16 
REV. S. H AUGHT ON ON THE TIDES OE THE ARCTIC SEAS. 
Substituting, in these equations, the values of A, B, A', B', we find 
0‘60 Mcos 2<£+0 , 86 S cos^=3 - l .... 
M (075 cos 2<H-0*64 sin 2<£)= 13‘3. . . . 
M (076 cos 2^— 0*61 sin 2<£)= — 10*2 . . . 
— 0*60 M sin 2</>+0’86 S sin ^=6*0 . 
M (0‘64 cos 2(f)— 0‘75 sin 2<£= — 8*2 . . . 
M (0’61 cos 2<£-f-0*76 sin 2<£)=12‘2 . . . 
From ( b ') and (c') we find, remembering that M must be positive, 
2<£=84°48', M=19*0 inches 
and from ( d ') and (/') we find 
2<f)=77° 10', M=14*0 inches. 
Mean values 
2<^>= 80° 59', M=15'5 inches. . . 
• («') 
• (V) 
■ M 
• m 
■ m 
■ (/') ( 53 ) 
(54) 
Hence we find, since, 
using the values of s' and c already given, 
i m =S7° 42'= 6 h 2 m 5 8 
(55) 
From (a') and (d') we find, using the mean values of M and 2 </>, 
S=5-9 inches. ^=41° 55'=2 h 48 m (56) 
We thus find, finally, from the present and former calculations, 
Semidiurnal Tidal Constants. 
-Hourly Observations. 
M=15 - 5 inches 
i m = 6 h 2 m l 
S=5’9 inches 
t,= 2 h 48 ta 
S_ 
M' 
:0-39 
II. — H. and L. W. Observations. 
M=17‘0 inches 
4=23 h 48 m 
S=7‘0 inches 
AV 
M 
