PART XX._Catechism of the locomotive

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:

================+===+===+===+===+===+====+====+====+====+====+==+====

per hour | 5 | 10| 15| 20| 25| 30 | 35 | 40 | 45 | 50 |60| 70

----------------+---+---+---+---+---+----+----+----+----+----+--+----

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.