# Circumference of elliptical rim in a tilted half-full cylindrical cup

I thought I could find the circumference of this water ellipse using Pythagoras, but I get a slightly higher number when I use online ellipse circumference calculators. I have stared down into my cup for an hour trying to figure out why, but the cup did unfortunately not offer me any answers... hopefully someone here can.

By Pythagoras I mean unfolding the cylinder into a rectangle - as half of the ellipse travels from the rim to the bottom I thought the ellipse would turn into the half-rectangle diagonal, i.e. the hypotenuse, to be solved with $$\pir$$ as the base and $$length$$/$$height$$ of the cylinder. But I get a slightly different answer using the ellipse's minor and major axis in online calculators. Should I not be getting the same answer?

Thank you

• When you unroll correctly you'll get a sine wave (with a particular amplitude related to the tilt). Also, the circumference of an ellipse is an example of a thing called an elliptical integral, and is generally held to not be expressible in an elementary way. Jun 27 at 23:19
• There is no simple formula for the perimeter of an ellipse.
– PC1
Jun 27 at 23:53
• The major axis of the ellipse is $\sqrt{4r^2+L^2}$ and it looks like the minor axis is equal to the diameter of the cylinder. Jun 28 at 0:38
• Thank you! Makes perfect sense now. Visually like this then: imgur.com/Fd7ADYY Jun 28 at 7:46

For a cylinder of radius $$r$$ and height $$h$$, the circumference of the half-tilted elliptical rim is $$2\int_0^1\sqrt{\frac{4r^2}{1-x^2}+ h^2} \ dx$$ On the other hand, the Pythagorean calculation yields $$2 \sqrt{\pi^2r^2+ h^2}$$, which is always shorter. Note that it is the limiting value of the integral for either $$r\ll h$$ or $$r\gg h$$.

• If you do not provide the closed form solution of the integral, may I suppose that it does not exist ? Cheers :-) Jun 28 at 3:31
• @ClaudeLeibovici - I suppose it is of an elliptical type, which has no elementary close form.. Jun 28 at 4:11
• I got it (simpler than expected) Jun 28 at 5:54
• It seems interesting to notice that $\frac {L_1}{L_2}$ has a maximum value of $1.03623$ for $k=1.94208$ corresponding to $h=3.88415r$ Jun 28 at 8:19

Starting from @Quanto's answer, let $$h=2 k r$$ and let us compare $$L_1=4r \int_0^1\sqrt{\frac{1}{1-x^2}+ k^2} \, dx\qquad \text{and} \qquad L_2=2r \sqrt{4 k^2+\pi ^2}$$

We have $$L_1=4r\sqrt{k^2+1}\, E\left(\frac{k^2}{k^2+1}\right)\implies \frac {L_1}{L_2}=\frac{2 \sqrt{k^2+1} }{\sqrt{4 k^2+\pi ^2}} \,E\left(\frac{k^2}{k^2+1}\right)$$ where $$E(t)$$ is the the complete elliptic integral of the second kind. As a series $$E(t)=\frac \pi 2 \sum_{n=0}^\infty \frac{2^{-4 n} ((2 n)!)^2 }{(1-2 n) (n!)^4}\,t^n$$

Using $$k=\sqrt{\frac{t}{1-t} } \quad \implies\quad \frac {L_1}{L_2}=\frac { 2 E(t)}{\sqrt{ \pi ^2-\left(\pi ^2-4\right) t} }$$

Some numerical values $$\left( \begin{array}{cc} k & \frac {L_1}{L_2} \\ 0.00 & 1.00000 \\ 0.25 & 1.00283 \\ 0.50 & 1.00991 \\ 0.75 & 1.01835 \\ 1.00 & 1.02578 \\ 1.25 & 1.03115 \\ 1.50 & 1.03440 \\ 1.75 & 1.03593 \\ 2.00 & 1.03621 \end{array} \right)$$