A drawing is the first and very important step in solving a geometric problem. What should be the drawing of a regular pyramid?

Let's remember first parallel design properties:

- parallel segments of the figure are depicted as parallel segments;

- the ratio of the lengths of segments of parallel lines and segments of one straight line is preserved.

Drawing of a regular triangular pyramid

First, draw the base. Since the angles and ratios of the lengths of non-parallel segments are not preserved in parallel design, the regular triangle at the base of the pyramid is represented by an arbitrary triangle.

The center of an equilateral triangle is the intersection point of the medians of the triangle. Since the medians at the intersection point are divided in a ratio of 2: 1, counting from the top, we mentally connect the top of the base with the middle of the opposite side, approximately divide it into three parts, and put a point at a distance of 2 parts from the top. Draw a perpendicular from this point upwards. This is the height of the pyramid. We draw the perpendicular so long that the side edge does not cover the image of the height.

Drawing of a regular quadrangular pyramid

The drawing of a regular quadrangular pyramid also starts from the base. Since the parallelism of the segments is preserved, but the magnitudes of the angles are not, the square at the base is depicted as a parallelogram. It is desirable to make the acute angle of this parallelogram smaller, then the side faces are larger. The center of a square is the intersection point of its diagonals. We draw diagonals, from the point of intersection we restore the perpendicular. This perpendicular is the height of the pyramid. We choose the length of the perpendicular so that the side edges do not merge with each other.

Drawing of a regular hexagonal pyramid

Since the parallel projection preserves the parallelism of the segments, the base of a regular hexagonal pyramid - a regular hexagon - is depicted as a hexagon, in which the opposite sides are parallel and equal. The center of a regular hexagon is the intersection point of its diagonals. In order not to clutter up the drawing, we do not draw diagonals, but we find this point approximately. From it we restore the perpendicular - the height of the pyramid - so that the side edges do not merge with each other.

The development of the surface of the pyramid is a flat figure, composed of the base and faces of the pyramid, aligned with a certain plane. In the example below, we will consider the construction of a sweep using the triangle method.

Pyramid SABC is crossed by the frontally projecting plane α. It is necessary to build a development of the SABC surface and draw an intersection line on it.

On the frontal projection S""A""B""C"" we mark the points D"", E"" and F"", in which the trace α v intersects with segments A""S"", B""S"" and C""S"" respectively. Determine the position of the points D", E", F" and connect them to each other. The intersection line is indicated in the figure in red.

Determining the length of the ribs

To find the natural values ​​of the lateral edges of the pyramid, we use the method of rotation around the projecting line. To do this, draw the i-axis through the vertex S perpendicular to the horizontal plane H. Rotating the segments SA, SB and SC around it, we move them to a position parallel to the frontal plane V.

The actual values ​​of the edges are equal to the projections S""A"" 1 , S"" 1 B"" 1 and S""C"" 1 . We mark points on them D "" 1, E"" 1, F"" 1, as shown by the arrows in the figure above.

Triangle ABC at the base of the pyramid is parallel to the horizontal plane. It is displayed on it in full size, equal to ∆A"B"C".

Sweep construction order

In an arbitrary place on the drawing, mark the point S 0. We draw a straight line n through it and set aside the segment S 0 A 0 = S "" A "" 1.

We build the face ABS = A 0 B 0 S 0 as a triangle on three sides. To do this, from points S 0 and A 0 we draw arcs of circles with radii R 1 \u003d S "" B "" 1 and r 1 \u003d A "B", respectively. The intersection of these arcs determines the position of the point B 0 .

The faces B 0 S 0 C 0 and C 0 S 0 A 0 are constructed similarly. The base of the pyramid, depending on the layout of the drawing, is attached to any of the sides: A 0 B 0, B 0 C 0 or C 0 A 0.

Let us draw on the development a line along which the plane α intersects with the pyramid. To do this, on the edges S 0 A 0 , S 0 B 0 and S 0 С 0 we mark the points D 0 , E 0 and F 0 respectively. In this case, the point D 0 is located at the intersection of the segment S 0 A 0 with a circle of radius S""D"" 1 . Similarly, E 0 = S 0 B 0 ∩ S""E"" 1 , F 0 = S 0 C 0 ∩ S""F"" 1 .

GENERAL CONCEPTS ABOUT THE DEVELOPMENT OF SURFACES

We will consider the surface as flexible inextensible shell. In this case, some surfaces can be combined with the plane by transformation no breaks or wrinkles . Surfaces that allow such a transformation are called deployable.

The figure obtained by combining a developable surface with a plane is called a development.

The construction of developments is of great importance in the design of products from sheet material (vessels, pipelines, patterns, etc.).

Surfaces that develop geometrically accurate : polyhedral, conical, torsos, cylindrical.

Of the curved surfaces, the unfolding ones include those ruled surfaces (conical, cylindrical, torsos), in which the tangent plane touches the surface along its rectilinear generatrix.

All other curved surfaces are non-developing, but you can build them if necessary. approximate sweeps.

To build a development of any curved surface, it is divided into such curvilinear sections, each of which can be approximated by some flat figure, which requires to determine its nature only measurements.

For example:

The cylinder is divided into rectangles (Figure 16-1a);

a right cone into isosceles triangles (Figure 16-1b);

elliptical cylinder - into parallelograms (Figure 16-1c);

elliptical cone - into triangles (Figure 16-1d);

sphere - on a trapezoid.

PYRAMID AND CONICAL SURFACE REVEALS

As examples, consider the construction of developments of only four surfaces: a pyramid, a cone, a prism, and a cylinder.

Surface development of the pyramid

The development of such a surface is a flat figure, which is obtained by combining all its faces with one plane.

Example 1. Build a development of the surface of the pyramid ABCS (Figure 16-2) and draw a line MN on it .

Since the side faces of the pyramid are triangles, in order to build a development it is necessary to find the natural form of these triangles, for which it is necessary to determine the true lengths of the sides - the edges of the pyramid.

The base of the pyramid lies in a horizontal plane, therefore, the actual size of the ribs AB, BC and AC is already on the drawing.

Rib SA is a frontal, so it is depicted in front view in full size.

The nature of the ribs SВ and SC is determined by the method of a right-angled triangle. One of its legs is the excess of the point S over the points B and C, and the second is the top view of the ribs SB and SC.

Then, on three sides, we build successively all the side faces of the pyramid.

To plot the line MN on the development, we first determine the true value of the segments AM and B1 and put them on the development on the corresponding edges.

To plot the point M, we draw a straight line S2 on the face SBC and find its position on the development, setting aside the segment B2 (measured in the top view) on the side BC. Then, in the front view, draw a segment 3-4 through point 4 parallel to the edge BC and find its position on the development, for which we set aside the segment C4 on the side SC and draw a straight line 3-4 parallel to the edge BC through the resulting point. At the intersection of lines S -2 and 3-4 we find the point N. By connecting the obtained points M, 1, N we get the desired line.

It is necessary to build a development of the faceted bodies and draw on the development the line of intersection of the prism and the pyramid.

To solve this problem in descriptive geometry, you need to know:

- information about the development of surfaces, methods of their construction and, in particular, the construction of developments of faceted bodies;

- one-to-one properties between a surface and its unfolding and methods for transferring points belonging to the surface to the unfolding;

- methods for determining the natural values ​​of geometric images (lines, planes, etc.).

Procedure for solving the Problem

The scan is called a flat figure, which is obtained by cutting and unbending the surface until it is completely aligned with the plane. All surface unfolds ( blanks, patterns) are built only from natural values.

1. Since the scans are built from natural values, we proceed to determine them, for which a tracing paper (graph paper or other paper) of A3 format is transferred task No. z with all points and lines of intersection of polyhedra.

2. To determine the natural values ​​of the edges and the base of the pyramid, we use right triangle method. Of course, others are possible, but in my opinion, this method is more intelligible for students. Its essence lies in the fact that “on the constructed right angle, the projection value of the straight line segment is plotted on one leg, and on the other, the difference in the coordinates of the ends of this segment, taken from the conjugate projection plane. Then the hypotenuse of the resulting right angle gives the natural value of this line segment..

Fig.4.1

Fig.4.2

Fig.4.3

3. So, in the free space of the drawing (Fig.4.1.a) making a right angle.

On the horizontal line of this angle, we set aside the projection value of the edge of the pyramid DA taken from the horizontal projection plane - lDA. On the vertical line of the right angle, we plot the difference in the coordinates of the points D’ AndA’ taken from the frontal projection plane (along the axis z down) - . Connecting the obtained points with a hypotenuse, we obtain the natural size of the edge of the pyramid | DA| .

Thus, we determine the natural values ​​\u200b\u200bof other edges of the pyramid D.B. And DC, as well as the base of the pyramid AB, BC, AC (fig.4.2), for which we construct the second right angle. Note that the definition of the natural size of the edge DC is made in those cases when it is given in projection on the original drawing. This is easily determined if we remember the rule: if a straight line on any projection plane is parallel to the coordinate axis, then on the conjugate plane it is projected in full size.

In particular, in the example of our problem, the frontal projection of the edge D’ C’ parallel to axis X, therefore, in the horizontal plane DC immediately expressed in natural size | DC| (fig.4.1).

Fig.4.4

4. Having determined the natural values ​​of the edges and the base of the pyramid, we proceed to the construction of a sweep ( fig.4.4). To do this, on a sheet of paper closer to the left side of the frame, we take an arbitrary point D considering that this is the top of the pyramid. Draw from a point D arbitrary straight line and set aside on it the natural size of the edge | DA| , getting a point A. Then from the point A, taking on the solution of the compass the full size of the base of the pyramid R=|AB| and placing the leg of the compass at the point A we make an arc. Next, we take on the solution of the compass the full size of the edge of the pyramid R=| D.B.| and placing the leg of the compass at the point D we make a second arc notch. At the intersection of arcs we get a point IN, connecting it with points A and D get the edge of the pyramid DAB. Similarly, we attach to the edge D.B. facet DBC, and to the edge DC- edge DCA.

To one side of the base, for example INC, we attach the base of the pyramid also by the method of geometric serifs, taking the size of the sides on the compass solution ABAndAWITH and making arc serifs from points BAndC getting a point A(fig.4.4).

5. Building a sweep prism is simplified by the fact that in the original drawing in the horizontal plane of projections the base, and in the frontal plane - 85 mm high, it set at full size

To build a sweep, we mentally cut the prism along some edge, for example, along E, having fixed it on the plane, we will expand the other faces of the prism until it is completely aligned with the plane. It is quite obvious that we will get a rectangle whose length is the sum of the lengths of the sides of the base, and the height is the height of the prism - 85mm.

So, to build a sweep of the prism, we proceed:

- on the same format where the pyramid sweep is built, on the right side we draw a horizontal straight line and from an arbitrary point on it, for example E, successively lay off segments of the base of the prism EK, KG, GU, UE, taken from the horizontal projection plane;

- from points E, K, G, U, E we restore the perpendiculars, on which we set aside the height of the prism, taken from the frontal projection plane (85mm);

- connecting the obtained points with a straight line, we obtain a development of the side surface of the prism and to one of the sides of the base, for example, GU we attach the upper and lower bases using the method of geometric serifs, as was done when building the base of the pyramid.

Fig.4.5

6. To build a line of intersection on the development, we use the rule that "any point on the surface corresponds to a point on the development." Take, for example, the edge of a prism GU where the line of intersection with the points 1-2-3 ; . Set aside on the development of the base GU points 1,2,3 by distances taken from the horizontal projection plane. Restore the perpendiculars from these points and plot the heights of the points on them 1’ , 2’, 3’ , taken from the frontal projection plane - z 1 , z 2 Andz 3 . Thus, we got points on the sweep 1, 2, 3, connecting which we get the first branch of the line of intersection.

All other points are transferred similarly. The constructed points are connected, getting the second branch of the line of intersection. Highlight in red - the desired line. Let us add that in case of incomplete intersection of faceted bodies, there will be one closed branch of the intersection line on the development of the prism.

7. The construction (transfer) of the intersection line on the development of the pyramid is carried out in the same way, but taking into account the following:

- since the sweeps are built from natural values, it is necessary to transfer the position of the points 1-8 lines of intersection of projections on the lines of edges of natural sizes of the pyramid. To do this, take, for example, the points 2 and 5 in the frontal projection of the rib DA we transfer them to the projection value of this right angle edge (fig.4.1) along communication lines parallel to the axis X, we get the required segments | D2| and |D5| ribs DA in natural values, which we set aside (transfer) to the development of the pyramid;

- all other points of the intersection line are transferred in the same way, including points 6 and 8 lying on the generators Dm And Dn why right angle (fig.4.3) the natural values ​​of these generators are determined, and then points are transferred to them 6 and 8;

- on the second right angle, where the natural values ​​\u200b\u200bof the base of the pyramid are determined, points are transferred mAndn intersections of generators with the base, which are subsequently transferred to the development.

Thus, the points obtained on natural values 1-8 and transferred to the development, we connect in series with straight lines and finally we get the line of intersection of the pyramid on its development.

Section: Descriptive geometry /

The development of the lateral surface of the pyramid (Fig. 16.3) consists of three triangles, representing in true form the lateral faces of the pyramid.

To construct a development, it is necessary to first determine the true lengths of the side edges of the pyramid. Turning these ribs around the height of the pyramid to a position parallel to the p 2 plane, on the frontal projection plane we obtain their true lengths in the form of segments and .

Having built on three sides and the face of the pyramid ASB (Fig. 16.4), we attach to it an adjacent face - the triangle BSC, and to the last face CSA. The resulting figure will be a development of the side surface of this pyramid.

To obtain a complete sweep, we attach the base of the pyramid to one of the sides of the base - the triangle ABC.

To build a line on the development along which the surface of the pyramid is intersected by plane a (Fig. 16.3), it is necessary to put on the edges SA, SB and SC, respectively, points 1, 2 and 3 at which this plane intersects the edges, determining the true lengths of the segments S1 , S2 and S3.

Rice. 16.3	Rice. 16.4

Control questions on the topic of the lecture:

1. What is called surface development?

2. What surfaces are called developable or non-developable. Give examples.

3. General rules for constructing the development of the surface of a prism, pyramid.