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



Moreover CO = sin 6= ^ a, which is the value of e when 



I + cos U 



-^0= — . With these conditions, we have between ^q, 0p <^2 ^^^ 



equations F(^, </>o) = F(^, <^i)-W. </>2), 



cos (^0= cos^i COS </>2+ sin ^, sin 4>ci^{0, ^q). 



We may remark that this theorem has not the one-sidedness 

 of Jacobi's. Indeed, as the angle <^ corresponds to 2i|r [twice ■^), 

 (f) may be measured either from A or from B. 



The last observation enables us to determine the auxiliary 

 circle without the help of Lagrange's scale. Draw VQ, pq, 

 through a and b perpendicular to A B, then 



PO A = AOQ =i.OB = B0(? = (/)i, 

 by which we understand an amplitude such that F(<^i-) = IF(^). 

 If therefore we suppose <^| to be knoAvn = w, it is easy to deter- 

 mine the auxiliary circle by marking off the points a and b, and 

 thence determining the other elements. 



Now suppose to be the centre of the sphere, and let us sup- 

 pose the last figure to be a stereographic projection from the 

 south pole on the plane of the equator. The longitudes will still 

 remain unaltered. The circles will still remain circles. The 

 straight lines passing through will be meridians, and the tan- 

 gent lines will be small circles passing through the south pole. 



If we pass to Mercator's chart, the tangent lines become me- 

 ridional curves, and the arc AMo=MiM2 in actual measurement 

 on the chart. This is the geometric meaning of the transcen- 

 dental equation r(^, </)o) = F(^, ct>2)-^{0, <^,). 



We have still to determine the elements of the auxiliary circle 

 from the value (supposed to be known) of &> or ^l. To do this 

 we shall return to the surface of the sphere. Here P Q S is the 

 circle which replaces the tangent line P Q, P N S is the longi- 

 tude ft), P A Q is the circle whose modulus is 6 ; and the auxi- 

 liary circle therefore passes through a. 



Let G be the centre of P S Q. Join G P, and draw G H_LS P; 



