Wave Resistance. 32 
(1, m, n), and M’ at (h’, kh’, —f’) in the direction (I’, m’, n’). We have simply 
to put the surface values of ¢ and 0¢/0z in the standard form, and evaluate 
the quantity XFG. The reduction need not be reproduced here; omitting 
terms in p. which make no contribution in the limit, we obtain 
TD EG . {( — Ky sec? 0)? + pu? sec? O}/icyx? sec? 0 
= [(l cos 8 sin P cos Q + msin 6 cos P sin Q — n cos P cos Q) Me“! 
+ (I cos 0 sin P’ cos Q’ + m’ sin 0 cos P’ sin Q’ — n cos P’ cos Q’) Me“ "2 
+ [(lcos 8 cos P cos Q — m sin 0 sin P sin Q +n sin P cos Q) Me7** 
-+ (U’ cos 6 cos P’ cos Q’— m’ sin 6 sin P’ sin Q’ +n’ sin P’ cos Q’) M’e~ "2 
+ [(L cos 0 sin P sin Q — m sin 0 sin 0 cos P cos Q — ncos P sin Q) Me~*! 
+ (I' cos 6 sin P’ sin Q’— m’ sin 0 cos P’ cos Q’— n’ cos P’ sin Q’) M’e~*F'2 
+ [(lcos 6 cos P sin Q + m sin 0 sin P cos Q + sin P sin Q) Me~*f 
+ (V’ cos 6 cos P’ sin Q’+ m’ sin 6 sin P’ cos Q’+-n’ sin P’ sin Q’) M’e~"F' 2, 
(30) 
where P = kh cos 0, Q = «ck sin 0, and similarly P’ and Q’. Carrying out the 
rest of the calculation for R, the wave resistance is given by 
1/2 ; 
R= 16px! | [iw cos” § + m? sin? 6 + n2) M2e~ of sect 6 
0 
++ (I? cos? 8 + m/’? sin? 6 + n’2) M’2e~2xoF’ sect é 
+ 2 {(ll’ cos? 8 + mm’ sin? 0 + nn’) cos A cos B 
— (Im’ + I’m) sin 6 cos 0 sin A sin B + (xm’ —n’m) sin 0 cos A sin B 
+ (nl’ — n'l) cos 8 sin A cos B} MM’e~" (+4) sec® ‘| sec’ 0d0, (31) 
where A = ky (h — h’) sec 0, B = ky (k — k’) sin 0 sec? 0. The various terms 
represent the contributions of the three components of each doublet and their 
mutual interference in pairs. 
Water of Finite Depth. 
7. For water of finite depth h, we shall consider only the simplest case of a 
horizontal doublet of moment M at depth f. It is clear that the same surface 
integral can be used for evaluating the wave resistance. 
We now assume the velocity potential in the form 
= —F| oe 0 a0 | {e-*f 4 eR ennetir (x cos O-+Y SiN 8) 1 dye 
ate J) — 0 
+ | cos a0 | F (0, «) cosh « (2 + h) ef (© 008 OY 8i0 9) 1. dye, (32) 
0 
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