PROPOSITION V. To construct the intersection of two given planes. Let 00, and 00, be their traces, and the scales 0,3,, 0.3, be figured: through 1, 1, draw parallels to the traces intersecting at 1: then 01 being drawn indefinitely, will be the projection of the line required, and in length equal to a unit of its figuring. For 11, is the projection of a line parallel to the trace at the distance unit from the plane of comparison, and hence is the projection of all the points in the plane (1) which are at a unit's distance from the plane of comparison. In like manner 1,1 is the projection of all the points in (2) which are at that distance from the plane of comparison. Whence their intersection 1 is the projection of the point common to both planes which is a unit distance from the plane of comparison. Also the projection of the intersection of the given planes is a straight line, and it passes through 0 and 1: and hence it may be figured from having its zero point, projection, and projected unit given. Cor. 1. If one of the planes, as (1), be perpendicular to the plane of comparison, its scale will disappear, and the plane will be represented by its trace only. In this case the trace of this plane is the actual projection of the line of intersection sought; and the scale of this Jine will be found by drawing lines 1,1, 2,2 ... from the points of the scale of the other plane parallel to its trace, to meet the trace of the perpendicular plane. Cor. 2. If the two planes have their traces parallel, their line of section will be parallel to them, and hence also its projection on the plane of comparison will be parallel to them both. The preceding method is therefore inapplicable to this case; but the following will obviate the difficulty : Let 0,4,, 0,4, be the scales of the te given planes ; conceive the eidograph cut by a profile, meeting the traces in av, Qa: from any point (as 12 3.) draw the parallel to the traces, and set off aßa, aßı, equal to the altitudes belonging to 3,, upon the scales of the planes. Then if a Big B, inter-sect in p, the parallel through p to the traces will be the projection required. Its figure is uniform in reference to either scale, though different, of course, in the two. PROPOSITION V). To find the intersection of a given line with a given plane. Let a,O be the projection of the given line: construct the line of section of the given plane with a plane through the given line having its trace parallel to the given trace (Prop. v. Cor. 2). Then the point of section of the line and plane is projected at some point in this line. It is also in the projection a,0 of the given line, and hence at their inter: section a. This may be figured as a point in the plane, or as a point in the line, according to the ulterior objects in view. PROPOSITION VII. To draw a line perpendicular to a given plane through a given point. Let a be the projection of the given point, and conceive a profile plane through it, cutting the trace of the given plane in 0 : draw aa, parallel to the trace; and in it take aa, the figure of the point, and aa, the figure of the scale of the plane corresponding to a. Join a, 0, and draw a, a perpendicular to Oag, meeting Oa in 0, and 0a, in a. Make 01 the unit of altitude, draw 11' parallel to Oa, and i'l, parallel to the trace. Also, draw aß parallel to 01. Then 0, is the zero point of the perpendicular, 0, 1, is the unit of its projection, a, a is the length of the perpendicular, and B is the projection of its point of intersection with the given plane. For since the line is perpendicular to the plane, its projection is perpendicular to the trace of the plane, which condition is in the first place fulfilled, since aß is perpendicular to the trace. In the next place, the profile plane being turned about Oa, the line of declivity becomes Oag, the position of the point becomes an, and the perpendicular to the plane is perpendicular to Oag. Whence 0, in which it meets Oa is the zero point, and the scale is obviously true. It also follows that we have a, a equal to the distance of the point from the plane; and B the projection of a is obviously the projection of the point of section. Cor. The declivity of the line and plane being complementary of each other, the figuring may be effected without quite so much constructive work. The method of effecting it is left as an exercise. VOL. II. 2 A PROPOSITION VIII. Through a given point to draw a plane perpendicular to a given line. The line being given, its projection and zero point are given : and the trace of the required plane will be perpendicular to that projection. Assume therefore a plane at pleasure, having its trace such as to fulfil that one condition ; and figure this plane. Then through the given point draw a plane parallel to this (Prop. 111.) and it is that required. This, manifestly, is an exact converse of the preceding proposition. PROPOSITION IX. Through a given point to draw a plane which shall be parallel to two given lines. Let 0, 0, be the zero points of the figured lines which are drawn through them, these being the given lines, and let a be the point through which the plane is to pass. Through 0, draw a line parallel to the other given line (Prop. III., Sec. 2.), and figure it to the same scale: then this will be a line parallel to (1). Hence the plane which contains these lines is parallel to the line (1), and therefore to the required plane. Join 1, 1,: then this is the projection of a line parallel to the trace of the required plane at a unit of altitude from the plane of comparison. Wherefore we can trace and figure the plane which contains these lines. Through a draw a figure scale parallel to this, and from a set off the given figure of a, to 0: the line through O parallel to the former trace is the trace required ; and this scale being figured, we have the plane determined. PROPOSITION X. To find the angle contained between two figured planes. Let (1), (2) be the given figured planes (the diagram is easily conceived), and Ob the figured line in which they intersect (found as in Prop. v.): through any convenient point in Ob as 1, draw a line la at right angles to Ob, and make it equal to the vertical figure of the point 1: join Oa, and draw ab perpendicular to Oa: through b draw cd perpendicular to Ob to meet the traces of the given planes in c, d, and in 50 make ba = ba: then drawing ca, ad, the angle cad is the inclination of the planes to one another. For, as was shown in the Descriptive Geometry, cad is the triangle revolved on the plane of comparison in which the plane through 1 cuts the two given planes and the plane of comparison. PROPOSITION XI. There is given the trace of a plane, and a figured line in it; it is required to draw through a given point another line, which shall lie in the traced plane and make a given angle with the given line. Let 03 be the figured line in the plane whose trace is 0,02, and let a be the projection of the given point: through a draw a), parallel to 03, and figure it: draw al, perpendicular to the trace of the plane, and make aA equal to the given figure of the given point : join 0, A, and make 0,c=0, A, and join 0,c: make the angle O, O, equal to the given angle and draw 03 a, and figure it by parallels to the trace 0,0,. This is the projection sought. For, obviously, 0, c0g is equal to the triangle whose projection is 0, alg; and hence all the conditions as to magnitude are fulfilled. PROPOSITION XII. Through a given line to draw a plane perpendicular to a given plane. Let 0, 2, be the scale of the given plane, its trace being 0,0s; let a be the projection of any point, taken at pleasure, in the given line Oa; draw aó, perpendicular to 0, 0s, and au parallel to it; make 's aa' the figure of the selected point, and ad the figure of the given plane for that point ; join aly, and draw a'p perpendicular to 0, a, meeting 0, a in 0g. Then 0, is the trace of this perpendicular (Prop. VII.), and the required plane will have its trace passing through this trace Og. The trace of the required plane also passes through o the trace of the given line; and hence 00, is the trace of the plane rest, quired. Draw pß parallel to 0, 0s: then this is the figure of a point in the plane drawn. If we draw the scale line po, perpendicular to the trace, and graduate this proportionally to pß, it will be the scale of the required plane, and the problem is solved. For this graduation different methods may be employed, one of which is the following : Join 30x, and make Bh a unit of altitude: 1 then hi, parallel to 80, meets the scale line in "T 1., the first division of the scale of the required plane. · PROPOSITION XIII. To find the intersection of a given line with a given plane. Let Oa be the projection of the line, AB the trace of the plane, both figured 0, 1, 2. Through O draw 00 perpendicular to Oa; then this is the trace of a plane through the given line whose declivity is the same as that of the given line ; and hence whose scale is Oa. Whence the intersection of the given line and given plane will be in the intersection of the two planes. Finding then the projection Op of this intersection (Prop. v.), we have the projections of two lines through the ^/ point required; and hence p, their intersection, is the projection sought. The height (or figure) is known at once from the scale op, formed during the preceding construction. SCHOLIUM. Problems may be proposed for solution in infinite variety ; but the general principles of the method may be sufficiently understood from those which have been here given. There is only one remark necessary: viz., that though for the sake of keeping up a view of the analogy between the Topographic Projection and the Descriptive Geometry, we have generally employed the traces of the line and plane as our points of departure ; yet in practice it is quite sufficient that we take corresponding figures (as 5, 5; 8}, 8; etc.), instead of the special unes 0, 0. The advantage gained by this consists in two things :- that it often saves the trouble of extending the scale back to 0 in the drawing; and that in consequence it requires less space and avoids damaging the paper. These details, however, belong more properly to the practical than to the theoretical branch of the subject, as do likewise the applications of the principles to actual contour operations, whether in respect to the surveys or the drawings. |