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, conversely, 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 rabatting a plane α 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 (see Projection, §§ 17 and 18) may therefore now be used. The trace α′ 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 perpendicular to the trace α′ and that corresponding lines meet on this trace. We also see that the correspondence is completely determined if we know for one point or one line in the plan the corresponding point or line in the rabatted plane.
Before, however, we treat of this we consider some special cases.
§ 13. To determine the distance between two points A, B given by their projections A1, B1 and A2, B2, or, in other words, to determine the true length of a line the plan and elevation of which are given.
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| Fig. 43. |
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| Fig. 44. |
Solution.—The two points A, B in space lie vertically above their plans A1, B1 (fig. 43) and A1A = A0A2, B1B = B0B2. The four points A, B, A1, B1 therefore form a plane quadrilateral on the base A1B1 and having right angles at the base. This plane we rabatt about A1B1 by drawing A1A and B1B perpendicular to A1B1 and making A1A = A0A2, B1B = B0B2. Then AB will give the length required.
The construction might have been performed in the elevation by making A2A = A0A1 and B2B = B0B1 on lines perpendicular to A2B2. Of course AB must have the same length in both cases.
This figure may be turned into a model. Cut the paper along A1A, AB and BB1, and fold the piece A1ABB1 over along A1B1 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 π1. Similarly if B2BAA2 be cut out and turned along A2B2 through a right angle we shall get AB in its true position relative to the plane π2. Lastly we fold the whole plane of the paper along the axis x till the plane π2 is at right angles to π1. In this position the two sets of points AB will coincide if the drawing has been accurate.
Models of this kind can be made in many cases and their construction cannot be too highly recommended in order to realize orthographic projection.
§ 14. To find the angle between two given lines a, b of which the projections a1, b1 and a2, b2 are given.
Solution.—Let a1, b1 (fig. 44) meet in P1, a2, b2 in T, then if the line P1T is not perpendicular to the axis the two lines will not meet. In this case we draw a line parallel to b to meet the line a. This is easiest done by drawing first the line P1P2 perpendicular to the axis to meet a2 in P2, and then drawing through P2 a line c2 parallel to b2; then b1, c2 will be the projections of a line c which is parallel to b and meets a in P. The plane α which these two lines determine we rabatt to the plan. We determine the traces a′ and c′ of the lines a and c; then a′c′ is the trace α′ of their plane. On rabatting the point P comes to a point S on the line P1Q perpendicular to a′c′, so that QS = QP. But QP is the hypotenuse of a triangle PP1Q with a right angle P1. This we construct by making QR = P0P2; then P1R = 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 (§ 12) and determine the angle between it and the given line. The complement 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. 45 the angle P1QR is the angle which the plane α makes with the horizontal plane.
§ 15. We return to the general case of rabatting a plane α of which the traces α′ α″ are given.
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| Fig. 45. |
Here it will be convenient to determine first the position which the trace α″—which is a line in α—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 therefore (fig. 45) P is any point in α″ originally coincident with its elevation P2, and if O is the point where α″ cuts the axis xy, so that O is also in α′, then the point P will after rabatting the plane assume such a position that OP = OP2. At the same time the plan is an orthographic projection of the plane α. Hence the line joining P to the plan P1 will after rabatting be perpendicular to α′. But P1 is known; it is the foot of the perpendicular from P2 to the axis xy. We draw therefore, to find P, from P1 a perpendicular P1Q to α′ and find on it a point P such that OP = OP2. Then the line OP will be the position of α″ when rabatted. This line corresponds therefore to the plan of α″—that is, to the axis xy, corresponding points on these lines being those which lie on a perpendicular to α′.
We have thus one pair of corresponding lines and can now find for any point B1 in the plan the corresponding point B in the rabatted plane. We draw a line through B1, say B1P1, cutting α′ in C. To it corresponds the line CP, and the point where this is cut by the projecting ray through B1, perpendicular to α′, is the required 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 without 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 π1 and π2 to be in their positions in space perpendicular to each other, we take a section of the whole figure by a plane perpendicular to the trace α′ about which we are going to rabatt the plane α. Let this section pass through the point Q in α′. Its traces will then be the lines QP1 and P1P2 (fig. 9). These will be at right angles, and will therefore, together with the section QP2 of the plane α, form a right-angled triangle QP1P2 with the right angle at P1, and having the sides P1Q and P1P2 which both are given in their true lengths. This triangle we rabatt about its base P1Q, making P1R = P1P2. The line QR will then give the true length of the line QP in space. If now the plane α be turned about α′ the point P will describe a circle about Q as centre with radius QP = QR, in a plane perpendicular to the trace α′. Hence when the plane α has been rabatted into the horizontal plane the point P will lie in the perpendicular P1Q to α′, so that QP = QR.
If A1 is the plan of a point A in the plane α, and if A1 lies in QP1, then the point A will lie vertically above A1 in the line QP. On turning down the triangle QP1P2, the point A will come to A0, the line A1A0 being perpendicular to QP1. Hence A will be a point in QP such that QA = QA0.
If B1 is the plan of another point, but such that A1B1 is parallel to α′, then the corresponding line AB will also be parallel to α′. Hence, if through A a line AB be drawn parallel to α′, and B1B perpendicular to α′, then their intersection gives the point B. Thus of any point given in plan the real position in the plane α, 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.
§ 16. It is required to draw the plan and elevation of a polygon of which the real shape and position in a given plane α are known.
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| 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
