F„.= (0.7) (0.5) (1025.2) (9.8) (0.3) (3)^ (0.71) = 6,741 N (1,515 lb) 



DM 



From equation (7-41), compute 



V ^ (2.0) (0.3) ^ 



C^H (0.7) (3) ^^ 



Interpolation between Figures 7-77 and 7-78 for (j) is required. Calculate 



H 3.0 



= 0.0031 



2 2 



gT (9.8) (10) 



and recall that 



d 



= 0.0046 



2 

 gT 



Find the points on Figures 7-77 and 7-78 corresponding to the computed 

 values of H/gT^ and d/gT^ and determine ^ (w = 10,047 N/m or 64 

 Ib/ft^) . '" 



Figure 7-77: W = 0.1 ; (j> = 0.35 



Interpolated Value: W = 0.29 ; (}> « 0.365 



Figure 7-78: W = 0.5 ; (}. = 0.38 



From equation (7-42), the maximum force is 



F = d) w C„ H D 



m m D 



F = 0.365 (10,047) (0.7) (3)^ (0.3) = 6,931 N (1,558 lb) 



m 



say 



F = 7,000 N (1,574 lb) 

 m 



To calculate the inertia moment component, enter Figure 7-73 with 



d 



= 0.0046 



2 

 gT 



and H = 0.91 H, (interpolate between H = H^ and H = 3/4 Hr) to find 



S . = 0.82 



Similarly, from Figure 7-74 for the drag moment component, determine 



7-128 



