Wave Resistance 462 
where 
BPS Ce = WP ap PPP GaP py 
np = (6s = Rae Soap Gir 
@ = (x —h)cos 0+ ysin 0. 
Using the theorem given in § 2, we may write down the total wave resis- 
tance for the body which is represented by the given distribution. It is 
given by 
Ree are | | o (h', f’) u dh’ df’, (3) 
taken over the distribution, u being the x-component of fluid velocity at 
the point (h’, 0, —f’). 
Consider the contribution of the various terms in (2) to the value of u 
in (3). We may omit the uniform stream c since the total source strength 
is zero, and also any contribution from the internal sources. Further, it 
is easily seen that there is no total horizontal force on the internal sources 
due to the image system represented in (2) by the term involving rz. 
Thus the only part of u which gives any integrated effect from (3) comes 
from the x-derivative of the last term in (2). Thus we obtain the expres- 
sion 
R= Acre | | a! dh’ df’ {Je dh df | "sec 0 d0 
8 ea SAS) +i (h’—h) COs 6 
x| de. (4) 
Ss}? —————_ 
0 K — Ky Sec” 0 + iu sec 0 
The integrations in 8 and « may be written as 
7/2 co ff eit (’—h) cos 6 
2 | sec 0 d0| )— 
0 9 le — ky sec? 0 + ip sec 6 
e7* (h’—h) cos @ : 
we oes eS) dx. (5) 
kK — Ky Sec? 0 — iv sec 8 
Regarding « as a complex variable we may transform the integrals by 
taking as contour an appropriate quadrant bounded by the positive half 
of the real axis and the positive or negative half of the imaginary axis 
according to the sign of h’ —h. Reducing the expressions and finally 
putting » zero, the integral with respect to « in (5) is equivalent to 
— 2i| Ko Sec” 0 sin m Gise i) — MCcos Mm Gi if ’) e7m (h’—h) cos om dam 
a m? + Kk, sect 0 z 
for h’ —h>0O, (6) 
