C as A does. The amount by which A will get ahead of B can be calculated easily. We begin with the wheels in the position shown in Fig. 223. A point, I, on A is exactly over the topmost point of C. For the sake of convenience we will first assume that instead of B running round C, B is revolved on its axis for one complete revolution in a clockwise direction, and that A and C move as in Fig. 224. If B has 10 teeth, C 30, and A 40, A will have been moved 10⁄40 = ¼ of a revolution in a clockwise direction, and C 10⁄30 = ⅓ of a revolution in an anti-clockwise direction.
Now, coming back to what actually does happen, we shall be able to understand how far A rotates round C relatively to the motion of B, when C is fixed and B rolls (Fig. 225). B advances ⅓ of distance round C; A advances ⅓ + ¼ = 7⁄12 of distance round B. The fractions, if reduced to a common denominator, are as 4:7, and this is equivalent to 40 (number of teeth on A): 40 + 30 (teeth on A + teeth on C.)
To leave the reader with a very clear idea we will summarize the matter thus:—If T = number of teeth on A, t = number of teeth on C, then movement of A: movement of B:: T + t: T.
Here is a two-speed hub. Let us count the teeth. The chain-ring (= A) has 64 internal teeth, and the central cog (= C) on the axle has 16 teeth. There are four cogs (= B) equally spaced, running on pins projecting from the hub-shell between A and C. How much faster