(V(f> +U ) • n = on the body surface (4) 



-»■ -> 



Here U (-U, 0, 0) is the steady forward velocity vector, and n is the unit normal 



vector of the body directed into the fluid domain. 



The solution of Equations (2)- (4) is given by Tuck as 



*1 = f log r + f log r l ~ U X (5) 



where a = — S' = the strength of the source 



S = area of the immersed cross section 

 S' = dS/dx 



2 2 1/2 

 r = [(y-n) +(z-C) ] 



and 



1/2 



r x = [(y-n) +(z+C) ] 



Here, (y,z) is the point where the potential is solved and (r),C) is the source point. 



TWO-DIMENSIONAL POTENTIAL OF OSCILLATION 



The two-dimensional velocity potential due to pure heave oscillation is given by 



<P 3 (S) = <j> 3 + ^ (6) 



The first potential satisfies the following conditions 



2 2 

 8 cj), 3 <j), 



+ r^ = when z < (7) 



2 2 



dy 9z 



\ 2 



* 3 = on z = (8) 



