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.
53. PROBLEM IV. Fig. 19. To find the perspective of a rectangular block resting upon a horizontal plane \" below the level of the eye, and turned so that the long side of the block makes an angle of 3o° with the picture plane.
The block is shown in plan and elevation at the left of the figure. The first step will be to make an auxiliary horizontal projection of the block on the plane of the horizon, showing the exact position of the block as it is to be seen in the perspective projection. This auxiliary horizontal projection is really a revolved plan of the object, and is called a Diagram. It is the general rule, in making a perspective projection, to place the object behind the picture plane with one of its principal vertical lines lying in the picture plane (21 h). HPP is usually drawn near the upper edge of the paper, leaving just room enough behind to place the auxiliary plan or diagram. In the figure the diagram is shown in the required position, i.e., with one of its long sides (abfe) making an angle of 80° with the picture plane. The vertical edge (ae) of the block is supposed to lie in the picture plane. VH may now be drawn parallel to HPP at any convenient distance from it, as indicated. VH1, the vertical trace of the plane on which the block is supposed to rest, should be assumed in accordance with the given data, i.e., 1" below VH (§ 44).
The position of the observer's eye should next be established. SPh is its horizontal projection, and shows by its distance from HPP the distance in front of the picture plane at which the observer is supposed to stand. SPV is its vertical projection, and must always be found in VH. In this problem the station point is 1 1/2" in front of the picture plane.
Note.- As a general rule, it is well to assume the station point on a vertical line half way between two lines dropped from the extreme edges of the diagram, as indicated. This is not necessary, but, as will be explained later,, it usually insures a more pleasing perspective projection.
Next find the vanishing points for the different systems of lines in the object (§ 12). There are three systems of lines in the block, formed by its three sets of parallel edges.
1st. A system formed by the four horizontal edges vanishing to the right: ab, ef, dc, and kg.
2d. A system formed by the four horizontal edges vanishing towards the left: ad, ek, be, nndfg.
3d. A system formed by the four vertical edges.
First find the vanishing point for the system parallel to ab by drawing through the station point a line parallel to ab and finding where it pierces the picture plane (24 g). AH drawn through SPH is the horizontal projection of such a line. Its vertical projection (Av), drawn through SPV, will coincide with VH, and its vanishing point will be found on VH at vab (§ 51). All lines in the perspective of the object that are parallel to ah will meet at vab (§ 24 a). In a similar manner find vad, which will be the vanishing point for all lines parallel to ad.
54. If the method for finding any vanishing point is applied to the system of vertical lines, it will be found that this vanishing point will lie vertically over SPV at infinity. That is to say, since all vertical lines are parallel to the picture plane, if a vertical line is drawn through the station point, it will never pierce the picture plane. Therefore (24 g), the perspective of the vanishing point of a vertical line cannot be found within any finite limits, but will be vertically over SPV, and at an infinite distance from it. In a perspective projection all vertical lines are drawn actually vertical, and not converging towards one another.
Note.- This is true of all lines in an object which are parallel to the picture plane. Thus, the perspective of any line which is parallel to the picture plane, will actually be parallel to the line itself; and the perspectives of the elements of a system of lines parallel to the picture plane, will be drawn parallel to, and not converging towards, one another. That this must be so, is evident, since, if the perspectives of such a system of lines did converge towards one another, they would meet within finite limits. But it has just been found that the perspective of the vanishing point of such a system is at infinity. The perspectives of the elements of any system can meet only at the perspective of their vanishing point, and must, therefore, in a system parallel to the picture plane, be drawn parallel to one another.
The direction's of the perspectives of all lines in the object have now been determined, and will he as follows: All lines parallel to ah will meet at vab. All lines parallel to ad will meet at vad. All vertical lines will be drawn vertical.
Since the point e is in the base of the object, it lies on the plane of the ground, and also, since the line ae lies in the picture plane, the point e must lie on the intersection of the plane of the ground with the picture plane. Therefore, the point e must lie in VH1, and must be vertically under the point e in the diagram. Since the point e lies in the picture plane, it will be its own perspective ; and ep will be found on VHl, vertically under e in the diagram, as shown in the figure. From ep the perspective of the lower edges of the cube will vanish at vab and wad, respectively, as indicated.
fp is the perspective of the point f, and will be found on the lower edge of the block, vertically under the intersection of HPP with the horizontal projection of the visual ray drawn through the point/ in the diagram.
- Similarly, kp is found on the lower edge of the block, vertically under the intersection of HPP and the visual ray drawn through the point k in the diagram.
Vertical lines drawn through fp, ep, and kp, will represent the perspectives of the visible vertical edges of the block.
The edge epap being in the picture plane will be its own perspective, and show in its true size (§ 24 h). Therefore, ap may be established by making the distance epap equal to evav as taken from the given elevation. From av two of the upper horizontal edges of the block will vanish at vab and vad, respectively, establishing the points bp and dv, by their intersections with the vertical edges drawn through kp and fp, respectively. Lines drawn through dp and bp, and vanishing respectively at vab and vad will intersect at ep, and complete the perspective of the block.
Before going farther in the notes, the student should solve the problems on Plate I.