24 



Fine Groin Sand Strip 7/8" Wide; 20/30 Sand 



/of Mean Diameter 0.028" 

 Sta.6 4 2 



V 



U =0.40 knot ^=5.8-10 U = 1.2 knots R, =5.7x10" 



^ = 0.079 % = 0.20 R = l7xl0 5 j±- = 0.238; |= =0.062; R = 52 xlO 5 



R K =6.1 - 7.3x10" U = 1.4 knots R x = 3.3 x I0 5 



jjU 0.099- 0.119; ^=0.17; R = 2l-26xlQ 5 j£ = 0.275; ^==0.03; R = 60xl0 5 



U = 0.5 -0.6 knot 

 S, 



-p =0.158; % = QH; R = 35xl0 5 4r = 0.317; -i = 0.03; R = 69 x I0 1 



U = 1.8 -3.5 knots R =4.2-8.3x10 



°r =0.198; -^ = 0.091; R=43xlO b =■ = 0.357- 0.693; ^ = 0.03; R = 77- 150 xlO 5 

 vT_ ■» vL S 



{Turbulent 



~7 L.W.L. at U = 



'H'ronsifionai v<; 



Bow Profile of TMB Model 4083 



Figure l8 - Boundary Layer Flow Conditions on Tanker Model 

 with a Sand Strip as Turbulence Stimulator 



Definitions of symbols are given in Figure 10. 



A simplified expression for the decrement in frictional resistance 

 may be obtained from the following exact expression: 



*m = 



(r: 



- r ) cos d> ds 

 m ^ 



[4 



f t' cos 6 ds 



IT 



J S 



where £ is the ratio of the decrement of the model frictional resistance to 

 m the total model turbulent frictional resistance, 



t is the laminar shearing stress, 

 m 



t' is the turbulent shearing stress on the model, 

 m 



ds is an element of model wetted area, 



<j> is the direction angle between the longitudinal centerline of the 

 model and the shear force, 



S-r is the hull area covered by a laminar boundary layer, and 



S is the wetted surface of the model at U = 0. 



