DR. T. R. ROBINSON ON THE REDUCTION OE ANEMOG-RAMS. 
421 
u xcos (6— 5)+m+«£Xcos (0 -\-b)=u cos 6 cos£+m+ u cos 6 cos b— ( u — u ) sin 6 sin b 
-b 0 0+b e-b 6 0+b \0+& 6-b) 
= ( u +m+ u ) cos 6— ( u + u \ cos 0 versine b— / u — u \ sin 6 sin b. 
\0-6 0 0+b) \0+6 0-b) \0+b 0-b) 
Developing the sum and difference of the ms, which gives 
u + u =2K+2A cos 6 cos 0+20 sin 0 cos 0+ 2B cos 2d cos 20+&c., 
0+b 0-b 
u — u = — 2A sin 0 sin 0+20 cos 0 sin0 — 2B sin 26 sin 2b &c., 
0+5 0-b 
we obtain the term 
= 3u cos 6—2 versine 0{K cos 0+A cos 2 0+0 sin 6 cos 0+&c.[ 
+2 sin 2 0{ A sin 2 0+0 sin0cos0+&c.f. 
Summing round the circle, calling Su 1 cos 6= F, and remembering that all except 
S cos 2 6 and S sin 2 6 vanish, that each of these = 4, and 12A=3F in ordinary cases, we 
have 
12A=3F— 8 A cos versine 0+8A sin 2 b , 
and ultimately 
A X 4 (1 — -f versine b)= F. 
O is given by the same formula, changing the cosines for sines in F. For higher 
orders, P, it is only necessary to use jpd and j)b. In the case of D, however, the 
formula must be modified ; for in this instance S cos 2 =8, S sin 2 =0, and the expression is 
D (4+- 3 - cos versine 40) = F. The values of the constants are : — 
A (3-9091)=^+ (d—d'j sin 45°. 
B (3-6428)=s-s. 
0 6 
C (3-2190 )=d-(d—d\ sin 45°. 
0 \ 3 9/ 
D (5-333) =s+s — ^s+s^. 
The suffixes here are the same as in 
0 is 90°. 
O (3-9091)=^+^+^ sin 45°. 
P(3-6428)=s— s. 
3 9 
B (3-2190)= -d+(d+dj sin 45°. 
S =0. 
the preceding formulae. Thus 0 is 45°, 
I have compared this formula with the observations of February and March, the 
most irregular of the whole set, and the results, along with those of the preceding- 
one, are given in the following Table. The numbers are the observed — the calculated 
values. 
