THE METHOD OF OPTIMAL PROJECTION AND THE AFFINE CORRESPONDENCE OF TWO SUPERPOSED PLANES

L. S. Skripov

(Presented by Prof. I. A. Balashev, Doctor of Technical Sciences)

The method of optimal projection onto a single plane was proposed by the author as one of the new methods for a simpler solution of metric problems, the construction of shadows, and axonometric images.

In the present article it is proposed to consider it together with the affine correspondence of two superposed planes, in order to clarify certain common properties of both methods, the convenience of their joint application, and the advantages of the former over the latter in particular cases.

Let us briefly recall the essence of the method of optimal projection onto one plane. In space, a certain plane is chosen and positioned with respect to the given object in such a way that, under orthogonal projection of the given object onto it, the required solution of the proposed problem is obtained on this plane at once.

The affine correspondence of two superposed planes, as is known from descriptive geometry, is determined by the following data:

1. The axis of affinity and a pair of corresponding points are given.
2. Two pairs of corresponding lines are given.
3. Three pairs of corresponding points not lying on one straight line are given.
4. Two corresponding triangles are given.

Orthogonal projection also establishes an affine correspondence between a plane figure and each of its projections. In this case the axis of affinity is the straight line of intersection of the plane of the given figure with one or another plane of projection, while the direction of projection is perpendicular to the axis of projections.

Under such projection there also exist properties of affinity between the projected figures and their projections, such as those given above.

In the theory of descriptive geometry it is proved that on the epure the horizontal and vertical projections of one and the same figure (a plane one) are figures mutually affine.

We shall make use of the stated properties of the first and second methods, and consider their application in a number of examples.

In Fig. 1 a certain triangle \(ABC\) is represented in three projections, lying in the profile-projecting plane \(Q_1\) and parallel to the axis \(OX\).

On the basis of the stated properties of affinity of superposed plane figures, we note that between the given triangle \(ABC\) and its horizontal projection there must exist an affine correspondence.

To find the axis of affinity, we superpose the plane \(Q_1\) with the plane \(H\) by rotation about the trace \(Q_{H1}\), then rotate the plane \(H\) about the axis.

\(OX\) until it is superposed with the plane \(V\), and we obtain the figure of the superposed given triangle in the form \(a_1b_1c_1\).

Extend its sides \(b_1a_1, b_1c_1, a_1c_1\) until they intersect, respectively, with the extensions of the corresponding sides of the horizontal projection of the triangle \(ba, bc, ac\) at the points \(c_0, a_0, b_0\), lying on the horizontal trace \(Q_H\) of the given plane,

Fig. 1

which in the present case is also the axis of affinity \(x_1x_1\), thereby confirming the previously indicated properties of affinity under orthogonal projection.

Let us now move the plane \(Q_1\) into a new position \(Q_2\), parallel to the original one, and project the given triangle \(ABC\) orthogonally onto this plane. The construction of the new projection is carried out on the profile plane in the form of the line \(c_2''a_2''b_2''\). On the new plane \(Q_2\), the given triangle will be represented in its true size; consequently, its figure will remain unchanged, and we may assert that between the new representation of the given triangle on the plane \(Q_2\) and the remaining unchanged horizontal projection \(abc\) the former affinity is preserved.

To find the new axis of affinity we proceed as in the preceding case: we bring the plane \(Q_2\) into coincidence with the plane \(H\) by rotation about its horizontal trace, then rotate the plane \(H\) until it coincides with \(V\) about the axis \(OX\), and construct the figure of the triangle \(a_3b_3c_3\), congruent to \(a_1b_1c_1\). We extend the sides of the triangle \(b_3a_3\) and \(b_3c_3\) until they intersect with the extensions of the corresponding sides of the horizontal projection \(ba\) and \(bc\) at the points \(c_{02}\) and \(a_{02}\), through which we draw the straight line \(x_2x_2\), thus obtaining a new axis of affinity, parallel to the preceding one \(x_1x_1\). The line \(x_1x_1\) will move into the new position \(x_2x_2\) by a distance equal to \(l_2\).

To confirm what has been set forth, another plane \(Q_3\), parallel to \(Q_1\), has also been taken; onto it the given triangle is orthogonally projected in the form \(c_4''a_4''b_4''\) on the plane \(W\), and the preceding arguments and constructions are repeated. A new axis of affinity \(x_3x_3\) is obtained as the result of shifting the axis \(x_1x_1\) by the amount \(l_3\).

Let us now move the plane of the triangle \(ABC\) into a new general position (Fig. 2). Such a displacement will not disturb the affinity between the triangle \(ABC\) and its horizontal projection. The axis of affinity in this case as well is the horizontal trace \(Q_H\), on which the extensions will intersect...

the continuation of the side of the triangle \(BC\) and its projection \(DC\), and any lines in its plane, for example \(A1\) and \(a1\).

The same constructions are also carried out on the plane of Fig. 3, where the given plane \(Q\) is brought into coincidence with the plane \(H\) by rotation about the trace \(Q_h\).

Fig. 2

In Fig. 3 the points \(a_0, b_0\), belonging to the axis of affinity \(x x\), are found by extending the sides \(a_1 c_1, b_1 c_1\) and \(ac, bc\). Fig. 3 also shows the finding of the axis of affinity by drawing other lines in the plane of the triangle, namely the frontal lines of the plane \(a_1 1_0, b_1 2_0, C_1 3_0\) and their horizontal projections \(a1_0, b2_0, c3_0\), passing through the vertices of the triangle \(ABC\).

Fig. 3

If the plane \(Q\) is placed in a position parallel to the plane of the triangle \(ABC\) (Fig. 4) and the triangle is projected orthogonally onto it, then it will be projected in its true size as \(a_1b_1c_1\), and its affinity with its former unchanged horizontal projection \(abc\) will be preserved, while the axis of affinity will move to a new position \(x_1x_1\), parallel to the line \(Q_H\) or \(xx\).

Figure 4 shows the superposed position \(a_2b_2c_2\) of the triangle \(a_1b_1c_1\) with the plane \(H\), and the axis of affinity \(x_1x_1\) has been found by extending the sides \(b_2c_2\) and \(bc\) of the triangles to their mutual intersection at the point \(a\).

Fig. 4

The second point \(l_0\) of the axis of affinity \(x_1x_1\) is determined by the intersection of the projections of some line arbitrarily taken in the plane of the triangle, for example, \(a1\) and \(a_2 1_2\). The same construction is carried out in the plane drawing (Fig. 5) by the method of the most advantageous projection of the triangle \(ABC\) onto the plane \(Q_1\), whose traces are drawn parallel to the horizontal \(B1\) and the frontal \(A2\) of the triangle \(ABC\).

After superposing the plane \(Q_1\) with the plane \(H\) and constructing, in the superposed position, the triangle \(a_2b_2c_2\), we extend the sides of the triangles \(a_2c_2\) and \(ac\), \(b_2c_2\) and \(bc\) to their mutual intersection at the points \(a_0\) and \(b_0\), through which the axis of affinity \(x_2x_2\) can already be drawn. Figure 5 also shows the projections of the frontals drawn through the vertices of the triangles \(a_2b_2c_2\) and \(abc\) to their mutual intersection at the points \(a_{02}\), \(b_{02}\), \(c_{02}\) on the axis of affinity \(x_2x_2\).

Having established the affine correspondence of two plane images of a certain triangle in the method of the most advantageous projection, let us consider a number of concrete examples of applying this affinity.

Example 1. Determine the distance between two parallel straight lines \(AB\) and \(CD\) (Fig. 6) and construct the projections of this distance on the planes \(H\) and \(V\).

Applying the method of the most advantageous projection, we place a certain plane \(Q\) in space perpendicular to the direction of the given straight lines, drawing the traces of the plane \(Q_H\) and \(Q_V\) at an arbitrary place in the drawing. We project the given lines orthogonally onto the chosen plane, drawing perpendiculars from the points of the lines \(A, B, C, D\) to the traces of the plane \(Q_H\) and \(Q_V\). We superpose the plane \(Q\) with the plane \(H\) by rotation about the trace \(Q_H\). Pere-

we transfer the points of intersection of the perpendiculars with the vertical trace \(Q_v\) to the superposed position of the trace \(Q_{v1}\). We continue drawing the perpendiculars to \(Q_{v1}\) on the superposed plane until they meet the continuations of the corresponding perpendiculars to the horizontal trace, and obtain the projections of the given

Fig. 5

lines on the chosen plane in the form of the points \(a_1 b_1\) and \(c_1 d_1\). The distance between them will be the required distance between the given lines \(AB\) and \(CD\), in the form \(5_1—6_1\).

Fig. 6

To construct this distance on the previous projections of the lines, we determine the location of points 5 and 6 on the horizontal and vertical projections by reverse reasoning. The line \(5—6\) is perpendicular to the given lines \(AB\) and \(CD\) and parallel to the selected plane \(Q\); consequently, in the plane \(Q\) one can draw a line \(mn\) parallel to \(5—6\).

We return the plane \(Q\) to its former position together with the line \(mn\) and construct its projections on \(H\) and \(V\), i.e., \(mn\) and \(m'n'\). Taking arbitrarily on the line \(CD\) a point \(5(5,5')\), we draw through it a line parallel to \(MN(mn',m'n')\), and obtain the second point \(6(6,6')\) on the line \(AB\). By joining the obtained points 5 and 6, we obtain the projections of a sought distance \(5—6\) between the given lines \((5—6,\ 5'—6')\).

The same solution could have been obtained by applying the affinity correspondence between the images of points 5 and 6 on the superposed plane \(Q\) and on the plane \(H\).

Suppose that points 5 and 6 \((5,6)\) have been found on the horizontal projections of the lines. Through these points we draw the projections of frontal lines, both on \(H\) and on the superposed plane \(Q\), until their mutual intersection at points \(5_0\) and \(6_0\). These points determine the position of the axis of affinity \(x_1x_1\), parallel to the line \(Q_h\), of the two superposed planes, on which the corresponding points \(5_1\) and \(5,\ 6_1\) and 6 are situated.

The distance between parallel lines may be measured at any place on the given lines; therefore, applying the laws of affinity correspondence, we proceed as follows to find this distance on the projections of the given lines.

We choose an arbitrary point \(3,3'\) on the line \(CD\). We draw, at an arbitrary place, the line of affinity, parallel to the trace \(Q_h\). From point 3 we draw the projection of a frontal line to meet the axis of affinity at point \(3_0\), from which we draw the corresponding frontal line, parallel to the trace \(Q_h\), until it meets the continuation of the perpendiculars from points \(c\) and \(d\) to the trace \(Q_h\). The obtained point \(3_1\) will be the point corresponding to point 3 on the line \(cd\). To find point 4, the end of the perpendicular measuring the distance between the given lines \(AB\) and \(CD\), we draw from the found point \(3_1\) a line parallel to \(5_1—6_1\) on the superposed plane, until it intersects the axis of affinity at point \(3_{10}\), which we then join with the previously selected point 3 on the line \(CD\). The required point 4 is found at the intersection of the line \(3—3_{10}\) with the projection of the line \(a—b\). Following such reasoning, one may take any points on the given lines and find the sought distance between the lines.

If we wished to find the distance between the lines by measuring it between the points of intersection of the given lines with the constructed plane \(Q\), then the trace of the plane \(Q_H\) should be taken as the axis of affinity, and from points \(1_1\) and \(2_1\) the projections of frontal lines should be drawn on the superposed plane \(Q\) until they intersect \(Q\) at points \(1_0\) and \(2_0\), and from them—the projections of frontal lines on the plane \(H\) until they meet the given lines at points 1 and 2. The distance \(12,\ 1'2'\) would be the required one on the projections of the given lines.

The same points 1 and 2 would have different corresponding points 1 and 2 on another superposed plane, if the line \(xx\) were taken as the axis of affinity.

The example given shows that, for solving the proposed question, one may use exclusively the method of most advantageous projection, without applying the laws of affinity of two superposed planes, which in this problem do not bring any substantial advantage in the sense of simplifying the solution.

Example 2. Determine the shortest distance between two skew lines \(AB\) and \(CD\) (Fig. 7) and its projections on the planes \(H\) and \(V\).

To determine the true magnitude of the sought shortest distance between the skew lines, we choose, at any place in the drawing, a plane \(Q\), placing it perpendicular to one of the given lines, ...

for example, \(CD\). We project both lines orthogonally onto the selected plane and then bring the plane \(Q\) into coincidence with the plane \(H\). The line \(CD\) is projected into the point \(c_1d_1\), and the line \(AB\) into a certain line \(a_1b_1\). Since the distance between straight lines is measured by the length of the perpendicular dropped from a point of one of the lines to the other line, let this be the point \(l\); then we drop from the point \(l_1\) (coincident in projection with \(c_1\) and \(d_1\)) a perpendicular to the line \(a_1b_1\) at the point \(k_1\), since the line \(l_1k_1\) is parallel to the plane \(Q\), and the right angle \(l_1k_1a_1\) is projected onto it in true size.

Fig. 7

To find the points \(l, l'\) and \(k, k'\) on the given projections of the lines \(AB\) and \(CD\), we draw in the plane \(Q\) the line \(mn_1\) parallel to \(l_1k_1\), and return the plane \(Q\) to its former position, constructing the projections of the line \(mn, m'n'\). Then, from the point \(k\), which is easily found on the horizontal projection of the line \(ab\), we draw a line parallel to the horizontal projection of the line \(mn\) selected on the plane until it meets the line \(cd\) at the point \(l\), and obtain the horizontal projection of the required shortest distance \(KL\), and from it also its vertical projection \(k'l'\).

It is possible to find the point \(l\) on the horizontal projection of the line as the corresponding point to \(l_1\) on the superposed plane \(Q\). For this, through the points \(k\) and \(k_1\) we draw projections of frontals and extend them until they meet at the point \(k_0\), through which we then draw the axis of affinity \(xx\), parallel to the horizontal trace of the plane \(Q_H\). From the point \(l_1\) we draw a frontal of the plane \(Q\) and extend it to the axis of affinity \(xx\) at the point \(l_0\), from which we draw a frontal on the plane \(H\) until it meets the line \(cd\) at the point \(l\), which will be the desired point.

In the proposed problem, the method of optimal projection and the affine correspondence are perhaps equivalent in convenience for obtaining the final result on the given projections.

Example 3. Construct the figure of the normal section of the triangular prism \(ABCABC\) by the plane \(Q\), and determine its true size (Fig. 8).

By the method of optimal projection, we determine the true size of the section figure by projecting the section orthogonally onto the given plane \(Q\), in the form of the triangle \(1_1 2_1 3_1\), which, on the superposed plane \(Q\), coincides with the projection onto this plane of the entire prism \(ABCABC\).

The construction of the projections of the section figure may be carried out by the reverse transfer of the points \(1_1, 2_1, 3_1\) onto the horizontal and vertical projections of the edges of the prism, by drawing through these points frontals first in the superposed plane \(Q\) to the trace \(Q_H\) at the points \(1_0, 2_0, 3_0\), then transferring these frontals to the original position of the plane \(Q\) from the points \(1'_0, 2'_0, 3'_0\) until they meet the edges of the prism at the points \(1', 2', 3'\), and, finally, already by constructing ...

horizontal projections of the points 1, 2, 3. Joining the points obtained, we construct the figure of the desired section in projections.

This same question can be solved by finding the corresponding points on the plane \(H\) from the found points of the vertices of the triangle \(1_1, 2_1, 3_1\). Since

Fig. 8

the figure of the section lies in the plane \(Q\), and its corresponding horizontal projection will lie on the plane \(H\), the axis of correspondence \(xx\), in this case, will be the trace of the plane \(Q_H\).

Through the points \(1_1, 2_1, 3_1\) we draw frontals on the superposed plane \(Q\) to the axis of correspondence, i.e., to the points \(1_0, 2_0, 3_0\). From these points we construct the lines corresponding to the frontals on the plane \(H\), i.e., lines parallel to the axis \(OX\), until they meet the corresponding projections of the edges of the prism at the points 1, 2, 3; joining these, we obtain the desired horizontal projection of the figure of the section 123, and then also the vertical projection.

In the given problem it is more advantageous to apply the laws of the corresponding relation of the superposed planes, since they require fewer constructions. To find the horizontal projection of the figure of the section it would have been sufficient to draw only one frontal, for example, through the point \(1_1\), and to find its corresponding point 1; the remaining points would be determined by extending the sides of the triangle \(1_1, 2_1, 3_1\) to the axis of correspondence at the points \(b_0\) and \(c_0\), from which it would be necessary to draw the corresponding lines \(b_0b_1\) and \(c_0c_1\), and, at their intersections with the edges of the prism, obtain the points 3 and 2.

Example 4. Through the given straight line \(AB(ab, a'b')\), draw a plane tangent to the given sphere (center of the sphere \(o,o'\))—(Fig. 9).

Using the method of most advantageous projection, we draw the plane \(Q\) at any place on the drawing, perpendicular to the given straight line \(AB\); we then project the straight line \(AB\) and the sphere onto this plane orthogonally and, finally, superpose the plane \(Q\) with the plane \(H\), constructing the superposed projection of the sphere and of the straight line \(AB\). The straight line \(AB\) is projected into the point \(a_1b_1\), and the sphere into a circle of the same radius with center at the point \(o_2\).

The problem will be solved if the point of tangency of the required plane with the sphere, passing through the given line \(AB\), is found, since a point and a line completely determine the position of a plane in space.

For this purpose, through the point \(a_1-b_1\) (into which the given line \(AB\) was projected) we draw two straight lines tangent to the circle (the projection of the sphere) at two points \(k_3\) and \(k_4\), since the problem has two solutions. The constructed lines \((a_1-b_1)k_3\) and \((a_1-b_1)k_4\) represent the projections of certain planes, and the points \(k_3\) and \(k_4\) are their points of tangency with the sphere. It remains to transfer the found points \(k_3\) and \(k_4\) onto the horizontal and vertical planes of projection; for this, we first draw from the points \(k_3\) and \(k_4\) lines perpendicular to the trace \(Q_H\). Then we draw the lines \(o_2k_3\) and \(o_2k_4\), as projections of the radii of the sphere from the center of the sphere to the points of tangency of the planes, and consequently perpendicular to the required planes or, in other words, parallel to the plane \(Q\).

In the superposed plane we draw two lines in an arbitrary place: \(mn\), parallel to \(o_2k_3\), and \(m_1n_1\), parallel to \(o_2k_4\). We return the plane \(Q\) to its original position, and with it the constructed lines, and construct their projections \(mn_1\), \(m'n'\) and \(m_1n_1\), \(m_1'n_1'\). From the horizontal projection of the center of the sphere \(o_1\) we draw lines parallel to the horizontal projections \(mn\) and \(mn_1\), until they meet the previously drawn lines from the points \(k_3\) and \(k_4\). The obtained points \(k_1\), \(k_2\) will be the required ones. To find the vertical projections of these points \(k_1'\) and \(k_2'\), lines are first drawn from the point \(o_1'\) parallel to the vertical projections of the lines \(m'n'\) and \(m_1'n_1'\), until they meet the vertical lines from \(k_1\) and \(k_2\). The same points \(k_1'\) and \(k_2'\) can also be found by the reverse construction of perpendiculars to the traces of the plane from the points \(k_3\) and \(k_4\).

Fig. 9

The required points of tangency of the planes with the sphere can be determined by using the affinity of the points \(k_3\) and \(k_1\), \(k_4\) and \(k_2\); for this, we first construct the axis of affinity by drawing through the projections of the center of the sphere \(o_2\) and \(o_1\) corresponding frontal lines until they meet each other at the point \(O_0\). Through the point \(O_0\) we draw the line of affinity \(xx\), parallel to the trace \(Q_H\). Then, from the points \(k_3\) and \(k_4\), we draw frontals in the plane \(Q\) until they meet the axis of affinity at the points \(k_{01}\) and \(k_{02}\), and from them the lines corresponding to them, parallel to the axis \(OX\), until they meet the perpendiculars to \(Q_H\) from the points \(k_3\) and \(k_4\).

In the present problem, a solution can be obtained somewhat more simply and quickly by the method of affinity.

Example 5. Construct the projections of the cast shadow on a sphere for the given direction of the light rays \(S(s_1 s')\) (Fig. 10).

Fig. 10

By the method of optimal projection, we choose a plane in any place on the drawing perpendicular to the direction of the light rays \((s, s')\). We project the sphere orthogonally onto the selected plane and obtain the superposed position of the projection of the sphere in the form of a circle with center \(O_2\).

Onto this same circle there is also projected the circle of the cast shadow on the sphere, as the geometric locus of the points of tangency of the light rays with the surface of the sphere.

We choose an arbitrary number of points on this circle (in our case 12 are taken) and transfer them to the horizontal and vertical projections of the sphere, using exclusively the method of affinity, as giving the simplest solution of the proposed problem.

First of all, we determine the axis of affinity by drawing the lines of corresponding frontals through the central points of the sphere \(O_2\) and \(O_1\) to their mutual intersection at the point \(O_0\), through which, parallel to the trace \(Q_H\), we draw the line of affinity \(xx\).

Using this axis, we draw, through the chosen points, frontals to the axis of affinity at the points \(1_0, 2_0, \ldots, 12_0\), and through them lines corresponding to the frontals on the plane \(H\), up to their intersection with the lines drawn from the points \(1_1, 2_1, 3_1, \ldots, 12_1\) perpendicular to the trace \(Q_H\). The obtained points \(1, 2, 3, 4, \ldots, 12\) on the horizontal projection are then joined by a smooth curve, taking its visibility into account. The points found for the outline of the shadow are transferred to the vertical projection by the usual reverse construction according to the method of optimal projection.

The example presented shows the advantageous application of the affine correspondence of two superposed planes in the method of optimal projection.

Example 6. Through the given three arbitrarily positioned lines in space \(AB\), \(CD\), \(EF\), draw a fourth line intersecting the three given ones (Fig. 11).

By the method of optimal projection, we choose some plane in space perpendicular to one of the given lines, for example \(CD\), and project all three lines orthogonally onto it; we then bring

Fig. 11

the plane \(Q\) into coincidence with the plane \(H\). The line \(CD\) is projected onto this plane as the point \(c_1d_1\), and the other two, \(AB\) and \(EF\), as the lines \(a_1b_1\) and \(e_1f_1\). Through the point \(c_1d_1\), or more precisely through some point \(2_1\) on the line \(CD\), we draw an arbitrary line \(2_1 3_1 1_1\), intersecting the other two lines at the points \(3_1\) and \(1_1\). We return the plane \(Q\) to its former position and transfer the points obtained, \(1_1, 3_1\), first onto the horizontal projections \(ab\) and \(ef\), and then onto the vertical projections \(a'b'\) and \(e'f'\), and then connect them by the straight lines \(1\!-\!3\) and \(1'\!-\!3'\) up to their intersection at the point \(2,2'\) with the line \(CD\).

The problem has an unlimited number of solutions.

The example presented shows the case in which, in order to solve the problem, it is impossible to resort to the laws of affine correspondence of two superposed planes, and one must use exclusively the laws of optimal projection.

Example 7. Construct a spatial image (axonometry) of a certain angle bracket, given by its orthogonal projections onto \(H\) and \(V\) in a specified direction of projection (Fig. 12).

We choose in space some plane \(Q\) perpendicular to the given direction of projection \((s, s')\). We bring the plane into coincidence with the plane \(H\) and, applying the method of optimal projection, construct the required spatial image of the angle bracket, which must be viewed along the direction of the projecting rays \(S\), rotating the drawing so that the trace of the plane \(Q_H\) assumes a horizontal position.

In the proposed problem, an exceptional application of the method of optimal projection is shown, which alone makes it possible to solve a problem of this kind easily and simply.

The last question is set forth in greater detail and with greater thoroughness in the author’s work Axonograph (1940).

The examples considered, of course, by no means exhaust all cases of the joint application of the method of optimal projection and the affine correspondence of two superposed planes; they show:

1. In some cases the joint application does not provide any special advantages in simplifying the solution of the proposed questions.

2. In other cases the joint application of the methods simplifies and accelerates the solution.

3. In still other cases the simplification obtained is indisputable (the shadow on a sphere).

4. In a fourth group of cases, only the method of optimal projection possesses exceptional simplicity and makes it possible to solve the proposed problem (Examples 5 and 6).

Fig. 12

In conclusion, it may be noted that the joint application of both methods will undoubtedly help to solve a number of problems in descriptive geometry more simply and quickly.

Noted Misprints