Wave Resistance of a Spheroid. 279 
in a vertical plane parallel to the direction of motion, to which previous work 
has been limited, we have 
‘0 } ‘020 co 7/2, 
Re L6mpr! | af | af’ | dh | a | M (h, f) M(h', f’) 
0 0 — —o 0 
xX e7*o (f +f’) sec? @ cos {ko (h — h’) sec 6} sec 0 d6, (18) 
where we have taken y = 0 as the plane of distribution. This expression can 
be written as 
[2 
= lewd | (P2 + Q®) sec 6 d0, (19) 
0 
where 
Ree Oe | af {a _M (hi, ff) « en Hof see? #bicgh seo 0. (20) 
0 —o 
When the distribution is in a plane perpendicular to the direction of motion, 
say the plane x = 0, it is edsily seen that we have the same expression (19) 
for R, but now 
P aL 1 = |v {a _M (kf). Em Kof Sec? @-+ixok sin @ sec? O° (21) 
0 —o 
If the doublets are distributed along a line, the suitable forms for R may 
readily be deduced from these expressions. 
Before proceeding to apply these results to spheroids, we may notice a simple 
case of (21). The first problem in wave resistance to be solved was that of a 
two-dimensional doublet corresponding to the motion of a circular cylinder 
with its axis horizontal and moving at right angles to the axis; the next 
problem was the three-dimensional doublet for the motion of a sphere. By 
means of (19) and (21) we may pass from the second problem to the first by 
integration. 
Write down the velocity potential of a uniform distribution of three-dimen- 
sional doublets of moment M per unit length over a straight line of finite length, 
the axes of the doublets being at right angles to this line ; evaluate the expres- 
sion in the limit when the length of the distribution becomes infinite, and we 
obtain the velocity potential of a two-dimensional doublet of moment 2M. 
Consider now the expression for the wave resistance for the same process ; 
if 21 is the length of the distribution, (21) gives 
cl 
Pp + 2Q) 4 | Me —rof sec? 6+ixyk sin 6 sec* 6 dk. (22) 
—l 
