Observations on the Comet of Halley. 215 



tance by .0001 ; the time of perihelion passage by .01 ; and the 

 other three elements, each by one minute. The resulting equa- 

 tions were : 



444= -2955- 150r+511*+45v-75! 



1016=-245<5 + 180r+418-7r'-284v+272. 



778=— 8405-|-3609r+1843^-146v + 135. 



960=- 16695+ 1086r+2451*-371v + 539. 



397=-802'3 + 16t+S57* + 150v-361. 



270= - 139H 720r+382^4-261v -879j 

 4241 =+3375 + 249r4-6797f+25v-176i 



706 = + 1 385 + 50r + 306* - 1 83v + 520» 

 8350 = +4475 + 175r+10807r-v-20ii 



334= + 955 - 3r+264*- 203v+753i 



Applying the method of minimum squares, the resulting values- 

 of the unknown quantities are; 5 = + 5,2755, t=— .9785, *= 

 +4.3792, v= -5.1624, i = — 2.8864. These corrections give the- 

 following final elements : 



Perihelion distance, .586544 



Perihelion passage, November 15.937837, Greenwich meatt 



time from noon. 

 Longitude of perihelion, 304° 31' 17'^ 

 Longitude of ascending node, 55 7 55 - 



Inclination of orbit, - 162 13 



With these elements, I recomputed the comet's place for eaclm 

 time of observation, and the result is shown in the following table : 



