Elipse mathematics?

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Yoshida
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Elipse mathematics?

Post by Yoshida »

Does anyone know how to calculate the widest part of an elipse? I have some formulae for calculating the circumferance and know how to draw one with some string tied to two points. but the radii of the elipse is on either ends of the figure and not in the middle. How do I find it out?
..l.,
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JT
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Post by JT »

Um... the standard forumla I have for an ellipse basically gives the information to you:

(x**2)/a + (y**2)/b = 1

The max width is 2*sqrt(a), and the max height is 2*sqrt(b).


What formula/description of the ellipse do you have?
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Halberds
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Post by Halberds »

If this is for a head band the standard ellipse is not adequate.
Humans heads are not shaped like that.

If this is for your geometry class then the textbook formulas are correct, read no further.
____________________________________________________________________

If it is for a helm head size the front to back and the side to side are the most important dimensions, next to the circumference. Get that right and shape it like a human head and the helm should fit. Remember that often more padding is on the forehead portion.
This allows the nose and faceplate to have better clearance.

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Yoshida
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Post by Yoshida »

I have area=pi*x*y
and perimeter=pi*sqrt of 2(Xsqr+Ysqr)-(X-Y)sqr/2.2
..l.,
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Post by olaf haraldson »

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Rhoetus
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Post by Rhoetus »

Okay, wait.. I know this one... take a liter of water... no .... too small... take a measured amount of water in a delineated container.... cross section your head... and ...no thats a three dimensional object....
Do what I did... guess.
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Post by raito »

Yoshida wrote:I have area=pi*x*y
and perimeter=pi*sqrt of 2(Xsqr+Ysqr)-(X-Y)sqr/2.2


You do realize that that's approximate, right?

I prefer Ramanujan's approximation:

c is about pi(3(a+b)-sqrt((3a+b)(a+3b))

http://en.wikipedia.org/wiki/Circumference
http://home.att.net/~numericana/answer/ellipse.htm#elliptic
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rameymj
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Post by rameymj »

The widest dimension of the ellispe is equal to the length of the string used to draw it.

The narrowest dimension is

sqrt[ (distance between the radii points / 2)^2 + (length of string / 2)^2]

Is that what you are asking? :?

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Yoshida
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Post by Yoshida »

If I have a measure of the circumferance, and the length, what's the widest part?
..l.,
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JT
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Post by JT »

Yoshida wrote:If I have a measure of the circumferance, and the length, what's the widest part?


Depends on which approximation of the circumference (perimeter) you've been told is "right."

Since you gave the circumference as

Code: Select all

perimeter=pi*sqrt of 2(Xsqr+Ysqr)-(X-Y)sqr/2.2


Just plug in the circumference(perimeter), and the length (pick X as being the length) and solve for the width (Y).

Code: Select all

P / PI = sqrt( 2(Xsqr + Ysqr) - (X-Y)sqr/2.2 )
(P / PI)sqr = 2(Xsqr + Ysqr) - (X-Y)sqr/2.2
(P / PI)sqr = 2*Xsqr + 2*Ysqr - (Xsqr - 2XY + Ysqr)/2.2
2.2*(P / PI)sqr = 3.4*Xsqr + 2XY + 5.4*Ysqr


Plus in all the known values, and you have a quadratic equation (with Y) to solve for.
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Mike F
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Post by Mike F »

You know, I actually need to study this for my calc class . . .

What do you need it for? With only a minimal amount of work I can get you the general and standard forms, but they're largely useless unless you want to do a focal point/axis dealie.
It's up to you now.
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Post by Yoshida »

Well, one thing I'd like to have it for is to get an idea about the size of some helms on ebay that are measured by widths.
Another thing I would like to have it for is making a possible Korean helm later on that looks like halves of elipses put together.
..l.,
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