With rectangular projection, the system of projection planes consists of two mutually perpendicular projection planes (Fig. 2.1). One agreed to be placed horizontally, and the other vertically.

The plane of projections, located horizontally, is called horizontal projection plane and denote sch, and the plane perpendicular to it frontal projection planel 2 . The system of projection planes itself is denoted p / p 2. Usually use abbreviated expressions: plane L[, plane n 2 . Line of intersection of planes sch and to 2 called projection axisOH. It divides each projection plane into two parts - floors. The horizontal plane of projections has an anterior and posterior floors, while the frontal plane has an upper and lower floor.

planes sch and p 2 divide the space into four parts called quarters and denoted by Roman numerals I, II, III and IV (see Fig. 2.1). The first quarter is called the part of space bounded by the upper hollow frontal and front hollow horizontal projection planes. For the remaining quarters of the space, the definitions are similar to the previous one.

All engineering drawings are images built on the same plane. On fig. 2.1 the system of projection planes is spatial. To move to images on the same plane, we agreed to combine the projection planes. Usually plane p 2 left motionless, and the plane P turn in the direction indicated by the arrows (see Fig. 2.1), around the axis OH at an angle of 90 ° until it is aligned with the plane n 2 . With such a turn, the front floor of the horizontal plane goes down, and the back one rises. After alignment, the planes have the form depicted

female in fig. 2.2. It is believed that the projection planes are opaque and the observer is always in the first quarter. On fig. 2.2, the designation of planes invisible after alignment is taken in brackets, as is customary for highlighting invisible figures in the drawings.

The projected point can be in any quarter of space or on any projection plane. In all cases, to construct projections, projecting lines are drawn through it and their meeting points are found with planes 711 and 712, which are projections.

Consider the projection of a point located in the first quarter. The system of projection planes 711/712 and the point BUT(Fig. 2.3). Two straight LINES are drawn through it, perpendicular to the PLANES 71) AND 71 2. One of them will intersect plane 711 at the point BUT ", called horizontal projection of point A, and the other is the plane 71 2 at the point BUT ", called frontal projection of point A.

Projecting lines AA" and AA" determine the plane of projection a. It is perpendicular to the planes Kip 2, since it passes through perpendiculars to them and intersects the projection planes along straight lines A "Ah and A" A x. Projection axis OH perpendicular to the plane oc, as the line of intersection of two planes 71| and 71 2 perpendicular to the third plane (a), and hence to any line lying in it. In particular, 0X1A "A x and 0X1A "A x.

When combining planes, the segment A "Ah, flat to 2, remains stationary, and the segment A "A x together with plane 71) will be rotated around the axis OH until aligned with the plane 71 2 . View of combined projection planes together with projections of a point BUT shown in fig. 2.4, a. After aligning the point A", A x and A" will be located on one straight line perpendicular to the axis OH. This implies that two projections of the same point

lie on a common perpendicular to the projection axis. This perpendicular connecting two projections of the same point is called projection line.

The drawing in fig. 2.4, a can be greatly simplified. The designations of the combined projection planes in the drawings are not marked and the rectangles conditionally limiting the projection planes are not depicted, since the planes are unlimited. Simplified point drawing BUT(Fig. 2.4, b) also called diagram(From French ?pure - drawing).

Shown in fig. 2.3 quadrilateral AE4 "A X A" is a rectangle and its opposite sides are equal and parallel. Therefore, the distance from the point BUT up to the plane P, measured by a segment AA", in the drawing is determined by the segment A "Ah. The segment A "A x = AA" allows you to judge the distance from a point BUT up to the plane to 2 . Thus, the drawing of a point gives a complete picture of its location relative to the projection planes. For example, according to the drawing (see Fig. 2.4, b) it can be argued that the point BUT located in the first quarter and removed from the plane p 2 to a shorter distance than from the plane ts b since A "A x A "Ah.

Let's move on to projecting a point in the second, third and fourth quarters of space.

When projecting a point AT, located in the second quarter (Fig. 2.5), after combining the planes, both of its projections will be above the axis OH.

The horizontal projection of the point C, given in the third quarter (Fig. 2.6), is located above the axis OH, and the front is lower.

Point D depicted in fig. 2.7 is located in the fourth quarter. After combining the projection planes, both of its projections will be below the axis OH.

Comparing the drawings of points located in different quarters of space (see Fig. 2.4-2.7), you can see that each is characterized by its own location of projections relative to the axis of projections OH.

In particular cases, the projected point may lie on the projection plane. Then one of its projections coincides with the point itself, and the other will be located on the projection axis. For example, for a point E, lying on a plane sch(Fig. 2.8), the horizontal projection coincides with the point itself, and the frontal projection is on the axis OH. At the point E, located on the plane to 2(Fig. 2.9), horizontal projection on the axis OH, and the front coincides with the point itself.

PROJECTION OF A POINT ON TWO PLANES OF PROJECTIONS

The formation of a straight line segment AA 1 can be represented as a result of moving point A in any plane H (Fig. 84, a), and the formation of a plane can be represented as a displacement of a straight line segment AB (Fig. 84, b).

A point is the main geometric element of a line and surface, so the study of the rectangular projection of an object begins with the construction of rectangular projections of a point.

In the space of the dihedral angle formed by two perpendicular planes - the frontal (vertical) plane of projections V and the horizontal plane of projections H, we place the point A (Fig. 85, a).

The line of intersection of the projection planes is a straight line, which is called the projection axis and is denoted by the letter x.

The V plane is shown here as a rectangle, and the H plane as a parallelogram. The inclined side of this parallelogram is usually drawn at an angle of 45° to its horizontal side. The length of the inclined side is taken equal to 0.5 of its actual length.

From point A, perpendiculars are lowered on the planes V and H. Points a "and a of the intersection of perpendiculars with the projection planes V and H are rectangular projections of point A. The figure Aaa x a" in space is a rectangle. The side aax of this rectangle in the visual image is reduced by 2 times.

Let us align the H plane with the V plane by rotating V around the line of intersection of the x planes. The result is a complex drawing of point A (Fig. 85, b)

To simplify the complex drawing, the boundaries of the projection planes V and H are not indicated (Fig. 85, c).

Perpendiculars drawn from point A to the projection planes are called projecting lines, and the bases of these projecting lines - points a and a "are called projections of point A: a" is the frontal projection of point A, a is the horizontal projection of point A.

Line a "a is called the vertical line of the projection connection.

The location of the projection of a point on a complex drawing depends on the position of this point in space.

If point A lies on the horizontal projection plane H (Fig. 86, a), then its horizontal projection a coincides with the given point, and the frontal projection a "is located on the axis. When point B is located on the frontal projection plane V, its frontal projection coincides with this point, and the horizontal projection lies on the x-axis. The horizontal and frontal projections of a given point C, lying on the x-axis, coincide with this point. A complex drawing of points A, B and C is shown in Fig. 86, b.

PROJECTION OF A POINT ON THREE PLANES OF PROJECTIONS

In cases where it is impossible to imagine the shape of an object from two projections, it is projected onto three projection planes. In this case, the profile plane of projections W is introduced, which is perpendicular to the planes V and H. A visual representation of the system of three projection planes is given in fig. 87 a.

The edges of a trihedral angle (the intersection of projection planes) are called projection axes and are denoted by x, y and z. The intersection of the projection axes is called the beginning of the projection axes and is denoted by the letter O. Let us drop the perpendicular from point A to the projection plane W and, marking the base of the perpendicular with the letter a, we will obtain the profile projection of point A.

To obtain a complex drawing, points A of the H and W planes are aligned with the V plane, rotating them around the Ox and Oz axes. A complex drawing of point A is shown in fig. 87b and c.

The segments of the projecting lines from point A to the projection planes are called the coordinates of point A and are denoted: x A, y A and z A.

For example, the coordinate z A of point A, equal to the segment a "a x (Fig. 88, a and b), is the distance from point A to the horizontal projection plane H. The coordinate at point A, equal to the segment aa x, is the distance from point A to the frontal plane of projections V. The x A coordinate equal to the segment aa y is the distance from point A to the profile plane of projections W.

Thus, the distance between the projection of a point and the projection axis determines the coordinates of the point and is the key to reading its complex drawing. By two projections of a point, all three coordinates of a point can be determined.

If the coordinates of point A are given (for example, x A \u003d 20 mm, y A \u003d 22 mm and z A \u003d 25 mm), then three projections of this point can be built.

To do this, from the origin of coordinates O in the direction of the Oz axis, the coordinate z A is laid up and the coordinate y A is laid down. segments equal to the x coordinate A. The resulting points a "and a are the frontal and horizontal projections of the point A.

According to two projections a "and a point A, its profile projection can be constructed in three ways:

1) from the origin O, an auxiliary arc is drawn with a radius Oa y equal to the coordinate (Fig. 87, b and c), from the obtained point a y1 draw a straight line parallel to the Oz axis, and lay a segment equal to z A;

2) from the point a y, an auxiliary straight line is drawn at an angle of 45 ° to the axis Oy (Fig. 88, a), a point a y1 is obtained, etc.;

3) from the origin O, draw an auxiliary straight line at an angle of 45 ° to the axis Oy (Fig. 88, b), get a point a y1, etc.

In this article, we will find answers to questions about how to create a projection of a point onto a plane and how to determine the coordinates of this projection. In the theoretical part, we will rely on the concept of projection. We will give definitions of terms, accompany the information with illustrations. Let's consolidate the acquired knowledge by solving examples.

Yandex.RTB R-A-339285-1

Projection, types of projection

For convenience of consideration of spatial figures, drawings depicting these figures are used.

Definition 1

Projection of a figure onto a plane- a drawing of a spatial figure.

Obviously, there are a number of rules used to construct a projection.

Definition 2

projection- the process of constructing a drawing of a spatial figure on a plane using construction rules.

Projection plane is the plane in which the image is built.

The use of certain rules determines the type of projection: central or parallel.

A special case of parallel projection is perpendicular projection or orthogonal projection: in geometry, it is mainly used. For this reason, the adjective “perpendicular” itself is often omitted in speech: in geometry they simply say “projection of a figure” and mean by this the construction of a projection by the method of perpendicular projection. In special cases, of course, otherwise can be stipulated.

We note the fact that the projection of a figure onto a plane is, in fact, the projection of all points of this figure. Therefore, in order to be able to study a spatial figure in a drawing, it is necessary to acquire the basic skill of projecting a point onto a plane. What we will talk about below.

Recall that most often in geometry, speaking of projection onto a plane, they mean the use of perpendicular projection.

We will make constructions that will enable us to obtain the definition of the projection of a point onto a plane.

Suppose a three-dimensional space is given, and in it - a plane α and a point M 1 that does not belong to the plane α. Draw a straight line through a given point M 1 a perpendicular to the given plane α. The point of intersection of the line a and the plane α will be denoted as H 1 , by construction it will serve as the base of the perpendicular dropped from the point M 1 to the plane α .

If a point M 2 is given, belonging to a given plane α, then M 2 will serve as a projection of itself onto the plane α.

Definition 3

is either the point itself (if it belongs to a given plane), or the base of the perpendicular dropped from a given point to a given plane.

Finding the coordinates of the projection of a point on a plane, examples

Let in three-dimensional space given: rectangular coordinate system O x y z, plane α, point M 1 (x 1, y 1, z 1) . It is necessary to find the coordinates of the projection of the point M 1 onto a given plane.

The solution obviously follows from the above definition of the projection of a point onto a plane.

We denote the projection of the point M 1 onto the plane α as H 1 . According to the definition, H 1 is the point of intersection of the given plane α and the line a through the point M 1 (perpendicular to the plane). Those. the coordinates of the projection of the point M 1 we need are the coordinates of the point of intersection of the line a and the plane α.

Thus, to find the coordinates of the projection of a point onto a plane, it is necessary:

Get the equation of the plane α (in case it is not set). An article about the types of plane equations will help you here;

Determine the equation of a straight line apassing through the point M 1 and perpendicular to the plane α (study the topic of the equation of a straight line passing through a given point perpendicular to a given plane);

Find the coordinates of the point of intersection of the line a and the plane α (article - finding the coordinates of the point of intersection of the plane and the line). The data obtained will be the coordinates of the projection of the point M 1 onto the plane α that we need.

Let's consider the theory on practical examples.

Example 1

Determine the coordinates of the projection of the point M 1 (- 2, 4, 4) onto the plane 2 x - 3 y + z - 2 \u003d 0.

Solution

As we can see, the equation of the plane is given to us, i.e. there is no need to compose it.

Let's write the canonical equations of the straight line a passing through the point M 1 and perpendicular to the given plane. For these purposes, we determine the coordinates of the directing vector of the straight line a. Since the line a is perpendicular to the given plane, then the directing vector of the line a is the normal vector of the plane 2 x - 3 y + z - 2 = 0. In this way, a → = (2 , - 3 , 1) – direction vector of the line a .

Now we compose the canonical equations of a straight line in space passing through the point M 1 (- 2, 4, 4) and having a direction vector a → = (2 , - 3 , 1) :

x + 2 2 = y - 4 - 3 = z - 4 1

To find the desired coordinates, the next step is to determine the coordinates of the point of intersection of the line x + 2 2 = y - 4 - 3 = z - 4 1 and the plane 2 x - 3 y + z - 2 = 0 . To this end, we pass from the canonical equations to the equations of two intersecting planes:

x + 2 2 = y - 4 - 3 = z - 4 1 ⇔ - 3 (x + 2) = 2 (y - 4) 1 (x + 2) = 2 (z - 4) 1 ( y - 4) = - 3 (z + 4) ⇔ 3 x + 2 y - 2 = 0 x - 2 z + 10 = 0

Let's make a system of equations:

3 x + 2 y - 2 = 0 x - 2 z + 10 = 0 2 x - 3 y + z - 2 = 0 ⇔ 3 x + 2 y = 2 x - 2 z = - 10 2 x - 3 y + z = 2

And solve it using Cramer's method:

∆ = 3 2 0 1 0 - 2 2 - 3 1 = - 28 ∆ x = 2 2 0 - 10 0 - 2 2 - 3 1 = 0 ⇒ x = ∆ x ∆ = 0 - 28 = 0 ∆ y = 3 2 0 1 - 10 - 2 2 2 1 = - 28 ⇒ y = ∆ y ∆ = - 28 - 28 = 1 ∆ z = 3 2 2 1 0 - 10 2 - 3 2 = - 140 ⇒ z = ∆ z ∆ = - 140 - 28 = 5

Thus, the desired coordinates of a given point M 1 on a given plane α will be: (0, 1, 5) .

Answer: (0 , 1 , 5) .

Example 2

Points А (0 , 0 , 2) are given in a rectangular coordinate system O x y z of three-dimensional space; In (2, - 1, 0) ; C (4, 1, 1) and M 1 (-1, -2, 5). It is necessary to find the coordinates of the projection M 1 onto the plane A B C

Solution

First of all, we write the equation of a plane passing through three given points:

x - 0 y - 0 z - 0 2 - 0 - 1 - 0 0 - 2 4 - 0 1 - 0 1 - 2 = 0 ⇔ x y z - 2 2 - 1 - 2 4 1 - 1 = 0 ⇔ ⇔ 3 x - 6y + 6z - 12 = 0 ⇔ x - 2y + 2z - 4 = 0

Let's write the parametric equations of the straight line a, which will pass through the point M 1 perpendicular to the plane A B C. The plane x - 2 y + 2 z - 4 \u003d 0 has a normal vector with coordinates (1, - 2, 2), i.e. vector a → = (1 , - 2 , 2) – direction vector of the line a .

Now, having the coordinates of the point of the line M 1 and the coordinates of the directing vector of this line, we write the parametric equations of the line in space:

Then we determine the coordinates of the point of intersection of the plane x - 2 y + 2 z - 4 = 0 and the line

x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ

To do this, we substitute into the equation of the plane:

x = - 1 + λ , y = - 2 - 2 λ , z = 5 + 2 λ

Now, using the parametric equations x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ, we find the values ​​of the variables x, y and z at λ = - 1: x = - 1 + (- 1) y = - 2 - 2 (- 1) z = 5 + 2 (- 1) ⇔ x = - 2 y = 0 z = 3

Thus, the projection of the point M 1 onto the plane A B C will have coordinates (- 2, 0, 3) .

Answer: (- 2 , 0 , 3) .

Let us dwell separately on the question of finding the coordinates of the projection of a point on the coordinate planes and planes that are parallel to the coordinate planes.

Let points M 1 (x 1, y 1, z 1) and coordinate planes O x y , O x z and O y z be given. The projection coordinates of this point on these planes will be respectively: (x 1 , y 1 , 0) , (x 1 , 0 , z 1) and (0 , y 1 , z 1) . Consider also the planes parallel to the given coordinate planes:

C z + D = 0 ⇔ z = - D C , B y + D = 0 ⇔ y = - D B

And the projections of the given point M 1 on these planes will be points with coordinates x 1 , y 1 , - D C , x 1 , - D B , z 1 and - D A , y 1 , z 1 .

Let us demonstrate how this result was obtained.

As an example, let's define the projection of the point M 1 (x 1, y 1, z 1) onto the plane A x + D = 0. The rest of the cases are similar.

The given plane is parallel to the coordinate plane O y z and i → = (1 , 0 , 0) is its normal vector. The same vector serves as the directing vector of the straight line perpendicular to the plane O y z . Then the parametric equations of a straight line drawn through the point M 1 and perpendicular to a given plane will look like:

x = x 1 + λ y = y 1 z = z 1

Find the coordinates of the point of intersection of this line and the given plane. We first substitute into the equation A x + D = 0 equalities: x = x 1 + λ, y = y 1, z = z 1 and get: A (x 1 + λ) + D = 0 ⇒ λ = - D A - x one

Then we calculate the desired coordinates using the parametric equations of the straight line for λ = - D A - x 1:

x = x 1 + - D A - x 1 y = y 1 z = z 1 ⇔ x = - D A y = y 1 z = z 1

That is, the projection of the point M 1 (x 1, y 1, z 1) onto the plane will be a point with coordinates - D A , y 1 , z 1 .

Example 2

It is necessary to determine the coordinates of the projection of the point M 1 (- 6 , 0 , 1 2) onto the coordinate plane O x y and onto the plane 2 y - 3 = 0 .

Solution

The coordinate plane O x y will correspond to the incomplete general equation of the plane z = 0 . The projection of the point M 1 onto the plane z \u003d 0 will have coordinates (- 6, 0, 0) .

The plane equation 2 y - 3 = 0 can be written as y = 3 2 2 . Now just write the coordinates of the projection of the point M 1 (- 6 , 0 , 1 2) onto the plane y = 3 2 2:

6 , 3 2 2 , 1 2

Answer:(- 6 , 0 , 0) and - 6 , 3 2 2 , 1 2

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

Projection of a point on three planes of projections of the coordinate angle begins with obtaining its image on the plane H - the horizontal plane of projections. To do this, through point A (Fig. 4.12, a) a projecting beam is drawn perpendicular to the plane H.

In the figure, the perpendicular to the H plane is parallel to the Oz axis. The point of intersection of the beam with the plane H (point a) is chosen arbitrarily. The segment Aa determines how far point A is from the plane H, thus indicating unambiguously the position of point A in the figure with respect to the projection planes. Point a is a rectangular projection of point A onto the plane H and is called the horizontal projection of point A (Fig. 4.12, a).

To obtain an image of point A on the plane V (Fig. 4.12, b), a projecting beam is drawn through point A perpendicular to the frontal projection plane V. In the figure, the perpendicular to the plane V is parallel to the Oy axis. On the H plane, the distance from point A to plane V will be represented by a segment aa x, parallel to the Oy axis and perpendicular to the Ox axis. If we imagine that the projecting beam and its image are carried out simultaneously in the direction of the plane V, then when the image of the beam intersects the Ox axis at the point a x, the beam intersects the plane V at the point a. Drawing from the point a x in the V plane perpendicular to the Ox axis , which is the image of the projecting beam Aa on the plane V, the point a is obtained at the intersection with the projecting beam. Point a "is the frontal projection of point A, i.e. its image on the plane V.

The image of point A on the profile plane of projections (Fig. 4.12, c) is built using a projecting beam perpendicular to the W plane. In the figure, the perpendicular to the W plane is parallel to the Ox axis. The projecting beam from point A to plane W on the plane H will be represented by a segment aa y, parallel to the Ox axis and perpendicular to the Oy axis. From the point Oy parallel to the Oz axis and perpendicular to the Oy axis, an image of the projecting beam aA is built and, at the intersection with the projecting beam, the point a is obtained. Point a is the profile projection of the point A, i.e., the image of the point A on the plane W.

The point a "can be constructed by drawing from the point a" the segment a "a z (the image of the projecting beam Aa" on the plane V) parallel to the Ox axis, and from the point a z - the segment a "a z parallel to the Oy axis until it intersects with the projecting beam.

Having received three projections of point A on the projection planes, the coordinate angle is deployed into one plane, as shown in Fig. 4.11, b, together with the projections of the point A and the projecting rays, and the point A and the projecting rays Aa, Aa "and Aa" are removed. The edges of the combined projection planes are not drawn, but only the projection axes Oz, Oy and Ox, Oy 1 (Fig. 4.13).

An analysis of the orthogonal drawing of a point shows that three distances - Aa", Aa and Aa" (Fig. 4.12, c), characterizing the position of point A in space, can be determined by discarding the projection object itself - point A, on a coordinate angle deployed in one plane (Fig. 4.13). The segments a "a z, aa y and Oa x are equal to Aa" as opposite sides of the corresponding rectangles (Fig. 4.12, c and 4.13). They determine the distance at which point A is located from the profile plane of projections. Segments a "a x, a" a y1 and Oa y are equal to segment Aa, determine the distance from point A to the horizontal projection plane, segments aa x, a "a z and Oa y 1 are equal to segment Aa", which determines the distance from point A to frontal projection plane.

The segments Oa x, Oa y and Oa z located on the projection axes are a graphic expression of the sizes of the X, Y and Z coordinates of point A. The point coordinates are denoted with the index of the corresponding letter. By measuring the size of these segments, you can determine the position of the point in space, i.e., set the coordinates of the point.

On the diagram, the segments a "a x and aa x are arranged as one line perpendicular to the Ox axis, and the segments a" a z and a "a z - to the Oz axis. These lines are called projection connection lines. They intersect the projection axes at points a x and and z, respectively.The line of the projection connection connecting the horizontal projection of point A with the profile one turned out to be “cut” at the point a y.

Two projections of the same point are always located on the same projection connection line perpendicular to the projection axis.

To represent the position of a point in space, two of its projections and a given origin (point O) are sufficient. 4.14, b, two projections of a point completely determine its position in space. Using these two projections, you can build a profile projection of point A. Therefore, in the future, if there is no need for a profile projection, diagrams will be built on two projection planes: V and H.

Rice. 4.14. Rice. 4.15.

Let's consider several examples of building and reading a drawing of a point.

Example 1 Determination of the coordinates of the point J given on the diagram by two projections (Fig. 4.14). Three segments are measured: segment Ov X (X coordinate), segment b X b (Y coordinate) and segment b X b "(Z coordinate). Coordinates are written in the following order: X, Y and Z, after the letter designation of the point, for example , B20; 30; 15.

Example 2. Construction of a point according to the given coordinates. Point C is given by coordinates C30; ten; 40. On the Ox axis (Fig. 4.15) find a point with x, at which the line of the projection connection intersects the projection axis. To do this, the X coordinate (size 30) is plotted along the Ox axis from the origin (point O) and a point with x is obtained. Through this point, perpendicular to the Ox axis, a projection connection line is drawn and the Y coordinate is laid down from the point (size 10), the point c is obtained - the horizontal projection of the point C. The coordinate Z (size 40) is plotted upwards from the point c x along the projection connection line (size 40), the point is obtained c" - frontal projection of point C.

Example 3. Construction of a profile projection of a point according to the given projections. The projections of the point D - d and d are set. Through the point O, the projection axes Oz, Oy and Oy 1 are drawn (Fig. 4.16, a). it to the right behind the Oz axis. The profile projection of the point D will be located on this line. It will be located at such a distance from the Oz axis, at which the horizontal projection of the point d is located: from the Ox axis, i.e. at a distance dd x. The segments d z d "and dd x are the same, since they determine the same distance - the distance from point D to the frontal projection plane. This distance is the Y coordinate of point D.

Graphically, the segment d z d "is built by transferring the segment dd x from the horizontal plane of projections to the profile one. To do this, draw a line of projection connection parallel to the Ox axis, get a point d y on the Oy axis (Fig. 4.16, b). Then transfer the size of the segment Od y to the Oy 1 axis , drawing from the point O an arc with a radius equal to the segment Od y, until it intersects with the axis Oy 1 (Fig. 4.16, b), get the point dy 1. This point can also be constructed, as shown in Fig. 4.16, c, by drawing a straight line at an angle 45 ° to the Oy axis from the point d y. From the point d y1 draw a projection connection line parallel to the Oz axis and lay a segment on it equal to the segment d "d x, get the point d".

The transfer of the value of the segment d x d to the profile plane of the projections can be carried out using the constant straight line of the drawing (Fig. 4.16, d). In this case, the projection connection line dd y is drawn through the horizontal projection of the point parallel to the Oy 1 axis until it intersects with a constant straight line, and then parallel to the Oy axis until it intersects with the continuation of the projection connection line d "d z.

Particular cases of the location of points relative to projection planes

The position of a point relative to the projection plane is determined by the corresponding coordinate, i.e., the value of the segment of the projection connection line from the Ox axis to the corresponding projection. On fig. 4.17 the Y coordinate of point A is determined by the segment aa x - the distance from point A to plane V. The Z coordinate of point A is determined by the segment a "a x - the distance from point A to plane H. If one of the coordinates is zero, then the point is located on the projection plane Fig. 4.17 shows examples of different locations of points relative to the projection planes.The Z coordinate of point B is zero, the point is in plane H. Its frontal projection is on the Ox axis and coincides with point b x. The Y coordinate of point C is zero, the point is located on the plane V, its horizontal projection c is on the x-axis and coincides with the point c x.

Therefore, if a point is on the projection plane, then one of the projections of this point lies on the projection axis.

On fig. 4.17, the Z and Y coordinates of the point D are equal to zero, therefore, the point D is on the projection axis Ox and its two projections coincide.

projection(lat. Projicio - I throw forward) - the process of obtaining an image of an object (spatial object) on any surface using light or visual rays (rays that conditionally connect the observer's eye with any point of a spatial object), which are called projecting.

There are two projection methods: central and parallel .

Centralprojection is to pass through each point ( A, B, C,…) of the depicted object and in a certain way selected projection center (S) straight line ( SA, SB, >… — projecting beam).

Figure 1.1 - Central projection

Let's introduce the following notation (Figure 1.1):

S– projection center (observer's eye);

π 1 - projection plane;

A, B, C

SA, SB- projecting straight lines (projecting rays).

Note: left mouse button can move the point in the horizontal plane, when you click on the point with the left mouse button, the direction of movement will change and you can move it vertically.

Central projection point the point of intersection of the projecting line passing through the projection center and the projection object (point) with the projection plane is called.

Property 1 . Each point in space corresponds to a single projection, but each point in the projection plane corresponds to a set of points in space that lie on the projecting line.

Let's prove this statement.

Figure 1.1: dot BUT 1 is the central projection of point A on the plane of projections π 1 . But all points lying on the projecting line can have the same projection. Take on the projecting line SA point FROM. Central projection point FROM(FROM 1) on the plane of projections π 1 coincides with the projection of the point BUT(BUT 1):

1. FROM∈ SA;
2. SC∩ π 1 = C 1 →C 1 ≡ A 1 .

The conclusion follows that by the projection of a point it is impossible to judge unambiguously about its position in space.

To eliminate this uncertainty, i.e. make a drawing reversible, we introduce one more projection plane (π 2) and one more projection center ( S 2) (Figure 1.2).

Figure 1.2 - Illustration of the 1st and 2nd properties

Let's construct projections of a point BUT on the plane of projections π 2 . Of all points in space, only a point BUT has its projections BUT 1 to the plane π 1 and BUT 2 to π 2 at the same time. All other points lying on the projecting rays will have at least one different projection from the projections of the point BUT(e.g. dot AT).

Property 2 . The projection of a straight line is a straight line.

Let's prove this property.

Connect the dots BUT and AT among themselves (Figure 1.2). We get a segment AB defining a straight line. triangle SAB defines a plane, denoted by σ. It is known that two planes intersect in a straight line: σ∩π 1 = BUT 1 AT 1 , where BUT 1 AT 1 - central projection of a straight line given by a segment AB.

The central projection method is a model of image perception by the eye, it is mainly used when making perspective images of building objects, interiors, as well as in film technology and optics. The method of central projection does not solve the main task facing the engineer - to accurately reflect the shape, dimensions of the object, the ratio of the sizes of various elements.

1.2. Parallel projection

Consider the method of parallel projection. We will impose three restrictions that will allow us, albeit to the detriment of the visibility of the image, to get a drawing more convenient for using it in practice:

1. Let's delete both projection centers to infinity. Thus, we will ensure that the projecting rays from each center become parallel, and, therefore, the ratio of the true length of any line segment and the length of its projection will depend only on the angle of inclination of this segment to the projection planes and do not depend on the position of the projection center;
2. Let's fix the projection direction relative to the projection planes;
3. Let's arrange the projection planes perpendicular to each other, which will make it easy to move from the image on the projection planes to the real object in space.

Thus, having imposed these restrictions on the central projection method, we have come to its special case - parallel projection method(Figure 1.3). Projection, in which the projecting rays passing through each point of the object are parallel to the selected projection direction P, is called parallel .

Figure 1.3 - Parallel projection method

Let's introduce the notation:

R– direction of projection;

π 1 - horizontal plane of projections;

A,B– projection objects – points;

BUT 1 and AT 1 - projections of points BUT and AT onto the projection plane π 1 .

Parallel point projection is the point of intersection of the projecting line parallel to the given direction of projection R, with the projection plane π 1 .

Pass through the dots BUT and AT projecting beams parallel to a given projection direction R. Projecting ray passing through a point BUT intersects the projection plane π 1 at the point BUT one . Similarly, a projecting ray through a point AT intersects the projection plane at a point AT one . By connecting the dots BUT 1 and AT 1 , we get a segment BUT 1 AT 1 is the projection of the segment AB onto the plane π 1 .

1.3. Orthographic projection. Monge method

If the projection direction R perpendicular to the plane of projections p 1 , then the projection is called rectangular (Figure 1.4), or orthogonal (gr. orthos- straight, gonia- angle) if R not perpendicular to π 1, then the projection is called oblique .

quadrilateral AA 1 AT 1 AT defines the plane γ, which is called the projecting plane, since it is perpendicular to the plane π 1 (γ⊥π 1). In what follows, we will use only rectangular projection.

Figure 1.4 - Orthographic projection Figure 1.5 - Monge, Gaspard (1746-1818)

The French scientist Gaspard Monge is considered the founder of orthogonal projection (Figure 1.5).

Before Monge, builders, artists and scientists possessed quite significant information about projection methods, and yet only Gaspard Monge is the creator of descriptive geometry as a science.

Gaspard Monge was born on May 9, 1746 in the small town of Beaune (Burgundy) in eastern France in the family of a local merchant. He was the eldest of five children, to whom his father, despite the low origin and relative poverty of the family, tried to provide the best education available at that time for people from the humble class. His second son, Louis, became a professor of mathematics and astronomy, the youngest, Jean, also a professor of mathematics, hydrography and navigation. Gaspard Monge received his initial education at the city school of the Oratory order. After graduating in 1762 as the best student, he entered the college of Lyon, also owned by the Oratorians. Soon Gaspard was entrusted with teaching physics there. In the summer of 1764, Monge drew up a plan of his native city of Beaune, remarkably accurate. The necessary methods and instruments for measuring angles and drawing lines were invented by the compiler himself.

While studying in Lyon, he received an offer to join the order and remain a college teacher, however, instead, having shown great abilities in mathematics, drafting and drawing, he managed to enter the Mézieres School of Military Engineers, but (due to origin) only as an auxiliary non-commissioned officer officer department and without a paycheck. Nevertheless, success in the exact sciences and an original solution to one of the important problems of fortification (the placement of fortifications depending on the location of enemy artillery) allowed him in 1769 to become an assistant (teaching assistant) in mathematics, and then in physics, and already with a decent salary at 1800 livres a year.

In 1770, at the age of 24, Monge held the position of professor at the same time in two departments - mathematics and physics, and, in addition, conducts classes in cutting stones. Starting with the task of accurately cutting stones according to given sketches in relation to architecture and fortification, Monge came to create methods that he later generalized in a new science - descriptive geometry, the creator of which he is rightfully considered. Given the possibility of using the methods of descriptive geometry for military purposes in the construction of fortifications, the leadership of the Mézières school did not allow open publication until 1799, the book was published under the title descriptive geometry (Geometry descriptive) (a verbatim record of these lectures was made in 1795). The approach to lecturing on this science and doing the exercises outlined in it has survived to this day. Another significant work of Monge - Application of analysis to geometry (L'application de l'analyse à la geometrie, 1795) - is a textbook of analytic geometry, in which special emphasis is placed on differential relations.

In 1780 he was elected a member of the Paris Academy of Sciences, in 1794 he became director of the Polytechnic School. For eight months he served as minister of the sea in the government of Napoleon, was in charge of the gunpowder and cannon factories of the republic, and accompanied Napoleon on his expedition to Egypt (1798–1801). Napoleon granted him the title of count, honored him with many other distinctions.

The method of depicting objects according to Monge consists of two main points:

1. The position of a geometric object in space, in this example a point BUT, is considered relative to two mutually perpendicular planes π 1 and π 2(Figure 1.6).

They conditionally divide the space into four quadrants. Dot BUT located in the first quadrant. The Cartesian coordinate system served as the basis for the Monge projections. Monge replaced the concept of projection axes with the line of intersection of projection planes (coordinate axes) and proposed to combine the coordinate planes into one by rotating them around the coordinate axes.

Figure 1.6 - Model for constructing point projections

π 1 - horizontal (first) projection plane

π 2 - frontal (second) projection plane

π 1 ∩ π 2 is the axis of projections (we denote π 2 / π 1)

Consider an example of projecting a point BUT onto two mutually perpendicular projection planes π 1 and π 2 .

Drop from point BUT perpendiculars (projecting rays) on the planes π 1 and π 2 and mark their bases, that is, the points of intersection of these perpendiculars (projecting rays) with the projection planes. BUT 1 - horizontal (first) projection of the point BUT;BUT 2 - frontal (second) projection of the point BUT;AA 1 and AA 2 - projecting lines. Arrows show the direction of projection on the plane of projections π 1 and π 2 . Such a system allows you to uniquely determine the position of a point relative to the projection planes π 1 and π 2:

AA 1 ⊥π 1

BUT 2 BUT 0 ⊥π 2 /π 1 AA 1 = BUT 2 BUT 0 - distance from point A to plane π 1

AA 2 ⊥π 2

BUT 1 BUT 0 ⊥π 2 /π 1 AA 2 \u003d A 1 A 0 - the distance from point A to the plane π 2

2. Let's combine the rotation around the axis of projections π 2 / π 1 of the projection plane into one plane(π 1 with π 2), but so that the images do not overlap, (in the α direction, Figure 1.6), we get an image called a rectangular drawing (Figure 1.7):

Figure 1.7 - Orthogonal drawing

Rectangular or orthogonal is called Monge diagram .

Straight BUT 2 BUT 1 called projection link , which connects opposite projections of the point ( BUT 2 - frontal and BUT 1 - horizontal) is always perpendicular to the projection axis (coordinate axis) BUT 2 BUT 1 ⊥π 2 /π 1 . On the diagram, the segments indicated by curly brackets are:

• BUT 0 BUT 1 - distance from the point BUT to the plane π 2 corresponding to the coordinate y A;
• BUT 0 BUT 2 - distance from the point BUT to the plane π 1 corresponding to the coordinate z A.

1.4. Rectangular point projections. Orthographic drawing properties

1. Two rectangular projections of a point lie on the same projection connection line perpendicular to the projection axis.

2. Two rectangular projections of a point uniquely determine its position in space relative to the projection planes.

Let's verify the validity of the last statement, for which we turn the plane π 1 to its original position (when π 1 ⊥ π 2). To build a point BUT needed from points BUT 1 and BUT 2 to restore the projecting rays, and in fact - the perpendiculars to the planes π 1 and π 2 , respectively. The intersection point of these perpendiculars fixes the desired point in space BUT. Consider an orthogonal drawing of a point BUT(Figure 1.8).

Figure 1.8 - Plotting a point

Let us introduce the third (profile) plane of projections π 3 perpendicular to π 1 and π 2 (given by the axis of projections π 2 /π 3).

Distance from the profile projection of a point to the vertical axis of the projections BUT‘ 0 A 3 allows you to determine the distance from the point BUT to the frontal projection plane π 2 . It is known that the position of a point in space can be fixed relative to the Cartesian coordinate system using three numbers (coordinates) A(X A ; Y A ; Z A) or relative to the projection planes using its two orthogonal projections ( A 1 =(X A ; Y A); A 2 =(X A ; Z A)). On an orthogonal drawing, using two projections of a point, you can determine its three coordinates and, conversely, using three coordinates of a point, build its projections (Figure 1.9, a and b).

Figure 1.9 - Plotting a point according to its coordinates

By the location on the projection diagram of a point, one can judge its location in space:

• BUT — BUT 1 lies under the coordinate axis X, and the front BUT 2 - above the axis X, then we can say that the point BUT belongs to the 1st quadrant;
• if on the plot the horizontal projection of the point BUT — BUT 1 lies above the coordinate axis X, and the front BUT 2 - under the axle X, then the point BUT belongs to the 3rd quadrant;
• BUT — BUT 1 and BUT 2 lie above the axis X, then the point BUT belongs to the 2nd quadrant;
• if on the diagram there are horizontal and frontal projections of the point BUT — BUT 1 and BUT 2 lie under the axle X, then the point BUT belongs to the 4th quadrant;
• if on the diagram the projection of a point coincides with the point itself, then it means that the point belongs to the plane of projections;
• a point belonging to the projection plane or projection axis (coordinate axes) is called private point.

To determine in which quadrant of space a point is located, it is enough to determine the sign of the coordinates of the point.

Dependences of the quadrant of the position of the point and the signs of the coordinates

	X	Y	Z
I	+	+	+
II	+	—	+
III	+	—	—
IV	+	+	—

An exercise

Construct orthogonal projections of a point with coordinates BUT(60, 20, 40) and determine in which quadrant the point is located.

Problem solution: along the axis OX set aside the value of the coordinate XA=60, then through this point on the axis OX restore the line of projection connection, perpendicular to OX, along which to set aside the value of the coordinate ZA=40, and down - the value of the coordinate YA=20(Figure 1.10). All coordinates are positive, which means that the point is located in the I quadrant.

Figure 1.10 - Solution of the problem

1.5. Tasks for independent solution

1. Based on the diagram, determine the position of the point relative to the projection planes (Figure 1.11).

Figure 1.11

2. Complete the missing orthogonal projections of points BUT, AT, FROM on the projection plane π 1 , π 2 , π 3 (Figure 1.12).

Figure 1.12

3. Build point projections:

• E, symmetric point BUT relative to the projection plane π 1 ;
• F, symmetric point AT relative to the plane of projections π 2 ;
• G, symmetric point FROM relative to the projection axis π 2 /π 1 ;
• H, symmetric point D relative to the bisector plane of the second and fourth quadrants.

4. Construct orthogonal projections of the point To, located in the second quadrant and remote from the projection planes π 1 by 40 mm, from π 2 - by 15 mm.