PROBLEM V. Fig. 105.

To find the curve of intersection of a cylinder and a surface of revolution.

GRF is the horizontal trace of the cylinder, and FH one of its generators; the vertical line through O is the axis of the surface of revolution, and o'a'p'q' the projection of that meridian which is parallel to the vertical plane of projection.

This problem might be solved by taking horizontal auxiliary planes, cutting the surface of revolution in circles and the cylinder in curves equal and similar to its horizontal trace. But the construction of these latter curves is, in general, too laborious, and the following method will be found preferable.

Construction. Take any plane section ABC of the surface of revolution, at right angles to its axis. Let this circle be taken as a directrix of a cylinder, having its generators parallel to those of the given cylinder. The horizontal trace of that auxiliary cylinder is the circle GDF, equal to ABC. As the horizontal traces of the given and auxiliary cylinders intersect at the points G and F, the two cylinders intersect along the two straight lines FH and GK, and the points H and K where these two lines meet ABC are evidently points on both the given surfaces, and consequently points on the required curve.

In a similar way as many points as desired may be found on the curve. It is clear there are two auxiliary cylinders which touch the given one, namely, those of which the horizontal traces touch GRF. These give the limiting points of the curve, but in general they can only be determined by trial.

Note. In finding the intersection of a cone and a surface of revolution, the auxiliary surfaces are to be cones having the same vertex as the given one, and circular sections of the surface of revolution for their directrices. The horizontal traces of these auxiliary cones will be circles, and the construction will be similar to that given above.

PROBLEM VI. Fig. 106.

To find the curve of intersection of two surfaces of revolution when the axes are parallel.

As the axes are parallel to one another, they are both at right angles to the same plane (Theorem VIII. Ch. I.), so that the auxiliary planes may be taken perpendicular to the axes, and consequently cutting both surfaces in circles.

Let one of the given surfaces be that generated by the circle IBP revolving about the axis Oo, that is to say, a ring of circular section; and let the other surface be the cone generated by the straight line VU revolving about the axis Vv. The horizontal plane is, as usual, taken at right angles to the axes.

Construction. Draw any line a'b' parallel to xy, and cutting the vertical projections of the two surfaces. This is the vertical trace of a horizontal plane which cuts the cone in the circle EFG and the ring in the two circles AEB, CGD. The circle EFG cuts AEB at the points E and F, and the circle CGD at G and H. Therefore the four points E, F, G, H are on the required curve. In a similar way as many points as are necessary for determining the curve may be found.

The limiting points of the curve are those points at which a circular section of one of the surfaces touches the circular section of the other. As the points of contact of the horizontal projections of these circles must always be on the line of centres ov, it follows that the limiting points of the curves are on the lines VL and VM, that is, the two generators of the cone which are in the plane of the axes, Hence the limiting planes may be found as follows:

Let the plane of the axes, together with its sections of the two given surfaces, be turned about Oo till it is parallel to the vertical plane of projection. The elevation of the section of the ring is the circle 'b'q', and the elevation of the section of the cone is the isosceles triangle l'v'm',. The limiting planes are those of which the vertical traces pass through the points p, q, ", j', and the limiting points are determined from them in the usual way. They are P1, I ̧, Q, and J.

In this figure the cone penetrates the ring, and there are two curves of intersection; the lower one has been omitted in the plan for the sake of clearness. The cone is supposed be removed out of the ring.

PROBLEM VII. Fig. 107.

To find the curve of intersection of two surfaces of revolution when their axes intersect.

Let Oo be the axis and PEC the meridian of one surface; OD the axis and PGQ the meridian of the second surface; it is required to find their common section. The vertical plane of projection is taken parallel to the axes, and the horizontal plane perpendicular to one of them.

When two surfaces of revolution have the same axis they must intersect in one or more circles; for as every point on either meridian describes a circle, the point of intersection of the two meridians describes a circle about the common axis, which is the intersection of the two surfaces. Hence, as any diameter of a sphere may be considered as its axis, when the centre of a sphere is on the axis of a surface of revolution, the sphere and that surface intersect in a circle which is a parallel of the given surface. By taking for auxiliary surfaces spheres having their common centre at the point of intersection of the axis, the given surfaces will be cut in circles by the spheres.

Construction. From the centre o' describe a circle ef'h' intersecting the elevations of the two given surfaces. Suppose e'f'h' to be the elevation of a sphere, it intersects the first surface in the parallel EKF, and the second surface in the parallel GKH. These two circles are in planes perpendicular to the vertical plane of projection, since the axes are parallel to that plane, and their common section is consequently the straight line through K perpendicular to the vertical plane; that is, the line KL. K and L are points on the curve required. In a similar way any number of points on the required curve may be determined.

In the figure the surface shown in dotted lines is supposed to be removed.