Divide C8 into any number of equal parts (8 in the fig.), and with centre C and radius CP draw a circle. The point P in which this

circle cuts AB produced will be the vertex of the required curve. Divide the semi-circumference of the circle into the same number of equal parts Pg, gf, &c., as has been chosen for the division of the path of the centre.

Draw 16 parallel to Cg and gG parallel to AX: their intersection G will be a point on the required curve. Similarly 2F parallel to Cf and fF parallel to AX will intersect in a point on the curve, and so on in succession. When B has come down to I the tracing point will evidently be at P, vertically below X on 8X produced so that 8P, CP; the tangent at P, is parallel to AX.

1

The construction is obvious from that of the cycloid.

In the figure a second trochoid is drawn generated by a point Q inside the rolling circle, to which the foregoing description applies exactly by the substitution of Q for P.

To draw the normal at any point of a trochoid. Consider for a moment the point F. When the tracing point is at F the centre of the rolling circle will be at 2 and the point of contact of the rolling circle and directing line will be H on AX vertically below 2; i. e. H will be the instantaneous centre, and therefore FH will be the normal at F, since the direction of motion of F must be perpendicular to FH. But FH is parallel to fA,

since the triangles F2H, ƒCA are in all respects equal and are similarly situated, and therefore the normal at any point R may be thus constructed :—

Through R draw a parallel to the directing line meeting the circle described with C as centre and CP as radius in the point r, and the normal RM will be parallel to the line joining r to A, the lowest point of the rolling circle when its centre is C.

To find the centre of curvature at any point R*.

Find K, the position of the centre of the rolling circle corresponding to R. (K will of course be vertically above M.) Join RK and draw MN perpendicular to RM meeting RK in N. Draw NS perpendicular to the guiding line meeting RM in S. be the required centre of curvature.

S will

DEF. The Epicycloid is the path described by a fixed point on the circumference of a circle rolling on the convex side of a fixed circle, both circles lying in the same plane.

PROBLEM 140. To describe an epicycloid, the radii of the rolling and directing circles being given (Fig. 145).

Let O be the centre of the directing circle, OA its radius, AC the radius of the rolling circle, C, on OA produced, its centre, and let B be the other extremity of the diameter through A. Suppose B to be one position of the tracing point. As the one circle rolls round the other let the point B come down to X ôn the one side of A and to Y on the other, X and Y being on the directing circle. The arc AX will necessarily be equal to the arc AY, and equal to the semi-circumference of the rolling circle.

These points may be thus determined :

Let the length of the semi-circumference AB be S,

S =T. AC,

being the circular measure of two right angles.

then

* The construction for the centre of curvature of this and the following roulettes was given by M. Savary in his Leçons des Machines à l'École Polytechnique, and is quoted by Williamson, Differential Calculus, 3rd ed., p. 345, where its proof is given.

Let be the circular measure of the angle subtended by the

arc AX (the length of which is S), at the centre of the directing circle; then

S=0.40=π. AC;

..:: AC: AO,

or if n is the number of degrees in the angle AOX,

[In the figure A03AC so that the angle AOX contains 60°.1

Draw the path of the centre of the rolling circle, i.e. an arc with centre O, and radius OC, and let OX produced meet it in 8. Divide up the arc C8 into any convenient number of equal parts (8 in the fig.) and draw the radii 01, 02... cutting the directing circle in a' b'.... Divide up the semi-circumference of the rolling circle into the same number of equal parts Aa, ab...:

As the one circle rolls on the other, the point a will evidently come down to the point a, b to b' and so on: when a has come to a', the centre of the rolling circle will be at the point 1, and the tracing point will evidently be on a line making with a'l an angle equal to the angle aCB which is equal to the angle ACg. Hence an arc described with centre 1, and radius CB, will intersect an arc described with centre 0, and radius Og, in a point G of the required curve, for the triangles G10 and gCO are equal in all respects:—i.e. G is the position of the tracing point corresponding to a', being the point of contact of the rolling and directing circles.

Similarly an arc described with centre 2, and radius CB will intersect an arc described with centre 0, and radius of in a point F of the required curve, and so on in succession for the points 3, 4, &c.

The arcs gG, fF, &c. will cut the corresponding arcs described with the successive centres 1, 2, &c. in two points, but it is evident by inspection which of the points must be taken, viz. that on the side of the corresponding radius 01, 02, &c. remote from OA.

The radius 0X8 is a tangent to the curve at the point X.

To draw the normal at any point P of an epi-cycloid. From P with the radius AC of the rolling circle describe an arc cutting the path of the centre in K. [It will do so in two points but the one lying within the angle POB must be taken.] This will be the position of the centre of the rolling circle when the tracing point is at P. Draw KO cutting the directing circle in M, the point of contact between the circles when the tracing point is at P: i.e. M is the instantaneous centre corresponding to P.

Therefore PM is the normal at P.

To find the centre and radius of curvature at any point P. From M the instantaneous centre draw MN perpendicular to PM meeting PK, the radius of the rolling circle when the tracing point is at P, in N. Then NO (O being the centre of the guiding

circle) will cut PM produced in S the required centre of cur

vature.

DEF. The Hypo-cycloid is the path described by a fixed point on the circumference of a circle rolling on the concave side of a fixed circle, both circles lying in the same plane.

PROBLEM 141. To describe a hypo-cycloid the radii of the rolling and directing circles being given (Fig. 145).

OA is the radius of the directing circle, and O its centre, AC is the radius of the rolling circle, and B' the tracing point when the centre is at C'. The construction is identical with that for the epi-cycloid. In the figure the radius AC' is equal to AC the radius of the epi-cycloid, and B' of course reaches the directing line at X and Y-the points F" and D' are the positions of the tracing point when the points b, and d, are the points of contact of the rolling and directing circles.

DEF. When, as in the epi-cycloid, a circle rolls on the convex side of another, any point in the plane of the rolling circle, but not on its circumference traces out a curve called an Epitrochoid.

PROBLEM 142. To describe an epi-trochoid, the rolling and guiding circles, and the position of the tracing point being given (Fig. 146).

[In the figure the tracing point is assumed outside the rolling circle; it might be inside it.]

Let O be the centre of the directing circle, OA its radius, AC the radius of the rolling circle; C, on OA produced, its centre; let B be the other extremity of the diameter through A, and P on CB produced be one position of the tracing point. As in the epi-cycloid determine an arc AX or AY of the guiding circle equal in length to the semi-circumference of the rolling circle, so that B comes down to X and Y as the circle rolls round: i.e. construct angles AOX and AOY each containing n degrees where