Theory of Wave Resistance. 345 
any continuous distribution of pressure p(x, y), we obtain by integration 
() ene tina 
poae f p(h, k) dS [sec 0.0 | xdx; (21) 
a 4x*co 0 K — Ky sec? 8 + iu sec 0 
where now we have 7 = (x — h) cos 0 + (y — k) sin 0. 
6. We obtain the corresponding wave resistance from the rate of dissipation 
of energy exactly as in the previous sections, and we use the formula (8). 
The surface values of ¢ and 0¢/0z are put into the form (11) and the calculation 
carried out as in (14). From the similarity of the forms for ¢ in the two cases, 
the result may be written down. We obtain 
roe, (i sec? 0 dO ( ae GE yee art On)) dk 
R = Lin 
u>0 477079 o (kK — Ky sec? 0)? + py? sec? 0 
Dace i (P22 pr O2-- @,2)sec> Old 0) (28) 
Teo Jo 
where the quantities P and Q are as in (13) with f zero and o replaced by p. 
We may also write this in the form 
2 (te 
R= (P2 + Q2) sec® 0 d0, (29) 
Pee) Aaa 
with 
ol = |p (a, y) nee {ky (ec cos 0 + ysin 6) sec? 6} dS, (30) 
the latter integrations extending over the given surface distribution of pressure. 
We may obtain an alternative form by integrating with respect to x in (30) ; 
provided the pressure distribution is continuous and is zero at its outer 
boundaries, we then have 
Lor 
== L. | (P2 + Q2) sec? 6.40, (31) 
ATC 0 J-in 
with 
ot = [2 ae {ky (x cos 0 + ysin 8) sec? 0} dS. (32) 
x 
We may compare (31) and (32) with the expressions (16) and (17) for a dis- 
tribution of sources on a surface within the liquid. Suppose we may neglect 
the depth of this latter surface at every point ; then without considering the 
actual surface elevation, which would require a closer examination, we may 
say that the wave resistance for the two cases would be the same with the 
connection between the source density and the pressure distribution given by 
4rgoo = c Op/da. 
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