Yet another condition may occur. Thus, suppose the countershaft is on hand, and that it has on it two pulleys, as a 12 and a 24-inch; then a 36 on the inner shaft will be three times as large as the 12, and a 12-inch on the machine will be twice as small; or, what is the same, one half as large as the 24.
When the principle is clearly understood the calculations can be performed mentally with ease so far as the required diameters to attain the necessary speed is concerned, but there are other considerations that claim attention.
Thus, for example, to multiply the rotations 6 times we may proportion the first pair as follows: Driver 48, driven 16; second pair, driver 30, driven 15 inches in diameter.
Or we may proportion them as follows: First pair: driver 36, driven 12; second pair: driver 28, driven 14 inches in diameter.
In the second arrangement of diameters the drivers are each 2 inches, and the driven each 1 inch less in diameter than in the first; hence their cost would be diminished, as would also be the wear of the journals, on account of the reduced weight of the pulleys; hence, if the driving capacity of each pulley is equal to the requirements the second arrangement would be preferable.
In considering this part of the subject, first let it be shown that although the horse-power transmitted by the two belts is equal whatever be the proportions of the pulleys (provided, of course, that the belts do not slip), yet the strain or wear and tear of the belts varies, and the requirements for one belt are therefore different from those for the other.
Fig. 2655.
In [Fig. 2655] let a represent a 36-inch pulley on the driving shaft, b a 12-inch, and c a 24-inch pulley on a countershaft, and d a 12-inch pulley on a machine shaft. Let the main shaft make 100 revolutions per minute, and the machine requires a force to move it equal to 50 pounds applied to the perimeter of d. Now the rotations of d will, with these pulleys, be six to one of the main shaft or a, which gives d 600 revolutions per minute, thus: 100 × 6 = 600. The circumference of d is about 37.69 inches, which, multiplied by 600 (the number of its revolutions), gives 22,614 inches, or 1,884.5 feet as its speed per minute. This multiplied by the 50 pounds it takes to move the machine at the perimeter of d, gives 94,225 as the foot pounds per minute required to drive the machine 600 revolutions per minute, and this, therefore, is the amount of power transmitted by each belt. On the second belt this is shown to be composed of 50 pounds moving 1,8841⁄2 feet per minute, hence we may now find how it is composed on the first belt, as follows:—