described in (a) above, and that f and c have been plotted against speed. From 

 these curves read value of speed corresponding to c = .10, and also value of f at 

 this speed. Suppose we find for the speed v = 30 ft. /sec. and f = 1.00 lb. /ft. 



(b') From Fig. 3 and table 3 compute T.., T.. sin $ , T.. cos $ for c = .10, 

 and hence at 30 ft. /sec. with f = 1.00 lb./ft. This yields table 4. Suppose now 

 that although the resistance of the float and its displacement when towed at 

 30 ft. /sec. are unknown, we can estimate that its resistance will lie between 500 

 and 1000 pounds. For values of T.. cos <j> in this range we see from table 4 that 

 the additional displacement T.. sin $ varies from 1726 lbs. to 1815 lbs. Take 1770 

 lbs. as a mean value of the additional displacement. Supposing that the weight of 

 the float itself is 300 lbs., we have 1770 + 300 = 2070 lbs. for the total dis- 

 placement of the float when towed by the cable at 30 ft. /sec. 



(c') By model tests, or otherwise, find the resistance of the float for a 

 speed of 30 ft. /sec. when its displacement is 2070 lbs. Suppose we find that the 

 resistance at this speed and displacement is 690 lbs., i.e. T.. cos $ = 690 lbs., 

 whence, from table 4, we have T- sin <j> = 1800 lbs. as a corrected value for the 

 additional displacement. The resistance of the float at 30 ft. /sec. could then be 

 corrected using 1800 + 300 = 2100 lbs. for the total displacement. Finally, we see 

 again from table 4 that when T.. cos 4> = 690 lbs., T 3 = 2930 lbs. which is the ten- 

 sion in the cable at the ship at this speed. 



