the figure in space, and hence Op is equal to the projector. It is also clear that the construction for 0.p might have been made upon any of the corresponding lines of the figure, should greater practical convenience result from such a change. Again, since the angle OaA is the inclination of XOY to XZY, it is found in the preceding construction to be Ojap. Lastly, the inclinations of the edges will be found by taking (for instance, that of OZ) Ap, perpendicular to AZ and equal to 0,p, and joining p.Z. Then p, ZA is the inclination of OZ to XYZ. This figure was proposed by the late Mr. Nicholson, which he called the directing diagram. We shall distinguish it as Nicholson's diagram ; and some applications of it to practice will be given in a future part of this work. The student is recommended in the present stage to vary the data so as to present new problems for exercise ; as a familiarity with the diagram will prove of great use to him in several branches of his future practice. PROPOSITION XV. If a point be projected upon three rectangular planes, and likewise upon any other plane which cuts them, and if the three planes be made to revolve as in the preceding proposition, and lines be drawn through the three projections, after revolution, perpendicular to the traces respectively; (1.) These lines will pass through the same point ; (2.) That point will be the projection of the original point on the plane which cuts the three rectangular planes. Let P be the point whose projection on XYZ is Q, and wliose projection on XOY is R; draw the plane through PQR, which will be perpendicular to both planes XZY and XOY, being drawn through the projectors PQ, PR. Whence QC, RC are both perpendicular to the trace XY; and when XOY is turned on the plane XYZ, QC, CR will be in one line perpendicular to XY. The projection R, then, is such that a perpendicular being drawn from it to the trace, it will pass through Q the projection of the point P upon XYZ. And similarly for the other planes. R X PROPOSITION XVI. The projections of a point on two given rectangular planes are given, to find its projection on a third given plane. This problem naturally divides itself into two general cases, and each of these into subordinate ones. The general cases are (a.) When the rectangular planes are given by their traces on the third plane. (6.) When the third plane is given by its traces on the two rectangular ones. OL Y X (a,. Let XY, YZ be the traces of the rectangular planes upon the third given plane, and a, a, the projections of the point on those planes, given with respect to any two lines in those planes respectively. Then the rectangular planes being made to revolve about the traces till they coincide with the third plane, draw a, a, aga perpendicular to the traces, meeting in a; and a will be the projection required. For by the preceding proposition, both these perpendiculars pass through the projection of the point on XYZ. (az.) When the traces are parallel, these lines coincide in direction, and the preceding construction fails in this case. We may, however, proceed thus : Draw a,a , 0qdq parallel to each other meeting the traces in ang as: then a,a,,a,a, will intersect in a the projection sought. (6.) Given the projections of a point an, Ag, on the two rectangular planes P, Pp, P,Pp; and the traces of the plane P upon which the projection is to be made; to find the projection, when the plane P is turned about its trace on one of the given rectangular planes (Fig, 3). Cut the system of planes by a profile to the given rectangular ones, and construct the triangle P,PP, (Prop. xi.), on the horizontal plane, suppose, and let it be PiPR. . x We may now consider that one of the rectangular planes is turned upon P,PR, ; and we have only to construct upon R,P the right angled triangle RSP equal to P,pP, and place a in it as a, is placed in P.P. Then we have, evidently, the problem reduced to the precise state of (a); and for similar reasons, it is only required to draw perpendiculars from a, to PP, and from a to PR,, intersecting in a the projection sought. În the actual construction of the problem, it is only necessary to construct PR, and a; the other parts of the figure being merely illustrative. (bx.) When the third plane is a profile plane, the steps are more simple. Draw açd, aja' parallel to the axis ; make Oa" = Oa', and draw a" a parallel to YZ, 1월 meeting aga in a: then a is the projection on the profile plane, when turned on the vertical plane. If the horizontal projection be required, reverse the operations with respect to a, х and Ayo 29 B R P (63.) Let BỊP be the axis, B.P2, B,P, the traces of the third plane parallel to the axis aP; and let an, az be the projections of a point on the rectangular planes: it is required to find its projection on the third plane. The projection will be on the line in which the profile plane through the given point cuts the plane of projection. Produce the line aja, to meet the traces in B,, B, and the axis in a; make aB' = aB, and join B,B', then B,B' is the intersection of the profile and third plane of projection, turned on the horizontal plave. Make a, A equal to ab, and parallel to ap: then A is the given point turned also on the horizontal plane. Draw Aa' perpendicular to B,B'; then a' is the projection on the horizontal plane. Make Bia Ba': then a is the point required on the plane P,P, turned about B,P. The trace of the third plane on the vertical plane, after revolution, is found by making Bib = B,B', and drawing the parallel to aP through b. PROPOSITION XVII. If a point be projected upon any three planes which meet, and these planes with the projections respectively upon them be turned on a fourth plane which cuts those three, about their traces, diculars from the three projections to the traces will all pass through one point, the projection of the given point upon the fourth plane. For it follows from Prop. xv. that each of the three projectors of the point upon the three given planes does so ; and hence it follows that all do so. SCHOLIUM This is, obviously, a more general theorem than that which forms the basis of Nicholson's theorem in the preceding propositions. All projections, indeed, by means of plan and section on rectangular planes, are capable of being made with facility by that method : still this more general form will sometimes simplify the actual then perpen, process to a considerable degree, since we may take these three planes to represent any trihedral angle of the figure to be projected. M PROPOSITION XVIII. To project a given angle. Project the containing sides : then the projections will, obviously, contain an angle which is the projection of the given angle. The construction of this problem has, in fact, been already virtually given : but as there are some facts in connexion between the given angle and its projection which it is important to state specifically, the problem is enunciated here for the purpose of noticing them at length. 1. If both lines AB, AC be parallel to the plane of projection, their projections a,b,, a,q, will be respectively parallel to themselves; and hence the lines and their projections will contain equal angles (Pls. $ Sols, 1. 7.). 2. Let both lines cut the plane of projection, as in B and C ; and draw As from the angular point A perpendicular to the plane of projection MN; and join Ba,, a,C. Then Ba,C is the projection of BAC. But (Pls. & Sols. II. 5) the angle Ba, C is greater than BAC. Wherefore, the projection of the angle formed by AB, AC and opposite to the trace of their plane on the plane of projection, is greater than the angle itself. The vertical angle EAD to BAC is projected into the angle ea,d vertical to Ba,C, and hence also the same holds true respecting these. But if one branch of the line be estimated towards the trace, and the other in the opposite direction, as CA, AD, then the projected angle Cad is less than the given angle CAD: these being the respective supplements of the angles to which the preceding demonstration was applied, and the greater angle having the less supplement. 3. Let one line be parallel to the plane of projection, and the two form a right angle; then the right angle is projected into a right angle. Let AB be parallel to the plane of projection and the angle CAB a right angle: then its projection will also be a right angle. Draw the plane CD through A perpendicular to AB: then it will be perpendicular to the plane MN (Pls. 8 Sols. 11. 19). It also passes through AC since all the perpendiculars through A to AB are in a plane perpendicular to AB. Whence CD is the projecting plane of AC, and a,c, is its projection on MN. Now the projection ab, of AB is pa N M D А E M rallel to AB itself, and AB is perpendicular to the plane CD: hence also ab, is perpendicular to CD, and therefore to the line aç, in it. Whence the projections c,a,, ab, of the sides of the right angle CAB form, themselves, a right angle ca, b,. 4. Conversely, if the projection be a right angle and one of the sides of the angles itself be parallel to the plane of projection, the angle itself which was projected is a right angle. 5. If one leg of the angle be parallel to the plane of projection, the angle is projected into an obtuse angle if itself be obtuse, and into an acute angle if itself be acute; and the projecting planes will form an obtuse or an acute angle in the two respective cases. Let AC be parallel to the plane MN of projection, CAB be an obtuse angle, and B the trace of the leg AB upon the plane MN; and let BAa, and Ac, be the projecting planes of BA, AC; and Ba,, a,c, the projections : then Ba,c, will be an obtuse angle, and the planes BA,, CAa,c, form an obtuse angle. For let the plane BAC be drawn: its trace on the projecting plane MN is a line parallel to AC and its projection a,c, (Pls. and Sols. 1. 2): also draw, in the plane BAC, the line AD at right angles to AC, meeting the trace of the plane BAC in D, and join Da,. Then, since Aa, is the intersection of the projecting planes of BA, AC it is perpendicular to MN; and hence Da, is the projection of DA, and (Case 3) Da,c, is a right angle. But Ba,c, is greater than Dac, ; and therefore the projection is an obtuse angle. Also, since Ba,, a,c, are perpendicular to the common section Aa, of the projecting planes, the angle Bac, is the measure of the angle formed by those planes; hence the inclination of the projecting planes to one another is obtuse. Again, produce CA, ca, to E and e, respectively; then since BAC, Bac, are obtuse angles, their supplements BAE, Ba,e, are acute; and Bage, is the projection of BAE. Also the angle Bae, is the measure of the inclination of the planes EAa,e,, Aa,B to each other, and hence the inclination of the projecting planes is acute. Both conclusions are, therefore, established. B H EXERCISES ON ORTHOGRAPHIC PROJECTION. 1. Given a triangle and the inclinations of two of its sides to the plane of projection, to find the inclination of the third side and of the plane which contains the triangle to the plane of projection. 2. An angle is given, and its containing sides make given angles with the horizon; it is required to construct its projection on the plane of the horizon, or “ to reduce the given angle to the horizon." 3. Given an angle BAC to construct a plane upon which the projection of that angle shall be a given angle, whilst the trace of that plane shall be a given line in the plane of the angle BAC. |