the changes of E, the speed curves are parallel to one another. If the induction factor is altered as well as the load, the speed curves will be differently inclined to the axis of speed (see Fig. 19).

We shall now consider the conditions of motion when the load is a function of the speed. Suppose that the resistance to motion increases with the speed, and that it can be represented by a curve, where vertical ordinates represent the speed in revolutions per second, and whose horizontal ordinates represent the torque resisting the motion.

For uniform speed the torque due to the current assisting the motion must be equal to the torque resisting the motion. Hence there will be a certain speed at which the motor must run when it is carrying a certain current. For, if M is constant, a given current means a given assisting torque; and there is only one speed at which the motor can run for that torque. If the speed is greater than that given by the curve, the resisting torque will be greater than the assisting torque due to the current, and the motor will be retarded, and if the speed is less than that given by the curve, the resisting torque will be less than the assisting torque, and the motor will accelerate.

If we place such a motor in series with a circuit in which the current varies, the speed of the motor will depend upon the current. The motor would then act as a current meter, and the total revolutions at the end of any time would depend upon the current that had been passing during that time. If the curve of torque and speed is straight and passes through the origin, the speed is proportional to the current. In practice it is not

possible to make the curve straight nor to make it pass through the origin.

If ty is the torque required to overcome the initial

friction, an initial current c·71 t will be required to

=

M

start the meter. If n1 and n, are the speeds for currents c, and c2, we shall have, assuming the torque curve to be straight,

For example, suppose that the curve is straight and inclined to the axis of current or torque at an inclination of 1.98, i.e. the ratio of any vertical ordinate, in terms of revolutions per minute, to the corresponding horizontal ordinate, measured up to the point where the curve cuts the current axis, in amperes, is 1.98. Suppose, also, that the starting current is 0.2 ampere. When the current is 3 amperes the speed is 5.55 r.p.m.; if the starting current were nothing it would be 5.95 r.p.m., the meter is therefore running 6.7 per cent. too slow. When the current is 15 amperes the speed is 29-3 r.p.m., it should be 29.7 r.p.m., the meter is now running 1.3 per cent. too slow. We see then that the accuracy increases with the total current.

Curves for three types of current meter are given in Fig. 17. The curves in the figure give (1) the speed, and (2) the ratio of the speed to the total current, both on a base of total current. In the Hookham meter the starting current is 0.2 ampere, and the maximum current 25 amperes. In the Perry meter the starting current is 0.25 ampere, the maximum current 20 amperes. In the

6

Thomson meter the starting current is 0.14 ampere, the maximum current 10 amperes. The curves are obtained from data given by Mr. G. W. D. Ricks, in a paper read before the British Association, August 1897; see the Electrician' of August 27, 1897.

If the torque curve is not straight, a further error will be introduced, tending to make the meter run slower or faster according as the curve is concave or convex to the current axis. In all these meters the curves are concave to the current axis, and for small speeds the error due to the starting torque is the greater of the two. For high speeds, when this error is very small, the error due to the bend of the curve becomes apparent, and the accuracy curve turns down again.

Hitherto we have supposed M to be constant. If M varies, the speed will depend on the product of M and c, so that ife is fixed and M increased, the speed will increase with M, or if both c and M increase, the speed will increase with their product.

If the magnetising coil is wound as a shunt across the main line terminals of the circuit in which the current is flowing, the current in the shunt will be proportional to the tension, and if the magnetic circuit contains no iron the induction curve will be a straight line, i.e. the value of M will be proportional to the tension, hence the speed will be proportional to the watts in the circuit. This is the principle of the Thomson watt-meter. For a given current the speed increases directly in proportion to the tension.

We shall now consider the case when two dynamos are mechanically coupled so that they rotate at the same speed, and have their main terminals connected in

parallel to the same line. The magnets should be shunt wound and separately excited, and provided with rheostats so that the induction factor of each dynamo can be adjusted to any desired value and kept constant. The connections will be made so that the dynamos tend to turn in the same direction when both are acting as motors.

We have further to suppose that by means of a pulley fitted on to one of the dynamo shafts and connected by a belt to a third dynamo, or by simply coupling a third dynamo direct on to the shaft line, we can produce a load of any required amount or of any required sign; in other words, we suppose that we can vary the mechanical torque on the shaft, making it either positive so that the dynamos have to act as motors, or negative so that they act as generators. Any arrangement will act provided the motion can be assisted or retarded at will.

The two dynamos may be distinguished by the letters A and B. Suppose that dynamo A has an induction factor of 6 and a resistance of 1·2 ohms, and that dynamo B has an induction factor of 4 and a resistance of 1.09 ohms. Let the tension of the line be 120 volts.

Set off ab in Fig. 18 equal to 100 amperes, the maximum current in A. Set off ad equal to 110 amperes, the maximum current in B. Let af represent 1,200 r.p.m., the speed of A when its induced tension is equal to the tension of the line. Let ag represent 1,800 r.p.m., the speed of B when its induced tension is equal to the tension of the line. Join fb and gd. These will be the speed curves of the two dynamos on a base of current in the

armature.

Since the dynamos are mechanically coupled, we can at once find from the diagram the current that each is