The two forms of train resistance, which, under some circumstances, are the greatest resistances to be overcome by the engine, are the grade and inertia resistances, and fortunately both of these resistances may be computed with mathematical precision. The problem may be stated as follows: What constant force P (in addition to the forces required to overcome the grade and various "frictional resistances, etc.) will be required to impart to a body a velocity of v feet per second in a distance of s feet? The required number of footpounds of energy is evidently Ps. But this work imparts a kinetic energy which may be expressed by Wʋ2/2g. Equating these values, we have Ps = Wʋ2/2g, or P=Wʋ2/2gs.........(2)
The force required to increase the velocity from v1 to v2 may likewise be stated as P=Wʋ2/2gs.(ʋ22 - v12). Substituting in the formula the values W = 2000 lbs. (one ton), g=32.16, and s=5280 feet (one mile), we have P = .00588 (ʋ22-ʋ12).
Multiplying by (5280ö3600)2 to change the unit of velocity to miles per hour, we have P = .01267 (ʋ22-V12).
But this formula must be modified on account of the rotative kinetic energy which must be imparted to the wheels of the cars. The precise additional percentage depends on the particular design of the cars and their loading, and also on the design of the locomotive. Consider, as an example, a box car, 60,000 lbs. capacity, weighing 33,000 lbs. The wheels have a diameter of 36", and their radius of gyration is about 13". Each wheel weighs 700 lbs. The rotative kinetic energy of each wheel is 4877 ft.-lbs. when the velocity is 20 miles per hour, and for the eight wheels it is 39,016 ft.-lbs. For greater precision (really needless) we may add 192 ft.-lbs. as the rotative kinetic energy of the axles. When the car is fully loaded (weight, 93,000 lbs.) the kinetic energy of translation for 20 miles per hour is 1,244,340 ft.-lbs.; when empty (weight, 33,000 lbs.) the energy is 441,540 ft.-lbs. The rotative kinetic energy thus adds (for this particular car) 3.15% (when the car is loaded) and 8.9% (when the car is empty) to the kinetic energy of translation. The kinetic energy which is similarly added, owing to the rotation of the wheels and axles of the locomotive, might be similarly computed. For one type of locomotive it has been computed to be about 8%. The variations in design, and particularly the fluctuations of loading, render useless any great precision in these computations. For a train of "empties" the figures would be high, probably 8 to 9%; for a fully loaded train it will not much exceed 3%. Wellington considered that 6% is a good average to use (actually used 6.14% for "ease of computation"), but considering (a) the increasing proportion of live load to dead load in modern car design, (6) the greater care now used to make up full train-loads, and (c) the fact that full train-loads are usually the critical loads, it would appear that 5% is a better average for the conditions of modern practice. Even this figure allows something for the higher percentage for the locomotive and something for a few empties in the train. Therefore, adding 5% to the coefficient in the above equation, we have the true equation:
P = .0133(V22-V12),.....(3) in which V2 and V1 are the higher and lower velocities respectively in miles per hour, and P is the force required per ton to impart that difference of velocity in a distance of one mile. If more convenient the formula may be used thus:
P1=70.224/S (V22-V12), .... (4) in which S is the distance in feet and P1 is the corresponding force.
As a numerical illustration, the force required per ton to impart a kinetic energy due to a velocity of 20 miles per hour in a distance of 1000 feet will equal:
P1=70.224 (400-0)/1000 = 28lbs., which is the equivalent (see § 117) of a 1.4% grade. Since the velocity enters the formula as V2, while the distance enters only in the first power, it follows that it will require four times the force to produce twice the velocity in the same distance, or that with the same force it will require four times the distance to attain twice the velocity.
As another numerical illustration, if a train is to increase its speed from 15 to 60 miles per hour in a distance of 2000 feet, the force required (in addition to that required for all the other resistances) will be P=70.224 (3600-225)/2000 = 118.50 lbs. per ton.
This is equivalent to a 5.9% grade, and shows at once that it would be impossible, unless there were a very heavy down grade, or that the train was very light and the engine very powerful.