This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

91. In the foregoing problems the perspective projection has been found from a diagram of the object. Another way of constructing a perspective projection is by the method of Perspective Plan. In this method no diagram is used, but a perspective plan of the object is first made, and from this perspective plan the perspective projection of the object is determined. The perspective plan is usually supposed to lie in an auxiliary horizontal plane below the plane of the ground. The principles upon which its construction is based will now be explained.

92. In Fig. 26, suppose the rectangle aHbhchdH to represent the horizontal projection of a rectangular card resting upon a horizontal plane. The diagram of the card is shown at the upper part of the figure. It will be used only to explain the construction of the perspective plan of the card.

First consider the line ad, which forms one side of the card. On HPP lay off from a, to the left, a distance (ae) equal to the length of the line ad. Connect the points e and d. ead is by construction an isosceles triangle lying in the plane of the card, with one of its equal sides (ae) in the picture plane. Now, if this triangle be put into perspective, the side ad, being behind the picture plane, will appear shorter than it really is; while the side ae, which lies in the picture plane, will show in its true length.

Let VH1 be the vertical trace of the plane on which the card and triangle are supposed to rest. The position of the station point is shown by its two projections SPH and SPV. The vanishing point for the line ad will be found at Vad in the usual manner. In a similar way, the vanishing point for the line ed, which forms the base of the isosceles triangle, will he found at ved, as indicated. ap will be found onVH, vertically under the point a, which forms the apex of the isosceles triangle ead. The line aPdp will vanish at vad. The point ep will he found vertically below the point e. epdp will vanish at ved, and determine by its intersection with apdp the length of that line. epapdp is the perspective of the isosceles triangle ead.

If the line ad in the diagram is divided in any manner by the Points t, s, and r, the perspectives of these points may be found on the line apdp in the following way. If lines are drawn through the points t, s, and r in the diagram parallel to the base; de of the isosceles triangle (ead~), these lines will divide the line ae in a manner exactly similar to that in which the line ad is divided. Thus, aw will equal at, wv will equal ts, etc. Now, in the perspective projection of the isosceles triangle, apep lies in the picture plane. It will show in its true length, and all divisions on it will show in their true size. Thus, on apep lay off ap wp, wpvp, and vpup equal to the corresponding distances at, ts, and sr, given in the diagram. Lines drawn through the points wp, vp, and up, vanishing at ved, will he the perspective of the lines wt, vs, and ur in the isosceles triangle, and will determine the positions of tp, sp, and rp, hy their intersections with apdp.

93. It will he seen that after having found vad and ved, the perspective of the isosceles triangle can he found without any reference to the diagram. Assuming the position of ap at any desired point on VH1, the divisions ap, wp, vp, up, ep may be laid off from ap directly, making them equal to the corresponding divisions aH, t h, sh, rh, dh, given in the plan of the card. A line through ap vanishing at vad will represent the perspective side of the isosceles triangle. The length of this side will be determined by a line drawn through ep, vanishing at ved. The positions of tp, sp, and rp may he determined by lines drawn through wp, vp, and uv, vanishing at ved.

94. It will be seen that the lines drawn to ved serve to measure the perspective distances ap tp, tp sp, sp rp, and rp dp, on the line apdp, from the true lengths of these distances as laid off on the line apep. Hence the lines vanishing at ved are called Measure Lines for the line apdp, and the vanishing point vedis called a Measure Point for ap dp.

95. Every line in perspective has a measure point, which may be found by constructing an isosceles triangle on the line in a manner similar to that just explained.

Note. - The vanishing point for the base of the isosceles triangle always becomes the measure point for the side of the isosceles triangle which does not lie in HPP.

96. All lines belonging to the same system will have the same measure point. Thus, if the line be, which is parallel to adt be continued to meet HPP, and an isosceles triangle (cku) constructed on it, as indicated by the dotted lines in the figure, the base (uc) of this isosceles triangle will be parallel to de, and its vanishing point will be coincident with vde.

97. There is a constant relation between the vanishing point of a system of lines and the measure point for that system. Therefore, if the vanishing point of a system of lines is known, its measure point may be found without reference to a diagram, as will be explained.

In constructing the vanishing points vad and ved,fh was drawn parallel to ad,fg was drawn parallel to ed, and since hg is coincident with HPP, the two triangles ead and fhg must be similar. As ae was made equal to ad in the small triangle, Af must be equal to hg in the large triangle; and consequently ved, which is as far from vad as g is from h, must be as far from vad as the point f is from the point h.

If the student will refer back to Figs. 8, 9, and 9a, he will see that the point h bears a similar relation in Fig. 26 to that of the point mH in Figs. 8, 9, and 9a, and that the point h in Fig. 26 is really the horizontal projection of the vanishing point vad. (See also § 32.) Therefore, as ved is as far from vad as the point h is from the point f, we may make the following statement, which will hold for all systems of horizontal lines.

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