20 U. S. COAST AND GEODETIC SURVEY 
c/d= (true parallax of moon)/(mean parallax of moon) 
=unity 
+e cos (s—p)+e? cos 2(s—p)___-_____- (elliptic inequality) 
+15/8 me cos (s—2h+p)___--_____- (evectional inequality) 
=m? COSi 248A). 2. ean (variational inequality) (52) 
in which 
s’=true longitude of moon in orbit (referred to equinox) 
s=mean longitude of moon 
h=mean longitude of sun 
p=mean longitude of lunar perigee 
e=eccentricity of moon’s orbit=0.0549 
m=ratio of mean motion of sun to that of moon=0.0748 
The clements e and m are small fractions of the first order and the 
square of either or the product of both may be considered as being of 
the second order. In the following development the higher powers of 
these elements will be omitted. 
63. Since k has been taken as the difference between the true and 
the mean longitude of the moon, we may obtain from (51) 
je—2 Sina (S—/p) 10) 40€- SUS 7p) 
+15/4 me sin (s—2h+p)+11/8 m? sin 2(s—h) (53) 
The value of k is always small, its maximum value being about 0.137 
radian. It may therefore be assumed without material error that the 
sine of k or the sine of 2k is equal to the angle itself. Then 
sin 2k=2k=4e sin (s—p)+5/2 e? sin 2(s—>p) 
+15/2 me sin (s—2h+p)+11/4 m? sin 2(s—h) (54) 
cos 2k=1—2 sin? k=1—2k? 
= 1—4e?+ 4e? cos 2(s—p) (55) 
terms smaller than those of the second order being omitted. 
64. Cubing (52) and neglecting the smaller terms, we obtain 
(c/d)?=1+3/2 e?+3e cos (s—p)+9/2 e cos 2(s—p) 
+45/8 me cos (s—2h+p)+3 m? cos 2(s—h) (56) 
Multiplying (54) and (55) by (56) 
(c/d)* sin 2k=4e sin (s—p)+17/2 e? sin 2(s—p) 
+15/2 me sin (s—2h+p)+11/4 m? sin 2(s—h) (57) 
(c/d)? cos 2k=1—5/2 e?+3 e cos (s—p)+17/2 e? cos 2(s—p) 
+45/8 me cos (s—2h+p)+3 m? cos 2(s—h) (58) 
65. From (56), (57), and (58), we may obtain the following general 
expressions applicable to the further development of formulas (48) 
to (50). Negative coefficients have been avoided by the introduction 
of 180° in the angle when necessary. 
(c/d)? cos (A—2k) = (ce/d)? cos 2k cos A+ (e/d)? sin 2k sin A 
= (1—5/2 e?) cos A 
+7/2 e cos (A—s+p)+1/2e cos (A+s—p+180°) 
+17/2 e cos (A—2s+2p) 
+105/16 me cos (git p)+ 15/16 me cos (A+s—2h+p+180°) 
+23/8 m? cos (A— 38+ 2h) 11/8 m? cos (A+2s-—2h) (59) 
