DISCUSSION : According to wave data reported by Thompson (1977), (H^in^m ^^ °^ 

 the order of one-half the average significant height for Point Mugu, while 

 (H . )f^ is considerably larger than the average significant wave height. As 

 stated, available evidence generally supports the use of iZ^)^ in field situ- 

 ations, and additional information for this site further indicates (Hm-jji^m ^^ 

 the appropriate calculated critical condition. 



Mugu Canyon extends close to the coastline, with a water depth at its head 

 of about 50 feet (15 meters) . Shepard and Dill (1966) state that Mugu Canyon 

 head presently receives a good supply of sediments, supporting the value of 

 (Hinin)m in that usual waves can thus agitate nearshore sands in the water 

 depths near this canyon head. Measurements of wave effects on the sea bottom 

 off southern California by Inman (1957) and Vernon (1965) confirm that waves 

 appreciably lower than (Hmin) t cause sand movement in water depths on the 

 order of 60 feet at comparable sites. 



With 



H = 1.27 ft (0.39 m) and (2Trd/L) = 0.98, (d/L) = 0.16 > 0.075, 



gHT^ _ (32.2) (1.27) (11)2 - 1 -^7 < ^n .r.A 

 ^2 - ^^2 - 1-37 < 30, and 



S = 54 = _H_ = 1.27 ^ Q^^^ ^ 



2d3 2d/d\2 (2) (60) (0.16)2 



:0 



so that equations (8), (9), and (10) are satisfied. However, the expected 

 bed forms in 60-foot water depths and the appreciable tidal range at this 

 site indicate that 11-second waves somewhat lower than 1.27 feet in height 

 might induce sand motion. 



*************** EXAMPLE PROBLEM 2*************** 



GIVEN: A laboratory movable-bed test is to be performed with sediment of median 

 diameter D = 0.00066 foot (0.020 centimeter), wave height H = 0.5 foot (15 

 centimeters), and wave period T = 2 seconds. Also, y' = 1.66 (quartz sand in 

 freshwater) and v = 1.08 x 10"^ square foot per second (0.010 square centi- 

 meter per second) (20° Celsius average water temperature) . 



FIND : Maximum water depth for sand motion in the given situation. 



SOLUTION : Equation (1) gives 



0.35 D°-25(y'g)0.75 0.35 (0.00066) ^ • 25 [ (i. 66) (32 .2) 1 ° -^^ 



(e")v ;;oT5 ^^yrr. o.63 £t 



(19.04 cm)/s 

 With 



[w H/2(5a))y] = [(^)(0. 5)72(0.63)] = 1.25 



