Step 5. Compute FL using equation (15) and FS using equation (16). 

 Since OS is zero, equation (15a) is used for FL, giving 



(4.92 ft/s) (1.0) 



FL = = 0.214 (dimensionless) 



[(32.2 ft/s 2 ) 16.40 ft] 1/2 



Since 6L is 180° (it rad) , equation (16a) is used for FS, giving 



-(4.92 ft/s) (1.0) 



FS = = -0.214 (dimensionless) 



[(32.2 ft/s 2 ) 16.40 ft] 1 / 2 



Step 6. In Figure 3, find the value of R^ corresponding to FL and 

 ftL. This value of R H is R H s given approximately by 



R„ c = 0.77 (dimensionless) 



Step 7. In Figure 3, find the value of Rj| corresponding to FS and 

 SIL. This value of R H is R p L , given approximately by 



R„ , = 1.75 (dimensionless) 

 ri , L 



Steps 1 through 7 solve part 1 under "Find." 



Step 8. To answer part 2 under "Find," note that R R s is less than 

 0.85 and R H l is greater than 1.15. As a result, the current cannot be 

 neglected for all wave periods between 6 and 15 seconds. 



Step 9. To solve part 3 under "Find," Figure 3 is first used to deter- 

 mine the factors ftA, fiB, fiC, and fiD. Invoking the definitions for 

 these parameters given in equation (19), it is found that approximately (see 

 Fig. 4) 



fiA = 0.80; ftB = MC = 0; HD = 0.59 (all dimensionless) 



Step 10. Compute US' using equation (18). The result is 



ftS' = MIN(0.80, 0.0, 0.0, 0.59) = 0.59 (dimensionless) 



Step 11. Compute TS 1 using equation (17) and the value of d~ T L 

 found in step 3. The "result is 



[2(3,14159)] / 23.28 ft \ 1/2 



TS' = ( ^ = 9.1 s 



0.59 \32.2 ft/s 2 / 



This value of TS' • shows that only for waves between 9.1 and 15 seconds 

 can the current be neglected in computing H from P. Waves between 6 and 

 9.1 seconds must have the influence of the current taken into account when 

 using bottom pressure measurements to compute wave heights. 



19 



