When constructing a point according to the given coordinates, it must be remembered that, in accordance with the drawing rules, the scale along the axis Oh decreases in 2 times compared to the scale along the axes OU and Oz.
1.Build points: A(2; 1; 3) x A = 2; y A = 1; z A = 3
a) usually, first of all, they build the projection of a point onto a plane Ohu. Mark points x A =2 and y A=1 and draw straight lines through them parallel to the axes Oh and OU. The point of their intersection has coordinates (2;1; 0) point built A 1 (2;1; 0.)
A(2; 1; 3)
0
y A=1
x A =2 at
A 1 (2; 1; 0) 0 y A=1at
X x A \u003d 2 A 1 (2; 1; 0)
X
b) further from the point A 1 (2; 1; 0) restore perpendicular to the plane Ohu (draw a line parallel to the axis Oz ) and lay a segment equal to three on it: z A = 3.
2.Build points: B(3; - 2; 1) x B = 3; y B = -2; Z B = 1
z
y B = - 2
B(3; -2; 1) O at
B 1 (3;-2) x B \u003d 3
X
3. Build a point C(-2; 1; 3 ) z C (-2; 1; 3)
X A \u003d -2; Y A = 1; Z A = 3
x C \u003d - 2 C 1 (-2; 1; 0)
y A =1 y
4.Dan cube. A ... D 1, whose edge is 1 . The origin is the same as the point AT, ribs VA, Sun and BB 1 coincide with the positive rays of the coordinate axes. Name the coordinates of all other vertices of the cube. Calculate the diagonal of a cube.
z
AB = BC = BB 1 BD 1 = =
B 1 (0; 0; 1) C 1 (0; 1; 1) = =
A 1 (1; 0; 1) D 1 (1; 1; 1)
В(0;0;0) С(0;1;0)
A(1;0;0) D(1;1;0)
5.Plot points A(1;1;-1) and B(1; -1; 1). Does the segment intersect the coordinate axis? coordinate plane? Does the line segment pass through the origin? Find the coordinates of the intersection points, if any. z The points lie in a plane perpendicular to the axis Oh.
The segment intersects the axis Oh
and plane hoy
at the point
B(1; -1; 1) 
0(0;0;0)
С(1;0;0)
A(1;1;-1)
6.Find the distance between two points: A(1;2;3) and B(-1;1;1).
a)AB = = = =3
b)С(3;4;0) and D(3;-1;2).
CD = = =
In space, to determine the coordinates of the middle of the segment, a third coordinate is introduced.
B (x B; y B; z B)
FROM( ; ; )
A(x A; y A; z A)
7.Find coordinates FROM midpoints of segments: a)AB, if A(3; - 2; - 7), B(11; - 8; 5),
x M = = 7; y M = = - 5; z M = = - 1; C(7; - 5; - 1)
8. Point coordinates A(x; y; z). Write the coordinates of points that are symmetrical to the given one with respect to:
a) coordinate planes
b) coordinate lines
in) origin
a) If point A 1
symmetrical to the given one with respect to the coordinate plane ho,
then the difference in
coordinates of points will only be in the sign of the coordinate z: A 1 (x; y; -z).
dot A 2 Ohz, then A 2 (x; -y; z).
dot A 3 symmetrical to the given one with respect to the plane Ouz, then A 2 (-x; y; z).
b) If point A 4
symmetrical to the given one with respect to the coordinate line Oh,
then the difference in
coordinates of points will be only in signs of coordinates at
and z: A 4 (x; -y; -z).
dot A 5 OU, then A 5 (-x; y; -z).
dot A 6 symmetrical to a given one with respect to a straight line Oz, then A 6 (-x; -y; z).
in) If point A 7 is symmetrical to the given one with respect to the origin, then A 6 (-x; -y; -z).
COORDINATE CONVERSION
The transition from one coordinate system to another is called coordinate system transformation.
We will consider two conversion cases coordinate systems, and derive formulas for the dependence between the coordinates of an arbitrary point of the plane in different systems coordinates. (The technique of transforming the coordinate system is similar to transforming graphs).
1.Parallel transfer. In this case, the position of the origin of coordinates changes, while the direction of the axes and the scale remain unchanged.
If the origin of coordinates goes to the point 0 1 with coordinates 0 1 (x 0; y 0), then for the point M(x; y) relationship between system coordinates x0y and x 0 0y 0 expressed by the formulas:
x \u003d x 0 + x "
y = y 0 + y"
The resulting formulas allow us to find old coordinates from known new ones. X" and at" and vice versa.
y M(x; y) M(x"; y")
0 1 (x 0; y 0), x "
x 0 x"
2.Rotation of coordinate axes. In this case, both axes are rotated by the same angle, while the origin and scale remain unchanged.
M(x; y)
y 1 x 1
Point coordinates M in the old system M(x; y) and M(x"; y") - in the new one. Then the polar radius in both systems is the same, and the polar angles are respectively equal + and , where - polar angle in new system coordinates.
According to the formulas for the transition from polar to rectangular coordinates, we have:
x = rcos( + ) x = rcos cos - rsin sin
y = rsin( + ) y = rcos sin + rsin cos
But rcos = x" and rsin = y", that's why
x \u003d x " cos - y "sin
y \u003d x "sin + y" cos
Answer the following questions in writing:
- What is a rectangular coordinate system in a plane? in space?
- What is the applicate axis? Ordinate? Abscissa?
- What is the notation for unit vectors on the coordinate axes?
- What is an ort?
- How is the length of a segment given by the coordinates of its ends calculated in a rectangular coordinate system?
- How are the coordinates of the middle of a segment given by the coordinates of its ends calculated?
- What is a polar coordinate system?
- What is the relationship between the coordinates of a point in rectangular and polar coordinate systems?
Complete the tasks:
1. How far from the coordinate planes is the point A(1; -2; 3)
2. How far is the point A(1; -2; 3) from coordinate lines a)OU; b) OU; in)Oz;
3. What condition is satisfied by the coordinates of points in space that are equally distant:
a) from two coordinate planes Ohu and Оуz; AB
b) from all three coordinate planes
4. Find the coordinates of a point M middle of the segment AB, A(-2; -4; 1); B(0; -1; 2) and name the point symmetrical to the point M, relatively a) axes Oh
b) axes OU
in) axes Oz.
5. Given a point B(4; - 3; - 4). Find the coordinates of the bases of the perpendiculars dropped from a point on the coordinate axes and coordinate planes.
6.On axle OU find a point equidistant from two points A(1; 2; - 1) and B(-2; 3; 1).
7. Flat Ohz find the point equidistant from three points A(2; 1; 0); B(-1; 2; 3) and C(0;3;1).
8. Find the lengths of the sides of the triangle ABC and its area , if the vertex coordinates : A (-2; 0; 1), B (8; - 4; 9), C (-1; 2; 3).
9. Find the coordinates of the projections of points A(2; -3; 5); In (3;-5; ); FROM(- ; - ; - ).
10. Points are given A(1; -1; 0) and B(-3; - 1; 2). Calculate the distance from the origin to the given points.
VECTORS IN SPACE. BASIC CONCEPTS
All quantities that are dealt with in physics, technology, everyday life are divided into two groups. The former are fully characterized by their numerical value: temperature, length, mass, area, work. Such quantities are called scalar.
Other quantities such as force, speed, displacement, acceleration, etc. determined not only by their numerical value, but also by direction. These quantities are called vector, or vectors. A vector quantity is geometrically represented as a vector.
Vector-this is a directed straight line segment, i.e. segment that has
specific length and direction.
Construct traces of the plane given by ∆BCD and determine the distance from point A to the given plane by the method right triangle (coordinates of points A, B, C and D see Table 1 of the Assignments section);
1.2. An example of completing task number 1
The first task is a set of tasks on the topics:
1. Orthographic projection, Monge plot, point, line, plane: by known coordinates of three points B, C, D construct horizontal and frontal projections of the plane given by ∆ BCD;
2. Traces of a straight line, traces of a plane, properties of belonging to a straight plane: build traces of the plane given by ∆ BCD;
3. General and particular planes, intersection of a line and a plane, perpendicularity of a line and a plane, intersection of planes, right triangle method: determine the distance from a point BUT to the plane ∆ BCD.
1.2.1. Known coordinates of three points B, C, D construct the horizontal and frontal projections of the plane given by ∆ BCD(Figure 1.1), for which it is necessary to build horizontal and frontal projections of the vertices ∆ BCD, and then connect the projections of the vertices of the same name.
It is known that trace plane is called a straight line obtained as a result of the intersection of a given plane with the plane of projections .
Near the plane general position 3 tracks: horizontal, frontal and profile.
In order to construct traces of a plane, it is enough to construct traces (horizontal and frontal) of any two lines lying in this plane and connect them to each other. Thus, the trace of the plane (horizontal or frontal) will be uniquely determined, since through two points on the plane (in this case, these points will be traces of lines) it is possible to draw a straight line, and moreover, only one.
The basis for this construction is property of belonging to a straight plane: if a line belongs to a given plane, then its traces lie on the traces of the same name of this plane .
The trace of a straight line is the point of intersection of this straight line with the plane of projections .
The horizontal trace of a straight line lies in the horizontal plane of projections, the frontal trace lies in the frontal plane of projections.
Consider the construction horizontal track straight D.B. for which you need:
1. Continue frontal projection straight D.B. to the intersection with the axis X, intersection point M 2 is the frontal projection of the horizontal trace;
2. From a point M 2 restore the perpendicular (line of projection connection) to its intersection with the horizontal projection of the straight line D.B. M 1 and will be a horizontal projection of the horizontal trace (Figure 1.1), which coincides with the trace itself M.
Similarly, the construction of a horizontal trace of the segment SW straight: dot M'.
To build frontal footprint segment CB direct, you need:
1. Continue the horizontal projection of the straight line CB to the intersection with the axis X, intersection point N 1 is a horizontal projection of the frontal trace;
2. From a point N 1 restore the perpendicular (projective connection line) until it intersects with the frontal projection of the straight line CB or its continuation. Intersection point N 2 and will be the frontal projection of the frontal trace, which coincides with the trace itself N.
By connecting the dots M′ 1 and M1 straight line segment, we get the horizontal trace of the plane απ 1 . Point α x of intersection απ 1 with the axis X called vanishing point . To construct the frontal trace of the plane απ 2, it is necessary to connect the frontal trace N 2 with trace vanishing point α x
Figure 1.1 - Construction of plane traces
The algorithm for solving this problem can be represented as follows:
- (D 2 B 2 ∩ OX) = M 2 ;
- (MM 1 ∩ D 1 B 1) = M 1 = M;
- (C 2 B 2 ∩ OX) = M′ 2 ;
- (M′ 2 M′ 1 ∩ C 1 B 1) = M′ 1 = M′;
- (CB∩ π 2) = N 2 = N;
- (MM') ≡ απ 1 ;
- (α x N) ≡ απ 2 .
1.2.2. To solve the second part of the first task, you need to know that:
- distance from point BUT to the plane ∆ BCD is determined by the length of the perpendicular restored from this point to the plane;
- any line is perpendicular to a plane if it is perpendicular to two intersecting lines lying in this plane;
- on the diagram, the projections of a straight line perpendicular to the plane are perpendicular to the oblique projections of the horizontal and frontal of this plane or the traces of the plane of the same name (Fig. 1.2) (see the Theorem on the perpendicular to the plane in the lectures).
To find the base of the perpendicular, it is necessary to solve the problem of the intersection of a straight line (in this problem, such a straight line is the perpendicular to the plane) with the plane:
1. Enclose the perpendicular in an auxiliary plane, which should be taken as a private plane (horizontally projecting or frontally projecting, in the example, horizontally projecting γ is taken as an auxiliary plane, that is, perpendicular to π 1, its horizontal trace γ 1 coincides with a horizontal projection of the perpendicular);
2. Find the line of intersection of the given plane ∆ BCD with auxiliary γ ( MN in fig. 1.2);
3. Find the point of intersection of the line of intersection of the planes MN with a perpendicular (point To in fig. 1.2).
4. To determine the true value of the distance from the point BUT up to a given plane ∆ BCD should take advantage right triangle method: the true value of the segment is the hypotenuse of a right triangle, one leg of which is one of the projections of the segment, and the other is the difference in distances from its ends to the projection plane in which the construction is being carried out.
5. Determine the visibility of the perpendicular segments using the competing points method. For example, dots N and 3 to determine visibility on π 1 , points 4 , 5 — to determine the visibility on π 2 .
Figure 1.2 - Construction of a perpendicular to the plane
Figure 1.3 - An example of registration of control task No. 1
Video example of completing task No. 1
1.3. Job options 1
| Option | Coordinates (x, y, z) of points | |||
|---|---|---|---|---|
| BUT | AT | FROM | D | |
| 1 | 15; 55; 50 | 10; 35; 5 | 20; 10; 30 | 70; 50; 40 |
| 2 | 80; 65; 50 | 50; 10; 55 | 10; 50; 25 | 75; 25; 0 |
| 3 | 95; 45; 60 | 130; 40; 50 | 40; 5; 25 | 80; 30; 5 |
| 4 | 115; 10; 0 | 130; 40; 40 | 40; 5; 25 | 80; 30; 5 |
| 5 | 55; 5; 60 | 85; 45; 60 | 100; 5; 30 | 50; 25; 10 |
| 6 | 55; 5; 60 | 70; 40; 20 | 30; 30; 35 | 30; 10; 10 |
| 7 | 60; 10; 45 | 80; 45; 5 | 35; 0; 15 | 10; 0; 45 |
| 8 | 5; 0; 0 | 35; 0; 25 | 20; 0; 55 | 40; 40; 0 |
| 9 | 50; 5; 45 | 65; 30; 10 | 30; 25; 55 | 20; 0; 20 |
| 10 | 60; 50; 35 | 40; 30; 0 | 30; 15; 30 | 80; 5; 20 |
| 11 | 65; 35; 15 | 50; 0; 30 | 20; 25; 25 | 5; 0; 10 |
| 12 | 75; 65; 50 | 45; 10; 35 | 60; 20; 10 | 10; 65; 0 |
| 13 | 95; 0; 15 | 85; 50; 10 | 10; 10; 10 | 55; 10; 45 |
| 14 | 45; 40; 40 | 80; 50; 10 | 10; 10; 10 | 55; 10; 45 |
| 15 | 80; 20; 30 | 55; 30; 60 | 15; 10; 20 | 70; 65; 30 |
| 16 | 75; 35; 35 | 55; 30; 60 | 25; 10; 20 | 70; 65; 30 |
| 17 | 75; 65; 50 | 45; 5; 55 | 5; 45; 10 | 70; 20; 0 |
| 18 | 65; 15; 20 | 40; 5; 60 | 0; 5; 25 | 60; 60; 20 |
| 19 | 70; 20; 10 | 45; 15; 60 | 5; 10; 20 | 60; 65; 10 |
| 20 | 20; 50; 45 | 10; 20; 10 | 55; 50; 10 | 80; 0; 60 |
| 21 | 0; 5; 50 | 50; 50; 40 | 5; 55; 10 | 45; 5; 0 |
| 22 | 55; 50; 65 | 45; 55; 5 | 0; 10; 45 | 70; 0; 40 |
| 23 | 65; 5; 15 | 40; 60; 10 | 0; 20; 5 | 60; 20; 60 |
| 24 | 50; 20; 45 | 45; 60; 30 | 5; 20; 10 | 60; 30; 5 |
| 25 | 55; 15; 40 | 40; 50; 25 | 5; 15; 10 | 50; 40; 10 |
| 26 | 15; 45; 40 | 10; 25; 5 | 20; 10; 30 | 65; 40; 35 |
| 27 | 70; 30; 30 | 55; 30; 60 | 20; 5; 15 | 65; 60; 25 |
| 28 | 90; 0; 15 | 80; 45; 10 | 10; 10; 10 | 50; 10; 45 |
| 29 | 110; 10; 0 | 120; 35; 30 | 35; 5; 20 | 70; 20; 5 |
| 30 | 45; 40; 40 | 80; 45; 10 | 10; 10; 10 | 55; 10; 40 |
|
Word form |
Graphic form |
|
1. Set aside on the X, Y, Ζ axes the corresponding coordinates of point A. We get points A x , A y , A z | |
|
2. Horizontal projection A 1 is located at the intersection of communication lines from points A x and A y drawn parallel to the X and Y axes |
|
|
3. Frontal projection A 2 is located at the intersection of communication lines from points A x and A z, drawn parallel to the axes X and z |
|
|
4. Profile projection A 3 is located at the intersection of communication lines from points A z and A y drawn parallel to the axes Ζ and Y |
|
3.2. Point position relative to projection planes
The position of a point in space relative to the projection planes is determined by its coordinates. The X coordinate determines the distance of the point from the P 3 plane (projection to P 2 or P 1), the Y coordinate - the distance from the P 2 plane (projection to P 3 or P 1), the Z coordinate - the distance from the P 1 plane (projection to P 3 or P 2). Depending on the value of these coordinates, a point can occupy both a general and a particular position in space with respect to the projection planes (Fig. 3.1).
Rice. 3.1. Point classification
Tpointsgeneralprovisions. The coordinates of a point in general position are not equal to zero ( x≠0, y≠0, z≠0 ), and depending on the sign of the coordinate, the point can be located in one of eight octants (Table 2.1).
On fig. 3.2 drawings of points in general position are given. An analysis of their images allows us to conclude that they are located in the following octants of space: A(+X;+Y; +Z( Ioctant;B(+X;+Y;-Z( IVoctant;C(-X;+Y; +Z( Voctant;D(+X;+Y; +Z( IIoctant.
Private position points. One of the coordinates of the point of particular position is equal to zero, so the projection of the point lies on the corresponding field of projections, the other two - on the axes of the projections. On fig. 3.3 such points are points A, B, C, D, G.A P 3, then the point X A \u003d 0; AT P 3, then the point X B \u003d 0; FROM P 2, then point Y C \u003d 0; D P 1, then point Z D \u003d 0.
A point can belong to two projection planes at once, if it lies on the line of intersection of these planes - the projection axis. For such points, only the coordinate on this axis is not equal to zero. On fig. 3.3, such a point is the point G(G OZ, then point X G =0, Y G =0).
3.3. Mutual position of points in space
Let us consider three options for the mutual arrangement of points depending on the ratio of the coordinates that determine their position in space.
On fig. 3.4 points A and B have different coordinates.
Their relative position can be estimated by the distance to the projection planes: Y A >Y B, then point A is located farther from the plane P 2 and closer to the observer than point B; Z A >Z B, then point A is located farther from the plane P 1 and closer to the observer than point B; X A On fig. 3.5 shows points A, B, C, D, in which one of the coordinates is the same, and the other two are different. Their relative position can be estimated by their distance to the projection planes as follows: Y A \u003d Y B \u003d Y D, then points A, B and D are equidistant from the plane P 2, and their horizontal and profile projections are located respectively on the lines [A 1 B 1 ]llOX and [A 3 B 3 ]llOZ. The locus of such points is a plane parallel to П 2 ; Z A \u003d Z B \u003d Z C, then points A, B and C are equidistant from the plane P 1, and their frontal and profile projections are located respectively on the lines [A 2 B 2 ]llOX and [A 3 C 3 ]llOY. The locus of such points is a plane parallel to П 1 ; X A \u003d X C \u003d X D, then points A, C and D are equidistant from the plane P 3 and their horizontal and frontal projections are located respectively on the lines [A 1 C 1 ]llOY and [A 2 D 2 ]llOZ . The locus of such points is a plane parallel to П 3 . 3. If the points have two coordinates of the same name, then they are called competing. Competing points are located on the same projecting line. On fig. 3.3 three pairs of such points are given, in which: X A \u003d X D; Y A = Y D ; Z D > Z A; X A = X C ; Z A = Z C ; Y C > Y A ; Y A = Y B ; Z A = Z B ; X B > X A . There are horizontally competing points A and D located on the horizontally projecting line AD, frontally competing points A and C located on the frontally projecting line AC, profile competing points A and B located on the profile projecting line AB. Conclusions on the topic 1. A point is a linear geometric image, one of the basic concepts of descriptive geometry. The position of a point in space can be determined by its coordinates. Each of the three projections of a point is characterized by two coordinates, their name corresponds to the names of the axes that form the corresponding projection plane: horizontal - A 1 (XA; YA); frontal - A 2 (XA; ZA); profile - A 3 (YA; ZA). Translation of coordinates between projections is carried out using communication lines. From two projections, you can build projections of a point either using coordinates or graphically. 3. A point in relation to the projection planes can occupy both a general and a particular position in space. 4. A point in general position is a point that does not belong to any of the projection planes, i.e., lies in the space between the projection planes. The coordinates of a point in general position are not equal to zero (x≠0,y≠0,z≠0). 5. A point of private position is a point belonging to one or two projection planes. One of the coordinates of a point of particular position is equal to zero, so the projection of the point lies on the corresponding field of the projection plane, the other two - on the axes of the projections. 6. Competing points are points whose coordinates of the same name are the same. There are horizontally competing points, frontally competing points, and profile competing points. Keywords Point coordinates General point Private position point Competing points Methods of activity necessary for solving problems – construction of a point according to the given coordinates in the system of three projection planes in space; – construction of a point according to the given coordinates in the system of three projection planes on the complex drawing. Questions for self-examination 1. How is the connection of the location of coordinates on the complex drawing in the system of three projection planes P 1 P 2 P 3 with the coordinates of the projections of points established? 2. What coordinates determine the distance of points to the horizontal, frontal, profile projection planes? 3. What coordinates and projections of the point will change if the point moves in the direction perpendicular to the profile plane of the projections П 3 ? 4. What coordinates and projections of a point will change if the point moves in a direction parallel to the OZ axis? 5. What coordinates determine the horizontal (frontal, profile) projection of a point? 7. In what case does the projection of a point coincide with the point in space itself, and where are the other two projections of this point located? 8. Can a point belong to three projection planes at the same time, and in what case? 9. What are the names of the points whose projections of the same name coincide? 10. How can you determine which of the two points is closer to the observer if their frontal projections coincide? Tasks for independent solution 2. Construct projections of points A and B according to their coordinates on a visual image and a complex drawing: A (13.5; 20), B (6.5; -20). Construct a projection of point C, located symmetrically to point A relative to the frontal plane of projections П 2 . 3. Build projections of points A, B, C according to their coordinates on a visual image and a complex drawing: A (-20; 0; 0), B (-30; -20; 10), C (-10, -15, 0 ). Construct point D, located symmetrically to point C with respect to the OX axis. An example of solving a typical problem Task 1. Given the coordinates X, Y, Z of points A, B, C, D, E, F (Table 3.3) To build an image of an object, first depict its individual elements in the form of the simplest elements of space. So, depicting a geometric body, one should build its vertices, represented by points; edges represented by straight and curved lines; faces represented by planes, etc. The rules for constructing images on drawings in engineering graphics are based on the projection method. One image (projection) of a geometric body does not allow us to judge its geometric shape or the form of the simplest geometric images that make up this image. Thus, one cannot judge the position of a point in space by one of its projections; its position in space is determined by two projections. Consider an example of constructing a projection of a point BUT, located in the space of the dihedral angle (Fig. 60). Let's place one of the projection planes horizontally, let's call it horizontal plane projections and denote by the letter P 1. Element projections spaces on it will be denoted with index 1: A 1 , a 1 , S 1 ... and call horizontal projections(points, lines, planes). We place the second plane vertically in front of the observer, perpendicular to the first one, let's call it vertical plane projections and denote P 2 . The projections of space elements on it will be denoted with the index 2: A 2, 2 and call front projections(points, lines, planes). The line of intersection of the projection planes is called projection axis. Let's project a point BUT orthogonally on both projection planes: AA 1 _|_ P 1 ;AA 1 ^P 1 =A 1 ; AA 2 _|_ P 2; AA 2 ^P 2 \u003d A 2; Projection beams AA 1 and AA 2 mutually perpendicular and create a projecting plane in space AA 1 AA 2, perpendicular to both sides of the projections. This plane intersects the projection planes along the lines passing through the projections of the point BUT. To get a flat drawing, we match the horizontal projection plane P 1 with the frontal plane P 2 rotation around the axis P 2 / P 1 (Fig. 61, a). Then both projections of the point will be on the same line perpendicular to the axis P 2 /P 1. Straight A 1 A 2, connecting the horizontal A 1 and frontal A 2 point projection is called vertical line of communication. The resulting flat drawing is called complex drawing. It is an image of an object on several combined planes. A complex drawing consisting of two orthogonal projections connected to each other is called a two-projection one. In this drawing, the horizontal and frontal projections of the point always lie on the same vertical connection line. Two interconnected orthogonal projections of a point uniquely determine its position relative to the projection planes. If we determine the position of the point a relative to these planes (Fig. 61, b) its height h (AA 1 =h) and depth f(AA 2 =f ), then these the values in the multidrawing exist as segments of the vertical connection line. This circumstance makes it easy to reconstruct the drawing, i.e., to determine the position of the point relative to the projection planes from the drawing. To do this, it is enough at point A 2 of the drawing to restore the perpendicular to the plane of the drawing (considering it to be frontal) with a length equal to the depth f. The end of this perpendicular will determine the position of the point BUT relative to the plane of the drawing. Image: Image: 4. What is the name of the distance that determines the position of a point relative to the plane of projections P 1, P 2? 7. How to build an additional projection of a point on a plane P 4 _|_ P 2 , P 4 _|_ P 1 , P 5 _|_ P 4 ? 9. How can I build a complex drawing of a point by its coordinates? To determine the position of a geometric body in space and obtain additional information on their images, it may be necessary to build a third projection. Then the third projection plane is placed to the right of the observer perpendicular to the simultaneously horizontal projection plane P 1 and the frontal plane of projections P 2 (Fig. 62, a). As a result of the intersection of the frontal P 2
and profile P 3
projection planes we get a new axis P 2 / P 3 ,
which is located on the complex drawing parallel to the vertical communication line A 1 A 2(Fig. 62, b). Third point projection BUT- profile - turns out to be connected with the frontal projection A 2 a new line of communication, which is called horizontal Rice. 62 Noah. The frontal and profile projections of a point always lie on the same horizontal line of communication. And A 1 A 2 _|_ A 2 A 1 and A 2 A 3 , _| _ P 2 / P 3. The position of a point in space in this case is characterized by its latitude- the distance from it to the profile plane of the projections P 3, which we denote by the letter R. The resulting complex drawing of a point is called three-projection. In a three-projection drawing, the point depth AA 2 is projected without distortion on the plane P 1 and P 2 (Fig. 62, a). This circumstance allows us to construct the third - frontal projection of the point BUT along its horizontal A 1 and frontal A 2 projections (Fig. 62, in). To do this, through the frontal projection of the point, you need to draw a horizontal line of communication A 2 A 3 _|_A 2 A 1 . Then, anywhere on the drawing, draw an axis of projections П 2 / П 3 _|_ A 2 A 3, measure the depth f of a point on a horizontal
projection field and set aside it along the horizontal line of communication from the axis of projections P 2 /P 3 . Get a profile projection A 3 points BUT. Thus, in a complex drawing consisting of three orthogonal projections of a point, two projections are on the same line of communication; communication lines are perpendicular to the corresponding projection axes; two projections of a point completely determine the position of its third projection. It should be noted that in complex drawings, as a rule, the projection planes are not limited and their position is set by the axes (Fig. 62, c). In cases where the conditions of the problem do not require this It turns out that projections of points can be given without depicting axes (Fig. 63, a, b). Such a system is called baseless. Communication lines can also be drawn with a gap (Fig. 63, b). Image: Image: The location of the projections of points in the complex drawing depends on the position of the point in the space of a three-dimensional angle. Let's consider some cases: Two points in space can be located in different ways. In a particular case, they can be located so that their projections on some projection plane coincide. Such points are called competing. On fig. 64, a a complex drawing of points is given BUT and AT. They are located so that their projections coincide on the plane P 1 [A 1 \u003d= B 1]. Such points are called horizontally competing. If the projections of the points A and B coincide on the plane P 2(Fig. 64, b) they're called frontally competitive. And if the projections of the points BUT and AT coincide on the plane P 3 [A 3 \u003d= B 3] (Fig. 64, c), they are called profile competitive. Competing points determine the visibility on the drawing. Horizontally competing points will see the one with a greater height, frontally competing ones - the one with more depth, and profile competing ones - the one with more latitude. Image: The properties of a three-projection drawing of a point allow, using its horizontal and frontal projections, to build a third onto other projection planes introduced instead of the given ones. On fig. 65 a showing dot BUT and its projections - horizontal A 1 and frontal A 2 . According to the conditions of the problem, it is necessary to replace the planes П 2 . Let's designate the new projection plane P 4 and place it perpendicularly P 1. At the intersection of planes P 1 and P 4 we get a new axis P 1 / P 4 .
New point projection A 4 will be located on
communication line passing through a point A 1 and perpendicular to the axis P 1 / P 4 .
Since the new plane P 4 replaces the frontal projection plane P 2 , point height BUT depicted equally in full size and on the plane P 2 and on the plane P 4 . This circumstance allows us to determine the position of the projection A 4 , in the system of planes P 1 _|_ P 4(Fig. 65, b) on the complex drawing. To do this, it is enough to measure the height of the point on the replaced plane sti projection P 2, put it on a new line of communication from the new axis of projections - and a new projection of the point A 4 will be built. If a new projection plane is introduced instead of the horizontal projection plane, i.e. P 4 _ | _ P 2 (Fig. 66, a), then in the new system of planes the new projection of the point will be on the same line of communication with the frontal projection, and A 2 A 4 _|_. In this case, the depth of the point is the same on the plane P 1, and on the plane P 4 . On this basis they build A 4(Fig. 66, b) on the line of communication A 2 A 4 at such a distance from the new axis P 1 / P 4 at what A 1 is located from the axis P 2 /P 1. As already noted, the construction of new additional projections is always associated with specific tasks. In the future, a number of metric and positional problems solved using the method of replacing projection planes will be considered. In tasks where the introduction of one additional plane will not give the desired result, another additional plane is introduced, which is denoted by P 5 . It is placed perpendicular to the already introduced plane P 4
(Fig. 67, a), i.e. P 5 P 4 and produce a construction similar to those previously considered. Now the distances are measured on the replaced second of the main projection planes (in Fig. 67, b on surface P 1) and deposit them on a new line of communication A 4 A 5, from the new projection axis P 5 /P 4 . In the new system of planes P 4 P 5, a new two-projection drawing is obtained, consisting of orthogonal projections A 4 and A 5 ,
connected by a communication line The position of a point in space can be specified by its two orthogonal projections, for example, horizontal and frontal, frontal and profile. The combination of any two orthogonal projections allows you to find out the value of all coordinates of a point, build a third projection, determine the octant in which it is located. Let's consider some typical tasks from the course of descriptive geometry. According to the given complex drawing of points A and B, it is necessary: Let us first determine the coordinates of point A, which can be written in the form A (x, y, z). The horizontal projection of point A is point A ", having coordinates x, y. Draw from point A" perpendiculars to the x, y axes and find, respectively, A x, A y. The x-coordinate for point A is equal to the length of the segment A x O with a plus sign, since A x lies in the region of positive x-axis values. Taking into account the scale of the drawing, we find x \u003d 10. The y coordinate is equal to the length of the segment A y O with a minus sign, since t. A y lies in the region of negative y-axis values. Given the scale of the drawing, y = -30. The frontal projection of point A - point A"" has x and z coordinates. Let's drop the perpendicular from A"" to the z-axis and find A z . The z-coordinate of point A is equal to the length of the segment A z O with a minus sign, since A z lies in the region of negative values of the z-axis. Given the scale of the drawing, z = -10. Thus, the coordinates of point A are (10, -30, -10). The coordinates of point B can be written as B (x, y, z). Consider the horizontal projection of point B - point B. "Since it lies on the x axis, then B x \u003d B" and the coordinate B y \u003d 0. The abscissa x of point B is equal to the length of the segment B x O with a plus sign. Taking into account the scale of the drawing, x = 30. The frontal projection of the point B - point B˝ has the coordinates x, z. Draw a perpendicular from B"" to the z-axis, thus finding B z . The applicate z of point B is equal to the length of the segment B z O with a minus sign, since B z lies in the region of negative values of the z-axis. Taking into account the scale of the drawing, we determine the value z = -20. So the B coordinates are (30, 0, -20). All necessary constructions are shown in the figure below. Points A and B in the P 3 plane have the following coordinates: A""" (y, z); B""" (y, z). In this case, A"" and A""" lie on the same perpendicular to the z-axis, since they have a common z-coordinate. In the same way, B"" and B""" lie on a common perpendicular to the z-axis. To find the profile projection of t. A, we set aside along the y-axis the value of the corresponding coordinate found earlier. In the figure, this is done using an arc of a circle of radius A y O. After that, we draw a perpendicular from A y to the intersection with the perpendicular restored from the point A "" to the z axis. The intersection point of these two perpendiculars determines the position of A""". Point B""" lies on the z-axis, since the y-ordinate of this point is zero. To find the profile projection of point B in this problem, it is only necessary to draw a perpendicular from B"" to the z-axis. The point of intersection of this perpendicular with the z-axis is B """. Visually imagining a spatial layout composed of projection planes P 1, P 2 and P 3, the location of octants, as well as the order of transformation of the layout into diagrams, you can directly determine that t. A is located in octant III, and t. B lies in the plane P 2 . Another option for solving this problem is the method of exceptions. For example, the coordinates of point A are (10, -30, -10). The positive abscissa x makes it possible to judge that the point is located in the first four octants. A negative y-ordinate indicates that the point is in the second or third octant. Finally, the negative applicate of z indicates that point A is in the third octant. The given reasoning is clearly illustrated by the following table. Point B coordinates (30, 0, -20). Since the ordinate of t. B is equal to zero, this point is located in the projection plane П 2 . The positive abscissa and the negative applicate of point B indicate that it is located on the border of the third and fourth octants. Using the frontal isometric projection, we built a spatial layout of the third octant. It is a rectangular trihedron, whose faces are the planes P 1, P 2, P 3, and the angle (-y0x) is 45 º. In this system, segments along the x, y, z axes will be plotted in full size without distortion. The construction of a visual image of point A (10, -30, -10) will begin with its horizontal projection A ". Having set aside the corresponding coordinates along the abscissa and ordinates, we find the points A x and A y. The intersection of perpendiculars restored from A x and A y respectively to the x and y axes determines the position of point A". Putting from A" parallel to the z axis towards its negative values the segment AA", whose length is equal to 10, we find the position of point A. A visual image of point B (30, 0, -20) is constructed in a similar way - in the P 2 plane, the corresponding coordinates must be plotted along the x and z axes. The intersection of the perpendiculars reconstructed from B x and B z will determine the position of point B.Chapter 6. PROJECTIONS OF A POINT. INTEGRATED DRAWING
§ 32. Complex drawing of a point
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7. Self-test questions
SELF-CHECK QUESTIONS
33. Elements of a three-projection complex drawing of a point
§ 33. Elements of a three-projection complex drawing of a point
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34. The position of a point in the space of a three-dimensional angle
§ 34. The position of a point in the space of a three-dimensional angle
35. Competing points
§ 35. Competing points
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36. Replacing projection planes
§ 36. Replacement of projection planes
Construction of projections of points
Determining the position of points in space
Octants
Coordinate signs
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Construction of a visual image of points in the system of planes P 1, P 2, P 3
