Construction of projections of points. Stage iv
To build images of a number of parts, it is necessary to be able to find the projections of individual points. For example, it is difficult to draw a top view of the part shown in Fig. 139, without building horizontal projections of points A, B, C, D, E, F, etc.
The problem of finding projections of points one at a time, given on the surface of an object, is solved as follows. First, the projections of the surface on which the point is located are found. Then, drawing a connection line to the projection, where the surface is depicted as a line, the second projection of the point is found. The third projection lies at the intersection of communication lines.
Let's look at an example.
Three projections of the part are given (Fig. 140, a). A horizontal projection a of point A, lying on the visible surface, is given. We need to find the rest of the projections of this point.
First of all, you need to draw an auxiliary line. If two views are given, then the place of the auxiliary line in the drawing is chosen arbitrarily, to the right of the top view, so that the view on the left is at the required distance from the main view (Fig. 141).
If three types have already been constructed (Fig. 142, a), then the place of the auxiliary line cannot be arbitrarily chosen; you need to find the point through which it will pass. To do this, it is enough to continue until the mutual intersection of the horizontal and profile projections of the axis of symmetry and through the obtained point k (Fig. 142, b) draw a line segment at an angle of 45 °, which will be an auxiliary straight line.
If there are no axes of symmetry, then continue until the intersection at point k 1 of the horizontal and profile projections of any face, projected in the form of straight line segments (Fig. 142, b).
Having drawn an auxiliary line, they begin to construct projections of the point (see Fig. 140, b).
Frontal a "and profile a" projections of point A should be located on the corresponding projections of the surface to which point A belongs. These projections are found. In fig. 140, b they are highlighted in color. Communication lines are drawn as indicated by the arrows. At the intersection of the lines of communication with the surface projections, there are the required projections a "and a".
The construction of projections of points B, C, D is shown in Fig. 140, in the lines of communication with arrows. The specified projections of the points are colored. Communication lines lead to the projection on which the surface is depicted as a line, and not in the form of a figure. Therefore, the frontal projection from point C is first found. The profile projection from point C is determined by the intersection of the communication lines.
If the surface is not represented by a line on any projection, then an auxiliary plane must be used to construct the projections of the points. For example, given a frontal projection d of point A, lying on the surface of the cone (Fig. 143, a). An auxiliary plane is drawn through a point parallel to the base, which will intersect the cone in a circle; its frontal projection is a straight line segment, and its horizontal projection is a circle with a diameter equal to the length of this segment (Fig. 143, b). Drawing a connection line to this circle from point a ", a horizontal projection of point A is obtained.
The profile projection a "of point A is found in the usual way at the intersection of communication lines.
In the same way, you can find the projection of a point lying, for example, on the surface of a pyramid or a ball. When the pyramid intersects with a plane parallel to the base and passing through a given point, a shape similar to the base is formed. The projections of this figure are the projections of the given point.
Answer the questions
1. At what angle is the auxiliary line drawn?
2. Where is the auxiliary line drawn if front and top views are given, but you need to build a left view?
3. How to determine the place of the auxiliary line in the presence of three types?
4. What is the method of constructing projections of a point using one given one, if one of the surfaces of an object is depicted by a line?
5. For what geometric bodies and in what cases are the projections of a point given on their surface found using an auxiliary plane?
Tasks for § 20
Exercise # 68
Write in workbook, what projections of the points indicated by numbers on the views correspond to the points indicated on the visual image by letters in the example indicated to you by the teacher (Fig. 144, a-d).
Exercise # 69
In fig. 145, a-b letters only one projection of some of the vertices is indicated. Find in the example given to you by the teacher, the rest of the projections of these vertices and designate them with letters. Construct in one of the examples the missing projections of points given on the edges of the object (Fig. 145, d and e). Highlight in color the projections of the edges on which the points are located. Perform the task on transparent paper, superimposing it on the page of the tutorial. It is not necessary to redraw Fig. 145.
Exercise # 70
Find the missing projections of points given by one projection on the visible surfaces of the object (fig. 146). Label them with letters. Highlight the specified projections of points. A visual image will help you solve the problem. The task can be completed both in a workbook and on transparent paper, overlaid on a page of the textbook. In the latter case, cross out Fig. 146 is not necessary.
Exercise # 71
In the example given to you by the teacher, outline three types (fig. 147). Construct the missing projections of points given on the visible surfaces of the object. Highlight the specified projections of points with color. Label all point projections. Use the construction line to construct projections of points. Complete a technical drawing and mark the specified points on it.
In this article we will find answers to questions about how to create a projection of a point onto a plane and how to determine the coordinates of this projection. In the theoretical part, we will rely on the concept of projection. We will give definitions of terms and accompany the information with illustrations. Let's consolidate the knowledge gained by solving examples.
Yandex.RTB R-A-339285-1
Projection, types of projection
For the convenience of examining spatial figures, drawings with the image of these figures are used.
Definition 1
Projection of a figure onto a plane- drawing of a spatial figure.
Obviously, there are a number of rules used to construct a projection.
Definition 2
Projection- the process of constructing a drawing of a spatial figure on a plane using construction rules.
Projection plane- this is the plane in which the image is built.
The use of certain rules determines the type of projection: central or parallel.
A special case of parallel projection is perpendicular or orthogonal projection: it is mainly used in geometry. For this reason, in speech, the adjective "perpendicular" itself is often omitted: in geometry they simply say "projection of a figure" and mean by this the construction of a projection by the method of perpendicular projection. In particular cases, of course, otherwise may be stipulated.
Note the fact that the projection of a figure onto a plane is essentially a projection of all points of this figure. Therefore, in order to be able to study a spatial figure in a drawing, it is necessary to acquire the basic skill of projecting a point onto a plane. What we will talk about below.
Recall that most often in geometry, speaking of projection onto a plane, they mean the use of perpendicular projection.
Let's make constructions that will give us the opportunity to get the definition of the projection of a point on a plane.
Suppose a three-dimensional space is given, and in it there is a plane α and a point M 1 that does not belong to the plane α. Draw a straight line through a given point M 1 but perpendicular to the given plane α. The point of intersection of the straight line a and the plane α will be denoted as H 1; by construction, it will serve as the base of the perpendicular dropped from the point M 1 onto the plane α.
If a point M 2 belonging to a given plane α is given, then M 2 will serve as a projection of itself onto the plane α.
Definition 3
Is either the point itself (if it belongs to a given plane), or the base of a perpendicular dropped from a given point onto a given plane.
Finding the coordinates of the projection of a point on a plane, examples
Let the following be given in three-dimensional space: a rectangular coordinate system O x y z, plane α, point M 1 (x 1, y 1, z 1). It is necessary to find the coordinates of the projection of the point M 1 on a given plane.
The solution follows in an obvious way from the definition of the projection of a point onto a plane given above.
Let's designate the projection of the point М 1 onto the plane α as Н 1. According to the definition, H 1 is the point of intersection of the given plane α and the straight line a drawn through the point M 1 (perpendicular to the plane). Those. the coordinates of the projection of the point M 1 we need are the coordinates of the point of intersection of the straight line a and the plane α.
Thus, to find the coordinates of the projection of a point onto a plane, it is necessary:
Get the equation of the plane α (if it is not specified). An article on the types of plane equations will help you here;
Determine the equation of the straight line a passing through the point M 1 and perpendicular to the plane α (study the topic of the equation of the straight line passing through a given point perpendicular to a given plane);
Find the coordinates of the point of intersection of the straight line a and the plane α (article - finding the coordinates of the point of intersection of the plane and the straight line). The obtained data will be the coordinates of the projection of the point M 1 on the plane α, we need.
Let's consider the theory with practical examples.
Example 1
Determine the coordinates of the projection of point M 1 (- 2, 4, 4) on the plane 2 x - 3 y + z - 2 = 0.
Solution
As we can see, the equation of the plane is given to us, i.e. there is no need to compose it.
Let us write down the canonical equations of the straight line a passing through the point М 1 and perpendicular to the given plane. For this purpose, we define the coordinates of the direction vector of the straight line a. Since the straight line a is perpendicular to the given plane, the direction vector of the straight line a is the normal vector of the plane 2 x - 3 y + z - 2 = 0. Thus, a → = (2, - 3, 1) is the direction vector of the straight line a.
Now we compose the canonical equations of a straight line in space passing through the point M 1 (- 2, 4, 4) and having a direction vector a → = (2, - 3, 1):
x + 2 2 = y - 4 - 3 = z - 4 1
To find the desired coordinates, the next step is to determine the coordinates of the point of intersection of the straight line x + 2 2 = y - 4 - 3 = z - 4 1 and the plane 2 x - 3 y + z - 2 = 0 . To this end, we pass from the canonical equations to the equations of two intersecting planes:
x + 2 2 = y - 4 - 3 = z - 4 1 ⇔ - 3 (x + 2) = 2 (y - 4) 1 (x + 2) = 2 (z - 4) 1 ( y - 4) = - 3 (z + 4) ⇔ 3 x + 2 y - 2 = 0 x - 2 z + 10 = 0
Let's compose a system of equations:
3 x + 2 y - 2 = 0 x - 2 z + 10 = 0 2 x - 3 y + z - 2 = 0 ⇔ 3 x + 2 y = 2 x - 2 z = - 10 2 x - 3 y + z = 2
And let's solve it using Cramer's method:
∆ = 3 2 0 1 0 - 2 2 - 3 1 = - 28 ∆ x = 2 2 0 - 10 0 - 2 2 - 3 1 = 0 ⇒ x = ∆ x ∆ = 0 - 28 = 0 ∆ y = 3 2 0 1 - 10 - 2 2 2 1 = - 28 ⇒ y = ∆ y ∆ = - 28 - 28 = 1 ∆ z = 3 2 2 1 0 - 10 2 - 3 2 = - 140 ⇒ z = ∆ z ∆ = - 140 - 28 = 5
Thus, the required coordinates of a given point M 1 on a given plane α will be: (0, 1, 5).
Answer: (0 , 1 , 5) .
Example 2
In a rectangular coordinate system O x y z of three-dimensional space, points A (0, 0, 2) are given; B (2, - 1, 0); C (4, 1, 1) and M 1 (-1, -2, 5). It is necessary to find the coordinates of the projection M 1 on the plane A B C
Solution
First of all, we write down the equation of a plane passing through three given points:
x - 0 y - 0 z - 0 2 - 0 - 1 - 0 0 - 2 4 - 0 1 - 0 1 - 2 = 0 ⇔ xyz - 2 2 - 1 - 2 4 1 - 1 = 0 ⇔ ⇔ 3 x - 6 y + 6 z - 12 = 0 ⇔ x - 2 y + 2 z - 4 = 0
We write the parametric equations of the straight line a, which will pass through the point M 1 perpendicular to the plane A B C. The plane x - 2 y + 2 z - 4 = 0 has a normal vector with coordinates (1, - 2, 2), i.e. vector a → = (1, - 2, 2) is the direction vector of the straight line a.
Now, having the coordinates of the point of the straight line M 1 and the coordinates of the direction vector of this straight line, we write the parametric equations of the straight line in space:
Then we determine the coordinates of the point of intersection of the plane x - 2 y + 2 z - 4 = 0 and the straight line
x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ
To do this, substitute into the equation of the plane:
x = - 1 + λ, y = - 2 - 2 λ, z = 5 + 2 λ
Now, using the parametric equations x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ, we find the values of the variables x, y, and z at λ = - 1: x = - 1 + (- 1) y = - 2 - 2 (- 1) z = 5 + 2 (- 1) ⇔ x = - 2 y = 0 z = 3
Thus, the projection of point M 1 onto the plane A B C will have coordinates (- 2, 0, 3).
Answer: (- 2 , 0 , 3) .
Let us dwell separately on the question of finding the coordinates of the projection of a point on the coordinate planes and planes that are parallel to the coordinate planes.
Let points M 1 (x 1, y 1, z 1) and coordinate planes O x y, O x z and O y z be given. The coordinates of the projection of this point on these planes will be, respectively: (x 1, y 1, 0), (x 1, 0, z 1) and (0, y 1, z 1). Consider also the planes parallel to the given coordinate planes:
C z + D = 0 ⇔ z = - D C, B y + D = 0 ⇔ y = - D B
And the projections of a given point M 1 onto these planes will be points with coordinates x 1, y 1, - D C, x 1, - D B, z 1 and - D A, y 1, z 1.
Let us demonstrate how this result was obtained.
As an example, let us define the projection of the point M 1 (x 1, y 1, z 1) onto the plane A x + D = 0. The rest of the cases are by analogy.
The given plane is parallel to the coordinate plane O y z and i → = (1, 0, 0) is its normal vector. The same vector serves as the direction vector of the straight line perpendicular to the plane O y z. Then the parametric equations of the straight line drawn through the point M 1 and perpendicular to the given plane will have the form:
x = x 1 + λ y = y 1 z = z 1
Let's find the coordinates of the point of intersection of this straight line and the given plane. First, we substitute in the equation A x + D = 0 the equalities: x = x 1 + λ, y = y 1, z = z 1 and we obtain: A (x 1 + λ) + D = 0 ⇒ λ = - DA - x one
Then we calculate the required coordinates using the parametric equations of the straight line at λ = - D A - x 1:
x = x 1 + - D A - x 1 y = y 1 z = z 1 ⇔ x = - D A y = y 1 z = z 1
That is, the projection of point М 1 (x 1, y 1, z 1) onto the plane will be the point with coordinates - D A, y 1, z 1.
Example 2
It is necessary to determine the coordinates of the projection of the point M 1 (- 6, 0, 1 2) on the coordinate plane O x y and on the plane 2 y - 3 = 0.
Solution
The coordinate plane O x y will correspond to the incomplete general equation of the plane z = 0. The projection of point М 1 onto the plane z = 0 will have coordinates (- 6, 0, 0).
The plane equation 2 y - 3 = 0 can be written as y = 3 2 2. Now it's easy to write down the coordinates of the projection of the point M 1 (- 6, 0, 1 2) on the plane y = 3 2 2:
6 , 3 2 2 , 1 2
Answer:(- 6, 0, 0) and - 6, 3 2 2, 1 2
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POINT PROJECTION.
ORTHOGONAL SYSTEM OF TWO PLANES OF PROJECTIONS.
The essence of the orthogonal projection method is that an object is projected onto two mutually perpendicular planes by rays orthogonal (perpendicular) to these planes.
One of the projection planes H is placed horizontally, and the second V is placed vertically. Plane H is called the horizontal projection plane, V - frontal. The H and V planes are infinite and opaque. The line of intersection of the projection planes is called the coordinate axis and is denoted OX. The projection planes divide the space into four dihedral angles - quarters.
Considering orthogonal projections, it is assumed that the observer is in the first quarter at an infinitely large distance from the projection planes. Since these planes are opaque, only those points, lines and figures that are located within the same first quarter will be visible to the observer.
When building projections, it must be remembered that point orthogonal projectionon the plane is called the base of the perpendicular dropped from a given pointonto this plane.
The figure shows the point BUT and its orthogonal projections a 1 and a 2.
Point a 1 are called horizontal projection points BUT, point a 2- her frontal projection... Each of them is the base of the perpendicular dropped from the point BUT respectively on the plane H and V.
It can be proved that point projectionalways located on straight lines, perpendicurly axesOH and crossing this axisat the same point. Indeed, the projecting rays BUTa 1 and BUTa 2 define a plane perpendicular to the projection planes and the line of their intersection - axes OH. This plane crosses H and V by direct a 1 ax and a 1 ax, which form with the axis OX and right angles to each other with apex at the point butx.
The converse is also true, i.e. if points are given on the projection planesa 1 and a 2 , located on straight lines intersecting axis OXat a given point at a right angle,then they are projections of somepoint A. This point is determined by the intersection of the perpendiculars retrieved from the points a 1 and a 2 to the planes H and V.
Note that the position of the projection planes in space may turn out to be different. For example, both planes, being mutually perpendicular, can be vertical. But in this case, the above-proved assumption about the orientation of opposite projections of points relative to the axis remains valid.
To get a flat drawing consisting of the above projections, the plane H combined by rotation around the axis OX with plane V as shown by the arrows in the illustration. As a result, the front half-plane H will be aligned with the lower half-plane V, and the back half-plane H- with the upper half-plane V.
A projection drawing, in which the projection planes with everything that is depicted on them, are aligned in a certain way with one another, is called plot(from French epure - drawing). The figure shows a plot of a point BUT.
With this method of aligning the planes H and V projections a 1 and a 2 will be located on the same perpendicular to the axis OX... In this case, the distance a 1 a x — from the horizontal projection of the point to the axis OX BUT to plane V and the distance a 2 a x — from the frontal projection of the point to the axis OX is equal to the distance from the point itself BUT to plane H.
Straight lines connecting opposite projections of a point on the plot, we agree to call projection communication lines.
The position of the projections of points on the plot depends on which quarter the given point is in. So if the point IN is located in the second quarter, then after aligning the planes, both projections will be lying above the axis OX.
If point WITH is in the third quarter, then its horizontal projection, after aligning the planes, will be above the axis, and the frontal projection will be below the axis OX. Finally, if the point D is located in the fourth quarter, then both projections of it will be under the axis OX. The figure shows the points M and N lying on the projection planes. In this position, the point coincides with one of its projections, while its other projection turns out to lie on the axis OX. This feature is reflected in the designation: near the projection with which the point itself coincides, a capital letter is written without an index.
Note also the case when both projections of a point coincide. This will be the case if the point is in the second or fourth quarter at the same distance from the projection planes. Both projections are aligned with the point itself, if the latter is located on the axis OX.
ORTHOGONAL SYSTEM OF THREE PLANES OF PROJECTIONS.
It was shown above that two projections of a point determine its position in space. Since each figure or body is a collection of points, it can be argued that two orthogonal projections of an object (in the presence of letter designations) completely determine its shape.
However, in practice, images building structures, machines and various engineering structures, it becomes necessary to create additional projections. They do this for the sole purpose of making the projection drawing clearer, more readable.
The model of three projection planes is shown in the figure. The third plane perpendicular to and H and V, denoted by the letter W and called profile.
The projections of points onto this plane will also be called profile, and denote them by capital letters or numbers with an index of 3 (ah,bh,cs, ...1h, 2h, 3 3 ...).
The projection planes, intersecting in pairs, define three axes: OX, OY and OZ, which can be considered as a system of rectangular Cartesian coordinates in space with the origin at point O. The system of signs indicated in the figure corresponds to the "right system" of coordinates.
Three projection planes divide the space into eight trihedral angles - these are the so-called octants... The numbering of octants is given in the figure.
To get a plot of a plane H and W rotate as shown in the figure until aligned with the plane V... As a result of rotation, the front half-plane H turns out to be aligned with the lower half-plane V, and the back half-plane H- with the upper half-plane V... When rotated 90 ° around the axis OZ front half-plane W will be aligned with the right half-plane V, and the back half-plane W- with the left half-plane V.
The final view of all aligned projection planes is shown in the figure. In this drawing, the axes OX and OZ, lying in a non-movable plane V, are shown only once, and the axis OY shown twice. This is explained by the fact that, rotating with the plane H, axis OY on the plot is aligned with the axis OZ, while rotating with the plane W, the same axis is aligned with the axis OX.
In the future, when the axes are designated on the diagram, the negative semiaxes (- OX, — OY, — OZ) will not be specified.
THREE COORDINATES AND THREE PROJECTIONS OF A POINT AND ITS RADIUS VECTOR.
Coordinates are numbers thatmatch the point for the definitionits position in space or onsurface.
In three-dimensional space, the position of a point is set using rectangular Cartesian coordinates x, y and z.
Coordinate NS are called abscissa, at— ordinate and z — applicate. Abscissa NS determines the distance from a given point to the plane W, ordinate y - to plane V and applicate z - to plane H... Having adopted the system shown in the figure for the reference point coordinates, we will compose a table of coordinate signs in all eight octants. Any point in space BUT, given by coordinates, will be denoted as follows: A(x, y,z).
If x = 5, y = 4 and z = 6, then the record will take the following form BUT(5, 4, 6). This point BUT, all coordinates of which are positive are in the first octant
Point coordinates BUT are at the same time the coordinates of its radius vector
OA relative to the origin. If i, j, k- unit vectors, respectively directed along the coordinate axes x, y,z(figure), then
OA =OA x i+ OAyj + OAzk , where OA X, OA U, OA g - vector coordinates OA
It is recommended to construct an image of the point itself and its projections on a spatial model (figure) using a coordinate rectangular parallelepiped. First of all, on the coordinate axes from the point O set aside segments, respectively equal 5, 4 and 6 units of length. On these segments (Oa x , Oa y , Oa z ), like on the edges, build rectangular parallelepiped... Its vertex, opposite to the origin, will determine the given point BUT. It is easy to see that to define a point BUT it is enough to construct only three edges of the parallelepiped, for example Oa x , a x a 1 and a 1 BUT or Oa y , a y a 1 and a 1 A and so on. These edges form a coordinate polyline, the length of each link of which is determined by the corresponding coordinate of the point.
However, the construction of a parallelepiped allows you to determine not only the point BUT, but also all three of its orthogonal projections.
Rays projecting a point on a plane H, V, W are those three edges of the parallelepiped that intersect at the point BUT.
Each of the orthogonal projections of the point BUT, being located on a plane, it is defined by only two coordinates.
So, the horizontal projection a 1 defined by coordinates NS and y, frontal projection a 2 - coordinates x andz, profile projection a 3 — coordinates at and z... But any two projections are defined by three coordinates. This is why specifying a point with two projections is equivalent to specifying a point with three coordinates.
On the diagram (figure), where all projection planes are aligned, projections a 1 and a 2 will be on the same perpendicular to the axis OX, and the projection a 2 and a 3 — on one perpendicular to the axis OZ.
As for projections a 1 and a 3 , then they are also connected by straight lines a 1 a y and a 3 a y , perpendicular to the axis OY. But since this axis occupies two positions on the diagram, the segment a 1 a y cannot be a continuation of the segment a 3 a y .
Point projection A (5, 4, 6) on the plot along the given coordinates, perform in the following sequence: first of all, a segment is laid on the abscissa axis from the origin of coordinates Oa x = x(in our case x =5), then through the point a x draw a perpendicular to the axis OX, on which, taking into account the signs, we postpone the segments a x a 1 = y(we get a 1 ) and a x a 2 = z(we get a 2 ). It remains to construct a profile projection of the point a 3 . Since the profile and frontal projection of the points must be located on the same perpendicular to the axis OZ , then through a 3 conduct a direct a 2 a z ^ OZ.
Finally, the last question arises: at what distance from the axis OZ should there be a 3?
Considering the coordinate parallelepiped (see figure), the edges of which a z a 3 = O a y = a x a 1 = y we conclude that the required distance a z a 3 equals at. Line segment a z a 3 laid to the right of the axis OZ, if y> 0, and to the left, if y
Let's see what changes will occur on the diagram when the point begins to change its position in space.
Let, for example, point A (5, 4, 6) will move in a straight line perpendicular to the plane V... With this movement, only one coordinate will change y, showing the distance from a point to a plane V... Coordinates will remain constant x andz , and the projection of the point defined by these coordinates, i.e. a 2 will not change its position.
As for projections a 1 and a 3 , then the first will begin to approach the axis OX, the second - to the axis OZ. In the figures, the new position of the point corresponds to the designations a 1 (a 1 1 a 2 1 a 3 1 ). The moment the point is on the plane V(y = 0), two of the three projections ( a 1 2 and a 3 2 ) will lie on the axes.
Moving from I octant in II, the point will start moving away from the plane V, coordinate at becomes negative, its absolute value will increase. The horizontal projection of this point, being located on the back half-plane H, on the plot will be above the axis OX, and the profile projection, being on the back half-plane W, on the plot will be to the left of the axis OZ. As always, the segment a za 3 3 = y.
In the subsequent plots, we will not denote by letters the points of intersection of the coordinate axes with the lines of the projection connection. This will simplify the drawing to some extent.
In the future, there will be diagrams without coordinate axes. This is done in practice when depicting objects, when only the image itself is essentialthe position of the object, and not its position, is relatedspecifically the projection planes.
In this case, the projection planes are determined with an accuracy only up to parallel translation (figure). They are usually moved parallel to themselves in such a way that all points of the object are above the plane. H and in front of the plane V... Since the position of the X 12 axis turns out to be undefined, the formation of the diagram in this case does not need to be associated with the rotation of the planes around the coordinate axis. When switching to a plot of a plane H and V are combined so that opposite projections of points are located on vertical lines.
Axleless plot of points A and B(drawing) notdetermines their position in space,but allows one to judge their relative orientation. So, the segment △ x characterizes the displacement of the point BUT with respect to the point IN in the direction parallel to the planes H and V. In other words, △ x indicates how much the point BUT located to the left of the point IN. The relative displacement of a point in the direction perpendicular to the plane V is determined by the segment △ y, i.e., the point And in our example is closer to the observer than the point IN, by a distance equal to △ y.
Finally, the segment △ z shows the elevation of the point BUT over point IN.
Supporters of axle-free study of the descriptive geometry course rightly point out that when solving many problems, you can do without coordinate axes. However, a complete rejection of them cannot be considered expedient. Descriptive geometry is designed to prepare the future engineer not only for the competent execution of drawings, but also for solving various technical problems, among which the problems of spatial statics and mechanics are not the last. And for this it is necessary to educate the ability to orient one or another object relative to the Cartesian axes of coordinates. These skills will be necessary in the study of such sections of descriptive geometry as perspective and axonometry. Therefore, on a number of plots in this book, we save images of the coordinate axes. Such drawings determine not only the shape of the object, but also its location relative to the projection planes.
The position of a point in space can be specified by two of its 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, and determine the octant in which it is located. Let's consider several typical problems from the descriptive geometry course.
According to a given complex drawing of points A and B, it is necessary:
Let us first determine the coordinates of point A, which can be written as A (x, y, z). Horizontal projection of point A - point A ", having coordinates x, y. Draw from point A" perpendiculars to axes x, y and find A х, A у, respectively. 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 values of the x axis. Taking into account the scale of the drawing, we find x = 10. The y coordinate is equal to the length of the segment A y O with a minus sign, since m. A y lies in the region of negative values of the y axis. Taking into account the scale of the drawing y = –30. Frontal projection of point A - point A "" has coordinates x and z. Let us 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. Taking into account 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 - m. B ". Since it lies on the x-axis, then B x = B" and the coordinate B y = 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. Frontal projection of point B - point B˝ has coordinates x, z. Let's draw a perpendicular from B "" to the z-axis, so we find 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 the necessary constructions are shown in the figure below.
Building projections of points
Points A and B in the plane П 3 have the following coordinates: A "" "(y, z); B" "" (y, z). In this case, A "" and A "" "lie in the same perpendicular to the z-axis, since they have a common z-coordinate. Similarly, B" "and B" "" lie on the common perpendicular to the z-axis. To find the profile projection of point A, we will postpone the value of the corresponding coordinate found earlier along the y-axis. In the figure, this is done using an arc of a circle of radius A y O. After that, draw a perpendicular from A y until it intersects with the perpendicular restored from point A "" to the z-axis. The intersection point of these two perpendiculars defines 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, you just need to draw a perpendicular from B" "to the z-axis. The intersection of this perpendicular with the z-axis is B "" ".
Determining the position of points in space
Visualizing a spatial layout made up of projection planes P 1, P 2 and P 3, the arrangement of octants, as well as the order of transformation of the layout into diagrams, one can directly determine that point A is located in the third octant, and point B lies in the plane P 2.
Another option for solving this problem is the method of exclusions. For example, the coordinates of point A are (10, -30, -10). The positive abscissa x allows us 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 octants. Finally, a negative applicate z indicates that m. A is located in the third octant. The above reasoning is clearly illustrated by the following table.
| Octants | Coordinate signs | ||
| x | y | z | |
| 1 | + | + | + |
| 2 | + | – | + |
| 3 | + | – | – |
| 4 | + | + | – |
| 5 | – | + | + |
| 6 | – | – | + |
| 7 | – | – | – |
| 8 | – | + | – |
Point B coordinates (30, 0, -20). Since the ordinate of m. B is equal to zero, this point is located in the plane of projections P 2. The positive abscissa and negative applicate of point B indicate that it is located on the border of the third and fourth octants.
Construction of a visual image of points in the system of planes P 1, P 2, P 3
Using a frontal isometric projection, we have built a spatial layout of the III octant. It is a rectangular trihedral, whose faces are the planes P 1, P 2, P 3, and the angle (-y0x) is 45 º. In this system, the segments along the x, y, z axes will be plotted in full size without distortion.
We will start constructing a visual image of point A (10, -30, -10) with its horizontal projection A ". Putting the corresponding coordinates along the abscissa and ordinate axes, we find the points A x and A y. Intersection of perpendiculars reconstructed from A x and A y respectively to the axes x and y determines the position of point A ". Setting aside from A "the segment AA", the length of which is equal to 10, parallel to the z-axis towards its negative values, we find the position of point A.
A visual image of point B (30, 0, -20) is constructed in a similar way - in the plane P2 along the x and z axes, you need to postpone the corresponding coordinates. The intersection of the perpendiculars reconstructed from B x and B z will determine the position of point B.
The projection of a point on three projection planes of the coordinate angle begins with obtaining its image on the H plane - the horizontal projection plane. To do this, through point A (Fig.4.12, a), a projection beam is drawn perpendicular to plane H.
In the figure, the perpendicular to the H plane is parallel to the Oz axis. The point of intersection of the beam with the plane H (point a) is chosen arbitrarily. The segment Aa determines at what distance point A is from the plane H, thereby indicating uniquely the position of point A in the figure in relation to the projection planes. Point a is a rectangular projection of point A onto the plane H and is called the horizontal projection of point A (Fig. 4.12, a).
To obtain an image of point A on the plane V (Fig. 4.12, b), a projection beam is drawn through point A perpendicular to the frontal plane of projections V. In the figure, the perpendicular to the plane V is parallel to the Oy axis. On the H plane, the distance from point A to the plane V is represented by a segment aa x parallel to the Oy axis and perpendicular to the Ox axis. If we imagine that the projection ray and its image are held simultaneously in the direction of the plane V, then when the image of the ray crosses the Ox axis at point a x, the ray will cross the plane V at point a. " , which is the image of the projection ray Aa on the plane V, at the intersection with the projection ray, point a "is obtained. Point a "is a frontal projection of point A, that is, its image on the plane V.
The image of point A on the profile plane of the projections (Fig.4.12, c) is built using a projection beam, perpendicular to the plane W. In the figure, the perpendicular to the plane W is parallel to the Ox axis. The projection ray from point A to the plane W on the plane H will be represented by a segment aa y parallel to the Ox axis and perpendicular to the Oy axis. From the point Oy parallel to the Oz axis and perpendicular to the Oy axis, an image of the projection ray aA is constructed and, at the intersection with the projection ray, point a "is obtained. Point a" is a profile projection of point A, that is, an image of point A on the plane W.
Point a "can be constructed by drawing from point a" segment a "a z (the image of the projection ray Aa" on the plane V) parallel to the Ox axis, and from point a z - segment a "a z parallel to the Oy axis until it intersects with the projection ray.
Having received three projections of point A on the projection planes, the coordinate angle is deployed into one plane, as shown in Fig. 4.11, b, together with the projections of the point A and the projection rays, and the point A and the projection rays Aa, Aa "and Aa" are removed. The edges of the aligned projection planes are not drawn, but only the projection axes Oz, Oy and Oy, Oy 1 are drawn (Fig. 4.13).
Analysis of the orthogonal drawing of the point shows that three distances - Aa ", Aa and Aa" (Fig. 4.12, c), characterizing the position of point A in space, can be determined by discarding the projection object itself - point A, on the coordinate angle unfolded into one plane (fig. 4.13). Segments a "a z, aa y and Oa x are equal to Aa" as opposite sides of the corresponding rectangles (Fig. 4.12, c and 4.13). They determine the distance at which point A is located from the profile plane of the projections. Segments a "a x, a" and y1 and Oa y are equal to the segment Aa, determine the distance from point A to the horizontal plane of projections, the segments aa x, and "a z and Oa y 1 are equal to the segment Aa", which determines the distance from point A to frontal projection plane.
Segments Oa x, Oa y and Oa z, located on the projection axes, are a graphical expression of the dimensions of the coordinates X, Y and Z of point A. The coordinates of the point are designated with the index of the corresponding letter. By measuring the size of these segments, you can determine the position of the point in space, that is, set the coordinates of the point.
On the diagram, the segments a "a x and aa x are located as one line perpendicular to the Ox axis and the segments a" a z and a "az - to the Oz axis. These lines are called projection connection lines. They intersect the projection axes at the points a x and and z respectively.The line of the projection connection connecting the horizontal projection of point A with the profile one turned out to be "cut" at the point a y.
Two projections of the same point are always located on the same line of the projection connection, perpendicular to the projection axis.
To represent the position of a point in space, two of its projections and a given origin of coordinates (point O) are sufficient. 4.14, b two projections of a point completely determine its position in space.According to these two projections, you can build a profile projection of point A. Therefore, in the future, if there is no need for a profile projection, the diagrams will be built on two projection planes: V and H.
Rice. 4.14. Rice. 4.15.
Let's consider several examples of building and reading a drawing of a point.
Example 1. Determination of the coordinates of the point J given on the diagram by two projections (Fig. 4.14). Three segments are measured: segment Ov X (coordinate X), segment b X b (coordinate Y) and segment b X b "(coordinate Z). The coordinates are written in the following row: X, Y and Z, after the letter designation of the point, for example , B20; 30; 15.
Example 2... Constructing a point based on specified coordinates. Point C is given by coordinates C30; 10; 40. On the Ox axis (Fig. 4.15) find a point with x, at which the line of the projection connection intersects the projection axis. To do this, along the Ox axis from the origin (point O), the X coordinate (size 30) is plotted and a point with x is obtained. Through this point, perpendicular to the Ox axis, a line of projection connection is drawn and the coordinate Y is laid down from the point (size 10), point c is obtained - the horizontal projection of point C. Upward from point c along the line of projection connection, the coordinate Z is laid down (size 40), point is obtained c "- frontal projection of point C.
Example 3... Creation of a profile projection of a point according to given projections. The projections of the point D - d and d "are set. The projection axes Oz, Oy and Oy 1 are drawn through point O. her to the right behind the Oz axis. On this line, the profile projection of point D will be located. It will be located at such a distance from the Oz axis, at which the horizontal projection of point d is located: from the Ox axis, that is, at a distance dd x. The segments d z d "and dd x are the same, since they define the same distance - the distance from point D to the frontal plane of projections. This distance is the Y coordinate of point D.
Graphically, the segment dzd "is constructed by transferring the segment dd x from the horizontal projection plane to the profile one. To do this, draw a line of projection connection parallel to the Ox axis, get the point dy on the Oy axis (Fig. 4.16, b). Then transfer the size of the Od y segment to the Oy 1 axis , drawing from point O an arc with a radius equal to the segment Od y, up to the intersection with the axis Oy 1 (Fig. 4.16, b), point dy 1 is obtained.This point can be constructed and, as shown in Fig. 4.16, c, drawing a straight line at an angle 45 ° to the axis Oy from the point dy. From the point d y1 draw a line of projection connection parallel to the axis Oz and lay on it a segment equal to the segment d "dx, get a point d".
The transfer of the value of the segment d x d to the profile plane of the projections can be carried out using a constant straight drawing (Fig. 4.16, d). In this case, the line of projection connection dd y is drawn through the horizontal projection of a point parallel to the axis Oy 1 until it intersects with a constant straight line, and then parallel to the axis Oy until it intersects with the continuation of the line of projection connection d "d z.
Special cases of the location of points relative to the projection planes
The position of a point relative to the projection plane is determined by the corresponding coordinate, that is, by the size of the segment of the projection connection line from the Ox axis to the corresponding projection. In fig. 4.17 the Y coordinate of point A is determined by the segment aa x - the distance from point A to the plane V. The Z coordinate of point A is determined by the segment a "and x is the distance from point A to the plane H. If one of the coordinates is zero, then the point is located on the projection plane Fig. 4.17 shows examples of different locations of points relative to the projection planes.The Z coordinate of point B is zero, the point is in the plane H. Its frontal projection is on the Ox axis and coincides with the point b x. The Y coordinate of point C is zero, the point is on the plane V, its horizontal projection c is on the Ox axis and coincides with the point c x.
Therefore, if a point is on the projection plane, then one of the projections of this point lies on the projection axis.
In fig. 4.17 coordinates Z and Y of point D are equal to zero, therefore, point D is located on the axis of projections Ox and its two projections coincide.