Falttnsen 
Pace sin@ v7 + ne 
where B,= 
l r sin @ 
A is chosen so that B, is sufficiently small. It was found 
that it was satisfactory touse B, = 0.0001. 
Each integral in (104) was evaluated by first locating the 
zeroes for the integrand. For the first integrand the zeroes are 
easily found to be u=0 and u = Spee gore + 1% m= 0, 1g 
and for the second integrand the zeroes are u = reos gms m =—Osmele 
2...Between each zero in an integrand we then used Simpson's for- 
mula. As is seen above the length of the interval between each zero 
depends on @ , and so the number of points used in Simpson integra- 
tion should depend on 6. When @ was close to 7/2, as many as 
50 points were needed in the Simpson integration. But when 6 was 
close to 0, it was only necessary touse 8 points. If A was less 
than u at the second zero of an integrand, then Simpson's formula 
was only used between 0 and A. 
9So was numerically evaluated in a way similar to that for 
= 
So. We have now explained how to obtain numerically the terms in 
the brackets of (96). We will refer to these terms as "Ursell's 
solution'' and denote them by @¢ ae So 
Os B. ¢ (105) 
(See (96) ).  ,, has been plotted in Figure 3 as a function of yr 
for different values of @ and in Figure 4 asa function of 6 for 
different values of pr. 
We now have to find B, in (105). By will of course be de- 
termined in the same way as we did in the previous chapters where 
we solved the zero- and the forward-speed problem. We prefer now 
to use the coordinate system shown in Figure l. 
If we take an outer expansion of (105), the term which is 
linear in y will be 
1804 
