I need the $x$ coordinates of two new points that split an elliptical arc between 2 given points in quadrant I into 3 equal pieces. How can I get these coordinates? I want to use Wolfram but I can also use Excel, C++ or Javascript. Quick-and-dirty is fine.

There are two parts to this problem, and I have already done the first.

First: Find the total arc length (easy). Second: Find the points that evenly partition the arc into three arcs of equal length.

My work so far:

I have a vertically-oriented ellipse whose equation is


were a and b are measured in meters. Then



$\frac{dy}{dx}=\frac{-ax}{b\sqrt{b^2 - x^2}}$

I checked online for arclength integral formulas and decided it made most sense to keep my equation in terms of x or y so my answers would be in meters. The formula I am using is

$L = \int \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$

So then I need $1/3$ of

$L = \int_{x_1}^{x_2} \sqrt{1 + \frac{a^2 x^2}{b^2 \left(b^2 - x^2\right)}} dx$

Now, for my use-case, $a = 14$ and $b = 8$. I have my endpoints approximated at

$\left(x_1,y_1\right) = \left(2.07052,13.523\right)$

$\left(x_2,y_2\right) = \left(7.43548,5.15728\right)$

(endpoints are fixed for this problem)

Using only the x-coordinates, Wolfram spits out the following for the required $1/3$ arc length:

$\frac{1}{3}L = \int_{2.07052}^{7.43548} \sqrt{1 + \frac{196 x^2}{64 \left(64 - x^2\right)}} dx \approx 3.41743 $

My Problem:

I would like to find 2 more x-coordinates between $x_1$ and $x_2$ such that each of the 3 arcs between them are about the same length as Wolfram just calculated. I attempted to see what Wolfram would say if I just told it to "solve for b:"

$3.41743 = \int_{2.07052}^b \sqrt{1 + \frac{196 x^2}{64 \left(64 - x^2\right)}} dx$

But Wolfram had no idea what I was talking about. I also tried using Wolfram's "approximation" of the integral and set that equal to the value I needed. I told wolfram to "solve" this:

$3.41743 = \frac{32 \sqrt{64 - x^2} \sqrt{1 - \frac{49 x^2}{16 \left(-64 + x^2 \right)}} E\left(sin^{-1}\left(\frac{x}{8}\right)|-\frac{33}{16}\right)}{\sqrt{1024 + 33 x^2}}$

but the answer Wolfram gave me was "approximately" an even nastier formula (I was hoping for a decimal!) You can feed this to Wolfram to see for yourself:

Solve[3.41743 == (32 Sqrt[64 - x^2] Sqrt[1 - (49 x^2)/(16 (-64 + x^2))] EllipticE[ArcSin[x/8], -33/16])/Sqrt[1024 + 33 x^2], {x}]

How can I find these points?

  • 1
    $\begingroup$ Call the arc length $\ell(b) = \int_{2.07\dots}^b \sqrt{\cdots}\,dx$. You want to find $b$ such that $\ell(b) = 3.41\dots$. Since $\ell$ is monotonic, you could do bisection manually by trying different values of $b$. In fact, since you know $\ell'(b) = \sqrt{\cdots}$, you could even do Newton's method. $\endgroup$ – user856 Apr 23 '19 at 7:38
  • $\begingroup$ Thank you, Rahul, testing different upper limit values ended up being much quicker than I anticipated $\endgroup$ – hatinacat2000 Apr 23 '19 at 7:56
  • $\begingroup$ Rahul, since the integral is not algebraic, how can I evaluate points using Newton's Method without using Wolfram to compute the integral (the integral is the function evaluated at a point, required since : $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$) ? I understand there should be a Taylor series or something but I don't know how to find it. Can you post a link to some instructions? $\endgroup$ – hatinacat2000 Apr 23 '19 at 8:31
  • 1
    $\begingroup$ You do need WolframAlpha to evaluate $\ell(b)$ in order to get the next Newton step. The advantage is just that Newton's method will get you close to the solution with fewer WolframAlpha calls than bisection will. $\endgroup$ – user856 Apr 23 '19 at 8:58
  • $\begingroup$ This question may be relevant. $\endgroup$ – Ng Chung Tak Apr 23 '19 at 15:04

I am sorry if this solution underwhelms anybody, but it was sufficient for my use-case to test different upper limit values on the arc length integral in Wolfram until the first 5 digits matched. This took about 10 minutes of adjustments and note-taking. The solution points are approximately

$2.31809, 13.523$ (left endpoint)

$4.676463, 11.3589282$ (first point determined)

$6.393351, 8.415491989$ (next point based on first)

$7.88893, 5.15728$ (right endpoint)

That is,

$\frac{1}{3}L = \int_{2.07052}^{7.43548} \sqrt{1 + \frac{196 x^2}{64 \left(64 - x^2\right)}} dx \approx 3.41743 $


$\int_{2.07052}^{4.676463} \sqrt{1 + \frac{196 x^2}{64 \left(64 - x^2\right)}} dx \approx 3.41743 ,$

$\int_{4.676463}^{6.393351} \sqrt{1 + \frac{196 x^2}{64 \left(64 - x^2\right)}} dx \approx 3.41743 $ and

$\int_{6.393351}^{7.43548} \sqrt{1 + \frac{196 x^2}{64 \left(64 - x^2\right)}} dx \approx 3.41743$


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