1861.] 
The great Comet of 1861. 
287 
L — 1= 3o 38' 54" 
but L = 2830 29' 15 ' 
I = 279o 50' 21" Again 
8" = 9.33S2650 r" = 0.0129150 
Sin L" — = 9. 9567378 Cos l" = 9.9622722 
9.2950028 9.9751872 
9.9751872 
9.3198156 = sin L" — l" 
L" — l" = 12o 3' 16" 
But L" = 293o i' 45" 
I" = 280o 58' 29". 
Therefore, because the heliocentric longitudes are increasing, the 
motion is dieect. 
I" = 280o 58' 29" 
l =279o 50' 21'' 
Z" + l 
4 the sum, or = 280» 24' 25" 
2 
Z" — l 
4 the diff., or = 0° 34' 4" 
2 
3. Jfind the node. Tang ( 
l + Z" 
sin b + b 1 ' 
\ 2 
Sin 5 + 5" = 33o 54' 2'' 
) sm 
= ~~r~ 
sin b — 
■Q m 
sin b" 
: 9.7464420 
■tans: 
Sin h" — Z> = 13o 10' 32" = 9.3578119 
l" — l 
tana 
tang 
r Z" + l 
( L F-“) 
0.3886301 
: 7.9960700 
: 8.3847001 
Z" + Z 
■ a = 10 23' 21" 
but 
l + l" 
: 280o 24' 25" 
Ascending Node = 279° 1' 4" 
tang b 
4. To find the inclination (</>) . Tang - 
tang b" 
l = 279o 50' 21" 
a = 279o 1' 4" 
sin Z — - fZ 
Z" = 280o 58' 
279o 1, 
sin Z" 
29" 
4" 
-O 
z — n : 
00 49' 17" 
Z" — n = 10 67' 25" 
