CH. VII ACCELERATION 165 



will have to accelerate 17 '5 tons. Using the Equation 

 84 we find the initial acceleration to be 0'463 f.p.s. per 

 second. As the speed when the rheostat is all out is 

 16 f.p.s., the acceleration will remain constant for 34'6 

 seconds. Plot the first portion of curve C showing the 

 speed in f.p.s. on a base of seconds. Let the point p l 

 represent the time when the rheostat is all out. Draw 

 a line at right angles to op 1? passing through the 

 point q v and cutting the speed base in s r Draw a 

 horizontal line passing through q l and jp t cutting the speed 

 base in t r Then q l t l is the accelerating torque at the 

 point PI on the acceleration curve, and the tangent at 

 the point p { is at right angles to the line drawn through 

 the points q l on the torque curve and s l on the speed 

 base. 



We have then a general construction for drawing the 

 acceleration curve. Take the point p^ where the accelera- 

 tion ceases to be constant, draw a horizontal line to cut 

 the torque curve in g, and the speed base in r Measure 

 t l s l along the speed base, join q r s v and from p^ draw a 

 line at right angles to g^s,, and continue this line for a 

 period during which the acceleration may be assumed to 

 be constant, say to the point p 2 . Now draw another 

 horizontal line through p 2 , cutting the torque curve in q v 

 and the speed base in 2 ; set off from 2 along the speed 

 base a distance 2 s 2 equal to t L s, ; join s z q 2 and draw a line 

 from p 2 at right angles to ^ 2 s 2 , extend this line to a point 

 p y and so on. In the triangle q^s^ s 2 2 = Sj 1} and is 

 a constant quantity, and q 2 t 2 is the accelerating torque, 

 it follows that the line drawn through _p 2 at right 

 angles to g- 2 s 2 will be the tangent to the curve at p 2 for 

 the tangent at any point of the acceleration curve represents 



