3
$\begingroup$

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.

enter image description here

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 $\pi$$r$ 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

$\endgroup$
4
  • 3
    $\begingroup$ 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. $\endgroup$
    – user24142
    Jun 27 at 23:19
  • 1
    $\begingroup$ There is no simple formula for the perimeter of an ellipse. $\endgroup$
    – PC1
    Jun 27 at 23:53
  • $\begingroup$ 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. $\endgroup$
    – John Douma
    Jun 28 at 0:38
  • 1
    $\begingroup$ Thank you! Makes perfect sense now. Visually like this then: imgur.com/Fd7ADYY $\endgroup$
    – Antman
    Jun 28 at 7:46

2 Answers 2

3
$\begingroup$

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 $r\gg h$.

$\endgroup$
4
  • 2
    $\begingroup$ If you do not provide the closed form solution of the integral, may I suppose that it does not exist ? Cheers :-) $\endgroup$ Jun 28 at 3:31
  • $\begingroup$ @ClaudeLeibovici - I suppose it is of an elliptical type, which has no elementary close form.. $\endgroup$
    – Quanto
    Jun 28 at 4:11
  • $\begingroup$ I got it (simpler than expected) $\endgroup$ Jun 28 at 5:54
  • $\begingroup$ It seems interesting to notice that $\frac {L_1}{L_2}$ has a maximum value of $1.03623$ for $k=1.94208$ corresponding to $h=3.88415r$ $\endgroup$ Jun 28 at 8:19
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.