108 REPORT—1883. 
In precisely the same manner we find the clearance for Sch sin li to be 
—[d(n)o(m+1) sin (An+1821) —Y(n)¢(m—1) sin (An—1827)]A 
+[w(n)g(m+1) cos (An +1821) —L(n)o(n—D) cos (An—182/)]B 
These coefficients may be written in a form more convenient for com- 
putation. For 
sin 2§5(mE1) 
S| f) a MIA ol ed 
ne) ae aD, 
=}cos 182(m+1) +4 sin 182(m+l) cot (ml). « (70) 
Then let 
K(n, )=d¢(m4+l) +¢(n— 
(n, D=o(m+1)+9(m—D | = 
Z(n, 1)=$(m+1)—d(m—) } 
Also let 
Aah ea sin 12n | yeaa ee 
¥(n) cos An=5), Scere cos An=C (i) er 
Y(n) sin An =S(n) } 
The functions K(n, 1), Z(n, 1), C(n), S(x) may be easily computed 
from (70), (71), (72). 
Then if we denote the additive clearance for Sc h cos li by 
TA, n, 1, cos|A+[B,’n, 7, cos]B, 
and that for 36h sin li by 
[A, 2, J, sm]A+[B, n, J, sin]B. 
We have 
[A, n, 1, cos]=—C(n)K(n, 1) cos 182174 S(n)Z(n, 1) sin 1821 
[B, n, 1, cos] = —S(n)K(n, 1) cos 1821—C(n)Z(n, 1) sin el 73) 
[A, n, 1, sin]=—S(n)Z(n, 1) cos 1821—C(n)K(n, 2) sin 1827 \ 
[B, n, 1, sinJ= C(nr)Z(n, 1) cos 182/—S(n)K(m, 1) sin 1821 
We must remark that if 4(m+1)=3860°, ¢(m-+1) is equal to 182°5. 
This case arises when J is the tide MSf of speed 2(o—n), and m the 
tide M, of speed 2(y—<), for m+ is then 24 x 2(y--)=720°. 
The clearance of the long-period tide J from the effects of the short- 
period tide requires the computation of these four coefficients. For — 
the clearance of the five long-period tides from the effects of the three tides 
M,, N, O, it will be necessary to compute 60 coefficients. 
If it shall be found convenient to make the initial instant or epoch for 
the tides of long period different from that chosen in the reductions of — 
those of short period, it will, of course, be necessary to compute the 
