How heavy an engine is needed to draw two hundred tons (including engine and tender) at twenty miles per hour over sixty feet grades?
The resistance on a level is
| 200 ×(20 × 20 171 + 8) = |
2,060 | pounds. |
| The resistance due to the grade | ||
| 200 × (60 5280 × 2240) = |
5,200 | pounds. |
| The resistance due to curves | ||
| 200 × 5= | 1,000 | pounds. |
| And the whole resistance, | 8,260 | pounds. |
| which multiplied by 6, is | 49,560 | pounds. |
or 22.1 tons, to which add 5 tons as the necessary load upon the truck, and the whole weight is 27.1 tons, which is the necessary weight of an engine to draw 200 tons over 60 feet grades, at 20 miles per hour.
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,
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,
nine tons nearly.
To which add five tons, and we have as the whole weight, fourteen tons.
Fig. 158.
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