9.

Different elliptic sections of a right cone are taken such that their minor axes are equal; shew that the locus of their centres is the surface formed by the revolution of an hyperbola about the axis of the cone.

[Consider a series of sections perpendicular to a principal section of the cone. Take any section parallel to the base and divide the diameter of that section, so that the product of the two parts = b2 where b is the semi-length of the constant minor axis; the corresponding elliptic section must pass through this point of division, and all these points lie on a hyperbola, the asymptotes of which are the generators of the principal section taken (Prop. 1, p. 160).]

10. Shew how to cut a right cone so that the section may be an ellipse whose axes are of given lengths.

[The centre of the section made by the plane perpendicular to any principal section must be the intersection of the ellipse and hyperbola in which such principal section cuts the surfaces referred to in examples 8 and 9.]

11. Shew how to cut from a right cone a section of given latus rectum.

be

[Any point F on a hyperbola described as in Ex. 9 may taken as focus, and the plane of section must be a tangent plane at F to the sphere inscribed in the cone, and passing through Fl

CHAPTER X.

CYCLOIDAL CURVES.

WHEN one curve rolls without sliding upon another, any point invariably connected with the rolling curve describes another curve, called a roulette. The curve which rolls is called the generating curve, and the curve on which it rolls is called the directing curve, or the base.

Only a few of the simpler examples of roulettes are here given, the first being the most simple of all, viz. the cycloid.

DEF. The cycloid is the path described by a point on the circumference of a circle, rolling upon a fixed right line, in one plane passing through the line.

In the construction this plane coincides with the plane of the paper.

PROBLEM 138. To describe a cycloid, the diameter of the circle being given (Fig. 143).

Let AB be the diameter of the given circle, C its centre, and suppose that the tracing point is the point B, and that at the moment A is the point of contact of the circle with the directing line. Draw the directing line XAY a tangent at A to the circle. The tracing point B will evidently reach the guiding line at points X and Y on opposite sides of A such that _AX = AY= the semicircumference AB, since each point of the semi-circumference comes down successively on a corresponding point of the line.

The following geometrical construction gives an exceedingly close approximation to the length of the circumference of a circle:-From C, the centre, draw a radius CH making an angle of 30° with the radius CB, and draw HK perpendicular to AB meeting it in K. At A, the extremity of the diameter through B, draw a

tangent to the circle and on it make AL=3. AB. KL will be

very nearly the circumference of the circle and its semi-length may be taken for the length AX or AY.

[In the figure L does not fall within the limits of the paper, but if AK is bisected in h and hk on a parallel to the tangent at A be made 3 times the radius of the circle, Kk may be taken as the semi-circumference.]

=

Divide up AX into any number of equal parts (say 8) as at a', b', c',... and divide the semi-circumference AB into the same number as at a, b, c,... Draw a line through C parallel to XAY, which will evidently be the path of the centre of the circle, i. e. as the circle rolls along AX the centre will always be on this line; and draw a'l, b'2, c'3... perpendicular to AX, the points 1, 2, 3, &c., being on the path of the centre. The point a will evidently come down to a, b to b', and so on; and when a has come to a', the centre of the circle will be at 1 and the tracing point will be on a line making an angle with a'l equal to the angle aCB, which is of course equal to ACg, since AagB. Draw 1G parallel to Cg and make 1GCg, the radius of the rolling circle. G will be a point on the required curve.

Similarly, when b has rolled down to b', the centre of the circle will be at 2 vertically above b', the tracing point will be on a line making with 6'2 an angle = the angle bCB, i. e. = the angle ACƒ, or

it will be on a line 2F parallel to Cf and at a distance from 2 equal to the radius of the circle.

Similarly for the remaining points c', d', &c.

It will be noticed that the lengths 1G, 2F, &c., may be determined without actual measurement by drawing through g, f, &c., parallels to AX meeting the corresponding lines through 1, 2, &c., in the points G, F, &c., the figures 1CgG, 2CfF are parallelograms and therefore in each case 1G=Cg, 2F Cf, and so on.

=

The curve should be drawn free-hand through the series of points thus found, and the half loop corresponding to the circle rolling on AY may be found by the same construction or may be put in by symmetry. The line X8 is a tangent to the curve at the point X.

The length AX may be determined arithmetically by multiplying the length of the radius AC by 3. 14... and may then be laid down by scale: the diagonal scales usually supplied with cases of mathematical instruments can conveniently be used for the purpose. In many works on geometry the length AX is determined by dividing up the semi-circle into any number of equal parts (say n) and laying off along AX the length of the chord of one of the parts repeated n times. This method is radically bad and should never be adopted: if the number of equal parts into which the semi-circle is divided is small it gives only a very rough approximation to the truth, while if the number is increased it is almost impossible to measure the length of the chord so accurately but that in repeating it n times an appreciable error will be introduced. A long length should in fact never be determined as the sum of a series of short ones.

To draw the normal at any point of a cycloid.

In all roulettes the normal at any point passes through the corresponding point of contact of the rolling and guiding curves. This point is called the Instantaneous Centre. The direction of motion of the tracing point will evidently at any moment be perpendicular to the line between it and the point about which the rolling curve is turning, i. e. the corresponding instantaneous

centre, and since the direction of motion at any point must coincide with the tangent at that point, the normal must pass through the instantaneous centre.

In the figure, when the tracing point is at E the centre is at 3 and c' is the instantaneous centre, so that Ec' is the normal at E; this is evidently parallel to eA, e being the point in which a parallel to AX through E meets the circle on AB as diameter, so that the normal at any point P may be thus constructed :—

Through P draw a parallel to the directing line AY meeting the circle on AB in the point Q. The normal at P will be parallel to AQ, and since the angle AQB is a right angle the tangent at P will be parallel to QB.

If the normal at P meet the directing line in M and PM be produced to S so that PS2PM, S will be the centre of curvature at the point P. The evolute of the cycloid is two equal semicycloids, the vertices being at X and Y and the cusp on BA produced at a distance from A = AB.

Let the tangent at P meet the tangent at the vertex in T, then the length of the arc BP of the cycloid is double the intercept TP of the tangent, i. e. double the chord BQ of the circle. Hence the whole length of the cycloid is 4 times the diameter of the generating circle.

DEF. If, as in the cycloid, a circle rolls along a straight line, any point in the plane of the circle but not on its circumference traces out a curve called a Trochoid.

PROBLEM 139. To describe a trochoid, the diameter of the circle and the distance of the tracing point from its centre being given (Fig. 144).

Let AB be the diameter of the given circle, C its centre, and CP the distance of the tracing point from the centre.

Draw XAY a tangent to the circle, and as in the last problem determine the length AX or AY equal to the semi-circumference of the circle AB. Draw C8, the path of the centre, through C parallel to XAY, and through X draw X8 perpendicular to XA.