8. To determine the centre and radius of a sphere, on the surface of which four given points are situated.

Care must be taken not to assume these points so that any three are in the same straight line : and the fourth point cannot be situated in the same plane with the other three, unless the four are in the same great circle of the sphere.

If the line joining any two points be bisected by a plane perpendicular to it, since every point in that plane is equidistant from the extremities of the line which it bisects, the plane must contain the centre of the sphere. For the same reason if two other planes be assumed, similarly bisecting any other two of the lines joining the given points, three planes will have been determined, each containing the required centre of the sphere. It will therefore be in their common intersection, that is, in the point in which the three planes intersect.

The length of the radius will be found by means of the right-angled triangle. If the distance from the plan of the centre to that of any point on the surface be taken as the base of a right-angled triangle, and the difference of level between the centre and this point as the perpendicular, the hypothenuse will be the radius required.

9. Draw the cylinder which would envelop a sphere of 15 inches radius and 1 inch above the ground. Let its axis be inclined 50°, and make in plan an angle of 30° with the ground line. Show circle of contact and horizontal trace of cylinder.

(1) Draw sphere and plan of axis of cylinder according to conditions.

(2) Use the plan of the axis as the ground line for a new elevation, in which draw a second elevation of the sphere. A line through the centre of sphere meeting the new ground line at the given inclination will be the elevation of the axis, and parallels to it, tangential to the circle which represents the sphere, will give the major axis of the ellipse in which the cylinder cuts the horizontal plane. The minor axis of the curve will be marked off by parallels to the axis of the cylinder in plan, tangential to the circle which is the plan of the sphere. From the second or auxiliary elevation, the elevation on the original vertical plane can be easily determined.

(3) The circle of contact is shown, in the auxiliary plane of elevation, as a straight line, and from this can readily be determined in plan, and from thence on the original vertical plane.

IO. A cone, with vertex 5 inches high, axis inclined 50o and in plan making an angle of 30° with the ground line, envelops a sphere of 1 inch radius and 1 inch high. Determine horizontal trace of cone and circle of contact.

As in the last problem use the plan of the axis as the ground line for an auxiliary elevation. In this auxiliary plane set up the elevation of the axis by drawing a line inclined 50° with xy through the point which is the elevation of the centre of the sphere, and determine vv the vertex of the cone 5 inches high. Tangents from v to the circle representing the elevation of the sphere will mark off on xy the line which is the elevation of the horizontal trace of the This same line is also the length of the major axis of the ellipse which is the horizontal trace of the cone.

cone.

To find the minor axis of the same ellipse, the point must be determined on the circle of contact, through which a line from the vertex to the extremity of the minor axis would pass. In the auxiliary elevation this line will be shown by joining the vertex and the centre of the major axis. The intersection of this line with the straight line representing the elevation of the circle of contact gives the elevation of the point required, through the plan of which point a line drawn from the plan of the vertex marks off the extremity of the minor axis sought.

But note that the straight line which in the auxiliary elevation will be the elevation of the circle of contact will not pass, as in the case of the cylinder, through the elevation of the centre of the sphere. It is the line joining the tangent points before determined.

The plan of the circle of contact can be readily determined by previous problems.

The traces on the horizontal plane would represent the shadows of spheres cast by rays of light which are parallel in the case of the cylinder, and which proceed from a fixed point in that of the cone.

The circles of contact would separate the dark and light parts of these spheres.

IX.

TANGENT PLANES TO CONES.

PROBLEM I.

To determine a tangent plane to a cone at a given point A on its convex surface.

Let

be the vertex of the cone. Join points A and V: the indefinite straight line AV is the generatrix of the conic surface. Determine the 'trace' of the line AV and that of the conic surface, the former will be a point in the latter. At this point determine a tangent to the trace of the conic surface; the plane of this tangent and AV is the tangentplane required.

PROBLEM II.

To determine a tangent-plane to a cone, which shall also pass through a point P which is external to the conic surface.

If V be the vertex of the cone: join points V and P and obtain the horizontal trace of the indefinite line VP; through this point H draw a tangent to the horizontal trace of the conic surface; the plane of VH and this tangent is the tangent-plane required.

PROBLEM III.

To determine the traces of a plane inclined degrees, the horizontal trace of the plane making an angle of m degrees with xy. See Fig. 29.

Fig. 29.

Assume a point aa' in the vertical plane; make this point the vertex of a right cone generatrix inclined degrees. Let cef be the circular base of the cone. Draw ct a tangent to cef at c, meeting xy at t, and making the angle xtc an angle of m degrees; join a't, then cta' is the required plane; being a tangent-plane to the cone, which it touches. in the line AC.