f f ti i Ax m 0.088 x 3 x 5,000 

 d 2 20. 5 2 " 



For the period, T = 3.2 seconds, and d = 20.5 feet, 



2sd a 2, (20.5) 91- 

 gT 2 32.2 (3.2) 2 



For 2nd/ (gT 2 ) = 0.391 (from Fig. 14) 



K f.01 = °- 996 for f f = °- 01 and £ f % Ax/d 2 = 0.357 



K^ a = 0.965 for ff = 0.088 and ff H^ Ax/d 2 = 3.14 . 



From equation (4) , 



1 " Kf.Ol i _ 0.996 0.004 _ ~ 



a = * = = = 114 * 



1 - Kf a 1-0.965 0.035 



from equation (5) , 



¥ a = a Ax = 0.114 (5,000) = 570 feet ; 

 from equation (8) , 



F = F e + F a = 4,000 + 570 = 4,570 feet . 



For d = 20.5 feet, U = 70 miles per hour, and F = 4,570 feet (from 

 Figs. 1, 2, or 6) 



Hf = 3.17 feet and T = 3.31 seconds 

 AH = 3.17 



This satisfies the requirements of equation (3), and the solution may 

 proceed to the next segment which is the remaining 5,000 feet of the 

 area, with the water depth varying from 18 to 13 feet. 



0.25 d i = 0.25 (18) = 4.5 feet . 



Since Ad = 18 - 13 = 5 feet > 0.25 d£ , which does not satisfy equation 

 (1), a shorter segment is required. For a 3,000-foot segment, assuming 

 a uniform depth variation, the depth will vary from 18 to 15 feet. 

 For the 15 -foot depth (using curve B in Fig. 13) 



f f = 0.120 

 £». = 0.095 at the 18-foot depth as previously shown. 



Lff = 0.120 - 0.095 = 0.025 = 0.25 f fi . 

 30 



