If we now consider that the tension at the top, T, , 
is limited, as specified in equation (4), we can calculate 
the limiting case of the maximum amount of cable, 5,, that 
can be towed at the critical angle with nothing on the bottom 
end , by simple geometry. These values and the corresponding 
values of x, and y, are given as follows: 
c lbs 
S, (Wsing, + 0.02 R) = 3.84 x 10° — a? (14a) 
Substituting from equations (2) and (9) 
3.84 x 10° £t2 (Se 
5s. = (14b) 
"(10 £t) ain 9, (=) + 0.024 sec 
Ym = Sq 5in 9 | (15) 
Xn = 8, COS Pe (16) 
The results of these calculations are given in Table II, 
TABLE IL 
d 
Pe v= Sry Yn Xn 
dege Beg" /ft as ee a ae, = ft. 
5 0.435 x 10 * 6,740. 586 6,700 
10 L74. x 107% 22,000 3,830 21,700 
15 4.00 x 107* 33,700 8,730 32,500 
20 7.15 x 10~* 36, 400 12,400 34,200 
25 Lia) a 2e* 35,100 14,800 31,800 
30 16:6 “* 167* 32,200 16,100 27,900 
35 23.0 x 167° 29, 400 16,900 24,100 
40 30.9 »x 107% 26,800 17,200 20,500 
45 40.5 x 107¢ 24,800 17,500 17,500 
50 51.5. * 197* 23,200 17,800 14,900 
55 68.5 x 107° 21,800 17,900 12,500 
60 88.5 x 107% 20,800 18,000 10, 400 
65 114 x 10°* 19,900 18,100 8,440 
70 149 x 
17 
