242 THE DIRECT-CURRENT MOTOR < II XI 



speed and finding the resistance in the circuit, and thru 

 multiplying this by the square of the current. If the 

 heat watts is divided by the tension of the line we obtain 

 the part of the total current that represents the loss due to 

 heat. In Figs. 58 and 59 this current has been set off from 

 a horizontal line passing through 45 amperes ; the points 

 obtained lie on a straight line passing through the origin 

 and a point a, where al represents the heat loss when the 

 current of 45 amperes is passing through the resistance 

 of the motor only. The area fabc represents the heat loss 

 during the period of constant acceleration. 



Beyond the point a the current used to make up the 

 heat loss is veiy small compared with the whole current, 

 and is neglected in the following discussion. 



The heat loss can be predetermined, since it is very 

 nearly one half of the area of the current curve up 

 to the point at which the starting rheostat is all out. 

 There is not much difference in the final speed of the two 

 motors, the kinetic energy is. therefore, nearly the same. 

 The energy expended in overcoming the train resistance is 

 the same ; this should be checked by comparing the areas 

 marked train resistance in the two diagrams. Hence the 

 difference in energy expenditure is nearly represented by 

 the difference of the heat areas. We thus see the import- 

 ance of setting back as far as possible the point at which the 

 starting rheostat is all out. This is what the series wind- 

 ing effects for us. 



If f j is the time in seconds during which the maximum 

 current c, is flowing, the energy, H, expended in heat 

 is nearly ^ Ec^t^ If s l is the speed at which the starting 



rheostat is all out, s, = 'j., and - 1 = A-,.W. being the 



-'/ ', 



