525 Prof. T. H. Havelock. Some Cases of 
sources and Xz is then a circle of sources on the inverse of this line with 
respect to the cylinder. We have now for the velocity potential to this stage 
Ae D+ 2ea? e-*U-9) {(A —4) sin xx + B cos xx} de 
0 
+ 2ea? |  enwf+ ntl? (A — 4) sin (402/72) +B cos (wax /72)} de. (23) 
0 
We have put A—4 for A so as to include under the integral sign the 
doublet previously denoted by Dy. 
The method could theoretically be carried on step by step; however, we 
stop at this stage because it is sufficient for obtaining the wave resistance 
R from the pressure equation to the same approximation as by the energy 
method. 
2rr 
We have Re | ap cos 6 dé ; (24) 
0 
pip = —cdp/dx—gy + up —} 0. (25) 
If we write (23) as 6 =D+Xi+Xz, and omit terms which obviously 
contribute nothing to the value of R, we have, when r = a, 
Dye, Geac, _ 10D 0(%, + X:2) 
pan Ca (Xi + X2) +4 (Xi + Xs) A ey 
= (2c/a) sin 60 (Xi+X2)/00+m(Xit Xe), (26) 
where we have used (22) and the value of D. From (23), omitting the 
doublets D and Dy, which will from symmetry give no contribution to R when 
pis zero, we have 
p = ca? [eevee in?{ 2c A sin Osin(p—0)+pAsing 
0 
+2x«cB sin 6cos(@—0)+pBeosp}dx, (27) 
where ¢ = xacos@. Substituting in (24) we have an expression for R. We 
may now change the order of integration and take first that with respect 
to 6; we can carry this out, after some transformation, by means of the 
integrals 
oe cos (1 sin @—n@) dé = rh"/T (n+1), 
0 
[ie °89 cos (isin +n) dé = 0, (28) 
0 
where 7 is a positive, odd integer. In fact the integration with respect to 6 
gives simply mxa(xcB+pA); hence we have 
R = 4arpeat | ” (eB pA)e-2 de, (29) 
0 
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