given by means of its projection a and projector Aa; and let YXZ be

the inclination, and aX the trace of the new plane upon which A is to be projected.

Since Aa is perpendicular to aZ, its projection on aY is perpendicular to the trace aX (Pls. 11. 18), and hence the projections of the points A, a, in it are in a line perpendicular to aX.

Whence, since a is given in the given plane aZ, its projection a, on aXY can be found by a former construction (Prop. 1.), and the line a,a in which the projection of A upon aXY is situated is known.

Also since Aa is given in magnitude and position, and that aa, is found, it will follow that AA' parallel to aa, can be constructed; thus giving A' as the required projection of A upon aXY, AA' as its required projector, and A'a, as the projection of the line Aa.

Moreover, the system may be turned about aa, to coalesce with aXY the new plane of projection, and all the parts of the figure will retain their relative positions as regards aa, and each other. Wherefore, the entire data and constructive lines might have been originally exhibited there, and the point A' and the magnitude of AA' the new projector have been determined without any use of the figure in space at all. In fact, the figure actually used in the construction of the problem in practice (which is exhibited in the margin) is precisely that which the turning round of the figure in space produces.

SCHOLIUM. This problem is one of great importance in projection, being the foundation of the method of projecting figures of three dimensions. The figures are supposed to be given standing with one of their faces resting on the given plane aZ, whilst their several points which are not in that plane are supposed to be given by means of their projections on it and their corresponding projectors. These projectors are sometimes actual lines in he tproposed figure, as in the case of a cube or any right prism, but more frequently they are lines having definite relations to the figure, and which it will be necessary to determine à priori when the figure is defined independently of them. As exercises or examples of this kind, the platonic bodies, or regular solids, have generally been employed.

It is too obvious to call for special remark that a line is projected when two points in it are projected, and that its inclination becomes known when the projectors of three points are found. The same applies to all rectilinear and curvilinear boundaries, and lines drawn upon a given plane or surface. The entire projection of all bodies is thus comprised in this problem. It would lead us beyond moderate

VOL. II.

S

bounds to give any considerable number of examples; but it still appears incumbent on us to offer a few instructive ones in this place.

EXAMPLE 1.

Given two points in a line in reference to a given plane, to find the projection and inclination of the line in reference to the plane of projection.

Let a, b be the projections of the points A, B on the given plane: project these in a, b, on the plane of projection; and project also A, B in a, B, and a B is the projection sought, and Aa', Bb' are the projectors on that plane.

Make aa', BB' perpendicular to aß and equal respectively to Aa', Bb': then a'B', aß being produced to meet in y, we have aya' the 'inclination of the given line to the plane of projection.

EXAMPLE 2.

A cube is given having one of its faces ABCD in a given position on a given plane, to construct its projection.

Project the square as before in a,b,c,d,. Project also the altitude A'A", which is equal to one of the sides, giving a,a,: then a,b, ad, being drawn parallel to a,b,, a,d, will give b, and d, in the perpendiculars Bb, Dd; and lastly dc, bc, parallel to dc, be, give c2. The figure is then projected, as was required.

The same process applies to any right prism, whose base is situated on the given plane: but in case of such prism being cut obliquely, each of the edges must be separately projected, instead of the upper face being projected by parallels, as has been done in this construction.

EXAMPLE 3.

To construct a triangular pyramid, having its base given in a given plane, and likewise given the perpendicular from the vertex to the base in magnitude and position.

Find the projections of the base and of the vertex, and draw lines from the projections of the base to that of the vertex; and the projection is complete.

EXAMPLE 4.

Given an oblique prism in a given position on a given plane, to construct its projection.

Project the base and one of the upper corners of the prism as in the preceding example; and then, as in the second, construct the projection of the upper face by parallels, and of the edges likewise.

EXAMPLE 5.

To project a cube on a plane at right angles to one of its diameters. (a.) All the three edges which meet in one extremity of that diameter are equally inclined to the diameter, and hence, also, to the plane of projection: they will hence be projected into equal straight lines. Also, the three planes which connect these edges two and two, having their inclinations complementary to the inclinations of the opposite edges, will be equally inclined to the plane of projection; and hence their contained (right) angles will be projected into equal angles. These three projected angles, making together the angular space about the projection of the solid angle, are equal to four right angles; and hence each of them is 120'.

D

Let these be AB, AC, AD, of the annexed figure. In the next place, the opposite solid angle of the cube is projected also in A (which we shall also mark A'), and the three edges which meet in this point are parallel to those already projected, but lying in the opposite side of the diagonal. These have the same inclination to the plane of projection that the former ones had and hence their projections will be of the same length as the former ones, and be situated in continuation of them. They will therefore be represented by A'B', A'C', A'D' respectively.

Lastly, the remaining edges being those which join the points already projected, we have only to join the adjacent projected points in succession, to obtain the projection of the entire cube.

It is obvious, therefore, that the projections of all the edges lie in the sides and diagonals of a regular hexagon; the edges which meet in the given diagonal forming the three diagonals, and the remaining edges the sides, of the hexagon.

This construction, however, merely gives the form, but not the magnitude of the projected figure, for any given length of edge. This relation is fixed, and is exhibited by a very simple construction.

Take two lines, ab, ac (Fig. 1), at right an-, gles, and make each equal to the edge of the cube, and join be: this will be the diagonal of one of the square faces. Make cd perpendicular to be and equal to ab or ac, and join bd; this will be the diagonal of the cube itself.

In Fig. 2, draw AA' perpendicular to the line b'b, and equal to the diagonal of the cube, bd. If, now, we conceive a plane to pass through the diagonal AA' and one of the edges AB, it will be the projecting plane of AB, and Ab will be its projection, as to position. Moreover this plane will pass through the opposite edge (since they are parallel and AA'joins them), and will cut the faces adjacent to these edges at the points A, A' in diagonals of those faces. Whence AB

is an edge, A'B the diagonal of a face, and AA' the diagonal of the cube itself. The three sides AB, BA', A'A of the triangle AA'B are therefore given, being equal to ba, be, bd of Fig. 1.

Wherefore, the entire construction will be:

Construct figures 1 and 2 as above; then, with any centre and radius Ab describe a circle; and in this circle describe a regular hexagon, as in the figure of the first analysis. This will be the projection sought.

(b.) The preceding construction is rather effected by special considerations belonging to the particular case, than by means of the general principle already developed, and to which the problem was proposed as an illustration. It may, therefore, be proper to give another which shall be more illustrative of the general principle.

In the first place, since each edge is perpendicular to the plane containing the other adjacent ones, that edge is perpendicular to the face which contains them: and hence the angle made by that plane with the plane of projection is the complement of that made by the edge with the plane of projection.

Let ABCD be a face of the cube; hAX the trace of the plane of

that face on the plane of projection. Make Ah the diagonal of the cube, Ak the diagonal of a face, and kh the edge of the cube into a triangle, and draw Ag perpendicular to the trace hX. Also, make Af equal to AF the semi-diagonal of the square face.

Then, as was shown in the concluding part of the former construction, kAg is the inclination of the face of the cube ABCD to the plane of projection; and by the general principle of the method, if we draw kc, and dfb, the figure Abcd will be the projection of the face ABCD.

Again, hk is perpendicular to Ak, and equal to the edge CC' of the cube which stands at right angles to ABCD at C; and hence the point C' is projected in A. Also, the edges of the opposite face are parallel, each to each, to those of ABCD; that is, AB to A'B', etc. Whence their projections will be parallel and equal to those of ABCD. We have, therefore, the parallalogram Ab'c'd' as the projection of the face A'B'C'D', equal and similar in all respects to Abcd.

Lastly, the projection of the edges joining these points two and two in order, will be the lines joining the projections of their extremities in the same order. Wherefore, drawing dd', bb', Ac, Ac', we have the entire projection sought.

All that was actually required by the problem to be done has been effected but it may be desirable still to investigate the character of the figure into which the cube has been projected-that investigation to be founded on the construction itself as here effected.

By the construction the figure becomes a hexagon, with lines drawn to its angular points from a point within it; we shall show that it is a regular hexagon, and that each of the lines from the point within it is the radius of its circumscribing circle.

Since the six edges of the cube that meet, three in A and three in C' are equally inclined to the diagonal AC', they are equally inclined to the plane of projection, which is, by hypothesis, perpendicular to AC'. Whence they are projected into equal lines; or Ab, Ab', Ac, Ac', Ad, Ad' are all equal, and are situated, therefore, in the circumference of the circle about A.

Again, since AD, C'B' are parallel and in the same projecting plane through the diagonal AC, they are projected in Ad, b'A which form one continuous line. In like manner bÀ, Ad' are one line, and c'A, Ac are one line. The diagonals of the hexagon are, therefore, diameters of the circle about A.

Finally, since dd', Ac', b'b and Ac are the projections of four parallel and equal edges of the cube, they are equal to each other; and hence the sides dd', bb' are each equal to the radius. In the same way each other pair of opposite sides may be proved respectively equal to the radius. Whence the figure is a regular hexagon.

If, as is sometimes the case, such a projection of the cube be given, and it is required to find the edge of the original cube, it may be easily effected as follows:

Take any one of the equilateral parallelograms of which the hexagon is composed, as Abcd; draw the diagonals Ac, bd meeting in f', and make fgfd: then dg is the edge of the original cube.

For, bd is the projection of the diagonal BD of the square face ABCD; and as this is parallel to the plane of projection, bd is equal to BD. Our construction, then, gives dg equal to the side of the square