86 



F II. Bigelow — Solution of the Aurora Problem. 



the existing non-homogeneous distribution. It may be hoped 

 that an accumulation of the observations proposed will throw 

 some light upon the pole of reference, as distinguished from 

 the principal magnetic pole. 



Pass a plane through the station of observation, tangent to 

 the sphere at A, intersecting the magnetic pole of reference 

 2. extended in B, and the aurora ray in 



C, the triangle ABC, being there- 

 fore in the tangent plane. Draw 

 AD at right angles to AB. Now if 

 planes be passed through these lines 

 and the center of the sphere, the 

 traces of intersection of the planes 

 with the surface of the earth are as 

 follows : 



the trace of AB is AP, 



" DB « DP, 



" " AC " AF. 



Connect the center of the 

 with A, B, C and D ; O C 

 the sphere at F. 



Let the ray spring from the 

 sphere at E, and reach the horizon 

 plane at C, the visible portion being 

 CH, upon which the measurements 

 are made. Since a ray of given 

 polar distance #, has a given inclina- 

 tion to the perpendicular to the 

 tangent plane at C, by measuring 

 the angle and interpreting it as pro- 

 jected on a plane perpendicular to 

 we have the means of describing its 

 of all the lines of force of similar 



sphere 

 pierces 



Fig. 

 of a 



2. — Showing the relation 

 ray with respect to the 



horizon plane. 



the line of vision AC 

 space 



For 



location in 



reference that the line of vision passes, there is only one that 

 can have the inclination due to a ray at the distance AC. This 

 can be easily perceived by considering these lines around a 

 sphere. Hence by standing at A the only question is how far 

 along the line AC must one go to meet the ray whose angle, 

 corrected for projection, will correspond with the one that is 

 measured. The critical angle of our theorem is therefore at 

 B, namely ABC, on the tangent plane, or the equivalent AOC, 

 at the center of the sphere. If we assume AOC and compute 

 thence to the measured angle, as will be explained, and find 

 the values to be the equal, we have chosen the correct value of 

 AOC. Thus by trial and error, may be found the true angle, 

 and the distance of the point C from the observer. 



