through which the cylinder advances is only half that through
which the upper rule advances, it follows that the _force_ which must
act on the upper rule is only half as great as that overcome in moving
the cylinder. The carter makes use of this principle when he puts his
hand to the top of a wheel to help his cart over an obstacle.

[Illustration: FIG. 223.]

[Illustration: FIG. 224.]

[Illustration: FIG. 225.]

Now see how this principle is applied to the change-speed gear. The
lower rule is replaced by a cog-wheel, C (Fig. 223); the cylinder by a
cog, B, running round it; and the upper rule by a ring, A, with internal
teeth. We may suppose that A is the chain-ring, B a cog mounted on a pin
projecting from the hub, and C a cog attached to the fixed axle. It is
evident that B will not move so fast round 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 = 1/4 of a revolution in a clockwise direction, and C 10/30
= 1/3 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 1/3 of distance round C; A
advances 1/3 + 1/4 = 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 than B
does A run round C? Apply the formula:--Motion of A: motion of B:: 64 +
16: 64. That is, while A revolves once, B and the hub and the
driving-wheel will revolve only 64/80 = 4/5 of a turn. T