250 TIIK DIRECT-CURRENT MOTOR CH. XI 



we can make use of the record of the current taken when 

 running at a uniform speed on a level with the existing 

 equipment. Taking the average of a number of tests, it 

 appears that when running on a level at a uniform speed, 

 the motors take 75 amperes. The induction curve of 

 these motors is given in Fig. 61, and for this current .M =30, 

 giving us a total torque on the motor shaft of 3,180 inch- 

 pounds, the speed being 15'1 miles per hour. Experi- 

 ments previously made with these motors showed that 

 with 75 amperes the torque available for useful effort was 

 2,310 inch-pounds, giving 73 "4 per cent, mechanical 

 efficiency. Now 2,340 inch-pounds is equivalent to 450 

 pounds of horizontal pull on 33-inch wheels, with r=3 - 18. 

 Hence the train resistance is 450 pounds per motor, or 10 

 pounds per ton. 



Assuming that the retarding resistances are the same 

 for all speeds, we see that the torque on the motor shaft 

 at full speed is 3,180, and since 3/=35 - 9 the current when 

 running at full speed is 63 amperes per motor. Since 

 the drop at full speed has been fixed at 5 volts, the 

 resistance of the motor must be 0'0795 ohm. 



We have thus found one point on the induction curve, 

 namely. .V = 35'9 for 63 amperes. Suppose now that 

 the limit of weight imposed gives the maximum value of 

 M at 72 and that we are not at liberty to increase r or 

 diminish d. With this value of M the current required 

 to overcome the retarding torque is 31 '5 amperes, so that 

 the total current at the start must be 226 + 31 '5, or say 

 257 amperes. This gives us a second point on the induction 

 curve, namely. M=72 for 257 amperes. These points are 

 plotted at a and b in Fig. 61. We will suppose that A is 



