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 c C g. 

 We have thus 



cos cCd = cos /S = 

 gC = R cos s; cos pC£ = cos /?' = 



B 



R + h' 

 B cose 



cos /S . cos e. 



B-j-h 



The geocentrical angular radius of the twilight arch is 

 £C5 = 9O°-0?' + r + e + £) = 9O° — (,?'+fc), when r + g + e^k. 



3r 3 



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 C6 or Z. The geo- 

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

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

 below the horizon of is 90°— 3 05, or 305 = 90° — «. We then have in 

 the spherical triangle 3 65 



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

 and in the plane triangle bOC 



sin (C — Z) sin C ■ ,^ rn a • r 



R -R + h'' sm(C--Z) = cos/JsmS 



In order to obtain the tine zenith-distance of the top of the twilight-arch, ^0, 

 we say e = 0, and hence 



sin(/S' + &) = sin(Z+a), ft' + k = Z-{-a, Z = fi' + k — a 

 sin (Co — Z) = sin ^0 cos Z — cos ^„ sin Z=cos fi sin ^0 



