f^ = Q-080 + 0.095 , 0^088 



Using Figure 3-21 for d = 6.25 meters, H = 0.9 meter, and U^ = 31.3 

 meters per second with 



B^. 9.8.6 25 ^ 0.0626 



U, (31.3) 

 A 



^ = ^^^A^= 0.0090 

 U^ (31.3) ^ 

 it is found that 



and 



= 38 



For ff = 0.01, 



U^ 2 



F =38 — = 38 ^^^'P = 3800 m 

 e g 9.8 



fy. H^ Ax ^ Q^Q^ ^ Q^g ^ J525 ^ ^ ^^^ 



d^ (6.25)^ 



and for f^ = 0.088, 



^/ ^i ^^ _ 0.088 X 0.9 X 1525 _ ^ ^q 

 d^ (6.25)^ 



For the period T = 2.6 seconds and d = 6.25 meters 



^= 2.(6.25) 0.593 

 gr 9.8 (2.6)^ 



Using Figure 3-38, for 2Tid/(gT^) = 0.593 



f^ H. Ax 

 K^ 01 " 0-9998 for f^ = 0.01 and ~ — | = 0.351 



fr. H. Ax 



K^ = 0.998 for f^ = 0.088 and — — ^ = 3.09 



fa f ^2 



From equation (3-45) 



^ " y.lO _ 1 - 0.9998 _ 0.0002 _ ^ in 



a = 



From equation ( 3-46) 



1 - K^ 1 - 0.998 0.002 

 fa 



F = a Ax = 0.10 (1525) = 152.5 m (500 ft) 



3-73 



