of Deposit of a Submarine Cable. 5 



5. Now x' = x-\-nt -{■¥.; therefore 



da^ _dx _ dx ds 

 dt dt ds dt 



But s'=s4-w^ + L: and, for a dcfiuite point of the cable, 6' is 



invariable with regard to time: therefore 0= Yt'^^^' ^^^ ^'^^^' 

 sequeutlv dx' dx 



from which 



d^x' _ £ (dx\ _ d_ fdx\ ds_ ^d^ 



Also 



dt ds dt ds dt^ di\ds) ds^' 



6. The equations of motion therefore become 



Or transposing and dividing by c/', 



«=i{(T-<')S}.- 



(T-„)^^=c, 



(T-o)'/^ = <' + ». 

 ' ds 



At the bottom, on the supposition that there is even the 



smallest conceivable tension, -f-=0 ; and s is : therefore c' is 0. 

 ds 



But c is (nearly) the tension at the bottom, and must be retained. 



7. Dividing the equation {T — a)-^ — c 



by the equation „ „\^!/-. 



^^ "^'ds ' 

 we find 



dx _ c 



ly-S' 

 the equation to the common catenary. Also T — « is given by 



Integrating, 



