1861.] 
The great Comet of 1861. 
281 
2. tang A = cos (f> — c) tang a 
cos <f> 
cos (</> — e) = 8.292358 
tang a — 9.856204 
8.148562 
cos <f> = — 9.579797 

tang A = — 8.568765 
A = geoc. long. == 177° 52' 42'' 
3. tang /3 = sin A tang ( <j> — e) 
sin A = 8.568455 
tang (<£ — e) = 1.707558 
tang — 0.276013 /3= .geoc. latitudes 62° 5' 32' 1 . 
We have next to find the earth’s longitude and radius-vector, 
which we denote by L and E. Still taking the 15th by way of ex- 
ample, we find in the Nautical Almanac ; 
July 15th. Sun’s long. Eadius Vector. 
112° 54' 7" 7 0.0070489 
Corr. for 3/i. 12 m. + 7 37 7 Corr. for 37<. 12m. — 40 
113 1 45 4 E = 0.0070449 
Add 180 
L = earth’s long. 293 1 45 4 
Eesults having been obtained, by the same process, for the 5th and 
1 0th, we may now proceed to more direct computation. We use the 
following additional symbols. 
Let 8, 8' 1 = comet’s curtate distances from the earth at the 1st and 
3rd observations. 
c — the chord joining the comet’s positions at 1st and 3rd 
observations. 
r, r" — radii vectores of the comet at the 1st and 3rd ob- 
servations. 
t, tf = intervals between the observations. 
T = whole time between 1st and 3rd. 
h, h" = heliocentric latitudes for 1st and 3rd. 
7, 7" = — longitudes. 
u, u" = arguments of latitude. 
