Now, to find the traction of the locomotive, we multiply the length of our crank, which is 12 inches, by .6366, which gives 7.64 as the length of the crank receiving constant pressure from the pistons. The aggregate steam-pressure has already been found to be for each piston, 18158.4 pounds. This is multiplied by 7.64; and the product is divided by the radius of the wheel, which is 34 inches, giving as a quotient 4080, being the power exerted on one side of the engine. That, multiplied by 2, gives 8160 pounds,—the same as was found by the other two methods.
HORSE-POWER OF LOCOMOTIVES.
When people wish to find the horse-power developed by a locomotive, the indicator is generally employed, and the calculations made from the diagram. Others figure out locomotive horse-power as they do that of other engines. By this method, the engine whose traction we have been investigating might have the horse-power calculated as follows:—
| 226.98 | square inches piston area. |
| 80 | pounds mean pressure. |
| ——— | |
| 18158.4 | piston pressure. |
| 4 | feet piston travel each revolution. |
| ——— | |
| 72633.6 | |
| 2 | cylinders. |
| ———— | |
| 145267.2 | |
| 145 | revolutions per minute. |
| ———— | |
| 20063944 | ÷ 33000 = 608 horse-power when running at a speed of 30 miles an hour. |
No allowance has been made for frictional losses in any of these calculations.
One horse-power is equivalent to the work performed in raising 33,000 pounds one foot high in one minute. One pound raised 33,000 feet high, or 330 pounds raised 100 feet high, would amount to the same thing. One horse-power is usually spoken of as 33,000 foot-pounds; and engineers in this country always calculate work by foot-pounds,—that is, so many pounds raised a certain number of feet. To indicate the capacity of any prime motor, the foot-pounds of work it is capable of raising in a given time must be stated. Although the work is often done without any thing being raised vertically, the power represented would be capable of raising the equivalent weight in the stated time.
FORMULAS OF TRAIN RESISTANCES.
The work which a locomotive performs in pulling a train is expended in overcoming the resistances due to wheel-friction, gradients, curves, and atmospheric or wind pressure. Formulas have been propounded for calculating all train resistances, but they are utterly untrustworthy for American railroad trains. The best known formula of this kind is that given by D. K. Clark in his Railway Machinery. One calculation will show how misleading its figures are when applied to American railroad train resistances. Figured by the Clark formula, the total resistance per ton of a passenger train running at a speed of fifty miles an hour on a straight level track, is 22.6 pounds. By accurate records with his dynagraph car, Professor P. H. Dudley found the total resistances of an express train running at a speed of fifty-one miles an hour, to be 11 pounds. The resistances are so much different under different conditions, that nothing closer than a loose approximation can be calculated of the work done by a locomotive, unless indicator or dynamometer tests are made.