DYNAMICS AND KINEMATICS OF SUBMARINE CABLE 



1183 



The singularity at u = 1 makes the numerical evaluation of the inte- 

 gral in (71) cumbersome. Therefore we consider the evaluation of A(l). 

 But for convenience of numerical calculation we write (70) as 



A(l) = -ntan 

 and integrate by parts to get 



a 





u'Y'd 



1 - u 



u 



). 



a 



A(l) =tan^((l -Wy^ -It, 



Jr 



— U 



u 



w^ 



(1 - u-yy^ 



du 



We note again that (1 — R^f ':=^ 1. Further, essentially all of contribu- 

 tion to the integral in this equation occurs near u = 1 because of the 

 large value of y. On the other hand, the values of To which are of in- 

 terest will normally be smaller than unity. Hence R, the lower limit, will 

 normally be less than one-half, and thus will be outside of the region of 

 significant contribution to the integral. Therefore, we can take the in- 

 tegral to be a constant for a given a. Denoting this integral by n and 

 combining these considerations we obtain 



A(l) = tan - — - n tan - To . 



Zi Zi Zi 



Finally solving for ^ and X from (68) and (69), we find 



1 



(73) 



tan a 



which are a dimensionless form of (24). For brevity we have written 



X = 



- - 



2\X 

 1 /2 



^-11-0" 



2\7L 



'-\\k-\ 



n 



«tii 9 - 9 ^ tan - 1 , 



ctn - + ^ n tan - 



Since the integral n and the constant 7 depend only on a, the constants 

 K and X are also functions of a only. We have evaluated n by numerical 

 integration and have plotted the resulting values of k and \ — k in Fig. 10. 



C.2 Recovery 



In the conventional recovery situation we have as boundary conditions 



