463 T. H. Havelock 
and 
2i | en S01) ln ey) C08 A AP) anemia agp 
‘ m + K,92 sec! 0 
— Ark i sec” Be~*0(I+1) se" 8 Cos {ey (h’ — h) sec 9}, 
for h’'—h<0, (7) 
the function defined by (6) and (7) being continuous at h’ —h=0. 
Writing —iF (f’, f’, h, f, ®) for this function, we have 
| 
R = See | o dh af’ || o dh df | F(W’,f’, h, f, 8) sec8d0. (8) 
J 0 
It is obvious from (6) and (7) that the part of F from the integrals in m 
will give zero result when integrated twice over the distribution; and 
we are left with 
R= 32n.2p {| o’ dh’ df’ | | o dh df 
wr /2 
x | sec? De" (F1) see’? Cas {xy (h’ — h) sec 8} d0, (9) 
0 
with h’ — h < 0, the integrand being zero for h’ — h > 0. 
This is the wave resistance expressed in a form which brings out more 
clearly than the usual forms the fact that the solution we require is one 
in which the regular waves trail aft from each element of the distribution. 
It is easily seen that the limitation h’ — h < 0 in (9) is equivalent to 
taking one-half the result of the repeated integration over the distribution 
without this limitation. Hence we obtain the result 
R = Gree [- (P? + Q2) sec? 6 d0, 
0 
with 
P = iQ = \| ceikok sec 6—xof sec? 6 dh df. (10) 
This agrees with the general result obtained from energy considerations 
in the paper already quoted, where the distribution was not necessarily 
confined to the plane y—O. There is no difficulty in extending the 
present method to more general cases, but that is left over until occasion 
arises for applying the results to some particular problem. 
5—To proceed to the case of two bodies, it is only necessary to suppose 
that the distribution of sources is divisible into two parts, each contained 
