Concrete example using elliptic integral of the second kind to calculate arc length?

Question

There's an identical question here but it was never answered fully and the link providing an essential component of the accepted "answer" is broken. The Keisan website presents a solution here (PDF) that shows the length of an arc going clockwise from the Y axis as $${s(x) = aE({x \over a},k)}$$

Consider an ellipse with a = 20, b = 5 (k = 0.9682) and an arc starting at (0,5) and ending at (7.3625, 4.6489). E(0.3681, 0.9682) = 0.3604, according to the Keisan integral calculator, yielding an arc length of 7.208. But using Keisan's own arc length calculator (the angle is 32.27°) yields a length of 7.3736. Close, but not exact.

For an arc ending at (18.8774, 1.6516): E(0.9439, 0.9682) = 0.8196 yields a length of 16.392 versus the calculated 19.3729 (angle of 5°). Considerably off.

Keisan certainly seems to have done something right in their arc length calculator! But I can't duplicate their results, so I'm clearly missing something in the calculation of the parameters and/or the application of the formula. Can anyone provide a clue, or suggest a concrete path for how to apply arc parameters such as these to the elliptic integral to calculate the arc length?

Answer 1

The formula should be $$\color{blue}{s(x)=a E\left(\sin ^{-1}\left(\frac{x}{a}\right)|1-\frac{b^2}{a^2}\right)}$$ So, for your case $(a=20,b=5,x=\frac{73625}{10000}=\frac{589}{80})$, the result is $$20 E\left(\sin ^{-1}\left(\frac{589}{1600}\right)|\frac{15}{16}\right) \approx 7.37381$$ while $$20 E\left(\frac{589}{1600}|\frac{15}{16}\right) \approx 7.20786$$ So, a typo !

Answer 2

The mismatch comes from a confusion in the definition of the elliptic integral.

In the PDF : $$E_{PDF}(X,k)=\int_0^X \sqrt{\frac{1-k^2t^2}{1-t^2}}dt$$ and for example, with $X=\frac{x}{a}\simeq\frac{7.3625}{20}=0.368125$ $$E_{PDF}(X,k)=E_{PDF}(0.368125 ,0.968246) =\int_0^X \sqrt{\frac{1-k^2t^2}{1-t^2}}dt\simeq 0.3687$$ $$s=aE_{PDF}(X,k)\simeq 20(0.3687)=7.3738$$ which is the correct approximate of the arc length.

The standard definition of the Incomplete Elliptic Integral definition is : $$E_{}(\phi,k)=\int_0^{\phi} \sqrt{1-k^2\sin^2(\theta)}d\theta=\int_0^{\sin(\phi)} \sqrt{\frac{1-k^2t^2}{1-t^2}}dt$$ where $t=\sin(\theta)$

With the above example $\phi=\sin^{-1}(X)\simeq \sin^{-1}(0.368125)=0.376992$ $$E(\phi,k)=E(0.376992,0.968246)= \int_0^{\phi} \sqrt{1-k^2\sin^2(\theta)}d\theta \simeq 0.3687$$ $$s=aE(\phi,k)\simeq 20(0.3687)=7.3738$$ This is the same result than above.

In a word, the mistake is in $E(0.368125 ,0.968246)\simeq 0.3601$ which should be $E(0.376992,0.968246)\simeq 0.3687$ .