Consider now equation (5a). The tension in the cable vj.ll vary 

 from T at the upper end to something of the order of Tq ± D sin © 



:o 



? 



at the other end. This drag contribution is less than 1 lb. in 

 1300 lb. In the case of neutrally buoyant cable = O.smd the tension 

 will increase with depth in proportion to 0, but the whole increase 

 will be only a pound or so. For the value given above for we can 

 neglect 0, so dT/dx becomes a constant; and the tension increase with 

 depth is now linear with depth and still very small. Thus on passing 

 to equation (5b) a good first approximation to the factor T is to 

 make it a constant Tq. 



Consider now equation (5b) with T = Tq. It can now be directly 

 integrated, but first we discuss it qualitatively. If is positive 

 (as for non- buoyant cable) the cosine ratio has a msiximum value 

 greater than one for = 0. All values of Q cwcurxliTig in our problem 

 will be small ccanpared to 0^ and the cosine ratio will be greater 

 than one for all 0. Therefore, the total offset Y will always be 

 greater than for the case of weightless cable. In the case of buoyant 

 cable (0 negative) the<j)posite is true. To find the numerical amount, 

 we put equation (5b) in the form 



cos d(Q-0) ^ D dx + d(in c) 

 cos (9-0) Tq 



where c is an arbitrary constant. 



Then 



Hence 



2D X 



X*UnM) ^ _ ^^^ . 



l-sin(©-iZ5) ^ 



The constant is to be determined from the condition that 0=©q for 

 x=0, thus 



c = 



Then 



l+sin(9o-0) 

 l-sin(©o-0) 



2Dx/TqCOS0 



^^°(^-^^ = - 2Dx/ToCOs0 • 

 1+c e " 



290 



