during the process of Sheathing. 175 



have fig. 4. And supposing balance were established, we have, 



Fig. 4. 



in order to use the above equation, to substitute 

 c=l—y, 



q = y - X) 

 where / represents the resistance of the whole cable, thus : — 

 (y-x) (<*F.-flF,)+ (F. + F,) { (W + x)d-a(l-y) \ = 0. 



Substituting ^r 



a and developing x, we have 



(1) 



which formula gives the position of one fault when the position 

 of the other is known as well as the resistances of the two faults 

 or the proportion of the two resistances only. 



Proposing in formula (1) a = 0, i.e. F, r =0 or F y = ao, which 

 is equivalent to having only one fault in the cable, we have 



.- al ~ dW (2) 



the known formula for the position of a fault if only one exists. 



But where ~F X and F y are both definite quantities and larger 

 than zero, we cannot put a = 0, and therefore formula (2) would 

 give quite another value of x than formula (1). Let us call this 

 value z, and see what it amounts to in a case where we have in 

 reality two faults. By comparing the two formulae, it will be 

 seen that as long as y > oc, z is always larger than x and smaller 

 than y, thus z indicates a position between the two faults o 1 and 



