ANALYSIS TO THE DYNAMICAL THEORY OF THE TIDES. 183 



On substituting the numerical values for the quantities on the right, we obtain 



C; = - 2-9242 3, C'i = 0-28546 *. 



The remaining constants may now be computed from the formulae Cg/C; = ftil/Tji, 

 g/Cg = Hg/Tjg &c., and we finally obtain 



= <[ 2-9242P; + 0-28546P; - 0'00733P^ - 0'000147Pi - 0'000007P? -....]. 



This makes the ratio of the height of the tide to that of the equilibrium tide at 

 the equator 



3-6690. 



The tide will evidently be in the main inverted, the longest period of free oscilla- 

 tion of the first class being in excess of twelve hours. 



As a further example, I have computed the series for when the depth is given 

 by the formula 



-!%. -I. -i-m-/9 

 4o>V~" 10 



that is, when the depth is 29,040 feet at the poles and shallows to 19,360 feet at the 

 equator. This series is found to be 



= *[ 2'366lP:i + 0-35649P; 0'03953P^ + 0'0037GPs 



- 0-000333Pi + 0-000029PJ, - 0'000002Pi4 + ....], 



making the ratio of the tide to the equilibrium tide at the equator 



- 3'4583. 



4 



If we put -- 3 = 3i>) and replace X by 2<a in the period-equation, regarding this 



*x(O Cv 



us an equation for K, the largest root is found to be 



K = 21,765. 



Thus there will be a period of free oscillation coinciding exactly with twelve hours 

 when 



h = (21,765 + 9680 shr<?) feet ; 



this formula makes the polar depth 21,765 feet, the equatorial depth 31,445 feet, and 

 the mean depth 28,222 feet. 



In like manner, when ;rf 3 = -fo, there will also be a period of free oscillation 



