166 THE DIRECT-CURRENT MOTOR ('II. VII 



the acceleration at that point, and the tangent at p% is 

 equal to the tangent of the angle </ 2 s 2 /2> ^ a ^ * s * sa y? ^ s 

 proportional to the accelerating torque, and therefore to 

 the acceleration at p z . 



In this way we can construct the whole of the 

 acceleration curve, taking care to keep the points p } p z 

 closer together as the curve becomes natter. The curve 

 will finally approximate to the speed of 21 ! 2 feet per 

 second. 



From the curve C we can obtain the distance travelled 

 at any time. Draw a vertical line through any given 

 time, cutting the acceleration curve ; the area enclosed 

 between this line, the curve, and the time base, will be 

 the distance travelled from the start. Find these distances 

 for successive seconds, and then plot them as a curve D on 

 the time base, taking vertical ordinates as distances 

 travelled. 



From the acceleration curve we can deduce the current 

 curve E, showing on a base of seconds the current taken 

 from the line. This will begin at 150 amperes, and will be 

 straight for 34'6 seconds, when it will rapidly slope down 

 and approximate to the uniform value 70 amperes. The 

 area of this curve up to any point on the time base will 

 give the energy drawn from the line. 



The curves in Fig. 41 record the results of tests made 

 on the South London E ail way with the motors previously 

 described. They are constructed from data given in the 

 ' Proceedings of the Institution of Civil Engineers,' 

 vol. cxii. p. 246. The track is not level, as shown by the 

 outline of the grades, which is drawn on a time base, so 

 that the position of the train on any grade can be seen at 

 once. The distances travelled in 80 and 120 seconds are 



