590 



H. MOHN. METEOROLOGY. 



[NORW. POL. EXP. 



phere of which the air is capable of reflecting visible light in the point 

 B, whose height above the earth is h. This terrestrial horizontal refraction 

 is a little less than the astronomical horizontal refraction. It is represented 

 in the figure by the angle e, or dcB, which is equal to the angle cCg. 

 We have thus 



cos c Cd = cos /? 





p I T. 

 gC = R cos e; cos gCB cos ft' = R . , = cos (i 



= cos 8 . cos e. 



The geocentrical angular radius of the twilight arch is 

 + e + e) = W (f+k), when r 



An observer in the point sees 

 the demarcation line between the twi- 

 light and the earth's shadow as an arch, 

 BB. To him, a point 6 in the arch 

 has the true zenith-distance 306, or , 

 r and an azimuth, S$b, or e reckoned 

 from the sun's meridian on the opposite 

 side of the sun. The geocentric zenith- 

 distance of b is 3 Cb or Z. The geo- 

 centric angular radius of the earth's shadow, or the twilight arch is 6 CS, 

 which is equal to B CS or 90 (/?'-{- k). The depression (a) of the sun's centre 

 below the horizon of is 90 3 CS, or 3 CS= 90 a. We then have in 

 the spherical triangle 3 bS 



sin (/?' + k) = cos e cos a sin Z + sin a cos Z, 

 and in the plane triangle 6 C 



S in(-Z) = sing pr sin(? _ z) = cos/SsinC 

 ft xi -f- n 



In order to obtain the true zenith-distance of the top of the twilight-arch, , 

 we say e = o, and hence 



sin ( Z) = sin cos Z cos sin Z = cos /? sin 



