is

the acceleration at that point, and the tangent at P2 equal to the tangent of the angle st, that is to say, is proportional to the accelerating torque, and therefore to the acceleration at P

P2

In this way we can construct the whole of the acceleration curve, taking care to keep the points Pil closer together as the curve becomes flatter. The curve will finally approximate to the speed of 21.2 feet per second.

From the curve C we can obtain the distance travelled at any time. Draw a vertical line through any given time, cutting the acceleration curve; the area enclosed between this line, the curve, and the time base, will be the distance travelled from the start. Find these distances for successive seconds, and then plot them as a curve Don the time base, taking vertical ordinates as distances travelled.

From the acceleration curve we can deduce the current curve E, showing on a base of seconds the current taken from the line. This will begin at 150 amperes, and will be straight for 34.6 seconds, when it will rapidly slope down and approximate to the uniform value 70 amperes. The area of this curve up to any point on the time base will give the energy drawn from the line.

The curves in Fig. 41 record the results of tests made on the South London Railway with the motors previously described. They are constructed from data given in the 'Proceedings of the Institution of Civil Engineers,' vol. cxii. p. 246. The track is not level, as shown by the outline of the grades, which is drawn on a time base, so that the position of the train on any grade can be seen at The distances travelled in 80 and 120 seconds are

once.

400 and 670 yards respectively, while the first 100 yards is covered in 36 seconds. The weight of the train is 35 tons.

The down grade at the station exit has been constructed to assist the locomotive in starting the train. In this case the acceleration is increased by about 20

per cent.

169

CHAPTER VIII

THE FORCE FACTOR

THE method of rating a motor in horse-power gives us no indication of its ability to accelerate, although the function of accelerating may be the most important that it is called upon to perform. We shall find it convenient to be able. to describe a motor in terms of a unit that shall be independent of speed and that will tell us to what extent a motor is able to accelerate under given conditions. This must be a force unit and not a power unit.

If a pulley of d centimetres diameter is placed on the shaft of a dynamo of induction factor M, carrying a current of c amperes, the tangential force at the rim of the pulley is given by

If d=107 centimetres, this may be written.

π

T=Mc dynes

.(91).

⚫(92).

The force of a dynamo may thus be defined as a force of Mc dynes at the rim of a pulley 107 centimetres in circumference. We shall call Mc the force factor of the dynamo.

Since M is equal to e, the induced volts, divided by n, the number of revolutions per second, it follows that

се

Mc= but ce

is the work done per second, hence Me is the

work done per revolution. If then we are given Mc and M, we can find the work done per second, that is the rate of working, or the power, by simply multiplying the product Me by the number of revolutions per second.

The ratio of the watts to the revolutions per second, sometimes called the mass factor,' has been used as a basis for the comparison of dynamos, its true significance not, however, being perceived. The fact seems to have been overlooked that the ratio of the induced volts to the revolutions per second is a constant, so long as the number of useful lines per pole remains unaltered, being, in fact, what we have termed the induction factor.

While the force factor and the so-called 'mass factor' are one and the same thing, the latter is expressed in a way involving the idea of speed, and consequently of power, while the former indicates the real nature of this ratio, showing that it is independent of speed, and therefore not a power unit, much less a mass unit, but a force

unit.

Example 42.-A four-pole dynamo with armature parallel connected, giving p=1, has 440 surface conductors, with 16·1 × 106 lines per pole. The induction factor is 77 and the force factor for 600 amperes is 77 × 600=46.2 kilodynes; the power at 450 r.p.m. is 46.2 x 7.5=346 kilowatts. The dynamo is a General Electric four-pole railway generator.

Example 43.--A ten-pole dynamo with armature parallel connected has 1,440 surface conductors and