172 TIIH DIRECT-CURRENT MOTOR ('II. VIII 



the current per motor in the former would be half that in 

 the latter, yet since the South London motors are in series, 

 the current from the line would be the same. 



Example 45. A motor that will draw a car at 

 thirteen miles an hour, with a total tractive effort of 800 

 pounds and a mechanical efficiency of 85 per cent., would 

 be rated as a 24 horse-power motor. Two such motors 

 would drive a 10-ton car up a 5 per cent, grade with 

 a train resistance of 20 pounds per ton. These motors, 

 however, would not be able to start the car on this 

 grade. They could drive the car when once started, 

 but if it should stop on the grade, they could not 

 start it again. In order to start up on the grade we do 

 not need greater power, but greater force. Let us find 

 the force factor required to start up in 10 seconds. 

 The final speed of 13 miles an hour is equivalent to 

 19 - 1 feet per second; to make up this speed in 10 

 seconds, the acceleration must be 19'1 f.p.s. per second, 

 assuming uniform acceleration throughout. From Equa- 

 tion 21 we find that if J?=500, <y = 478, and d = 33, M 

 must be 41 '7 and the final current 47 amperes. 

 Inserting these values in Equation 83, remembering that 

 each motor has to accelerate half the car, we find the 

 accelerating current to be twenty amperes, so that the 

 total current from the line at the start is 67 amperes. 

 The motors then must be capable of carrying this current^ 

 so that the force factor must be 2,790 dynes. Thus, to 

 drive the car up the grade we need a force factor of 1,960 

 dynes, and in order to be able to start on the grade we 

 have to increase the force factor by 42 per cent. 



Example 46. A motor has to be designed for a lift 

 which will raise an unbalanced weight of 2,000 pounds 



