The article reveals the concept of perpendicularity of a straight line and a plane, gives a definition of a straight line, a plane, graphically illustrated and shows the designation of perpendicular lines and a plane. Let us formulate a sign of perpendicularity of a straight line with a plane. Consider the conditions under which a straight line and a plane will be perpendicular with the given equations in the plane and three-dimensional space. Everything will be shown with examples.

Yandex.RTB R-A-339285-1 Definition 1

The line is perpendicular to the plane when it is perpendicular to any line lying in this plane.

It is true that the plane is perpendicular to the line, as well as the line to the plane.

Perpendicularity is indicated by "⊥". If the condition specifies that the line c is perpendicular to the plane γ , then the notation is c ⊥ γ .

For example, if the line is perpendicular to the plane, then it is possible to draw only one line, due to which two adjacent walls of the room will intersect. The line is considered perpendicular to the plane of the ceiling. The rope located in the gym is considered as a straight line segment that is perpendicular to the plane, in this case semi.

If there is a perpendicular line to the plane, the angle between the line and the plane is considered right, that is, equal to 90 degrees.

Perpendicularity of a straight line and a plane - a sign and conditions of perpendicularity

To find the detection of perpendicularity, it is necessary to use a sufficient condition for the perpendicularity of a line and a plane. It guarantees that the line and the plane are perpendicular. This condition is considered sufficient and is called a sign of perpendicularity of a line and a plane.

Theorem 1

For a given line to be perpendicular to a plane, it is sufficient that the line be perpendicular to two intersecting lines that lie in this plane.

A detailed proof is given in the geometry textbook of grades 10-11. The theorem is used to solve problems where it is necessary to establish the perpendicularity of a line and a plane.

Theorem 2

Provided that at least one of the lines is parallel to the plane, it is considered that the second line is also perpendicular to this plane.

The sign of perpendicularity of a straight line and a plane has been considered since school, when it is necessary to solve problems in geometry. Let us consider in more detail one more necessary and sufficient condition under which the line and the plane will be perpendicular.

Theorem 3

In order for the line a to be perpendicular to the plane γ, a necessary and sufficient condition is the collinearity of the directing vector of the line a and the normal vector of the plane γ.

Proof

For a → = (a x , a y , a z) being a vector of the line a , for n → = (n x , n y , n z) being a normal vector of the plane γ to fulfill perpendicularity, it is necessary that the line a and the plane γ belong to the fulfillment of the condition of collinearity of the vectors a → = (a x , a y , a z) and n → = (n x , n y , n z) . Hence we get that a → = t n → ⇔ a x = t n x a y = t n y a z = t n z , t is a real number.

This proof is based on the necessary and sufficient condition of perpendicularity of the line and the plane, the directing vector of the line and the normal vector of the plane.

This condition is applicable to prove the perpendicularity of a straight line and a plane, since it is enough to find the coordinates of the directing vector of the straight line and the coordinates of the normal vector in three-dimensional space, and then perform calculations. It is used for cases when a straight line is defined by an equation of a straight line in space, and a plane by an equation of a plane of some kind.

Example 1

Prove that the given line x 2 - 1 = y - 1 2 = z + 2 2 - 7 is perpendicular to the plane x + 2 2 + 1 y - (5 + 6 2) z .

Decision

The denominators of the canonical equations are the coordinates of the direction vector of the given line. Hence we have that a → = (2 - 1 , 2 , 2 - 7) is the directing vector of the line x 2 - 1 = y - 1 2 = z + 2 2 - 7 .

In the general equation of the plane, the coefficients in front of the variables x, y, z are the coordinates of the normal vector of the given plane. It follows that n → = (1 , 2 (2 + 1) , - (5 + 6 2)) is the normal vector of the plane x + 2 2 + 1 y - (5 + 6 2) z - 4 = 0

It is necessary to check the fulfillment of the condition. We get that

2 - 1 \u003d t 1 2 \u003d t 2 (2 + 1) 2 \u003d t (- (5 + 6 2)) ⇔ t \u003d 2 - 1, then the vectors a → and n → are related by the expression a → = ( 2 - 1) n → .

This is the collinearity of vectors. it follows that the line x 2 - 1 \u003d y - 1 2 \u003d z + 2 2 - 7 is perpendicular to the plane x + 2 (2 + 1) y - (5 + 6 2) z - 4 \u003d 0.

Answer: line and plane are perpendicular.

Example 2

Determine if the line y - 1 = 0 x + 4 z - 2 = 0 and the plane x 1 2 + z - 1 2 = 1 are perpendicular.

Decision

To answer the question of perpendicularity, it is necessary that the necessary and sufficient condition be satisfied, that is, first you need to find the vector of the given line and the normal vector of the plane.

From the straight line y - 1 = 0 x + 4 z - 2 = 0, it can be seen that the direction vector a → is the product of the normal vectors of the plane y - 1 = 0 and x + 4 z - 2 = 0 .

Hence we get that a → = i → j → k → 0 1 0 1 0 4 = 4 i → - k → .

The coordinates of the vector a → = (4 , 0 , - 1) .

The equation of the plane in the segments x 1 2 + z - 1 2 = 1 is equivalent to the equation of the plane 2 x - 2 z - 1 = 0 , whose normal vector is equal to n → = (2 , 0 , - 2) .

You should check for collinearity of the vectors a → = (4 , 0 , - 1) and n → = (2 , 0 , - 2) .

To do this, we write:

4 = t 2 0 = t 0 - 1 = t (- 2) ⇔ t = 2 t ∈ R ⇔ t ∈ ∅ t = 1 2

From this we conclude that the directing vector of the straight line is not collinear with the normal vector of the plane. So y - 1 = 0 x + 4 z - 2 = 0 is a straight line not perpendicular to the plane x 1 2 + z - 1 2 .

Answer: line and plane are not perpendicular.

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Outline of a lesson in geometry in grade 10 on the topic "Perpendicularity of a line and a plane"

Lesson Objectives:

educational

introduction of a sign of perpendicularity of a straight line and a plane;

to form students' ideas about the perpendicularity of a straight line and a plane, their properties;

to form the ability of students to solve typical problems on the topic, the ability to prove statements;

developing

develop independence, cognitive activity;

develop the ability to analyze, draw conclusions, systematize the information received,

develop logical thinking;

develop spatial imagination.

educational

education of the culture of speech of students, perseverance;

instill in students an interest in the subject.

Lesson type: Lesson of studying and primary consolidation of knowledge.

Forms of student work: front poll.

Equipment: computer, projector, screen.

Literature:"Geometry 10-11", Textbook. Atanasyan L.S. and etc.

(2009, 255p.)

Lesson plan:

Organizational moment (1 minute);

Updating knowledge (5 minutes);

Learning new material (15 minutes);

Primary consolidation of the studied material (20 minutes);

Summing up (2 minutes);

Homework (2 minutes).

During the classes.

Organizational moment (1 minute)

Greeting students. Checking the readiness of students for the lesson: checking the availability of notebooks, textbooks. Checking absenteeism.

Knowledge update (5 minutes)

Teacher. Which line is called perpendicular to the plane?

Student. A line perpendicular to any line lying in this plane is called a line perpendicular to this plane.

Teacher. How does the lemma about two parallel lines perpendicular to a third sound?

Student. If one of two parallel lines is perpendicular to a third line, then the other line is also perpendicular to this line.

Teacher. Theorem on the perpendicularity of two parallel lines to a plane.

Student. If one of two parallel lines is perpendicular to a plane, then the other line is also perpendicular to that plane.

Teacher. What is the inverse of this theorem?

Student. If two lines are perpendicular to the same plane, then they are parallel.

Checking homework

Homework is checked if students have difficulty solving it.

Learning new material (15 minutes)

Teacher. You and I know that if a line is perpendicular to a plane, then it will be perpendicular to any line lying in this plane, but in the definition, the perpendicularity of a line to a plane is given as a fact. In practice, it is often necessary to determine whether the line will be perpendicular to the plane or not. Such examples can be given from life: during the construction of buildings, piles are driven in perpendicular to the surface of the earth, otherwise the structure may collapse. The definition of a straight line perpendicular to the plane cannot be used in this case. Why? How many lines can be drawn in a plane?

Student. There are infinitely many straight lines that can be drawn in a plane.

Teacher. Correctly. And it is impossible to check the perpendicularity of a straight line to each individual plane, since it will take an infinitely long time. In order to understand whether a line is perpendicular to a plane, we introduce the sign of perpendicularity of a line and a plane. Write in your notebook. If a line is perpendicular to two intersecting lines lying in a plane, then it is perpendicular to that plane.

Notebook entry. If a line is perpendicular to two intersecting lines lying in a plane, then it is perpendicular to that plane.

Teacher. Thus, we do not need to check the perpendicularity of a line for each straight plane, it is enough to check the perpendicularity for only two lines of this plane.

Teacher. Let's prove this sign.

Given: p and q- straight, p ∩ q = O, a⊥ p, a⊥ q, p ϵ α, q ϵ α.

Prove: a⊥ α.

Teacher. And yet, for proof, we use the definition of a straight line perpendicular to the plane, how does it sound?

Student. If a line is perpendicular to a plane, then it is perpendicular to any line lying in that plane.

Teacher. Correctly. Draw any line m in the plane α. Draw a line l ║ m through the point O. On the line a mark points A and B so that the point O is the midpoint of the segment AB. Let's draw the line z in such a way that it intersects the lines p, q, l, the points of intersection of these lines will be denoted by P, Q, L, respectively. Connect the ends of the segment AB with points P, Q and L.

Teacher. What can we say about the triangles ∆APQ and ∆BPQ ?

Student. These triangles will be equal (according to the 3rd criterion for the equality of triangles).

Teacher. Why?

Student. Because lines p and q are perpendicular bisectors, then AP = BP , AQ = BQ , and side PQ is common.

Teacher. Correctly. What can we say about the triangles ∆APL and ∆BPL ?

Student. These triangles will also be equal (according to 1 sign of equality of triangles).

Teacher. Why?

Student. AP = BP, PL- common side APL =  BPL(from the equality ∆ APQ and ∆ BPQ)

Teacher. Correctly. So AL = BL . So what will be ∆ALB ?

Student. So ∆ALB will be isosceles.

Teacher. LO is the median in ∆ALB, so what will it be in this triangle?

Student. So LO will also be the height.

Teacher. Hence the straight linelwill be perpendicular to the linea. And since the straight linelis any line belonging to the plane α, then by definition the linea⊥ a. Q.E.D.

Proven with presentation

Teacher. But what if the line a does not intersect the point O, but remains perpendicular to the lines p and q? If the line a intersects any other point of the given plane?

Student. It is possible to construct a line 1 , which will be parallel to the line a, will intersect the point O, and by the lemma on two parallel lines perpendicular to the third, we can prove thata 1 ⊥ p, a 1 ⊥ q.

Teacher. Correctly.

Primary consolidation of the studied material (20 minutes)

Teacher. In order to consolidate the material we have studied, we will solve the number 126. Read the task.

Student. Line MB is perpendicular to sides AB and BC of triangle ABC. Determine the type of triangle MBD, where D is an arbitrary point of the straight line AC.

Picture.

Given: ∆ ABC, MB⊥ BA, MB⊥ BC, D ϵ AC.

Find: ∆ MBD.

Decision.

Teacher. Can you draw a plane through the vertices of a triangle?

Student. Yes, you can. The plane can be drawn at three points.

Teacher. How will the lines BA and CB be located relative to this plane?

Student. These lines will lie in this plane.

Teacher. It turns out that we have a plane, and there are two intersecting lines in it. How does the line MW relate to these lines?

Student. Direct MW⊥ VA, MV ⊥ BC.

Writing on the board and in notebooks. Because MV⊥ VA, MV ⊥ VS

Teacher. If a line is perpendicular to two intersecting lines lying in a plane, then the line will belong to this plane?

Student. The straight line MB will be perpendicular to the plane ABC.

⊥ ABC.

Teacher. Point D is an arbitrary point on the segment AC, so how will the line BD relate to the plane ABC?

Student. So BD belongs to the plane ABC.

Writing on the board and in notebooks. Because BD ϵ ABC

Teacher. What will be the lines MB and BD relative to each other?

Student. These lines will be perpendicular by the definition of a line perpendicular to the plane.

Writing on the board and in notebooks. ↔ MV⊥ BD

Teacher. If MB is perpendicular to BD, then what will be the triangle MBD?

Student. Triangle MBD will be right angled.

Writing on the board and in notebooks. ↔ ∆MBD – rectangular.

Teacher. Correctly. Let's solve number 127. Read the task.

Student. In a triangleABC sum of angles A and Bequals 90°. StraightBDperpendicular to the planeABC. Prove that CD⊥ AC.

The student goes to the blackboard. Draws a drawing.

Write on the board and in a notebook.

Given: ∆ ABC,  A +  B= 90°, BD⊥ ABC.

Prove: CD⊥ AC.

Proof:

Teacher. What is the sum of the angles of a triangle?

Student. The sum of the angles in a triangle is 180°.

Teacher. What is angle C in triangle ABC?

Student. Angle C in triangle ABC will be 90°.

Writing on the board and in notebooks. C = 180° - A- B= 90°

Teacher. If angle C is 90°, how do lines AC and BC lie relative to each other?

Student. Means AC⊥ Sun.

Writing on the board and in notebooks. ↔ AC⊥ Sun

Teacher. Line BD is perpendicular to plane ABC. What follows from this?

Student. So BD is perpendicular to any line from ABC .

BD⊥ ABC ↔ BDperpendicular to any lineABC(a-priory)

Teacher. In accordance with this, how will direct BD and AC be related?

Student. So these lines are perpendicular.

BD⊥ AC

Teacher. AC is perpendicular to two intersecting lines lying in the plane DBC, but AC does not pass through the intersection point. How to fix it?

Student. Draw a line through point B and parallel AC. Since AC is perpendicular to BC and BD, then a will also be perpendicular to BC and BD by the lemma.

Writing on the board and in notebooks. Draw a line through point B a ║AC ↔ a⊥ BC, and ⊥ BD

Teacher. If the line a is perpendicular to BC and BD, then what can be said about the relative position of the line a and the plane BDC?

Student. This means that the line a will be perpendicular to the plane BDC, and hence the line AC will be perpendicular to BDC.

Writing on the board and in notebooks. ↔ a⊥ bdc↔ AC ⊥ bdc.

Teacher. If AC is perpendicular to BDC, then how will the lines AC and DC be located relative to each other?

Student. AC and DC will be perpendicular by the definition of a line perpendicular to the plane.

Writing on the board and in notebooks. Because AC⊥ bdc↔ AC ⊥ DC

Teacher. Well done. Let's solve number 129. Read the assignment.

Student. StraightAMperpendicular to the plane of the squareABCD, whose diagonals intersect at point O. Prove that: a) the lineBDperpendicular to the planeAMO; b)MO⊥ BD.

A student comes to the board. Draws a drawing.

Write on the board and in a notebook.

Given:ABCD- square,AM⊥ ABCD, AC ∩ BD = O

Prove:BD⊥ AMO, MO⊥ BD

Proof:

Teacher. We need to prove that theBD⊥ AMO. What conditions must be met for this to happen?

Student. It is necessary that the direct BD is perpendicular to at least two intersecting lines from the plane AMO.

Teacher. The condition says that BD perpendicular to two intersecting lines AMO?

Student. No.

Teacher. But we know that AM perpendicular ABCD . What conclusion can be drawn from this?

Student. Means what AM perpendicular to any line from this plane, i.e. AM perpendicular B.D.

AM⊥ ABCD ↔ AM⊥ BD(a-priory).

Teacher. One line is perpendicular BD there is. Pay attention to the square, how the lines will be located relative to each other AC and BD?

Student. AC will be perpendicular BD by the property of the diagonals of a square.

Write on the board and in a notebook. BecauseABCD- square, thenAC⊥ BD(by the property of the diagonals of a square)

Teacher. We have found two intersecting lines lying in a plane AMO perpendicular to the line BD . What follows from this?

Student. Means what BD perpendicular to the plane AMO.

Writing on the board and in notebooks. BecauseAC⊥ BDandAM⊥ BD ↔ BD⊥ AMO(by sign)

Teacher. Which line is called the line perpendicular to the plane?

Student. A line is said to be perpendicular to a plane if it is perpendicular to any line in that plane.

Teacher. How are the lines related to each other? BD and OM?

Student. Means BD perpendicular OM . Q.E.D.

Writing on the board and in notebooks. ↔BD⊥ MO(a-priory). Q.E.D.

Debriefing (2 minutes)

Teacher. Today we studied the sign of perpendicularity of a line and a plane. How does it sound?

Student. If a line is perpendicular to two intersecting lines lying in a plane, then this line is perpendicular to this plane.

Teacher. Correctly. We have learned to apply this feature in solving problems. Who answered at the blackboard and helped from the place, well done.

Homework (2 minutes)

Teacher. Paragraph 1, paragraphs 15-17, learn: lemma, definition and all theorems. No. 130, 131.

In order for a straight line in space to be  of the plane, it is necessary and sufficient that on the diagram the horizontal projection of the straight line be  of the horizontal projection of the horizontal, and the frontal projection to the frontal projection of the front of this plane.

Determining the distance from a point to a plane(Fig. 19)

1. From the point, lower the perpendicular to the plane (for this, in the plane

hold h, f);

2. Find the point of intersection of the straight line with the plane (see Fig. 18);

3.Find n.v. perpendicular segment (see Fig. 7).

The second section Method for replacing projection planes

(to tasks 5, 6.7)

This geometric figure is left motionless in the system of projection planes. New projection planes are set so that the projections obtained on them provide a rational solution to the problem under consideration. Moreover, each new system of projection planes must be an orthogonal system. After projecting objects on the plane, they are combined into one by rotating them around the common straight lines (projection axes) of each pair of mutually perpendicular planes.

For example, let the point A be set in the system of two planes P 1 and P 2. Let's supplement the system with one more plane P 4 (Fig. 20), P 1 P 4. It has a common line X 14 with the plane P 1 . We build the projection A 4 on P 4.

AA 1 \u003d A 2 A 12 \u003d A 4 A 14.

On fig. 21, where the planes P 1, P 2 and P 4 are brought into alignment, this fact is determined by the result A 1 A 4 X 14, and A 14 A 4 A 2 A 12.

The distance of the new point projection to the new projection axis (A 4 A 14) is equal to the distance from the replaced point projection to the replaced axis (A 2 A 12).

A large number of metric problems of descriptive geometry are solved on the basis of the following four problems:

1. Transformation of a general position line into a level line (Fig. 22):

a) P 4 || AB (X-axis 14 || A 1 B 1);

b) A 1 A 4 X 14; B 1 B 4 X 14;

c) A 4 A 14 \u003d A 12 A 2;

V 4 V 14 = V 12 V 2 ;

A 4 B 4 - present

2. Transformation of a straight line in general position into a projecting one (Fig. 23):

a) P 4 || AB (X 14 || A 1 B 1);

A 1 A 4 X 14;

B 1 B 4 X 14;

A 14 A 4 \u003d A 12 A 2;

14V 4 = 12V 2 ;

A 4 B 4 - n.v.;

b) P 5 AB (X 45 A 4 V 4);

A 4 A 5 X 45;

B 4 B 5 X 45;

A 45 A 5 \u003d B 45 B 5 \u003d A 14 A 1 \u003d B 14 B 1;

3. Transformation of a plane of general position into a projecting position (Fig. 24):

A plane can be brought into a projecting position if one straight line of the plane is made projecting. Let's draw a horizontal line (h 2 ,h 1) in the ABC plane, which can be made projective in one transformation. Let's draw a plane P 4 perpendicular to the horizontal; it is projected onto this plane by a point, and the plane of the triangle is projected by a straight line.

4. Transformation of a generic plane into a level plane (Fig. 25).

Make the plane a level plane using two transformations. First, the plane must be made projecting (see Fig. 25), and then P 5 || A 4 B 4 C 4, we get A 5 B 5 C 5 - n.v.

Task #5

Determine the distance from point C to a straight line in general position (Fig. 26).

The solution comes down to the 2nd main problem. Then the distance along the diagram is defined as the distance between two points

A 5  B 5  D 5 and C 5.

Projection С 4 D 4 || X 45.

Task #6

Determine the distance from ()D to the plane given by points A, B, C (Fig. 27).

The problem is solved using the 2nd main problem. The distance (E 4 D 4), from () D 4 to the straight line A 4 C 4 B 4, into which the plane ABC was projected, is the natural value of the segment ED.

Projection D 1 E 1 || X 14;

E 2 E X12 = E 4 E X14.

Build your own D 1 E 1.

Build your own D 2 E 2.

Task #7

Determine the actual size of the triangle ABC (see the solution of the 4th main problem) (Fig. 25)

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umbrellas. Segment KL determines the direction of the projections of the line of intersection of two given planes.

2.8 Perpendicularity of a line and a plane, two planes

The condition of perpendicularity of a line and a plane and perpendicularity of two planes is based on the right angle projection theorem. Adapting the theorem to solve metric problems for determining the distance from a point to a plane, determining the distance from a point to a straight line, or for constructing a plane parallel to a given one at a certain distance, we formulate the condition of perpendicularity of a straight line and a plane.

The line l (l1 ,l2 ) is perpendicular to the plane , if it is perpendicular to two intersecting level lines (for example, horizontal and frontal) belonging to the given plane.

l 1h 1

l 2f 2

Consider examples of solving typical metric problems on the application of the condition of perpendicularity of a straight line and a plane.

Example 1. Determine the distance from the point N to the plane Q(mIIn) (Figure 2.35).

Algorithm for solving the problem:

1. Analyze the condition of the problem. (The shortest distance from a point to a line is determined by the perpendicular dropped from point N to plane Q.)

2. In order to fulfill the condition of perpendicularity of a straight line and a plane, it is necessary first to construct a horizontal h (h 1, h 2 ) and a frontal f (f 1 , f 2 ) in the plane, and then construct a line l (l 1 , l 2 ) perpendicular to the plane Q ( figure 2.35).

Figure 2.35 - Line perpendicular to the plane

3. Find the base of the perpendicular, i.e. the point of intersection of the constructed line l(l 1 , l 2 ) with a given plane Q. To construct a point K, we conclude, for example, the frontal projection of the line l 2 into the frontally projecting plane Σ. We determine the projections of the line of intersection of the line l with the corresponding projection of the line of intersection of two planes (Q∩∑). We determine the position of the projections of the point K1 and K2.

4. Determine the actual size of the segment NK as the hypotenuse of a right triangle (Figure 2.36).

Figure 2.36 - Projections of the distance from a point to a plane

Example 2. Determine the distance from point A to line n. Algorithm for solving the problem:

1. Analysis of the conditions of the problem. After analyzing the condition of the problem, we state that the shortest distance from a point to a line is measured by a perpendicular dropped from point A to line n. Since the given line n (n 1 , n2 ) is a line in general position, then to solve the problem it is necessary to perform additional constructions.

2. Through the projections of the point A(A 1 ,А2 ) we construct a plane Σ (h ∩ f) perpendicular to the line n (n1 , n2 ).

3. Determine the point of intersection of the given line n(n 1 , n2 ) with the plane Σ (h ∩ f) and find the projections of the line segment A1 B1 and A2 B2 as projections of the distance from point A to line n.

4. We build the natural value of the distance from point A to the straight line n (Figure 2.37).

Figure 2.37 - Distance from point A to straight line n

Example 3. Construct a plane Θ, parallel to the plane Σ (ΔABC), at a distance of 25 mm from it.

Algorithm for solving the problem:

1. Analysis of the conditions of the problem. The plane will be built at a distance of 25 mm from the plane Σ (ΔABC). Therefore, you need to build a perpendicular to the plane.

2. To construct a line perpendicular to the plane, we set the level lines in the plane - the horizontal h(h 1 , h2 ) and frontal f(f1 , f2 ) and build a line l(l 1 , l 2 ) perpendicular to the plane Σ (ΔАВС) (Figure 2.38).

Figure 2.38 - Position of point L

3. Find the base of the perpendicular, i.e. point K (K1, K2) of the intersection of the line l (l 1, l 2) with the plane Σ (ΔABS).

4. Choose on line l(l 1 , l 2 ) arbitrary point N(N1 ,N2 ) and determine the distance from the chosen point to the plane (N1 Kº).

5. We find on a straight line l(l 1 , l 2 ) the position of the point L(L1 , L2 ) having a distance from the plane of 25 mm.

6. Through the point L(L 1 , L2 ) we construct a plane Θ(m∩n) parallel to the given plane Σ (ΔАВС) (Figure 2.39)

Figure 2.39 - Plane parallel to the given one at the required distance

Questions for self-control on topic 2:

1. What is the position of the point relative to the straight line?

2. When does a point belong to a straight line?

3. How can straight lines be arranged relative to each other?

4. What points are called competing?

5. Continue the sentence: A right angle is projected onto the frontal projection plane without distortion if it is formed by two intersecting straight lines, one of which is a straight line in general position, and the second ...... ..

6. How to determine the natural size of a line segment in general position?

7. What is the condition for a straight line and a plane to be perpendicular?

8. What is the condition for two planes to be perpendicular?

9. When is a line parallel to a plane?

10. When are two planes parallel?

11. What is the condition for a straight line to belong to a plane?

12. When does a point belong to a plane?

13. What is the algorithm for finding the point of intersection of a straight line with a plane?

14. What is the essence of the method of auxiliary planes of intermediaries when finding the line of intersection of two planes?

15. What is the projection plane?

3 PROJECTION CONVERSION

3.1 Essence and main ways of converting a drawing

The solution of positional and metric problems in descriptive geometry is greatly simplified if straight and flat figures occupy the position of projecting straight lines and planes, or straight lines and level planes.

A necessary condition for simplifying the solution of problems is the construction of new additional projections, which make it possible to obtain either degenerate projections of individual elements, or these elements in full size. The construction of additional projections is called drawing transformation.

The conversion can be done in the following ways:

1. Change (replacement) of the projection planes with the condition that the object in question or its elements will occupy one of the particular positions relative to the new system of projection planes;

2. Rotation of geometric objects in space around the projecting axis so that they occupy any particular position relative to the projection planes.

3. Plane-parallel movement of the object, in which, the method of rotation around the projecting axis and the movement of the object, achieve a transition from an object of general position to an object of particular position;

4. The rotation of geometric objects in space around the level line so that they occupy the position of either a level line or a level plane.

3.2 Theory and algorithms for solving basic positional and metric problems

The essence of the method of changing projection planes is the transition from a given system of projection planes to a new one. In this case, line segments and flat figures retain their position, and their new projections are obtained by introducing additional projection planes.

When changing the projection planes, the mutual perpendicularity of the two projection planes - new and non-replaceable - is necessarily preserved.

Consider the mechanism for changing projection planes using the example of a transformation with a point (Figure 3.1.).

Figure 3.1 - The mechanism for changing the plane of projections P2 to P4

In the diagram, this transformation is shown in Figure 3.2. We set two projections of the point A (A1, A2) in the system of projection planes P1 and P2. Let us introduce the position of the plane P4. From the irreplaceable projection of the point A - A1

we draw a communication line perpendicular to the trace line of the plane P4. Since the height of the point does not change when moving from the system of projection planes P1 - P2 to the system of planes P1 - P4, this height is measured on the field P2 and deposited on the field P4 from the line of intersection of the planes in the direction of the new communication line.

Figure 3.2 - The mechanism of transition from the system P1 - P2 to P1 - P4 on the diagram

Replacing one of the projection planes does not always lead to the final solution of the problem, therefore, we will sequentially consider the mechanism of transition from the system of projection planes P1 - P2 to P1 - P4, and then to P4 - P5. (Figure 3. 3).

To obtain the projection of point A on the plane of projections P5, it is necessary to sequentially transfer the point first to the plane P4, and then to the plane P5. To perform the construction, we replace the plane P2 with the plane P4.

Figure 3.3 - The mechanism of transition from the system P1 - P2 to P4 - P5 on the diagram

The projection of the point A4 is obtained as follows: from the non-replaceable projection of the point A1 we draw a connection line perpendicular to the line of intersection of the planes P1 - P4 and set aside from it the distance measured from the replaced projection of the point to the line of intersection of the planes P1 - P2. During the transition to the system of projection planes P4 - P5, the plane P1 is replaced by P5. From the irreplaceable projection of the point A4, we draw a communication line perpendicular to the line of intersection of the planes P4 - P5. From this line we postpone the distance measured from the replaced projection of the point A1 to the line of intersection of the planes P1 - P4. As a result, we construct the projection of the point A5.

Another way to transform a drawing is the rotation method. It consists in the fact that the given system of projection planes remains unchanged, and the figure is rotated around a fixed axis until it takes a particular position relative to the projection planes, in particular, it becomes parallel or perpendicular to one of the projection planes.

tions. Rotation is performed around axes perpendicular or parallel to the projection planes.

Let us dwell on the mechanism of point rotation around the projecting axis. Let point A rotate around a horizontally projecting axis i. In this case, the point will describe a circle with a center passing through the axis of rotation i (i 1 ,i 2 ). During rotation, the trajectory of point A is a circle, the plane of which is parallel to the horizontal projection plane (Figure 3. 4).

Figure 3. 4 - Rotation around a horizontally projecting axis

On the diagram, the process of point rotation is depicted as follows. Choose the axis of rotation i (i1 , i2 ). On the horizontal plane of projections, this axis is projected to the point i1. From the center i1, the projection of the point A1 describes a circle, turning at any angle until it takes the positions A1 ". The frontal projection of the point A2 then moves along a horizontal straight line to the new position of the point A2 ".

Thus, when rotating around the horizontal

projecting axis, the horizontal projection of the point moves along a circle, and the frontal projection moves along a straight line perpendicular to the projection of the axis of rotation (Figure 3.5).

Figure 3.5 - Algorithm of rotation around a horizontally projecting axis

When a point rotates around a frontally projecting axis, the point describes a trajectory in the form of a circle, the plane of which is parallel to the frontal projection plane (Figure 3. 6).

Figure 3.6 - Rotation around the front projecting axis

When rotating around a frontally projecting straight line, the frontal projection of the point describes a circle, and the horizontal one moves along a straight line perpendicular to the axis of rotation. The algorithm for rotating a point around a frontally projecting axis is shown in Figure 3.7.

Figure 3.7 - Algorithm of rotation around the front projecting axis

3.3. The method of changing projection planes. Solution of the main tasks

No matter how the drawing is converted, the main tasks of the conversion can be reduced to the following:

1. A transformation in which a generic straight line becomes a level straight line.

2. A transformation in which the level line becomes a projecting line.

3. A transformation in which a generic plane becomes a projection plane.

4. A transformation in which the projection plane becomes a level plane.

Let's consider the solution of the main tasks of converting a drawing by changing the projection planes.

In order for a general position line to be a level line, it is necessary to introduce a new projection plane П4 that would be parallel to it. Let's replace, for example, the plane P2 with the plane P4 (figure 3.8).

The plane P4 is located parallel to the irreplaceable projection of the straight line segment A1 B1. The resulting projection of the line segment A4 B4 is a level line, therefore, this projection is the natural size of the segment. The solution of this problem allows you to determine the angle of inclination of the straight line segment AB to the horizontal projection plane -α.

Figure 3.8 - Transformation of a general position line into a level line

In order for a straight level line to become a projecting one (that is, to be projected onto some projection plane by a point), the new projection plane must be perpendicular to it.

Perpendicularity in the complex drawing is preserved only to the level line. Therefore, a new projection plane P4 is chosen perpendicular to the corresponding projection of the level line, i.e. to the natural size of the segment AB (Figure 3.9).

Figure 3.9 - Converting a direct level into a projecting one

In order for a plane in general position to be projective, it is necessary that the new system of projection planes be perpendicular to it. A plane will be perpendicular to a given plane if it is perpendicular to any level line of this plane. Therefore, in order to choose the position of the new plane P4, it is necessary to decide which of the projection planes will be replaced. For example, let's replace the P2 plane with the P4 plane (Figure 3.10). In the horizontal projection plane, the horizontal is projected without distortion.

the umbrella projection of the horizontal h1, so we build the plane P4 perpendicular to it.

On the plane P4, the triangle ABC occupies a projecting position

Figure 3.10 - Transformation of a general position plane into a projecting plane

In order for the given plane to turn out to be a level plane, it is necessary to place the P4 plane parallel to it (Figure 3.11).

Figure 3.11 - Transformation of the projecting plane into a level plane

In order to convert a general position plane into a level plane, it is necessary to perform two transformations: first, transform the general position plane into a projecting one, and then, by introducing another plane П5, transform the projecting plane into a level plane.

3.4 Method of rotation around the projecting axis. Solution of the main tasks

Task 1. Transform a general position line into a level line

In order to solve the problem, it is necessary to choose the position of the axis of rotation. Let us choose, for example, a horizontally projecting line as the axis of rotation. In this case, the rotation will be carried out in the horizontal projection plane. The angle of rotation of the straight line is determined by the condition of the problem: the straight line must be rotated to the position of the level line, in this case, to the position of the frontal level line (Figure 3.12).

Figure 3.12 - Transformation of a general position line into a level line by rotation

Task 2. Transform the level line into a projecting line.

When performing a rotation, you must select the position of the axis of rotation. In this case, a horizontally projecting axis should be chosen as the axis of rotation and the angle of rotation of the straight line should be determined. The angle of rotation is determined by the condition of the problem (Figure 3.13).

Figure 3.13 - Transformation of the level line into a projecting line by the rotation method

Task 3. Transform the plane of general position into a projected

The solution of the problem begins with the choice of the axis of rotation. Let us choose, for example, a horizontally projecting line as the axis of rotation. In this case, the rotation should be performed in the horizontal projection plane. The angle of rotation of the plane of the triangle around the horizontally projecting axis will set the horizontal projection of the horizontal lying in the given plane (Figure 3.14).

Figure 3.14 - Transformation of a plane of general position into a projective one by the rotation method

Task 4. Convert the projecting plane to a level plane.

Let's choose the position of the axis of rotation. In this case, you should select a horizontally projecting axis of rotation. The rotation angle of the object determines the rotation of the specified plane to the position of the frontal plane of the level (Figure 3.15).

Figure 3.15 - Transformation of the projecting plane into a level plane by the rotation method

3.5 Plane-parallel movement method

The method of plane-parallel movement consists in that the projection planes remain unchanged, and the object is rotated around the projecting axis until it takes a particular position relative to the projection planes and is moved. Depending on the conditions of the tasks, the object should be transformed so that it is located perpendicularly or parallel to the projection planes.

Task 1. Transform a generic plane into a level plane.

Figure 3.16 - Method of plane-parallel movement

Questions for self-control on topic 3:

1. What is the essence of the method of changing projection planes?

2. Can a generic line be transformed into a level line using a single transformation?

3. How is the projection direction chosen to transform a generic plane into a projection plane?

4. What is the difference between the method of changing projection planes and the method of plane-parallel movement?

5. How many times should a line in general position change its position relative to the projection planes П 1 , P2 to become a front-projecting straight line?

6. What is the essence of the method of rotation around the projecting line?

4 POLYHEDA

4.1 General information about polyhedra. Specifying polyhedra in a multidrawing

Polyhedra, representing the simplest geometric shapes, are fundamental in the design of engineering structures. Polyhedral forms are widely used in the design of machine parts and mechanisms in technology, as well as in various architectural structures.

Of the greatest practical interest are prisms, pyramids and convex uniform polyhedra, all faces of which are regular and equal polygons - Plato's solids (tetrahedron - 4, octahedron - 8, icosahedron - 20 regular triangles; hexahedron (cube - 6 regular rectangles); dodecahedron - 12 regular pentagons). A polyhedron is called convex if it is located on one side of the plane of any of its faces.

A polyhedron is a body bounded by flat polygons. These polygons are called edges (Figure 4.1).

Figure 4.1 - Examples of polyhedra

The totality of all the faces of a polyhedron is called its surface

Faces intersect along straight lines called edges. Edges intersect at points called vertices.

Drawings of polyhedra must be reversible. This can be achieved if certain conditions for the location of the edges of the polyhedron in the projections are met.

In the drawing, polyhedra are depicted as projections of their vertices and edges. In Figure 4.2, a straight tetrahedral prism ABCDKLMN and a trihedral pyramid SABC are given. A prism is called straight if its side faces and edges are perpendicular to the base. A right prism is called regular if its base is a regular polygon.

Figure 4.2 - Specifying polyhedra on the diagram

4.2 Intersection of polyhedra by a plane and a straight line

The line of intersection of the polyhedron with the plane is a flat polygon (Figure 4.3).

Figure 4.3 - Intersection of a polyhedron by a plane

The cut line of a polyhedron by a plane can be constructed in two ways.

First way. Find the vertices of the desired polygon as a result of the intersection of the edges of the polyhedron with the cutting plane.

The second way. Find the sides of the desired polygon as a result of the intersection of the faces of the polyhedron with the cutting plane.

In the first case, one has to repeatedly solve the problem of constructing a point of intersection of a straight line with a plane, in the second case, of constructing a line of intersection of two planes. In cases where the cutting plane or surface is in a particular position, the task is greatly simplified, since on one of the projection planes the projection of the section line will coincide either with the projection of the cutting plane (Figure 4.4), or with the degenerate projection of the surface of the polyhedron (Figure 4.5).

To construct a line of intersection of a trihedral pyramid with a frontally projecting plane, it is necessary to find the intersection points of each edge of the SABC pyramid with the frontally projecting plane ∑. As a result of the construction, we get the triangle DFE. If a generic surface is intersected by a frontally projecting plane, then the frontal projection of the section line (triangle) will coincide with the frontal projection of the cutting plane ∑2. Frontal projections of the vertices of the section line (D2 , F2 , E2 ) are defined as the result of the intersection of each edge of the pyramid with the cutting plane. By projecting the points that define the section line onto the horizontal plane of projections onto the projections of the corresponding edges, we obtain the horizontal projection of the desired section line (D1, F1, E1).

Figure 4.4 - Intersection of the pyramid by the projecting plane

To construct a section of a straight prism ABCD by a generic plane Q(a||b), you need to construct the sides of the desired polygon

KLMN as a result of the intersection of the faces of the polyhedron with the plane Q(a||b) (Figure 4.5). To do this, we draw an auxiliary cutting plane Θ through the projection of the face B1 C1. This plane will intersect the given plane Q(a||b) along a straight line passing through the points 11 , 21 . We construct the projection of the section line of two planes in the frontal plane of projections (12, 22) and find the intersection points of this segment with the edges B and C - L and M. Similarly, we construct the line of intersection of the face AD ​​with the plane Q - the segment KN. In the frontal plane of projections, we connect the projections of the segments of the polygon K2 L2 M2 N2, taking into account the visibility of the faces

– segment projection is visible if the face is visible in the given projection, not visible – if the face projection is not visible. In addition, it is necessary to establish the mutual visibility of the edges of the prism and the cutting plane.

Figure 4.5 - Intersection of the projecting prism by a plane of general position

Consider the construction of a section of a pyramid in general position by a plane in general position (Figure 4.6).

Figure 4.6 - Intersection of the pyramid by a plane in general position

To construct the line of intersection, we will define the vertices of the section as the result of the intersection of each edge of the pyramid with the plane of general position ∑(a||b). To find the point of intersection of the edge SA with the plane ∑(a||b), it is necessary to enclose the edge in the secant plane Q and find the line of intersection of the two planes Q and ∑ - the segment 12 22 ;11 21 . The vertex K is constructed as a result of the intersections of the corresponding projections of the projections of the edge SA and the segment 1,2. The vertices L and N are found according to the same algorithm as the results of the intersections of the edges SB and SC with the plane ∑(a||b).

Definition tasks points of intersection of a polyhedron with a straight line solved on the basis of the method of auxiliary cutting planes. In this case, one of the projections of a given straight line is enclosed in a projecting secant plane. Find the line of intersection of the auxiliary

cutting plane with a polyhedron. The projections of the points of intersection of a straight line with a polyhedron are found as a result of the intersection of the constructed section line and another projection of a given straight line and the subsequent determination of their position in both projection planes. Find the points of intersection of the pyramid with a straight line in general position (Figure 4.7).

Figure 4.7 - The intersection of a straight line with a pyramid

Let us conclude, for example, the frontal projection of the given straight line l 2 into the frontally projecting plane Q2 and construct a line of section of the pyramid by this plane. We construct the points of intersection of the pyramid with the line l as a result of the intersection of the triangle of the section first with the horizontal projection of the line l 1 - K1 and L1, and then we get their frontal projections (K2, L2).

Let us determine the mutual visibility of the line l (l 1 ,l 2 ) with the pyramid SABC. The tasks of determining the intersection points of polyhedra with lines are simplified if one of the elements is in a particular position.

For example, when determining the points of intersection of a line in general position with a projecting prism, the problem is reduced to determining the points of intersection of a line with degenerate projections of the prism faces (Figure 4.8).

Figure 4.8 - The intersection of a straight line with a straight prism

When finding the intersection points of the pyramid with the projecting line, the horizontal projections of the intersection points (K1, N1) are determined on the degenerate projection of the straight line, and then their frontal projections are lined up (K2, N2) and their mutual visibility is established (Figure 4.9).

Figure 4.9 - The intersection of the pyramid with the projecting line

4.3 Construction of developments of polyhedra

If the surfaces are given the properties of flexibility and inextensibility, then some of them can be combined with the plane without the formation of folds and breaks, i.e., to obtain a surface development.

A development of a polyhedron is a flat figure obtained by combining all the faces of a polyhedron with one plane in a certain order.

To build a development of a prism or pyramid, it is necessary to determine the actual size of their edges and bases, and then build a development of surfaces (Figures 4.10 and 4.11).

The construction of the development of the pyramid is reduced to the repeated construction of the natural size of the triangles that limit its surface.

Let's build a full development of a trihedral pyramid (Figure 4.10). To do this, we determine the actual size of each edge using the right-angled triangle method. The edge SC is the front line of the level, so its projection S2 C2 is natural. The base of the pyramid is a horizontal level plane, so the horizontal projection of the triangle ABC is a natural value.

Figure 4.10 - Development of the pyramid

The construction of scans of inclined prisms is reduced to the construction of the natural values ​​of the faces of the polyhedron. These builds can be done in the following ways:

1. The method of normal section, in which the width of each face is determined using a cutting plane perpendicular to the edges of the prism;

2. The rolling method, which is based on the sequential combination of all the faces of the prism with the plane, by rotating around the level line;

3. A triangulation method based on dividing rhombuses by diagonals into triangles and determining the natural values ​​​​of the sides of the triangles.

Let us dwell in more detail on the consideration of the essence of the normal section method. Let's set the position of the prism in such a way that its edges are, for example, in the position of the fronts (Figure 4.11).

Figure 4.11 - Scanning a prism using the normal section method

Let us intersect the given prism with an auxiliary plane perpendicular to the edges of the prism, i.e. determine the width of each face of the prism. Let us determine the natural value of this normal section and construct a development of the surface of the prism. The construction of the development begins with the construction of a horizontal line, on which we set aside the segments that determine the width of each face along its normal section.

Through the points that determine the lengths of the segments, we draw lines perpendicular to them, on which we plot the lengths of the segments of the ribs enclosed between the section line and the bases of the prism.

The development of the lateral surface of the prism is obtained after connecting the ends of the constructed segments with straight lines. To build a complete sweep of the prism, it is necessary to complete the natural values ​​​​of the bases of the prism.

4.4 Mutual intersection of polyhedra

The result of the intersection of two polyhedra is a spatial polygonal closed line running along the lateral surface of both polyhedra.

Its links are defined as the result of the intersection of the faces of one polyhedron with the faces of another, and the vertices are defined as the points of intersection of the edges of each polyhedron with the faces of another. Thus, the problem of constructing a line of mutual intersection of two polyhedra can be reduced to solving the problem of the intersection of two planes, or to the intersection of a line with a plane.

The line of intersection of polyhedra can break up into two or more branches, which can be both closed spatial polygonal lines and flat polygons. The intersection line can be within the common part of the projections of both intersecting surfaces.

Let's build a line of intersection of the KLMN prism with the SABC pyramid.

To build an intersection line, we first find the intersection points, for example, the edges of a prism with the faces of a pyramid (Figure 4.12). It can be seen from the drawing that the edges M, N, L are outside the overlapping area of ​​the two polyhedra, therefore, they do not intersect with the pyramid. The edge K is located in the area of ​​superposition of the projections of the two faces of the pyramid CSA and CSB (determined by the horizontal projections of the faces C1 S1 A1 and C1 S1 B1 and the edge K1 ), so we determine the points of intersection of the edge K with these faces.

Figure 4.12 - Finding the intersection points of the edges of the prism with the faces of the pyramid

For construction, we will use auxiliary straight lines (S1 11 , S1 21 ), which we draw in the faces CSB and CSA through the projections of the intersection points of the edge K with the faces - points 3 and 4 (first we determine their horizontal projections 31 and 41 ). Let us construct the frontal projections of points 3 and 4 at the intersection of the projections of the edge K2 with the projections of the auxiliary lines S2 12 , S2 22 .

We find the points of intersection of the edges of the pyramid with the faces of the prism. We will start constructing these points from the horizontal plane of projections, since the prism occupies a horizontally projecting position. The projection of the edge S1 A1 intersects two faces of the prism K1 L1 and L1 N1 at points 51 and 61 . Let's project these points into the frontal plane of projections onto the projection of the edge S2 B2 and construct projections 52 and 62 .

Arguing similarly, we construct projections of the intersection points of the edges SA and SC with the faces of the prism KL, KN and KM (7,8, 9, 10) (Figure 4.13) .

Figure 4.13 - Finding the intersection points of the edges of the pyramid with the faces of the prism

Connect successively the projections of the points of intersection by segments of straight lines that belong simultaneously to the faces of the prism and the pyramid. For example, the projections of points 7-5-4-9-3-7 are sequentially connected, connecting the segments of the line of intersection of two polyhedra in the entry area and points 8, 6 and 10 in the exit area of ​​two polyhedra.

The last stage of construction is to determine the visibility of sections of the constructed intersection line. The projection of the intersection line segment is considered visible if the segment is in the visible projections of the pyramid face and the prism face. If at least one of the projections of the faces is not visible, then the projection of the considered section of the intersection line is not visible. Let's connect sections of the intersection line and stroke the drawing, taking into account the visibility of the faces (Figure 4.14).

Figure 4.14 - Mutual intersection of polyhedra

Questions for self-control on topic 4:

1. What is a polyhedron?

2. What defines the surface of a polyhedron in a complex drawing?

3. What methods are used to construct a section of a polyhedron by a plane?

4. How are entry and exit points constructed when a polyhedron intersects with a straight line?

5. What is the essence of the normal section method when constructing a sweep of a prism?

6. What method is used to construct a pyramid sweep?

5 CURVES AND SURFACES

5.1 Curved lines

Curved lines are used in the design of various surfaces, in the theory of machines and mechanisms, in modeling and marking business, in the construction of state diagrams of multicomponent systems.

A curved line is a set of successive positions of a point moving in space.

Curved lines, all points of which belong to the same plane, are called flat, for example, a straight line, a circle, an ellipse, a parabola, a hyperbola, a sinusoid, graphs of functions of one variable, graphs of equations with two unknowns, other curved lines - spatial, for example, helical lines.

Each curve includes geometric elements that make up its determinant, i.e. a set of independent conditions that uniquely determine this curve.

There are the following ways to define curves:

1. Analytical - the curve is given by a mathematical equation;

2. Graphical - the curve is set only graphically;

3. Tabular - the curve is specified by the coordinates of a successive series of its points.

Any curved line can be obtained by moving a point in space, as a result of the intersection of curved surfaces by a plane and as a result of the mutual intersection of surfaces, at least one of which is a curve.

The points of a flat curved line are divided into ordinary (tangent point A) and special (inflection point B - at the inflection point, the curvature changes sign - from

on one side of this point, the curve is convex, on the other, concave; cusps C - cusps of the 1st kind (point F of the cycloid refers to cusps of the 1st kind), D - cusps of the 2nd kind; point E is a double point of the strophoid, at this point the curve has two different tangents m1 and m2) (Figure 5.1).

Figure 5.1 - Ordinary and singular points of the curve

Regular curved lines are divided into algebraic (circle, parabola) and transcendental (sinusoid).

When studying a flat curved line, it often becomes necessary to determine its order. The order of a flat curved line is determined by the largest number of points of its intersection with a straight line, or the degree of its equation. The line of the first order is a straight line. Curved lines of the second order - an ellipse (its particular form is a circle), a parabola, a hyperbola.

A circle is a closed curve, all points of which are at the same distance from some point O lying in this plane, called the center. Circle equation: x 2 +y 2 =R 2 .

An ellipse is a set of all points in a plane, the sum of the distances to two given points F1 and F2, called foci, is a constant value (2a). Ellipse equation: x 2 / a 2 + y 2 / b 2 =1.

Figure 5.2 - Lines of the second order: circle and ellipse

The parabola is defined by the equation y 2 = 2px . A parabola has one improper point, has one axis of symmetry.

The hyperbola is defined by the equation x2 /a2 – y2 /b2 =1. A hyperbola has a center and two axes of symmetry, and has two improper points.

Figure 5.3 - Lines of the second order: parabola and hyperbola

Of the spatial curved lines, cylindrical and conical helical lines are of the greatest practical interest.

Cylindrical helix - this is a line described by a point with uniform motion along a straight line, with uniform rotation of its rotation around an axis parallel to it.

Figure 5.4 - Helix

The height to which point A rises in one complete revolution is called helix pitch.

The frontal projection of a cylindrical helical line is a sinusoid, the horizontal projection is a circle.

5.2 Formation of curved surfaces

A curved surface is a set of successive positions of a certain line moving in space according to a certain law.

Surfaces can be defined in a drawing in the following ways:

1. Kinematic - the surface is considered as a continuous set of positions of a line moving in space according to a certain law.

The moving line is called the generatrix of the surface, and the line

along which the generatrix moves is called the guide (Figure 5.5).

Figure 5.5 - Kinematic way of defining surfaces

2. Wireframe - if it is impossible to describe mathematically, the surface is set by a sufficiently dense network of lines belonging to these surfaces. The surface skeleton can consist of three-dimensional curves or families of plane sections (figure 5.6).

Figure 5.6 - Defining the surface with a frame

3. Analytical - the surface is considered as a continuous two-dimensional set of points. The coordinates of the points of this set satisfy some equation F(x,y,z) = 0.

4. A determinant is a set of conditions necessary and sufficient for a unique assignment of a surface. Surface qualifier

consists of geometric and algorithmic parts D = [G] Λ [A] . For example, the surface of a cylinder of revolution can be defined by rotating a straight line a around a fixed axis i using the determinant: D = Λ [A]. The geometric part of the determinant is represented by frontal projections of the axis and generatrix. In the algorithmic part, the “surface of revolution” should be written (Figure 5.7).

Figure 5.7 - Defining the surface with a determinant

5. Outline - the boundary of the visible part of the surface on the corresponding projection plane. This method is the most visual in solving problems of descriptive geometry. For example, the surface of a right circular cylinder can be represented by projections of its horizontal and frontal outlines (Figure 5.8).

Figure 5.8 - Defining a surface with a sketch

A large variety of surfaces, various ways of their formation, the complexity of geometric characteristics create difficulties in attempts to classify surfaces.

All curved surfaces, depending on the type of generators, are divided into ruled surfaces, in which the generatrix is ​​a straight line, and non-ruled, in which the generatrix is ​​a curve.

Separate ruled surfaces, if they are given the physical properties of flexibility and inextensibility, can be expanded to coincide with the plane without wrinkles or breaks. Such surfaces are called deployable. Those ruled surfaces that do not meet the specified requirements, as well as non-ruled surfaces, are called non-deployable.

5.3 Surfaces: rotations, ruled, helical, cyclic

5.3.1 Surfaces of revolution

A surface of revolution is a surface described by a curve (or straight line) generatrix when it rotates around a fixed axis.

Each point of the generator describes during its rotation a circle centered on the axis. These circles are called parallels. The parallel of the largest radius is called the equator, the smallest - the throat (Figure 5.9).

The curves obtained in the section of the body of revolution by planes passing through the axis are called meridians. The meridian parallel to the frontal projection plane is called the main one.

Figure 5.9 - Surface of rotation

The surfaces formed by the rotation of a straight line include the following surfaces:

1. Cylinder of rotation - is formed by the rotation of a straight line around the i-axis parallel to it.

2. Cone of rotation - formed by the rotation of a straight line around the i-axis intersecting with it.

3. A one-sheeted hyperboloid of revolution is obtained by rotating a straight line around the i-axis intersecting with it.

A hyperboloid of revolution can also be obtained by rotating a hyperbola around its imaginary axis.

The named surfaces are also ruled surfaces (Figure 5.10).

Figure 5.10 - Surfaces of revolution: cylinder, cone, hyperboloid

Surfaces of revolution formed by the rotation of a circle include:

1. Sphere - a surface formed by the rotation of a circle around its diameter;

2. Torus - a surface formed by the rotation of a circle around an axis lying in the plane of this circle, but not passing through its center;

3. Ring - a surface formed by the rotation of a circle around an axis lying outside the circle.

The torus is a surface of the fourth order.

Any surface is considered given if it is possible to determine the position of any point on its surface. To build points on the surface

the sphere or torus, it is necessary to use the parallels and meridians of these surfaces (Figure 5.11).

Figure 5.11 - Surfaces of revolution: sphere, torus, ring

The surfaces of revolution formed by the rotation of an ellipse, parabola and hyperbola are called respectively: ellipsoid of revolution, paraboloid of revolution, one-sheeted hyperboloid of revolution (Figure 5.12).

Figure 5.12 - Surfaces of revolution: ellipsoid, paraboloid, hyperboloid

5.3.2 Ruled surfaces

The surface formed by the movement of a straight line is called a ruled one.

A ruled surface formed by the movement of a rectilinear generatrix that constantly passes through some point S and in all cases intersects some guide curve is called conic.

A ruled surface formed by the movement of a generatrix parallel to a certain direction and intersecting a guide is called a cylindrical surface.

Ruled surfaces include surface with cusp- is formed by moving a straight line along a certain spatial curve, and the generatrix of the straight line remains at each point tangent to the curvilinear guide (Figure 5.13).

Figure 5.13 - Ruled surfaces: conical, cylindrical, surface with a return edge

5.3.3 Helical surfaces

The helical surface is formed by the helical movement of some generating line (Figure 5.14).

Helical surfaces with generating straight lines are called helicoids.

A helicoid is called straight if the generating straight line makes a right angle with the z-axis of the surface. In other cases, the helicoid is called oblique or oblique.

Figure 5.14 - Straight and oblique helicoids

5.3.4 Cyclic surfaces

A surface is called cyclic if it is described by a circle of constant or variable radius during its arbitrary motion.

An example of a cyclic surface can be any surface of revolution. In addition, they include channel and tubular surfaces.

The channel surface is formed by moving a circle of variable radius along a curved guide.

A tubular surface is formed by moving a circle of constant radius along a curved guide (Figure 5.15).

Figure 5.15 - Cyclic surfaces: channel and tubular

5.4 Generalized positional problems

5.4.1 Intersection of curved surfaces by a plane

When a curved surface is intersected by a plane, in the general case, a plane curve (ellipse, circle) is obtained. When crossing ruled surfaces with a plane, straight lines can also be obtained, in a particular case, if the secant plane is directed along the generators or passes through one point (cylinder or cone).

To construct a line of intersection of a curved surface by a plane, the method of auxiliary cutting planes is used. The auxiliary plane is chosen so that it intersects the given plane along a straight line, and the surface along a graphically simple line (circle or straight line). The intersection points of these lines will be the desired points belonging to the surface and the cutting plane.

The construction of the projections of the line of the section of the surface by the plane is greatly simplified if the cutting plane occupies the projecting position

zhenie. In this case, one of the projections of the section line is already on the drawing: it coincides with the projection of the plane. The task is reduced only to constructing another projection of this line.

Consider the construction of a section line of a cylinder by a projecting plane (Figure 5. 16).

Figure 5.16 - Intersection of the cylinder by the projecting plane

The cylinder is intersected by the plane Σ along an ellipse. Since the cylinder occupies a horizontally projecting position, the ellipse degenerates onto the horizontal projection plane into a circle coinciding with the horizontal outline of the cylinder. Since the cutting plane ∑ occupies a frontally projecting position, the frontal projection of the ellipse degenerates into a straight line segment 12 22 .

Consider the construction of a section line of a right circular cylinder by a plane in general position (Figure 5.17).

Construction algorithm:

1. Analyze the condition of the problem. Since the cylinder occupies a horizontally projecting position, the horizontal projection of the section ellipse degenerates into a circle, and the front projection is projected into an ellipse.

Viewpoints A and B are points dividing the frontal projection of the section ellipse into visible and invisible parts. The projections A2 and B2 are determined using the auxiliary secant plane Q (the level frontal plane) drawn through the projections A1 and B1.

Near and far points C and D are determined using cutting planes of the frontal level drawn through the projections C1 and D1 and intersecting the cylinder along the near and far generators, and the given plane - along the corresponding fronts. The projections of points C2 and D2 are found at the intersection of the corresponding projections of the lines.

Figure 5.17 - Intersection of the cylinder by a plane of general position

The highest and lowest points of the section K and L are on the slope line drawn through the axis of the cylinder perpendicular to the horizontal of a given plane. The segment KL determines the position of the major axis of the ellipse.

Minor axis of the ellipse MN is located perpendicular to the major axis, perpendicular to it and passes through the axis of the cylinder.

3. Determine the position of random points. Spend auxiliary secant planes of the frontal level and determine the position of the projections of random points on the horizontal and frontal planes of the projections.

4. Set the visibility of the ellipse in the frontal projection plane. Set in the projections the mutual visibility of the cylinder and the cutting plane.

AT as a result of the intersection of a right circular cone by planes, lines can be obtained, the nature of which can be foreseen depending on the location of the cone and the secant plane. These lines can be: a circle, an ellipse, a parabola, a hyperbola, and if the cutting plane passes through the top of the cone, a pair of straight lines (Figure 5.18).

Let's build a section line of a right circular cone by a projecting plane (Figure 5.19).

Construction algorithm:

1. Analyze the condition of the problem.

The cutting plane is in a frontally projecting position, therefore, the frontal projection of the section ellipse degenerates in the frontal projection into a straight line segment AB.

2. Determine the position of the reference points: the upper and lower points of the section A and B determine the position of the major axis of the ellipse. The position of the near and far points (C and D) is determined on the minor axis of the ellipse, which is perpendicular to the major axis and is located in the middle of the segment AB.

3. Determine the position of random points: K,L and M,N. For their construction, auxiliary cutting planes of the level are used, which

rye intersect the surface of the cone along the circles of the corresponding radii, and the plane - along the frontally projecting straight lines.

Figure 5. 18 - Conic sections (conics)

Figure 5.19 - Intersection of a cone by a frontally projecting plane

5.4.2 Intersection of a curved surface with a straight line

The result of the intersection of a curved surface with a straight line is a pair of points.

A pair of points of intersection of a straight line with a curved surface is conditionally called entry and exit points. To construct these points, the method of auxiliary cutting planes is used.

Construction algorithm:

1. Any projection of a given straight line is enclosed in a cutting plane. (Usually, projecting planes are chosen as the auxiliary plane.)

2. Build projections of the line section of the surface by a plane.

3. Determine the intersection points of the resulting line with a given straight line

4. Determine the mutual visibility of a straight line and a surface. Consider various cases of constructing intersection points of curves

straight line surfaces.

Problem solving is simplified if one of the elements (a line or a surface) is in a particular position (Figure 5.20). In this case, in one of the projections, the position of the projections of the points of intersection of the straight line with the curved surface is determined.

By enclosing the frontal projection of the given straight line in the projecting secant plane in the section of the cylinder, we obtain an ellipse, which is projected onto the horizontal projection plane in the form of a circle coinciding with the horizontal outline of the cylinder surface. The points of intersection of the projecting cylinder with the straight line are determined on the horizontal projection plane at the intersection of the horizontal contour of the cylinder with the projection of the straight line. Mutual visibility of a straight line and a cylinder is established.

When finding the points of intersection of a line of particular position with the surface of a cone in general position, one can use the construction of generators belonging to the surface of the cone. Construct the intersection points of M and N and establish the mutual visibility of the line and the cone.

Figure 5.20 - Particular cases of intersection of surfaces with straight lines

Consider the general case of the intersection of a curved surface with a straight line in general position using the example of the intersection of a cone with a straight line (Figure 5.21). We will solve this problem in two ways.

In the first case, the frontal projection of the straight line AB is enclosed in a plane passing through the top of the cone (plane ABS). This plane will intersect the cone along the lines S1 and S2. To construct these lines, the line DC of the intersection of the plane ABS with the plane of the base of the cone and points 1 and 2 of its intersection with the circle of the base of the cone are found. The intersection points K and N of line AB with the surface of the cone are found as a result of the intersection of line CD with lines S1 and S2. Determine the mutual visibility of a straight line and a cone.

In the second case, the line AB is enclosed in a frontally projecting plane that intersects the cone in an ellipse. The intersection points K and N are found as a result of the intersection of the constructed ellipse with a straight line

AB and determine the mutual visibility of a straight line and a cutting plane.

The first way to solve the problem is the most rational.

Figure 5.21 - Intersection of a cone with a straight line in general position

To solve the problem of determining the points of intersection of a sphere with a straight line in general position (Figure 5.22), it is more rational to use the method of changing projection planes. In this case, for example, a horizontal projection of a given straight line AB into a horizontally projecting plane is concluded. In the section of the sphere by this plane, a circle is obtained, which is projected onto the P4 plane without distortion in the form of a circle,

and the line segment A4 B4 - in its natural size. The intersection points C and D are determined at the intersection of the circle and the straight line in the plane P4, and then their projections on the planes P1 and P2 are determined. Set the visibility of the projections of a straight line and a sphere in accordance with the visibility of the constructed section line.

Figure 5.22 - Intersection of a sphere with a straight line in general position

5.4.3 Methods for constructing lines of intersection of curved surfaces

Two curved surfaces intersect in the general case along a spatial curved line (Figure 5.23).

Figure 5.23 - Mutual intersection of curved surfaces

The line of intersection of two curved surfaces is built on its individual points. These points are determined with the help of auxiliary intermediary surfaces. Intersecting the given surfaces with some auxiliary surface, section lines are obtained, at the intersection of which they find points that belong simultaneously to both surfaces and, therefore, to the desired section line.

Planes or spheres are most often chosen as intermediary surfaces. The use of these surfaces is determined by the type and location of the specified surfaces.

5.4.3.1 Auxiliary cutting plane method

The method of auxiliary cutting planes is used when both surfaces can be intersected along graphically simple lines (circles or straight lines) by a certain set of projecting planes or level planes (Figure 5.24).

Figure 5.24 - Intersection of a cone and a cylinder

Consider the application of the method of auxiliary cutting planes of the level on the example of the problem of constructing a line of intersection of a cylinder and a cone (Figure 5.25).

Figure 5.25 - Cutting plane method: intersection of a cylinder and a cone

Let's start the construction by defining reference points (upper, lower, right and left points of the section and visibility points). Since the surface of the circular cylinder is in a frontally projecting position, these points are located on the frontal outline of the surface - the circle into which the cylinder is projected.

The section line itself in the frontal plane of the projections will coincide with the frontal outline of the cylinder and is determined by the area of ​​superposition of the projections of the two surfaces.

The construction of projections of the upper and lower points of the section will begin with the definition of their frontal projections 12 and 22 . Let's build them on the mountains

the umbrella plane of projections onto the projections of the main meridian and find the horizontal projections of points 11 and 21 .

To construct horizontal projections of the rightmost and leftmost points of the section, we will use the method of cutting level planes. We choose the position of the auxiliary plane in such a way that it simultaneously intersects both surfaces along graphically simple lines - along circles or straight lines. An auxiliary cutting plane - a horizontal level plane - will be drawn through the frontal projections of points 3 and 4. In this case, the surface of a circular cylinder will be intersected by it in straight lines, and the surface of a circular cone - in a circle. The horizontal projections of points 31 and 41 will be obtained at the intersection of the horizontal projections of the section lines.

Points 3 and 4 are at the same time the points of view for the horizontal projection of the section line, i.e. delimit this projection into visible and invisible parts.

All other points belonging to the section line will be auxiliary and their choice is random. The number of random points is determined by the accuracy of the construction: the more there are, the more accurately the solution is made.

Let us dwell on the construction of a pair of random points 5 and 6. To do this, we select a pair of competing points in the frontal projection plane and use the auxiliary secant plane of the horizontal level to determine their horizontal projections.

Connecting the constructed projections of points with a smooth curved line, we obtain a horizontal projection of the section line of two surfaces. In this case, in the horizontal projection plane, we will take into account the position of the visibility points. The section of the section line above points 3 and 4,

will be visible, and below them - invisible. The frontal projection of this line coincides with the frontal outline of the cylindrical surface, and, being symmetrical, will be visible.

Thus, to build a line of intersection of surfaces, it is necessary:

1. Determine which surfaces intersect and whether there is a projection of the intersection line in the condition of the problem.

2. Determine the position of anchor points.

3. Select the position of the auxiliary cutting planes.

4. Find the position of the rest of the reference and random points using the selected cutting planes.

5. Draw projections of the desired section line.

6. Define visibility.

To construct a line of intersection of surfaces that do not have a common plane of symmetry, use the method of secant planes (Figure 5.26). To determine the position of points 1 and 2, through the axis of symmetry of the cone we draw the frontal plane of the level Σ, which intersects the cone along the main meridian, and the sphere along the circumference. The frontal projections of points 12 and 22 are determined, and then projections 11, 21.

The position of the highest and lowest points (3 and 4) is determined using the secant plane Q, passing through the centers of the cone and sphere and being the plane of symmetry of the two surfaces. To determine the projections of points 32 , 42 and 31 , 41, the method of rotation of the obtained sections (meridians of both surfaces) around the axis passing through the axis of symmetry of the cone was used.

Figure 5.26 - Intersection of a cone and a sphere - the method of cutting planes

Points of view for the horizontal plane of projections (5.6) are determined using the plane Θ drawn through the equator of the sphere.

The position of random points is determined using cutting planes of the horizontal level.

The viewpoints for the frontal projection plane will be on the main meridian of the sphere. If we draw a cutting plane through the main meridian of the sphere, then in the section of the sphere there will be a circle, and in the section of the cone - a hyperbola. Let us determine the approximate position of these

points after constructing a common section line of surfaces.

We connect the projections of the constructed points, taking into account the visibility in the corresponding projection planes.

5.4.3.2 Auxiliary cutting spheres method

The use of the auxiliary secant spheres method is based on a property inherent in surfaces of revolution. It consists in two

any coaxial surfaces of revolution intersect along circles passing through the points of intersection of the meridians of the surfaces.

In this case, the planes of the circles of the section are perpendicular to the axis of rotation, and the centers of the circles belong to this axis. Therefore, if the axes of the surfaces of revolution are parallel to the plane of projections, then on this plane the circles of the section are projected into segments of straight lines perpendicular to the projections of the axes of the surfaces of revolution, and on the other plane - in the form of circles.

As an auxiliary secant surface of revolution, it is convenient to use a spherical surface, the center of which should belong to the axis of the surface of revolution (Figure 5.27).

Figure 5.27 - Property of cutting spheres

AT Depending on the relative position of the surfaces, there are two possible options for solving problems using the method of secant spheres:

1. The axes of both surfaces are parallel to the projection plane.

2. Intersecting surfaces have a common symbol plane

AT in the first case, the method of concentric secant spheres is used (Figure 5.28), in the second case, eccentric secant spheres.

Figure 5.28 - Method of concentric secant spheres: intersection of cones

Let us dwell in more detail on using the method of concentric secant spheres to solve the problem of constructing a line of intersection of two cones (Figure 5.29).

The construction of the line of intersection begins with determining the position of the projections of the reference points. Projections of points 12 , 22 and 32 , 42 are the highest and lowest points in the entry area of ​​the cone surfaces and in the area of ​​their exit. Their horizontal projections 11 , 21 , 31 , 41 are obtained by projecting onto the axis of symmetry in the horizontal projection plane.

To obtain the remaining points of the line of intersection of surfaces, the method of concentric secant spheres is used. The center of the secant spheres is chosen in the frontal projection plane at the intersection of the symmetry axes of the surfaces. The construction begins with determining the minimum radius of the secant sphere - the value of the larger of the two perpendiculars, lowered from the center of the spheres to the generatrix surfaces of the cones.

Figure 5.29 - Method of concentric cutting spheres

Let's construct the points belonging to the line of intersection of the surfaces as a result of the intersection of two chords (space circles along which the auxiliary sphere intersects the cones).

Let's build random points belonging to the intersection line - points 5 and 6, using a secant sphere, the radius of which is chosen from the range: greater than the minimum and less than the maximum (from the center to the projection of point 22).

We connect the projections of the section line, taking into account their visibility in the corresponding projections.

Consider using the method of eccentric cutting planes to solve the problem of determining the intersection of a cone and a sphere that have a common plane of symmetry (Figure 5.30).

Figure 5.30 - Coaxial cone and sphere

We begin the construction of the intersection line by determining the position of the upper and lower points of the section (12, 22) at the intersection of the frontal sketches of the surfaces and determine their horizontal projections 11 and 21 (Figure 5.31). The remaining points are determined using secant spheres drawn from one or different centers lying on the axis of symmetry of the cone.

Figure 5.31 - Intersection of a cone and a sphere - the way of spheres

Pairs of points 3.4 and 5.6 are determined first in the frontal plane of projections at the intersection of chords from the corresponding sections of the auxiliary sphere of the given surfaces. Then they build their horizontal projections. The visibility of the line of intersection is determined in the horizontal projection plane using a cutting plane passing through the equator of the sphere. In the frontal projection plane, the section line, being symmetrical, is projected into a visible smooth curve.

The method of eccentric secant spheres is used when constructing the line of intersection of an open torus and a truncated cone (Figure 5.32). The upper and lower points of the section A and B are in the plane of the main meridian of both surfaces and therefore are determined by their frontal projections at the intersection of the outlines of the surfaces. Then their horizontal projections A1 and B1 are built.

Figure 5.32 - Method of eccentric spheres: intersection of a torus and a cone

The remaining points are constructed using secant spheres that intersect the surface of the ring along its meridional circles. To find the centers of the secant spheres, secant planes are drawn that pass through the center of the ring. A tangent is drawn through the intersection point of this plane and the axis of the torus until it intersects with the axis of the cone - this point will be the center of the secant sphere common to both the torus and the cone. The projections of the point C2 and D2 are determined at the intersection of chords (space circles) on the surfaces of the torus and the cone. The position of the generators is determined and the projections C1 and D1 are built on the corresponding projections of the generators of the torus.

The viewpoints for the horizontal projection of the section line are determined on the axis of symmetry of the truncated cone in the frontal plane of the projections (a horizontal level plane is drawn) and the horizontal projections of the viewpoints (L1 and N1) are determined. In the frontal projection plane, the line is projected as a visible curve.

5.5 Tangent lines and planes to surfaces

A straight line lying in the same plane as a curve can intersect it at two or more points. Such a line is called a secant. If the secant is moved so that the length of the arc AB between the two intersection points approaches zero, then in the limit position the secant will take position t and will be called a tangent (Figure 5.33).

The tangent indicates the direction of movement along the curve at each tangent point.

A plane tangent to a surface has a point in common with this surface, a straight line or a flat curved line. A plane can touch a surface in one place and intersect it in another. The line of contact can simultaneously be the line of intersection of the surface with the plane.

Figure 5.33 - Tangent to the curve

In general, the plane tangent to the surface is a set of straight lines tangent to any curves belonging to

pressing the surface and passing through a given point of this surface.

To set a tangent plane to any surface, it is sufficient to draw curves belonging to the surface through a point given on the surface and construct a tangent line to each of them passing through one point. These straight lines will define the tangent plane. The plane tangent to the surface is the limiting position of the secant plane.

A straight line passing through the tangent point and perpendicular to the tangent plane is called the surface normal at that point. The surface normal at a given point determines the direction of the plane tangent to the surface at that point (Figure 5.34).

It is not possible to construct a tangent plane at every point on the surface. At some points, the tangent plane cannot be defined or is not unique. Such points are called special points of surfaces, for example, the points of the edge of the return of the surface of the torso, the vertex of the conical surface, the points of the surface of revolution, where the meridian and the axis do not intersect at right angles, etc.

Figure 5.34 - Tangent plane

The task of constructing tangent planes passing through a given point on the surface is reduced to the following:

1. Any two secants are drawn through a point on a curved surface

planes.

2. Find the lines of section of the surface by these planes.

3. Build tangents at a given point to the section lines.

Two tangents define the desired plane. When choosing cutting planes, they tend to get the simplest kind of section - a straight line or a circle.

Consider the case of constructing a tangent plane through point A, which belongs to the surface of the cone of revolution (Figure 5.35).

To construct two necessary sections, one cutting plane is drawn through a given point A and the top of the cone. This plane will intersect the surface of the cone along the generatrix serving as the line of tangency, and therefore is one of the straight lines that define the tangent plane. The second straight line m, tangent to the circumference of the section of the cone by a horizontal level plane drawn through point A. The tangent could also be drawn to the circumference of the base of the cone.

Figure 5.35 - Tangent plane to the surface of the cone

5.6 Surface developments

A surface development is a flat figure formed by combining a surface with a plane.

From the geometric properties of surface elements that are preserved during unfolding, it can be noted that the surface line passes into the unfolded line and that the lengths of lines, the values ​​of plane angles and areas bounded by closed lines remain unchanged.

Not all surfaces can be exactly flattened. Therefore, the surfaces are divided into developable and non-developable. Developable surfaces include ruled surfaces: cylinders, cones and torsos, since adjacent generators are parallel or intersect, i.e. form a plane.

To build a sweep of a right circular cylinder, you need to build a rectangle with a base 2πR, where R is the radius of the base circle. The height of the rectangle is equal to the height of the cylinder (Figure 5.36).

2. What lines are obtained when planes intersect a cylinder of revolution?

3. What curves are obtained when the planes intersect the cone of revolution?

4. What are the extreme points of the curved section line?

5. In what cases is it recommended to use the method of auxiliary cutting planes or the method of auxiliary cutting spheres to construct a line of intersection of two curved surfaces?

6 COMPUTER GRAPHICS

6.1 Computer graphics and its place in computer-aided design

Computer graphics studies the methods and means of creating and processing images using software and hardware systems.

Computer graphics includes a complex of various software tools used to form, convert and display information in visual form on display devices (displays, graph plotters).

Among the hardware are specialized devices and general purpose devices.

The first are inputs such as light pen, digital tablets and output means - plotters(Figure 6.1).

Figure 6.1 - Specialized devices

To the second - Input Devices- "mouse" and "joystick" manipulators, and output devices-bitmap graphic displays, printers, keyboards(Figure 6.2).

The software is focused on the following main types of graphics: business, illustrative, scientific, design (for CAD), cartographic (architectural and land management CAD), fine arts and advertising.

Computer graphics developed in line with the general development of computer technology and software. Initially, programs were created for displaying graphs as part of application packages as part of high-level languages. For example, the GRAFOR package was created as part of FORTRAN language application packages.

Figure 6.2 - General purpose devices

AT further, the creation of graphic programs stood out as an independent direction of software.

AT Depending on the method of image formation, computer graphics are divided into:

∙ raster graphics;

∙ vector graphics;

∙ fractal graphics.

The image element in raster editors is a dot. A point can have several parameters: coordinates, color, tone, transparency. The image is made by systematization of points. In this case, there is an indicator of image resolution - the number of dots per unit area of ​​the image. Modern engineering graphics tools allow you to create images with a resolution of 2540 dpi (dots per inch) or more. Each point requires addressing for storage on media. A significant amount of data being processed, as well as the data required to save images, is a significant drawback of raster graphics.

A common disadvantage of raster editors is that when the image is scaled up, the points increase accordingly, so when the image is enlarged, its resolution and, as a result, accuracy are lost; inability to work with elements (enlarged images) - pixelation.

Since the image element is a point, the line will already require the systematization of points. From this we can conclude that the creation of two-dimensional and three-dimensional objects significantly complicate the description of the image, increasing the amount of processed and stored data.

Raster editors include Paint, Adobe Photoshop, etc. They are designed to create images such as artistic drawings, illustrations, graphics (Figure 6.3).

Figure 6.3 - Examples of using raster graphics

In vector graphics, the basic element is the line. The line is described mathematically as a single object, and therefore the amount of data for displaying an object in vector graphics is significantly lower than in raster graphics.

All considered graphic editors are either the simplest editors, for example, Paint, or a wide range of editors.

Three main blocks: a simulator, a calculation block and an expert system - perform all the main procedures that may be necessary during design work.

The calculation block can execute any program from the application package, which contains all the necessary programs used by developers. The call of a particular program is carried out at the request of either the simulator or expert system, or the constructor itself.

Database

Task formation block

User

Figure 6.5 - Typical diagram of CAD B Task formation block the designer introduces the technical

the design brief, which specifies all the goals to be achieved in the design and all the constraints that cannot be violated.

Unit for preparation of technical documentation allows the designer to prepare the necessary documents for the last two stages of creating new products.

Specific systems may deviate from this typical scheme.

Consider specific examples of CAD and engineering graphic editors and CAD / CAM / CAE systems

6.3 Functionality of 2D-3D Modeling Modules

The AutoCAD graphics system is the de facto standard in engineering graphics systems. The latest versions of AutoCAD are modern 32-bit Windows applications for engineers and CAD users. AutoCAD provides an efficient work environment and thus allows designers to concentrate more on projects and spend less time entering parameters from the keyboard.

Features such as the Multiple Design Environment, AutoCAD DesignCenter, Intellimouse support, and more support a natural, intuitive, efficient work environment.

SOLIDCAM is a product of CADTECH Ltd. - powerful

a tool for obtaining control programs for CNC machines when processing parts containing complex

surface or solid geometry. SOLIDCAM provides 2.5 and 3-axis milling with guaranteed

tired absence of "undercuts", turning

bodies of revolution, visualization of the cutting process with imitation of material removal.

Figure 6.6 - Using the program SOLIDCAM in production

The bCAD system was developed for a wide range of applications, so its functionality is quite universal (Figure 6.7).

The bCAD system is designed and developed as a universal designer's workstation, which allows to carry out a wide range of work in the “end-to-end” mode - from a drawing to a three-dimensional model or, conversely, from a three-dimensional representation to flat projections. At the same time, it is possible to produce technical documentation in accordance with the requirements of standards, obtain realistic images, and prepare data for settlement systems.

Figure 6.7 - Window of the bCAD system

Raster images prepared in bCAD can be written in GIF, TGA, BMP, JPG, TIFF or PCX formats and used in publishing or illustration packages.

Recently, when developing design documentation in the educational process of technical universities, the KOMPAS-3D system developed by the Russian company ASCON is widely used.

The drawing and design editor KOMPAS-3D contains sufficient drawing tools for making drawings of any level of complexity with full support for Russian standards. The simple and understandable interface of this program is successfully combined with the flexibility of a professional system when constructing, selecting, deleting drawing objects, typing according to GOST, setting dimensions of all types, shape tolerances and location of surfaces, positions, bases, etc.

KOMPAS-3D is designed specifically for the MS Windows operating environment and makes full use of all its features and benefits, providing the user with maximum efficiency and convenience in work.

The following graphical objects are supported in KOMPAS-3D.

Geometric objects:						straight line segment
	circular arc,				polygon,		broken line,
bezier Curve,		NURBS curve,		hatching,		equidistant curve,
macronutrient.
		linear size,			angle size,		radial size
	diametrical size,			height size.
Special and technological designations:							multiline

text inscription, base designation, shape and location tolerance,

Drawing design objects: technical requirements, main inscription (stamp), designation of roughness of unspecified surfaces.

The main documents in the KOMPAS-3D system are:

drawing, fragment, text document, specification, assembly and detail.

The main task solved with the help of any drawing system is the creation and release of various graphic documentation (Figure 6.10).

Figure 6.10 - Fragment of the detail drawing in KOMPAS-3D

The simplest and most understandable way of constructing is direct pointing to the input field with the cursor. For example, when creating a segment, its start point is sequentially fixed, and then the end point.

Another way is to specify the exact values ​​of the coordinates to move to the desired point and then fix it. To display and enter coordinates, special X and Y fields are provided, displayed on the right side of the Current Status Bar.

And, finally, the Object Parameters Bar allows you to implement the widest possibilities for managing drawing objects.

You can move drawing or fragment objects either with the mouse or using menu commands.

The basic methods of work are: moving objects with the mouse; copying objects with the mouse; simple removal of graphic objects; editing characteristic points of objects; editing object parameters.

The KOMPAS-3D system is capable of generating three-dimensional models of a part in order to transfer geometry to various design parameters or to packages for developing control programs for CNC equipment, as well as to create design documentation for the developed parts (Figure 6.11).

Figure 6.11 Example of work in KOMPAS-3D

The main tasks that KOMPAS-3D solves are the formation of a three-dimensional model of a part in order to transfer geometry to various calculation packages or to packages for developing control programs for

CNC ruding, as well as the creation of design documentation for the developed parts.

The generally accepted procedure for modeling a rigid body is the sequential execution of Boolean operations (union, subtraction and intersection) on solid elements (spheres, prisms, cylinders, cones, pyramids, etc.). An example of such operations is shown in Figure 6.12.

Figure 6.12 - An example of performing Boolean operations

Boolean operations on solid elements: a) cylinder; b) combination of a cylinder and a prism; c) prism subtraction; d) subtraction of the cylinder.

In KOMPAS-3D, to set the shape of three-dimensional elements, such a displacement of a flat figure in space is performed, the trace of which determines the shape of the element (for example, the rotation of a circular arc around an axis forms a sphere or a torus, a displacement of a polygon - a prism, etc.). Formation of volumetric elements: a) a prism, b) a torus, c) a kinematic element (Figure 6.12).

Figure 6.12 - Formation of volumetric elements

A flat figure, on the basis of which a body is formed, is called a sketch, and the shaping movement of a sketch is called an operation.

Questions for self-control on topic 6:

1. What does the term "computer graphics" include?

2. What belongs to computer graphics hardware?

3. List the main types of graphics.

4. According to the method of image formation, computer graphics are divided into ……….. What is their difference?

5. What is the basic element of fractal graphics?

6. What is the basic element of vector graphics?

7. What are the elements of a typical CAD system?

8. Name engineering graphic systems known to you.

9. What operations are used to model a rigid body?

Bibliography

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