F. H. Bigelow — Solution of the Aurora Problem. 87 



Let u = polar distance of station A to the pole P. 

 " v = angular distance from station to the point D on the 



plane. 

 u io = angular distance from station to the point C. 

 The ray lies in the plane, C, PEFD. 

 AD = r tan v. 



AB = r tan u. tan B = tan v cot u. 

 In the triangle ABC, 



A = the measured azimuth from magnetic meridian. 

 c — AB = ftanw. 



B = the value obtained by assuming v. 

 Compute, 



b = distance of observer to base of ray on horizon. 

 C = 180 —(A + B), which varies with v. 

 b = c sin B cosec C = r tan w. 

 OC = r sec w. 



FC = r secw— r — r (sec w — 1), the height of C above the 

 surface of the earth, measured along a radius. 

 At C this radius makes with the perpendicular to the plane, 

 the angle 10, lying in the vertical plane AOC. 



Furthermore in the plane triangle BOC, we have, 

 OB = r sec u = c', 

 OC = r secw=. b'. 



BC = a may be computed from the formula, 



c. sin A 

 a — • / 1 — ^t • Hence 

 sin(A + B) 



tan^=r j/ ( S ~^^~ C '\ 6' being the angle BOC, or the 



magnetic polar distance of the point C on the ray. 



Now we have by Gauss' Theorem, 

 2 cot 6' — cot I, 



where £ is the angle that the ray at the point C makes with the 

 radius of the earth to that point. This is the angle which is 

 seen by the observer in its projected position. 

 ^ If we assume the true length in space of the measured por- 

 tion of the ray to be s, we have by fig. 3, the projections of s 

 on the radius CK, and CI at right angles to the radius in the 

 plane of the ray, 



CK = s cos £, CI = s sin I. 



The projection of CK on CJST the normal to the plane, drawn 

 at the point C, is CM = s cos I cos w ; and of CI on CL per- 

 pendicular to CA, is CL = ssin I sin (A+B). Therefore, rep- 

 resenting by z/A the difference in azimuth of the two points 

 of the ray, and by Ah the difference in altitude of the same, 

 we find, 



