NO. n.] THE METEOROLOGICAL PERIOD S IN THE ARCTIC SEA. 591 



, ^ cos {^' -}-k — o) — cos /? 



^°' ^'' sin(^' + fc — a) ~ 



sin (/?' + /? — «) 



The apparent zenith-distance is ^o — «• I have taken £ from the table 

 given below. 



In order to find the azimuth, reckoned from the antisolar vertical of the 

 sun, Bo, of the point where the arch cuts the horizon of 0, we say ^ = 90° -|- e, 

 and get cos {Z — e) = cos /3 cos £ = cos /?', or Z = /?'+£. Hence 



_ sin (l^' + fc) - cos ili' + £) sin a 

 sm (p + £) cos a 



cos (^^ +-^+2+ 2J «'" U~2+2J 



2e 

 or tan -^ 



sm 



f ., . k . £ a\ (s k a\ 



(/J + 2 + 2-2J <^os(2-2-2J 



For computation I assume that R = 6398-147 km., log B = 3-80605 

 (gp = 83°), /i = 53 km., r = 16', ^ = 40' (Mean temperature of the dark season 

 = — 30°) and £ = 30'. The value for £ I have taken from a table given by 

 Prof. Fearnley in "Forhandlinger i Videnskabsselskabet i Christlania, Aar 

 1859", p. 137. This table, of which the argument is the apparent zenith- 

 distance, I have transformed into the following table with the argument true 

 zenith-distance ^. 



^==0° 45° 60° 70° 75° 80° 85° 86° 87° 88° 89° 90° 90°27'-4 

 e O'-O 0'-8 l'-5 2'-2 3'-0 4'-5 8'-3 9'-4 ll'-3 14' 17'-9 23'-0 27'-4 



The table is applicable to ordinary temperatures. For the low temper- 

 atures prevailing at the Fram's station, I have taken the horizontal refraction 

 £ to be 30'. 



We get thus 



/? = 7° 20', /}' = 7° 21'-5, ^-^ = 7° 21'; fc = 16' + 40' + 30' = 86'; 



2 



I = 43';-; = 15', 



