My Math Forum Find the length of parts of an ellipse

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April 26th, 2016, 04:38 AM   #1
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Find the length of parts of an ellipse

Hi,

I have an ellipse, where radius 1 and radius 2 are known variables.
I know an angle alpha, which intersects the ellipse at point T. From that point T, I would like to draw a tangent line, spanning by known distance L from both sides of the point T. Ends of this tangent line will cut the ellipse at points S1 and S2.

Is it possible to calculate the lengths of S1_T and S2_T ellipse parts (colored in green on photo attached below)?

So what I have:
alpha
L

What I would like to calculate is the length of ellipse parts:
S1_T
S2_T

I would appreciate any kind of help.

Thank you.
Attached Images
 ellipse_sections_length.jpg (24.4 KB, 28 views)

 April 26th, 2016, 11:49 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,624 Thanks: 2077 Will approximate values suffice? They can be obtained by use of calculus.
April 27th, 2016, 12:59 AM   #3
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Joined: Jun 2012

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Quote:
 Originally Posted by skipjack Will approximate values suffice? They can be obtained by use of calculus.
Thank you for the reply skipjack.

Is it not possible to get the precise values?
If it's not possible, then I guess I have no other option but to use those approximate values.

 April 27th, 2016, 07:26 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,624 Thanks: 2077 It's not possible except in very special cases, such as when the ellipse is a circle.
 April 27th, 2016, 10:25 AM #5 Member   Joined: Jun 2012 Posts: 73 Thanks: 2 Thank you skipjack. I am interested in ellipse example where radius 1 and radius 2 are not equal. Can you show me the calculation of those approximate values, in case radius 1 and radius 2 are not equal?
April 30th, 2016, 02:26 PM   #6
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@ skipjack:

Did I misunderstand you when you said:
Quote:
 Will approximate values suffice? They can be obtained by use of calculus.
The approximate values, were values obtained from a circle instead of an ellipse?
Is this correct?

Last edited by stgeorge; April 30th, 2016 at 02:44 PM.

 April 30th, 2016, 03:38 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,624 Thanks: 2077 No, an ellipse. You will need an equation for the ellipse. Thanks from stgeorge
 May 1st, 2016, 08:40 AM #8 Member   Joined: Jun 2012 Posts: 73 Thanks: 2 Hi @skipjack. Can you present that approximate solution please? Thank you.
 May 5th, 2016, 01:46 AM #9 Member   Joined: Jun 2012 Posts: 73 Thanks: 2 Hi @skipjack, Did I do something wrong? Let's assume that radius 1 (semi-major axis) length is 8, and that radius 2 (semi-minor axis) length is 5. I assume that an ellipse equation would like like: x^2/8^2 + y^2/5^2 = 1 Can you show me how would those those approximate values obtained through calculus look like? Thank you in advance. Last edited by skipjack; May 5th, 2016 at 04:31 AM.
 May 8th, 2016, 04:52 PM #10 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Hi stgeorge, I would like to take the liberty of answering your question. I limited myself to quadrant 1 (I think) so be aware of that. Let's work an example with the numbers for $radius_1$ and $radius_2$ that you chose and arbitrary (I chose) $alpha$ and $L$ : \displaystyle \large { \begin{align*} radius_1 &=8 \\ radius_2 &=5 \\ alpha &= \frac{\pi}{5} \\ L &=2 \end{align*} } It is easiest (in my opinion) to work in parametric equations with the ellipse defined as: $\displaystyle \large{ x=radius_1 \cdot \cos(t) \\ y=radius_2 \cdot \sin(t)}$ as referenced here in equations (13) and (14). Note that the parameter $t$ is not the same as the angle from the origin to a point on the ellipse. However, we can calculate the relationship between t and an arbitrary alpha, $aplha_{arb}$ as follows: \displaystyle \large{ \begin{align*} \tan(alpha_{arb})&=\frac{y}{x} \\ \tan(alpha_{arb})&=\frac{radius_2 \cdot \sin(t)}{radius_1 \cdot \cos(t)} =\frac{radius_2}{radius_1} \cdot \tan(t) \\ \tan(t)&=\frac{radius_1}{radius_2} \cdot \tan(alpha_{arb}) \\ t&=\arctan\left (\frac{radius_1}{radius_2} \cdot \tan(alpha_{arb}) \right) \end{align*}} So at point $T$ : \displaystyle \large{ \begin{align*} t_{alpha}&=\arctan \left( \frac{8}{5} \cdot \tan \left( \frac{\pi}{5} \right) \right) \approx 0.86038812 \quad [1] \\ x_T&=8 \cos(t_{alpha}) \approx 5.217146282 \\ y_T&=5 \sin(t_{alpha}) \approx 3.790478649 \end{align*}} Now for a little calculus. The derivative of the curve, which gives the slope of the line tangent to the curve is: \displaystyle \large{ \begin{align*} \frac{dy}{dt}&=5 \cos(t) \\ \frac{dx}{dt}&=-8 \sin(t) \\ \frac{dy}{dx}&=\frac{dy/dt}{dx/dt}=-\frac{5 \cos(t)}{8 \sin(t)}=-\frac{5}{8}\ \cot(t) \end{align*}} So: \displaystyle \large{ \begin{align*} T_{slope}&=-\frac{5}{8} \cot(0.86038812) \approx -0.537649188 \\ x1_{L_{S1}}&=5.217146282-2\cos(\arctan(-0.537649188 )) \approx 3.455607093 \\ y1_{L_{S1}}&=3.790478649-2\sin(\arctan(-0.537649188 )) \approx 4.737568763 \end{align*}} Note that we don't have to calculate the actual coordinates of S1. But we need $t_{S1}$ . Fortunately we have the tangent of the angle wrt the origin as $y1_{L_{S1}}/x1_{L_{S1}}$ . So we have: $\displaystyle \large{ t_{S1}=\arctan \left( \frac{8}{5} \cdot \frac{4.737568763}{3.455607093} \right) \approx 1.143064833 \quad [2] }$ Similarly for $S2$ : \displaystyle \large{ \begin{align*} x2_{L_{S2}}&=5.217146282+2\cos(\arctan(-0.537649188 )) \approx 6.978685471 \\ y2_{L_{S2}}&=3.790478649+2\sin(\arctan(-0.537649188 )) \approx 2.843388535 \\ t_{S2}&=\arctan \left( \frac{8}{5} \cdot \frac{2.843388535}{6.978685471} \right) \approx 0.577711407 \quad [3] \end{align*}} ----------------------------------------------- Now for more calculus to find the arc lengths. If, for ease of notation, we let: $\displaystyle \large{ a=radius_1 \\ b=radius_2}$ The differential arc length, $\Delta s$, and arc length $s$, are given by: \displaystyle \large{ \begin{align*} \Delta s &= \sqrt{\Delta x^2+\Delta y^2} \\ s &= \int_{t1}^{t2} \sqrt{a^2 \sin^2t+b^2 \cos^2t} \ \ dt \\ s &= \int_{t1}^{t2} \sqrt{b^2 \sin^2t+b^2 \cos^2t + \left(a^2-b^2 \right)\sin^2t} \ \ dt \\ s &= \int_{t1}^{t2} \sqrt{b^2 + \left(a^2-b^2 \right)\sin^2t} \ \ dt \\ s &= \int_{t1}^{t2} \sqrt{b^2 + b^2 \frac{\left(a^2-b^2 \right)}{b^2}\sin^2t} \ \ dt \\ s &= b \ \int_{t1}^{t2} \sqrt{1 + \frac{\left(a^2-b^2 \right)}{b^2}\sin^2t} \ \ dt \\ s &= b \ \int_{t1}^{t2} \sqrt{1 - \frac{\left(b^2-a^2 \right)}{b^2}\sin^2t} \ \ dt \\ s &= b \ \int_{t1}^{t2} \sqrt{1 - \left(1-\frac{a^2}{b^2} \right) \sin^2t} \ \ dt \\ \end{align*}} This is an elliptic integral and fortunately, W|A can provide the answers. For S1_T, using values [2] and [1], look here For S2_T, using values [3] and [1], look here. Note that in these equations, I used $1-\frac{a^2}{b^2}$ in the EllipticE function while equation (57) in the W|MW reference given above, erroneously, I think, uses the square root of this value. I tried the square root and came up with unreasonable values. I am quite sure that the results using the non square root value are correct as I used two methods of integration to verify the answers. It is easy to make mistakes so I wrote a program to calculate these values and do the integration. I will post the code shortly. It is in Ruby but one should be able to convert it to other languages. Thanks from greg1313 and stgeorge Last edited by greg1313; May 10th, 2016 at 05:55 PM.

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