APPENDIX B 

 SAMPLE CALCULATIONS 



In the derivation of the long-term distributions of ship motions and bending moments, it 

 is necessary to fit a log-normal distribution to the truncated histograms computed in Tables 8 

 through 11. This requires the calculation of the mean value and the standard deviation from the 

 truncated data. 



The method and tables of Reference 7 are applied in making these calculations as in- 

 dicated below. In the calculations, the symbols used are: a for standard deviation, y and z 

 for the parameters needed to enter Table IX of Reference 7, z being an estimate of the point of 

 truncation. 



LONG-TERM DISTRIBUTION OF VARIATION IN PITCH ANGLE 



The mean value and standard deviation of the long-term distribution of the variation in 

 pitch angle are computed from the data given in Table 8. These data are truncated at a pitch 

 angle of 1/2 deg (6) = 5). 









tog,„ d 





Percent of 









'Og,n ^ 



•Oglr, Q 



lU 1 



Measured 



s^ = 



Variations 







Variation 



deg X 10 



"10 



at End of 

 Class 



^10 



at Center 

 of Class 



from 

 Point of 



(h-hj,)^ 



Falling 

 within 

 Class 



Nx^ 



Nx 





Interval 



Interval 



Truncation 









e 





h 



X = h—hj< 





A' 











— OO 















5 



0.6990=Aj, 



0.8495 



0.1505 



0.0227 



^5.4 



0.577 





10 



1.0000 



1.1505 



0.4515 



0.2039 



36.2 



7.381 





20 



1.3010 



1.4516 



0.7526 



0.5664 



17.5 



9.912 





40 



1.6021 



1.6902 



0.9912 



0.9825 



5.6 



5.502 





60 



1.7782 



1.8407 



1.1417 



1.3035 



2.96 



3.858 





80 



1.9031 



1.9516 



1.2526 



1.5690 



1.28 



2.008 





100 



2.0000 



2.0731 



1.3741 



1.8882 



0.608 



1.148 





140 



2.1461 



2.2007 



1.5017 



2.2551 



0.0571 



0.129 





180 



2.2553 









89.605 



30.515 



44.792 



In accordance with the procedure outlined on page 29 of Reference 7, we have: 



y _ S/Va;2XJV _ 30.515(89.61) ^ p fisifi 

 2(lNxf 2(44.79)2 



z = -1.087 (from Table IX of Reference 7, corresponding to y = 0.6815) 

 g{z) = 0,7798 (from Table IX of Reference 7) 



a = s = M£ g(z) = 44.79(0.7798) ^ 0.3898 

 2N 89.61 



35 



