[Math] Parameterizing the path of a point on a circle rolling on another circle

parametricplane-curves

Problem:

A wheel of radius $a$ rolls on the outside of a circle with radius $b$ (see figure).

Fig. 1

Find the parameterization for the curve a point on the wheel follows. You may choose freely how you position the coordinate system, and the starting point.

My attempt:

Fig. 2

I'm not very well versed in parametrizing curves, so I tried first determining the curve that the center of the small circle would follow, and came up with $$P_s = ((a+b)\cos t, (a+b)\sin t)$$

Then I added to that the curve that the point $m$ would have relative to the center $s$ of the tiny wheel, $$P_m = P_s + (a\cos t, a\sin t) \\ = ((2a+b)\cos t, (2a+b)\sin t)$$

That is my answer, but unless I'm missing some fundamental algebra, this is not equal to the real answer.

The real answer:

With the origin at the center of the large circle, and with starting point $(b, 0)$, we get: $$x(t) = (a+b)\cos t – a\cos(\frac{a+b}{a}t) \\ y(t) = (a+b)\sin t – a\sin(\frac{a+b}{a}t)$$

Best Answer

enter image description here

Imagine the smaller circle has rolled along the bigger circle as shown above. The distance it has rolled along the larger circle is equal to $bt$ (shown as the green path). The distance that point $m$ on the smaller circle has moved is equal to $as$ (shown as the pink path). These two distances must be equal, therefore:$$as=bt$$$$\therefore s=\frac{bt}{a}$$Now look at the magnified section below the larger circle. If we drop a vertical line down from the center of the smaller circle then the angle made between the red line and this vertical line must be $\frac{\pi}{2}-t$. Therefore the angle marked as $p$ is given by:$$p=s-\left(\frac{\pi}{2}-t\right)=\frac{bt}{a}-\left(\frac{\pi}{2}-t\right)=\frac{(a+b)t}{a}-\frac{\pi}{2}$$Hopefully this should give you enough information to solve from here...