Intersection of planes defined by triangles online. Straight as the line of intersection of planes

One of the fundamental tasks of descriptive geometry is the task of constructing the line of intersection of two planes general position. The cases of specifying planes are different, but in any case you will encounter a task in which you will need to construct a line of intersection of two planes defined by triangles (or other flat geometric shapes). I propose to consider an algorithm for solving such a problem now.

So, two planes are given, defined by triangles ABC and DEF. The method boils down to alternately finding two points of intersection of two edges of one triangle with the plane of another. By connecting these points we get the line of intersection of two planes. The construction of the point of intersection of a straight line with a plane was discussed in more detail in the previous lesson, let me remind you only of the mechanical actions:

Let us enclose the straight line AC in the frontally projecting plane and transfer along the communication lines to the horizontal projection the points of intersection of this plane with the straight lines DE and DF - points 1 and 2

On the horizontal projection, we connect the projections of points 1 and 2 and find the point of intersection of the resulting line with the horizontal projection of the straight line that we included in the frontally projecting plane, in this case with the straight line AC. We got the M point.

Let us enclose the straight line BC in the frontally projecting plane and transfer along the communication lines to the horizontal projection the points of intersection of this plane with the straight lines EF and DF - points 3 and 4

Let's connect their horizontal projections and get the point of intersection of this line with straight line BC - point N.

By connecting points M and N we get the line of intersection of the planes defined by the triangles. In fact, the intersection line has already been found. - All that remains is to determine the visibility of the edges of the triangles. This is done using the competing point method.

With the help of the most attentive site visitors, we were able to find an inaccuracy in determining the visibility of the planes. Below is a drawing in which the visibility of the lines limiting the planes on the horizontal plane has been corrected

17. Method of replacing projection planes.

Changing the relative position of the object being studied and the projection planes is achieved by replacing one of the planes P 1 or P 2 new planes P 4 (rice. 148 ). The new plane is always selected perpendicular to the remaining projection plane.

To solve some problems, double replacement of projection planes may be required (Fig. 149 ). Consecutive transition from one system of projection planes to another must be carried out by following the following rule: the distance from the new projection of the point to the new axis must be equal to the distance from the replaced projection of the point to the replaced axis.

Problem 1 : Determine the natural size of the segment AB straight line of general provisions (Fig. 148 ). From the property of parallel projection it is known that a segment is projected onto a plane in full size if it is parallel to this plane.

Let's choose a new projection plane P 4 , parallel to the segment AB and perpendicular to the plane P 1 . By introducing a new plane, we move from the system of planes P 1 P 2 into the system P 1 P 4 , and in the new system of planes the projection of the segment A 4 IN 4 will be the natural value of the segment AB .

3.6. Constructing the line of intersection of two planes

The straight line obtained by the mutual intersection of two planes is determined by two points, each of which simultaneously belongs to both planes.

In Fig. 3.37 the general position plane defined by triangle ABC intersects the frontally projecting plane defined by triangle DEF. Since triangle DEF is projected onto plane V in the form of a straight line D "F", then the frontal projection of the line of intersection of both planes is a segment K 1 "K2". We find its horizontal projection and determine visibility.

Fig.3.37 Fig.3.38

The horizontally projecting plane a intersects the plane triangle ABC(Fig. 3.3 8), The horizontal projection of the line of intersection of these planes is a segment M "N", which is determined on the trace of the axis.

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If the planes are defined by traces on the projection planes, then the currents that define the line of intersection of the planes) should be selected at the intersection points of the same plane traces(Fig. 3.39); the straight line passing through these points common to both planes is their line of intersection. Therefore, to construct projections of the line of intersection of planes  and  it is necessary:

1) find point M" at the intersection of traces н" and н" and point N" at the intersection of    and   , and along them the projections M" and N".

2) draw straight lines MN and M"N".

The intersection points of plane traces of the same name aretraces of the line of intersection of these planes.

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is.3.40

In Fig. 3.40 planes  and  intersect. Plane  is a general plane, Plane  is a horizontal plane. To construct an intersection line you must:

1) find point N" at the intersection of traces  and v;

2) draw a straight line through this point, based on the position

planes and their traces.

Figures (3.40 - 3.42) show cases when the direction of the intersection line is known. Therefore, it is enough to have only one point from the intersection of the traces and, then, draw a straight line through this point, based on the position of the planes and their traces.

3.7.Intersection of a straight line with a general plane

The construction of the point of intersection of a straight line with a general plane is performed using the following algorithm:

1) draw some auxiliary plane () through this line (MN);

2) construct a straight line (ED), the line of intersection of this plane (ABC) and the auxiliary plane ();

3) determine the position of the point (K) of intersection of this straight line (MN) and the constructed intersection line (ED);

4) determine the visibility of the straight line (MN) relative to the H and V planes.

In Fig. 3.43, straight line MN intersects the plane defined by triangle ABC. Through the straight line MN we draw

nickname ABC. Through the straight line MN we draw

G horizontally projecting plane . Since the auxiliary plane  is horizontally projecting, then the horizontal projection of the plane  and triangle ABC is the straight line E "D". Finding its frontal projection E"D". Then we construct K", in which E" D" intersects M"N" and determine its horizontal projection K". We determine the visibility of the segments MK and

K

N using competing points

Fig.3.44 Fig.3.45 3.46

In Fig. 3.44 straight line AB intersects plane a in general position. We draw a horizontal projection plane  through straight line AB and find the line of intersection of plane a and plane  (MN).

We define point K" as the intersection point of M"N" and A"B". We find point K" and determine visibility.

In Fig. 3.45 plane a is defined by traces. The straight line intersecting the plane  is horizontal, Through straight line AB we draw a horizontal plane (||Н). Plane p intersects plane a horizontallyN.K., belonging to the plane Then we determine visibility. In Fig. 3.46 plane a is given by traces; the straight line AB intersecting the plane a is horizontally projecting, onto the plane H it is projected to a point and, therefore, the horizontal projection of the point of intersection of the straight line AB and the plane (K) is located at this point.

A"=B=K", Position K" is determined using the horizontal.

3.8. Intersection of two generic planes

Let's consider the general case of constructing a line of intersection of two planes (Fig. 3.47).

One of the intersecting planes () is defined by two intersecting straight lines (AB  BC). The second plane () is defined by two parallel lines (DE FG). As a result of the mutual intersection of planes  and , the straight line K 1 K 2 (== K 1 K 2) is obtained. To determine the position of points K 1 and K 2, we take two auxiliary frontally projecting planes  1 and  2 intersecting both plane  and plane . When plane  1 intersects plane , a straight line is formed with projections 1"2" and 12". When plane  1 intersects plane , a straight line is formed with projections 3"4" and 3"4". The intersection of lines 12 and 34 determines the first point K 1 line of intersection of planes  and .

By introducing the front-projection plane  2 , at its intersection with planes  and  we obtain straight lines with projections 5 "b", 5 "b" and 7 "8", 7 "8". These straight lines located in the plane  2 , V

at their intersection, they determine the second point K 2 of the intersection line  and . Having obtained the projections K 1 " and K 2 ", we find on the traces  1 v" and  2 v" the projections K 1 " and K 2 ". The projections K 1 "K 2  and K 1 "K 2 " are projections of the desired line of intersection of the planes  and .

3.9. Constructing a line of intersection of two planes using the intersection points of straight lines with the plane

This method consists in finding the intersection points of two straight lines belonging to one of the planes with another plane. Therefore, it is necessary to be able to construct the point of intersection of a straight line with a general plane (Fig. 3.43).

N and fig. 3.48 shows the construction of the line of intersection of two triangles ABC and DEF. Straight line K 1 K 2 is constructed at the points of intersection of sides AC and BC of triangle ABC with the plane of triangle DEF. Auxiliary frontal projecting plane  1 drawn through AC intersects triangle DEF along a straight line with projections 1."2" and 1"2"; at the intersection of projections A "C" and 1 "2" we obtain the horizontal projection of point K 1 " - intersection

straight AC and triangle DEF. Then we build the frontal projection K 1 //

The auxiliary frontal projection plane  2, drawn through BC, intersects the triangle DEF in a straight line with projections 3"4" and 3"4", At the intersection of projections 3"4" and BC" we obtain a horizontal projection of point K 2 - the intersection of the straight line BC and triangle DEF. Then we build a frontal projection of point K2. Visibility in the drawing is determined by the method of competing points (see, Fig. 3.36),

4. METHODS OF CONVERTING A DRAWING

Specifying straight lines and flat figures in particular positions relative to projection planes, it significantly simplifies the construction and solution of problems, allows you to obtain an answer either directly from a given drawing, or using simple constructions. Such a particular mutual arrangement of straight lines, flat figures and projection planes can be ensured by transforming the drawing. This is achieved:

1) by introducing additional projection planes so that a straight line or flat figure, without changing its position in space, finds itself in some particular position in the new system of projection planes (a method of changing projection planes).

2) changing the position of a straight line or a flat figure, by rotating around a certain axis so that the straight line or figure is in a particular position relative to an unchanged system of projection planes (the method of rotation and a special case of its method of alignment).

3) changing the position of a straight line or a flat figure by moving them to a particular position so that the trajectories of movement of their points are in parallel planes with a constant system of projection planes (method of parallel movement).

4.1 Method of changing projection planes

The essence of the method of changing projection planes is that the position of points, straight lines, flat figures, surfaces in space remains unchanged, and the system H, V is supplemented by planes that form with H or with V, or between themselves systems of two mutually perpendicular planes, taken as projection planes,

Each new system is selected in such a way as to obtain the most advantageous position for performing the required construction.

4.1.1. Introduction to the H, V system of one additional projection plane

In most cases, an additional plane in the system , V is introduced according to a certain condition that meets the purpose of the construction. An example is plane V 1 in Fig. 4.1.

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Since it was necessary to determine the size of the segment AB and the angle between AB and the plane H, then the plane Vi is located perpendicular to the plane H (the system H, V 1 was formed) and parallel to AB

WITH
Therefore, in the system H, V 1, the segment AB is the frontal (A "B" || axis X 1) and the value A 1 "B 1" is equal to the natural value of the segment AB, the angle  1 is equal to the angle of inclination ka AB to the plane H.

Fig.4.2 Fig.4.3

In Fig. 4.2, the choice of plane H 1 is also subordinate to the goal: to determine the angle between straight line CD and plane V, as well as the natural size of the segment CD. Therefore, the plane H 1 is chosen perpendicular to V and at the same time parallel to the segment CD (axis H 1 /V || C "D") Therefore, in the system V, H 1 the segment CD is horizontal

(C"D" axis V/H 1), the value of C 1 "D 1" is equal to the natural value

segment CD, and the angle φ 2 is equal to the angle of inclination of the segment CD to the plane V.

In the case shown in Fig. 4.3, the choice of plane H 1 completely depends on the task.

It is necessary to determine the natural appearance of triangle ABC. Since in this case the plane defined by the triangle is perpendicular to the plane V, then to image it without distortion it is necessary to introduce into the system H 1, V an additional plane that meets two conditions: H 1, V (to form systems V,H 1) and H 1 || ABC (H 1 || A"B"C"), which makes it possible to depict triangle ABC on the Hi plane without distortion. New V/H axis 1 || A"B"C". To construct the projection A" 1 B 1 "C" 1 from the new axis, we lay off segments equal to the distances of points A, B, C" from the V/H axis. The natural appearance of triangle ABC is expressed by its new projection A" 1 B" 1 C" 1 .

The introduction of an additional projection plane makes it possible to transform the drawing in such a way that the general position plane given V system H, V, become perpendicular to the additional projection plane.

Fig.4.4 Fig.4.5

In Fig. 4.4, the general position plane, defined by the triangle ABC in the H, V system, becomes perpendicular to the additional projection plane V 1. To do this, a horizontal line AD is drawn in triangle ABC. The plane perpendicular to AD is perpendicular to ABC and at the same time to the plane H (since ADH). This corresponds to plane V 1 and triangle ABC is projected onto it into the segment B" 1 C" 1. Angle f 1 corresponds to the angle of inclination of triangle ABC to plane H.

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If we take the plane H 1 (Fig. 4.5), perpendicular to the plane V and the plane defined by the triangle ABC (for which it is necessary to draw the axis V/H 1 perpendicular to the front of the triangle ABC), then we determine the angle φ 2 - the inclination of the plane of the triangle ABC to plane V.

4.1.2.Introduction to the systemH. Vtwo additional projection planes

Consider the following example (Fig. 4.b, 4.7): a straight line in general position AB, defined in the system H , V must be positioned perpendicular to the additional projection plane.

In this case, we adhere to the following scheme:

1) from the system H,V we move to the system H, V 1 in which the additional plane V 1  H and V 1 || AB,

2) from the system H,V 1 we move to the system V 1 H 1 where H 1 V 1 and H 1 AB. The solution comes down to the sequential construction of projections A 1  and A 1 "points. A, B 1  and B 1 "points B.

Straight AB, general position in H,V system, becomes parallel to the plane V 1 in the H, V 1 system and is projected to a point on the H 1 plane in the V 1, H 1 system i.e. AB H 1 ,

Figure 4.8 gives an example of constructing a natural type of triangle ABC.

The solution to this problem is carried out according to the following scheme:

1) from the H,V system we move to the H,V 1 system, in which V 1  H and V 1  AD (AD is the horizontal of the triangle ABC), V 1  ABC .

2) from the system H, V 1 we move to the system Vi, Hi, in which H 1 1 V 1 and H 1 || ABC,

In the first part of the problem, the additional plane V 1 is perpendicular to the plane of the triangle ABC . This construction repeats that shown in Fig. 4.4.

In the second part, the construction in Fig. 4.8 is reduced to drawing the axis V 1 /H 1  C" 1 "A 1 "B 1" i.e. plane H 1 is drawn parallel to plane ABC, which leads to the definition of a natural view expressed by the projection C" 1 "A 1 "B 1".

4.2.Method of rotation around an axis perpendicular to the projection plane

When rotating around some fixed straight line i (rotation axis), each point of the rotated figure moves in a plane perpendicular to the rotation axis (rotation plane). In this case, the point moves along a circle, the center of which is at the point of intersection of the axis with the plane of rotation (pricemr rotation). The radius of the circle is equal to the distance from the rotated point to the center (this is the radius of rotation). If any point of a given system is located on the rotation axis i, then when the system rotates, this point is considered stationary. The rotation axis can be specified and selected. If the axis of rotation is selected, then it is advantageous to place it perpendicular to one of the projection planes, since this simplifies the construction.

4.2.1.Rotation around a given axis

Fig.4.9 Fig.4.10

Let point A rotate around axis i, perpendicular to plane H (Fig. 4.9). During rotation, point A describes a circle of radius R, the plane of which is in the plane  and perpendicular to the plane V, and, therefore, parallel to the plane H. The value of the radius R is expressed by the length of the perpendicular drawn from point A to the axis of rotation. A circle described in space by point A is projected onto plane H without distortion. Since plane a is perpendicular to V, the projections of points of the circle onto plane V will be located on v", i.e. on a straight line perpendicular to the frontal projection

axis of rotation. Figure 4.9 on the right: the circle described by point A when rotating it around the i axis is projected without distortion onto the plane H. From the center O a circle of radius R=OA is drawn. On the plane V this circle is projected in the form of a line segment equal to 2R,

Figure 4.10 shows the rotation of point A around axis i, perpendicular to plane V. The circle described by point A is projected without distortion onto plane V. From point O a circle of radius R==OA is drawn." On plane H this circle is represented by a line segment equal to 2R.

It follows from this that when a point rotates around an axis perpendicular to any of the projection planes, one of the projections of the rotated point moves along a straight line perpendicular to the projection of the rotation axis.

4.2.2.Rotation around the selected axis

In some cases, the axis of rotation can be selected. Moreover, if the axis of rotation is chosen to pass through one of the ends of the segment, then the construction is simplified, since the point through which the axis passes will be stationary and to rotate the segment it will be necessary to construct a new position of the projection of only one point - the other end of the segment.

Fig.4.11 Fig.4.12

In Fig. 4.11 it is necessary to determine the natural size of the segment AB and its angle of inclination to the plane H. The axis of rotation i is chosen perpendicular to the plane H and passes through point A. Rotating the segment AB around the axis i we move it to the position

parallel to the plane V (i.e. AB becomes frontal). The value A B is equal to the natural value of the segment AB, and the angle A // B // B // is equal to the angle of inclination of the segment AB to the plane H.

The natural size of the segment CD and its angle of inclination to the plane V are determined similarly (Fig. 89). The rotation axis i is chosen perpendicular to the plane V and passes through point C. By rotating the segment CD around the i axis, we move it to a position parallel to the plane H (i.e. CD becomes horizontal). The value C D is equal to the natural size of the segment CD and the angle C / D is equal to the angle of inclination of the segment CD to the plane V.

In Fig. 4.13, it is necessary to determine the natural appearance of the triangle ABC and its angle of inclination to the plane H. Since the plane of the triangle ABC is a general plane, we solve this problem according to the scheme:

1 Rotation around axis i, perpendicular to plane H and passing through point C , we transfer triangle ABC from the general position to the position of the frontally projecting plane.

2. By rotating around the i 1 axis, perpendicular to plane V and passing through point A, we transfer triangle ABC from the position of the frontally projecting plane to the position of a plane parallel to plane H.

In order to transfer the ABC triangle to the position of the frontal-projecting plane, in the plane of the ABC triangle we draw a horizontal plane of the SC plane, Its frontal projection C // K // is parallel to the X axis. The horizontal projection C / K / is equal to the natural size of the segment SC. We select the rotation axis i perpendicular to H and draw it through point C. The ABC plane becomes the position of the frontal-projecting plane if the horizontal of a given triangle (TC) takes a position perpendicular to the plane V and, therefore, the segment SC becomes perpendicular to the X axis, and the frontal projection C // K // is projected to a point. From the center i / С / with a radius equal to C / K / , we draw an arc and build a new projection K.T.k. When any point rotates around an axis perpendicular to the projection plane, the trajectory of the point is located in a plane perpendicular to the axis of rotation, then the projection K // is located on the straight line K // K // parallel to the X axis.

Using the serif method we find B/ and A/. Frontal projection B"" lies on line B // B // and parallel to the axis X, the frontal projection A / lies on the straight line A / A / parallel to the X axis. As a result of this rotation, the ABC plane has become frontally projecting and the angle (p is equal to the angle of inclination of the ABC plane to the H plane.

We select the axis of rotation i 1 perpendicular to V and draw it through point A. We rotate point K and point C with radius A K, point B with radius A B until plane ABC takes a position parallel to plane H and, therefore, segment A 1 / / K 1 // B 1 // parallel to the OX axis. Because movement trajectories points C, B and K were projected onto the horizontal plane H c into straight lines parallel

C / lies on the straight line C / C / ,

B / 1 lies on the straight line B / B / 1,

K/1 lies on the straight line K/K/1.

Projection A / B / C / determines the natural appearance of triangle ABC.

The straight line obtained by the mutual intersection of two planes is completely determined by two points, each of which belongs to both planes. Thus, the straight line K 1 K 2 (Fig. 163), along which the plane defined by the triangle ABC and pl. β, given by the lines DE and DF, passes through the points K 1 and K 2; but at these points the lines AB and AC of the first plane intersect the square. β i.e. points K 1 and K 2 belong to both planes.

Hence, in the general case, to construct the line of intersection of two planes, it is necessary to find any two points, each of which belongs to both planes; these points define the line of intersection of the planes.

To find each of these two points, it is usually necessary to perform special constructions. But if at least one of the intersecting planes is perpendicular to the projection plane, then the construction of projections of the intersection line is simplified. Let's start with this case.

In Fig. 164 shows the intersection of two planes, one of which (defined by the triangle DEF) is located perpendicular to the square. π 2. Since the triangle DEF is projected onto the square π 2 in the form of a straight line (D "F"), the frontal projection of the straight line segment along which both triangles intersect is a segment K " 1 K " 2 on the projection D "F". Further construction is clear from the drawing.

Another example is given in Fig. 165. The horizontally projecting plane α intersects the plane of triangle ABC. The horizontal projection of the line of intersection of these planes - the segment M"N" - is determined on the trace α".

Now let's consider general case of constructing the line of intersection of two planes. Let one of the planes, β, be defined by two intersecting lines, and the other, γ, by two parallel lines. The construction is shown in Fig. 166. As a result of the mutual intersection of the planes β and γ, the straight line K 1 K 2 is obtained. Let's express this by writing: β × γ = K 1 K 2.

To determine the position of points K 1 and K 2, we take two auxiliary frontally projecting planes (α 1 and α 2) intersecting each of the planes β and γ. When planes β and γ intersect with plane α 1. we obtain straight lines with projections 1"2", 1"2" and 3"4", 3"4". These straight lines, located in the square. α 1, at their intersection determine the first point, K 1, the intersection line of the planes β and γ.

Having obtained the projections K" 1 and K" 2, we find on the traces both α" 1 and α" 2 the projections K" 1 and K" 2. This determines the projections K" 1 K" 2 and K" 1 K" 2 of the desired straight line of intersection of the planes β and γ (the projections are drawn by a dash-dot line).

When constructing, you can keep in mind the following: since the auxiliary cutting planes α 1 and α 2 are mutually parallel, then, having constructed projections 1"2" and 3"4", one should take one point each for projections 5"6" and 7"8" , at least 5 and 8, since 5"6"||1"2" and 7"8"||3"4".

In the construction considered, two frontal-projecting planes were taken as auxiliary. Of course, it was possible to take other planes, for example, two horizontal or one horizontal, the other frontal, etc. The essence of the constructions does not change from this. However, such a case may occur. Let us assume that two horizontal planes were taken as auxiliary and those obtained when they intersected

the horizontal planes β and γ turned out to be mutually parallel. But rice. 167 shows that β and γ intersect each other, although their horizontal lines are parallel. Consequently, having received mutually parallel horizontal projections of the horizontals AB and CD and knowing that the planes are not necessarily parallel, but can intersect (along a common horizontal for them), it is necessary to test the planes β and γ using at least a horizontally projecting plane (see Fig. 167); if the straight lines along which this auxiliary plane σ intersects β and γ would also turn out to be parallel to one another, then the planes β and γ do not intersect, but are parallel to one another. In Fig. 167 these lines intersect at point K, through which the line of intersection of planes β and γ passes parallel to straight lines BA and CD.

If the planes are defined by their traces on the projection planes, then it is natural to look for the points that define the line of intersection of the planes at the points of intersection of the same traces of the planes (Fig. 168): the straight line passing through these points is common to both planes, i.e., their line intersections.

The scheme for constructing the line of intersection of two planes (see Fig. 166) can, of course, be extended to the case of specifying planes by their traces. Here the role of auxiliary cutting planes is played by the projection planes themselves:

α × π 1 =h" 0α ; β× π 1 =h" 0β ; h" 0α × h" 0β =M;

α × π 2 =f" 0α ; β× π 2 =f" 0β ; f" 0α × f" 0β =N.

The intersection points of traces of planes of the same name are traces of the line of intersection of these planes. Therefore, to construct projections of the line of intersection of planes α and β (Fig. 168), it is necessary: ​​1) find point M" at the intersection of traces h" 0α and h" 0β

and point N" at the intersection of f" 0α and f" 0β, and along them - projections M" and N"; 2) draw straight lines M"N" and M"N",

In Fig. 169-171 show cases where the direction of the intersection line is known. Therefore, it is enough to have only one point from the intersection of the traces and then draw a straight line through this point, based on the position of the planes and their traces.

Questions for §§ 22-24

1. What relative position can the two planes occupy?
2. What is the sign of parallelism of two planes?
3. How are the frontal traces of two parallel frontally projecting planes mutually located?
4. How are the horizontal traces of two parallel horizontally projecting planes mutually located?
5. How are the tracks of the same name of two planes parallel to each other mutually located?
6. Is the intersection of at least one pair of their tracks of the same name a sign of the mutual intersection of two planes?
7. How to establish the relative position of a straight line and a Plane?
8. How is the point of intersection of a straight line with a plane perpendicular to one or two projection planes constructed?
9. Which point from among those located on a common perpendicular to a) pl. π 1 b) pl. π 2 is considered visible respectively on π 1, on π 2?
10. How to construct the line of intersection of two planes, at least one of which is perpendicular to the square. π 1 or to pl. π 2?
11. What is the general method for constructing the line of intersection of two planes?

Constructing the point of intersection of a straight line with a projecting plane comes down to constructing a second projection of a point on a diagram, since one projection of a point always lies on the trace of the projecting plane, because everything that is in the projecting plane is projected onto one of the traces of the plane. In Fig. 224, A shows the construction of the point of intersection of a straight line E.F. with a frontally projecting plane of a triangle ABC(perpendicular to the plane V) To the plane V triangle ABC is projected into the segment a"c" straight line and point To" will also lie on this line and be at the intersection point e"f With a"c". A horizontal projection is constructed using a projection connection line. Visibility of a straight line relative to the plane of the triangle ABC determined by the relative position of the triangle projections ABC and straight E.F. on surface V. Viewing direction in Fig. 224, A indicated by an arrow. That section of the straight line, the frontal projection of which is above the projection of the triangle, will be visible. To the left of the point To" the projection of the line is above the projection of the triangle, therefore, on the plane N this area is visible.

In Fig. 224, b straight E.F. intersects the horizontal plane R. Frontal projection To" points TO- points of intersection of a line E.F. with plane P - will be at the intersection point of projection e" f"with a trace of the plane Pv, since the horizontal plane is a frontally projecting plane. Horizontal projection k points TO found using a projection communication line.

Constructing the line of intersection of two planes comes down to finding two points common to these two planes. To construct an intersection line, this is enough, since the intersection line is a straight line, and a straight line is defined by two points. When a projecting plane intersects a generic plane, one of the projections of the intersection line coincides with the trace of the plane located in the projection plane to which the projecting plane is perpendicular. In Fig. 225, A frontal projection t"p" intersection lines MN matches the trace Pv frontal projection plane R, and in Fig. 225, b horizontal projection kl coincides with the trace of the horizontally projecting plane R. Other projections of the intersection line are constructed using projection link lines.

Constructing the point of intersection of a line with a general plane(Fig. 226, A) performed using an auxiliary projection plane R, which is drawn through this straight line E.F. Constructing an intersection line 12 auxiliary plane R. with a given triangle plane ABC, get in plane R two straight lines: E.F.- given straight line and 12 - constructed intersection line that intersect at a point K.

Finding projections of a point TO shown in Fig. 226, b. Constructions are carried out in the following sequence.

Via direct E.F. draw an auxiliary horizontal projection plane R. Her trail Rн coincides with the horizontal projection ef straight E.F.

Constructing a frontal projection 1׳2" intersection lines 12 plane R with a given triangle plane ABC using projection communication lines, since the horizontal projection of the intersection line is known. It coincides with the horizontal trace R H plane R.

Determine the frontal projection To" the desired point TO, which is located at the intersection of the frontal projection of this line with the projection 1"2" intersection lines. The horizontal projection of a point is constructed using a projection link line.

Visibility of a straight line relative to the plane of the triangle ABC determined by the method of competing points. To determine the visibility of a straight line on the frontal plane of projections (Fig. 226, b) compare the coordinates Y points 3 And 4, whose frontal projections coincide. Coordinate Y points 3, lying on a straight line sun, less coordinate Y points 4, lying on a straight line E.F. Therefore, the point 4 is closer to the observer (the direction of view is indicated by the arrow) and the projection of the straight line is depicted on the plane V visible. The straight line passes in front of the triangle. To the left of the point TO' the straight line is covered by the plane of the triangle ABC. Visibility on the horizontal projection plane is shown by comparing the Z coordinates of the points 1 And 5. Because Z 1 > Z 5, point 1 visible. Therefore, to the right of the point 1 (to the point TO) straight E.F. invisible.

To construct the line of intersection of two general planes, auxiliary cutting planes are used. This is shown in Fig. 227, a. One plane is defined by a triangle ABC, the other - parallel lines E.F. And MN. Specified planes (Fig. 227, A) intersected by a third auxiliary plane. For ease of construction, horizontal or frontal planes are taken as auxiliary planes. In this case, the auxiliary plane R is a horizontal plane. It intersects given planes in straight lines 12 And 34, which at the intersection give a point TO, belonging to all three planes, and therefore to two given ones, that is, lying on the line of intersection of the given planes. The second point is found using the second auxiliary plane Q. Found two points TO And L determine the line of intersection of two planes.

In Fig. 227, b auxiliary plane R given by the frontal trace. Frontal projections of intersection lines 1"2" And 3"4" plane R with given planes coincide with the frontal trace Rv plane R, since the plane R perpendicular to the plane V, and everything that is in it (including the lines of intersection) is projected onto its frontal trace R v . Horizontal projections of these lines are constructed using projection connection lines drawn from the frontal projections of points 1", 2", 3", 4" until they intersect with the horizontal projections of the corresponding lines at points 1, 2, 3, 4. The constructed horizontal projections of the intersection lines are extended until they intersect each other at the point k, which is the horizontal projection of the point K, belonging to the line of intersection of two planes. The frontal projection of this point is on the trace R v .

Two planes in space can be parallel or intersecting; a special case of intersecting planes are mutually perpendicular planes.

Constructing the line of intersection of planes is one of the main tasks of descriptive geometry, which has a large practical significance. It belongs to the so-called positional tasks.

Positional are called problems to determine common elements various mating geometric shapes. These include tasks for belonging geometric elements and at the intersection geometric objects, for example, the intersection of a line and a plane with a surface, the intersection of two surfaces and, in particular, the problem of the intersection of two planes.

The line of intersection of two planes is a straight line that simultaneously belongs to both intersecting planes. Therefore, to construct a line of intersection of planes, it is necessary to determine two points of this line or one point and the direction of the intersection line.

Let's consider special case intersection of planes when one of them is projecting. In Fig. 3.6 shows a plane in general position, defined by the triangle ABC and the horizontally projecting P. The two common points belonging to both planes are points D and E, which determine the line of intersection.

To determine these points, the intersection points of sides AB and BC with the projecting plane were found. Constructing points D and E both on the spatial drawing (Fig. 3.6, a) and on the diagram (Fig. 3.6, b) does not cause difficulties, because based on the collective property of projecting traces of planes discussed above.

By connecting the same projections of points D and E, we obtain the projections of the line of intersection of the plane of the triangle ABC and the plane P. Thus, the horizontal projection D 1 E 1 of the line of intersection of the given planes coincides with the horizontal projection of the projecting plane P - with its horizontal trace.

Let's consider general case intersection when both planes are in general position. In Fig. 3.7. shows two generic planes defined by a triangle and two parallel lines. To determine two common points of the line of intersection of planes, we draw two auxiliary (horizontal) level planes R and T. The auxiliary plane R intersects the given planes along two horizontals h and h 1, which at their intersection define point 1, common to planes P and Q, so how they simultaneously belong to the auxiliary cutting plane R. The second plane - the mediator T also intersects each of the given planes along the horizontals h 2 and h 3, which are parallel to the first two horizontals. At the intersection of the horizontal lines we obtain the second common point of 2 given planes. By connecting the projections of these points of the same name on the diagram (Fig. 3.8,b), we obtain the projections of the line of intersection of the planes.

In Fig. Figure 3.8 shows two planes defined by traces. The common points of the planes are the intersection points of M and N of the same traces. By connecting the projections of these points with the same name with a straight line, I obtained the projections of the line of intersection of the planes.

If the intersection points of the same traces are outside the drawing field (see example 5), as well as in cases where the planes are defined not by traces, but by other geometric elements, then to determine the line of intersection of the planes you should use auxiliary level planes– horizontal or frontal. It should be noted that when constructing the line of intersection of the planes specified by the traces, the role of auxiliary cutting planes is played by the projection planes P 1 and P 2.

In Fig. Figure 3.9 shows the case of intersection of two planes, when the direction of the intersection line is known, because plane P is the level plane (P||P 1). Therefore, it is enough to have only one intersection point of the traces and then draw a straight line through this point, based on the position of the planes and their traces. In our case, the intersection line is the common horizontal NA of the P and T planes.