day. The corresponding duration, as read 

 off from the top or bottom scale, represents 

 the time correction to be added to the time 

 interval between the maps. 

 Example: To determine correction for du- 

 ration time (map interval 12 hours) . 

 Given C/ = 29 knots 

 (i?) = 8 feet. 

 From plate V the time needed for a 29- 

 knot wind to raise 8-foot waves = 4 hours. 

 Therefore, corrected duration time of ?7 = 

 16 hours. 



Wind Waves 



Either fetch or duration can be the determin- 

 ing factor in the production of wind waves. To 

 determine which is critical plate III has been 

 devised. It translates fetch into equivalent 

 duration time so that a comparison between the 

 two — fetch in terms of time and actual duration 

 time — can be made. The lesser of the two is 

 used for all computations involving the dura- 

 tion of the wind. 



Plates IV and V give wave height and pe- 

 riods at the end of the fetch corresponding to 

 the duration time and wind speed. 



When the actual duration is short plate V is 

 used for computation of wave height and pe- 

 riod at the end of the fetch. 



Example A: To find Hp and Tp (imin. greater 

 than to). 



Given C7 = 30 knots 



F = 600 nautical miles 

 ta = 32 hours 

 From plate III, imin. = 52 hours; therefore 

 use 32 hours for t^. 



From Plate IV, H^ = 18 feet 



Tp = 8.7 seconds. 

 Example B: To find Hp and Tp (^min. less 

 than ta). 



Given U = S0 knots 



F ^ 60 nautical miles 

 ti = 12 hours. 

 From plate III, t,nin. = 9.6 hours; therefore 

 use 9.6 hours for ta. 



From plate V, H^ = 13 feet 



Tp = 5.6 seconds. 

 When the wind over the generating area de- 

 creases from its value on the preceding map, 

 special problems arise if the resultant wind is 

 insufficient to raise waves of the height which 

 are already known to exist. In such cases, if 



the wind decreases to a value less than two- 

 thirds of the original value use the following 

 wind procedure outlined below (p. 20). 



Otherwise proven techniques are lacking, but 

 the following suggestions give reasonable re- 

 sults : 



On plate IV use for Hp the wave height for 

 the lower wind with a duration time of 60 

 hours. The height probably adjusts to this 

 value rather quickly. 



In computing the wave period, the value 

 obtained from plate IV by using the lower 

 wind and the computed duration time 

 (greater than 60 hours) will generally be too 

 high, although the period will continue to in- 

 crease even with lowered wind velocity. In 

 this case use the increase in period which 

 the waves computed for the previous map 

 would undergo at the lowered wind velocity 

 in the additional duration time (map in- 

 terval) . 



Example: To find Hp and Tp [resultant 

 wind U insufficient to cause previously exist- 

 ing wave height (H) ] . 



Given Tp = 6.3 seconds 

 (H) = 13 feet 

 U = 22 knots. 

 From plate IV, when U = 22 knots, maxi- 

 mum wave height =11 feet = Hp. 



From plate IV, time for a 22-knot wind to 

 produce wave of 6.3 seconds = 22 hours. 

 Then, 22 hours + 12 hours (map interval) 

 = 34 hours. 



For C7 = 22 knots and to. = 34 hours, Tp 

 = 7.3 seconds. 



It may happen that the lower wind speed will 

 not produce waves of the period known to exist 

 even with 60 hours of duration time. In this 

 case use the increase of period which would oc- 

 cur if the energy front were to decay through 

 the distance obtained by multiplying the group 

 velocity of the initial period by the actual dura- 

 tion of the lowered wind speed. 



Example: To find Tp (resultant wind U in- 

 sufficient to cause previously existing wave 

 period Tp) . 



Given previously existing Tp ^ 9 seconds 



C/ = 20 knots. 

 Group velocity of a 9-second wave is 13.5 

 knots (one-half of wave velocity). Distance 

 traveled in 12 hours is 162 nautical miles. 



From plate VI, 9-second waves increase to 10 

 seconds in this decay distance. 



19 



