OF THE TIDES IN THE PORT OF LONDON. 25 



Table XIX. freed from the Term 1 1 — 132 sin' 5 = 132 (sin* A — sin'S). 



From the changes of magnitude and sign, it appears that each vertical column of 

 differences may be represented nearly by a term A sin 2 ((p — y), p being the hour- 

 angle of the first column, and y a certain other angle. Also it appears that in each 

 horizontal line, A passes from positive to negative, and vice versd, when the declination 

 passes through its mean value. Hence there is a factor § — A, A being the mean 

 value of the declination. 



For declinations less than the mean, the maximum values of the correction are 

 about the hours of transit O*' 0" and 7^ 0™. This would give for y the value 3'' 30°^. 



For declinations greater than the mean, the maximum values of the correction 

 would occur nearly when the hour of transit is I** 30" or 7^ 30™. This would give 

 for y, 4^ 30™ ; the mean of this and the other value is 4'^ 



Hence the formula for the above residual quantities will be 



D ( A - S) sin 2 (^ ■- y) ; 

 where, however, instead of A — 5, we may have other functions, as sin A — sin \ 

 sin* A — sin* h. Hence the whole correction for lunar declination appears to be 



132 (sin* ^ - sin* A) + D (S - A) sin 2 {p - y), 

 which will be simplified if we put sin* S — sin* A for § - A ; the expression then becomes, 



MDCCCXXXIV. E 



