Page:Encyclopædia Britannica, Ninth Edition, v. 19.djvu/826

Rh 802 PROJECTION when, as in the above case, some lines must meet on the axis, or if three points must lie in a line. A careful draughtsman will always avail himself of these checks. 40. To draw a plane through a given point parallel to a given plane a, we draw through the point two lines which are parallel to the plane a, and determine the plane through them; or, as we know that the traces of the required plane are parallel to those of the given one ( 37), we need only draw one line I through the point parallel to the plane and find one of its traces, say the vertical trace I; a line through this parallel to the vertical trace of o will be the vertical trace |&quot; of the required plane 0, and a line parallel to the horizontal trace of o meeting ft&quot; on the axis will be the horizontal trace /3. Let Aj A 2 (fig. 15) be the given point, a a&quot; the given plane, a line /i through Aj parallel to a and a horizontal line 1 2 through A 2 will be the projec tions of a line /through A parallel to the plane, because the horizontal plane through this line will cut the plane a in a line c which has its horizontal projection Cj parallel to a. 41. We now come to the metrical proper ties of figures. A line is perpendicu lar to a plane if the projections of the line are perpendicular to the traces of the plane. We prove it for the horizontal projection. If a line p is perpendicular to a plane a, every plane through _p is per pendicular to a; hence also the vertical plane which projects the line p to p v As this plane is perpendicular both to the horizontal plane and to the plane a, it is also perpendicular to their intersection that is, to the horizontal trace of o. It follows that every line in this projecting plane, therefore also p l} the plan of^, is perpendicu lar to the horizontal trace of a. Q.E.D. To draw a plane through a given point A perpendicular to a given linep, we first draw through some point in the axis lines y, y&quot; perpendicular respectively to the projections p l and j 2 of the given line. These will be the traces of a plane y which is perpendicular to the given line. We next draw through the given point A a plane parallel to the plane y; this will be the plane required. Other metrical properties depend on the determination of the real size or shape of a figure. In general the projection of a figure differs both in size and shape from the figure itself. But figures in a plane parallel to a plane of projection will be identical with their projections, and will thus be given in -their true dimensions. In other cases there is the problem, constantly recurring, either to find the true shape and size of a plane figure when plan and elevation are given, or, con versely, to find the latter from the known true shape of the figure itself. To do this, the plane is turned about one of its traces till it is laid down into that plane of projection to which the trace belongs. This is technically called rabatting the plane respectively into the plane of the plan or the elevation. As there is no difference in the treatment of the two cases, we shall consider only the case of rabatt ing a plane a into the plane of the plan. The plan of the figure is a parallel (orthographic) projection of the figure itself. The results of parallel projection ( 17 and 18) may therefore now be used. The trace a will hereby take the place of what formerly was called the axis of projection. Hence we see that corresponding points in the plan and in the rabatted plane are joined by lines which are perpen dicular to the trace a and that corresponding lines meet on this trace. We also see that the correspondence is completely deter mined if we know for one point or one line in the plan the corre sponding point or line in the rabatted plane. Before, however, we treat of this we consider some special cases. 42. To determine the distance between two points A, B given by their projec tions Aj, B! and Aj, B 2. Solution. The two points A, B in space lie vertically above their plans Aj, B! (fig. 16) and A 1 A = A A 2 , BjB = B B2. Thefourpoints A, B, A t, Bj therefore form a plane qua&amp;lt; trilateral on the base AjBj and having right angles at the base. This plane we rabatt about A^ by drawing A,A and B,B jxTpendicular to A 1 B 1 and making A^Y^A^Y^ B 1 B = B B 2. Then AB will give the length rcfinired. The construction might have been performed in the elevation by making A 2 A = A A! and B 2 B = B B 1 on lines perpendicular to A 2 B.|, Of course AB must have the same length in both cases. This figure may be turned into a model. We cut the paper along A 1 A, A13, and BBj and fold the piece A 1 ABB 1 over along AjBj till it stands upright at right angles to the horizontal plane. The points A, B will then be in their true position in space relative to TT^ Similarly if B 2 BAA 3 be cut out and turned along A 2 B 2 through a right angle we shall get AB in its true position relative to the plane ir. Lastly we fold the whole plane of the paper along the axis x till the plane TT, is at right angles to TTJ. In this position the two sets of points A~B will coincide n the drawing has been accurate. Models of this kind can be made in many cases and their construc tion cannot be too highly recommended in order to realize ortho graphic projection. 1 43. To find the angle between two given lines a, b of which l/m projections a lt b l and a. 2, b. 2 are given. Solution. Let a v b l (fig. 17) meet in P l5 a 2, b 2 in T, then if the line P X T is not perpendicular to the axis the two lines will not meet. In this case we draw a line paral lel to b to meet the line a. This is easiest done by drawing first the line PjP-j perpendicular to the axis to meet a. 2 in P 2, and then drawing through P 2 a line c. 2 parallel to b.,; then b lt c. 2 will be the projections of a line c which is parallel to b and meets a in P. The plane a which these two lines determine we rabatt to the plan. We determine the traces a rig. 17. and c of the lines a and c; then a c is the trace a of their plane. On rabatting the point P comes to a point S on the line PjQ per pendicular to a c, so that QS = QP. But QP is the hypothenuse of a triangle PPjQ with a right angle Pj. This we construct by making QR = P P 2; then PjR = PQ. The lines a S and c S will therefore include angles equal to those made by the given lines. It is to be remembered that two lines include two angles which are supplementary. Which of these is to be taken in any special case depends upon the circumstances. To determine the angle between a line and a plane, we draw through any point in the line a perpendicular to the plane ( 41) and determine the angle between it and the given line. The comple ment of this angle is the required one. To determine the angle between two planes, we draw through any point two lines perpendicular to the two planes and determine the angle between the latter as above. In special cases it is simpler to determine at once the angle between the two planes by taking a plane section perpendicular to the intersection of the two planes and rabatt this. This is especially the case if one of the planes is the horizontal or vertical plane of projection. Thus in fig. 18 the angle P : QR is the angle which the plane a makes with the horizontal plane. 44. We return to the general case of rabatting a plane o of which the traces a a&quot; are given. Here it will be convenient to determine first the position which the trace o&quot; which is a line in a assumes when rabatted. Points in this line coincide with their elevations. Hence it is given in its true dimension, and we can measure off along it the true distance between two points in it. If there fore (fig. 18) P is any point in a&quot; originally coincident with its elevation P 2, and if O is the point where a&quot; cuts the axis x, so that is also in a , then the point P will. - after rabatting the plane assume such a position that OP = OP 2. At the same time the plan is an orthographic projection of the plane a. Hence the line joining P to the plan Pj will after rabatting be perpendicular to a. But P! is known; it is the foot of the perpendicular from P a to the axis x. We draw therefore, to find P, from P x a perpen dicular PjQ to a and find on it a point P such that OP = OPg^ 1 In order to make a sharp crease along A, 15,, it is well to place a straight edge along this line, and then to turn the piece AiABB, up against it.