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Construction of sections of polyhedra. Research work on the topic "Methods for constructing sections of polyhedra"

The task itself usually goes like this: "build a natural view of the section figure". Of course, we decided not to leave this question aside and try, if possible, to explain how the oblique section is constructed.

In order to explain how an oblique section is built, I will give a few examples. Of course, I will start with the elementary, gradually increasing the complexity of the examples. I hope that after analyzing these examples of section drawings, you will understand how this is done and will be able to complete your learning task yourself.

Consider a "brick" with dimensions of 40x60x80 mm by an arbitrary inclined plane. The cutting plane cuts it along the points 1-2-3-4. I think everything is clear here.

Let's move on to the construction of a natural form of the sectional figure.
1. First of all, let's draw the axis of the section. The axis should be drawn parallel to the section plane - parallel to the line into which the plane is projected on the main view - usually it is on the main view that the task is set for construction of an oblique section(Further on, I will always mention the main view, bearing in mind that this is almost always the case in training drawings).
2. On the axis, we set aside the length of the section. In my drawing, it is designated as L. The size L is determined in the main view and is equal to the distance from the point where the section enters the part to the point where it exits.
3. From the resulting two points on the axis perpendicular to it, we set aside the section widths at these points. The width of the section at the point of entry into the part and at the point of exit from the part can be determined in the top view. IN this case both segments 1-4 and 2-3 are equal to 60 mm. As you can see from the picture above, the edges of the section are straight, so we simply connect our two resulting segments, getting a rectangle 1-2-3-4. This is - the natural view of the figure of the section of our brick with an inclined plane.

Now let's complicate our detail. Let's put a brick on the base 120x80x20 mm and add stiffeners to the figure. Let's draw a cutting plane so that it passes through all four elements of the figure (through the base, brick and two stiffeners). In the picture below you can see three views and a realistic image of this part.

Let's try to build a natural view of this slanted section. Let's start again with the section axis: draw it parallel to the section plane indicated on the main view. On it we set aside the length of the section equal to A-E. Point A is the entry point of the section into the part, and in a particular case, the entry point of the section into the base. The exit point from the base is point B. Let's mark point B on the axis of the section. Similarly, we mark the entry-exit points to the edge, to the "brick" and to the second edge. From points A and B perpendicular to the axis, we set aside segments equal to the width of the base (on each side of the axis, 40, only 80 mm). Connect extreme points- we get a rectangle, which is a natural view of the section of the base of the part.

Now it's time to build a piece of section, which is a section of the edge of the part. From points B and C, we set aside perpendiculars of 5 mm in each direction - we will get segments of 10 mm. Connect the extreme points and get the cross section of the rib.

From points C and D we set aside perpendicular segments equal to the width of the "brick" - completely similar to the first example of this lesson.

Having set aside the perpendiculars from points D and E equal to the width of the second edge and connecting the extreme points, we obtain a natural view of its section.

It remains to erase the jumpers between the individual elements of the resulting section and apply hatching. You should get something like this:

If, according to a given section, we divide the figure, then we will see the following view:

I hope you are not intimidated by the tedious paragraphs of the description of the algorithm. If you have read all of the above and still do not fully understand, how to draw a cross section, I strongly advise you to take a piece of paper and a pencil in your hands and try to repeat all the steps after me - this will almost 100% help you learn the material.

Once I promised the continuation of this article. Finally, I am ready to present you with a step-by-step construction of an oblique section of a part, closer to the level of homework. Moreover, the oblique section is defined in the third view (the oblique section is defined in the left view)

or write down our phone number and tell your friends about us - someone is probably looking for a way to make drawings

or create a note about our lessons on your page or blog - and someone else will be able to master the drawing.

Yes, everything is fine, but I would like to see how the same thing is done on a more complex part, with chamfers and a cone-shaped hole, for example.

Thank you. But aren't the stiffeners hatched on the cuts?
Exactly. It is they who do not hatch. Because they are general rules making cuts. However, they are usually hatched when making cuts in axonometric projections - isometry, dimetry, etc. When performing inclined sections, the area related to the stiffener is also shaded.

Thank you, very accessible. Can you tell me if the oblique section can be done in the top view, or in the left view? If so, I would like to see the simplest example. Please.

It is possible to make such cuts. But unfortunately I don't have an example at hand right now. And there's another one interesting point: on the one hand, there is nothing new there, but on the other hand, in practice, it is really more difficult to draw such sections. For some reason, everything starts to get confused in the head and most students have difficulties. But don't give up!

Yes, everything is fine, but I would like to see how the same thing is done, but with holes (through and non-through), otherwise they never turn into an ellipse in my head

help me with a complex problem

It's a pity that you wrote here. We would write in the mail - maybe we could have time to discuss everything.

You explain well. What if one of the sides of the part is semicircular? Also, there are holes in the part.

Ilya, use the lesson from the section on descriptive geometry "Section of a cylinder by an inclined plane". With it, you can figure out what to do with holes (they are also cylinders in fact) and with a semicircular side.

I thank the author for the article! Brief and understandable. About 20 years ago I myself gnawed at the granite of science, now I help my son. I forgot a lot, but your article returned a fundamental understanding of the topic. I’ll go with the inclined section of the cylinder to deal with)

Add your comment.

Axioms of planimetry:

In various textbooks, the properties of lines and planes can be presented in different ways, in the form of an axiom, a consequence from it, a theorem, a lemma, etc. Consider the textbook Pogorelov A.V.

The straight line divides the plane into two half-planes.

From any half-line to a given half-plane, one can lay off an angle with a given degree measure, less than 180 0 , and only one.

Whatever the triangle, there exists an equal triangle at a given location with respect to the given half-line.

Through a point not lying on a given line, at most one line can be drawn in the plane parallel to the given line.

Axioms of stereometry:

Whatever the plane, there are points that belong to this plane, and points that do not belong to this plane, and points that do not belong to it.

If two different planes have a common point, then they intersect along a straight line passing through this point.

If two different lines have a common point, then a plane can be drawn through them, and moreover, only one.

Whatever the line, there are points that belong to this line, and points that do not belong to it.

Through any two points you can draw a line, and only one.

Of the three points on a line, one and only one lies between the other two.

Each segment has a certain length greater than zero. The length of a segment is equal to the sum of the lengths of the parts into which it is divided by any of its points.

A straight line belonging to a plane divides this plane into two half-planes.

Each angle has a certain degree measure greater than zero. The straight angle is 180 0 . The degree measure of an angle is equal to the sum of the degree measures of the angles into which it is divided by any ray passing between its sides.

On any half-line from its starting point, you can put off a segment of a given length, and only one.

From a half-line on the plane containing it, an angle with a given degree measure less than 180 can be plotted into a given half-plane 0 , and only one.

Whatever the triangle, there is an equal triangle in the given plane at a given location relative to the given half-line in that plane.

In a plane, through a given point not lying on a given line, at most one line parallel to the given line can be drawn.

cross section

In space, two figures, for our case, a plane and a polyhedron can have the following mutual arrangement: do not intersect, intersect at a point, intersect in a straight line and the plane intersects the polyhedron along its interior (Fig. 1), and at the same time form the following figures:

a) an empty figure (do not intersect)

b) point

c) cut

d) polygon

If there is a polygon at the intersection of a polyhedron and a plane, then this polygonis called a section of a polyhedron with a plane .

fig.1

Definition. cross section a spatial body (for example, a polyhedron) is a figure obtained at the intersection of a body with a plane.

cutting plane polyhedron Let's call any plane, on both sides of which there are points of a given polyhedron.

We will consider only the case when the plane intersects the polyhedron along its interior. In this case, the intersection of this plane with each face of the polyhedron will be a certain segment.

If the planes intersect in a straight line, then the straight line is calledfrom one of these planes to another.

In the general case, the secant plane of a polyhedron intersects the plane of each of its faces (as well as any other secant plane of this polyhedron). It also intersects each of the lines on which the edges of the polyhedron lie.

The line along which the secant plane intersects the plane of any face of the polyhedron is calledfollowing the cutting plane on the plane of this face, and the point at which the secant plane intersects the line containing any edge of the polyhedron is calledfollowing the cutting plane onthis straight line. This point is also the trace of a straight line on the cutting plane. If the cutting plane directly intersects the face of the polyhedron, then we can talk about the trace of the cutting plane on the face, and, similarly, abouttrace of a cutting plane on an edge of a polyhedron, that is, the trace of an edge on a cutting plane.

Since a straight line is uniquely determined by two points, to find the trace of a secant plane on any other plane and, in particular, on the plane of any face of a polyhedron, it suffices to construct two common points of the planes

To construct a trace of a secant plane, as well as to construct a section of a polyhedron by this plane, not only the polyhedron, but also the secant plane must be specified. And the construction of the section plane takes place depending on the assignment of this plane. The main ways to define a plane, and in particular a secant plane, are as follows:

three points not lying on one straight line;

a straight line and a point not lying on it;

two parallel lines;

two intersecting lines;

a point and two intersecting lines;

There are other ways to define the cutting plane.

Therefore, all methods for constructing sections of polyhedra can be divided into methods.

Methods for constructing sections of polyhedra

The method of sections of polyhedra in stereometry is used in construction problems. It is based on the ability to build a section of a polyhedron and determine the type of section.

There are three main methods for constructing sections of polyhedra:

Axiomatic method:

trace method.

Combined method.

coordinate method.

Note that the method of traces and the method of auxiliary sections are varietiesAxiomatic method for constructing sections.

We can also distinguish the following methods for constructing sections of polyhedra:

construction of a section of a polyhedron by a plane passing through a given point parallel to a given plane;

construction of a section passing through a given line parallel to another given line;

construction of a section passing through a given point parallel to two given skew lines;

construction of a section of a polyhedron by a plane passing through a given line perpendicular to a given plane;

construction of a section of a polyhedron by a plane passing through a given point perpendicular to a given straight line.

The main actions that make up the methods for constructing sections are finding the point of intersection of a straight line with a plane, constructing a line of intersection of two planes, constructing a straight line parallel to a plane perpendicular to the plane. To construct a straight line of intersection of two planes, two of its points are usually found and a straight line is drawn through them. To construct the point of intersection of a line and a plane, find a line in the plane that intersects the given one. Then the desired point is obtained at the intersection of the found line with the given one.

Consider separately listed by usmethods for constructing sections of polyhedra:

trace method.

trace method is based (operated) on the axioms of stereometry, the essence of the method is to construct an auxiliary line, which is the image of the line of intersection of the cutting plane with the plane of any face of the figure. It is most convenient to build an image of the line of intersection of the cutting plane with the plane of the lower base. This linecalled the main trace of the cutting plane . Using the trace, it is easy to construct images of the points of the cutting plane located on the side edges or faces of the figure. Consistently connecting the images of these points, we obtain the image of the desired section.

Note that when constructing the main trace of the secant plane, the following statement is used.

If the points belong to the secant plane and do not lie on one straight line, and their projection (central or parallel) onto the plane chosen as the main one are, respectively, the points then the points of intersection of the corresponding lines, that is, the points and lie on the same line (Fig. 1, a, b).

fig.1.a fig.1.b

This line is the main trace of the cutting plane. Since the points lie on the main trace, it suffices to find two of these three points to construct it.

Method of auxiliary sections.

This method of constructing sections of polyhedra is sufficiently universal. In cases where the desired trace (or traces) of the cutting plane is outside the drawing, this method even has certain advantages. At the same time, it should be borne in mind that the constructions performed using this method often turn out to be “crowded”. Nevertheless, in some cases the method of auxiliary sections turns out to be the most rational.

Combined method

The essence of the combined method for constructing sections of polyhedra is the application of theorems on the parallelism of lines and planes in space in combination with the axiomatic method.

Coordinate method for constructing sections.

The essence of the coordinate method is to calculate the coordinates of the intersection points of edges or a polyhedron with a secant plane, which is given by the equation of the plane. The equation of the section plane is calculated based on the conditions of the problem.

Note that this method of constructing a section of a polyhedron is acceptable for a computer, since it is associated with a large amount of calculations and therefore it is advisable to implement this method using a computer.

Our main task will be to construct a section of a polyhedron with a plane, i.e. in constructing the intersection of these two sets.

Construction of sections of polyhedra

First of all, we note that the section of a convex polyhedron is a convex flat polygon, whose vertices in the general case are the points of intersection of the secant plane with the edges of the polyhedron, and the sides with its faces.

Section creation examples:

There are many different ways to define a section. The most common of these is the method of specifying a cutting plane by three points that do not lie on one straight line.

Example 1 For box ABCDA 1 B 1 C 1 D 1 . Construct a section passing through the points M, N, L.

Solution:

Connect the points M and L lying in the plane AA 1 D 1 D.

Intersect the line ML (belonging to the section) with the edge A 1 D 1 1 D 1 D. Get point X 1 .

The point X1 lies on the edge A 1 D 1 , and hence the planes A 1 B 1 C 1 D 1 , connect it with a point N lying in the same plane.

X 1 N intersects edge A 1 B 1 at point K.

Connect the points K and M lying in the same plane AA 1 B 1 b.

Find the line of intersection of the section plane with the plane DD 1 C 1 C:

Intersect the line ML (belonging to the section) with the edge DD 1 , they lie in the same plane AA 1 D 1 D, get point X 2 .

Let us intersect the line KN (belonging to the section) with the edge D 1 C 1 , they lie in the same plane A 1 B 1 C 1 D 1 , we get the point X3;

Points X2 and X3 lie in the plane DD 1 C 1 C. Draw a line X 2 X 3 , which intersects the edge C 1 C at the point T, and the edge DC at the point P. And let's connect the points L and P, which lie in the plane ABCD.

Thus, the problem is considered solved if all the segments along which the plane intersects the faces of the polyhedron are found, which we did. MKNTPL - desired section.

Note. The same task for constructing a section can be solved using the property of parallel planes.

From the above, we can compose an algorithm (rule) for solving problems of this type.

Rules for constructing sections of polyhedra:

Example 2 DL, M

We solve by the axiomatic method:

Draw an auxiliary planeDKM, which intersects the edges AB and BC at points E andF(the course of the solution is in Fig. 2.). Let's build a "trace" of the CM of the section plane on this auxiliary plane, find the intersection point of the CM and EF- point P. Point P, as well asL, lies in the plane ABC, and it is possible to draw a straight line along which the section plane intersects the plane ABC ("trace" of the section in the plane ABC).

Example 3 On the edges AB and AD of the pyramid MABCD, we set the points P and Q, respectively, the midpoints of these edges, and on the edge MC, we set the point R. Let's construct a section of the pyramid by a plane passing through the points P, Q, and R.

The solution will be carried out by a combined method:

1). It is clear that the main trace of the plane PQR is the line PQ.

2). Find the point K at which the MAC plane intersects the line PQ. Points K and R belong to both the PQR plane and the MAC plane. Therefore, by drawing the straight line KR, we get the line of intersection of these planes.

3). Let's find the point N=AC BD, draw the line MN and find the point F=KR MN.

4). Point F is common point planes PQR and MDB, that is, these planes intersect along a straight line passing through the point F. At the same time, since PQ is the midline of triangle ABD, then PQ is parallel to BD, that is, the line PQ is also parallel to the plane MDB. Then the plane PQR passing through the line PQ intersects the plane MDB along the line parallel to the line PQ, that is, parallel to the line BD. Therefore, in the plane MDB through the point F we draw a line parallel to the line BD.

5). Further constructions are clear from the figure. As a result, we get the polygon PQD"RB" - the required section

Consider the sections of the prism for simplicity, that is, the convenience of logical thinking, consider the sections of the cube (Fig. 3.a):

Rice. 3.a

Sections of the prism by planes parallel to the side edges are parallelograms. In particular, diagonal sections are parallelograms (Fig. 4).

Def. diagonal section a prism is a section by a plane passing through two side edges that do not belong to the same face.

The polygon resulting from a diagonal section of a prism is a parallelogram. Question about the number of diagonal sectionsn-angular prism is more difficult than the question of the number of diagonals. There will be as many sections as there are diagonals at the base. We know that a convex prism has convex polygons at its bases, while a convexn-gon of diagonals. And so we can say that there are half as many diagonal sections as there are diagonals.

Note: When constructing sections of a parallelepiped in the figure, one should take into account the fact that if the cutting plane intersects two opposite faces along some segments, then these segments are parallel “by the property of the parallelepiped, i.e. Opposite faces of a parallelepiped are parallel and equal.

We provide answers to frequently asked questions:

What polygons are obtained in the section of a cube by a plane?

"triangle, quadrilateral, pentagon, hexagon".

Can a plane cross section of a cube produce a heptagon? And the octagon?

"can not".

3) The question arises, what is the largest number of sides of a polygon obtained by a section of a polyhedron with a plane?

Largest number sides of the polygon obtained in the section of the polyhedron by the plane is equal to the number of faces of the polyhedron .

Example 3 Construct a section of prism A 1 B 1 C 1 D 1 ABCD by a plane passing through three points M, N, K.

Consider the case of location of points M, N, K on the surface of a prism (Fig. 5).

Consider the case: In this case it is obvious that M1 = B1.

Building:

Example 4 Construct a section of the parallelepiped ABCDA 1 B 1 C 1 D 1 a plane passing through the points M, N, P (the points are indicated on the drawing (Fig. 6)).

Solution:

Rice. 6

Points N and P lie in the plane of the section and in the plane of the lower base of the parallelepiped. Let's construct a line passing through these points. This line is the trace of the secant plane on the plane of the base of the parallelepiped.

Let us continue the line on which the side AB of the parallelepiped lies. The lines AB and NP intersect at some point S. This point belongs to the section plane.

Since the point M also belongs to the section plane and intersects the line AA 1 at some point x.

Points X and N lie in the same plane of face AA 1 D 1 D, connect them and get the line XN.

Since the planes of the faces of the parallelepiped are parallel, it is possible to draw a straight line through the point M in the face A 1 B 1 C 1 D 1 parallel to line NP. This line will intersect side B 1 WITH 1 at point Y.

Similarly, we draw the line YZ, parallel to the line XN. We connect Z with P and get the desired section - MYZPNX.

Sections of the pyramid by planes passing through its top are triangles. In particular, diagonal sections are triangles. These are sections by planes passing through two non-adjacent side edges of the pyramid.

Example 4 Construct a section of the pyramid ABCDa plane passing through the points K,L, M.

Solution:

Let's solve the same example differently. .

Let us assume that for points K,L, and М the sectionKLNM(Fig. 7). Denote byFthe intersection point of the diagonals of the quadrilateralKLNM. Let's draw a straight lineD.F.and denote byF 1 its point of intersection with face ABC. DotF 1 coincides with the point of intersection of lines AM and SK (F 1 simultaneously belongs to the planes AMDAndDSC). pointF 1 easy to build. Next we build a pointFas a point of intersectionD.F. 1 AndLM. Next we find the pointN.

The method considered is calledinternal design method . (For our case we are talking about central design. quadrilateralKISA is the projection of a quadrilateralKMNLfrom a pointD. In this case, the point of intersection of the diagonalsKMNL- dotF- goes to the point of intersection of the diagonals of the quadrilateralKISA - dotF 1 .

Sectional area of a polyhedron.

The problem of calculating the cross-sectional area of a polyhedron is usually solved in several stages. If the problem says that the section is built (or that the cutting plane is drawn, etc.), then at the first stage of the solution, the shape of the figure obtained in the section is found out.

This must be done in order to select the appropriate formula for calculating the cross-sectional area. After the form of the figure obtained in the section is clarified and the formula for calculating the area of this figure is chosen, they proceed directly to the computational work.

In some cases, it may turn out to be easier if, without finding out the form of the figure obtained in the section, we proceed immediately to calculating its area using the formula that follows from the theorem.

The theorem on the area of the orthogonal projection of a polygon: the area of the orthogonal projection of a polygon onto a plane is equal to the product of its area and the cosine of the angle between the plane of the polygon and the projection plane: .

A valid formula for calculating the cross-sectional area is: where is the area of the orthogonal projection of the figure obtained in the section, and is the angle between the secant plane and the plane on which the figure is projected. With such a solution, it is necessary to construct an orthogonal projection of the figure obtained in the section and calculate

If the condition of the problem says that the section needs to be built and the area of the obtained section should be found, then at the first stage it is reasonable to build the given section, and then, naturally, determine the shape of the figure obtained in the section, etc.

We note the following fact: since sections of convex polyhedra are constructed, the section polygon will also be convex, so its area can be found by dividing it into triangles, that is, the section area is equal to the sum of the areas of the triangles from which it is composed.

Task 1.

– correct triangular pyramid with a base side equal and equal to a height Construct a section of the pyramid by a plane passing through the points, where is the midpoint of the side, and find its area (Fig. 8).

Solution.

The cross section of the pyramid is a triangle. Let's find its area.

Since the base of the pyramid is an equilateral triangle and the point is the midpoint of the side, then it is the height and then, .

The area of a triangle can be found:

Task 2.

Side rib of a regular prism is equal to the side of the base. Construct sections of a prism by planes passing through a pointA, perpendicular to the line If you find the area of the resulting section of the prism.

Solution.

Let's construct the given section. Let us do this from purely geometrical considerations, for example, as follows.

In a plane passing through a given line and a given point, we draw a line through this point perpendicular to the line (Fig. 9). Let us use for this purpose the fact that in the triangle that is, its median is also the height of this triangle. Thus, a straight line.

Through the point we draw another line perpendicular to the line. Let us draw it, for example, in a plane passing through a straight line. It is clear that this line is a line

So, two intersecting lines, perpendicular to the line, are constructed. These lines define a plane passing through a point perpendicular to the line, that is, a secant plane is given.

We construct a section of the prism by this plane. Note that since the line is parallel to the plane. Then the plane passing through the line intersects the plane along a line parallel to the line, that is, the line. Draw a straight line through the point and connect the resulting point with a dot.

Quadrilateral given section. Let's determine its area.

It is clear that a quadrilateral is a rectangle, that is, its area

rice. 9

CONSTRUCTION OF SECTIONS AND SECTIONS ON DRAWINGS

The drawing of the part is formed by sequentially adding the necessary projections, cuts and sections. Initially, a custom view is created with a user-specified model, and the model orientation is set to best suit the main view. Further, the necessary cuts and sections are created for this and the following types.

The main view (front view) is selected in such a way that it gives the most complete idea of the shapes and dimensions of the part.

Sections in drawings

Depending on the position of the cutting plane, the following types of cuts are distinguished:

A) horizontal, if the cutting plane is parallel to the horizontal projection plane;

B) vertical, if the cutting plane is perpendicular to the horizontal projection plane;

C) inclined - the cutting plane is inclined to the projection planes.

Vertical sections are divided into:

· frontal - the cutting plane is parallel to the frontal projection plane;

· profile - cutting plane is parallel to the profile projection plane.
Depending on the number of cutting planes, the cuts are:

· simple - with one cutting plane (Fig. 107);

· complex - with two or more cutting planes (Fig. 108)
The standard provides for the following types of complex cuts:

· stepped, when the secant planes are parallel (Fig. 108 a) and broken lines - the secant planes intersect (Fig. 108 b)

Fig.107 Simple cut

A) b)

Fig.108 Complex cuts

Designation of cuts

In the case when in a simple section the secant plane coincides with the plane of symmetry of the object, the section is not indicated (Fig. 107). In all other cases, the sections are indicated by capital letters of the Russian alphabet, starting with the letter A, for example, A-A.

The position of the cutting plane in the drawing is indicated by the section line - a thickened open line. With a complex cut, strokes are also carried out at the inflections of the section line. Arrows indicating the direction of view should be placed on the initial and final strokes, the arrows should be at a distance of 2-3 mm from the outer ends of the strokes. On the outside of each arrow indicating the direction of view, the same capital letter is applied.

The same button is used to designate cuts and sections in the KOMPAS system Section line located on the Legend page (fig.109).

Fig.109 Section line button

Connecting Half View to Half Section

If the view and section are symmetrical figures (Fig. 110), then you can connect half of the view and half of the section, separating them with a dash-dotted thin line, which is the axis of symmetry. Part of the section is usually placed to the right of the axis of symmetry separating the part of the view from the part of the section, or below the axis of symmetry. Hidden contour lines on the connected parts of the view and section are usually not shown. If the axial line separating the view and the section coincides with the projection of some line, for example, the edge of a faceted figure, then the view and the section are separated by a solid wavy line drawn to the left of the symmetry axis if the edge lies on the inner surface, or to the right if the edge is outer .

Rice. 110 Connecting part of a view and a section

Building cuts

We will study the construction of sections in the KOMPAS system using the example of constructing a drawing of a prism, the task for which is shown in Fig. 111.

The drawing sequence is as follows:

1. By given dimensions let's build a solid model of a prism (Fig. 109 b). Let's save the model in the computer's memory in a file named "Prism".

Fig.112 Lines panel

3. To build a profile section (Fig. 113) draw a line section A-A on the main view using the button Cut line.

Fig.113 Construction of a profile section

The direction of view and the text of the designation can be selected on the control panel with the command at the bottom of the screen (Fig. 114). The construction of the section line is completed by pressing the Create object button.

Fig.114 Control panel for the command for constructing cuts and sections

4. On the Associative Views panel (Fig. 115), select the Cut line button, then specify the cut line with the trap that appears on the screen. If everything is done correctly (the cut line must be drawn in the active view), then the cut line will turn red. After specifying the cut line A-A, an image phantom will appear on the screen in the form of an overall rectangle.

Fig.115 Associative views panel

With the help of the Cut/section switch on the Property bar, the image type is selected - Cut (Fig. 116) and the scale of the displayed cut.

Fig.116 Control panel for the command for constructing cuts and sections

The profile section will be built automatically in the projection connection and with a standard notation. If necessary, the projection connection can be turned off by the switch Projection connection (Fig. 116). To set hatching parameters that will be used in the created section (section), use the controls on the Hatching tab.

Fig.117 Construction of a horizontal section B-B and section C-C

If the selected cutting plane when constructing the cut coincides with the plane of symmetry of the part, then in accordance with the standard, such a cut is not indicated. But if you simply erase the section designation, then due to the fact that the view and the section in the computer's memory are interconnected, the entire section will be erased. Therefore, in order to remove the designation, you must first destroy the connection between the view and the section. To do this, click the left mouse button to select the section, and then click the right mouse button to open the context menu, from which the Destroy view item is selected (Fig. 97). The section symbol can now be deleted.

5. To construct a horizontal section, let's draw a B-B section line through the lower plane of the hole in the front view. The front view must first be made current by two clicks of the left mouse button. Then a horizontal section is built (Fig. 117).

6. When constructing a frontal section, a part of the view and a part of the section are compatible, because they are symmetrical figures. The outer edge of the prism is projected onto the line separating the view and the cut, so we delimit view and section of a solid thin wavy line drawn to the right of the axis of symmetry, because outer rib. The button is used to draw a wavy line. Bezier curve located on the Geometry panel drawn with the For clipping line style (Fig. 118). Sequentially specify the points through which the Bezier curve should pass. To finish the command execution, click the Create object button.

Fig.118 Selecting a line style for a break

Sectioning

A section is an image of an object that is obtained by mentally dissecting an object with a plane. The section shows only what is located in the cutting plane.

The position of the cutting plane, with which the section is formed, is indicated in the drawing by the section line, just as for sections.

Sections, depending on their location in the drawings, are divided into extended and superimposed. The removed sections are most often located on the free field of the drawing and are outlined by the main line. The superimposed sections are placed directly on the image of the object and outlined with thin lines (Fig. 119).

Fig.119 Construction of sections

Consider the sequence of constructing a drawing of a prism with an extended oblique section B-B(Fig. 117).

1. Make the front view active by double-clicking the left mouse button on the view and draw a section line with the button cutting line . Let's select the text of the inscription В-В.

2. Using the Cut line button located on the Associative Views panel (Fig. 115), which appears as a trap, indicate the secant line planes B-B. Using the Cut/section switch on the Property bar, select the image type - Section (Fig. 116), the scale of the displayed section is selected from the Scale window.

The constructed section is located in a projection relationship, which limits its movement in the drawing, but the projection relationship can be turned off using the button projection connection.

On the finished drawing, draw centerlines, if necessary, enter the dimensions.

The tasks for constructing sections of polyhedra take significant place as a school geometry course for high school, and at exams at various levels. The solution of this type of problems contributes to the assimilation of the axioms of stereometry, the systematization of knowledge and skills, the development spatial representation and constructive skills. Difficulties that arise in solving problems on the construction of sections are well known.

From the early childhood we are faced with cuts. We cut bread, sausage and other products, cut a stick or pencil with a knife. The secant plane in all these cases is the plane of the knife. Sections (sections of pieces) are different.

The section of a convex polyhedron is a convex polygon, the vertices of which, in the general case, are the points of intersection of the cutting plane with the edges of the polygon, and the sides are the lines of intersection of the cutting plane with the faces.

To construct a line of intersection of two planes, it is enough to find two common points of these planes and draw a line through them. This is based on the following statements:

1. if two points of a straight line belong to a plane, then the whole line belongs to this plane;

2. if two different planes have a common point, then they intersect along a straight line passing through this point.

As I have already said, the construction of sections of polyhedra can be carried out on the basis of the axioms of stereometry and theorems on the parallelism of lines and planes. At the same time, there are certain methods for constructing plane sections of polyhedra. The following three methods are the most effective:

trace method

Internal design method

Combined method.

In the study of geometry and, in particular, those sections of it where images are considered geometric shapes, images of geometric shapes help the use of computer presentations. With the help of a computer, many geometry lessons become more visual and dynamic. Axioms, theorems, proofs, tasks for construction, tasks for constructing sections can be accompanied by successive constructions on the monitor screen. Computer-generated drawings can be saved and pasted into other documents.

I want to show a few slides on the topic: "Construction of sections in geometric bodies»

To construct the point of intersection of a line and a plane, find a line in the plane that intersects the given line. Then the desired point is the point of intersection of the found line with the given one. Let's see it on the next slides.

Task 1.

Two points M and N are marked on the edges of the tetrahedron DABC; M GAD, N b DC. Pick the point of intersection of the line MN with the plane of the base.

Solution: in order to find the point of intersection of the line MN with the plane

base we will continue AC and segment MN. Let us mark the point of intersection of these lines through X. The point X belongs to the line MN and the face AC, and AC lies in the plane of the base, which means that the point X also lies in the plane of the base. Therefore, the point X is the point of intersection of the line MN with the plane of the base.

Let's consider the second problem. Let's complicate it a little.

Task 2.

Given a tetrahedron DABC of points M and N, where M € DA, N C (DBC). Find the point of intersection of the line MN with the plane ABC .

Solution: The point of intersection of the line MN with the plane ABC must lie in the plane that contains the line MN and in the plane of the base. We continue the segment DN to the point of intersection with the edge DC. We mark the point of intersection through E. We continue the line AE and MN to the point of their intersection. Note X. The point X belongs to MN, so it lies on the plane that contains the line MN and X belongs to AE, and AE lies on the plane ABC. So X also lies in the plane ABC. Hence X is the point of intersection of the line MN and the plane ABC.

Let's complicate the task. Consider a section of geometric figures by planes passing through three given points.

Task 3

Points M, N and P are marked on the edges AC, AD and DB of the tetrahedron DABC. Construct a section of the tetrahedron by the plane MNP.

Solution: construct a straight line along which the plane MNP. Intersects face plane ABC. Point M is a common point of these planes. To build another common point, we continue the segment AB and NP. We mark the intersection point through X, which will be the second common point of the plane MNP and ABC. So these planes intersect along the straight line MX. MX intersects the edge BC at some point E. Since E lies on MX and MX is a line belonging to the plane MNP, it follows that PE belongs to MNP. The quadrilateral MNPE is the required section.

Task 4

We construct a section of a straight prism ABCA1B1C1 by a plane passing through the points P , Q,R, where R belongs to ( AA 1C 1C), R belongs IN 1C1,

Q belongs to AB

Solution: All three points P,Q,R lie in different faces, so we cannot yet construct a line of intersection of the secant plane with any face of the prism. Let's find the intersection point of PR with ABC. Let us find the projections of the points P and R onto the base plane PP1 perpendicular to BC and RR1 perpendicular to AC. Line P1R1 intersects line PR at point X. X is the point of intersection of line PR with plane ABC. It lies in the desired plane K and in the plane of the base, like the point Q. XQ is a straight line intersecting K with the plane of the base. XQ intersects AC at point K. Therefore, KQ is the segment of the intersection of the plane X with the face ABC. K and R lie in the X plane and in the plane of the AA1C1C face. Draw a line KR and mark the point of intersection with A1Q E. KE is the line of intersection of the plane X with this face. Find the line of intersection of the X plane with the plane of the faces BB1A1A. KE intersects with A1A at point Y. The line QY is the line of intersection of the secant plane with the plane AA1B1B. FPEKQ - desired section.

Tasks for constructing sections of a cube by a plane, as a rule, are simpler than, for example, tasks for sections of a pyramid.

We can draw a line through two points if they lie in the same plane. When constructing sections of a cube, one more option for constructing a trace of a cutting plane is possible. Since the third plane intersects two parallel planes along parallel straight lines, then if a straight line has already been built in one of the faces, and there is a point in the other through which the section passes, then we can draw a straight line through this point parallel to the given one.

Consider on concrete examples how to construct sections of a cube by a plane.

1) Construct a section of the cube by a plane passing through points A, C and M.

Problems of this type are the simplest of all problems for constructing sections of a cube. Since points A and C lie in the same plane (ABC), we can draw a line through them. Its trace is segment AC. It is invisible, so we depict AC with a stroke. Similarly, we connect the points M and C, which lie in the same plane (CDD1), and the points A and M, which lie in the same plane (ADD1). Triangle ACM is the required section.

2) Construct a section of the cube by a plane passing through the points M, N, P.

Here, only the points M and N lie in the same plane (ADD1), so we draw a straight line through them and get the trace MN (invisible). Since the opposite faces of the cube lie in parallel planes, the cutting plane intersects the parallel planes (ADD1) and (BCC1) along parallel lines. We have already built one of the parallel lines - this is MN.

Through the point P we draw a line parallel to MN. It intersects the edge BB1 at the point S. PS is the trace of the secant plane in the face (BCC1).

We draw a straight line through the points M and S, which lie in the same plane (ABB1). Got the MS trace (visible).

Planes (ABB1) and (CDD1) are parallel. There is already a line MS in the plane (ABB1), so through the point N in the plane (CDD1) we draw a line parallel to MS. This line intersects the edge D1C1 at point L. Its trace is NL (invisible). Points P and L lie in the same plane (A1B1C1), so we draw a straight line through them.

The pentagon MNLPS is the required section.

3) Construct a section of the cube by a plane passing through the points M, N, P.

Points M and N lie in the same plane (BCC1), so a straight line can be drawn through them. We get the trace MN (visible). The plane (BCC1) is parallel to the plane (ADD1), so through the point P lying in (ADD1) we draw a line parallel to MN. It intersects the edge AD at point E. We got the trace PE (invisible).

There are no more points lying in the same plane, or a line and a point in parallel planes. Therefore, one of the already existing lines must be continued in order to obtain an additional point.

If we continue the line MN, then, since it lies in the plane (BCC1), we need to look for the intersection point of MN with one of the lines of this plane. There are already intersection points with CC1 and B1C1 - these are M and N. The lines BC and BB1 remain. We continue BC and MN to the intersection at point K. The point K lies on the line BC, which means that it belongs to the plane (ABC), so we can draw a line through it and the point E lying in this plane. It intersects edge CD at point H. EH is its trace (invisible). Since H and N lie in the same plane (CDD1), a straight line can be drawn through them. We get the trace HN (invisible).

Planes (ABC) and (A1B1C1) are parallel. One of them contains the line EH, the other contains the point M. We can draw a line through M parallel to EH. We get the trace MF (visible). We draw a straight line through points M and F.

Hexagon MNHEPF is the required section.

If we continued the line MN to the intersection with another line in the plane (BCC1), with BB1, then we would get a point G belonging to the plane (ABB1). This means that through G and P it is possible to draw a line whose trace is PF. Further, we draw straight lines through points lying in parallel planes, and we arrive at the same result.

Working with straight line PE gives the same cross section MNHEPF.

4) Construct a section of the cube by a plane passing through the point M, N, P.

Here we can draw a straight line through points M and N lying in the same plane (A1B1C1). Her footprint is MN (visible). There are no more points lying in the same plane or in parallel planes.

We continue the line MN. It lies in the plane (A1B1C1), so it can only intersect with one of the lines in this plane. There are already intersection points with A1D1 and C1D1 - N and M. Two more lines of this plane are A1B1 and B1C1. The intersection point of A1B1 and MN is S. Since it lies on the line A1B1, it belongs to the plane (ABB1), which means that a line can be drawn through it and the point P, which lies in the same plane. The line PS intersects the edge AA1 at the point E. PE is its trace (visible). Through points N and E, lying in the same plane (ADD1), it is possible to draw a straight line, the trace of which is NE (invisible). There is a line NE in the plane (ADD1), and a point P in the plane parallel to it (BCC1). Through the point P we can draw a line PL parallel to NE. It intersects the edge CC1 at point L. PL is the trace of this line (visible). Points M and L lie in the same plane (CDD1), which means that a straight line can be drawn through them. Her footprint is ML (invisible). The pentagon MLPEN is the required section.

It was possible to continue the line NM in both directions and look for its points of intersection not only with the line A1B1, but also with the line B1C1, which also lies in the plane (A1B1C1). In this case, we draw two straight lines through the point P at once: one in the plane (ABB1) through the points P and S, and the second one in the plane (BCC1), through the points P and R. After that, it remains to connect the points lying in the same plane: M c L, E - with N.