Dr. Hirst on Equally Attracting Bodies. 323 



easily find the following relations between the new coordinates 

 ^i, ^j of the projection, and the former ones r and d : — 



cos 2^. tan^ = tanf ^ — ^A 



TT 



rcos-^ 

 o 



'&-".)• 



By eliminating r and 6 from these and the equation (24), we 

 obtain, after a few simplifications, the following as result : — 



,.2 



12fl2 



' l+2cos2^i" 

 This is easily recognized to be the equation of a hyperbola, whose 

 transverse and conjugate semiaxes are respectively 2« and 2a -v/S; 

 the former of these axes coincides with the prime radius, and the 

 generating line of the cone diametrically opposed to the prime 

 radius is the asymptote of the hyperbola. Thus the curve of 

 double curvature into which the right line has been transformed 

 may be regarded as the intersection of the cone and a hyperbolic 

 cylinder, whose diametral and asymptotic planes touch the cone 

 along two of its generating lines diametrically opposite each 

 other. The plane of the equally attracting circular section of 

 the cone cuts the hyperbolic cylinder along the line which foi-ms 

 its vertex. 



We will briefly indicate the general solution of the present 

 problem. Let 



Aip,^,^)=of ^^^ 



be the polar equations of the given curve of double curvature 

 which attracts the pole, — ^ being the angle between the prime 

 radius and the projection of p on the polar plane, and ■^ the 

 inclination of p to that plane. Then dO being the infinitesimal 

 angle between two radii vectores to the curve (25), and ds the 

 element of its arc, we have on the one hand 

 ds'-=pW^ + dp^, 



whilst on the other 

 hence 



ds^ = dp^ + p^df"- + p V</)2 COS^ ^{r ■ 

 de^=dyjr^ + d<J>^COS^yfr. 



If we now agree that shall vanish when p=po, then by inte- 

 gration 



Theoretically, at least, the expression under the radical sign can 



Z 2 



