26 U. S. COAST AND GEODETIC SURVEY 
Then 
F439 /g=3/2 UA/2—3/2 sin?Y) 2 fC cos E (81) 
Fi [g=3/2 U sn 2Y > fC cos FE (82) 
F 32 [g=3/2 U cos’?Y 2 fC cos E (83) 
Latitude factors for each degree of Y are given in table 3. The 
column symbol in this table is Y with annexed letter and digits corre- 
sponding to those in the designation of the tidal forces. Thus, Yo9 
represents the latitude factor to be used with force Fy, its value 
being equal to the function (1/2—3/2 sin?7Y). Taking the numerical 
value for the basic factor U from table 1, the general coefficient 3/2 U 
is found to be 0.8373 X 107". 
HORIZONTAL COMPONENTS OF FORCE 
80. The horizontal component of the principal part of the tide- 
producing force as expressed by formula (25), page 14, is in the direc- 
tion of the azimuth of the tide-producing body. This component 
may be further resolved into a north-and-south and an east-and-west 
direction. In the following discussion the south and west will be 
considered as the positive directions for these components. Now let 
F 3 /g=south component of principal tide-producing force 
F 3 /g=west component of principal tide-producing force 
=azimuth of moon reckoned from the south through the west. 
From formula (25), we then have 
F 3 /g=3/2 (M/E) (a/d)? sin 22 cos A (84) 
F 3 /g=3/2 (M/E) (a/d)? sin 22 sin A (85) 
81. Referring to figure 3, page 16, the angle P’PM equals A, the 
azimuth of the moon. Now, keeping in mind that the angle MPC 
is the supplement of A, the angle PCM equals t, and the arcs MC and 
PC are the respective complements of D and Y, we may obtain from 
the spherical triangle 1/PC the following relations: 
sin zg cos A=—cos Y sin D+ sin Y cos D cos t (86) 
sin z sin A=cos D sin t (87) 
Multiplying each of the above equations by the value of cos z from 
formula (31), the following equations may be derived: 
sin 2z cos A=2 sin z cos z cos A 
=3/4 sin 2Y (2/3—2 sin’D) 
—cos 2Y sin 2D cos t 
+1/2 sin 2Y cos?D cos 2¢ (88) 
sin 22 sin A=2 sin z cos z sin A 
=sin Y sin 2D sin t¢ 
+eos Y cos*D sin 2¢ (89) 
82. Substituting in (84) and (85) the quantities from equations (88) 
and (89), we have 
F 3 /g =9/8 (M/E) (a/d)* sin 2Y (2/83—2 sin?D)_______- F 39 /g 
—3/2 (M/E) (a/d)? cos 2Y sin 2D cos ¢______---- Fy |g 
+3/4 (M/E) (a/d)? sin 2Y cos?D cos 2¢_______-_-- F302 |g (90) 
Fis /g—3/2, (ME) \(G/d)esimp acme 2 Dicimuias 2 ee Fs: |g 
+3/2 (M/E) (a/d)* cos Y cos?D sin 2t________---- Fuso [g (91) 
