;C0 



AN ESSAY ON CYCLOIDAL CURVES. 



the space desrribed by P, while NO is described by O : it 

 is obvious, from prop. 1, that the intersection of NA and 

 OP must be the centre of curvature. Let QF be perpendi- 

 cular to PO, and TA parallel to QN ; then, by prop. 3, 

 NO : VX or QPXPO : PY, but by similar triangles QP: 

 Or::PY:PZ; therefore NO: QFV.PO : PZ, and by 

 division, NO : AOI'.OP : OZ, and by similar triangles;: 

 OW : or or OP. 



Corolla BV 1. When Z coincides Vifith O or M, OW is 

 infinite ; therefore whenever PZY intersects the describing 

 ciiele, the cpittochoid vsill have a point of contrary flexure 

 at the same distance from C as this intersection ; and the 

 circle PZY is given when the basis and generating circle are 

 given, whatever the magnitude of the describing circle 

 may be. If the basis be a straight line, PY will be equal 

 toPK. 



CoHOLLiRY 2. By means of this proposition, we may 

 find the curve which will produce any given curve by toll- 

 ing on a given basis ; or having the two curves, we may 

 find the basis. When the basis is given, supposing NO a 

 small portion of the given curve, of which W is the centre 

 of curvature, VP being the circle of equal curvature with 

 the basis at the point P, if we take OZ a third proportional 

 to OW and OP, draw the perpendiculars PY and ZY, and 

 take CP— PY : PY::PY : YKj then K will be the centre 

 of curvature of the generating curve ; for by addition CP : 

 PY::PK:YK, CP:PK::PY:YK, and CP:CK::PY: 

 PK, as before. When the basis is a right line, Y is the 

 centre of curvature of the curve required. 



Scholium. Hence we may easily find the curvature 

 necessary in the tooth of a wheel for impelling a pallet 

 without friction, by determining the curve which will ge- 

 nerate, by rolling on the face of the pallet, a circle passing 

 through the axis of the wheel ; but the tooth could never 

 be disengaged from the pallet, without an escapement in- 

 troduced for the purpose. 



Proposition v. Problem. (Plate 7. 

 Fig. 60.) To find the evolute of an epicycloid. 



In the epicycloid SM, the point M being in the circum- 

 ference of PIVIR, PZ will be to PM in the constant ratio of 

 PY to PR, and MZ to PM as RY to PR, and PM to MW 

 in the same ratio; hence PM:PW::RY : PYIICR: CP, 

 therefore the point W is always in a circle PWH of which 

 the radius is to PK in that proportion, and which touches 

 SP in P. On the centre C describe a circle AS© touch- 

 ing PWH in S; then, since CR : CP::PR : PH, we have 

 by division CR ; CPX.CP ; VS, and the circle PWS being 

 to A20 as PMR to SP, the arc PM being equal to SP, the 

 similar arc PW will be equal to AS, and taking AS0=: 



PWH, 3Q will be always equal to SW, and W in a curve 

 ©WS similar to SM, of which it is the evolute. 



Proposition vi. Problem. (Plate 7. 

 Fig. Ol ) To find the area of an epitrochoid. 



On the centre C describe a circle touching the epitrochoid 

 in S, take GFI to GC as PR to PC, and let the circle G*!! 

 describe on the basis SG the epicycloid S*. Then taking 

 GM always to G* as GL to Gn, M will be in the epitro- 

 choid SM ; for the angular motion of the chord G<I>, is the 

 same as that of GM in the primary epitrochoid. Let SJi 

 be the evolute of S<t>, and GWS its generating circle. On 

 diameters equal to SG, SL, and Sn, describe three circles, 

 AD, AE, and AF, touching the light line AB in A ; 

 let the angl» BAD be always equal to GHO, and it is evi- 

 dent that AD, AE, and AF, will be equal respectively to 

 WG, WM, and W*. But the angular motion of WG oft 

 W being equal to the sum of the angular motions of GM 

 on G and CG on C, is to that of AF, or of GM, or half 

 that of KM, in the ratio of CIl to CG, or CR to CP ; 

 therefore the fluxions of the areas SWG, SWM and SW* • 

 are to those of the segments AD, AE, and AF, in the 

 same ratio ; and that ratio being constant, the whole 

 areas, and their differences, are also respectively to eax;h 

 other as CR toCP. 



Scholium. The quadrable spaces of Halley are those 

 which are comprehended between the arc of the epitro- 

 choid, that of the describing circle, and that of a circle 

 concentric with the basis, and cutting the describing circle 

 at the extremities of its diameter. 



Proposition VII. Problem. (Plate 7. 

 Fig. 62.) To find a central ecjuation for the 

 epic^'cloid. 



Let CT be perpendicular to RT, the tangent at the point 

 M, then PMR will be a right angle, and PM parallel to 

 CT. On the centre C describe through M the circle MNO, 

 and let MG be perpendicular to RO. Then the rectangle 

 OQN=PQil, OQ : PG::GR : GN, by addition OQ : PQ 

 ::0R : PN ; hence by division OP : PQ::IR : PN, and 



PQ ^2m-p^. But PM,=PR X PQ=^^X INP : 

 IK. IK. 



and by similar triangles CT : CR : : PM : PR, whence Cfj 

 Let MZ and RY be tan- 



CR(7 „ INP 



^£gxPM,=CR,Xj^. 



gents to SP, then INP=MZ^, and IRP^RYj, CT=:CR 



MZ 



RY 



and CT will be to MZ in the constant ratio of CR 



to RY. Putting CP=a, CR=b, CMzzs, CT=u, thea uit 



as — aa 



=i:b 



I'L'—aa 



