174 THK DIRECT-CURRENT MOTOR CH. VIII 



We cannot alter I/ so long as the final speed is 

 specified, so we must double the 'accelerating current, 

 making the total current at the start 1 74 amperes. The 

 motor would then be described as having an induction 

 factor of 4'45, with an armature capable of carrying 174 

 amperes safely in other words, the force factor must be 

 775 dynes , or 15 per cent, greater than that required to 

 start up in two seconds. 



Example 47. A train and locomotive weighing 

 780 tons have to mount a grade of - 8 per cent, at 10*7 

 miles per hour. Friction and other retarding forces 

 amount to nine pounds per ton. The tension of the line 

 is 625 volts. Four gearless motors are used, permanently 

 connected in series, each having an internal resistance 

 of 0-0209 ohrn. Each motor has to move 195 tons at a 

 tension of 156-2 volts. The draw-bar pull per motor for the 

 grade is 3,490 pounds ; for friction, 1,755 pounds; allow- 

 ing 95 per cent, mechanical efficiency, we get 5,500 

 pounds as the required tractive effort per motor ; insert- 

 ing this value together with those for E, R and 8 in 



Equation 21, we find the value of " to be 2'32, v being 



(L 



unity. Take Jf=144, d = 62 inches. When running 

 at full speed, each of these motors would be doing 

 work at the rate of 158 horse-power, so that the total 

 horse-power of the locomotive would be 632. This 

 gives us, however, no indication of the ability of the loco- 

 motive to accelerate. If a start has to be effected on the 

 grade in 40 seconds, the final speed being 10- 7 miles 

 an hour, or 15'7 feet per second, the acceleration must 

 be 0'393 f.p.s. per second. Inserting this value in 

 Equation 83, we find the current required for acceleration 



