DYNAMICS AND KINEMATICS OF SUBMARINE CABLE 



1147 



In Appendix C it is further shown that for small a 



>S = L + kT^/w, 



X = L cos a + \T,/w. 



(24) 



Here S and X are the distance along the cable and the horizontal dis- 

 tances respectively from the touchdown point to the ship (Fig. 11), 

 L and L cos a are the corresponding distances for straight-line laying at 

 the same ship speed, and k and X are functions of the critical angle a 

 which are plotted in Fig. 10. To illustrate the use of (24) we consider the 

 following. 



Example: Cable No. 2 is being laid without slack onto a rough bottom 

 from a ship moving at six knots. If the pay-out rate is decreased so the 

 slack is 1 per cent negative, what is the subsequent rise of the tension 

 with time at the ship? 



This is really a transient problem. However, we shall try to get an idea 

 of the average behavior of the cable by assuming it passes through a se- 

 quence of stationary configurations. Also, we assume that because of 

 the rough bottom there is no slippage of the cable along the ocean floor. 



If d is the amount of negative slack and V the ship speed, then in a 

 time t an amount V{1 — 8)toi cable will have been paid out. This amount 

 plus the inclined length L will equal the amount contained in the curve 

 AOC (Fig. 11). We then have 



L -f V(l - 8)t ^ S + Vt - {X - L cos a). 



(25) 



Substituting (24) into this equation and solving for To we find To = 

 (w/(X — K))8Vt and that by (21) the tension at the ship is given by 



Ts = wh -\- 



w 



8Vt. 



0.4 



0.3 



0.2 



0.1 



0.016 



0.012 



I 0,008 



0.004 



5 10 15 20 



a IN DEGREES 



25 



10 15 20 



a IN DEGREES 



25 



Fig. 10 — Variation of k and X with the critical angle. 



