§ 15. We return to the general case of rabatting a plane α of which the traces α′ α″ are given.
| 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.