86 General Method for detcrmhibig 



Suppose, for example, that we employ the two formnlce 

 (e) and [e') : it is visible that the tangent of" j may equally 

 belong to the two angles j, and 180° + j, j, being the" 

 smallest of the positive angles to which it can belong. In 

 order to determine which of the two to choose, we shall 

 observe that <p and tang (p ought to be positive, and thus, 

 sin (?"—/) ought to be of the same sign with tang sr"; 

 this condition will delerminc the angle j, and this angle 

 will be the position of the ascending node if the motion of 

 the comet be direct ; but if this motion be .etrograde we 

 must add ISO' to it, in order to have the position of this node. 



The hvpothenuse of the rtctangiilar spherical triangle of 

 which tf' —j and ot" are the sides, is the distance from the 

 comet to its ascendino node in the third observation ; and 

 the difference between that hvpothenuse and v" is the in- 

 terval between the node and the perihelion, reckoned on 

 the orbit. 



Let us apply these results to the second comet of 1781, 

 whose perihelion distance and instant of passage by this point 

 we have already determined by a first approximation. For 

 this purpose, we shall use 'iome observations of the Qth of 

 October J 781, and 17th of November and 20lh of Decen<- 

 her of the same year : these observations give 



Mean time at Paris. 

 J 78 1. 9th Oct. at \&^ 50' 0", a =124° 27' 42", 



9 = 0= ir4o', 



1 7lh Nov. at ■ 8'' 29' 44", oi — 306° 37' 32", 



9'= 44° 17' 12", 



29th Dec. at 6'> 6' 30", a"= 306M7' 39", 



5"= 17° 34' 23'', 



We have moreover 



C = 197" 13' 44", log i? = 9,998864, 



C = 235° 35' 43", log R' = 9^99'3602, 



C" = 269" 20' 35", log R' = 9,992748. 



This being done ; we shall form a first hypothesis, in 



which the perihelion distance will be, as we have found 



above, equal tf) 0,958339, and the instant of the perihelion 



passage took place on the 29ih of November, at 18'^ 10' 34" j 



we shall find in this hypothesis 



V = _ 61° 18' 3", v'=z — IS" 7' 12", y = 29° 8' 15", 

 which gives 



U = 43° JO' 41", V = 90° 26' 18"i 

 we shall afterwards find 

 e = 77" 5' 50", r = 37° 26' 37" , £■" = 346° 49' 32", 

 Ts = 0° 10' 34', a-: =18° 6' 32"i, TO-" = 27" 12' 37", 

 from which we extract 



F= 42° 33' 2", F' = 90° 9' 22', 

 dividing ?« = 17' 49'', n = 16' 5G". 



