CH. XI DESIGN OF RAILWAY MOTORS 229 



We now find that the acceleration is 0-421 f.p.s. per 

 second, and that the motor will accelerate for 53*4 

 seconds ; the distance travelled in this time is 600 

 feet; full speed is given either by 53'4 times 0*421, 

 or by three halves of Djt, both giving 22*5 feet per 

 second, or 15 '3 miles per hour, and the distance covered 

 at this speed in the remainder of the 80 seconds is COO 

 feet. 



The frictional current is 22 - 5 amperes : knowing the 

 final speed we can determine the drop, which is 54 volts ; 

 hence the resistance of the motor must be 2'4 ohms. This 

 is more than it need be. By taking c,, = 38 amperes, we 

 find the resistance to be 1*4 ohms. We may of course 

 make it anything we please. 



Taking the accelerating current then as 38 amperes, 

 the value of M will be 164-4, and c f will be 21-4 amperes ; 

 the starting rheostat must be arranged to carry 59*4 

 amperes, and the motor must be designed to carry this 

 current for the whole period of acceleration. 



Suppose, now, that instead of using the best diameter 

 we had taken tZ=40 inches, M being the same as before. 

 Full speed would not be reached, and the car would take 

 87 seconds to cover the given distance. If, on the other 

 hand, we had used a motor with a higher induction factor 

 than the best, say 3/=200, the acceleration would be 

 0'540 f.p.s. per second, but the final speed would fall to 

 18- 7 feet per second, or 12 -8 miles per hour, and the car 

 would take 81 -6 seconds to cover the 400 yards, so that 

 there would be no gain by increasing the weight of the 

 motor. 



We have seen that the best arrangement is that in 

 which one-half the given distance is covered at full speed 



