My Math Forum Ellipse: semi-major axis (a) as a function of semi-minor (b)

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 April 23rd, 2011, 03:54 AM #1 Member   Joined: Aug 2008 Posts: 44 Thanks: 0 Ellipse: semi-major axis (a) as a function of semi-minor (b) I want to build a simple green house with a single piece of a perforated metal sheet. I would put each end of the sheet inside the ground, as in the picture below: If I simplify and use a model of an ellipse, how can I get the height ($a$) as a function of width ($b$). In short, what is $a(b)$ My attempts: The "semi-latus rectum" ($l$) passes through the focus and is (according to Wikipedia) $\displaystyle{ l=\frac{b^2}{a} }$. Of course, I know the length of the metal sheet ... but this is still not enough to get $a(b)$ function. Or is it simpler to first get height $a$ using isosceles triangle as a model ... and then correct the over-estimated height $a$ by multiplying it with the ratio $\frac{\text{triangle side}}{\text{ellipse arc length}}$?
 April 23rd, 2011, 04:09 PM #2 Global Moderator   Joined: May 2007 Posts: 6,628 Thanks: 622 Re: Ellipse: semi-minor axis (b) as a function of semi-major The circumference of an ellipse for given axes is a "complete elliptic integral". Therefore there is no simple expression for the elliptic arc length that you are trying to use.
 April 23rd, 2011, 11:42 PM #3 Member   Joined: Aug 2008 Posts: 44 Thanks: 0 Re: Ellipse: semi-major axis (a) as a function of semi-minor But I know the metal sheet length (or half the ellipse circumference ) ... let us say it's 6 meters (from one ground point to another ground point). Had I started the other way around, I'd differentiate the explicit function for an ellipse $y=b\sqrt{1-\frac{x^2}{a^2}}$ and the put the derivative, $\frac{dy}{dx}$, into the formula for arc length $\int\sqrt{1+(\frac{dy}{dx})^2}dx=\int\sqrt{1+\frac {b^2x^2}{a^4-a^2x^2}}dx$. Now, the task of extracting the relationship $a(b)$ from this integral overwhelms me.
 April 24th, 2011, 12:55 AM #4 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Ellipse: semi-major axis (a) as a function of semi-minor You'll simply have to define your own inverse function to determine a. You know the arc length, which is 6. You know what b is. Now you must solve for a, which is actually impossible. Therefore, it must be approximated. You could use Newton's method of approximating the nearest zero given an initial starting point. I'd recommend a graphing calculator or a computer to approximate this. You could also use a scientific calculator that can perform numeric integrals, but it'll take longer to get your desired accuracy. Then you have a method of determing a(b,s) where s is the arc length of half the ellipse. The integral will be easier if you cut the half ellipse in half again, having bounds from 0 to b, and cutting the arc length in half. There will be less calculations performed and you'll still determine a. This won't be pretty to approximate, but it will work.
 April 24th, 2011, 03:28 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 The circumference of an ellipse ? 2??((aČ + bČ)/2).
April 24th, 2011, 05:14 PM   #6
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Re: Ellipse: semi-major axis (a) as a function of semi-minor

Quote:
 Originally Posted by courteous I want to build a simple green house with a single piece of a perforated metal sheet. I would put each end of the sheet inside the ground, as in the picture below: If I simplify and use a model of an ellipse, how can I get the height ($a$) as a function of width ($b$). In short, what is $a(b)$ My attempts: The "semi-latus rectum" ($l$) passes through the focus and is (according to Wikipedia) $\displaystyle{ l=\frac{b^2}{a} }$. Of course, I know the length of the metal sheet ... but this is still not enough to get $a(b)$ function. Or is it simpler to first get height $a$ using isosceles triangle as a model ... and then correct the over-estimated height $a$ by multiplying it with the ratio $\frac{\text{triangle side}}{\text{ellipse arc length}}$?
If you were to take the length and divide it into half for the ninety degree referring to the hundred and eighty degrees of the ground you will be half way. Then you divide the height by the length and then subtract the outer degrees of the intergral angles.

Basically you divide the height by two...

April 25th, 2011, 02:17 PM   #7
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Re:

Quote:
 Originally Posted by johnny The circumference of an ellipse ? 2??((aČ + bČ)/2).
Not true!

 April 25th, 2011, 02:30 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 According to from Kepler's lower bound to Muir's lower bound, p = 2.
 April 26th, 2011, 09:42 AM #9 Member   Joined: Aug 2008 Posts: 44 Thanks: 0 Re: Ellipse: semi-major axis (a) as a function of semi-minor It seems that Ramanujan's approximation for ellipse circumference C will do: $C\approx \pi\left[3(a+b)-\sqrt{(3a+b)(a+3b)}\right]$ I use Mathematica to solve this for given $C= 12$ and particular $b$ points. Code: b = 1 Solve[12 == Pi (3 (a + b) - Sqrt[(3 a + b) (a + 3 b)]), a] Now, how would you plot graph for a range of $b$ values?
 April 26th, 2011, 02:43 PM #10 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Ellipse: semi-major axis (a) as a function of semi-minor Mathematica has something called implicit plot. Here is the link to the documentation on it: http://tinyurl.com/3r6bgoj

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### use the grid markings to estimate the ratio of the semi major axis to the semimajor axis of mercury

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