the machine is at rest, shows that the force produced by a current in the armature is independent of the motion of the armature.

In Fig. 4 will be found the results of a torque test made with an Edison 12 K.W. dynamo. This machine has 200 conductors on the armature, and is designed to carry 96 amperes at 125 volts, running at 1,500 revolutions per minute. The conductors are laid on the surface of the iron core. The torque is plotted on a base of

amperes in the armature.

The current in the magnets was kept constant at 4.5 amperes. The leverage was 30 inches. A pull of 1.4 pounds due to friction was observed with no current in magnets or armature; this was not increased when the magnets were excited, showing that the hysteresis torque was inappreciable. The induction factor is obtained from Equation 5.

After the experiment a current of 78 amperes was passed through the armature, and a fall of potential of 3.74 volts observed, giving the resistance from brush to brush as 0.048 ohm.

In Fig. 5 let ab represent the length πr, where r is the mean distance of the axes of the conductors from the centre of the shaft; ab is thus equal to one half of the armature circumference measured round the circle of radius r.

1

A

Set off along ab points 1, 2, 3, 4 ... to represent the positions of the conductors. The distance from point to point will be of the complete circumference, assuming for the present that the conductors are laid side by side. At each point draw a vertical line, to represent by its length the number of lines of force per square centimetre

entering the armature across the cylinder of radius, on the surface of which the axes of the conductors are supposed to be lying, as shown in Fig. 2. The ends of these vertical ordinates will form a curve, the shape of which will depend upon the variation of the intensity of magnetisation from point to point round the armature. The curve will generally be quite irregular and follow no definite law; methods of obtaining it experimentally will be given later on. We shall call such a

curve the

magnetisation curve, and the points a, b, the neutral points; these points will be a distance π apart if the field is symmetrical.

The area of each element of the curve being the product of the intensity at any point into the distance over which the intensity may be assumed to be constant, will represent dn, the number of lines of force entering the armature between two adjacent conductors, and the whole area of the curve will represent N, the whole number of lines of force entering the armature from one pole.

In Fig. 6 magnetisation curves are given for a high tension arc light dynamo. This machine has a ring armature with 80 sections and 47 turns per section, making A=3,760. The areas of the curves for 3, 6, and 9 amperes in the magnets are as 100, 159, and 178. The corresponding values of N as obtained by a method to be presently described are 1.05 x 106, 1.68 x 106, and 1.89 × 106.

Amperes

6 Amperes

Let us suppose that the armature is turned through an

360
A

angle equal degrees. Each conductor will then sweep

through the space that lies between it and the next conductor. The number of lines of force occupying this space is represented by the area of the strip of the curve between the two conductors; hence each conductor will cut a number of lines of force represented by the area of the strip next to it. Let dn be the area of any such strip, and dt the time occupied in turning the armature through

1

Α

dn

dt

of a revolution, then the rate of cutting lines of force is

; there will therefore be an induced tension at the ends

dn of each conductor equal to c.g.s units of tension, or dt

dn10-8 volts, so that if each conductor were separate from

dt

all the other conductors, we should observe a reading of dn10-8 volts on a voltmeter connected to the ends of

dt

any one conductor, dn depending upon the position of the conductor.

In practice, if the brushes are placed at the neutral points ab, each conductor is in series with all the other conductors round one side of the armature between the brushes, so that the effect is as if we had one long conductor cutting lines at different rates along its length; the total number of lines cut by this imaginary long conductor in dt seconds is represented by the whole area of the curve;

hence the whole tension induced is 10-8 volts.

N. dt

If the armature is rotating uniformly at n revo

lutions per second, dt=

1

nA

; hence the tension induced

in all the conductors in series between the neutral points, and observed on a voltmeter connected by brushes to these points, will be given by the equation

If the dynamo has more than two poles, and if the conductors between two adjacent neutral points are placed in series with these between two other neutral points, p

representing as before the number of polar divisions of the armature thus connected in series, we may write :

We have assumed that the conductors are spaced evenly round the armature. A similar proof would hold good if the conductors lay two or more deep, with equal spaces between those in each layer.

If in Equation 8 we insert M in place of pAN 10−8, we get

e=Mn

..(9).

From this we see that the induced tension in a dynamo is given by multiplying the induction factor by the number of revolutions per second. The induction factor may thus be defined as the induced volts divided by the number of revolutions per second.

We must not however suppose that because M is thus defined it depends on the motion of the armature. Equation 6 shows that M depends only upon p, A, and N, and does not in any way involve the speed.

We have here a second method for finding the value of the induction factor-namely, to drive the dynamo as a generator, and observe the induced tension at the terminals of the machine, and the revolutions per second.

In making this test we must be careful to insure that the tension recorded is the true induced tension, expressed by Equation 7. The voltmeter will only indicate this tension provided that the brushes are placed precisely at the neutral points, for in this position only will all the conductors between two brushes be generating a tension of like sign. Any forward or backward lead will place two sets of conductors with opposing tensions between two