j\Ir. W. Gravatt on Ihe Atlantic Cable. 



35 



of a rough cylinder moving sideways in sea water is 'Sdv^ pretty 

 nearly (d being the diameter in feet, and v the velocity in feet 

 per second). 



fUU COS OL 



We must have therefore lo cos « — •8<//^ = 0, or t=\/ .q , -• 



1 1 Tr ^ 1 /w cos « , , 



We must also have V = -. — = -. — \/ — 5-5—^ and also 

 T -TV sin a sma V 'oa 



L — D coseca. 



Now, taking as an example tv = 'S2 and d='05, as is pretty 



nearly the case in the present Atlantic cable, we get 



log t = l (log 8 4- log cos a), 



log V = 5 (log 8 + log cos «) — log sin a. 



From whence, taking «= successively 10°, 20°, &c., we get 



immediately the following Table, which we can easily extend : — 



Note. — The slip is obviously V vers a in feet per second, giving a re- 

 sistance (as will be afterwards shown) of '012 ("GV vers a)'"* pounds per 

 fathom, or •0i2 ('fiV versa)' * L pounds ou the whole, which ought in 

 strictness to be subtracted from the Aveight of D, but which is too small a 

 quantity to he noticed in practice, at least in any cable similar to this. 



Now, taking the greatest depth of the sea to which the cable 

 in question is to be exposed at 2000 fathoms, and each fathom 

 to weigh in water 6x';^2 = l*92 lb., say 2 lbs. per fathom, we 

 have 4000 lbs. for the strain on the cable, whatever may be the 

 velocity of the ship ; or as the cable is said to break with about 

 4 J tons or 9500 lbs., the gi'catest necessary strain in smooth 

 water is about four-tenths the breaking weight. 



But if the cable be ])aid out at a greater velocity than the hori- 

 zontal velocity of the ship, the cable will be (so to speak) slipping 

 down the supposed inclined ])laiK', and we must, in that case, find 

 the resistance due to that slipping velocity. 



Now this resistance is not exactly as the square of the velocity, 

 nor is it indeed exactly as the diameter of the cable. 



D2 



