The width of the diameter DE on plan No. 2 is the same as the width of diameter DE on plan No. 1, and is found by parallels to xy drawn from D and E, No. 1, to meet a projector dropped from F", No. 2. It will now be seen that the plan of the circle No. 1 at an angle of 45°, as shown in No. 2, is an ellipse, of which BC and DE are the conjugate and transverse diameters. The points through which the curve of the ellipse passes are thus projected. On plan No. 1, mark off any points, e.g., aa, bb, cc, dd, on each side of BC, and at equal distances from it. Join aa, bb, cc, dd, and produce the lines as projectors to B'C' in the points a', b', c', d'.

Now let these points be transferred to B'C', the elevation of the diameter in No. 2, and from these transferred points drop projectors

to intersect the parallels to xy drawn from the corresponding points in plan No. 1. Then the curve of the ellipse passes through the points of intersection, which curve must be traced by hand.

The plan of the apex A' is found by projecting A'G on to the parallel BC produced. Tangents drawn from G to the curve of the ellipse, representing the sloping edges of the cone, will complete the plan of the whole cone.

It will be seen that the tangents do not meet the curve in D and E.

Problem 44.

To construct the projections of the tetrahedron.

In this case we assume the given solid to rest on one of its faces on the horizontal plane.

Let AB be one of the edges resting on the horizontal plane, and inclined to the ground line at any given angle q. On AB describe the equilateral triangle ABC. Find O, the centre of the circum

scribing circle, and join AO, BO, CO; then O will be the plan of the vertex, ABC will be the plan of the base, and AO, BO, ĈO, will be the plans of the edges meeting in 0.

From O, draw OD at right angles to OB; then with B as centre, and BA as radius, describe an are cutting OD in D; OD will be the height of the pyramid. Draw AA', BB', CC', and 00' perpendicular to xy; make E'O' equal to OD; then join A'O', B'O', and C'O', and the figure thus formed will be the elevation of the solid.

Problem 45.

To construct the projections of the octahedron when its axis is vertical.

In this case, the axis is assumed to be vertical; let AB be one

edge, and let it make any angle with xy. On AB describe the square ABCD, and draw the diagonals AC, BD, intersecting in 0; then ABCD will be the plan of the octahedron, O being that of the

vertex.

Next, from O draw 00' at right angles to xy, and make O'O' equal to the diagonal of the square ABCD; bisect O'O' in E', draw D'B parallel to xy, and DD', CC', AA', and BB' perpendicular to xy, then

O', O', A', B, C, D', will be the elevations of the angular points of the solid, whose elevation will be formed by joining O'D', OʻA', O'B', O'D', O'A', and O'B'.

NOTE. The elevations O'C' and O'C' would be unseen, and are consequently represented as dotted.

Problem 46.

To construct the projections of the octahedron when it lies on its face on the horizontal plane.

In this case, the solid lies on one of its faces on the horizontal plane. Let ABC be the face in the horizontal plane, and xy parallel to the axis passing through E; then if AB' and CC are drawn at right angles to xy, A'C' is the elevation of the face ABC.

Make A'G' equal to an edge of the solid; G'C' equal to B'C'; draw C'E' perpendicular to A'G', and bisected in 0; join A'E and G'E', thus completing the elevation.

Next, in order to project the plan, we proceed thus. From D', drop a projector perpendicular to xy to BD, AG, drawn parallel to the axis EC in D and G.

meet the lines Join DE, GE; Q

then EGD will be the plan of the surface ED'. Next join AE, EB, DC, CG, to complete the required plan.

Problem 47.

To construct the plan and elevation of an octahedron when resting on one of its faces, and when one edge of this face makes an angle of 15° with the vertical plane.

Draw CA inclined to xy at an angle of 15°, and upon it describe an equilateral triangle ACE. Describe a circle about the triangle, and in it inscribe a regular hexagon, ABCDEF. Join FD, DB, and BF, which completes the plan of the solid.

Now as the face ACE rests on the horizontal plane, the points A, C, E, will be projected on xy in A', C', E'. We now proceed to

find the height of the given solid.

Since FA is the projection of a line of which FB is the real length, draw AG at right angles to AF, and with centre F and radius FB, describe an arc intersecting AG in G; then AG is the height of the solid.

Make H'F' equal to AG, and draw a line parallel to the ground line. Cut this line in the points F', B', D', with perpendiculars from F, B, D. Then join F'A', F'E'; B'A', B'C'; and D'C', D'E'; which completes the elevation of the solid.