Descriptive geometry . n the plane Q, to havebeen revolved about HQ into H. The figure is left to the stu-dent. The path of the revolving point is the arc of a circle,which will project on Fas a straight line perpendicular toHQ, and on any plane perpendicular to HQ in its true shapeand size. Referring to the various solutions of Problem 21,§ 138, it is seen that the solution given in Fig. 220 shows thesetwo projections of the path of the revolving point. The con-struction of Fig. 220 may therefore be reversed to give thesolution of the present problem. Construction (Fig. 247). The plane Q is g
Descriptive geometry . n the plane Q, to havebeen revolved about HQ into H. The figure is left to the stu-dent. The path of the revolving point is the arc of a circle,which will project on Fas a straight line perpendicular toHQ, and on any plane perpendicular to HQ in its true shapeand size. Referring to the various solutions of Problem 21,§ 138, it is seen that the solution given in Fig. 220 shows thesetwo projections of the path of the revolving point. The con-struction of Fig. 220 may therefore be reversed to give thesolution of the present problem. Construction (Fig. 247). The plane Q is given by its tracesHQ and VQ, and ar is given as the position of a point which 164 XVII, § 147] COUNTER-REVOLUTION OF PLANES 165 has been revolved about HQ into H. Assume a secondaryground line GXLX perpendicular to HQ. Assume a point, f,in VQ; project to th in GL. Find the secondary projection,txv, of t, and through this point draw the trace and edge viewVXQ (§ 70). Project ar to GXLX. With o as center, revolve this. Fig. 247. point to axv in VXQ. Then axv is the secondary projection ofthe point a. Project from axv perpendicular to GXLX, and fromar perpendicular to HQ. Both of these lines must pass throughah, which is thus determined. To find a, we now have given a*as one projection of a point lying in the plane Q. Through ahdraw Mh parallel to HQ; this is one projection of a horizontalprincipal line of Q. The other projection, M, is parallel toGL; o lies in M (Prob. 16, § 133). Check. The distance of av from GL equals the distance ofaxv from GxIj^ (§ 67). A Second Result. The point a, thus found, lies above given revolved position may also he the revolved positionof a point b, lying below II. To obtain this result, utter pro-jecting from the given revolved position to GXLU revolve abouto to bxv, which lies in V{Q produced below GXLX. Then pro-ceed as for the point a. As before, there is a cheek on theconstruction ; the distance from b to GL equals the distancefrom
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