Goals:

• Studying the rules for constructing projections of points on the surface of an object and reading drawings.
• Develop spatial thinking, the ability to analyze the geometric shape of an object.
• To cultivate diligence, the ability to cooperate when working in groups, and interest in the subject.

DURING THE CLASSES

STAGE I. MOTIVATION OF LEARNING ACTIVITIES.

STAGE II. FORMATION OF KNOWLEDGE, ABILITIES AND SKILLS.

HEALTH-SAVING PAUSE. REFLECTION (MOOD)

STAGE III. INDIVIDUAL WORK.

STAGE I. MOTIVATION OF LEARNING ACTIVITIES

1) Teacher: Check your workplace, is everything in place? Is everyone ready to go?

TAKE A DEEP BREATH, HOLD YOUR BREATH AS YOU EXHALE, BREATH OUT.

Determine your mood at the beginning of the lesson according to the diagram (this diagram is on everyone’s desk)

I WISH YOU GOOD LUCK.

2)Teacher: Practical work on this topic " Projections of vertices, edges, faces” showed that there are guys who make mistakes when projecting. They are confused which of two coinciding points in the drawing is a visible vertex and which is an invisible one; when the edge is parallel to the plane, and when it is perpendicular. It's the same with edges.

To avoid repetition of mistakes, complete the necessary tasks using the counseling card and correct errors in practical work (by hand). And as you work, remember:

“EVERYONE CAN MAKE A MISTAKE, ONLY A CRAZY ONE STAYS WITH THEIR ERROR.”

And those who have mastered the topic well will work in groups with creative tasks (see. Annex 1 ).

STAGE II. FORMATION OF KNOWLEDGE, ABILITIES AND SKILLS

1)Teacher: In production there are many parts that are attached to each other in a certain way.
For example:
The desktop cover is attached to the vertical posts. Pay attention to the table you are at, how and with what are the lid and racks attached to each other?

Answer: Bolt.

Teacher: What is needed for a bolt?

Answer: Hole.

Teacher: Really. And in order to make a hole, you need to know its location on the product. When making a table, a carpenter cannot contact the customer every time. So, what do you need to provide a carpenter with?

Answer: A drawing.

Teacher: Drawing!? What do we call a drawing?

Answer: A drawing is an image of an object rectangular projections in projection connection. Using the drawing, you can imagine the geometric shape and design of the product.

Teacher: We have completed rectangular projections, what next? Will we be able to determine the location of the holes from one projection? What else do we need to know? What to learn?

Answer: Construct points. Find the projections of these points in all views.

Teacher: Well done! This is the goal of our lesson and topic: Construction of projections of points on the surface of an object. Write the topic of the lesson in your notebook.
You and I know that any point or segment on the image of an object is a projection of a vertex, edge, face, i.e. each view is an image not from one side (head view, top view, left view), but the entire object.
In order to correctly find the projections of individual points lying on the faces, you must first find the projections of this face, and then, using connection lines, find the projections of the points.

(We look at the drawing on the board, work in a notebook, where 3 projections of the same part are made at home).

– Opened the notebook with the completed drawing (Explanation of the construction of points on the surface of an object with guiding questions on the board, and students fix it in their notebooks.)

Teacher: Consider the point IN. Which plane is the face of this point parallel to?

Answer: The edge is parallel to the frontal plane.

Teacher: We define the projection of a point b' on the frontal projection. Swipe down from the point b' vertical communication line to the horizontal projection. Where will the horizontal projection of the point be located? IN?

Answer: At the intersection with the horizontal projection of the face that was projected into the edge. And it is located at the bottom of the projection (view).

Teacher: Profile projection of a point b'' , where will it be located? How will we find her?

Answer: At the intersection of a horizontal communication line from b' with a vertical edge on the right. This edge is the projection of the face with the point IN.

THOSE WHO WANT TO CONSTRUCT THE NEXT PROJECTION OF THE POINT ARE CALLED TO THE BOARD.

Teacher: Point projections A are also located using communication lines. Which plane is the face with the point parallel to? A?

Answer: The edge is parallel to the profile plane. We define a point on the profile projection A'' .

Teacher: In which projection was the face projected into the edge?

Answer: On the front and horizontal. Let's draw a horizontal connection line until it intersects with the vertical edge on the left on the frontal projection, we get a point A' .

Teacher: How to find the projection of a point A on a horizontal projection? After all, communication lines from the projection of points A' And A'' do not intersect the projection of the face (edge) on the horizontal projection on the left. What can help us?

Answer: You can use a constant straight line (it determines the location of the view on the left) from A'' draw a vertical communication line until it intersects with a constant straight line. A horizontal connection line is drawn from the intersection point until it intersects with the vertical edge on the left. (This is the face with point A) and denotes the projection with a point A .

2) Teacher: Everyone has a task card on the table with tracing paper attached. Examine the drawing, now try on your own, without redrawing the projections, to find the given projections of points in the drawing.

– Find the picture in the textbook, page 76. 93. Test yourself. Those who did it correctly – score “5”; one mistake – ‘’4’’; two – ‘’3’’.

(Grades are given by the students themselves on the self-control sheet).

– Collect cards for verification.

3)Work in groups: Time limited: 4 min. + 2 min. checks. (Two desks with students are combined, and a leader is selected within the group).

Each group is given tasks in 3 levels. Students choose tasks by level (as they wish). Solve problems involving constructing points. Discuss the formation under the supervision of the leader. Then the correct answer is displayed on the board using an overhead projector. Everyone checks that the projection of the points is done correctly. With the help of the group leader, grades are given on assignments and on self-control sheets (see. Appendix 2 And Appendix 3 ).

HEALTH-SAVING PAUSE. REFLECTION

“Pharaoh Pose”– sit on the edge of a chair, straighten your back, bend your arms at the elbows, cross your legs and place them on your toes. Inhale, tense all the muscles of the body while holding your breath, exhale. Do it 2-3 times. Close your eyes tightly, until they reach the stars, and open them. Mark your mood.

STAGE III. PRACTICAL PART. (Individual assignments)

There are task cards available to choose from at different levels. Students choose their own option according to their abilities. Find the projections of points on the surface of the object. Works are submitted and graded for the next lesson. (Cm. Appendix 4 , Appendix 5 , Appendix 6 ).

IV STAGE. FINAL

1) Homework assignment. (Instruction). Performed by levels:

B – understanding, at “3”. Exercise 1 Fig. 94a p. 77 – according to the assignment in the textbook: complete the missing projections of points on these projections.

B – application, at “4”. Exercise 1 Fig. 94 a, b. complete the missing projections and mark the vertices on the visual image in 94a and 94b.

A – analysis, “5”. (Increased difficulty.) Ex. 4 Fig.97 – construct the missing projections of the points and label them with letters. There is no visual image.

2)Reflexive analysis.

1. Determine the mood at the end of the lesson, mark it with any sign on the self-control sheet.
2. What new did you learn in class today?
3. What form of work is most effective for you: group, individual, and would you like it to be repeated in the next lesson?
4. Collect self-control sheets.

3)“The Mistaken Teacher”

Teacher: You have learned to construct projections of vertices, edges, faces and points on the surface of an object, following all the rules of construction. But they gave you a drawing that contains errors. Now try yourself as a teacher. Find the errors yourself, if you find all 8–6 errors, then the score is “5”; 5–4 errors – “4”, 3 errors – “3”.

Answers:

With rectangular projection, the system of projection planes consists of two mutually perpendicular projection planes (Fig. 2.1). They agreed to place one horizontally and the other vertically.

The projection plane located horizontally is called horizontal projection plane and denote sch, and the plane perpendicular to it is frontal plane of projectionsl 2. The system of projection planes itself is denoted p/p 2. Usually abbreviated expressions are used: 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 projection plane has front and rear, and the frontal plane has upper and lower floors.

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

All mechanical engineering drawings are images built on the same plane. In 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 flat n 2 left motionless, and the plane P rotate in the direction indicated by the arrows (see Fig. 2.1) around the axis OH at an angle of 90° until it aligns with the plane n 2. With this rotation, the front floor of the horizontal plane goes down, and the back goes up. After combining the planes they look like

married to fig. 2.2. It is believed that the projection planes are opaque and the observer is always in the first quarter. In Fig. 2.2 the designation of planes that are invisible after combining the floors is taken in brackets, as is customary for highlighting invisible figures in drawings.

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

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

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

When combining planes, a segment A "Ah, flat to 2, remains motionless, and the segment A "A x together with plane 71) will be rotated around the axis OH until aligned with plane 71 2. View of combined projection planes along with point projections A shown in Fig. 2.4, A. After combining the point A", Ax 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 communication line.

Drawing in Fig. 2.4, A can be greatly simplified. The designations of the combined projection planes are not marked in the drawings and the rectangles that conventionally limit the projection planes are not depicted, since the planes are unlimited. Simplified point drawing A(Fig. 2.4, b) also called diagram(from the French ?pure - drawing).

Shown in Fig. 2.3 quadrilateral AE4 "A H A" is a rectangle and its opposite sides equal and parallel. Therefore, the distance from the point A to 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 A to 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 A located in the first quarter and away from the plane n 2 at a smaller distance than from the plane 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 IN, 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 point C, specified in the third quarter (Fig. 2.6), is located above the axis OH, and the front one is lower.

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

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

In special 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 axis of projections. 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 one is on the axis OH. At the point E, located on a plane to 2(Fig. 2.9), horizontal projection on the axis OH, and the front one coincides with the point itself.

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 define the terms and provide information with illustrations. Let's consolidate the acquired knowledge by solving examples.

Projection, types of projection

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

Definition 1

Projection of a figure onto a plane– 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- this is the plane in which the image is constructed.

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

A special case of parallel projection is perpendicular projection or orthogonal: 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 by this they mean constructing a projection using the method of perpendicular projection. In special cases, of course, something else may be agreed.

Let us note the fact that the projection of a figure onto a plane is essentially a 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.

Let us recall that most often in geometry, when speaking about projection onto a plane, they mean the use of a perpendicular projection.

Let us make constructions that will give us the opportunity to obtain a definition of the projection of a point onto a plane.

Let's say a three-dimensional space is given, and in it there is a plane α and a point M 1 that does not belong to the plane α. Let's draw through given point M 1 straight A perpendicular given planeα. We denote the point of intersection of straight line a and plane α as H 1; by construction, it will serve as the base of a perpendicular lowered from point M 1 to 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

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

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

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

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

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

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

Obtain the equation of the plane α (if it is not specified). An article about the types of plane equations will help you here;

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

Find the coordinates of the point of intersection of the straight 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 we need for the projection of point M 1 onto the plane α.

Let's look at the theory with practical examples.

Example 1

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

Solution

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

Let us write down the canonical equations of a 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 line a is perpendicular to a given plane, the direction vector of line a is the normal vector of the plane 2 x - 3 y + z - 2 = 0. Thus, a → = (2, - 3, 1) – direction vector of straight line a.

Now let’s compose the canonical equations of a 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 required coordinates, the next step is to determine the coordinates of the intersection point of the straight line x + 2 2 = y - 4 - 3 = z - 4 1 and the plane 2 x - 3 y + z - 2 = 0 . For these purposes, we move 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 create 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 let's 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 required coordinates of a given point M 1 on a given plane α will be: (0, 1, 5).

Answer: (0 , 1 , 5) .

Example 2

IN rectangular system coordinates O x y z of three-dimensional space are given points A (0, 0, 2); B (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 down 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 - 6 y + 6 z - 12 = 0 ⇔ x - 2 y + 2 z - 4 = 0

Let us write down the parametric equations of the 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 = 0 has a normal vector with coordinates (1, - 2, 2), i.e. vector a → = (1, - 2, 2) – direction vector of straight line a.

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

Then we determine the coordinates of the intersection point of the plane x – 2 y + 2 z – 4 = 0 and the straight 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 for λ = - 1: x = - 1 + (- 1) y = - 2 - 2 · (- 1) z = 5 + 2 · (- 1) ⇔ x = - 2 y = 0 z = 3

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

Answer: (- 2 , 0 , 3) .

Let us separately dwell on the issue of finding the coordinates of the projection of a point onto coordinate planes and planes that are parallel to 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 coordinates of the projection of this point onto these planes will be, respectively: (x 1, y 1, 0), (x 1, 0, z 1) and (0, y 1, z 1). Let us also consider 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 a given point M 1 onto 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 point M 1 (x 1, y 1, z 1) onto the plane A x + D = 0. The remaining 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 direction vector of the line perpendicular to the O y z plane. Then the parametric equations of a straight line drawn through the point M 1 and perpendicular to a given plane will have the form:

x = x 1 + λ y = y 1 z = z 1

Let's find the coordinates of the intersection point of this line and the given plane. Let us first substitute the equalities into the equation A x + D = 0: x = x 1 + λ , y = y 1 , z = z 1 and get: A · (x 1 + λ) + D = 0 ⇒ λ = - D A - x 1

Then we calculate the required coordinates using the parametric equations of the straight line with 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 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 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 incompletely general equation plane z = 0. The projection of point M 1 onto the plane z = 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 down the coordinates of the projection of 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 an error in the text, please highlight it and press Ctrl+Enter

To construct images of a number of parts, you must be able to find the projections of individual points. For example, it is difficult to draw a top view of the part shown in Fig. 139, without constructing horizontal projections of points A, B, C, D, E, F, etc.

The problem of finding projections of points one at a time, given on the surface of an object, is solved as follows. First, the projections of the surface on which the point is located are found. Then, by drawing a connection line to the projection, where the surface is represented by a line, the second projection of the point is found. The third projection lies at the intersection of communication lines.

Let's look at an example.

Three projections of the part are given (Fig. 140, a). A horizontal projection a of point A lying on the visible surface is given. We need to find the remaining projections of this point.

First of all, you need to draw an auxiliary straight line. If two views are given, then the location of the auxiliary line in the drawing is chosen arbitrarily, to the right of the top view, so that the view on the left is at the required distance from the main view (Fig. 141).

If three views have already been constructed (Fig. 142, a), then the location of the auxiliary line cannot be chosen arbitrarily; you need to find the point through which it will pass. To do this, it is enough to continue the horizontal and profile projections of the symmetry axis until they mutually intersect and through the resulting point k (Fig. 142, b) draw a straight line segment at an angle of 45°, which will be the auxiliary straight line.

If there are no axes of symmetry, then the horizontal and profile projections of any face, projected in the form of straight segments, are continued until they intersect at point k 1 (Fig. 142, b).

Having drawn an auxiliary line, they begin to construct projections of the point (see Fig. 140, b).

The frontal a" and profile a" projections of point A must be located on the corresponding projections of the surface to which point A belongs. These projections are found. In Fig. 140, b they are highlighted in color. Draw communication lines as indicated by the arrows. At the intersections of the communication lines with the surface projections, the desired projections a" and a" are located.

The construction of projections of points B, C, D is shown in Fig. 140, in communication lines with arrows. The given point projections are colored. The connection lines are drawn to the projection on which the surface is depicted as a line, and not as a figure. Therefore, first find the frontal projection from point C. The profile projection from point C is determined by the intersection of communication lines.

If the surface is not represented by a line on any projection, then an auxiliary plane must be used to construct projections of points. For example, given a frontal projection d of point A lying on the surface of a cone (Fig. 143, a). An auxiliary plane is drawn through the point parallel to the base, which will intersect the cone in a circle; its frontal projection is a straight segment, and its horizontal projection is a circle with a diameter equal to the length of this segment (Fig. 143, b). By drawing a connection line to this circle from point a, a horizontal projection a of point A is obtained.

The profile projection a" of point A is found in the usual way at the intersection of communication lines.

Using the same technique, you can find the projections of a point lying, for example, on the surface of a pyramid or ball. When a pyramid is intersected by a plane parallel to the base and passing through a given point, a figure similar to the base is formed. On the projections of this figure lie the projections of a given point.

Answer the questions

1. At what angle is the auxiliary straight line drawn?

2. Where do you draw the auxiliary straight line if front and top views are given, but you need to construct a view on the left?

3. How to determine the location of an auxiliary line if there are three types?

4. What is the method for constructing projections of a point based on one given point, if one of the surfaces of an object is represented by a line?

5. For what geometric bodies and in what cases are the projections of a point given on their surface found using an auxiliary plane?

Assignments for § 20

Exercise 68

Write to workbook, what projections of the points indicated by numbers on the views correspond to the points indicated on the visual image by letters in the example indicated to you by the teacher (Fig. 144, a-d).

Exercise 69

In Fig. 145, a-b letters indicated by only one projection of some of the vertices. In the example given to you by your teacher, find the remaining projections of these vertices and label them with letters. In one of the examples, construct the missing projections of the points specified on the edges of the object (Fig. 145, d and e). Highlight in color the projections of the edges on which the points are located. Complete the task on transparent paper, placing it on the textbook page. There is no need to redraw Fig. 145.

Exercise 70

Find the missing projections of points defined by one projection on the visible surfaces of the object (Fig. 146). Label them with letters. Highlight the given projections of points in color. A visual image will help you solve the problem. The task can be completed either in a workbook or on transparent paper, overlaying it on a textbook page. In the latter case, redraw the figure. 146 is not necessary.

Exercise 71

In the example given to you by your teacher, redraw the three views (Fig. 147). Construct the missing projections of the points specified on the visible surfaces of the object. Highlight the given projections of points in color. Label all projections of points with letters. To construct projections of points, use an auxiliary straight line. Complete a technical drawing and mark the specified points on it.

Let's consider the projections of points onto two planes, for which we take two perpendicular planes (Fig. 4), which we will call horizontal frontal and planes. The line of intersection of these planes is called the projection axis. We project one point A onto the considered planes using a plane projection. To do this, it is necessary to lower the perpendiculars Aa and A from a given point onto the considered planes.

The projection onto the horizontal plane is called horizontal projection points A, and the projection A? on the frontal plane is called frontal projection.

Points to be projected are usually denoted in descriptive geometry using capital letters A, B, C. Small letters are used to indicate horizontal projections of points a, b, c... Frontal projections are indicated in small letters with a stroke at the top a?, b?, c?…

Points are also designated by Roman numerals I, II,... and for their projections - by Arabic numerals 1, 2... and 1?, 2?...

By rotating the horizontal plane by 90°, you can get a drawing in which both planes are in the same plane (Fig. 5). This picture is called diagram of a point.

Through perpendicular lines Ahh And Huh? Let's draw a plane (Fig. 4). The resulting plane is perpendicular to the frontal and horizontal planes because it contains perpendiculars to these planes. Therefore, this plane is perpendicular to the line of intersection of the planes. The resulting straight line intersects the horizontal plane in a straight line ahh x, and the frontal plane – in a straight line a?a X. Straight aah and a?a x are perpendicular to the axis of intersection of the planes. That is Aahaha? is a rectangle.

When combining horizontal and frontal projection planes A And A? will lie on the same perpendicular to the axis of intersection of the planes, since when the horizontal plane rotates, the perpendicularity of the segments ahh x and a?a x will not be broken.

We get that on the projection diagram A And A? some point A always lie on the same perpendicular to the axis of intersection of the planes.

Two projections a and A? of a certain point A can unambiguously determine its position in space (Fig. 4). This is confirmed by the fact that when constructing a perpendicular from projection a to the horizontal plane, it will pass through point A. In the same way, a perpendicular from projection A? to the frontal plane will pass through the point A, i.e. point A is simultaneously on two specific straight lines. Point A is their point of intersection, that is, it is definite.

Consider a rectangle Aaa X A?(Fig. 5), for which the following statements are true:

1) Point distance A from the frontal plane is equal to the distance of its horizontal projection a from the axis of intersection of the planes, i.e.

Huh? = ahh X;

2) point distance A from the horizontal plane of projections is equal to the distance of its frontal projection A? from the axis of intersection of the planes, i.e.

Ahh = a?a X.

In other words, even without the point itself on the diagram, using only its two projections, you can find out at what distance a given point is located from each of the projection planes.

The intersection of two projection planes divides space into four parts, which are called in quarters(Fig. 6).

The axis of intersection of the planes divides the horizontal plane into two quarters - the front and rear, and the frontal plane - into the upper and lower quarters. The upper part of the frontal plane and the anterior part of the horizontal plane are considered as the boundaries of the first quarter.

When receiving the diagram, the horizontal plane rotates and is aligned with the frontal plane (Fig. 7). In this case, the front part of the horizontal plane will coincide with the bottom part of the frontal plane, and the back part of the horizontal plane will coincide with the top part of the frontal plane.

Figures 8-11 show points A, B, C, D, located in different quarters of space. Point A is located in the first quarter, point B is in the second, point C is in the third and point D is in the fourth.

When the points are located in the first or fourth quarters of them horizontal projections are on the front part of the horizontal plane, and on the diagram they will lie below the axis of intersection of the planes. When a point is located in the second or third quarter, its horizontal projection will lie on the back of the horizontal plane, and on the diagram it will be located above the axis of intersection of the planes.

Frontal projections points that are located in the first or second quarters will lie on the upper part of the frontal plane, and on the diagram they will be located above the axis of intersection of the planes. When a point is located in the third or fourth quarter, its frontal projection is below the axis of intersection of the planes.

Most often, in real constructions, the figure is placed in the first quarter of space.

In some special cases, the point ( E) can lie on a horizontal plane (Fig. 12). In this case, its horizontal projection e and the point itself will coincide. The frontal projection of such a point will be located on the axis of intersection of the planes.

In the case when the point TO lies on the frontal plane (Fig. 13), its horizontal projection k lies on the axis of intersection of the planes, and the frontal k? shows the actual location of this point.

For such points, a sign that it lies on one of the projection planes is that one of its projections is on the axis of intersection of the planes.

If a point lies on the axis of intersection of the projection planes, it and both of its projections coincide.

When a point does not lie on the projection planes, it is called dot general position . In what follows, if there are no special marks, the point in question is a point in general position.

2. Lack of projection axis

To explain how to obtain projections of a point on a model perpendicular to the projection plane (Fig. 4), it is necessary to take a piece of thick paper in the shape of an elongated rectangle. It needs to be bent between projections. The fold line will represent the axis of intersection of the planes. If after this the bent piece of paper is straightened again, we will get a diagram similar to the one shown in the figure.

By combining two projection planes with the drawing plane, it is possible not to show the fold line, i.e., not to draw the axis of intersection of the planes on the diagram.

When plotting on a diagram, you should always place projections A And A? point A on one vertical line (Fig. 14), which is perpendicular to the axis of intersection of the planes. Therefore, even if the position of the axis of intersection of the planes remains uncertain, but its direction is determined, the axis of intersection of the planes can only be located on the diagram perpendicular to the straight line huh?.

If there is no projection axis on the diagram of a point, as in the first Figure 14 a, you can imagine the position of this point in space. To do this, draw anywhere perpendicular to the straight line huh? projection axis, as in the second figure (Fig. 14) and bend the drawing along this axis. If we restore perpendiculars at points A And A? before they intersect, you can get a point A. When changing the position of the projection axis, different positions of the point relative to the projection planes are obtained, but the uncertainty in the position of the projection axis does not affect mutual arrangement several points or figures in space.

3. Projections of a point onto three projection planes

Let's consider the profile plane of projections. Projections onto two perpendicular planes usually determine the position of a figure and make it possible to find out its real size and shape. But there are times when two projections are not enough. Then the construction of the third projection is used.

The third projection plane is drawn so that it is perpendicular to both projection planes simultaneously (Fig. 15). The third plane is usually called profile.

In such constructions, the common straight line of the horizontal and frontal planes is called axis X , the common straight line of the horizontal and profile planes – axis at , and the common straight line of the frontal and profile planes is axis z . Dot ABOUT, which belongs to all three planes, is called the origin point.

Figure 15a shows the point A and three of its projections. Projection onto the profile plane ( A??) are called profile projection and denote A??.

To obtain a diagram of point A, which consists of three projections a, a, a, it is necessary to cut the trihedron formed by all the planes along the y-axis (Fig. 15b) and combine all these planes with the plane of the frontal projection. The horizontal plane must be rotated about the axis X, and the profile plane is about the axis z in the direction indicated by the arrow in Figure 15.

Figure 16 shows the position of the projections huh, huh? And A?? points A, obtained by combining all three planes with the drawing plane.

As a result of the cut, the y-axis appears in two different places on the diagram. On a horizontal plane (Fig. 16) it takes a vertical position (perpendicular to the axis X), and on the profile plane – horizontal (perpendicular to the axis z).

There are three projections in Figure 16 huh, huh? And A?? points A have a strictly defined position on the diagram and are subject to unambiguous conditions:

A And A? should always be located on the same vertical line, perpendicular to the axis X;

A? And A?? should always be located on the same horizontal straight line, perpendicular to the axis z;

3) when carried out through a horizontal projection and a horizontal straight line, and through a profile projection A??– a vertical straight line, the constructed straight lines will necessarily intersect on the bisector of the angle between the projection axes, since the figure Oa at A 0 A n – square.

When constructing three projections of a point, you need to check whether all three conditions are met for each point.

4. Point coordinates

The position of a point in space can be determined using three numbers called its coordinates. Each coordinate corresponds to the distance of a point from some projection plane.

Determined point distance A to the profile plane is the coordinate X, wherein X = huh?Huh(Fig. 15), the distance to the frontal plane is coordinate y, and y = huh?Huh, and the distance to the horizontal plane is the coordinate z, wherein z = aA.

In Figure 15, point A occupies the width rectangular parallelepiped, and the measurements of this parallelepiped correspond to the coordinates of this point, i.e., each of the coordinates is presented in Figure 15 four times, i.e.:

x = a?A = Oa x = a y a = a z a?;

y = а?А = Оа y = а x а = а z а?;

z = aA = Oa z = a x a? = a y a?.

In the diagram (Fig. 16), the x and z coordinates appear three times:

x = a z a?= Oa x = a y a,

z = a x a? = Oa z = a y a?.

All segments that correspond to the coordinate X(or z), are parallel to each other. Coordinate at represented twice by an axis located vertically:

y = Oa y = a x a

and twice – located horizontally:

y = Oa y = a z a?.

This difference appears due to the fact that the y-axis is present on the diagram in two different positions.

It should be taken into account that the position of each projection is determined on the diagram by only two coordinates, namely:

1) horizontal – coordinates X And at,

2) frontal – coordinates x And z,

3) profile – coordinates at And z.

Using coordinates x, y And z, you can construct projections of a point on a diagram.

If point A is given by coordinates, their recording is defined as follows: A ( X; y; z).

When constructing point projections A the following conditions must be checked:

1) horizontal and frontal projections A And A? X X;

2) frontal and profile projections A? And A? must be located at the same perpendicular to the axis z, since they have a common coordinate z;

3) horizontal projection and also removed from the axis X, like profile projection A away from the axis z, since projections ah? and eh? have a common coordinate at.

If a point lies in any of the projection planes, then one of its coordinates is equal to zero.

When a point lies on the projection axis, two of its coordinates are equal to zero.

If a point lies at the origin, all three of its coordinates are zero.