220 THE DIRECT-CURRENT MOTOR (II. X 



But -? = l - i ; hence we find that b<j = bh. 

 oh k. 2 P 



The given distance will then be covered in the shortest 

 time when the distance travelled during the process of 

 acceleration is equal to that travelled at full speed, the 

 time of acceleration being two-thirds of the whole time. 



Substituting for /r, and A' 2 their values as given by 

 Equations 93 and 99, we get 



We thus see that when a train of weight W tons has 

 to be started from rest and moved through a distance of 

 D feet, the tension of the line being E volts, the accelerating 

 current c,, amperes, and the internal drop when running 

 at full speed c f R volts, the time occupied in covering the 

 given distance is least when the equipment is so de- 

 signed that half the distance is covered during the 

 period of acceleration. In order to secure that this 



may be so, the ratio - must be that given by Equation 

 Mv 



102. If the ratio is greater or less than that given by 



this equation, the time occupied will be greater than it 

 need be, and to reduce the time we shall have therefore 

 to use a greater accelerating current. 



In Fig. 55 time curves for different distances are 

 drawn on a base of seconds. The dotted lines give the 

 acceleration curves for wheels of different diameters and 

 equal values of M and v. Thus we see that when D is 200 

 yards, a 40-inch wheel will cover the distance in the shortest 

 time, while a 20-inch wheel will take about 10 seconds 

 longer. The condition assumed is, that with a 40-inch 



