200 Mr. C. W. Merrifield on the Geometry of 



of rectification of a curve in the projection is 



^ VCOS A,/ 



(1) In the first case, taking for simplicity the case of the 

 great circle (in which ^=k ^nd /x = 0), the complement of the 

 longitude [^ —<}>)> ^^^ *^^ latitude X, are the sides of a right- 

 angled triangle. The great circle itself forms the hypothenuse, 

 and 6 (the declination or modulus) is the angle opposite X. 

 Therefore tan A, = tan 6cos(f>, whence we easily obtain 



^=(l-,sin2^sin2 0)-*- 



If we change the variable to t, the angle which the circle 

 makes with the parallel, we have sin t = sin ^ sin ^, and conse- 

 quently 



(3) In the third case, we take for simplicity the case where 



TT 



a= —, and therefore /x=0. The oval is symmetrical to the 



equator, and the radius of the circle is 0. The radius 9 is the 

 hypothenuse of a right-angled triangle, of which the latitude 

 and longitude are the other sides. Therefore cos 6= cos A. cos <f), 

 which gives 



ds _ sin 6 



If we change the variable to t, the angle which the circle 

 makes with the parallel, we have 



sin d) , ds sin 6 



sin T= -. — ~ and -=- = . 



sm6» dT (l-sin^^sin^T)^ 



The inverse correlation between these cases saves us from the 

 necessity of considering more than one of them. We shall choose 

 the first, because it is not encumbered with the constant factor 

 sin 0. 



(2) The second case may be derived from either the first or 



the third by making sin 0=1, or sin a = sin /S. The mean lati- 



cos S 



tude merges in the pole, since = + 1 . In this critical case, 



° '^ cos« — ' 



therefore, the mean latitude and modulus do not determine the 



circle. We shall revert to this hereafter. 



