The diameter of the first driver is 36 inches, and its circumference 113.09 inches, or 9.42 feet; this, multiplied by its revolutions per minute, will give its speed, thus: 9.42 × 100 = 942 feet per minute. To obtain the necessary amount of pull for this first belt, we must divide this speed into the number of foot pounds it takes to drive the machine, thus: 94,225 ÷ 942 = 100.02. The duties of the two belts are therefore as follows:—
| First belt, | weight of pull | 100.02 |
| „ | speed per minute | 942 feet. |
| Second belt, | weight of pull | 50.00 |
| „ | speed per minute | 1884.5 feet. |
The duty in foot pounds being equal, as may be shown by multiplying the feet per minute by the force or weight of the pull, leaving out the fractions, thus:—
| 942 | × | 100 | = | 94,200. |
| 1884 | × | 50 | = | 94,200. |
The difference in the requirements is, then, that the first belt must have as much more weight or force of pull than the second as its speed is less than that of the second.
It is obvious that in determining the proportions of the pulleys this difference in the requirements should be considered, and the manner in which this should be done depends entirely upon the conditions.
Thus, in the case we have considered, the speed was increased, but the object of the countershaft may be to decrease the speed, and in that case the conditions would be reversed, inasmuch as though the foot pounds transmitted by both belts would still be equal, yet the speed would be greatest and the strain or pull the most on the second belt instead of on the first.
It is obvious, then, that the proportions of the pulleys being determined the actual diameters must be large enough to transmit the required amount of power without unduly straining the belt.