p = the effective mean pressure on the piston in pounds per square inch.
T = the equivalent tractive force at the rails in pounds.
This, given in plain terms, reads: Square the diameter of piston in inches; multiply by the length of stroke in inches; multiply by the mean pressure of steam per square inch, and divide by the diameter of the drivers in inches.
We will apply the calculations to the case of the standard Buchanan passenger engine; the cylinders being 17 by 24 inches, the drivers 68 inches in diameter, and the effective pressure on pistons about 80 pounds. The problem is worked thus:—
| 17 | inches diam. of cylinders. | |
| 17 | ||
| —— | ||
| 289 | = square of diameter. | |
| 24 | inches stroke of piston. | |
| —— | ||
| 6936 | ||
| 80 | lbs. per square inch. | |
| ——— | ||
| Diam. of drivers, 68) | 554880 | (8160 |
This gives 8,160 pounds as the pressure exerted to turn the drivers, which may be accepted as a close approximation to the truth. In using this formula, the mistake has frequently been made of taking the quotient to represent the power developed by one cylinder, when, in fact, it gives the power of both.
A method of calculating the locomotive traction that is a good deal followed by our engineers, is to ascertain the foot-pounds of work the engine is doing during each revolution of the drivers. By dividing the total thus found by the circumference of the drivers in feet, the force exerted through each foot that the engine moves is found. Taking the same engine, the pistons 17 inches diameter, give 226.98 square inches area. This is multiplied by the mean effective pressure of steam, giving 226.98 × 80 = 18158.4 pounds pressing on each piston through the whole stroke of four feet.
| 18158.4 | |
| 2 | pistons. |
| ———— | |
| 36316.8 | |
| 4 | feet of stroke. |
| ———— | |
| 145267.2 |
These figures show that 145267.2 foot-pounds are exerted during each turn of the wheel, whose circumference is 17.8024; therefore, 145267 ÷ 17.8024 = 8160, the number of pounds exerted through each foot moved.
There is still another way of figuring out tractive power by calculating the rotational force and the leverage through which it is applied to the locomotive wheels. The pressure of steam on the piston is productive of two strains on the crank-axle. When the crank is on the dead center, the pressure upon the crank-pin forces the axle against the axle-box without causing any tendency to rotate: but, when the crank-pin is on the quarter, the full leverage of the crank is available for rotation; the leverage increasing gradually as the crank gets away from the center. The mean effort upon the crank-pin during one revolution of the crank is to the effort of the piston as .6366 is to 1.