Ns" and NT, it is evident that the part of the perpendicular at s' which is projected into s"T", falls without the conic surface; the remaining parts of it falling within the upper and lower divisions of the conic surface.

If we produce q'k to meet D"H" in a", the point e" will be the vertical projection of the point in which the perpendicular from a to the horizontal plane, meets the upper division of the conic surface. In like manner, if R'm be produced to

A', we have the vertical projection of the point in which the perpendicular at R' meets the slant side of the cone passing through : and, because мR" and KQ" are equal, as is evident from the construction, in follows that R" and Q" are the vertical projections of two points diametrically opposite in a circular section of the upper conic surface parallel to the base.

If we take any point F' in 'Q' produced, and draw D'F', OE" at right angles to the ground line AB, it is plain that D' and E" are the vertical projections of the points in the upper and lower divisions of the conic surface through which a straight line passes, that is, perpendicular to the horizontal plane at F', so that oF', OD" are the horizontal and vertical ordinates of the points of intersection in the upper division, and DF', OE" in the lower. Also that part of the perpendicular at ', that is represented by D"E", falls without the conic surface and the remaining parts above D" and below E" fall within the upper and lower divisions of the conic surface.

PROBLEM IL

To find the point in which a given plane is cut by the given slant side of a cone.

Let EK be the ground line; r' the centre of the base T'R'Q' on the horizontal plane; p" the vertical projection of the vertex of the cone, e' any given point in the circumference of the base, and R'Q', PL the horizontal and vertical projections of the slant side passing through a'; also, let EG', Er" be the horizontal and vertical traces of the given plane.

2.

Since the slant side is given by its horizontal and vertical projections P'q', P"L, and the plane by its traces, we have only to find the projections of the required point by prob. 17, chap. The operation is as follows: Produce 'R' to G' and H; draw G'v and HF" at right angles to EH ; join vF' cutting PL in s", and draw s'"s' at right angles to EH cutting EH and r'Q' in N and s', and Ns', NS" will be the horizontal and vertical projections of the point required.

In a similar manner we find the horizontal and vertical projections D' and D" of the point in which the given plane is cut by the slant side which passes through R' the other extremity of the diameter Q'R'.

If the point E be at an infinite distance, the traces EG' and EF" become parallel to the ground line; and this circumstance produces a variation in the method of construction for some points that may require farther illustration.

Let AB be the ground line; Ka'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 PL, PM 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 'z', and through y" draw s"r" parallel to AB; then s's' and 'r' being drawn perpendicular to AB, will give the horizontal projections of the required points; and s", r' 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 xv intersecting c'o 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 xx'.

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 x ; let r' be the centre of the base, and at the same time the horizontal projection of the axis, and vertex; and KP" the vertical 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 to 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 KA equal to KK", 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 Q' and R' draw Q'L, R'м perpendicular to AB, and PL, P'M will be the projections of the slant side passing through Q' and R': let c'A meet p'o' in R', and having made кY" equal P'R', draw H′′Y"J" parallel to AB, and J, H'H' parallel to P'P' and 'H' will be the points required in q'r'.

Το

To find the point in which the curve meets any other radius r'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 r's' to meet A'v' and AD in v' and T; draw v'в, TT" parallel to r'r', and join BT", cutting PN in E"; draw E'E' parallel to P'P' and E' will be the point in which r's' is intersected by the curve.

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

When the points v, and T become too remote to be consequently 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;

VOL. II.

79