Or, generally,
| Let W | = | Weight of engine, tender, and train, in tons, |
| Let V | = | Speed in miles per hour, |
| Let a b | = | Fraction expressing the grade, |
| Let c | = | Resistance, in pounds per ton due to the sharpest curve, which, assume as 5 lbs., as we have no reliable data, |
and we have, as the weight of the engine,
[W(V2
171 + 8) + a
b × 2240 + 5]6/2240 =
weight of engine exclusive of weight on truck.
If we assume the adhesion as one fourth of the weight on the drivers, and load 150 tons, speed twenty miles per hour, and grade forty feet per mile, the above formula becomes,
[150(20 × 20
171 + 8) + (40
52802240) + 5]4/2240 =
nine tons nearly.
To which add five tons, and we have as the whole weight, fourteen tons.
