Minimum Wave Resistance Problem Solution fil 
A ee a ee +1 +1 oo 4) 1- 2 
peur Ye | hs ye el ey nen «(=F 
Li 1 
f(é,) 
0 0 
i MK T( CHL) LK 4 dr 
x e s 
co - —___—., 
1 2 - 1 
In order to obtain a simple integral expression, we neglect the dependence of g on € and 
assume infinite depth. An estimate of the validity of this procedure is obtained by consid- 
ering the case, where the cross sections are rectangular: 
B(e) = 1, OPS a 
&u)=0, w>il. 
Then 
ECS) 
ViLes 
bvi-€? \ rere titers caamsiciens 
at 8 dt = e dC 
RCE). S A 
-\2KT£(E) 
= i 1 => 3 V1 = 2 
2 KT 2 
which integral is simply replaced by 1/A°xT. 
If necessary, the method can be refined, yielding more complicated integrals which can 
be treated by numerical methods. The simplified expression for R is 
2p2y7 2,2 . ™ cos ¥(6=€,)r 
——— im «fo Be oe 
1 2-1 
or, using the formula 
V[y(é-6))] = - 2( —— dn, 
pic?B2L 2,2 +1 +1 
ps eee ee { dé | a, M2) 4.) Y, [v.(E- £1] | 
