the Elliptic Equation. 207 



hence 



cos ir (cos ir)^ 



and by combining these with the value of tau |<^, we have 



,, • 2fl • 2 \ cos^((;J) + t) .cosl(0 — t) 



(1 — sin^ 6 sm^ an = ^^ , ■, ..o" -• 



(cos i</))2 



We may also put the equation between the moduli under the 



more succinct form 



.J. cos 10 . „ 

 sm c= T^ . sm 0. 



COS^T 



I now proceed to a discussion of the critical case, in which 

 a=/3 or the circle passes through a pole, and its projection is 

 the meridional curve. 



One remarkable property is derived from the equation 

 sin T= sin ^ sin (f). Since sin ^ = 1 in the critical case, the angle 

 which the curve makes with the parallel is always the same as 

 the longitude measured from its axis. By the term axis I mean 

 the meridian passing through the centre of the circle. 



The angle <u is therefore obtained more simply from 0. In 



fact, the equation sin co= —- becomes sin (M = tan |0, whence 



. -2 cos <j) 



[COS coy = -. r^. 



(cos |<^)2 



It is e\Hdent that my theorem for the comparison of elliptic 



amplitudes applies also to the mei'idional curve ; but since the 



modulus is constant, and the mean latitude nugatory, in this 



case, we must make a=yS in the expression for the latitude 



where the curve crosses the axis. This gives 



fir <y \ 

 tan( — — -^ l = tana. cos to, or tan^= tana, cos «, 



since the auxiliary circle also passes through the pole. 



In the stereographic projection, the circle passes through the 

 centre of the sphere, and its polar equation is /3 = 2« .cos 0. 

 Now TT— 2a is the latitude of the point opposite the pole; and 



if \ be the current latitude, we have2a = tan a and p = tan( j + ^ ), 

 x ±H) = tanacos0. Taking the Napierian loga- 

 rithm, we have for the equation on the chart, 



log tan f — + — j = log tan a + log cos <^, 

 + log tan ( T "^ y ) — ^^D *^" "■ + ^^S ^^^ ^- 



