186 THE DIRECT-CURRENT MOTOR CH. IX 



0*4 ohm, the resistance of the rheostat on the first step of 

 the controller must be 4-6 ohms. The current used in 

 overcoming the friction which we suppose to be constant 

 at all speeds, is 29-6 amperes. The current available 

 for acceleration is therefore 70*4 amperes. Inserting this 

 in Equation 83 we obtain the initial acceleration namely, 

 0'415 f.p.s. per second. If the resistance is taken out as 

 the car speeds up, the current being kept constant, the 

 acceleration will be constant until the resistance is all out. 

 The speed is then given by the consideration that at that 

 moment the motor is taking 100 amperes, and the 

 resistance in circuit is that of the motor only, that is to 

 say, 0'4 ohm. Inserting these values in the speed 

 equation, we get 41 - 5 f.p.s., or 288 revolutions per minute. 

 The time from the start is 100 seconds, and the distance 

 travelled is 2,075 feet. These results are shown in Fig. 46. 

 Throughout this step the current per motor remains equal 

 to 100 amperes. 



STEP II. Starting rheostat all out ; motors speeding 

 up according to law of Equation 69 ; acceleration 

 gradually diminishing. 



From Equation 87 we find the time factor to be 6'1 

 seconds. Since the final speed is 306 r.p.m., and the 

 speed when the resistance is all out is 288 r.p.m., a is 

 0'058, so that the speed at 6'1 seconds on Step II. is 300 

 r.p.m., and the current per motor has fallen to 52 '5 amperes. 

 Now to find the distance travelled during this step, say 

 in 80 seconds from the point when the rheostat was all 

 out. We see that if the motor had attained its full speed 

 immediately on Step II., the distance travelled would 

 have been 3.520 feet; but we must deduct from this an 

 amount ^ar, or 15 feet, so that the whole distance 



