286 
[No. 3, 
The great Comet of 186J. 
1st approx. 8 = .1176 gives 1\ = 10.10115. Error Oi. 10115 
2nd .1160 T 2 = 9.9676 0.032 
3rd .11639 T ; , = 10.00325 |- 0.00325 
4th .11635 T 4 = 9.99964 — 0.00035 
comparing the last 2, we find the difference of the natural numbers 
.11639 and .11635 to be 4, and the sum of the errors 360 : 
.•. 360 : 35 = 4 : 04 nearly, 
or 8 = .116354 of which the log is 9.0657813 ; 
and M = 0.2724837 
.-. Log 8" = 9.3382650 ; for 8" = M 8. 
In like manner we find Log r — 9.9699588 
and Log r" = 0.0129150 
. This last value of 8 will give a time, which will not differ from 
T more than one or two seconds. 
We are now prepared for the direct computation of the comet’s 
elements. 
1. To find the heliocentric latitudes b, b>'. 
8 tang /3 S' 7 tang /I'* 
The formula are Sin b = and sin 8" = ■ 
Log of 5 = 9.0657813 Log of 5" = 9.3382650 
tang 0 = 55o 16' 15'' = 0.1591490 tang. 0" = 62» 5' 32" = 0.2760130 
9.2249303 
Log r = 9.9699588 
9.6142780 
Log r" = 0.0129150 
Sin b = 9.2549715 
Cos b — 9.9928580 
Tang b = 9.2621135 
.-. b = 10° 21' 
and b" = 23» 32' 
Sin l" — 9.6013630 
Cos b" = 9.9622722 
Tang b" — 9.6390907 
fn// ^ ] Ilelioeentrio latitudes. 
b -f b" = 33o 54' 2" 
b" — b =13o 10' 32" 
2 . 
To find the heliocentric longitudes l, l". 
8 sin L — X 8" sin L" — X" 
Sin L — l = and sin L" — l" — 
r cos b 
r h cos 8" 
5 = 9.0657813 r = 9.9692588 
149o 55' 15 Sin L — \ = 9.7000077 Cos b = 9.9928580 
8.7657890 9.9621168 
9.9621168 
Sin L — l 8.8036722 
