PROJECTION 803 Then the line OP will be the position of o" when rabatted. This line corresponds therefore to the plan of o" that is, to the axis x, corresponding points on these lines being those which lie on a per pendicular to a . AVe have thus one pair of corresponding lines and can now find for any point Bj in the plan the corresponding point B in the rubatted plane. We draw a line through B 1( say B,Pj, cutting a in C. To it corresponds the line CP, and the point where this is cut by the projecting ray through Bj, perpendicular to a, is the re quired point B. Similarly any figure in the rabatted plane can be found when the plan is known ; but this is usually found in a different manner with out any reference to the general theory of parallel projection. As this method and the reasoning employed for it have their peculiar advantages, we give it also. Supposing the planes ir l and ir 2 * be in their positions in space peipendicular to each other, we take a section of the whole figure by a plane perpendicular to the trace a about which we are going to rabatt the plane a. Let this section pass through the point Q in a . Its traces will then be the lines QP X and PjP 2 (fig. 18). These will be at right angles, and will therefore, together with the section QP._, of the plane a, form a right-angled triangle QP^ with the right angle at Pj, and having the sides PjQ and PjP 2 which both are given in their true lengths. This triangle we rabatt about its base PjQ, making PjK, = PiP 2 . The line QR will then give the true length of the line QP in space. If now the plane a be turned about a the point P will describe a circle about Q as centre with radius QP = QR, in a plane perpendicular to the trace o . Hence when the plane a has been rabatted into the horizontal plane the point P will lie in the perpendicular P X Q to a , so that QP = QR. If Aj is the plan of a point A in the plane a, and if Aj lies in QPj, then the point A will lie vertically above A l in the line QP. On turning down the triangle QPxPo, the point A will come to A , the line AjA,) being perpendicular to QP r Hence A will be a point in QP such that QA = QA . If Bj is the plan of another point, but such that A : Bj is parallel to a, then the corresponding line AB will also be parallel to o . Hence, if through A a line AB be drawn parallel to a , and B a B perpendicular to a, then their intersection gives the point B. Thus of any point given in plan the real position in the plane a, when rabatted, can be found by this second method. This is the one most generally given in books on geometrical drawing. The first method explained is, however, in most cases preferable as it gives the draughtsman a greater variety of constructions. It requires a somewhat greater amount of theoretical knowledge. If instead of our knowing the plan of a figure the latter is itself given ; then the process of finding the plan is the reverse of the above and needs little explanation. We give an example. 45. PROBLEM. It is required to draw the plan and elevation of a polygon of which the real shape and position in a given plane a are known. Solution. We first rabatt the plane a. (fig. 19) as before so that P x comes to P, hence to OP. Let the given polygon in o be the figure ABODE. We project, not the vertices, but the sides. To project the line AB, we produce it to cut in F and OP in G, and draw GGj perpendicular to a! ; then G] corresponds to G, therefore FG l to FG. In the same manner we might project all the other sides, at least those which cut OF and P in convenient points. It will be best, however, first to produce all the sides to cut OP and a and then to draw all the projecting rays through A, B,C . . . perpendicular to a , and in the same direc tion the lines G,G 1( &c. By drawing FG we get the points A lf Bj on the projecting ray through A and B. We then join B to the point M where BC produced meets the trace o . This gives Cj. So we go on till we have found E r The line Aj E x must then meet AE in a , and this gives a check. If one of the sides cuts o 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 corre sponding diagonals must meet in the trace a . Having got the plan we easily find the elevation. The elevation of G is above G x in a", and that of F is at F 2 in the axis. This gives the elevation F 2 G 2 of FG and in it we get A 2 B 2 in the verticals through A x and B x . As a check we have ~OG = OG f Similarly the elevation of the other sides and vertices are found, 46. We have now obtained the ABC of descriptive geometry, and proceed to give some applications to the representation of solids and of the solution of problems connected with them. PROBLEM. 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. Solution. (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 perpen dicular to the plane of the section. Let then (fig. 20) ABCD be the base of the pyramid, Y! the plan of the ver tex, then the elevations of A, B, C, D will be in the axis at A 2 , B 2 , C 2 , D 2 , and the vertex at some point V 2 above V] at a known distance from the axis. The lines VjA, VjB, &c., will be the plans and the lines V 2 A 2 , V 2 B.,, &c., the elevations of the edges of the pyramid, of which thus plan and eleva tion 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 43. 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 V X Q 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 a whose section we have to determine has its hori zontal 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 tAvo 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 o is perpendicular to the vertical plane, the trace a" contains the projection of every figure in it ; the points E. 2 , F 2 , G 2 , H 2 where this trace cuts the elevations of the edges will there fore be the elevations of the points where the edges cut a. From
these we find the plans E 1( F 1; G 1; H 1; and by joining them the