| Fig. 46. |
We first rabatt the plane α (fig. 46) as before so that P1 comes to P, hence OP1 to OP. Let the given polygon in α be the figure ABCDE. We project, not the vertices, but the sides. To project the line AB, we produce it to cut α′ in F and OP in G, and draw GG1 perpendicular to α′; then G1 corresponds to G, therefore FG1 to FG. In the same manner we might project all the other sides, at least those which cut OF and OP in convenient points. It will be best, however, first to produce all the sides to cut OP and α′ and then to draw all the projecting rays through A, B, C ... perpendicular to α′, and in the same direction the lines G, G1, &c. By drawing FG we get the points A1, B1 on the projecting ray through A and B. We then join B to the point M where BC produced meets the trace α′. This gives C1. So we go on till we have found E1. The line A1 E1 must then meet AE in α′, and this gives a check. If one of the sides cuts α′ or OP beyond the drawing paper this method fails, but then we may easily find the projection of some other line, say of a diagonal, or directly the projection of a point, by the former methods. The diagonals may also serve to check the drawing, for two corresponding diagonals must meet in the trace α′.
Having got the plan we easily find the elevation. The elevation of G is above G1 in α″, and that of F is at F2 in the axis. This gives the elevation F2G2 of FG and in it we get A2B2 in the verticals through A1 and B1. As a check we have OG = OG2. Similarly the elevation of the other sides and vertices are found.
§ 17. We proceed to give some applications of the above principles to the representation of solids and of the solution of problems connected with them.
Of a pyramid are given its base, the length of the perpendicular from the vertex to the base, and the point where this perpendicular cuts the base; it is required first to develop the whole surface of the pyramid into one plane, and second to determine its section by a plane which cuts the plane of the base in a given line and makes a given angle with it.
1. As the planes of projection are not given we can take them as we like, and we select them in such a manner that the solution becomes as simple as possible. We take the plane of the base as the horizontal plane and the vertical plane perpendicular to the plane of the section. Let then (fig. 47) ABCD be the base of the pyramid, V1 the plan of the vertex, then the elevations of A, B, C, D will be in the axis at A2, B2, C2, D2, and the vertex at some point V2 above V1 at a known distance from the axis. The lines V1A, V1B, &c., will be the plans and the lines V2A2, V2B2, &c., the elevations of the edges of the pyramid, of which thus plan and elevation are known.
We develop the surface into the plane of the base by turning each lateral face about its lower edge into the horizontal plane by the method used in § 14. If one face has been turned down, say ABV to ABP, then the point Q to which the vertex of the next face BCV comes can be got more simply by finding on the line V1Q perpendicular to BC the point Q such that BQ = BP, for these lines represent the same edge BV of the pyramid. Next R is found by making CR = CQ, and so on till we have got the last vertex—in this case S. The fact that AS must equal AP gives a convenient check.
2. The plane α whose section we have to determine has its horizontal trace given perpendicular to the axis, and its vertical trace makes the given angle with the axis. This determines it. To find the section of the pyramid by this plane there are two methods applicable: we find the sections of the plane either with the faces or with the edges of the pyramid. We use the latter.
As the plane α is perpendicular to the vertical plane, the trace α″ contains the projection of every figure in it; the points E2, F2, G2, H2 where this trace cuts the elevations of the edges will therefore be the elevations of the points where the edges cut α. From these we find the plans E1, F1, G1, H1, and by joining them the plan of the section. If from E1, F1 lines be drawn perpendicular to AB, these will determine the points E, F on the developed face in which the plane α cuts it; hence also the line EF. Similarly on the other faces. Of course BF must be the same length on BP and on BQ. If the plane α be rabatted to the plan, we get the real shape of the section as shown in the figure in EFGH. This is done easily by making F0F = OF2, &c. If the figure representing the development of the pyramid, or better a copy of it, is cut out, and if the lateral faces be bent along the lines AB, BC, &c., we get a model of the pyramid with the section marked on its faces. This may be placed on its plan ABCD and the plane of elevation bent about the axis x. The pyramid stands then in front of its elevations. If next the plane α with a hole cut out representing the true section be bent along the trace α′ till its edge coincides with α″, the edges of the hole ought to coincide with the lines EF, FG, &c., on the faces.
§ 18. Polyhedra like the pyramid in § 17 are represented by the projections of their edges and vertices. But solids bounded by curved surfaces, or surfaces themselves, cannot be thus represented.