When a train starts from rest and acquires its normal velocity, it overcomes not only the usual track resistances (and perhaps curve and grade resistances), but also performs work in accumulating a large amount of kinetic energy. Such work is not necessarily lost. In fact there need not be the loss of a single foot-pound of such energy, provided it is not necessary to dissipate the energy by the application of brakes. If for a moment we consider that a train runs without any friction, then, when running at a velocity of v feet per second, it possesses a kinetic energy which would raise it to a height of h feet, when h = v2/2g,, in which g is the acceleration of gravity which equals 32.16 feet per second in a second. Still ignoring friction, the train would climb a grade until it had attained an elevation of h feet above the point where its velocity was v. When it had climbed a height of h' feet (less than h) it would have a velocity V1=\/[2g(h-h')]. As an illustration assume that v = 30 miles per hour =44 feet per second.

Then h=v2/2g=30.1 feet. Still assuming that there is no friction, the kinetic energy in the train would carry it up a grade until it had attained an elevation of 30.1 feet, or it would carry it for two miles up a grade of 15 feet per mile or half a mile up a grade of 60 feet per mile. When the train had climbed 20 feet there would still be 10.1 feet left of velocity head, and its velocity would be v = \/[2g(10.1)] = 25.49 feet per second = 17.4 miles per hour. But these figures must be slightly modified, on account of the revolving-wheels, of the train, as already discussed in § 120. When train velocity is being acquired part of the work done is spent in imparting the energy of rotation to the driving-wheels and various truck-wheels of the train. Since these wheels run on the rails and must turn as the train moves, their rotative kinetic energy is just as effective (so far as it goes) in being transformed back into useful work. The proportion of this rotative energy to the kinetic energy of translation has already been computed in § 120, in which the corrective value of 5% has been adopted.

Since v equals 5280/3600 V = 1.4667 V, in which v equals the velocity in feet per second and V equals the velocity in miles per hour, and since v2 equals 2.151 V2 we may write Velocity head v2 in ft. per sec. 2.151V2 in m. per h. = = 64.32 = 64.32 0.03344 V2

Adding 5% for the rotative kinetic energy of the wheels..............................=0.00167V2

The correct velocity head therefore..........=0.03511V2

On account of the great usefulness of these values, as explained later, the velocity-head for velocities varying from 10 to 50 miles per hour have been computed as shown in Table XX. Part of these figures were obtained by interpolation, and the final hundredth may be in error by one unit, but it may readily be shown that the final hundredth is of no practical importance. It is also true that the chief use made of this table is with velocities much less than 50 miles per hour.