AN ESSAY ON CYCLOIDAL CURVES. 



S59 



describing poinf, and the point of contact 

 of the generating curve and the basis, is al- 

 ways perpendicular to the curve described. 



If may by some be deemed sufficient to consider the ge- 

 nerating curve as a rectilinear polygon, of an infinite num- 

 ber of sides ; since, in this point of view, the proposition 

 requires no further demonstration ; and, indeed, Newton 

 and others have not scrupled to lake it for granted : but it 

 is presumed, that a more rigid proof will not be considered 

 as superfluous. Let M be the describing point, and P the 

 point of contact : and let LO, MP, and NQ, be succes. 

 sive positions of the same chord of the generating curve at 

 infinitely small distances : then it is obvious, and easily de- 

 monstrable, that the arcs OP and PQ,. described by the 

 point P of the generating curve, in its passage from O to P, 

 and from P to Q, will be perpendicular to the basis at P, 

 and will therefore touch each other. Let the arcs L, LMK, 

 and N, be described with the radius PM, on the centres O, 

 P, and S. Then the curve described by M will touch 

 IMK ; for since O and Q lie ultimately in the same direc- 

 tion from P, if L be above IMK, N will also be above it, 

 since these points must be in the circles L and N, and 

 infinitely near to M ; and if L be below IMK, N, for the 

 same reason, must be below it ;, and M is common to the 

 circle and the curve, therefore the curve touches the circle 

 IMK at M, and is perpendicular to the radius PM. 



Proposition ii. Problem. To draw a 

 tangent to a cycloidal curve at aay given 

 point. 



On the given point, as a centre, describe a circle equal to 

 the desaribing circle of the curve, and from the intersection 

 of this circle, with the line described by the centre of the 

 generating circle, let fall a perpendicular on the basis ; the 

 point thus found will be the point of contact, and the tan- 

 gent will be perpendicular to the right line joining this 

 point of contact and the given point, by the first proposi- 

 tion. It will be obvious, from inspection, which of the two 

 intersections of the circle to be described, with the track of 

 the centre, is to be taken as the place of that centre cor- 

 responding to the given point. 



Scholium. The tangents of cycloids and epicycloids 

 may be diawn from any given point without them bj means 

 of curves, which are described by the intersection of two 

 lines revolving on given points, with proportionate angular 

 velocities, and in the case of the bicuspidate epicycloid, the 

 curve becomes an equilateral hyperbola. 



Proposition hi. Pkobleru (Plate 7. 

 Fig. 38.). To find the length of an ephro- 

 choid. 



Let C be the centre of the basis VP, K that of the rotat. 

 ing circle PR, and of the describing circle GL, P the point 

 of contact, and M the describing point. Then joining 

 MXC, and supposing VX to be an element representing the 

 motion of the point P, in cither the basis or the generating 

 circle, draw the arc MN on the centre C, and join CVN : 

 then NM will represent the motion of the point M as far as 

 it is produced by the revolution round the centre C : take 

 MO to VX as GR to PK, then MO will be the motion of 

 M arising from the revolution round K, and NO will be the 

 element of the curve produced by the joint motion. Let 

 CII be parallel to PM, then CX or CP : CM::VX : MN, 

 and PK:MK::CP:HM::VX:M0, therefore, CM; HM 

 :!MN:MO, and these lines being perpendicular to CM, 

 HM, the triangle NMO is similar to CMH, and MN r NO 

 ::CM:CH, hence CP.:CH::VX: NO. Take PY to CP 

 asPK to CK, thenCHtCP::PM:PY::NO:VX. On L 

 describe the circle PfB, and draw IMLF: let FD be per- 

 pendicular to PRB, take DE to DF as PG to PL, and E will 

 be always in the ellipsis BEP : let AE and AF be tangent* 

 to the ellipsis and circle at E and F ; then the increment of 

 the arc BF will be to MO as PL to GL, and to VX as PL to 

 PR. Join GM, and parallel to it draw PI ; then PIL is a 

 light angle, and IL1'=AFD, and IM : ILXPG: PLXDE ; 

 DF, by construction ; therefore the figure IPML is similar 

 to DAEF, and as PL to PM, so is AF to AE, and so is the 

 increment of the arc BF to that of BE ; but the increment 

 cf BF Is to VX as PL to PR, therefore the increment of BE 

 is to VX as PM to PR. Now it was proved that NO: VX. 

 ::P.M : PY ; therefore the increment of BE is to NO as PY 

 to PR, or as CP to 2 CK ; and the whole elliptic arc BE it 

 to the whole SM as the radius of the basis to twice the dl''_ 

 tance of the centres. 



C0K01.1AUY 1. The fluxion of every cycloidal arc Is 

 proportional to the distance of the describing point from 

 the point of contact. 



CoHOLLARY 2. Fot the epicycloid, the ellipsis coincides 

 with its axis HP, and the arc BE with BD, which is double 

 the versed sine of half the arc GM, in the describing or 

 generating circle ; therefore the length of the curve is to 

 this versed sine as four times the distance of the centres to 

 the radius of the basis. 



Proposition iv. Problem. (Plate 7. 

 Fig. 58, 59.) To HikI the centre of cur- 

 vature of an epitrochoid. 



Let PY be, as in the last proposition, to CP as PK to CK, 

 and on the diameter PY describe the circle PZY, cutting 

 PO In Z : take OW a third proportional to OZ and OP, and 

 W will be the centre of curvature.. For let QP::zVX be 



