If we divide the total tractive power by the gross rating tonnage, we will have the available power in pounds per ton. If we subtract from this the tractive resistances, including grade resistance and the resistance due to curvature, the remainder will be the force which is available for acceleration. If these resistances are more than the available power per ton, the difference will be negative and will indicate a retarding force.
On the same principle. as used in computing gravity resistance (§ 117), we may say that a given accelerative force per ton will be able to alter the potential head (or velocity head) by a proportional amount in a given distance. If P = the accelerating force per ton, then P:2000::(h2-h1):s, in which h2 and h1 are the two velocity heads and s the distance. If we say that h=0.03511 V2 (see §124), and substitute for V12 and V22 in Eq. 4 (§120) the values h1/.03511 and h2/.03511, we will have P= 2000/8 (h2-h1), or s= 2000/P (h2-h1), .(14) which is the same as the above equation. If our accelerating force is, say, 10 pounds per ton, then the distance required for a change of one foot of velocity head will be 2000 10 or 200 feet.
Assume that the consolidation locomotive which we have been discussing is loaded with a train-load of 2452 rating tons behind the tender, which is the capacity of that locomotive running for an indefinite distance at 7 miles per hour up a 0.4% grade. This loading (2452 rating tons) might be made up of an indefinite number of combinations of empty cars, loaded cars, or loaded and empty cars. The train having been started out from the terminal with this loading, we wish to compute its probable behavior on other grades and at different velocities. We will first study its behavior on a level grade. Since it is working in a manner which would carry the train up a 0.4% grade at 7 miles per hour, it will evidently gain velocity on the level grade. The total weight of train and engine is 2660 rating tons, on which the resistance due to traction is only 2660x2.6=6916 pounds. The tractive power of this engine at various velocities is as shown in the diagram Fig. 26. The tractive power for each velocity in miles per hour is as given in Table XXII. For example, at 10 miles per hour the tractive power is 26,400 pounds. Subtracting from this the power required for tractive resistance, which we will call 6900 for a round number, we will have 19,500 pounds available for acceleration. The velocity head at 10 miles per hour (as taken from Table XX) is 3.51; for 9 miles per hour it is 2.84. The difference is .67 foot. The total weight of our train in rating pounds equals 5,320,000. Therefore the distance required to increase the velocity from 9 to 10 miles per hour equals (see Eq. 14)
5,320,000 x .67 = 183, S=. 19,500 which means that the tractive power of the engine is such that it would increase the velocity of that train from 9 to 10 miles per hour in a distance of 183 feet. This figure, 183, is found in Table XXII in column 6, opposite the velocity of 10 miles per hour in column 1. The other numbers of column 6 are similarly obtained. The numbers of column 7 are obtained in each case by adding the sum of all the distances from zero, as given in column 6, and give the total distance from the origin that must be traveled before the locomotive working in that manner will attain the velocity as given in column 1. Considering the numbers in column 7 as ordinates, we may plot the acceleration curve marked "level" in Fig. 27.
To determine the behavior of this train on other grades, we will apply this same method to determine the other acceleration curves as given in Fig. 27. For example, the tractive power required of the locomotive at 7 miles per hour on a 0.4% grade is 28,200 pounds. At 5 miles per hour the tractive power of the engine is 29,100 pounds, and therefore the surplus is only 900 pounds (as given in column 8). The distance required for the difference of velocity head between 4 and 5 miles per hour, computed as before, equals about 1890 feet. Working out the other distances similarly, we would have the other numbers in column 9, and adding these partial distances we have the sum totals as given in column 10. It should be remembered, however, that this curve has but little value, since the computations depend on the assumption of a uniform resistance per ton, which is far from being true with very low velocities. The form of the curve, however, is very instructive, since it shows that when it is running up the 0.4% grade with a velocity less than 7 miles per hour its surplus accelerative power is very small; it must necessarily run a long distance before its velocity will materially increase, and theoretically it will require an infinite time to quite reach the velocity of 7 miles per hour. Speaking mathematically, the curve (+0.4) is asymptotic to the vertical line over 7.
In Table XXII the ordinates for the lowest of the acceleration curves ( - 1.0) has been worked out in columns 11, 12, and 13. In this case each rating ton assists the tractive power by a force of (20.0 - 2.6), or 17.4 pounds per ton. This makes a total added power of 46,284 pounds, which we call 46,300 for round numbers. Adding this constant to the tractive power of the locomotive, we have the net tractive power available for acceleration as shown in column 11. Applying these values similarly, we obtain the comparative short distances required to change the velocity heads by the amounts given in column 4, and in column 13 we have the total distance required to attain the given velocities.