192 The N.Z. Journal of Science and Technology. [Aug. 
The values of the mean anomaly are derived from Kepler’s equation, 
which is ,, , m n • -n 
M = n (t — T) = E — e sm E, 
where M is the mean anomaly, n is the mean daily motion, T is the time 
of perihelion passage, t is the time corresponding to M, E is the eccentric 
anomaly, e is the eccentricity of the ellipse. 
For M, n, E, and e in seconds of arc the equation is 
M = 611-565" (t - T) = E — 139826-91" sin E. 
The first step is to find the polar co-ordinates of the comet for given 
values of E. The selected values of E are 0°, 5°, 10°, and 15°. The 
calculations when E = 5° are given here. For E = + 15°, M = 5813'29" ; 
hence t — T = + 9*506 d. ; but T = June 12-95, hence t = June 3*444, 
and t — June 22-456. 
The radius vector r is obtained from r — a — ae cos E = 1-04833. 
The true anomalv w is obtained from tan 4 w = \/ C-— 3 tan 4 E or 
tan i w = 2-28238 tan J E ; w = 11° 22' 53*6". 
The values of r and w are checked by the following equations : 
r sin w — b sin E, r cos w = a cos E — ae, and give 
r sin w = 0-206879 \ r cos w = 1-027711 
b sin E = 0-206879) a cos E — ae — 1-02771 j 
In this way the following polar co-ordinates of the comet in the plane 
of the orbit and with the Sun in the focus are obtained :— 
E. 
r . 
w . 
Dates, 1921. 
- 15° 
1-1146 
- 33° 
26' 
54" 
May 
14-830. 
- 10° 
1-0732 
- 22° 
35' 
06" 
24-788. 
- 5° 
1-0483 
- 11° 
22' 
54" 
June 
3-445. 
0° 
1-0400 
0° 
0' 
0" 
12-95. 
+ 5° 
1-0483 
+ 11° 
22' 
54" 
22-455. 
+ 10° 
1-0732 
+ 22° 
35' 
06" 
July 
2-112. 
+ 15° 
1-1146 
+ 33° 
26' 
54" 
12-070. 
The next step is to calculate the heliocentric equatorial co-ordinates of 
the comet. 
When a number of positions is to be calculated considerable labour is 
saved by first calculating the constants for the equator. The constants for 
the equator are derived from the following equations :— 
sin a sin A cos S3, 
sin a cos A. = — cos i sin S3, 
sin b sin B — cos e G sin S3, 
sin b cos B = cos e D cos i cos S3 — sin sin i, 
sin c sin G = sin e 0 sin S3, 
sin c cos C = sin e o cos i cos S3 ~f cos sin i, 
where e G - the obliquity of the ecliptic = 23° 27'. 
The solutions of these equations give 
A = 188° 17' 17", log. sin a = 9-97658, sin a = + 0-94749, 
B = 105° 12' 29", log. sin b = 9-97396, sin b = + 0-94179, 
C = 58° 10' 12", log. sin c = 9-66651, sin c = + 0-40400. 
The constants are checked by the following equation :— 
sin b sin c sin (C — B) = sin a cos A tan i 
sin b sin c sin (C — B) -- 0-31979) 
sin a cos A tan i — 0*31979 j 
This affords a sufficient check. 
