# Finding coordinates of a point on a rolling circle

Ok, here's the situation:

I'm imagining a circle of radius A, and a circle of (a smaller) radius B. Circle A is centred at the point (0,0) and circle B starts centred at the point (b, 0). On circle B is a point P, which starts at (a+b, 0). The centre of circle B then travels along the circumference of circle A at a given rate (the rate itself is unimportant); simultaneously, P travels along the circumference of circle B, again at some arbitrary rate.

Now for the problem: given A, B, their respective rates of rotation, and the time elapsed, how do I calculate the new x and y coordinates for P? I've tried drawing countless diagrams, yet I just can't crack it. I suspect there's something missing in my intuition, or else I'm just overthinking everything. In any case, any help would be welcome.

Here's a rough diagram of what I'm trying to do:

enter image description here

You can ignore the bottom circle, as that was just me trying to figure something else out.

• Draw a picture and add it to your question. Sep 25, 2018 at 17:05
• In your pic, I guess (not good) circle A is the big one and circle B is the little one. Where is point (b,0), the start for circle B? Where is point P? What is variable "a"? Please try to make this easier. Sep 25, 2018 at 18:52
• Do you mean something like this: geogebra.org/m/hq5rpunj Sep 25, 2018 at 19:17
• @Narlin - yes, that's exactly it! Sep 26, 2018 at 1:32

You wrote that at $$t=0$$ the center of $$B$$ is at $$(b,0)$$ and the point $$P$$ at $$(a+b,0)$$. I'm assuming both of these indicate the rightmost possible position, so circle $$A$$ would have radius $$b$$ and circle $$B$$ would have radius $$a$$. Confusing. May I switch names? Let $$a$$ be the radius of circle $$A$$ and $$b$$ be the radius of circle $$B$$. Also let $$\alpha$$ be the rotational speed (in angle per time) along the circumference of $$A$$, and $$\beta$$ be the speed around $$B$$. Then the position of $$P$$ is

$$P(t)=\begin{pmatrix}a\cos(\alpha t)+b\cos(\beta t)\\a\sin(\alpha t)+b\sin(\beta t)\end{pmatrix}$$

because you simply have

$$\begin{pmatrix}a\cos(\alpha t)\\a\sin(\alpha t)\end{pmatrix}$$

as the position of the center of $$B$$ on $$A$$, and you have

$$\begin{pmatrix}b\cos(\beta t)\\b\sin(\beta t)\end{pmatrix}$$

as the offset of $$P$$ relative to the center of $$B$$. These two add up using regular vector addition. Done.

$$x=a\cos\omega_at+b\cos\omega_bt,\\y=a\sin\omega_at+b\sin\omega_bt.$$
Note that $$\omega_b$$ is the angular speed wrt the fixed frame. If you want to express it wrt the first rotating frame, the relative speed is $$\omega_b'=\omega_b-\omega_a$$.
If the circle $$b$$ is rolling on $$a$$ without gliding, when the contact point moves by $$a\omega_a\Delta t$$ units, the rolling circle rotates so that the contact point moves by $$b\omega_b'\Delta t$$ units in the revolving frame. Hence,
$$\frac{b\omega_b'}{a\omega_a}=\frac{a}{b}.$$