Figures 3-12 and 3-13 allow conversion of the fastest mile to the average 

 windspeed. The procedures for using these figures are illustrated by an 

 example problem . 



*************** EXAMPLE PROBLEM 3*************** 

 GIVEN : Fastest mile windspeed, U^ = 29 m/s (65 mph) . 

 FIND : Twenty-five-minute average windspeed, Uj._25 min. 



SOLUTION : 



3600 3600 ., , ,^. ^ ^ 1 1 -1 N 



t = rr—, TV = v v = 55.4 s (time to travel 1 mile) 



U,(mph) 65 



1609 1609 .. , 

 and 



U^ = Uj.^55.4 s " ^^ ™''^ ^^^ "P*^^ 



From Figure 3-13 for t = 55 seconds, -rr^ — = 1.25, and the 1-hour average 

 windspeed is 3600 



V3600 s= /u7T:\= T^Ii = 23.2 Ws or "^^ = 52.0 mph 



"t=25 min 

 Using Figure 3-13 again, find j for t = 25 minutes or 1500 



3600 seconds 



""=2^ "^" = 1.015 

 3600 



Solving for U^^25 min 



U. o. • = (tT^^^-^ 1 U-^Ann = 1-015 (23.2) = 23.5 m/s (52.8 mph) 

 t=25 mm \ U_ / 3600 



*************************************** 



If the fastest mile windspeed observations (or any duration windspeed 

 observations that can be converted by the procedure just outlined) are 

 available at 1-hour increments, the procedure may be used to compute hourly 

 average winds or some fraction thereof. If a duration of more than 1 hour is 

 needed, the hourly average values may then be averaged to achieve the desired 

 duration. If the hourly averages vary considerably (say more than 3 to 5 

 meters per second), then the assumption of constant wind made in the use of 

 wave growth formulas is not valid and the accuracy of the wave predictions is 

 questionable. If wind observations are available on a 3-hour basis, the same 



3-27 



