5 2 



SMITHSONIAN MISCELLANEOUS COLLECTIONS 



VOL. 51 



ra, from the focal points, by the distances D and D' of the two pairs 

 from each other (see fig. 5), then the sum of the quotients is unity 

 and we have therefore 



D D' 



(13) 



hence from equation (12) 

 2F 



D 



= 1, D =2F, D' =2F'. 



From these we at once find the distances of the points of symmetry 

 from the (upper or lower) boundaries of the atmosphere, namely, 



ab'=2F-R-H 

 a' c' = 2 F' + R 



and substituting 



F = 22 924 900 km. R = 6366 .7 km. 



F' = 22 918 400 km. H = 200 km. 



there results 



a V = 45 843 233 km (14) 



a' c' = 45 843 167 km. 



Hence the points of symmetry are at approximately equal distances 

 from the boundaries of the atmosphere, and since a = a' the points 

 of entrance and exit, b and c, of the corresponding pencil of rays 



FIG. 6. 



are at equal distances from the axis so that be is parallel to am. By 

 the zenith point of the pencil I understand the point of intersection 

 d (fig. 6) of its initial direction with the prolonged radius of the place 

 of observation. 



