What is locus of a fixed point on a circle of radius $r$ rolling over the curve $y=\sin x$?

Question

What is locus of a fixed point on a circle of radius $r$ rolling over the curve $y=\sin x$? I have been struggling on the problem for many days, but I could not solve it. Geometrically, the locus seems to be a series of cycloid-looking figures mounted on the exterior of the sine curve, but analytically it seems very difficult to find the equation of the locus.

Answer 1

It involves an elliptic integral of the second kind.

Let $M = M(\theta) = (\theta, \sin \theta)$ be the contact point. The tangent at this point is $T = (1, -\cos \theta)$ so the center $I$ of the circle is such that $\vec{MI} = (\frac {r \cos \theta} {1+\cos ^2 \theta}, \frac {r} {1+\cos ^2 \theta})$.

The fixed point $P$ on the rolling circle is obtained for $M$ by rotation around $I$ of an angle $s$ equal to the length of $OM$ along the curves, that is $s = \int _0^\theta \sqrt{1 + \cos^2 \theta} d\theta$. Explicitly, $IP= (\frac {r \cos \theta} {1+\cos ^2 \theta} \cos s + \frac {r} {1+\cos ^2 \theta} \sin s , \frac {-r \cos \theta} {1+\cos ^2 \theta} \sin s + \frac {r} {1+\cos ^2 \theta} \cos s) $.

So the only step that makes the curve non-rational is the integral, which indeed is an elliptic function, impossible to express with elementary functions alone.