as MR. RUMKER'S OBSERVATIONS 



Then is the reduction to the meridian R = 11 S — ^2 cotang. z (,« — h). 



Demonstration. — Be ^ the zenith distance corresponding to the middle time T; 

 z' z" z'" .... the different unknown zenith distances envelopped in the arc 

 run through, and z their mean, which is known. Call Z, — z' = a, Z, ^ z" 



= ft &c. &c Then is 



cos M — cos ? = 2 r sin' ^ t = r A sin 1" 



subtract 



COS M — COS 2* = 2 r sin' ^ / = r A' sin l" 



cos 5 — cos 3* = r ( A' — A ) sin 1" 



« • w/ PN r(A'- A)sinl" r(A'-A) , r{A"-A)„ „ 



2 Sm H^ - C) = -— roy- ^= sLiK + ia) ^ = sin(?+x,) &C.&c. 



2jp(A'- A) _ 2a 2 * / 2p(A'-A) 1 _ 



cot ? sin 1" ~ cot ? sin 1" ■'" " ' V cot ? sin 1" "*" cot* t sin* 1" — " "T 



cot ? sin 1" ~ cot ? sin 1" ~ " ' V cot ? sin 1" "•" cot* ? sin* 1" — " "T cot ? sin 1" 

 If we now call 2 ^ cot ^ sin l" = 5^, we obtain 



ten ? y — n rr tan t 



which resolved according to the binomial theorem gives 



_ tonz f (A'- A) 2 (A'- A)' , 3q'{A'- A? 3.5q*{A'- A)* , ,\ 



"— slnl"|? 2 ~9 2T4 ' 2T4T6 2.4.6.8 "T •" / 



Placing now for q its value in the two first parts, and considering that » 



A' sin 1" 



— 2 — = h according to the construction of Delambbe's Tables, 



/A( AN o v/vi .V A, A • ,iK ■ tan?3o*(A'— A)^ tan ?3 . 5 .o*( A' — A)' 



a=:p{A' - A) -p^COt^^ +S - A'Asml")+ ,inV.2.4.6 " sinl".2.4.6.8 



And in the same manner 

 and c =;) (A'"- A) - ;)2 cot. i {l"'+t- A'" A sin l") + &c. &c. &c. . . . 



I+4 + C + + 



adding up and taking a mean. 



C is therefore the quantity to be added to the mean z of the zenith distances, 

 in order to have the zenith distance ^ corresponding to the mean of the times. 

 If the change of altitude were proportional to the change of time, C would be • 



