Image of spatial figures

Kvant 1970, No. 12 p. 2, 4, 5, 6, 7, 8, 10, 12, 13, 14 Tag 0MDK

Rough automatic draft, low quality: expect mistakes and awkward wording. It will be replaced by a better translation. Original →

Parallel Projection Method

Unfortunately, there is no “spatial pencil”, the tip of which would leave a mark in what has arisen. With such a pencil it was possible to “draw” a real cube by tracing its edges. A flat image cannot be an exact copy of a spatial figure. A problem arises: what rules should be used to construct an image so that it is similar to the original, that is, so that when viewing the image, a visual impression arises that is close to that which occurs when viewing the original? Most often, he uses design methods for this purpose. The simplest of them are the method of central projections and the method of parallel projections.

Fig. 1.

The method of central projections is clear from figure 1. A fixed point $S$ (the center by the projection) and a fixed plane l (the plane by the projection) not passing through $S$ are taken. Through any point $A'$ of space a line $SA'$ (projecting line) is drawn. The point $A$ of intersection of this line with the plane l is called the project point $A'$ . figure 1 shows the projections of two points: $A'$ and $B'$ .

Original points lying on the same projection line have the same image.

The method of central projections gives the most straightforward images, because the described construction reproduces the process of vision: the rays coming from the original point ( $A', B', \dots$ ) or from their project ( $A, B, \dots$ ) coincide. Thus, the eye doesn’t care whether it looks at the original or its projection onto the plane. (True, we are simplifying somewhat here: after all, the original is viewed with two eyes.)

Artists mainly use the method of central projections.

The method of parallel projections differs from the method of central projections only in that the projecting lines do not pass through a fixed point, but are parallel to a fixed direction (Fig. 2). The image in parallel projection resembles an object viewed from afar: with an unlimited

MI, you can start any triangle $ABC$ and claim that it is similar to the image of the given triangle $A'B'C'$ .

Proof. We are given two triangles: the original $A'B'C'$ and the sample $ABC$ . On side $A'B'$ we construct a triangle $A'B'C_1$ (Fig. 4), similar to triangle $ABC$ and located in some plane $π$ , different from the plane of triangle $A'B'C'$ . Let us further take $C'C_1$ as the design direction. Then the projection of triangle $A'B'C'$ onto the plane $π$ turns out to be triangle $A'B'C_1$ , similar to triangle $ABC$ .

Theorem 2. If a given image of a triangle is $A'B'C'$ , then the image of each point lying in the plane of this triangle is thereby uniquely determined.

Proof. Let $A'B'C'$ be the original (Fig. 5, a), and $ABC$ be the image (Fig. 5, b). Let us take an arbitrary point $D'$ in the plane of the triangle $A'B'C'$ and connect it with any vertex of the triangle, for example with $A'$ . Let $E'$ be the point of intersection of line $A'D'$ with the opposite side $B'C'$ (it makes no difference whether $E'$ is inside the segment $B'C'$ or on its extension). The image of the point $E'$ can be found: it must divide the segment $BC$ in the same ratio in which $E'$ divides the segment $B'C'$ , that is

$\frac{BE}{EC} = \frac{B'E'}{E'C'}$

From the two theorems proved, a practical rule for constructing images of plane figures follows: one must select any three points in the figure $F'$ that do not lie on the same straight line, and depict them as arbitrary three points of general position. This is where the arbitrariness ends: images of all other points must be constructed. In other words, a given triangle can be drawn arbitrarily, but a given quadrangle cannot be drawn.

Example 1. Image of a regular hexagon.

Let us denote this regular hexagon (original) by $A'B'C'D'E'F'$ , and its center by the letter $O'$ (for convenience, the original is shown at figure 6, a). Let us now construct an image of this hexagon. Let us draw the triangle $A'B'C'$ arbitrarily (Fig. 6, b). Point $G'$ is the midpoint of segment $A'C'$ . Therefore, point $G$ must also be the midpoint of segment $AC$ . Now you can draw a straight line $BG$ . On it you can construct points $O$ and $E$ , use correspondences $BO=2BG$ , $BE=4BG$ . Next, $CD||BG$ and $AF||BG$ . Point $D$ can be found in different ways: 1) putting aside $CD=BO$ , 2) drawing

The Polke-Schwartz theorem and the representation of spatial figures

We know that when depicting flat figures, the triangle plays a special role. A similar situation arises when depicting spatial figures, only a tetrahedron is used instead of a triangle. Before sorting the corresponding theorem, we make the following definition.

A complete quadrilateral is a plane figure consisting of four points in general position (that is, no three of them lie on the same line) and six segments connecting these points in pairs.

Theorem 3 (Polke-Schwarz theorem)* Any given method can be represented, up to similarity, by an arbitrary complete quadrilateral (at figure 7, and

*) Polke proved this theorem in 1853 for an equilateral right tetrahedron (with equal lateral edges and right plane angles at the vertex). Schwartz proved it in 1864 for an arbitrary tetrahedron.

Fig. 7, a. Fig. 7, b.

ED||AB, 3) by drawing straight line AO. Point F is constructed in the same way.

As another example, prove on your own that a square can be represented by an arbitrary parallelogram.

Let us note in conclusion that the image of a circle is an ellipse.

Fig. 8, a. Fig. 8, b,

the quadrilateral ABCD is convex, at figure 7, b - non-convex).

We will accept this theorem without proof.

The dotted line is used in the depiction of spatial figures to indicate invisible lines (it is assumed that the edges of the tetrahedron are opaque).

Theorem 4. If an image of the metrahedron A’B’C’D’ is given, then this determines the image of each point in space.

Proof. Let A’B’C’D’ be the original signal (Fig. 8, a) ABCD be its image on the plane (Fig. 8, b). Suppose, in addition, that an arbitrary point M’ is given in space. We’ll show you how to construct its image. Let’s connect M’ with some vertex of the tetrahedron, for example, with A’, and mark the intersection point E’ of the straight line A’M’ with the opposite face B’C’D’ (the point E’ should not at all indicate being inside the tetrahedra B’C’D’).

Theorel will move to the image. An image of point E’ can be constructed: after all, this point lies in the plane of triangle B’C’D’, the image of which is given. Next, draw a straight line AE and construct a point M on it that satisfies the condition

$\frac{AM}{ME} = \frac{A'M'}{M'E'}$

Example 2. Image of a cube. When the edges of the cube extending from

Fig. 9.

one point define a tetrahedron. According to the Polke-Schwartz theorem, it can be represented by an arbitrary quadrilateral. Thus, depicting the cube (Fig. 9), we can mark the points $A_1, B_1, D_1$ and $A_2$ arbitrarily. The rest of the image is achieved based on the fact that the parallel edges of the cube and in the drawing should be parallel.

Most people think that figure 9 cannot always serve as an image of a cube and that the angles of its constituent parallelograms and the ratios of the segments must be chosen in a special way. This is explained by the fact that for us the most graftable are orthogonal projections obtained in the case when the projecting straight lines are perpendicular to the projection plane. Here we are talking about arbitrary (“oblique”) parallel projections.

Example 3. Representation of a regular quadrangular parallelogram. The base (square) can be depicted as an arbitrary parallelogram. In addition, according to the Polke-Schwartz theorem, one can draw arbitrarily one side edge, that is, mark an arbitrary parallelogram. If you need to check the height, you should connect the top to the intersection point of the base diagonals.

Fig. 10, $a$ . Fig. 10, $b$ .

In textbooks, it is common practice to depict a plane (more precisely, a “piece of a plane”) as a parallelogram, while implicitly assuming that the original is a rectangular piece of a plane. In this case, part of the arbitrariness at our disposal is involuntarily consumed. If we forget about this and further use the Polke-Schwartz theorem, then we arrive at groups of errors. Here’s an example. At figure 10, $a$ a regular quadrangular parallelogram is depicted. Standing on a plane. The individual SABCD image is constructed correctly, but figure 10, $a$ as a whole is grossly erroneous. Having depicted a rectangular piece of the plane as a parallelogram KLMN, we have already partially used the arbitrariness represented by the Polke-Schwartz theorem, and we do not have the right to depict the square $A'B'C'D'$ as an arbitrary parallelogram. On figure 10, $a$ $\rightarrow$ LKN and $\rightarrow$ DAB depict right angles. $AB \parallel KL$ should therefore be $AD \parallel KN$ . Since this is not the case, figure 10, $a$ is wrong.

Since the straightness of a piece of a plane does not matter at all when studying the properties of this parallel on it, there is no point in binding oneself to the condition that $\rightarrow$ LKN is an image

Sections of polyhedra

Now we will consider images encountered when studying stereometry.

Example 4. At figure 11, a the flat shape of a triangular feature is depicted.

It is “constructed” very simply: arbitrary points $A_3$ , $B_3$ , $C_3$ are taken on the edges of the feature and connected by segments of the feature. At figure 11, $\delta$ the same “method” is applied to the construction of a section of a quadrangular feature.

Figure 11 is correct, but on figure 11, $\delta$ there was a mistake (a very common one!). In fact, the plane is determined by the third point. Therefore, when depicting the plane shape of a triangular feature, we can take points on its edges arbitrarily. You cannot do this with a quadrangular sign. Noting the pro-

arbitrarily the points $A_3$ , $B_3$ , $C_3$ , we have already defined the cutting plane, and the point of its intersection with the four edges of the recognition cannot be taken arbitrarily.

We will show you how to correctly find the point of intersection of a cutting plane with four edges.

The plane passing through the edges $^*$ ) $A'_1A'_2$ and $C'_1C'_2$ intersects the plane section $A'_3B'_3C'_3D'_3$ along the diagonal $A'_3C'_3$ , and the plane passing through the edges $B'_1B'_2$ and $D'_1D'_2$ , along the diagonal $B'_3D'_3$ . The intersection of these two planes is parallel to the edge of the plane. Consequently, in all plane sections of the line the points of intersection of the diagonal line lie on a straight line parallel to the edge of the line. At figure11, $\delta$ the points $P_1$ and $P_2$ of intersection of the diagonal bases are marked; The yama $P_1P_2$ is parallel to the edge of the yam. Point $P_3$ does not lie on this line -

*) Do not forget that the elements of the original are indicated by letters with dashes.

Fig. 13.
therefore figure is incorrect. This means that the quadrilateral $A_3'B_3'C_3'D_3'$ depicted on it is not flat. A person with a trained eye, for example, an artist, will notice this immediately, without any construction.

This property allows you not only to check the correctness of the drawing, but also to construct a planar view of a quadrangular feature. Let us arbitrarily mark the points $A_3$ , $B_3$ , $C_3$ (Fig. 12). We construct points $P_1$ and $P_2$ and draw a straight line $P_1P_2$ (it turns out to be parallel to the straight line $A_1A_2$ ). We carry out a straight line $A_3C_3$ . We find the point $P_3$ of intersection $A_3C_3$ and $P_1P_2$ . We are holding $B_3P_3$ . The intersection point $B_3P_3$ with $D_1D_2$ is the desired one.

Depicting flat sections of polyhedra requires the use of only two rules:

If planes $\beta'$ and $\gamma'$ intersect along line $l'$ , and plane $\alpha'$ intersects their corresponding lines $b'$ and $c'$ , then $b'$ and $c'$ either intersect at a point lying on $l'$ or are parallel to $l'$ .
If planes $\beta'$ and $\gamma'$ are parallel, and plane $\alpha'$ intersects their corresponding lines $b'$ and $c'$ , then $b'$ and $c'$ are parallel.

Example 5. Image of a flat section of a parallelped.

We draw images of linear sections $RQ$ and $QP$ . The point of their intersection must lie on the line $A_2B_2$ , otherwise the lines $PQ$ and $QR$ are arbitrary. Next we carry out $RS \parallel QP$ and $PS \parallel QR$ .

On figure 13 the lines $PQ$ and $QR$ are drawn so that the point $Q$ falls outside the segment $A_2B_2$ . Mark the points $T \equiv PQ \times A_2D_2$ and $U \equiv QR \times A_1A_2$ and connect $U$ and $T^*$ ). We carry out $RS \parallel QP$ and $PS \parallel QR$ . It is clear that the point $S$ must lie on the line $C_1D_1$ . If this point falls on the continuation of the segment $C_1D_1$ , then we connect the points $V \equiv RS \times B_1C_1$ and $W \equiv PS \times C_1C_2$ . Analyze on your own the case when the point $Q$ lies inside the segment $A_2B_2$ .

Example 6. Image of a plane section of a quadrangular paramid (Fig. 14).

We carry out links $PQ$ and $QR$ . From point $R$ it is necessary to draw a section line along the rear edge. The planes of the front and rear faces intersect along a straight line $S'X'$ . The image of this line can be easily obtained by constructing the point $X \equiv AB \times CD$ . The traces of the cutting plane $\alpha'$ on the front and back faces must intersect at $S'X'$ . We find point $T \equiv PQ \times SX$ and draw $TR$ . We mark the segment of this straight line $RU$ , which is the interior of the triangle $SCD$ . And $AB$ represents the line of intersection of the front and bottom edges. We build point $Y \equiv AB \times PQ$ and find point $V \equiv AD \times UY$ .

*) The notation $M \equiv AB \times CD$ means: “M is the point of intersection of the lines $AB$ and $CD$ ”, and $m \equiv A \times B$ - “m is the line defining the points $A$ and $B$ ”.

Fig. 17.
Each ellipse shows two diameters: in the original they were redistributed, and in the image they are conjugate (see the article by I. N. Bronshtein “Ellipse”, Quantum No. 9).

The base of the cone is also depicted as an ellipse. In school practice, the following mistake is often encountered: the extreme generatrices are used to depict the main section, and it is assumed that they touch the ellipse at the ends of the major axis (Fig. 18, a). This is not laziness. If you draw tangents to the ellipse from point S and connect tangency points A and B, then straight line AB will not pass through the center of the ellipse. However, some

Fig. 19.
The “draftsmen” take her through the center of the violent. Figure 18, b - correct. One of the points of tangency, for example A, is connected to the center O, and point C is marked on the ellipse, diametrically opposite to A. Triangle SAC is an image of its own section.

To depict a master section, it is not at all necessary to use

Fig. 18, a.
Fig. 18, b.
Fig. 18, p.

Fig. 23, a.

The date and equator will be depicted as a segment (Fig. 22).

The correct image of the ball is constructed as follows. The equator and meridians are depicted as ellipses, and all meridians pass through two points N and S. It remains only to find out the connection between the image of the equator and the position of points N and S.

Let’s imagine the original. Let’s draw an equatorial section (circle A’B’C’D’) and a diameter N’S’ (earth’s axis) perpendicular to it. Let us further imagine a plane β’ perpendicular to A’C’ and (for convenience) passing outside the ball (Fig. 23, a). The sphere, together with the equatorial section and the N’S’ axis, is projected orthogonally onto the β’ plane. The ball is projected into a circle, and the system (A’B’C’D’, N’S’) into two perpendicular diameters of this circle.

If the figure on the left (Fig. 23, a) is rotated around A’C’, then the circle on the plane β’ will remain in place, and the cross of its two perpendicular diameters will rotate around the center.

Figure 23, b gives clues to the question of how the image of the equator is related to the images of the stripes. The diameter PQ determines the minor axis of the ellipse, and the perpendicular

Fig. 25.

connecting through the points $N$ and $S$ . Parallels are depicted by similar ellipses touching the contour of the shaar.

When solving stereometric problems, the drawing plays a supporting role, and there is no point in spending more work on drawing it than on solving the problem itself. Therefore, we recommend that the reader draw mainly by eye. However, we will give some general recommendations, the implementation of which will help you avoid mistakes.

The image of a ball is more complex than that of polyhedra, a cylinder and a cone. Therefore, it is recommended to carry out drawings taking into account the ball, starting with the ball, and then attaching the remaining figures to it.

If a ball is inscribed in a cylinder (Fig. 25), then the lateral surface of the cylinder touches the ball in a large circle, for example along the equator. Point $B$ is on the equator, $AB = BC$ , $AC = NS$ . All three ellipses (equatorial shape and base of the cylinder) are the same.

If a cylinder is inscribed in a ball (Fig. 26), then it should be remembered that its bases are parallel circles of equal size.

If the ball is inscribed in the recognition (Fig. 27), then the larger circle (for example, eq

Fig. 28.

To depict a prism inscribed in a ball (Fig. 28), you must first of all inscribe a polygon in some parallel circle. The rest is clear.

If the ball is inscribed in a paramid (Fig. 29), then the points of contact of the lateral faces are at equal distances from the top of the paramid and, therefore, lie on one parallel circle, the plane of which is perpendicular to $T'O'$ (O’ is the center of the ball, $T'$ is the top of the paramid). You can draw in the following sequence: draw a ball, then take

Fig. 29.

some parallel circle and describe a polygon around it (for example, $A_1B_1C_1D_1$ ). After this, draw tangents to the meridians passing through the tangent points $L, M, N, P$ at these points (you can do it by eye). They will intersect at one point $T$ , which represents the vertex of the paramid. Let the plane of the lower base be parallel to the plane $A_1'B_1'C_1'D_1'$ (if this is not so, then the problem is a condition, and we will not consider it in this article), then it remains to complete the paramid, guided by the proportions

$\frac{TS}{TO_1} = \frac{TA}{TA_1} = \frac{TB}{TB_1} = \frac{TC}{TC_1} = \frac{TD}{TD_1}$

( $S'$ - Southern stripe, $O'$ - center of the parallel circle inscribed in $A_1'B_1'C_1'D_1'$ ).

If a ball is described near a paramid (Fig. 30), then the base of the paramid is inscribed in some parallel circle, and the vertex is any point on the sphere.

From the editor. In 1971, in the series “Popular lectures on mathematics”, N. M. Beskin’s brochure “Image of spatial figures” was published, in which these issues will be presented in more detail.