Ytukyg

Suppose we have a moon $M$ following a circular orbit (with radius $R_M$) around a planet $P$, which is in turn following a circular orbit (with radius $R_P$) around a star $S$. The angular velocity of $M$ around $P$ is given to be $omega_M$ and the angular velocity of $P$ around $S$ is given to be $omega_P$. Then $M$ will trace out an epicycle $$p(t)=leftlangle R_Pcosleft(omega_P,tright)+R_Mcosleft(omega_M,tright),,R_Psinleft(omega_P,tright)+R_Msinleft(omega_M,tright)rightrangle$$

I am trying to represent this epicycle as an epitrochoid (EDIT: by which I mean that $M$ does not need to lie on the circumference of the rolling circle) (assuming both angular velocities to be positive), where the fixed circle centered at $S$ has radius $R$ and the rolling circle centered at $P$ has radius $r$. So I must relate $R$ and $r$ to the given values. Clearly, $R=R_P-r$.

Suppose a time $t$ as passed, and $P$ has revolved through an angle $theta$ around $S$ and $M$ has revolved through an angle $phi$ around $P$. By the construction of an epitrochoid (rolling without slipping), we have that $Rtheta=rphi$. Now, since $displaystyle t=frac{theta}{omega_P}=frac{phi}{omega_M}$, we have that $Romega_P=romega_M$. So $$displaystyle R=frac{R_Pomega_M}{omega_M+omega_P},;r=frac{R_Pomega_P}{omega_M+omega_P}$$
A similar working out can be done if one of the angular velocities is negative and the other is positive, though the curve would be a hypotrochoid.

Except, this doesn't work. When I tested this out in GeoGebra I found that the arc lengths "rolled" did not equal each other, that is, there was slippage while rolling. The actual epitrochoid corresponding to $R$, $r$, and $R_M$ did not match the epicycle. What have I done wrong?

$omega_P=frac{pi}{3}$ and $omega_M=frac{7pi}{6}$, GeoGebra calculated $R=frac{28}{9}$ and $r=frac{8}{9}$. The measured arc lengths are unequal.">

UPDATE: I have been messing around in GeoGebra further and found that the actual value for $R$ in the above example (where $R_P=4$, $R_M=1$, $omega_P=frac{pi}{3}$, and $omega_M=frac{7pi}{6}$) is about 2.8572. I have tried further decimal approximations but could not find a fraction with the value.

Furthermore when $R_P=4$, $R_M=1$, $omega_P=1$, and $omega_M=4$, then $R$ is exactly 3.

Unfortunately I cannot find a formula which produces both of these answers using the given values.

The issue is that $displaystyle tnot=frac{phi}{omega_M}$. Because the rolling circle is also revolving around the center of the fixed circle, we have to account for both effects, so $displaystyle t=frac{phi}{omega_M-omega_P}$. This means that $displaystyle Romega_P=rleft(omega_M-omega_Pright)$, so $$displaystyle R=frac{R_Pleft(omega_M-omega_Pright)}{omega_M},,r=frac{R_Pomega_P}{omega_M}$$