219] ON SOME FORMULAE RELATING TO THE VARIATION OF A PLANET’S ORBIT. 517 
the latter of which is at once seen to be true, since joining the points C and P, 
the two sides are respectively equal to cos CP. To verify the former one, write 
ZPCM — C X , ZPCN—C 2 , so that C = C 1 — C 2 . Then, since cos CP = cos PM cos CM = 
cos PN cos (717, sin PM = sin CP sin C\, sin PN = sin CP sin C 2 , the equation becomes 
cos CP (tan CM — tan CN) = — tan C sin CP (sin C 1 + sin C 2 ), or since tan CM = tan CP cos C x , 
tan CN — tan CP cos C 2 , this is 
cos C x — cos C 2 = — tan | C (sin C x + sin C 2 ), 
which is identically true, in virtue of the equation C — C 1 — C 2 ; and, conversely, we 
have the original two equations. 
Suppose that XM is the ecliptic, X being the origin of longitudes, DP the 
instantaneous orbit, D the departure-point therein, and P the planet, DD 0 the orthogonal 
trajectory of the successive positions of the orbit; and writing 
J?, the departure of the planet, 
v, the longitude of ditto, 
y, the latitude of ditto, 
6, the longitude of node, 
er, the departure of ditto, 
<fi, the inclination ; 
then, in the figure, DP = ]?, XM = v, PM — y, XA — 0, DA = ar, Z A — $>. 
The quantities 0 O , cr n , </>„, might be considered as altogether arbitrary; but to fix 
the ideas it is better to assume at once that they denote 
0 O , the longitude of node, 
<r 0 , the departure, 
</>„, the inclination, 
for the initial position of the orbit, viz., in the figure XB o =0 o , D 0 B 0 —cr 0 , Z P 0 = <£o- 
Take DB = ar 0 , Z B = </>,,, BX o = 0 o , this determines a travelling orbit of reference 
X 0 N, and origin of longitudes X 0 therein; such that, with respect to this travelling 
orbit, the -position of the planet’s orbit is determined by 
0 O , the longitude of node, 
cr 0 , the departure of node, 
<f) 0 , the inclination. 
We have in the triangle ABC, AB = a — a 0 , ZB = cj) 0 , ZA = 180° — <f); and if the other 
parts of the triangle are represented by 
BC — o), 
AC=0 o -0 + co + T, 
ZC = A>;