422 T. H. HAVELOCK 
Hence, for the quantities P, we have the infinite set of equations 
os t co 
P,(t) = ca?—1a? | F(«, t)ke—**F dx —gta? ( E(k, 7) dr | L*xke—"F dic, 
0 0 i) 
qrqen 
P(t) = — al | F(t, there2F die + 
0 
pr+1 7k 720 e 
a i eras | Lit tte2F dc. (15) 
4. We shall only attempt an approximate solution of these equations 
as far as the second order, that is, up to m = 2. It may be noted that the 
condition at the free surface is satisfied exactly, but the condition on the 
circular boundary is only satisfied approximately to the order indicated. 
For the forces (X, Y) on the cylinder we use the general expression suitable 
for axes moving with the cylinder (2), which is in this case 
X—1Y = tpi {Gy dz-+-mpa?é— pis, | w* dz*. (16) 
We shall find it convenient to take the corresponding resistance in two 
parts; thus, to the present order, 
. [ [dw\? 
i = Re{ —4pi | (a) az| 
Pa —mpaé-+Re pic, | w* dar} = ee 2p Re| apy. (18) 
8 Re {P, P*} (17) 
Further, from (15), P, and P, are oe by, 
a2 
B= oa? FP watt me FRO APE ie ae te-2S dic 
(19) 
4 4 : 
_ tat, 3a Palen at } £ clo—2Kf 
P= — FP iat i (P(r) Plo a |B oF dic. 
i (20) 
If we neglect gravity, we have approximately 
P, = ca?(1—a?/4f?); P, = —ica’/8f?. (21) 
From (17) and (18), A, is zero and 
R, = mpa*é(1—a?/2f2), (22) 
the coefficient of zpa?é being, to this order, the effective inertia coefficient 
for a free surface, neglecting gravity. The next step is to use these first 
approximations for P, and P, in the integrals in (19) and (20) and so 
obtain the next approximation. 
548 
