438 Dr. Glazebrook on the Practical Application of the 

 and our equation of condition becomes 



l*24xLx 



1 



io- 5 <s. 



y s/a 2 + 4y 2 



From this we obtain the rule : — 



Calculate in absolute units the leak in a rail 1 mile long 

 when carrying the maximum current required to drive the 

 maximum number of cars in the mile. Let the result be 

 L c.G.s. units of current. Then a must not exceed the value 

 given by the equation 



l*24xLx 



1 



V V(a z -\-±f) 



io- 5 =s 



or if L be the leak in amperes, 



Hence 



124Lx 



L=10L. 

 1 



io- 5 =s. 



y s/{a z + 4f} 

 If we take $ = j><y (a usual case in practice), this reduces to 



25xLx — 



and 



y ^(a' + fy 2 ) 

 With the numbers already assumed, 

 X/^ 2 = 60 amperes, 



= = 1. 



L=7*5 amperes = 5 per cent, of current. 



If we know the resistance of a mile of the track we can 

 calculate the potential-difference required to drive the current 

 through, for it is clear from Mr. Parry's curves and from 

 the theory that for a track 1 mile in length the drop required 

 is given by the product of the current and resistance to a 

 sufficient degree of approximation. 



Now for a rail weighing 92 lbs. to the mile we may take 

 as the resistance of a mile of continuous rail the value 

 0*0525 ohm. Thus, allowing for the bonds, the rails being 

 in 36-feet lengths, about one-fortieth of the total resistance, 

 we get as the resistance of a mile of single rail 0*0538 ohm. 

 Hence the resistance of a double track will be 0*0135 ohm, 

 and the potential-difference required to produce a current of 

 150 amperes will be about 2 volts. This, then, is the value 

 of v in the above formula. 



