltage may prevent the motor from getting its load started or keeping it running at rated speed. A 10% increase in supply voltage, on the other hand, increases the starting torque by 20%. This increased torque may cause damage during startup. A conveyor, for example, may lurch forward at startup. A voltage variation will cause similar changes in the motors starting amps, full-load amps, and temperature rise.
%Change In Motor Performance
Below +20 +15 +10 +5 0 -5 -10 -15 -20 -15 -10 -5 0 +5 +10 +15 % Voltage Variation 0 Above
54
Frequency
A variation in the frequency at which the motor operates causes changes primarily in speed and torque characteristics. A 5% increase in frequency, for example, causes a 5% increase in full-load speed and a 10% decrease in torque.
Frequency Variation +5% -5% % Change Full-Load Speed +5% -5% % Change Starting Torque -10% +11%
Altitude
Standard motors are designed to operate below 3300 feet. Air is thinner and heat is not dissipated as quickly above 3300 feet. Most motors must be derated for altitude. The following chart gives typical horsepower derating factors, but the derating factor should be checked for each motor. A 50 HP motor operated at 6000 feet, for example, would be derated to 47 HP providing the 40°C ambient rating is still required. ,
Altitude 3300 - 5000 5001 - 6600 6601 - 8300 8301 - 9900 9901 - 11,500 Derating Factor 0.97 0.94 0.90 0.86 0.82
50 HP X 0.94 = 47 HP
The ambient temperature may also have to be considered. The ambient temperature may be reduced from 40°C to 30°C at 6600 feet on many motors. A motor with a higher insulation class may not require derating in these conditions.
Ambient Temperature (°C) 40 30 20 Maximum Altitude (Feet) 3300 6600 9900
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AC Motors and AC Drives
Many applications require the speed of an AC motor to vary. The easiest way to vary the speed of an AC induction motor is to use an AC drive to vary the applied frequency. Operating a motor at other than the rated frequency and voltage has an effect on motor current and torque. Volts per hertz A ratio exists between voltage and frequency. This ratio is referred to as volts per hertz (V/Hz). A typical AC motor manufactured for use in the United States is rated for 460 VAC and 60 Hz. The ratio is 7.67 volts per hertz. Not every motor has a 7.67 V/Hz ratio. A 230 Volt, 60 Hz motor, for example, has a 3.8 V/Hz ratio.
460 = 7.67 V/Hz 60
230 = 3.8 V/Hz 60
Flux (F), magnetizing current (IM), and torque are all dependent on this ratio. Increasing frequency (F) without increasing voltage (E), for example, will cause a corresponding increase in speed. Flux, however, will decrease causing motor torque to decrease. It can be seen that torque (T = kFIW) is directly affected by flux (F). Torque is also affected by the current resulting from the applied load, represented here by IW. Magnetizing current (IM) will also decrease. A decrease in magnetizing current will cause a corresponding decrease in stator or line (IS) current. These decreases are all related and greatly affect the motors ability to handle a given load.
Φ≈
E F
T = kΦIW
IM =
56
E 2πFLM
Constant torque
AC motors running on an AC line operate with a constant flux (F) because voltage and frequency are constant. Motors operated with constant flux are said to have constant torque. Actual torque produced, however, is determined by the demand of the load.
T = kΦIW
An AC drive is capable of operating a motor with constant flux (F) from approximately zero (0) to the motors rated nameplate frequency (typically 60 Hz). This is the constant torque range. As long as a constant volts per hertz ratio is maintained the motor will have constant torque characteristics. AC drives change frequency to vary the speed of a motor and voltage proportionately to maintain constant flux. The following graphs illustrate the volts per hertz ratio of a 460 volt, 60 Hz motor and a 230 volt, 60 Hz motor. To operate the 460 volt motor at 50% speed with the correct ratio, the applied voltage and frequency would be 230 volts, 30 Hz. To operate the 230 volt motor at 50% speed with the correct ratio, the applied voltage and frequency would be 115 volts, 30 Hz. The voltage and frequency ratio can be maintained for any speed up to 60 Hz. This usually defines the upper limits of the constant torque range.
460 230
230
115
0 0 30 60
0 0 30 60
460 = 7.67 V/Hz 60 230 = 7.67 V/Hz 30
230 = 3.8 V/Hz 60 115 = 3.8 V/Hz 30
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Constant horsepower
Some applications require the motor to be operated above base speed. The nature of these applications requires less torque at higher speeds. Voltage, however, cannot be higher than the rated nameplate voltage. This can be illustrated using a 460 volt, 60 Hz motor. Voltage will remain at 460 volts for any speed above 60 Hz. A motor operated above its rated frequency is operating in a region known as a constant horsepower. Constant volts per hertz and torque is maintained up to 60 Hz. Above 60 Hz the volts per hertz ratio decreases, with a corresponding decrease in torque. Frequency 30 Hz 60 Hz 70 Hz 90 Hz V/Hz 7.67 7.67 6.6 5.1
Flux (F) and torque (T) decrease:
Φ≈
E F
T = kΦIW
Decreasing V/Hz
460
230
Constant Torque
Constant Horsepower
0 0 30 60 90
Horsepower remains constant as speed (N) increases and torque decreases in proportion. The following formula applies to speed in revolutions per minute (RPM).
HP (remains constant)=
T (decreases) x N (increases) 5250
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Reduced voltage and frequency starting
A NEMA B motor that is started by connecting it to the power supply at full voltage and full frequency will develop approximately 150% starting torque and 600% starting current. AC drives start at reduced voltage and frequency. The motor will start with approximately 150% torque and 150% current at reduced frequency and voltage. The torque/speed curve shifts to the right as frequency and voltage are increased. The dotted lines on the torque/speed curve illustrated below represent the portion of the curve not used by the drive. The drive starts and accelerates the motor smoothly as frequency and voltage are gradually increased to the desired speed. An AC drive, properly sized to a motor, is capable of delivering 150% torque at any speed up to speed corresponding to the incoming line voltage. The only limitations on starting torque are peak drive current and peak motor torque, whichever is less.
225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed 2 10 20 30 40 Frequency - Hz 50 60
Some applications require higher than 150% starting torque. A conveyor, for example, may require 200% rated torque for starting. If a motor is capable of 200% torque at 200% current, and the drive is capable of 200% current, then 200% motor torque is possible. Typically drives are capable of producing 150% of drive nameplate rated current for one (1) minute. If the load requires more starting torque than a drive can deliver, a drive with a higher current rating would be required. It is appropriate to supply a drive with a higher continuous horsepower rating than the motor when high peak torque is required.
59
Selecting a motor
AC drives often have more capability than the motor. Drives can run at higher frequencies than may be suitable for an application. Above 60 Hz the V/Hz ratio decreases and the motor cannot develop 100% torque. In addition, drives can run at low speeds, however, self-cooled motors may not develop enough air flow for cooling at reduced speeds and full load. Each motor must be evaluated according to its own capability before selecting it for use on an AC drive. Harmonics, voltage spikes, and voltage rise times of AC drives are not identical. Some AC drives have more sophisticated filters and other components designed to minimize undesireable heating and insulation damage to the motor. This must be considered when selecting an AC drive/ motor combination. Motor manufacturers will generally classify certain recommended motor selections based on experience, required speed range, type of load torque, and temperature limits.
Distance between drive and motor
Distance from the drive to the motor must also be taken into Consideration. All motor cables have line-to-line and line-toground capacitance. The longer the cable, the greater the capacitance. Some types of cables, such as shielded cable or cables in metal conduit, have greater capacitance. Spikes occur on the output of AC drives because of the charging current in the cable capacitance. Higher voltage (460 VAC) and higher capacitance (long cables) result in higher current spikes. Voltage spikes caused by long cable lengths can potentially shorten the life of the AC drive and motor. When considering an application where distance may be a problem, contact your local Siemens representative. A high efficiency motor with a 1.15 service factor is recommended when used on an AC drive. Due to heat associated with harmonics of an AC drive, the 1.15 service factor is reduced to 1.0.
Service factor on AC drives
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Matching AC Motors to the Load
One way to evaluate whether the torque capabilities of a motor meet the torque requirements of the load is to compare the motors speed-torque curve with the speedtorque requirements of the load.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
Motor
To find the torque characteristics a table, similar to the partial one shown below, can be used. NEMA publication MG 1 is one source of typical torque characteristics.
Load Torque as % FullLoad Drive Torque Breakaway Actuators: Screw-down (rolling mills) Positioning Agitators Liquid Slurry Blowers, centrifugal: Valve closed Valve open Blowers, positive displacement, rotary, bypassed Calenders, textile or paper 200 150 100 150 30 40 40 75 Accelerating 150 110 100 100 50 110 40 110 Peak Running 125 100 100 100 40 100 100 100
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The most accurate way to obtain torque characteristics of a given load is to obtain them from the equipment manufacturer. A simple experiment can be set up to show how the torque of a given load can be calculated. In the following illustration a pulley is fastened to the shaft of a load that a motor is to drive. A cord is wrapped around the pulley with one end connected to a spring scale. The torque can be calculated by pulling on the scale until the shaft turns and noting the reading on the scale. The force required to turn the shaft, indicated by the scale, times the radius of the pulley equals the torque value. It must be remembered that the radius is measured from the center of the shaft. If the radius of the pulley and shaft were 1 foot, for example, and the force required to turn the shaft were 10 pounds, the torque requirement is 10 Lb-Ft. The amount of torque required to turn the connected load can vary at different speeds.
Scale
Force (F)
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Centrifugal pump
When a motor accelerates a load from zero to full-load speed the amount of torque it can produce changes. At any point during acceleration and while the motor is operating at fullload speed, the amount of torque produced by the motor must always exceed the torque required by the load. In the following example a centrifugal pump has a full-load torque of 600 Lb-Ft. This is equivalent to 200 HP The centrifugal pump . only requires approximately 20% of full-load torque to start. The torque dips slightly after it is started and then increases to full-load torque as the pump comes up to speed. This is typically defined as a variable torque load.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
600 Lb-Ft
120 Lb-Ft
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A motor has to be selected that can start and accelerate the centrifugal pump. By comparing a 200 HP NEMA B motor curve to the load curve, it can be seen that the motor will easily start and accelerate the load.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
600 Lb-Ft
120 Lb-Ft
Screw down actuator
In the following example a screw down actuator is used. The starting torque of a screw down actuator is approximately 200% of full-load torque. Comparing the loads requirement with the NEMA design B motor of equivalent horsepower, it can be seen that the loads starting torque requirement is greater than the motors capability. The motor, therefore, will not start and accelerate the load.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed
Motor
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One solution would be to use a higher horsepower NEMA B motor. A less expensive solution might be to use a NEMA D motor of the same horsepower requirements as the load. A NEMA D motor would easily start and accelerate the load.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100 % Synchronous Speed Load Motor
The motor selected to drive the load must have sufficient torque to start, accelerate, and run the load. If, at any point, the motor cannot produce the required torque the motor will stall or run in an overloaded condition. This will cause the motor to generate excess heat and typically exceed current limits causing protective devices to remove the motor from the power source. If the overload condition is not corrected, or the proper motor installed, the existing motor will eventually fail.
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Review 7
1.
A motor rated for 460 VAC operating on an a supply of 437 VAC (-5%) will have a ____________ % change in motor performance. Using the altitude derating table the Derating Factors section, a 200 HP motor operated at 5500 feet would be derated to ____________ HP . The volts per hertz ratio of a 460 Volt 60 Hz motor is ____________ V/Hz. When applying an AC motor to an AC drive a motor with a ____________ service factor is recommended. If the radius of a pulley and shaft were 2 feet, and the force required to turn the shaft were 20 pounds, the amount of torque required to turn the load is ____________ Lb-Ft.
2.
3. 4. 5.
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Enclosures
Recall that the enclosure provides protection from contaminants in the environment in which the motor is operating. In addition, the type of enclosure affects the cooling of the motor. There are two categories of enclosures: open and totally enclosed. Open drip proof (ODP) Open enclosures permit cooling air to flow through the motor. The rotor has fan blades that assist in moving the air through the motor. One type of open enclosure is the drip proof enclosure. The vent openings on this type of enclosure prevent liquids and solids falling from above at angles up to 15° from vertical from entering the interior of the motor and damaging the operating components. When the motor is not in the horizontal position, such as mounted on a wall, a special cover may be necessary to protect it. This type of enclosure can be specified when the environment is free from contaminates.
Vents
Vents
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Totally enclosed non-ventilated (TENV)
In some cases air surrounding the motor contains corrosive or harmful elements which can damage the internal parts of a motor. A totally enclosed motor enclosure restricts the free exchange of air between the inside of the motor and the outside. The enclosure is not airtight, however, and a seal at the point where the shaft passes through the housing keeps out water, dust, and other foreign matter that could enter the motor along the shaft. The absence of ventilating openings means all heat dissipates through the enclosure by means of conduction. Most TENV motors are fractional horsepower. TENV motors are used, however, for larger horsepower special applications. For larger horsepower applications the frame is heavily ribbed to help dissipate heat more quickly. TENV motors can be used indoors and outdoors.
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Totally enclosed fan cooled (TEFC)
The totally enclosed fan-cooled motor is similar to the TENV except an external fan is mounted opposite the drive end of the motor. The fan provides additional cooling by blowing air over the exterior of the motor to dissipate heat more quickly. A shroud covers the fan to prevent anyone from touching it. With this arrangement no outside air enters the interior of the motor. TEFC motors can be used in dirty, moist, or mildly corrosive operating conditions. TEFC motors are more widely used for integral HP applications.
Fan
Explosion proof (XP)
The explosion proof motor enclosure is similar in appearance to the TEFC, however, most XP enclosures are cast iron. The application of motors used in hazardous locations is subject to regulations and standards set by regulatory agencies such as the National Electrical Code® and Underwriters Laboratories for XP motors used in the United States.
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Hazardous environments
Although you should never specify or suggest the type of location, it is important to understand regulations that apply to hazardous locations. It is the users responsibility to contact local regulatory agencies to define the location as Division I or II and to comply with all applicable codes. There are two divisions. Hazardous materials are normally present in the atmosphere. A division I location requires an explosion proof motor. Atmosphere may become hazardous as result of abnormal conditions. This may occur if, for example, a pipe breaks that is the conduit for a hazardous chemical. Once the location is defined as hazardous the location is further defined by the class and group of hazard. Class I, Groups A through D are chemical gases or liquids such as gasoline, acetone, and hydrogen. Class II, Groups E, F, and G include flammable dust, such as coke or grain dust. Class III is not divided into groups. It includes all ignitable fibers and lints such as clothing fiber in textile mills.
Division I Division II
Classes and groups
Class I Groups A - D Gases and Liquids Gasoline Acetone Hydrogen Ethyl
Class II Groups E - G Flammable Dust Coke Dust Grain Dust Metallic Dust
Class III Ignitible Fibers Rayon Jute
In some cases it may be necessary for the user to define the lowest possible ignition temperature of the hazardous material to future assure the motor complies with all applicable codes and requirements.
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Mounting
NEMA dimensions
NEMA has standardized frame size motor dimensions. Standardized dimensions include bolt hole size, mounting base dimensions, shaft height, shaft diameter, and shaft length. Existing motors can be replaced without reworking the mounting arrangement. New installations are easier to design because the dimensions are known. Letters are used to indicate where a dimension is taken. For example, the letter C indicates the overall length of the motor. The letter E represents the distance from the center of the shaft to the center of the mounting holes in the feet. The actual dimensions are found by referring to a table in the motor data sheet and referencing the letter to find the desired dimension.
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NEMA divides standard frame sizes into two categories: fractional and integral. Fractional frame sizes are designated 48 and 56 and include primarily horsepower ratings of less than one horsepower. Integral or medium horsepower motors are designated by frame sizes ranging from 143T to 445T. A T in the motor frame size designation of integral horsepower motors indicates the motor is built to current NEMA frame standards. Motors built prior to 1966 have a U in the motor frame size designation, indicated they are built to previous NEMA Standards.
143 T = Current NEMA Standards 326 U = Previous NEMA Standards
The frame size designation is a code to help identify key frame dimensions. The first two digits, for example, are used to determine the shaft height. The shaft height is the distance from the center of the shaft to the mounting surface. To calculate the shaft height divide the first two digits of the frame size by 4. In the following example a 143T frame size motor has a shaft height of 3½ inches (14 ÷ 4).
Shaft Height
Frame 143T 14 ÷ 4 = 3½"
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The third digit in the integral T frame size number is the NEMA code for the distance between the centerlines of the mounting bolt holes in the feet of the motor.
Frame 14 3 T Distance Between Bolt Holes
The dimension is determined by matching the third digit in the frame number with a table in NEMA publication MG-1. It can be seen that the distance between the centerlines of the mounting bolt holes in the feet of a 143T frame is 4.00 inches.
Frame Number Series 140 160 180 200 210 220 250 280 320 Third/Fourth Digit In Frame Number 3 4 1 2 4.00 4.50 5.00 5.50 5.50 6.25 7.00 8.00 9.00 4.50 5.00 5.50 6.50 6.25 6.75 8.25 9.50 10.50
D
5 4.50 5.00 5.50 6.50 6.25 6.75 8.25 9.50 10.50
4.00 4.50 5.00 5.25 5.50 6.25 7.00 8.00
3.50 4.00 4.50 4.50 5.00 5.50 6.25 7.00
4.00 4.50 5.00 5.00 5.50 6.25 7.00 8.00
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IEC dimensions
IEC also has standardized dimensions which differ from NEMA. Many motors are manufactured using IEC dimensions. IEC dimensions are shown in the following drawing.
Mounting positions
The typical floor mounting positions are illustrated in the following drawing, and are referred to as F-1 and F-2 mountings. The conduit box can be located on either side of the frame to match the mounting arrangement and position. The standard location of the conduit box is on the left-hand side of the motor when viewed from the shaft end. This is referred to as the F-1 mounting. The conduit opening can be placed on any of the four sides of the box by rotating the box in 90° steps.
Conduit Box
Shaft F-1 Position (Standard) F-2 Position
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With modification the foot-mounted motor can be mounted on the wall and ceiling. Typical wall and ceiling mounts are shown in the following illustration. Wall mounting positions have the prefix W and ceiling mounted positions have the prefix C.
Assembly W-1 Assembly W-2 Assembly W-3 Assembly W-4
Assembly W-5 Assembly W-6 Assembly W-7 Assembly W-8
Assembly C-1
Assembly C-2
Mounting faces
It is sometimes necessary to connect the motor directly to the equipment it drives. In the following example a motor is connected directly to a gear box.
Gear Box
Motor
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C-face
The face, or the end, of a C-face motor has threaded bolt holes. Bolts to mount the motor pass through mating holes in the equipment and into the face of the motor.
D-flange
The bolts go through the holes in the flange of a D-flange motor and into threaded mating holes of the equipment.
G N R A IN
Through Bolt Holes
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Review 8
1.
A type of open enclosure that prevents liquids and solids falling from above at angles up to 15° from vertical from entering the interior of the motor is an ____________ ____________ ____________ . A type of enclosure that is closed and uses a fan mounted on the shaft to supply cooling is referred as ____________ ____________ ____________ ____________ . Gasoline is defined as a Class ____________ hazard. The NEMA dimension from the center of the shaft to the mounting surface is designated by the letter ____________ . The letter ____________ in the motor frame size designation indicates a motor is built to current NEMA standards. The shaft height can be determined by dividing the first two digits of an integral frame designation by ____________ . A motor intended to be mounted on the wall with the conduit box facing up and the shaft facing left is an Assembly ____________ . A ____________ motor has threaded bolt holes to mount a motor to another piece of equipment.
2.
3. 4.
5.
6.
7.
8.
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Siemens Motors
Siemens manufactures a wide range of AC motors. The following motors provide only an introduction to these motors. Contact your local Siemens representative for more information on any of the motors discussed or other Siemens AC motors. Medallion motors Medallion motors represent the newer family of Siemens enclosed motors. Standard efficiency Medallion motors are available from 1 to 300 HP When used on a 230/460 VAC, 3. phase, 60 Hz power supply standard efficiency Medallion motors are wound for either 900, 1200, 1800, or 3600 RPM. Premium efficiency Medallion motors are available from 1 to 400 horsepower at 460 volts, 3-phase, 60 Hz. Premium efficiency Medallion motors are wound for 900, 1200, 1800, or 3600 RPM. Medallion motors are also available in standard and premium efficiency from 1 to 200 HP for use on 575 volt systems. Other voltages and frequency designs are also available. Contact your Siemens representative for information and lead times.
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Medallion motors are available with longer shafts for belt driven applications and vertical mounting for applications such as pumps. Various protective devices, such as thermocouples and thermistors, can be installed as an option. These devices are wired into the motor controller and shut down the motor if temperature becomes excessive. Space heaters can also be used to keep the temperature of the motor above the dew point in areas that are damp or humid. Space heaters are turned off when the motor is running and on when the motor is stopped. Medallion motors are also available in two-speed configurations. A two-speed motor often has two sets of windings, each wound with a different number of poles. The windings are brought out to an external controller. The motor can be run at either speed. Typical speed selections are 900 or 1200 RPM at low speed, and 1800 RPM at high speed.
L1 1 L2 L3 2 12 T3 13 T1 11 T11
Low Speed: Close 1, 2, 3 Open 11, 12,13 High Speed: Close 11, 12, 13 Open 1, 2, 3
3
T2
T12
Speed =
T13
120 x Frequency - Slip Number of Poles
PE-21 Plus motors
PE-21 Plus mottors are premium efficiency motors available from 1 to 500 HP Premium efficiency motors typically cost . slightly more than standard efficiency motors, but payback is in energy savings.
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The following example is used to show energy savings available over the life of a premium efficiency motor. A 25 HP standard motor with an efficiency of 86.5% costing \$590, and a PE-21 Plus with an efficiency of 93% costing \$768 are compared. Using a process life of 60,000 hours at \$.08 a kilowatt hour, the PE-21 Plus will save \$7,055 in total operating cost.
CL = PI +
.746 x HP x TO x RU E 100
Example: 25 HP 1800 RPM, TEFC , Std. F/L Eff. Std. Motor PI 86.5% \$590
CL = Lifetime motor operating cost PI = Initial price of motor HP = Motor horsepower TO = Lifetime hours of operation RU = Local utility rates - (\$/KWH) E = Motor nameplate efficiency
PE-21 Plus Eff. 93.0% PE-21 Plus PI \$768 Std. Motor .746 x 25 x 60,000 x .08 86.5 100
CL = 590 +
CL = \$104,081 PE-21 Plus Motor .746 x 25 x 60,000 x .08 CL = 768 + 93.0 100 CL = \$97,026
Savings = \$7,055
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Vertical pump motors
Vertical hollow shaft pump motors are designed for vertical pump applications. The motors are squirrel cage induction type with NEMA design B torque and current characteristics. Motors are rated from 25 to 250 HP and 1800 RPM. Vertical pump motors are designed for 460 volt, 3-phase, 60 Hz systems. Thermostats and space heaters are optional. These motors have NEMA standard P flange mounting shaft with a hollow shaft which accomodates the driven shaft to extend through the rotor. The coupling for connecting the motor shaft to the driven shaft is located in the top of the motor.
In addition to hollow shaft, more conventional solid shaft motors are supplied where the motor shaft is coupled to the driven shaft below the P flange face. Vertical solid shaft motors designed for in-line pump applications are available from 3 - 100 HP at 3600 RPM, and 3 - 250 HP at 1200 and 1800 RPM.
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IEC motors
IEC motors are manufactured to meet specifications of the International Electrotechnical Commission, IEC 34. Standard voltages at 50 Hz are 220, 400, 500, or 660 volts. Standard voltage at 60 Hz is 460 volts. Siemens IEC motors are available with 2, 4, 6, and 8 poles. IEC motors are also available for multispeed applications. Siemens IEC motors are available from 600 W (1.24 HP) to 630 KW (840 HP).
While mounting flange dimensions, shaft height, shaft extensions, and other performance standards clearly differ with comparable NEMA motors, a closer comparison will show remarkably similar characteristics. Perhaps the greatest obstacle to working with IEC motors is familiarity with unique terminology and the ability to correlate with more familiar NEMA standards.
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Above NEMA Motors
Motors that are larger than the NEMA frame sizes are referred to as above NEMA motors. These motors typically range in size from 200 to 10,000 HP Some above NEMA motors . manufactured by Siemens may also be considered Medallion motors. There are no standardized frame sizes or dimensions because above NEMA motors are typically constructed to meet the specific requirements of an application. Siemens offers large motors in seven basic frame sizes: 30, 500, 580, 680, 708, 800, and 1120 frames. For each frame size Siemens has standard frame dimensions similar to NEMA dimensions. For specific application information contact your local Siemens representative.
Yoke Width AB AC N XD
C
W
N-W
Keyway Length XW HT
XL AF X AA Pipetap D
E 2E
2F B
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The customer typically supplies specifications for starting torque, breakdown torque, and full-load torque based on speed-torque curves obtained from the driven equipment manufacturer. There are, however, some minimum torques that all large AC motors must be able to develop. These are specified by NEMA. Locked Rotor Torque ³ 60% of Full-Load Torque Pull-Up Torque ³ 60% of Full-Load Torque Maximum Torque ³ 175% of Full-Load Torque Above NEMA motors require the same adjustment for altitude and ambient temperature as integral frame size motors. When the motor is operated above 3300 feet a higher class insulation should be used or the motor should be derated. Above NEMA motors with class B insulation can easily be modified for operation in an ambient temperature between 40° C and 50° C. Above 50° C requires special modification at the factory. Enclosures Environmental factors also affect large AC motors. Enclosures used on above NEMA motors look differently than those on integral frame size motors. The open drip proof enclosure provides the same amount of protection as the integral frame size open motor. This provides the least amount of protection for the motors electrical components. It is typically used in environments free of contaminants.
Open drip proof (ODP)
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Horizontal drip proof weather protected I (Type CG)
The weather protected I enclosure is an open enclosure that has ventilating passages designed to minimize the entrance of rain, snow, and airborne particles that could come into contact with the electrical and rotating parts of the motor. All air inlets and exhaust vents are covered with screens. It is used on indoor applications when the environment has a small amount of moisture in the air. This enclosure is available to 10,000 horsepower.
Weather protected II (Type CGII)
Weather protected II enclosures are open enclosures with vents constructed so that high velocity air and airborne particles blown into the motor can be discharged without entering the internal ventilating passages leading to the electrical parts of the motor. The intake and discharge vents must have at least three 90° turns and the air velocity must be less than 600 feet per minute. It is used outdoors when the motor is not protected by other structures. This enclosure is available through 10,000 horsepower.
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Totally enclosed fan cooled (Type CGZ)
The totally enclosed fan cooled motor functions the same as the TEFC enclosure used on integral frame size motors. It is designed for indoor and outdoor applications where internal parts must be protected from adverse ambient conditions. Type CGZ utilizes cooling fins on all around the yoke and housing. Available up to 900 HP on 580 frames and 2250 HP on 708-880 frames.
Totally enclosed air to air cooled (Type TEAAC)
Type TEAAC totally enclosed motor utilizes air to tube type heat exchangers for cooling. Available through 4500 HP .
Totally enclosed water-toair cooled (Type RGG)
There comes a point when the motor frame cannot adequately dissipate heat, even with the help of a fan. This enclosure is designed to cool the motor by means of an water-to-air heat exchanger. This type of enclosure requires a steady supply of water. Available through 10,000 HP .
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Totally enclosed fan cooled explosion proof (Type AZZ)
Large AC motors are also used in hazardous environments. This enclosure meets or exceeds all applicable UL requirements for hazardous (Division 1) environmental operation. Available through 1750 HP .
Review 9
1. 2. 3.
Standard efficiency Medallion motors are available from 1 to ____________ horsepower. Premium efficiency Medallion motors are available from 1 to ____________ horsepower. An advantage of a premium efficiency motor over a standard efficiency motor is a savings in ____________ over the operating life of the motor. Vertical pump motors have NEMA design ____________ torque and current characteristics. IEC motors are built to IEC ____________ standards. Above NEMA motors are available from ____________ to ____________ HP . Locked rotor torque of an above NEMA motor is ³ ____________ % of full-load torque.
4. 5. 6. 7.
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Review 1 Review 2 Review 3 Review 4 Review 5 Review 6 Review 7 Review 8 Review 9
1) force; 2) 15; 3) torque; 4) 80; 5) inertia; 6) Speed; 7) revolutions per minute; 8) Acceleration 1) Work; 2) Power; 3) voltage; 4) true; 5) watts, volt amps; 6) 25 1) A. enclosure, B. stator, C. rotor; 2) stator and rotor; 3) stator; 4) rotor; 5) squirrel cage 1) north, south; 2) A. attract, B. repel, C. repel D. attract; 3) magnetic field; 4) D 1) 2, north; 2) synchronous; 3) 1800; 4) slip; 5) 4.2 1) 34.5; 2) 105 °C; 3) 150; 4) Breakdown 1) -10; 2) 188; 3) 7.67; 4) 1.15; 5) 40 1) open drip proof; 2) totally enclosed fan cooled; 3) I; 4) D; 5) T; 6) 4; 7) W-2; 8) C-face 1) 300; 2) 400; 3) energy; 4) B; 5) 34; 6) 200 to 10,000; 7) 60
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Final Exam
The final exam is intended to be a learning tool. The book may be used during the exam. A tear-out answer sheet is provided. After completing the test, mail the answer sheet in for grading. A grade of 70% or better is passing. Upon successful completion of the test a certificate will be issued. 1. ____________ is a twisting or turning force that causes an object to rotate. a. Torque b. Friction 2. c. Inertia d. Acceleration
If 50 pounds of force were applied to a lever 3 feet long, the torque would be ____________ Lb-Ft. a. 16.7 b. 53 c. 47 d. 150
3.
The rate of doing work is called ____________ . a. inertia b. speed c. power d. energy
4.
A motor with a rating of 60 KW would have an equivalent rating of ____________ HP . a. 45 b. 80 c. 65 d. 120
5.
The three basic parts of an AC motor are the ____________ . a. b. c. d. rotor, stator, and enclosure shaft, housing, and connection box cooling fan, rotor, and stator end brackets, bearings and cooling fan
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6.
A four-pole motor operating at 50 Hz has a synchronous speed of ____________ RPM. a. 1500 b. 3000 c. 1800 d. 3600
7.
A motor with a synchronous speed of 900 RPM and a rotor speed of 850 RPM has ____________ % slip. a. 3 b. 9.4 c. 5.5 d. 20
8.
____________ is an indication of how much electrical energy is converted into mechanical energy. a. Service factor b. Efficiency c. Temperature rise d. RPM
9.
____________ torque is also referred to as starting torque. a. Pull up b. Accelerating c. Breakdown d. Locked rotor
10. A NEMA B motor that is started by connecting it to the power supply at rated voltage and frequency has a typical starting current of ____________ %. a. 100 b. 150 c. 200 d. 600
11. The temperature rise of a motor with Class F insulation is ____________ ° C with a 10° C hot spot. a. 60 b. 105 c. 80 d. 125
12. The volts per hertz ratio of a 460 VAC, 60 Hz motor is ____________ V/Hz. a. 3.8 b. 7.67 c. 5.1 d. 9.2
13. A motor operated within a speed range that allows a constant volts per hertz ratio is said to be ____________ . a. constant hp b. variable torque 90 c. constant torque d. variable flux
14. A +5% variation in frequency can have a __________ % change in starting torque. a. +5% b. -5% c. -10% d. +10%
15. The following graph represents a NEMA ____________ motor.
300 275 250 225 200 175 150 125 100 75 50 25 0 0 10 20 30 40 50 60 70 80 90 100
a. A b. B
c. C d. D
16. A ____________ motor enclosure uses vent openings to prevent liquids and solids falling from above at angles from up to 15° from vertical from entering the interior of the motor. a. TENV b. XP c. TEFC d. ODP
17. Grain dust is in a hazardous location Class __________ . a. II b. III c. A d. C
18. The letter ____________ in the motor frame size designation of an integral horsepower motor indicates the motor is built to current NEMA standards. a. C b. U c. T d. N
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19. The shaft height of a 449 frame integral horsepower motor is ____________ inches. a. 3½ b. 9 c. 4.4 d. 11
20. PE-21 Plus motors are available from 1 to ____________ HP . a. 500 b. 1200 c. 800 d. 2500
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https://buildingspeed.org/blog/2014/06/24/knowing-pit-road-speed-in-the-rain/ | 1,579,861,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00007.warc.gz | 359,407,085 | 21,421 | # Knowing Pit Road Speed in the Rain
This weekend, we learned that the real weather challenge for the NASCAR Nationwide Series isn’t rain. It’s not enough rain. It wasn’t raining hard enough to put on rain tires, but it wasn’t quite dry enough to safely race on slicks. (I’ve written before about why racing in the rain is hard.) But they managed to pull it off, put on a great show and @Brendan62 finally got that long-sought-after win.
When the teams switched to rain tires, the crew chiefs had to remind the drivers that their tach readings for maintaining pit road speed would be different . I’ve written quite a bit about tach readings and speed limits:
If you know the gear ratios in the transmission and the rear end, you can convert the engine speed (in revolutions per minute) to the rotation rate at the wheels. How far the car moves each revolution of the axle depends on the distance around: the bigger the tire, the further the car moves. This is why your odometer and speedometer get screwed up if you don’t use the right size tire on your car. You can check this out yourself using two tumblers with different sizes. If you lie them on their sides and start at the same place, they roll them each one revolution, the larger tumbler will have gone further.
Rain tires have thicker tread and a larger circumference. The car moves further along with each rotation of the tires. If you don’t compensate for this in your tach readings for pit road speed, you’ll end up with a speeding penalty, even though you would have been fine at the same tach readings with the slicks on.
I couldn’t find the difference in the circumferences, but remember that teams are always pushing to get as close to the pit road speed limit as possible, so even a relatively small change in the relationship between speed and circumference can put you at the tail end of the longest line.
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# If A = [(5, -1, 4), (2, 3, 5), (5, -2, 6)] , find A^(-1) and use it to solve the following system of equation 5x - y + 4z = 5, 2x + 3y +5z = 2, 5x - 2y + 6z =-1
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# Solutions to Problems
P13-1. LG 2: CCC
Basic
a. OC = Average age of inventories
+ Average collection period
= 90 days + 60 days
= 150 days
b. CCC = Operating cycle Average payment period
= 150 days 30 days
= 120 days
c. Resources needed = (total annual outlays 365 days) CCC
= [\$30,000,000 365] 120
= \$9,863,013.70
d. Shortening either the AAI or the ACP, lengthening the APP, or a combination of these can
reduce the CCC.
Chapter 13 Working Capital and Current Assets Management 257
## P13-2. LG 2: Changing CCC
Intermediate
a. AAI = 365 days 8 times inventory = 46 days
OC = AAl + ACP
= 46 days + 60 days
= 106 days
CCC = OC APP
= 106 days 35 days = 71 days
b. Daily cash operating expenditure = total outlays 365 days
= \$3,500,000 365
= \$9,589
Resources needed = daily expenditure CCC
= \$9,589 71
= \$680,819
c. Additional profit = (daily expenditure reduction in CC)
financing rate
= (\$9,589 20) 0.14
= \$26,849
P13-3. LG 2: Multiple changes in CCC
Intermediate
a. AAI = 365 6 times inventory = 61 days
OC = AAI + ACP
= 61 days + 45 days
= 106 days
CCC = OC APP
= 106 days 30 days
= 76 days
Daily financing = \$3,000,000 365
= \$8,219
Resources needed = Daily financing CCC
= \$8,219 76
= \$624,644
b. OC = 56 days + 35 days
= 91 days
CCC = 91 days 40 days
= 51 days
Resources needed = \$8,219 51
= \$419,169
258 Gitman Principles of Managerial Finance, Brief Fifth Edition
## c. Additional profit = (daily expenditure reduction in CCC)
financing rate
= (\$8,219 26) 0.13
= \$27,780
d. Reject the proposed techniques because costs (\$35,000) exceed savings (\$27,780).
## P13-4. LG 2: Aggressive versus conservative seasonal funding strategy
Intermediate
a.
Total Funds Permanent Seasonal
Month Requirements Requirements Requirements
January \$ 2,000,000 \$2,000,000 \$ 0
February 2,000,000 2,000,000 0
March 2,000,000 2,000,000 0
April 4,000,000 2,000,000 2,000,000
May 6,000,000 2,000,000 4,000,000
June 9,000,000 2,000,000 7,000,000
July 12,000,000 2,000,000 10,000,000
August 14,000,000 2,000,000 12,000,000
September 9,000,000 2,000,000 7,000,000
October 5,000,000 2,000,000 3,000,000
November 4,000,000 2,000,000 2,000,000
December 3,000,000 2,000,000 1,000,000
## Average permanent requirement = \$2,000,000
Average seasonal requirement = \$48,000,000 12
= \$4,000,000
b. (1) Under an aggressive strategy, the firm would borrow from \$1,000,000 to \$12,000,000
according to the seasonal requirement schedule shown in part a at the prevailing short-
term rate. The firm would borrow \$2,000,000, or the permanent portion of its
requirements, at the prevailing long-term rate.
(2) Under a conservative strategy, the firm would borrow at the peak need level of
\$14,000,000 at the prevailing long-term rate.
c. Aggressive = (\$2,000,000 0.17) + (\$4,000,000 0.12)
= \$340,000 + \$480,000
= \$820,000
Conservative = (\$14,000,000 0.17)
= \$2,380,000
d. In this case, the large difference in financing costs makes the aggressive strategy more
attractive. Possibly the higher returns warrant higher risks. In general, since the conservative
strategy requires the firm to pay interest on unneeded funds, its cost is higher. Thus, the
aggressive strategy is more profitable but also more risky.
Chapter 13 Working Capital and Current Assets Management 259
## P13-5. LG 3: EOQ analysis
Intermediate
(2 S O) (2 1,200,000 \$25)
a. (1) EOQ = = = 10,541
C \$0.54
(2 1,200,000 0)
(2) EOQ = =0
\$0.54
(2 1,200,000 \$25)
(3) EOQ = =
\$0.00
EOQ approaches infinity. This suggests the firm should carry the large inventory to minimize
ordering costs.
b. The EOQ model is most useful when both carrying costs and ordering costs are present. As
shown in Part (a), when either of these costs are absent the solution to the model is not
realistic. With zero ordering costs the firm is shown to never place an order. (Assuming the
minimum order size is one, Tiger Corporation would place 2.3 orders per minute.) When
carrying costs are zero the firm would order only when inventory is zero and order as much
as possible (infinity).
## P13-6. LG 3: EOQ, reorder point, and safety stock
Intermediate
(2 S O ) (2 800 \$50)
a. EOQ = = = 200 units
C 2
200 units 800 units 10 days
b. Average level of inventory = +
2 365
= 121.92 units
(800 units 10 days) (800 units 5 days)
c. Reorder point = +
365 days 365 days
= 32.88 units
d. Change Do Not Change
(2) carrying costs (1) ordering costs
(3) total inventory cost (5) EOQ
(4) reorder point
260 Gitman Principles of Managerial Finance, Brief Fifth Edition
## P13-7. LG 3: Marginal costs
Challenge
Jimmy Johnson
Marginal Cost Analysis
Purchase of V-8 SUV vs. V-6 SUV
V-6 V-8
MSRP \$30,260 44,320
Engine (liters) 3.7 5.7
Ownership period in years 5 5
Depreciation over 5 years 17,337 25,531
Financing over 5 years.* 5,171 7,573
Insurance over 5 years 7,546 8,081
Taxes and fees over 5 years 2,179 2,937
Maintenance/repairs over 5 years 5,600 5,600
a
Total true cost for each vehicle over the 5-year period \$37,833 \$49,722
Average miles per gallon 19 14
Miles driven per year 15,000 15,000
Cost per gallon of gasoline over the 5-year ownership period 3.15 3.15
b
Total fuel cost for each vehicle over 5-year ownership period \$12,434 \$16,875
If Jimmy decides to buy the V-8, he will have to pay Marginal cost \$11,889
c
\$11,889 more than the cost of the smaller V-6 SUV Marginal fuel cost 4,441
over the 5 year period. Additionally, Jimmy will spend d
Total marginal costs \$16,330
\$4,441 more on fuel for the V-8 SUV. The total
marginal costs over the 5-year period, associated with
purchasing the V-8 over the V-6, are \$16,330.
## *Accumulated Finance Charges V-6 V-8
Cost of SUV \$30,260.00 \$ 44,320
Assumed annual discount rate 5.50% 5.5%
Term of the loan (years) 5 5
PV inters factor of the annuity (PVIFA) 4.2703 4.2703
Annual payback to be made over five years 7,086.2 \$10,378.7
Total interest and principals paid back over 5 years 35,431 \$ 51,893
Less: Cost of the SUV 30,280 \$ 44,320
Accumulated finance charges
over the entire 5 year period \$ 5,171 \$ 7,573
e. The true marginal cost of \$16,330 is greater than the simple difference between the costs of
the two vehicles.
Chapter 13 Working Capital and Current Assets Management 261
## P13-8. LG 4: Accounts receivable changes without bad debts
Intermediate
a. Current units = \$360,000,000 \$60 = 6,000,000 units
Increase = 6,000,000 20% = 1,200,000 new units
Additional profit contribution = (\$60 \$55) 1,200,000 units
= \$6,000,000
total variable cost of annual sales
b. Average investment in accounts receivable =
turnover of A/R
365
Turnover, present plan = = 6.08
60
365 365
Turnover, proposed plan = = = 5.07
(60 1.2) 72
Marginal investment in AR:
Average investment, proposed plan:
(7,200,000 units* \$55)
= \$78,106,509
5.07
Average investment, present plan:
(6,000,000 units \$55)
= 54,276,316
6.08
Marginal investment in AR = \$23,830,193
*
Total units, proposed plan = existing sales of 6,000,000 units + 1,200,000 additional units.
c. Cost of marginal investment in accounts receivable:
Marginal investment in AR \$23,830,193
Required return 0.14
Cost of marginal investment in AR \$ 3,336,227
d. The additional profitability of \$6,000,000 exceeds the additional costs of \$3,336,227.
However, one would need estimates of bad debt expenses, clerical costs, and some
information about the uncertainty of the sales forecast prior to adoption of the policy.
## P13-9. LG 4: Accounts receivable changes and bad debts
Challenge
a. Bad debts
Proposed plan (60,000 \$20 0.04) \$48,000
Present plan (50,000 \$20 0.02) 20,000
b. Cost of marginal bad debts \$28,000
c. No, since the cost of marginal bad debts exceeds the savings of \$3,500.
d. Additional profit contribution from sales:
10,000 additional units (\$20 \$15) \$50,000
Cost of marginal bad debts (from Part (b)) (28,000)
Savings 3,500
Net benefit from implementing proposed plan \$25,500
262 Gitman Principles of Managerial Finance, Brief Fifth Edition
This policy change is recommended because the increase in sales and the savings of \$3,500 exceed
the increased bad debt expense.
e. When the additional sales are ignored, the proposed policy is rejected. However, when all the
benefits are included, the profitability from new sales and savings outweigh the increased
cost of bad debts. Therefore, the policy is recommended.
## P13-10. LG 4: Relaxation of credit standards
Challenge
Additional profit contribution from sales = 1,000 additional units (\$40 \$31) \$9,000
Cost of marginal investment in AR:
11,000 units \$31
Average investment, proposed plan = \$56,055
365
60
10,000 units \$31
Average investment, present plan = 38,219
365
45
Marginal investment in AR \$17,836
Required return on investment 0.25
Cost of marginal investment in AR (4,459)
Cost of marginal bad debts:
Bad debts, proposed plan (0.03 \$40 11,000 units) \$13,200
Bad debts, present plan (0.01 \$40 10,000 units) 4,000
Cost of marginal bad debts (9,200)
Net loss from implementing proposed plan (\$4,659)
The credit standards should not be relaxed since the proposed plan results in a loss.
## P13-11. LG 5: Initiating a cash discount
Challenge
Additional profit contribution from sales = 2,000 additional units (\$45 \$36) \$18,000
Cost of marginal investment in AR:
42,000 units \$36
Average investment, proposed plan = \$124,274
365
30
40,000 units \$36
Average investment, present plan = 236,713
365
60
Reduced investment in AR \$112,439
Required return on investment 0.25
Cost of marginal investment in AR 28,110
Cost of cash discount = (0.02 0.70 \$45 42,000 units) (26,460)
Net profit from implementing proposed plan \$ 19,650
Since the net effect would be a gain of \$19,650, the project should be accepted.
Chapter 13 Working Capital and Current Assets Management 263
## P13-12. LG 5: Shortening the credit period
Challenge
Reduction in profit contribution from sales = 2,000 units (\$56 \$45) (\$22,000)
Cost of marginal investment in AR:
10,000 units \$45
Average investment, proposed plan = \$44,384
365
36
12,000 units \$45
Average investment, present plan = 66,576
365
45
Marginal investment in AR \$22,192
Required return on investment 0.25
Benefit from reduced
Marginal investment in AR \$ 5,548
Cost of marginal bad debts:
Bad debts, proposed plan (0.01 \$56 10,000 units) \$ 5,600
Bad debts, present plan (0.015 \$56 12,000 units) 10,080
Reduction in bad debts 4,480
Net loss from implementing proposed plan (\$11,972)
This proposal is not recommended.
## P13-13. LG 5: Lengthening the credit period
Challenge
Preliminary calculations:
(\$450,000 \$345,000)
Contribution margin = = 0.23333
\$450,000
Variable cost percentage = 1 contribution margin
= 1 0.233
= 0.767
a. Additional profit contribution from sales:
(\$510,000 \$450,000) 0.23333 contribution margin \$14,000
b. Cost of marginal investment in AR:
\$510,000 0.767
Average investment, proposed plan = \$64,302
365
60
\$450,000 0.767
Average investment, present plan = 28,368
365
30
Marginal investment in AR (\$35,934)
Required return on investment 0.20
Cost of marginal investment in AR (\$ 7,187)
264 Gitman Principles of Managerial Finance, Brief Fifth Edition
## c. Cost of marginal bad debts:
Bad debts, proposed plan (0.015 \$510,000) \$7,650
Bad debts, present plan (0.01 \$450,000) 4,500
Cost of marginal bad debts (3,150)
d. Net benefit from implementing proposed plan \$3,663
The net benefit of lengthening the credit period is a surplus of \$3,663; therefore the proposal
is recommended.
P13-14. LG 6: Float
Basic
a. Collection float = 2.5 + 1.5 + 3.0 = 7 days
b. Opportunity cost = \$65,000 3.0 0.11 = \$21,450
The firm should accept the proposal because the savings (\$21,450) exceed
the costs (\$16,500).
## P13-15. LG 6: Lockbox system
Basic
a. Cash made available = \$3,240,000 365
= (\$8,877/day) 3 days = \$26,631
b. Net benefit = \$26,631 0.15 = \$3,995
The \$9,000 cost exceeds \$3,995 benefit; therefore, the firm should not accept the lockbox
system.
## P13-16. LG 6: Zero-balance account
Basic
Current average balance in disbursement account \$420,000
Opportunity cost (12%) 0.12
Current opportunity cost \$ 50,400
Zero-balance account
Compensating balance \$300,000
Opportunity cost (12%) 0.12
Opportunity cost \$ 36,000
+ Monthly fee (\$1,000 12) 12,000
Total cost \$ 48,000
The opportunity cost of the zero-balance account proposal (\$48,000) is less than the current
account opportunity cost (\$50,400). Therefore, accept the zero-balance proposal.
Chapter 13 Working Capital and Current Assets Management 265
## P13-17. LG 6: Management of cash balance
a. Alexis should transfer her current savings account balances into a liquid
marketable security
Current savings balance \$15,000
Yield on marketable security @ 4.75% \$712.50
Interest on savings account balance @ 2.0% (\$300.00)
b.
Increase in annual interest earnings \$412.50
c. Alexis should transfer monthly the \$500 from her checking account to the liquid
marketable security
Monthly transfer \$500.00
Yield on marketable security @ 4.75% \$ 23.75
Interest on savings balance @ 2.00% (\$ 10.00)
Increase in annual earnings on monthly transfers \$ 13.75
d. Rather than paying bills so quickly, Alexis should pay bills on their
due date
Average monthly bills \$ 2,000
Total annual bills (\$2,000 12) \$24,000
Daily purchases (24,000 365 days) \$ 65.75
Additional funds invested (\$65.75 9) \$591.78
Marketable security yield 4.75%
Annual savings from slowing down payments (\$591.78 0.0475) \$ 28.11
Summary
Increase from investing current balances \$412.50
Increase from investing monthly surpluses 13.75
Savings from slowing down payments 28.11
Increase in Alexis annual earnings \$454.36
## P13-18. Ethics problem
Intermediate
Management should point out that what it is doing shows integrity, as it is honest, just, and fair.
The ethics reasoning portrayed in the ethics focus box could be used.
Case
Assessing Roche Publishing Companys Cash Management Efficiency
Chapter 13s case involves the evaluation of a furniture manufacturers cash management by its treasurer.
The student must calculate the OC, CCC, and resources needed and compare them to industry standards.
The cost of the firms current operating inefficiencies is determined and the case also looks at the decision
to relax its credit standards. Finally, all the variables are consolidated and a recommendation made.
266 Gitman Principles of Managerial Finance, Brief Fifth Edition
1. Roche Publishing:
OC = AAI
+ ACP
= 120 days + 60 days
= 180 days
CCC = OC APP
= 180 days 25 days
= 155 days
Total annual outlays
Resources needed = CCC
365 days
\$12,000,000
= 155 = \$5,095,890
365
2. Industry
Industry OC = 85 days + 42 days
= 127 days
Industry CCC = 127 days 40 days
= 87 days
\$12,000,000
Industry resources needed = 87 = \$2,860,274
365
## 3. Roche Publishing resources needed \$5,095,890
Less: Industry resources needed 2,860,274
\$2,235,616
Cost of inefficiency: \$2,235,616 0.12 = \$268,274
4. To determine the net profit or loss from the change in credit standards we must evaluate the three
factors that are impacted:
a. Changes in sales volume
b. Investment in accounts receivable
c. Bad-debt expenses
Changes in sales volume:
Total contribution margin of annual sales:
Under present plan = (\$13,750,000 0.20) = \$2,750,000
Under proposed plan = (\$15,000,000 0.20) = \$3,000,000
Increase in contribution margin = \$250,000 (\$3,000,000 \$2,750,000).
Investment in accounts receivable:
Turnover of accounts receivable:
365 365
Under present plan = = = 6.08
ACP 60
365 365
Under proposed plan = = = 8.69
ACP 42
Chapter 13 Working Capital and Current Assets Management 267
## Under present plan =
( \$13,750,000 0.80 ) = \$11,000,000 = \$1,809,211
6.08 6.08
## Under proposed plan =
( \$15,000,000 0.80 ) = \$12,000,000 = \$1,380,898
8.69 8.69
Cost of marginal investment in accounts receivable:
Average investment under proposed plan \$1,380,898
Average investment under present plan 1,809,211
Marginal investment in accounts receivable 428,313
Required return on investment 0.12
Cost of marginal investment in AR \$51,398
Cost of marginal bad debts:
Bad debt would remain unchanged as specified in the case.
Net profits from implementation of new plan:
Additional profit contribution from sales:
(\$1,250,000 0.20) 250,000
Cost of marginal investment in AR:
Average investment under proposed plan 1,380,898
Average investment under present plan 1,809,211
Marginal investment in AR 428,313
Cost of marginal investment in AR
(0.12 428,313) 51,398
\$198,602
5. Total cost incurred
Cost of 40-day payment period \$ 53,000
Cost of 85-day inventory period 150,000
Incremental cost of accounts receivable 51,398 \$254,398
Savings from plan implementation
Payment period extensiona \$ 59,178
Inventory period reductionb 138,082
Additional profit contribution on salesc 250,000 447,260
Annual savings \$192,862
a
(40 25) \$12,000,000/365) 0.12 = \$59,178
b
(120-85) \$12,000,000/365) 0.12 = \$138,082
c
\$1,250,000 0.20 = \$250,000
6. Roche Publishing should incur the cost to correct its cash management inefficiencies and should also
soften the credit standards to increase net, before-tax cash flow by \$192,862.
268 Gitman Principles of Managerial Finance, Brief Fifth Edition
Spreadsheet Exercise
The answer to Chapter 13s Eboy Corporation accounts receivable management spreadsheet problem is
located in the Instructors Resource Center at www.prenhall.com/irc.
## A Note on Web Exercises
A series of chapter-relevant assignments requiring Internet access can be found at the books Companion
Website at http://www.prenhall.com/gitman. In the course of completing the assignments students access
information about a firm, its industry, and the macro economy, and conduct analyses consistent with those
found in each respective chapter. | 5,342 | 17,818 | {"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} | 3.15625 | 3 | CC-MAIN-2019-35 | latest | en | 0.785255 |
http://www.jiskha.com/display.cgi?id=1334436827 | 1,498,400,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320532.88/warc/CC-MAIN-20170625134002-20170625154002-00273.warc.gz | 551,220,177 | 4,219 | # PSY315
posted by .
Based on the information given for each of the following studies, decide whether to reject the null hypothesis. For each, give (a) the Z-score cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should be rejected,(b) the Z score on the comparison distribution for the sample score, and (c) your conclusion. Assume that all populations are normally distributed.
Population
Study ì ó Sample Score p Tails of Test
A
100.0
10.0
80
.05
1 (low predicted)
B
100.0
20.0
80
.01
2
C
74.3
11.8
80
.01
2
D
16.9
1.2
80
.05
1 (low predicted)
E
88.1
12.7
80
.05
2
18. A researcher predicts that listening to music while solving math problems will make a particular brain area more active. To test this, a research participant has her brain scanned while listening to music and solving math problems,and the brain area of interest has a percentage signal change of 58. From many previous studies with this same math problems procedure (but not listening to music), it is known that the signal change in this brain area is normally distributed with a mean of 35 and a standard deviation of 10. (a) Using the .01 level, what should the researcher conclude? Solve this problem explicitly using all five steps of hypothesis testing, and illustrate your answer with a sketch showing the comparison distribution, the cutoff (or cutoffs),and the score of the sample on this distribution. (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores). | 381 | 1,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-26 | latest | en | 0.912056 |
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