299 T. H. Havelock 
With this value of ¢ in (8) we obtain the rate of propagation of energy outwards, 
and this must be equal to RhQ with R the steady wave resistance. Finally, replacing 
M by 4a%h, we obtain for the wave resistance of the sphere 
4r*pabQ* 2 
gh 
It is of interest to examine the limiting form of this expression as hoo, Q>0, 
hQ->c. The series then becomes an integral, and by using appropriate asymptotic 
expansions for Bessel functions of large order and large argument, it is found that 
(12) reduces to ia 
R = 4rpa’k§ Al sec 6 exp[—2k,fsec? 6] df, (13) 
0 
with ky = g/c?, and this is the wave resistance for a sphere in steady rectilinear motion 
with velocity c. 
R= Bim dnk (n?.Q2h/g) exp [ — 2n?Q?f/g]. (12) 
4. Returning to the general expression (6) we may evaluate the resultant fluid 
pressure on the sphere and so obtain both the radial force and the tangential force 
or wave resistance. The effective part of the pressure comes from pd¢/et, and we 
notice from (6) that the terms divide into two groups, (i) those symmetrical in the 
angle 0—t, (ii) those anti-symmetrical in that angle. Obviously the resultant 
radial force on the sphere comes from the terms in group (i), while the tangential 
force is due to those in group (ii). It is necessary to note that, in using this method, 
the expression for the velocity potential must be carried to a further degree of 
approximation, because the boundary condition at the surface of the sphere must 
be satisfied to the same stage. Let (r,a, /) be spherical polar co-ordinates referred 
to the centre of the sphere so that 
mw cos (9— Qt) = h+rsina cos f, 
o@ sin (9— Qt) = rsina sin £, | (14) 
z2=—f+rcosa. 
The first term in (6) is the doublet D giving the correct normal velocity at the surface 
of the sphere. The remaining terms in (6) may be expanded in the neighbourhood of 
the sphere in spherical harmonics so that we have (6) in the form 
o= D+ 3 ria)" S,,(%, 8). (15) 
The required extension is then 
fa) r \ 2 nN a n+1 
O= p+3 (7) + (6) | s,(aA). (16) 
Taking the tangential resultant force, the effective terms in p0¢/0t from (6) are 
(2mpa%Q?/g) ¥ n4J,(k,) Jy(k,,h) exp [—K,(f—2)] sin n(0— 2). (17) 
1 
In this, we put 
J,(K,@) exp [—K,(f—2)] sin n(8— Qt) 
= ( a exp [|—«,(f— aif” exp [ix,,w cos (9 — Qt —u)] sin nudu 
ak — exp[— ZF exp [ik,, r(sina cos # cosu+sin« sin £ sin u—1 cosa) 
—ik, heosu]sinnudu. (18) 
556 
