Weird Al revolution Observe, Weird Al on a Thing;
http://imgur.com/gallery/LBg2rYR I tried posting this as an image, but it's a .webm file.
This motion is also found in coins or tires or other circular objects as they roll in place, before they settle.
Assume all measurements are variable. I don't know how to define a revolution in mathematics terms.
The way I've figured it, there are two circles, a larger one and a smaller one, flat on the plane, that the edge of the larger one is tracing. I want to know what the formula is for finding out how much Weird Al's legs (a given point) has turned relative to the static plane for every revolution of the smaller circle.
 A: Imagine you took a cylinder, a wheel, let's say, and you stood it on its side and gave it a push so that it rolled. It would go in a straight line, right? How about if the wheel was shorter on one side than the other? That is, what if the diameter of the circle on one side was smaller than that of the other? It would roll in a circle if you laid it one its side and gave it a push, right? The way you determine how many revolutions the wheel would have to make before it came back to where it started would be to take the width of the wheel ($w$), the radius of the smaller circular face ($r_1$), and the radius of the larger circular face ($r_2$) and solve for the radius of the circular path through which the wheel will travel using the fact that the radius of the circular path will be smaller for the smaller side (by the width of the wheel) than it is for the larger side, yet the number of revolutions (times the wheel turns for every circuit, henceforth $m$) they each make in one circuit around their path (of radius $R$) is the same. That is: $$\frac{2 \pi m r_1}{2 \pi m r_2}=\frac{r_1}{r_2}=\frac{R}{R+w}$$ We can then use this relationship to solve for $m$ once we know $R$ from $r_1,r_2,$ and $w$ using the fact that $2mr_1\pi=R$. Hope that helps you understand why he's spinning like that, the uproarious bafoon.
A: You have a cone frustum, large radius $R$, small radius $r$, and altitude $h$.  This gives full cone altitude $H=Rh/(R-r)$, and full distance from the apex of the cone to the outer edge $\ell=\sqrt{H^2+R^2}$.
now, as it goes around, the outer edge traces out a circle of radius $\ell$, which should have a circumference of $2\pi\ell$... but the outer edge of the cone only has a circumference of $2\pi R$, which is smaller!
So, following Al's feet, he slowly loses ground - in the case shown, he's got $R\approx\frac{5}{6} \ell$ (I'm guessing here - it looks like he gets about halfway around in three cycles), so a full cycle of his feet going up and down only gets the low edge of the cone $5/6$ of the way around the circle -- so he's rotated $1/6$ of the way in the opposite direction of the cone's rotation. 
A: Lets call the radius of the horizontal disc that touches the ground $r$, so the circumference is $2\pi r$.
The other distance we need is the diagonal distance from the edge of the disc to the point where the axis of rotation perpendicular to the ground hits the ground. Lets call that $d$. (If this is hard to accurately measure we could make a right triangle and find it based on the height of the disc.)
So then the circle that the edge traces out on the ground has a circumference of $2\pi d$. So then every time Al comes back to the same point, the seat has rolled $2\pi d / 2\pi r = d/r$ times on the ground, so then he has made $d/r-1$ of a rotation relative to the static plane.
