I 



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 

 1016 

 778 

 960 

 397 

 270 



295<$-150r+511*+45v-75« 

 2456 + 1 80r + 4 1 8* - 284v -f 272< 

 8405 + 3609r + 1 843* - 1 46* + 1 35. 

 16695+ 1086r + 2451* -371v+539< 

 8025 4- 1 6*- + 857* + 1 50v - 36 1 , 

 13954 720r+382*r+261v - 



879 



424 1 = + 3376 + 249r 4- 679* 4- 25v - 1 76i 



706 = 4- 1 386 + 50^4-306* - 1 83v 4- 520. 



8350 = 4- 4475 4- 1 75r 4- 1 080* 



201. 



334= 4-955 - 3^4-264* - 203v 4- 753. 



Applying the method of minimum squares, the resulting values 



of the unknown quantities are; 6 = + 5.2755, «" 



.9785, 



1C 



+4.3792, v 



5.1624, i 



2.8864. These corrections give the 



following final elements : 



Perihelion distance, .586544 



Perihelion passage, November 15.937837, Greenwich mean; 



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 each 

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





