Yim 



inO<Zj<H,Xj = y=o and 



Ml = bH 



(37b) 



inH<Zi<cOjXi = y=o, this combination of singularities will not produce any 

 regular bow waves (2). Likewise, it is easy to see that the x component of per- 

 turbation velocity minus the local disturbance (this may be called the regular 

 part of the perturbation velocity) is zero in < x < a , > z > -H . It is known 

 that the local disturbance of 0^ does not contribute to the force (3) for any 

 source distribution. Therefore, it is evident that Eq. (34) in this case is zero. 

 However, Eq. (35) in this case is exactly the same as the wave resistance of the 

 doublet distribution itself, due to following reasons. When we consider a doublet 

 line (Eqs. (37)) located at x = -b, y = (b - 0), the regular 4>^ on the doublet line 

 due to the source distribution (Eq. (36)) is zero because the regular wave does 

 not propagate forward. Thus, 0xx('''^) i^ Eq. (35) is due to the doublet line it- 

 self only, without any contribution from the source distribution (Eq. (36)). In 

 fact, this wave resistance of the doublet line (Eqs. (37)) is exactly the same as 

 the bow (or shoulder) wave resistance. Namely, when we consider the pressure 

 distribution on the bulbous bow of the wedge ship corresponding to Eqs. (36) plus 

 (37), the bulb has eliminated the pressure on the ship hull, making the drag due 

 to the ship hull equal to zero, but is carrying its own drag, which is exactly the 

 same as the shoulder wave drag. 



For a symmetric ship, we consider, as an example, the source distribution 

 of Eq. (15), 



N 

 m(Xi) = X] ^n''" 



mo<x,<l, 0<z. 



H. The X component of the regular velocity can be ob- 



tained exactly the same way as for Eq. (16): 



4k r"^ ^ 

 <^^ = -^ I Ai(0,z,^) sin (kox sec 6) 



+ Bi(O,z,0) cos (kgx sec 6) - Bi(x,z,0) d9 , 



(38) 



where 



Ai(x,z,^) 



kgzsec 8 kQ(z— H)sec 



N n ( 2n) 



2.n .-. (_1) m^ (X) 



E 



k2"-i sec^" 6 



(39) 



and 



Bi(x,z,0) = 



kgZ sec 6 kQ(z-H) sec 



? - e 



E 



n+ 1 ( 2n+ 1 ) 

 (-1) m' \x) 



(40) 



, 2n 2n+ 1 



n=o k sec 



Because of symmetry 



694 



