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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.

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First, two cartoons:

curtate elliptic epitrochoid

curtate elliptic hypotrochoid


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.)

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  • $\begingroup$ I reused your animation in this question. $\endgroup$ – Joseph O'Rourke Mar 9 '17 at 0:57
  • $\begingroup$ @Joseph, I'm glad it was useful! $\endgroup$ – J. M. is a poor mathematician Mar 18 '17 at 16:18
  • $\begingroup$ 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 ? $\endgroup$ – asmaier May 3 '17 at 20:59
  • $\begingroup$ @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. $\endgroup$ – J. M. is a poor mathematician May 10 '17 at 21:53
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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:

enter image description here

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.

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    $\begingroup$ 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… $\endgroup$ – J. M. is a poor mathematician Dec 11 '15 at 12:21

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