# Circle Rolling on Ellipse

I've gotten interested in describing a circle rolling on an ellipse; specifically, the curve traced out by a point on the circumference of the circle. I want a symbolic solution to the general case, radius $r$, axes $a$ and $b$. I've written nine polynomial equations in terms of various angles and lengths.

Exactly what "solution" means is subject to debate. Let $(u,v)$ be the point on the circle. Similar to the cycloid, I would like an equation for $u$ in terms of a "natural" angle in the problem. Similarly, an equation for $v$.

Perhaps it is necessary to have a differential equation, so maybe $du/dt$, $u$, and $t$, where $t$ is an angle in the problem.

I would have thought this was known, but I can't find it anywhere.

First, two cartoons:

Since I have already outlined the general derivation for the circle-roulette with respect to a fixed curve in this answer, I will not repeat the derivation for this case, and instead present the necessary formulae for the "elliptic trochoid" corresponding to the ellipse $(a\cos t\quad b\sin t)^\top$.

Letting

$$\mathfrak{s}(t)=a\left(E\left(t+\frac{\pi}{2}\mid 1-\frac{b^2}{a^2}\right)-E\left(1-\frac{b^2}{a^2}\right)\right)$$

be the arclength function for the ellipse (where $E(\phi\mid m)$ is the incomplete elliptic integral of the second kind, and $E(m)=E\left(\frac{\pi}{2}\mid m\right)$ the complete elliptic integral of the second kind, both with parameter $m$), the general equation for the "elliptic trochoid" (in matrix-vector format) is

$$\begin{pmatrix}a\cos t\\ b\sin t\end{pmatrix}+\frac1{\sqrt{a^2\sin^2 t+b^2\cos^2 t}}\begin{pmatrix}a\sin t&b\cos t\\-b\cos t&a\sin t\end{pmatrix}\begin{pmatrix}h r\sin\left(\frac{\mathfrak{s}(t)}{r}\right)\\r-h r\cos\left(\frac{\mathfrak{s}(t)}{r}\right)\end{pmatrix}$$

where $|r|$ is the radius of the rolling circle, and $h$ gives the fraction of the distance of the tracing point from the rolling circle's center. Positive $r$ corresponds to the "elliptic epitrochoid" (the first cartoon), and negative $r$ corresponds to the "elliptic hypotrochoid" (the second cartoon). Letting $h=1$ gives the cycloid case, while $h < 1$ gives the "curtate" case (the case depicted in the given cartoons), and $h > 1$ gives the "prolate case", where the tracing point juts outside the rolling circle.

(The Mathematica notebook that generated these cartoons is available upon request.)

• I reused your animation in this question. Mar 9, 2017 at 0:57
• @Joseph, I'm glad it was useful! Mar 18, 2017 at 16:18
• Why is the arclength of the ellipse here not simply $s(t) = a E(t|1-b^2/a^2)$ as given here: mathworld.wolfram.com/Ellipse.html ? May 3, 2017 at 20:59
• @asmaier, formula 56 from MathWorld does not give the correct results, and formula 57 is usable only if your computing system supports imaginary moduli/negative parameters. Most systems require $m$ or $k$ to be in $[0,1)$, and this is the appropriate formula in that circumstance. May 10, 2017 at 21:53

Complicated! I can't even write out the parametric equation for this kind of epicycloid. But I was able to program a picture for an ellipse with semimajor and semiminor axes of $3$ and $1$, with a circle of radius $1$ rolling outside:

It looks like a closed curve, but in fact it is not quite closed. I will try to give more details about how I made this image when I have the time, but suffice it to say, it's not something that is trivially done.

• As I noted in the comments, the elliptic integral of the second kind is necessary here, since it pops up in the expression for the arclength of the ellipse. So yes, complicated… Dec 11, 2015 at 12:21