80 T. H. Havelock. 
4. Suppose the ship to be symmetrical fore and aft, and take the origin 
at the midship section. To simplify the calculations we assume, as in previous 
studies, the ship to be of infinite draught and to be of constant horizontal 
section, as shown in fig. 1. 
The length of parallel middle body is 2k, and 1 is the length of entrance or 
run ; the curved surface is of parabolic form, the equation of AB being 
y = b{1—(a—k)"/P}. (6) 
Substitute from (5) and (6) in (3). Since the normal velocity is zero over the 
parallel middle body, and since the ship is symmetrical, the integrations in 
h and h’ simplify considerably ; also, we may carry out the integrations in 
f and f’ and so obtain 
2 
Re sectn=* | J cos dé, 
0 
where 
= = | sin |S (h+-k) sec | dh. (7) 
Evaluating J, we find after some reduction 
7 
a) [ E cos? f+ = cos’ d— ; cos! ¢ sin (E sec é) 
, é) cos (eae d sec a) 
— = cos! ¢ sin pa (be), ecg ti “cost ¢ sin an 08 (2k-+}) secs 
64gob7l 
Tw 
R= 
pis = cos® ¢ cos (@ sec é) +4 (cos! o— 
cA 
4 a cos? ¢ cos {eee sec 4| is aD cos? $ cos (Po sec $)| ch ®) 
217 
