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 cCd ^ COS ^ ■ 



B 



B + h- 

 gC = B cos e; cos gCB = cos §' = v ii, = cos /S . cos e. 



The geocentrical angular radius of the twilight arch is 

 BCS ^W - (^' + r + Q -\- e) = 90° — (^' -\-k), when r-{-Q + e = k. 



^B 



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 b in the arch 

 has the true zenith-distance 3 06, 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 6 is 3 0& or .^. The geo- 

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

 which is equal to BCS or 90°— (/?'+ft). The depression (or) of the sun's centre 

 below the horizon of is 90°— 5 OS, or 5CS=90°—a. We then have in 

 the spherical triangle 3&S 



sin (/S' + fc) = cos e cos « sin ^ + sin a cos Z, 



and in the plane triangle bOC 



5HL%=^ = ^HLL or sin(S-.Z) = cos|SsinS 

 B B -\- "> 



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



we say e = 0, and hence 



sia{^' + k) = sm(Z+a), §'-\-k = Z+a, Z = fi'-\-k — a 

 sin {t,o — Z) = sin ^0 cos Z— cos t:„ sin ^ = cos /S sin ?„ 



