1174 THE BELL SYSTEM TECHNICAL JOURNAL, SEPTEMBER 1957 



20 



16 



|< 12 



z 8 

 < 



0.020 



0.040 



0.060 



0.080 



h/h 



0.10 



0.12 



0.14 



016 



Fig. 29 — Distance behind the ship at which the cable enters the lower stra- 

 tum. 



Example: A ship is laying cable No. 2 at a depth of 6.000 ft at a re- 

 sultant ground speed of 6 knots due east. There is a one knot cross-cur- 

 rent 600 feet deep rinming 30° east of north. Find e and d. 



Here jS = 60° and we obtain from (50) by a simple computation 

 V = 5.57 knots. Also since for cable No. 2 H = 70 degree-knots, 



h'/h = 600/6000 = 0.1, 



a' = 70/5.57 = 12.3 degrees. 



By interpolation, we find from Figs. 28 and 29 



1 e 



f 



h' 



= 2.7, 



1 



--)- = 4.-. 

 \tan a' h' / Aa 



Equation (51) yields in turn (p = 0.156 radians, and Aa is given by 



Act = — — — , = —0.6 degrees = —0.010 radians. 



Hence, we have 



e = 3.6 h'<p = 2.7 X 600 X 0.156 = 253 ft, 

 1 



d = h' 



tan a' 



4.7A 



a 



= 6000 [4.59 -f 4.7 X 0.010] = 2800 ft. 



