The question is to find the vertical rise, consuming an amount of power equal to that expended upon the horizontal unit of length. This has been estimated by engineers all the way from twenty to seventy feet. For simple comparison it does not matter much what number is used if it is the same in all cases; but to find the equivalent horizontal length to any location, regard must be had to the nature of the expected traffic.
The elements of the problem are, the length, the inclination or the total rise and fall, and the resistance to the motion of the train upon a level, which latter depends upon the speed and the state of the rails and machinery.
From chapter XIV. we have the following resistances to the motion of trains upon a level:—
| Velocity, in miles, per hour. | Resistance, in lbs. per ton. |
|---|---|
| 10 | 8.6 |
| 15 | 9.3 |
| 20 | 10.3 |
| 25 | 11.6 |
| 30 | 13.3 |
| 40 | 17.3 |
| 50 | 22.6 |
| 60 | 27.1 |
| 100 | 66.5 |
The power expended upon any road is of course the product of the resistance per unit of length, by the number of units. Calling R the resistance per unit upon a level, and R′ the resistance per unit on any grade, and designating the lengths by L and L′, that there shall be in both cases an equal expenditure of power, we must have
RL = R′L′,
whence the level length must be
L = L′R′
R.
Thus assuming the resistance on a level as twenty lbs. per ton, that on a fifty feet grade is
20 + 50
5280 of 2240, or 20 + 21.2 or 41.2,