CH. IX CONTROL 191 



the resistance 0'4 ohm. We find from Equation 21 that in 

 order to comply with the specification the motors must 

 have an induction factor of 44. The current required to 

 overcome friction will be 64' 4 amperes. Since the motors 

 are in series each will now carry the full current from the 

 line, namely, 200 amperes ; hence the current available 

 for acceleration is 135'6 amperes. 



STEP I. Motors in series. Starting rheostat of 1-7 

 ohms in circuit. Considering one motor and half the 

 weight of the car, we get from Equation 83 an initial 

 acceleration of 0'367 f.p.s. per second. The speed when 

 the rheostat is all out is 232 r.p.m., the time occupied is 

 91 seconds, and the distance travelled 1,518 feet. 



STEP II. Motors still in series. Rheostat all out. 

 Acceleration diminishing according to Equation 69. We 

 have T = 29'l seconds, ,= 0'24. Hence in fifteen seconds 

 the speed is 262 r.p.m., and the current is 145 amperes. 

 At 40 seconds the speed is 280 . r.p.m. and the current 

 110 amperes; at 90 seconds the speed is 302 '6 r.p.m. 

 and the current is 70 amperes. These results are also 

 shown in Figs. 46 and 47. 



We see that this method compares badly with that of 

 Case I., both in respect of the distance travelled in a given 

 time and of the energy expended. The rheostat is all out 

 at a lower speed, and the high time factor causes the 

 curve to bend over as the car speeds up. The final 

 current for two motors is 64'4 as compared with 59 '2 

 amperes in Case I. 



If we consider the car as ' started ' when the motors 

 have attained a speed 300 r.p.m., or 29*5 miles per hour, 

 we can find with a planimeter the area of the curve 

 giving 1 the work done when this speed is attained. The 



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