actual facts, of which we are still in ignorance: ================+===+===+===+===+===+====+====+====+====+====+==+==== Velocity of | | | | | | | | | | | | trains in miles | | | | | | | | | | | | per hour | 5 | 10| 15| 20| 25| 30 | 35 | 40 | 45 | 50 |60| 70 ----------------+---+---+---+---+---+----+----+----+----+----+--+---- Resistance on | | | | | | | | | | | | straight line in| | | | | | | | | | | | lbs. per ton (of| | | | | | | | | | | | 2,000 lbs.) |6.1|6.6|7.3|8.3|9.6|11.2|13.1|15.3|17.8|20.6|27|34.6 ================+===+===+===+===+===+====+====+====+====+====+==+==== Now, if we want to get the resistance at 30 miles an hour of a train of ten cars weighing each 20 tons, the calculation would be 10 × 20 × 11¹⁄₄ = 2,250 lbs. This will give the resistance on a level and straight track. On an ascending grade the resistance is greater than that given above, because, besides pulling the car horizontally, it is necessary to raise it vertically a distance equal to the ascent of the grade. Thus if we have a grade with a rise of forty feet in a mile, the amount of energy required to simply raise the weight of a car would be equal to its weight in pounds multiplied by the vertical height of the ascent. Thus, supposing a car which weighs 40,000 lbs. to be run one mile on a grade of forty feet ascent in that distance, then the energy expended in simply raising the car will be equal to 40,000 × 40 = 1,600,000 foot-pounds. Now, if it was necessary to raise that weight by a direct vertical lift or pull, it would require a force equal to or a little greater than the load to do it. But in pulling a car or train up a grade, which is an inclined plane, the force, which is the locomotive, instead of being exerted through the vertical distance is exerted through the horizontal distance, which in this case is one mile, or 5,280 feet. Therefore, if we divide the number of foot-pounds of energy required by the distance through which the power is exerted, it will give us the force exerted through one foot. That is, 1,600,000 --------- = 151.5 lbs. 5,280 The resistance due to the ascent alone of a train on a grade or incline can therefore be calculated by MULTIPLYING THE WEIGHT OF THE TRAIN IN POUNDS BY THE ASCENT IN ANY GIVEN DISTANCE IN FEET AND DIVIDING THE PRODUCT BY THE HORIZONTAL DISTANCE IN FEET. Thus in the above example the rate of the ascent is given in so many feet per mile; we therefore multiply by 40 and divide by 5,280, which is the number of feet in a mile. If the rate of the gradient had been given, as it sometimes is, as 1 in 132, we would simply have divided the weight of the train by the latter number. If we want to get the resistance per ton of train we substitute for its weight that of one ton in pounds; thus: 2,000 × 40 ---------- = 15.1 lbs. 5,280