416 T. H. Havelock 
These may be compared with the corresponding values at 9 = 0 for the 
circular cylinder. When xa is small, we obtain from (23) the approximate 
value 1 + (Ka/2)? for the ratio of the amplitude to that of the incident waves. 
We may, possibly, use this to give an upper limit for the resultant amplitude 
at the bow of a ship if we regard the front half of the ship as a parabola with 
its vertex at the bow. For instance, consider the model examined in § 3. 
Instead of a wedge-shaped bow, suppose it is rounded off into a parabola 
with its vertex at the bow and joining on to the parallel middle body at 
a distance of 5-81 ft. from the bow, the beam at that point being 1-5 ft. 
With these data, and taking the same wave-length of 16 ft., we find that 
ka = 0:019. From the approximate formula, this gives a relative amplitude 
at the bow of 1-17. Comparing with the previous calculations, this seems 
a reasonable estimate, in spite of the various assumptions; the ratio would, 
of course, be greater for smaller wave-lengths. 
THE PRESSURE OF WATER WAVES 
6. For the resultant pressure upon the obstacle, the first order effect 
is a purely periodic force with zero mean value; this was the effect considered 
in the previous paper (1937) and applied to a ship among waves. To obtaina 
steady mean force different from zero we have to proceed to second order 
terms; although much work was done at one time on the pressure of vibra- 
tions, water waves do not seem to have been considered in this connexion. 
We begin with plane waves, and the only general result we need is that 
given by Rayleigh (1915), that the usual first order expression for the 
velocity potential is also correct to the second order, the next term being 
of the third order; this was shown to be the case both for progressive waves 
and for stationary oscillations. There are, however, second order terms in 
the surface elevation. 
Consider plane waves incident directly upon the plane x = 0 as a fixed 
boundary. We take 
@ = (2gh/c) e cos kx sin ot, (24) 
€ = 2h cos xx cos ot + 2kh? cos 2xx cos? at, (25) 
with o? = gx. We have 0¢/éx = Oat x = 0; for the pressure we have 
2 ge pee — lf C2V (GF 5 
p= F(t)—gpz+p Ar ~40l(=) +( I (26) 
It may be verified that, with 
P = — 2gpxh? cos 20t — gpz + 2gphe cos Kx cos ot — 2gpKhe?*? sin? ot, (27) 
477 
