280 
[No. 3, 
The great Comet of 1SG1. 
A P = 70" 5' 40" P V = 51° 20' 32" 
A V == 59° 2' 16" p Y A = 88° 22' 0 
A P Y = 65° 43' 50" P A V = 56° T 0 
During the first half of July the comet was without the triangle : 
during the latter half within it. The observations of the 15th for 
3 h. 12m. Greenwich mean time (97*. 5 m. 33s. mean time at Chinsurah) 
will show how the R A and Dee. were determined. A C, measured 
by the sextant, corrected for index error, was 35° V 48" and C V 
was 44° 3' 0". Then P, C is evidently the comet’s N P distance, 
and A P C, V P C, its differences of R A from Arcturus and Yega. 
I. In C A V (3 sides given) find C A V = 52° 45' 10" 
P A V = 56° 7' 0 
P A C = 3“ 21' 50" 
2. In P A C (2 sides and included find PC = 35° 9' 28" 
.■. Comet’s Dec. = 54° 50' 32" North. 
3. In the same A, find A P C = 3° 21' 11J" = Oh. 13»*. 25s. 
Right Asc. of Arcturus 147*. 9m. 22 s. 
Comet’s AR = 147*. 22»*. 47s. ; 
and Comet’s Dec. N. 54° 50' 32". 
Comet’s distance from Eta Urs® Majoris. Measured 7° 51' 20" ; 
and computed 7° 52' 30". 
The next step is to reduce the right ascensions and declinations to 
Geocentric longitudes and latitudes. This is done by help of the 
following formula ; in which 
a == Right ascension. A = longitude. ' 
8 = Declination. f3 = latitude, 
f = Obliquity of Ecliptic. 4> = auxiliary angle. 
I select again the 15th as an example. 
tang 8 
1. tang — tang 8 = + 0.152230 tang <£ = (1) 
sin a 
Sin a == — 9.765896 (3rd quadrant.) 
<£ = 112° 20' 4" = — 0.386334 = tang <f> 
£ = 23 27 28 
</,— £= 88 52 36 
