CH. XI DESIGN OF RAILWAY MOTORS 243 



acceleration; hence ^ = A . ,, and 7/=^ '-I . c \ In- 

 serting the values of k l and k 2 we have 



TT Q.OQ /I? ,, T?\ 1 /I 1 "H 



XI = O O ( Ji CtJl) - VT^ (Ill), 



c a v 2 M 2 



where c l is the total starting current, and e,, the current 

 available for acceleration. Hence H varies inversely as 

 c,,!T 2 . Now the larger we make M at the start, the larger 

 will be the current c u , since the retarding torque is constant. 

 It is thus of great importance to increase M at the start. 



In the previous example the ratio of the induction 

 factors is 2 to 1 , and the ratio of the accelerating currents 

 is 1-3 to 1, giving 5-2 as the ratio of the heat losses. By 

 actual measurement of the diagrams this ratio is 5'25. 



When the current representing the heat loss has been 

 deducted from the total current at any instant, the 

 remainder represents the expenditure of energy in pro- 

 ducing acceleration and overcoming train resistance. The 

 proportion of these two can be obtained from the curve of 

 total torque, since that tells us how much is being used 

 for accelerating and how niuch for overcoming train 

 resistance at any speed. The curves od in Figs. 58 and 59 

 have been constructed in this way, thus dividing the 

 remaining area of the current curve into one portion, 

 shaded in the figures representing the energy expended 

 in acceleration, and a second portion representing the 

 expenditure of energy in overcoming the train resistance. 

 The latter of these two areas is equal in the two figures, 

 while the acceleration energy is a little greater for the 

 constant induction motor, the final speeds being 25 and 

 23'2 feet per second respectively. 



E 2 



