In order to construct the hyperbola (Fig. 72), reciprocal of the original hyperbola (Fig. 71) relatively to the external pole O, the lines drawn in the first instance were (Fig. 71) the diameter u, the conjugate chord v, the conjugate diameter n, and the polar m of the pole O. Next a polygon of forces was formed by drawing the tangents at the six points on

the curve where it is cut by the lines mnv, any arbitrary tangent 4, and the two asymptotes, which in all constitute a closed force-polygon 1, 2, 3, 4-9 of nine sides, eight of which meet two and two in the four points of intersection of the line u with the four parallel lines vnm∞.

Then in Fig. 72 choose a pole O, draw ON parallel to the direction of the lines vnm and set off along this line

the points NMV, so as to fulfil the condition of constancy (eq. 3, § 52); or such that the punctuated range NMO may be similar to the range u)mnv, made by the intersections of the lines mnv with the diameter u.

In Fig. 71 the four lines 59mu meet in a point; in Fig. 72 the reciprocal points 5'9'MU lie upon the straight line which is reciprocal of the point of intersection in Fig. 71.

In Fig. 71 the points of contact 3 and 8 lie upon the straight line n; therefore in Fig. 72 the tangents through 3′ and 8', reciprocal of those points, must meet in the point N reciprocal of line n. Since, also, the tangents at 3' and 8' must be parallel to the rays 03 and 08, the points 3' and 8' must lie at the intersections of N3' and N8' with OU, drawn through O parallel to the tangents 3 and 8 (Fig. 71).

Finally, the tangents Ol' and 06′ (Fig. 72) are parallel to the asymptotes in Fig. 71; and, corresponding to the limiting tangents 7 and 2 which traverse the pole O in Fig. 71, we have the asymptotes M2' and M7', passing through the centre M (Fig. 72).

All the above reciprocal relations are exhibited in the following scheme :

To find the point of application 4', draw 04' (Fig. 72) parallel to tangent 4 (Fig. 71), and M4' parallel to Om1· Then the point required will coincide with point of intersection of lines 04' and M4'.

In order to find the reciprocal parabola (Fig. 74), take the pole O" (Fig. 71) upon the hyperbola, at the point of contact 8 of the tangent through the vertex. Through the

corresponding pole O (Fig. 74) draw a ray parallel to the tangents at 3 and 8, and find the reciprocal points 3′ and 8' so as to fulfil the condition of constancy (§ 52). The point 3' may be located anywhere; but, once fixed, its position determines the positions of all other points. The parallel drawn through the point 3' to the chord n is the tangent at the vertex of the parabola. The parallels 01' and 06' to the asymptotes are tangents, the common abscissa of whose points of contact is equal to 03'. The points reciprocal of the tangents 2, 4, 5, 9 can be directly determined by means of the intersections of their projections from the centres O and 3′ (Fig. 74), of which the first ray (02') must be drawn parallel to the corresponding tangent (2), and the second ray (3'2') parallel to the corresponding polar ray (023"), by which, in Fig. 71, the point of intersection of the tangent with tangent 3 is projected from O".

Thirdly, to obtain the reciprocal ellipse, pitch a pole O' within the curve, draw the conjugate chord 99, through O', and find its pole. Next, draw 08'3' (Fig. 73) parallel to the tangents at 8 and 3 and determine the points 3' and 8' so as to fulfil the condition of constancy (§ 52). As in the case of the parabola make O and 3′ centres of projection from which to draw lines parallel respectively to tangents (9) and rays (O').

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Lastly, corresponding to the internal pole O' (Fig. 73), construct the funicular ellipse (Fig. 75) according to the same principle and method.

It is worthy of remark that when the pole lies on the curve it is projective of any series of points on the curve, that the tangent at the pole forms a projective series with the tangents at the other points (§§ 7, 11).

54. Both Poles at the Curve-Centres.-Since any tangent 9 in Fig. 71 is the reciprocal of the corresponding point 9′ on the ray 09' (Fig. 72), and the tangent at the point 9' is the reciprocal of the point of contact 9 (Fig. 71); it follows

that when the pole of one curve lies at the centre, the ray 09' becomes a diameter conjugate to the direction of the tangent at 9' or to the polar ray 09 (Fig. 71). Similarly, in the original curve, the ray to any point of contact and the tangent at that point would be conjugate. In this case, therefore, the poles of both curves would lie at their centres. When both pole-centres lie within their respective curves, the allied curves are similar ellipses; on the contrary, when both centres lie without their curves, both the allied curves are hyperbolas, situate in the supplemental angles of their parallel asymptotes. Imagine two such reciprocal curves, the force-polygon being inscribed in one. and the funicular circumscribed to the other. Then, since the polar rays of the first curve are parallel to the sides of the circumscribed funicular, it follows that the stress in any side of the funicular is represented by the half diameter drawn parallel to it in the first curve. Also, each of the forces applied at the joints of the funicular will be represented by the chord of the first curve lying between the polar rays drawn parallel to the sides of the funicular, meeting at the point of application.

This principle is illustrated by Fig. 76, which represents the case of a number of central pressures distributed over an elliptic vault. The circuit of the inner curve of

force is divided into segments which, measured at right angles to the directions of the forces, are equal. Imagine now the magnitude of any force to be represented by the area of the lamina resting upon its segment. Then the mean depths 654. . . of the laminæ must be proportional (here made equal) to the chords 654 . . ., representing the forces and joining the extremities of the diameters conjugate respectively to the diameters bounding the laminæ. Finally, the stress in the elliptic envelope at the end of any diameter is represented by the half conjugate diameter. 55. The Funicular Pole at Infinity.—If the pole of the funicular pass to infinity, the force-polygon is reduced

to a straight line; since then the directions of all the forces are parallel, as in the case of a series of vertical forces applied to a girder. If in the same case the funicular be an ellipse or a parabola, the line of forces is unlimited; for, if polar rays be drawn from the pole of the line of