To find the number of revolutions of a driven pulley in a given time, its diameter and the diameter and revolutions of the driver being given:

Rule.—Multiply the diameter of the driver by the number of its revolutions in the given time, and divide by the diameter of the driven, and the quotient will be the number of revolutions of the driven in the given time.

Suppose, however, that the speed of the shaft only is given, and we require to find the diameter of both pulleys, as, for example, suppose a shaft makes 150 revolutions per minute, and we require to drive the pulley on a machine 600 revolutions per minute. Here we have two considerations: first, the relative diameters of the two pulleys, and secondly, the diameter of pulley and width of belt necessary to transmit the amount of power necessary to drive the machine at the speed required. Leaving the second to be discussed hereafter in connection with the driving power of belts, we may proceed to determine the first as follows: The pulley on the machine must be as much smaller than that on the main shaft, as the speed of the pulley on the machine requires to run faster than does the main shaft, hence we divide the 600 by 150 and get four, which is the number of times smaller than the driver that the driven pulley must be. Suppose then the driver is made a 24-inch pulley, then the driven must be a 6-inch one, because 24 ÷ 4 = 6; or we may make the driver 36, and the driven 9, because 36 ÷ 4 = 9; or the driver being 48 inches in diameter, the driven must be 12, because 48 ÷ 4 = 12. To reverse the case, suppose the shaft to make 200 revolutions per minute, and the machine pulley to make 50, then since 200 ÷ 50 = 4, the driven (or machine pulley) must have a diameter four times that of the driver, and any two pulleys of which one is four times the diameter of the other may be used, as say: Pulley on line shaft 10 inches in diameter, pulley on machine 40 inches in diameter; or, pulley on line shaft 20 inches in diameter, pulley on machine 80 inches in diameter.

Now, in nearly all cases that are met with in practice, it would be inconvenient to have so large a pulley as 80 inches in diameter to drive a machine, and again in most cases a driving pulley of 10 inches in diameter would be too small. So likewise in cases where the machine pulley requires to run faster than the line shaft, a single pair of pulleys will be found to give, where great changes of revolution are required, too great a disproportion in the diameter of the pulleys; thus in the case of a shaft making 150, and the machine requiring to make 600, we may use the following pairs of pulleys:—

On Main Shaft.On Machine Shaft.
First32inch diameter 8inch diameter.
Second40„„10„„
Third48„„12„„
Fourth60„„15„„

But the machine may require so much power to drive it, that with the width of belt it is desired to employ, a pulley larger than either of these is necessary, as, say, one 20 inches in diameter. Now, with a 20-inch driven pulley, the driver would require to be 80 inches in diameter, because 20 × 4 = 80. But there may not be room between the shaft and the ceiling for a pulley of so large a diameter, or such a large pulley may be too heavy to place on the shaft, or it may be too costly, and to avoid these evils, countershafts are used.

By the employment of a countershaft we simply obtain—with two pairs of pulleys and by means of small pulleys—that which could be obtained in a single pair, providing the great difference in their diameters (necessary to obtain great changes of rotation), were not objectionable; all that is necessary, therefore, is to accomplish part of the required change of rotation in one pair, and the remainder in the other. In doing this, however, while the velocity of each driver and driven will be equal (as was explained with reference to a single pair), notwithstanding the difference in their diameters, yet the velocity of one pair will necessarily differ from that of the other, so that the pulley on the machine will vary in its velocity as well as in its rotation from that of the first driver. The first driver is that on the main or driving shaft, and the pulley it drives is the first driven. The second driver is the second pulley on the countershaft, and the second driven is the one it drives or that on the machine. Suppose, then, a driving shaft makes 100 revolutions per minute, and the machine requires to make 600, then the speed of rotation requires to be increased six times. Now we may effect this change of six times in several ways; thus: Suppose we increase the rotations three times in the first pair, then the second pulley will make 300 rotations, or three times those of the main shaft, and all we have to do is to make the second driven one-half the diameter of the second driver, and its rotations will be double those of the second driver, which will give the required speed of 600 revolutions. Suppose, however, we change the speed four times in the first pair, and the 100 of the shaft becomes 400 on the countershaft, and to increase this to 600 on the second driven, all that is required is to make its diameter one-half less than that of the second driver, because 600 is one-half more than 400. From this it will be perceived that the number of changes or amount of increase or decrease of speed being given, the proportion of diameters for both pairs of pulleys will be represented by any two numbers which, multiplied together, will give a sum equal to the number of increased revolutions required. Having found the proportions for each pair, it remains to determine their actual diameters, and they will be found to vary under different conditions.

Suppose, for example, we have the following conditions: Main shaft runs 100; machine must run 600. The pulley on the line shaft is 36 inches in diameter; required, the diameters for the other three pulleys.

To make three changes in the first pair, the first driven must be 1⁄3 the diameter of the first driver, which is 12 inches. Now the second pair we may make any diameters that are two to one; and since the second driver is to be the smallest, we may select as small a pulley as will answer for the machine, and make its driver twice its diameter.

But suppose it is the diameter of the pulley on the machine that is fixed, and the diameter of the other three require to be found. Let the diameter of the second driven be 12; then its driver on the countershaft must be 24. The other two must have diameters 3 to 1 as before, any suitable wheels being selected.