THEORY OF POLYGON IC PROJECTIONS. 101 



Since the three points A, D, U' are in a straight line, 



the angle at A of the triangle OAD is v^v^aal to ^? and it 



results, in this triangle and the triangle OAD' , that OB = 



tan ^, and OD' = cot |-. We thus have OD x OD' = 1, as 



it ought to be, since the tangent OU is the mean propor- 

 tional between OD and 0D\ 



The constant ratio of the distances of any point of the 

 projection of a parallel to the projections P and P' of 

 the two poles will be 



UP 



7= tan PP'J[7=tan^. 



UP' ^"" " " " ~ ^"-^^ 2 



Let us now consider the meridians. The longitude will 

 be reckoned as starting from that meridian the projec- 

 tion of which is the straight line PP\ and we shall define 

 the modified longitude of a meridian the angle at which 

 its projection intersects the projection of the central 

 meridian, an angle which we shall denote by X'; this angle 

 is also half the angle at which, from the center of the 

 projection of the meridian, we should see the line of 

 poles of the map. Therefore, for the meridian projected 

 into PGP' J y wul be the angle which PP' makes with the 

 tangent at P to the arc PGP', or, what amounts to the 

 same thing, to the angle OTP. The projection can vary 

 without the arc PGP' ceasing to be the projection of a 

 meridian; that which will vary will be the position of this 

 meridian upon the earth or, inversely, the expression of 

 X' as a function of X. Whatever this expression may be, 

 if we call R the radius TG or TP or TMoi the projection 

 of the meridian, and S the distance OT oi its center from 

 the center of the map, the right-angled triangle TP will 

 give 



P = cosec X' 



AS' = cotX' 



and the triangles OPG and OPG' will give 

 OG = tsin^,OG' = cot\' 



