6 
ME. Ct. I. TAYLOK ON TIDAL FRICTION IN THE IRISH SEA. 
Ijet ds be an element of length of the curve, .s, 
V = velocity of current at any point on s, 
0 = angle between element ds and direction of current, 
D = depth of bottom below mean sea-level, 
h = height of tide above mean sea-level, 
p = density of sea-water, 
g = acceleration due to gravity. 
First consider the rate at which energy is communicated to tlie portion of the sea 
which, at time, f was enclosed by the surface, S. Let S' be the moving surface 
which encloses this water. 
t' 
Fig. 1. l)iagram showing .succe.ssive positions of a surface, S, formed the vertical lines through 
the curve, s as that surface moves with the current. 
The mean hydrostatic pressure on a vertical strip of S, of height I) + /t and 
width ds is pg^{^ + ^i)- Its area is (I)d-A)d.$. The work done Iw hydrostatic 
pressure on the portion of the sea originally enclosed in tlie surface is therefore 
I Pd ( ) ('’ ^ (II + .(f f) 
J \ ^ ! 
This then is the amount of energy which has flowed during the time, dt, througli 
the surface. S', which originally coincided with the fixed surface, S, but which moves 
with the fluid. 
To find the amount of energy which has crossed the original surface, S, during the 
time, dt, it is necessary to take account of the energy contained in the fluid which 
has actually crossed the fixed surface, S. The gravitational potential energy of a 
vertical column of fluid, of height T) + Ji and horizontal cross-section csinOdfd.s, is 
evidently pg (D + h) 
A-D' 
sin 0 dt ds, mean sea-level being regarded as the surface 
of zero potential. The kinetic energy of the same column is 
^pv^ (D + A) 0 sin B dt ds. 
