Wave Resistance 466 
By the method already described, this is transformed into 
ali woe OE (3 Ko sec? 0 sin cli m COS 2mf . —mae0s 93 dm, 
| m? + Kk 92 sect 6 
for x >0, (18) 
a i aes Od) le Ky sec? 9 sin 2mf — m cos 2mf ome cos 8 3 din 
0 Jo m? + Kk, sect 6 
pr/2 
— 8rK,? | sec? De **F 8°"? Cos (gx sec 0) dO, 
0 
or ae<@, ils) 
the two expressions having the same limiting value as x tends to zero. 
Writing Ry, for the resistance of either doublet if existing alone, and 
using these expressions in (16), we get at once the known result 
cm] 2 
Ry = 16xpx,2 M2 | sec? He-2*of sect? JA), (20) 
0 . 
Considering R,, the contribution from the doublet M at B and the image 
doublet —M is easily found to be —R’ where 
ee pI EES) 
R’ = 24n0M tao oe ie (21) 
It may be noted that if we put M = 45%c, the first term in (21) gives 
6zeb® c?//*, which is the usual approximate value of the repulsion between 
two equal spheres moving in the line of their centres in an infinite liquid. 
Finally, for R,, there is the term which comes from (18) and (16); we 
denote this by —R”, with 
kK, sec? 0 sin 2mf— mcos 2mf onmlcos. 
- m dm. 
0 m? + kK," sect 0 
7/ 
R” = 16x eM? | 
) 
wens 0 d0 [. 
(22) 
If we calculate R;, remembering that A is in advance of B, the forces R’ 
and R” are reversed, and we have in addition the effect of the second term 
in (19). We obtain finally 
R, = R, — R’ — R” (23) 
7/2 
[Reg == TR qe TRY Se IR Se 32pxy!M | sec® Def se? COs (Kol sec 0) dO. 
0 
(24) 
The sum of Ry, and R, is the result which would be given by energy 
methods for the two parts regarded as one system. In (23) and (24) 
we have the separate resistances with the wave-interference part assigned 
414 
