Mr. T. T. P. B.Warren on Prof. Fleeming Jenkin's Formula. 171 



one minute, will require 13*2 seconds to lose 5 per cent. ; and 

 these values of t will be proportional to the Napierian logarithms 



of — , and therefore to their resistances : for 

 c 



~ ^,r^ = -p 1 ^ 001 , = 2449 millions, 



•5 x -04908 -5 x -22314 ' 



and 



13-2 , 60 .... 



"E mnno = ~r nooi a =538 millions. 



•5 x -04908 -5 x -22314 



From this it is evident that, for reliable comparisons between 

 cores or cables which have different rates of loss, the times should 

 be noted when they each lose the same proportion of charge. It 

 is incorrect to state the comparison by percentages of loss for the 

 same time. 



In the formula K is a function of length and thickness of di- 

 electric (log— ), so that, log -j being constant, R will vary as 



the length of the cable or core; consequently if K represent the 

 total capacity of a long length of cable, or that of a single mile 

 of the same cable, R will be the resistances of these respective 

 lengths. 



Suppose a loss of 5 per cent, in these cases ; and first, the 

 length being one mile, K being the same for each unit of length 

 and equal to *5 farad, t being one minute, 



R =^rS908= 2449milUons - 



Next } let the length be 1000 miles; 



and 



2-449 x 1000 = 2449 millions per mile. 



Hence in the formula K enters as the function of length in terms 

 of the capacity per unit of length. 



The next proposition is to show that whatever interval t be 

 taken, if the correct loss corresponding to that time be known, 

 the value for R is constant. 



The 1866 Atlantic Telegraph cable fell to half tension in 66 

 minutes 40 seconds. K= *3535 farad per nautical mile. Length 

 = 1896 nautical miles. 



U = - 3535 xlB96x -69314 = 9 millionS (near1 ^ 

 9 x 1896=17,000 millions per nautical mile. 



E= - 5xl000x-04908 = 2 ' 449 millions ' 



