iy means of Two known Stars. 195 
immediately after he observes the other, and after that he 
again observes the first;—few minutes are sufficient for 
these three observations. Daring this interval, it may be 
supposed that the first star has raised by an uniform mo- 
tion; and by means of simple proportional parts, the alu- 
tude of the first star is reduced to the moment w when. the 
second was observed. 
This being laid down, we shall, in order to save a figure, 
express in general formule Mr. Calkoen’s solution. . 
Let D be the distance between the two stars, we shall 
have cos D = cos ae AX) cosd cos & ++ sin o sind’,..(1). 
ef sin a _ sin %’ cos D sin 3” 
Make cot E = sinésinD _ = ( Tre Mee ae D) 
1 
“sin D eee 1(2)z 
E will be the continuation of the distance D till it reaches 
the equator (line.) 
: sin é 
Sin A = ane 77773) ° 
A will be the angle by which the continuation E inter- 
sects the equator. : 
Make cot F = 
l 
sin D ae ie) .(4). 
F will be the continuation of the distance D to the 
horizon. 
sin h’—sin hcosD _ ee h 
sin 2 sin D sin h 
— cos D) 
B will be the angle by which the continuation F will 
interséct the horizon. E and F will be in the same plane. 
Make tang « = cos (EZ — F) tang B......(6). 
Si age B cos (A—2) 
in ® cei isc vole a UE) 
Then (90° — ¢), (90 — 8), and (g0—h) will be the three 
sides of a spherical triangle, and the angle opposite to 
(go — A) will be the hurary angle of the star whose dechi- 
nation is 9. 0 
Thus eight formula give the solution of the problem, 
which is founded on a simple and | ingenious construction. 
The demonstration of the formula is easy to find, it is 
perceived on the simple inspection of the figare, The 
solution requires 33 logarithms. actually diferent; or, with 
tables of natural and logarithmic sines, 30 only; but in 
all cases it is longer than rtbe preceding solution. 
N 2 - The 
