described on the diagonal bd; and which is therefore equal to the side of the square on the diagonal BD; that is, to the edge of the cube.

The same figure suggests also a ready method of finding the side of the hexagon when the edge of the cube is given, viz. :

Let dg be the given side; on which describe an isosceles rightangled triangle df'g; and make the angle f'de equal to 30°: then de is the required length of the projection.

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For, since Abed is an equilateral parallelogram, db bisects the angle d, and hence, f'de 30°; and the construction is obviously true. This projection is the foundation of the Isometric method of which further discussion will be given hereafter in the APPLICATIONS.

PROPOSITION XI.

Given the trace and inclination of a plane to a horizontal plane, to find its trace upon, and inclination to, a given vertical plane, and to construct the angle made by the traces of the plane upon the horizontal and vertical planes.

Let X'X be the given trace of the vertical plane upon the horizontal, and PP, the trace of the given inclined plane upon the horizontal.

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(1.) Take any point p, in X'X, draw p2 P, perpendicular to PP1, and set off pp, on X'X (on either' side of p.) equal to p, P1, and make P2 PP equal to the given inclination of the inclined plane to the horizontal, meeting the perpendicular from p, in P2: then P, is a point in the trace of the inclined plane on the vertical plane through X'X. Also, Pis another point;

X'

and hence, PP, is the required trace on the vertical plane.

1

For, since in the vertical plane, PP, is perpendicular to the trace in the horizontal plane, the line PP, is itself perpendicular to the horizontal plane; and hence the plane P,p,P, through it is perpendicular to the horizontal plane. Also since p,P, is perpendicular to the trace of the oblique plane, and P2p, perpendicular to the horizontal plane, the line PP, is perpendicular to PP, (Pls. and Sols. II. 13), and PP、P2 is the inclination of the oblique plane to the horizontal one. But in the triangles Pap.P, PPP, the angles Pp.P, and PP1P; are right angles, and the side Pap, common and p.P1, p,p, equal; and hence the angles PPP, P, PP, are equal. Also Papp was made equal to the given inclination; and hence the plane whose traces are PP and PP, has the given inclination to the horizontal plane. That is, PP, is the vertical trace required.

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(2.) To find its inclination to the vertical plane. Take any point p, in X'X (Fig. 2), and draw p,P, perpendicular to X'X to meet the horizontal trace in P, and PP, perpendicular to the vertical trace; make PP2 equal to pP, and join p.P; then the angle p,p,P, is the inclination of the plane to the vertical plane.

For, as in the preceding reasoning, it may be shown that p'P, P, is the inclination, and that p, p,P,

is equal to it.

(3.) To find the angle made by the traces themselves.

Conceive in Fig. 1., that the oblique plane is turned about ✰' PP, to coincide with the horizontal plane: then as PP, is perpendicular to PP, it will during this revolution always be perpendicular to PP1; and hence when it coincides with the

horizontal, will coincide with the perpendicular p,P,. Make, therefore, PR, equal to P, P1, and join R,P. Then, evidently R,PP, is the angle made by the traces of the plane.

PROPOSITION XII.

Given the position of a line with respect to the horizontal plane, to find its position with respect to any vertical plane.

Let X'X be the trace of the vertical plane upon the horizontal one, aß, the projection of the given line

of which a, is the trace of the line on the horizontal plane, and B, the intersection of the projecting plane with the trace X,X.

(1.) Then the projecting plane being also perpendicular to the horizontal one, its intersection BB with the vertical will be perpendicular to X'X; and B,B, will contain the trace of the given line upon the vertical plane.

Make B equal to Ba1, and the

angle Baß, equal to the given inclination: then B, the intersection, is the trace of the given line upon the vertical plane.

For evidently we have only turned the projecting plane about BB to coincide with the vertical plane; and then turned the compound system about X'X to coincide with the horizontal plane.

(2.) Since the projecting plane of the given line upon the vertical is perpendicular to the vertical, its trace upon the horizontal will be perpendicular to X'X.

Whence, draw a,a, perpendicular to X'X, which gives the projection of the point a; and we have B, already found, which being the trace of the line on the vertical plane, the line aß, is the projection of the given line on the vertical plane.

(3.) To find its inclination to the vertical plane. Make aß equal to aß, and join Ba,: then aẞa, is the inclination of the given line to the vertical plane.

For a Ba is evidently the inclination in question, and aßa, is equal to it; the triangles a,,, and aẞa, being in all respects equal.

PROPOSITION XIII.

A point is given in reference to the horizontal plane by means of its projection and projector; it is required to find its projection and projector in reference to a given vertical plane.

Let X'Y be the horizontal plane, a, the projection of the given point A, and Aa, its projector on that plane;

also let X'Z be the given vertical plane, of which XX' is the trace.

Now conceive the projector on the vertical plane to be Aa, and a, the projection of the point A.

a

X

Then since Aa, and Aa, are perpendicular to the planes of projection respectively, the plane through them is perpendicular to each and to their common section XX' (Pls. 11. 16, 18); and a,a,a,a are perpendicular to the planes X'Y and X'Z. Whence Aa, and aga are parallel, as are likewise Aa, and aa. Also, since Aa, is perpendicular to X'Y it is perpendicular to a,a in it: and the figure Aa, aa, is a rectangle, having a,a equal to Aa, and aa, equal to Aa,.

The projection, then, of A upon the vertical plane is in a plane through A perpendicular to the trace of the two planes of projection; its projector is equal to the given distance of a, from the trace, and its height above the trace is equal to the given projector upon the horizontal plane.

If now we turn the vertical plane about XX' to coincidence with the horizontal, aa and aa, will form one straight line at right angles to XX'; and if we retain only these × planes and the lines internally in them, the figure will be as in the margin.

We see then, that a, being given, we have only to draw a a perpendicular to XX', and make aa, equal to the given projector.

SCHOLIUM. We have already seen that when a figure is given for projection on any plane, as the horizontal, for instance, the plane to which that figure is referred being also given, we do actually find the lengths of the projectors themselves. (See Props. 1.-IV.) We see by this proposition that the projections on a given vertical plane, when turned on the horizontal, are in lines from the horizontal projections perpendicular to the trace of the vertical plane; and that their altitudes above that line on the drawing are equal to the projectors of those points. Whence it becomes easy to construct simultaneously with the horizontal projection of a figure its projection on any given vertical plane.

PROPOSITION XIV.

Let the three faces of a trihedral right angle be projected on any plane which also cuts those faces, and let those three triangular faces be turned about their traces on the plane of projection: then,

(1.) The projections of the three edges will be respectively perpendicular to the traces of the opposite faces;

(2.) They will respectively pass through the vertices (after revolution) of the opposite faces;

(3.) The projections of the three edges will meet circles described on the three opposite traces in points which are the vertices of the three right-angled triangles, after revolution.

Let O be the position of the vertex of the trihedral right angle, and XY, YZ, ZX the traces of its faces on

the plane of projection XYZ. Let OA be perpendicular to XYZ, then A is the projection of the trihedral angular point O.

(1.) Then, since O is a trihedral right angle, OZ is perpendicular to each of the lines OX, OY, and therefore to x the plane XOY which contains them. Also the projecting plane of ZO passes through AO, and is hence perpendicular to XYZ; and it passes through OZ,and is hence perpendicular to XOY. (Pls. 11. 16.) It is, therefore, perpendi

cular to ZXY and OXY; and hence to their common section XY, and its traces ZB, BO are perpendicular to XY. The projection of OZ upon XYZ is therefore perpendicular to the trace of the face XOY upon XYZ, which is the first part of the proposition.*

(2.) In the next place, since BO is perpendicular to XY, it will remain so in its revolution on either side, as XO'Y or XO"Y; and O will hence fall either in BZ or its prolongation, as O" or O'. Whence the projections of the edges of the trihedral angle will pass through the vertices of the opposite faces when they are turned on th plane of projection.

(3.) The angle XOY being a right angle, its vertex will fall in the circle on XY; and so of the others. This establishes the third part of the proposition.

SCHOLIUM. In contemplating this figure in space, and after revolution, several properties of it, having important practical uses, will present themselves to our notice.

*This conclusion might be more simply deduced from Prop. IX., that since OZ is perpendicular to the plane XOY, the projection of OZ is perpendicular to the trace of XOY. In fact, the steps through which we have here passed are but a repetition of those by which that proposition was established; and they are employed here chiefly with the view of more distinctly impressing that important though simple theorem on the student's mind.

We shall remark that in this figure there are concerned, besides the

projection of O (which comprises, in fact, two data), the following triads :

1. The three traces.

2. The three angles formed by them.

3. The lengths of the three edges.

4. The inclinations of the three edges.

5. The projections of the three edges.

6. The inclinations of the three facial triangles.

7. The three pairs of acute angles of the facial triangles.

Now, it requires but little acquaintance with the principles of plane geometry to perceive that if any five of these are given (except there be more than three in the first pair of triads) all the rest can be found by very simple constructions. In the annexed figure, the revolutions are performed for the three facial right-angled triangles, O,, O2, 039 being the positions of the vertex O, when all fall without the triangle, and w1, w, w, those when all fall within it; A is the projection of O, and XA, YA, ZA those of the edges respectively; and the circles have their second sections a, b, c in the traces XY, YZ, ZX, since the lines Xb, Yc, Za are perpendicular to the traces.

Let us take an example:

Given two traces XY, YZ, the angle XYZ, and the projections of two of the edges OX, OY.

With the first three of the data form the triangle XYZ; and with base XY and the given projections, the triangle XAY. On the three sides of XYZ describe circles and draw the other lines as indicated in the figure.

Then XO,Y, YO,Z, ZOX, or X,Y, YwZ, ZwX will be the three facial triangles turned on the plane of XYZ; and ZA will be the projection of the third edge.

The projections of the lines in which the projecting planes cut the facial planes, will be Aa, Ab, Ac.

The length of the projector may be thus found :-on aO, describe a semicircle ap1, and in it set off ap Aa; then, obviously, the triangle apO, of the figure is in all respects equal to the triangle aOA of

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