280 T. H. Havelock. 
Evaluating the integral and using (19) we have 
LA pn o 2 
R= Btrpxg?ME| sin” (xgl sin 0 sec” 8)  — Prof sec? 6 JO. (23) 
0 sin? 0 cos 0 
The wave resistance for the corresponding two-dimensional doublet is for 
unit length perpendicular to the plane of motion, and should be given by 
lim (R/2l) asl > 2. From (23), this is 
Lim 3279)? M2 an sin? (qu V1 + u2/I?) eof? yw V/T 2/2 du 
l>o 0 
= 16r?oK,? M@e-7" | (24) 
and this is the known expression for the wave resistance of a two-dimensional 
doublet of the corresponding moment 2M. 
5. We proceed now to the wave resistance of a submerged spheroid, taking 
im each case the axis of the spheroid to be horizontal and at a depth f below 
the surface of the water. 
Prolate Spheroid in Direction of Axis.—From (7) and (20) we have 
(P — 1Q)/ Auer "oF see ess ic (ae? ue h?) etkoh sec 6 Jp, 
—ae 
= (8rraFe3/cg? sec? 0)? Jap (Kgve sec 8), (25) 
where J denotes the usual Bessel function. Hence from (19), 
mr [2 sures 
R= 128xigpatea? | EW Prot 82° 8 LT 15 (ky ae sec )}? sec? 0 dO, (26) 
0 
a result which was obtained previously by a different method.* For purposes 
of numerical calculation it is convenient to change the variable in the integration 
from 6 to tan 8; we then have 
R = 128rgpate?A%e? | e™ {Taio (gae V1 + #2)}? dt, (27) 
0 
where p = 2x) f = 2qf/u?, and A is given in (8). 
Oblate Spheroid in Direction of Axvis.—Here we have a surface distribution 
given by (10), and remembering that the centre of the circular distribution is 
at a depth f, (21) gives 
(P ak 1Q) /Bue="F sec? 9 __ {| (b?e’? ak. y? eats EPpy ekoe Sec? 6+7%xay sin @ sec* 6 dy dz, (28) 
taken over the circle y? + 22 = be’2. 
* “Proc. Roy. Soc.,’ A, vol. 95, p. 365 (1919). 
