Suppose the weight to have fallen 32.2 inches, and we have 10 lbs. moving through 32.2 inches, this power it will have transmitted to pulley b.

To find what this becomes at the perimeter of c, we must reduce the number of lbs. in the same proportion that the perimeter of c moves faster than does that of b; hence we divide the circumference of one into the other, and with the sum so obtained divide the amount of the weight; thus, 64.4 (circumference of c) ÷ 32.2 (circumference of b), = 2; and 10 lbs. ÷ 2 = 5 lbs., which, as the circumference of c is twice that of b, will move twice as fast as the 10 lbs. at b, hence for c we have 5 lbs. moving through 64.4 inches.

Now c communicates this to d by means of the belt h, hence we have at d the same 5 lbs. moving through 64.4 inches.

Now e moves twice as fast as d, because its circumference is twice as great, and both are fast upon the same shaft, hence the 5 lbs. at d becomes 21⁄2 lbs. at e, but moves through a distance equal to twice 64.4, which is 128.8 inches. To recapitulate, then, we have as follows:

The weight gives 10 lbs. moving through 32.2 inches.
Pulleyb„10 „„„32.2„
„c„5 „„„64.4„
„d„5 „„„64.4„
„e„21⁄2„„„128.8„

That the amount of power is equal in each case, may be shown as follows:

For c, 5 lbs. moving through 64.4 inches is an equal amount of power to 10 lbs. moving through 32.2 inches, because if we suppose the first pair of pulleys to be revolving levers, whose fulcrum is the centre of the shaft, it will be plain that one end of the lever being twice as long as the other, its motion will be twice as great, and the 5 at 101⁄4 inches just balances 10, at 51⁄8 inches from the fulcrum, as in the common lever.

In the case of d we have the same figures both for weight and motion as we have at c, because d simply receives the weight or force and the motion of c. In the case of e, we have the motion of the weight multiplied four times; for the distance e moves is 128.8 inches, which, divided by 4, gives 32.2 inches, which is the amount of motion of the weight, hence the 10 lbs. of the weight is decreased four times, thus 10 lbs. ÷ 4 = 21⁄2 lbs., hence the 21⁄2 lbs. moving through 128.8 inches is the same amount of power as 10 lbs. moving 32.2 inches, and we may concentrate or convert the one into the other, by dividing 128.8 by 4, and multiplying the 21⁄2 lbs. by 4, giving 10 lbs. moving 32.2 inches.

If, therefore, we make no allowance for friction, nothing has been lost and nothing gained.

Thus far, we have taken no account of the time in which the work was done, more than as one wheel is caused to move by the other, and all of them by the motion of the weight, they must all have begun and also have to move at the same time. Suppose, then, that the time occupied by the weight in falling the 32.2 inches was one minute, and the amount of power obtained may be found by multiplying the lbs. of the weight by the distance it moved through in the minute, thus 10 lbs. moving 32.2 inches in a minute gives 32.2 inch lbs. per minute, being the amount of power developed by the 10 lb. weight in falling the 32.2 inches.