( 421 ) 



2. Solution for Ai ^0. 

 In this (!ase we find : 



AM = — 9"0461 

 A^ =z — 2'32"41 

 and for the remaining errors : 



1. A«=: + 0^185 Ad— - — 3"18 



2. +0.089 +2.80 



3. — 0.226 + 3. 32 



3. Solution lüith 3 unknovm quantities: 

 Tlie results are : 



AM = — 5"3045 

 Ai r= + 7.32 

 A^=:- 1'2.90 



and according to the equations of condition there remain the following 

 differences Obs. — Comp. 



1. A«=r + 0^088 A(f= — 0"23 



2. + 0.095 + 1.34 

 3 —0.181 —1.01 



As we see the solution with AcTl = and that with A/=:0 

 satisfy the observations fairly well, the first one somewhat better, 

 especially in right ascension. Stili we cannot deny that in the values 

 Obs. — Comp. of Ö in both solutions, there exists a systematic varia- 

 tion. On account of that I prefer for the present the solution with 

 3 unknown quantities, where such a systematic variation doesnot 

 appear. I therefore take the following elements as the most probable 

 for the return in 1906 : 



Epoch 1906 January 16.0 M.T. Greenw. 



M, = 1266456"838 



=:351°47'36"838 

 jLt = 517"447665 

 log a =0.5574268 



T = 1906 March 14.09401 

 (f =24°20'25"55 

 e =0.4121574 

 t = 20°49' 0"62 



.T =346 2 31.63 1906.0 

 SI =331 4437.85 ) 



