HARMONIC ANALYSIS OF TIDAL OBSERVATIONS. 107 
If the daily means have been cleared by the use of the tide-predicter as 
above described, these ten equations are to be solved by successive 
approximation, and we are then furnished with the two component semi- 
amplitudes, say A,, B, of the five long-period tides. But the initial 
instant of time is the first 115 30™ in the year instead of the first noon. 
Hence if as before we put R?=A,?+B,’, and tan ¢;,=B,/A,, we must, in 
order to reduce the results to the normal form in which noon of the first day 
is the initial instant of time, add to ¢, the increment of the corresponding 
argument for 115 30™, according to method (i), or for 12 hours accord- 
ing to methods (ii) or (a1). 
If, however, the daily means have not been cleared, then before solu- 
tion of the final equations corrections for clearance will have to be applied, 
which we shall now proceed to evaluate. 
For this process we still suppose method (i) to be adopted. 
Let » be the speed of a short-period tide in degrees per m. s. hour, 
\ 1 Sin lan 
and let f (n)=3, i 
to dh,, the mean height of water at 11? 30™ of the (i+1)™ day, will be 
—w(n)R cos [n {24¢4+115} —2]. 
Then we have already seen that the clearance 
If we write m=24m (so that m is the daily increase of argument of the 
tide of short period), and S=n x 114—4, this becomes 
—d(n)R cos (mi+/). 
Hence the clearance for 6h; cos li is 
— I(n)R {cos [((m+)i+]+cos [(m—]i+_p}}, 
and for 6h; sin li is 
—t(n)R {sin [((m+)i+6]—sin [((m—D]i+/]}. 
Summing the series of 365 terms we find that the additive clearance 
for 36h cos li is 
—Ri(~) {o(m+B) cos [182(m+2)+5]+4(m—l) cos [182(m—1)+/]}, 
where as before 
es a Se eA 
sin 52 
If An denotes the increase of the argument nt in 182% 11 30, this 
may now be written 
—Ry(n) {6(m+1) cos [An +1821—¢]+ 4(m—1) cos [An—1821—Z]}, 
Tf therefore R cos =A, R sin (=B, so that A and B are the component 
semi-ranges of the tide » as immediately deduced from the harmonic 
en for the tides of short period, we have for the clearance to 
20h cos li 
—[Y(x)o(m+1) cos (An+1827) +(n)o(m—1) cos (An—1821)]A 
—[¥(n)o(m+1) sin (An +1827) +Y(n)¢(m—1) sin (An—1821)]B 
