438 PHILOSOPHICAL TRANSACTIONS. [aNNO 1708. 



semiaxis = bo X de. For because of the equiangular triangles, gbo, gla, 

 GAK, GPS, GOE, it wili be, 



sp : SG :: bo : go 



8G : gd :: bg : lg :: go : ga 



DG : DE :: ga : ak, 



hence sp : de :: bo : ak ; and sp X ak = de X bo = ^-l x sb. 



Hence, if a body move in an ellipsis, with a centripetal force tending to its 



centre, that force will be directly as the distance. For it is iL2iJ^i^ = a 



SA 



given quantity, because sp x ak is given. Therefore the force, as — 



will be as the distance sa. 



In fig. 3, from the other focus p drawing pi perpendicular to the tangent ; 



then by the equiangular triangles sap, pai, it will be sa : sp :: fa : pi = ^-L^ill ; 



hence it will be sp X fi = = the square of the less semi-axis. So 



that, if the greater axis be called b, and the less 2d, it will be 



., d* X SA J J . SA 



sp2 — _j and SP = rf/7 . 



6— SA ' ^ 6— SA 



SA 



But in the hyperbola it is sp = d^^rr — • 



And in the parabola it is sp = ^d X sa, putting 4d for its latus rectum. 

 Because ta** : to'* :: ap'* ; sp* :: sa' — sp^ : sp'* :: sa' — 



i^ .d^..SA-^ : r-^ :: ^SA - sa' - d' : rf", it will he 



b—SA 6— 8A 6 — SA 6 — SA 



^ (bsA. — 8a' — (f ) : rf :: ta : to; and since ta = sa, it will be 



(fSA 

 ''^ "" </(68A-SA»-rfO* 



Now let QAO, fig. 7, be any curve, the element of the arc being ao, to 

 which AP and op are tangents, also ar the radius of curvature, and sp, sp 



perpendiculars to the tangents : then will ar = ^^^^. For, by similar tri- 

 angles, it is /p : AO :: PA : ba, 



and AO : ta :: sa : pa; hence ex aequo it 

 will be /p : ta or sa :: sa : ra; 



SA ^ SA 



but /p = sp, therefore it will be ra = . 



Hence if the distance sa be drawn into its fluxion, and divided by the fluxion 

 of the perpendicular, it will give the radius of curvature. By which theorem 

 the curvature is easily determined in radial curves or spirals. For example. 



