Wave Resistance. 28 
Horizontal Doublet. 
3. To consider three-dimensional fluid motion, take first a horizontal doublet 
of moment M at the point (0,0, —f). Assume that the velocity potential can 
be expressed in the form 
Tr 00 
db ee =| | Ke * +f) + tx (x cos 6+y sin 6) cos () d9 dk 
a i | KE (0, x) e* J tHe (2 008 +950) eo3 @ dO dx, (14) 
0 
=o 
where real parts are to be taken, and where the first term is the velocity potential 
of the given doublet in a form valid for z+ f > 0. 
The surface condition is equation (2) as before; applying this, we obtain 
uM k + ky sec® 0 + tu sec 0 
= 5 15 
ea) Qn K — Ky sec? @ + ip sec 0 ) 
Hence from (14) and (15) the surface values of ¢ and 0¢/0z are 
9 Cd i) —xf tix (x cos 6+y sin 6) 
| | —_—*n sec 8 d0 dk, (16) 
tT J—zJo K — Ky sec” 0 + tp sec 0 
‘ 1 C2 —«ftix (x cos 6+y sin 4) 
as =| | g : Ale Ankes am O0bde (Om) 
Oz T 0 K — Ky sec” 0 + ip sec 0 
ip 
Taking real parts of these expressions we obtain 
‘= | : i {48 (O¥c)icosl (ier cog 6) eos) (reyisinl6) 
ae + F, (0, «) sin (Kx cos 0) cos (ky sin 0)} « dk d0, (18) 
and a similar form for 0¢/0z with G instead of F, with 
F, = M«yue” sec? 0./D 
F, = — Mky (k — ky sec? 8) e~” sec 0./D 
G, = Muxyxe~“ sec? 0./D 
G, = —M {« (k — kp sec® 0) + 2 sec? 6} xe"! cos 0./D 
D = zx {(k — ky sec? 6)? + p? sec? 6}. (19) 
We now apply a theorem in two dimensions corresponding to that given in (10). 
The Fourier integral theorem is 
Fay = 5] du | av | ( F (s, t) cos u(x — s) cos v (y — t) ds dt. 
282 
