184f Mr. Galbraith on the Determination of Distances 



By combining this with the former, there results 



first const, log 7*6170579 + log of 0*1904-762 = 6*8968986 

 = the second constant logarithm. 



Hence 



log a" = 1st const, log 7*6170579 + log K . (10.) 

 log a." = 2nd const.log 6-8968986 + log K, . (11.) 

 which two results being substituted in formula (4.), and the 

 operation repeated, will give K sufficiently correct, when the 

 distance is only a small, or even a moderate part of that of 

 the visible horizon. 



In practice it will be found convenient and sufficiently ac- 

 curate to use the mean quantities in the first and second ap- 

 proximations, and the true quantities under given circum- 

 stances for the last approximation only. 



3. Again, let the angle A H C be denoted by H, and sin H 



= 2 sin £ H cos £ H = 2 sin ^ H (1 — sin 2 \ H)*, therefore 



2 sin £ H (1 - sin 2 \ H)* = £™ cos D. 

 Squaring both sides, and there results 



4 sin 2 iH-4 sin 4 | H = (*— ) cos 2 D, 

 or, by arranging, 



4 sin 4 \ H - 4 sin 2 iH=- {l±J^ cos 2 D. 

 Add 1 to each side of the equation, and 



4sin 4 ^H-4sin 2 |H + I = 1 - (t±J^j cos 2 D, 



or 4sin 4 ^H-4sin 2 £H + 1 = (l + ^—J cos D 



x (l_L±i)cosD. 



Multiplying and dividing by If- ) , and 



4 sin 4 $ H - 4 sin 2 i H + 1 



But = sec D /5 D y being the depression of the horizon, 



4sin 4 |H— 4sin 2 iH + 1 rrsec^/cosD^ cosD)(cosD / — cosD). 

 Since (cos D, + cos D) (cos D, — cos D) = sin(D + D,) sin(D — D,) 

 4sin 4 |H-4sin 2 ±H + l=sec 2 I> y {sin (D + D,)sin(D-D,)}. 



