2. To project the slant side of the cone, take any point o' in the circumference of the base, through which and the centre r' draw Q'r'R' a diameter of the base; also, from q' draw Q'L at right angles to the ground line AB; and join P"L; then will P'Q' and PL be the horizontal projections of the slant side of the cone which passes through the point q'.

This construction of the slant side is evident, because L being the vertical projection of the horizontal point q'; there. fore P' and o' are the horizontal projections of two points of the slant side, and P"L, the corresponding vertical projections; and consequently, P'q', P"L are the required projections of the slant side passing through o'.

If we make a similar construction for the slant side passing through R' we have the construction of the two slant sides of the cone, in which the curve surface of the cone is intersected by a vertical plane passing through the axis of the cone. Thus PL and PM are the vertical projections on the opposite slant sides passing through the extremities of the diameter Q'R' of the base, and the opposite radii P'q', P'R' are the corresponding horizontal projections.

3. To find the vertical projection of any point of the surface corresponding to any given horizontal projection. Let s' be any given horizontal projection of a point of the curve surface of the cone; draw the radius r's'a', and having constructed the slant side by its projections P'o', PL, draw s'ns" at right angles to the ground line AB, meeting the vertical projection PL in s", and s" will be the vertical projection of that point of the conic surface which has s' for its horizontal projection.

In the preceding construction we have considered only that part of the whole conic surface which is between the vertex and base; but as the conic surface may be extended indefi. nitely downwards below the base, and upwards above the vertex, it is plain that the horizontal projection R'r'q' of the opposite slant sides as well as the vertical projections P′′K, P′′M, ^ should be produced indefinitely both ways; that is, r'Q' towards F' and '; and P"K, P"M towards E", R"; H", D".

Now, the vertical projections R"E", D"H" being both in a ver. tical plane passing through r's', if we produce s's" to meet D"H" in T", we shall have r" for the vertical projection of the point in which a perpendicular to the horizontal plane at s' meets the slant side which passes through the point R of the base; this perpendicular therefore meets the conic surface in two points, of which s" and T" are the vertical projections, the horizontal projections being coincident in the points s'. Also, the vertical ordinates of these two points being

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

R", we have the vertical projection of the point in which the perpendicular at n' meets the slant side of the cone passing through o' and, because MR" and κQ" are equal, as is evident from the construction, it follows that R" and o" 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 in R'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 or', 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 F', 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 II.

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; p' the centre of the base T'R'Q' on the horizontal plane; p" the vertical projection of the vertex of the cone, q' any given point in the circumference of the base, and R'q', P'L the horizontal and vertical projections of the slant side passing through a'; also, let EG', EF" be the horizontal and vertical traces of the given plane.

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. 2. The operation is as follows: Produce q'R' to G' and ; draw G'v and HF" at right angles to EH; join VF' cutting "L in s", and draw s's' at right angles to EH cutting EH and P'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; 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, 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 p'z', and through y" draw s"r" 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'p 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 KX'.

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; Ka'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 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 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 Q' and R'draw Q'L, R'M perpendicular to AB, and P"L, 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 KY" equal P'R', draw a"Y"J" parallel to AB, and J", "H' parallel to P'r' and 'H' will be the points required in q'R'.