H.. (m) = m(x) e dx 

 H (m) = u(x) e 1 m x dx 



(268) 



One can evaluate the above integral provided that the densities of sources 

 and dipoles are known. 



Considerable data of computation of added resistance in head sea waves 



37 

 by means of the above formula have been reported. As for the case of 



oblique seas, numerical results so far obtained are rather infrequent be- 

 cause of the complication of ship motions. Unfortunately the above formula 

 has failed to strike the fancy of engineers of a practical mind, because 

 the integral in the infinite interval necessitates much calculation. It 



will be found that the formula can be simplified if high frequencies such 



—1 / 2 3ft 1/9 



as 0) = 0(e ) are assumed. If we assume first U = 0(e ) and 



-1/2 -1 



a) = 0(e ), then K = 0(e ) and Q = 0(1). Omitting terms of 0(e), the 



formula of Equation (266) is reduced to 



.p J 



AR = 4 it p (m-K cos a) 



4 

 |H*(m)| 2 +^r |H*(m)| 2 



dm (269) 



Substitute Equation (268) in the above and apply the Fourier integral 

 theorem. Now we assume that m(x) and y(x) vanish at both ends of the ship. 

 Then, by integration by parts, we have 



f , * i m x , . f dm(x) i 



m(x) e m dx = - l — -J— — e 



dx 



The added resistance due to the longitudinal disturbance is then repre- 

 sented by 



00 



AIL = 4 7T p (m-K cos a) IH-Cm) | dm 



•Loo 



= - 8 T\ 2 p Re J |k cos a|m(x)| 2 +im(x) dm ^ 1 dx (270) 

 95 



