PART XX.
THE RESISTANCE OF TRAINS.
QUESTION 407. _What is meant by the resistance of trains or cars?_
_Answer._ It is the power required to move them on the track. Thus if a rope, fig. 219, was attached to a car at one end, and the other passed over a pulley, _a_, and a sufficiently heavy weight, _W_, was hung on the end of the rope, it would move the car. The weight _W_ would then be equal to the resistance of the car.
[Illustration: Fig. 219.]
QUESTION 408. _How can the resistance of cars under different circumstances be determined?_
_Answer._ It has been found that it takes a force of about 6 lbs. per ton (of 2,000 lbs.) to move a car slowly on a level and straight track after it is started. That is, if a car weighs 20 tons and a rope, fig. 219, is attached to it at one end and the other passed over a pulley, _a_, with a weight, _W_, suspended to it, it will require a weight equal to 20 × 6 = 120 lbs. to keep the car moving slowly. If two cars of the above weight were coupled together, it would require twice 120 or 240 lbs., and if three were attached to each other, three times 120, or 360 lbs., and so on. In other words, MULTIPLYING THE TOTAL WEIGHT OF THE CARS IN TONS (OF 2,000 LBS.) BY 6 WILL GIVE US THEIR RESISTANCE, OR THE FORCE REQUIRED TO KEEP THEM MOVING ON A LEVEL AND STRAIGHT TRACK AT A SLOW SPEED AFTER THEY ARE STARTED. The resistance is represented by the weight above, and the locomotive must exert a force equal to that weight to keep the train moving. As the speed increases the resistance increases, as is shown by the following table. It should be stated here, however, that our knowledge regarding this whole subject of the resistance of American cars and trains is exceedingly inaccurate and imperfect, and the data given in the books are nearly all based on experiments made in Europe, with cars of a different construction from those used here. There is reason for believing, however, that the resistance of American cars is less than that of European cars, and we have assumed it to be 6 lbs. per ton on a level at very slow speed, which is less than the resistance which is usually given, but the following figures should be regarded merely as an approximation to the 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
If, now, we have the resistance which is due to the _ascent or gravity alone_, we must add to this the resistance on a straight and level track, at the speed at which the train runs, in order to determine the total resistance on the grade. On a level road at a speed of 5 miles per hour it would be 6.1 lbs. per ton, so that on a grade of forty feet to a mile at that speed the resistance would be 6.1 + 15.1 = 21.2 lbs., per ton, and at 10 miles it would be 6.6 + 15.1 = 21.7 lbs., and at 30 miles per hour on the grade the resistance would be 11.2 + 15.1 = 26.3 lbs. per ton. To get the total resistance on a grade for any speed, we ADD THE RESISTANCE FOR THAT SPEED ON A STRAIGHT AND LEVEL LINE TO THE RESISTANCE DUE TO THE ASCENT ALONE. The resistances for various rates of speed and grades has been calculated, and is given in the table in the appendix.
The top horizontal row of figures of that table gives the rates of speed. The left-hand vertical row gives the rise of grade in feet per mile. The resistance for any given grade and speed is given where the vertical row of figures under the rate of speed and the horizontal row opposite the rise of the grade intersect each other. Thus, for a grade of 30 feet per mile and a speed of 45 miles per hour, we follow the vertical column under the 45 downward, and the horizontal column opposite 30 to the right, and where the two intersect the resistance, 29.1 lbs. is given.
QUESTION 409. _What effect do curves have on the resistance of trains?_
_Answer._ They increase the resistance, but in what proportion or to what degree is not known accurately. European authorities say that the resistance is increased, over what it would be in a straight and level line, about 1 per cent. for every degree of the curve occupied by a train. It is probable, however, that the resistance of American cars, which nearly all have double trucks, is not so great on curves as that of European cars, which nearly all have long and rigid wheel-bases, and whose wheels therefore can not adjust themselves so easily to the curvature of the track as they can when the American system of double trucks is used.
QUESTION 410. _What is meant by a degree of a curve?_
_Answer._ In order to measure circles, they are all supposed to be divided into 360 equal parts, which are called degrees. One degree of a curve is therefore ¹⁄₃₆₀ of a complete circle; but if the curve has a long radius, one degree of such a curved track will be longer than one degree of a curve with a short radius, but each will have the same amount of “bend” or curvature. It is this latter which increases the resistance of trains, and the greater the number of degrees of a curve occupied by a train of cars, the greater will be the “bend” of the track, and therefore the greater the resistance.
QUESTION 411. _What other causes affect the resistance of trains?_
_Answer._ The condition of the track and the force and direction of the wind. On a rough track the resistance is very much greater than on a smooth one, and a strong head-wind makes it much more difficult to pull a train than it is in calm weather.