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 (5280รถ3600)2 to change the unit of velocity to miles per hour, we have P = .01267 (ʋ22-V12).