Let AB be the ground line; KQ'c'R' the base of the cone on the horizontal plane, p' its centre, KP" the vertical projection of the axis of the cone; and EF, G"H" the traces of the given plane, which are parallel to AB.

Suppose the horizontal projection Q'R' of two opposite slant sides to be the diameter of the base parallel to AB; and therefore P"L, P′′M the vertical projections of those sides. To determine the points in which the plane meets those slant sides we may proceed as follows:

Make KD equal to Ko, join c'D cutting Q'R' in z'; make KY“ equal to p'z', and through y" draw s"T" parallel to AB; then s's and TT being drawn perpendicular to AB, will give the horizontal projections of the required points; and s", T" the corresponding vertical projections.

Again, to determine the intersection of the plane and slant side passing through K: make P'v' equal to KP" which is the altitude of the cone; draw Kv' intersecting c'D in w', make w'x' parallel to AB, and x' will be the horizontal projection of the required point, and x'w will be equal to the vertical ordinate, the horizontal ordinate being кx'.

PROBLEM III.

To construct the horizontal projection of the curve made by the intersection of a given plane with a given conic surface.

Let AB be the ground line; Ko'c' the circumference of the base on the horizontal plane, touching the ground line in K; let p' be the centre of the base, and at the same time the horizontal projection of the axis, and vertex; and KP" the ver tical projection of the axis.

Suppose the given intersecting plane to be parallel to the ground line, or which is the same in effect, let the horizontal and vertical traces A'V', T"B" of the given plane be parallel to AB, and suppose the horizontal trace a'v' touch the base of the cone in c'.

To find the axis of the projection, make кA equal to Kк", and join c'a in P'R'o' parallel to AB: take P'o' equal to KP"; join ko' intersecting ac' in u', and from u' draw u'D' parallel to AB, and c'D' will be the axis of the projection.

Again, to find the points in which the curve to be projected cuts the diameter Q'R' parallel to AB; from o' and R' draw QʻL, R'M perpendicular to AB, and P"L, PM will be the projections of the slant side passing through Q' and R': let c'a meet p'o'in R', and having made KY" equal P'R', draw "y"" parallel to AB, and JJ, н"н' parallel to r'r' and 'H' will be the points required in q'R'.

To find the point in which the curve meets any other radius P's', draw s'N at right angles to AB; join P'N which is the vertical projection of the slant side passing through s': produce p's' to meet A'v' and AB in v' and T; draw v'B, TT" pa. rallel to P'r", and join BT" cutting P'N in E"; draw E'E' paraldel to P'P', and E' will be the point in which p's' is intersected by the curve.

In a similar manner we may find any number of points in the required section C'E'D'F'.

When the points v and T become too remote to be conse. quently used in the construction, we may find the required points of the curve by the method used in determining the intersection of the plane by the slant side passing through K.

In this example, in which the cone is divided by the plane into upper and under parts of the conic section, is called an

ellipse; the curve c'E'D'F' is therefore the projection of an ellipse, and consequently, C'E'D'F' is also an ellipse.

PROBLEM IV.

To construct the ellipse of which the curve c'E'D'r' in the preceding problem is the horizontal.

This problem is readily solved by finding by prob. 12, chap. 11. the positions of the points of the curve on the horizontal plane by the revolution of the intersecting plane to a coincidence with the horizontal plane.

Or, we may proceed as follows, which is nearly equivalent. Because KO' is the position on the horizontal plane of the slant side passing through H, and KC'A the angle of elevation of the given plane; it is plain, that cu' is the transverse axis

of the ellipse section required. Make c'ô equal to c'u', and. will be the position on the horizontal plane of the vertex determined by u'.

In like manner, take c' equal to CR' and make the perpendicular equal to P'P', and will be a point in the required section. In a similar manner, we may find any number of points in the circumference of the required ellipse dec'p.

It is evident, from this construction, that the ellipse SEC'o is derived from C'E'D'F' by elongating each abscissa from c' as cr' in the constant ratio of c'D' to c'v'; so that cr' is to c'r as c'D' to c'd, while the semiordinate remains the same as P'H'.

And as the curve dec' is by the definitions of conic sections an ellipse, it is manifest from the constant ratio of the abscissas cp' and c' having a common semiordinate p'n' or , that the projection C'E'D'F' is also an ellipse.

PROBLEM V.

To construct the section of a cone by a plane parallel to the axis of the cone.

Let AB be the ground line; p' the centre of the circular base KQ'R' of the cone touching the ground line AB in к: produce the radius P'K to P', and take KP" equal to the axis of the cone which is supposed to be at right angles to the plane of its base, and consequently to the horizontal plane; then P"K is the vertical projection of the axis, and p" of the vertex.

Suppose the cutting plane to be parallel to the vertical plane, and to intersect the horizontal plane in the straight line R'vo', which is therefore parallel to AB, and consequently perpendicular to the diameter CK.

Draw any radius P's'r of the base, meeting R'q' in s”, and the circumference of the base in T. Find by prob. 1, chap. IV, the vertical projection p'v of the slant side passing through T, the corresponding horizontal projection of this slant side being P'T: through s' draw s'ss" at right angles to AB, and meeting VP" in s", and Ns" is the altitude of the conic surface at s', because NS', Ns" are evidently co-ordinates of a point of the slant slide passing through T.

And, since the cutting plane which passes through 'a' is perpendicular to the horizontal plane, it is evident that Ns', Ns" are the co-ordinates of the point in which the slant side

terminating in r penetrates the cutting plane; if therefore we make s'L equal to Ns", it is plain that L will be the position on the horizontal plane of the point denoted by s', s", by the revolution of the cutting plane about the intersection R' Q'.

By a similar construction, we may determine any number of points in the curve Q'LXR', which will be the section required.

The curve required may be obtained still more simply by merely finding the perpendiculars NS"H, and describing the curve through L, s", M, &c. without determining the corresponding points in Q'LXR'.

It is evident that the plane meeting the base at right angles in p'yo' must also meet the upper division of the conic surface, and produce another section equal and similar to QʻXR'. The curve determined by this construction is an hyperbola.