12 
BEY. S. HAUGHTON ON THE TIDES OF THE 
or, from (38) and (39), 
Aj= {0'69S cos^+O'llMsin <£} 
+M { — 0 - 35 cos ^>+0 , 29 sin <£} cos u 
+M{ — 0’34 cos (f>— 0’30 sin </>}sin u (40) 
Bj= {0‘69S sin^— 0'4lM cos <£} 
+M{0 , 30 cos 0'S5 sin (f)}cosu 
+ M{ — 0’30 cos </>+0’34 sin <£}sin u (41) 
We have, therefore, from Tables III. and IV., 
0'69Scos* s +0 , 4lMsin(/)= — 11 ’7 (a) 
M (—0\35 cos <£+0-29 sin <£) = + 8‘4 (6) 
M (— 0‘34 cos cf)— 0’30 sin <£) = — 0‘5 (c) 
0'69Ssin^— 0‘4lMcos </>= — 13 - 2 ( d ) 
+M (0-30 cos <H-0-35 sin <£)= + l-4 (e) 
+ M ( — 0‘30 cos ^>+^’34 sin ^>) = + 8‘0 (/) 
From ( b ) and (c) we find 
M= 19‘4 inches, </>= 128° 5', 
choosing that value of </> which makes M positive in the equations ( b ) and (c). 
From (e) and (f) we find 
M=l7-6 inches, <f>= 129° 30'. 
These values of M and <£ agree very well, and taking means, we have 
M=18’5 inches 
<£=128° 48' (42) 
but 
$=— c-f 6— i m , 
and Conjunction happened at June 29 d 0 h 
and ascending node . . . July 19 d I7 h 
Diff. 20°-7 
Hence 
and s' increases at the rate of a degree per day, or s'=20° 42'. 
Therefore 
i m =—c+s'—<f> 
= —360° 36'= — 0° 36' 
or, converting the arc into time 
i„ n =— 2 m> 48 . . , . 
(43) 
