461 T. H. Havelock 
force on it may be taken as the vector — 4remq, where q is the resultant 
fluid velocity at that point due to all the other sources, in which the remain- 
ing internal sources may be included as their actions and reactions do 
not affect the final result. 
It is true that for a solid of given form an important and difficult part 
of the problem is the complete determination of the internal sources so 
as to satisfy all the required conditions. However, assuming this has 
been done, we can proceed to calculate the resultant forces. Further, 
in certain problems results of some value may be obtained by using dis- 
tributions of internal sources which satisfy the conditions approximately. 
3—Take the origin O in the free surface of deep water which is stream- 
ing with uniform velocity c in the negative direction of Ox, and take Oz 
vertically upwards. Let there be a source of strength m in the fluid at 
the point (0, 0, —f). The velocity potential of the fluid motion is given* 
by 
mm «m7 2 ea (2) riko, 
BS eee. Se sec? 0 d0 | pai Sea) 
Pi IPs we Ja 0 K — Ky Sec? 0 + iu sec 0 
where the limit is to be taken as the positive quantity » tends to Zero, 
and 
r= tY EEL; re=e+yt@— fy: 
wm = xcos 6+ ysin 0; Rey = eC, 
The second term on the right of (1) is the given source, the third term 
represents an equal sink at the image point above the free surface, while 
the last term could be interpreted as a certain continuous distribution 
of sources lying in the plane z—f=0. The expression (1) may be 
generalized by summation and integration for the velocity potential of 
any given distribution of sources in the liquid. We shall assume that 
this distribution is such that it represents a solid body in the stream, the 
total source strength being therefore zero. 
4—Consider in the first place a continuous distribution over a finite 
part of the vertical plane y = 0, the surface density of source strength 
being co at a point (h, 0, —f). The velocity potential is 
b= ex+{{(t —1)odhaf 
Ty Ps 
‘CO e—* (f-2) Hike 
= {| o dh af |’ sec? 0. d0 | (2) 
up 
SS ed 
0 K — Ky Sec? 0 + iu sec 0 
* Havelock, ‘ Proc. Roy. Soc.,’ A, vol. 138, p. 340 (1932), 
409 
