Application 



Let us apply the above theory to the following problem: A wire cable 3600 

 ft. in length is towed through water at a speed of 20 knots by two towboats 1000 ft. 

 apart. What is the tension in the cable at the towboats when the cable diameter is 



1/2" 



5/8"; 



Assume 



3/4" ? 



R* = .27 V 8 d lb. /ft (11) 



F* = .006 V 2 d lb./ft (12) 



where d = cable diameter in inches 



V = speed of towboats in knots. 

 Then R/F = .27/. 006 = 45 and y/s = 500/1800 = .278. Fig. 2 is a plot of 

 Fs/T as computed from (8) against corresponding values of y/s as computed from (9) 

 and (10). From Fig. 2, when y/s = .278, Fs/T = .210. Hence T = 1800 F/.210 = 

 8570 F. Taking V = 20 knots, and computing F from (12), we have the following table: 



d 



1/2" 



5/8" 



3/4" 



F 



1.2 



1.5 



1.8 lb./ft 



T 



10300 



12850 



15400 lb. 



* From early U.S.E.M.B. experiments. 



.7 





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.6 





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WHE 



:n n 



DRM* 



I TC 



STP 



FAM 







\ 





F = RE3ISTANCE PER UNO - LENGTH OF CABLE WHEN PARALLEL TO STREAM 



-»=HALF LENGTH OF CABLE 



T- CABLE TENSION AT SHIP 



5 





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T " 























































































































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R\l 





4£ 





































































































































































































































.3 .4 .5 .6 



SCALE FOR -^lAy 



FIG. 2 



CABLE TENSION AT SHIP 



