A flat surface of a cone is a flat figure obtained by aligning the side surface and base of the cone with a certain plane.

Scanning options:

Flattened circular cone

The sweep of the lateral surface of a straight circular cone is a circular sector, the radius of which is equal to the length of the generatrix of the conical surface l, and the central angle φ is determined by the formula φ \u003d 360 * R / l, where R is the radius of the circumference of the base of the cone.

In a number of problems of descriptive geometry, the preferred solution is the approximation (replacement) of a cone with a pyramid inscribed in it and the construction of an approximate sweep, on which it is convenient to draw lines lying on the conical surface.

Construction Algorithm

1. We fit a polygonal pyramid into the conical surface. The more side faces of the inscribed pyramid, the more accurate the correspondence between the actual and the approximate scan.
2. We build a development of the side surface of the pyramid using the triangles method. We connect the points belonging to the base of the cone with a smooth curve.

Example

In the figure below, a regular hexagonal pyramid SABCDEF is inscribed in a straight circular cone, and an approximate sweep of its lateral surface consists of six isosceles triangles - the faces of the pyramid.

Consider a triangle S 0 A 0 B 0. The lengths of its sides S 0 A 0 and S 0 B 0 are equal to the generator l of the conical surface. The value A 0 B 0 corresponds to the length A'B '. To construct a triangle S 0 A 0 B 0 in an arbitrary place of the drawing, set aside the segment S 0 A 0 \u003d l, after which from points S 0 and A 0 we draw circles with a radius of S 0 B 0 \u003d l and A 0 B 0 \u003d A'B ' respectively. We connect the point of intersection of the circles B 0 with the points A 0 and S 0.

The faces S 0 B 0 C 0, S 0 C 0 D 0, S 0 D 0 E 0, S 0 E 0 F 0, S 0 F 0 A 0 of the SABCDEF pyramids are built similarly to the triangle S 0 A 0 B 0.

We connect the points A, B, C, D, E and F lying at the base of the cone with a smooth curve - an arc of a circle whose radius is equal to l.

Oblique Cone Sweep

Consider the procedure for constructing a sweep of the lateral surface of an inclined cone by the approximation (approximation) method.

Algorithm

1. We inscribe hexagon 123456 into the circle of the base of the cone. Connect points 1, 2, 3, 4, 5 and 6 with the vertex S. The pyramid S123456, constructed in this way, with some degree of approximation is a replacement for the conical surface and is used in this capacity in further constructions.
2. We determine the natural values \u200b\u200bof the edges of the pyramid, using the method of rotation around the projecting line: in the example, the i-axis is used, perpendicular to the horizontal plane of the projections and passing through the vertex S.
   So, as a result of the rotation of the edge S5, its new horizontal projection S'5''1 takes a position at which it is parallel to the frontal plane π 2. Accordingly, S''5''''1 is the actual size S5.
3. We build a development of the side surface of the pyramid S123456, consisting of six triangles: S 0 1 0 6 0, S 0 6 0 5 0, S 0 5 0 4 0, S 0 4 0 3 0, S 0 3 0 2 0, S 0 2 0 1 0. Each triangle is constructed on three sides. For example, △ S 0 1 0 6 0 length S 0 1 0 \u003d S''1 '' '0, S 0 6 0 \u003d S''6' '' 1, 1 0 6 0 \u003d 1'6 '

The degree to which the approximate scan corresponds to the actual one depends on the number of faces of the inscribed pyramid. The number of faces is selected based on the ease of reading the drawing, the requirements for its accuracy, the presence of key points and lines that need to be transferred to the scan.

Transferring a line from a cone surface to a flat pattern

Line n, lying on the surface of the cone, is formed as a result of its intersection with a certain plane (figure below). Consider the algorithm for constructing the line n on the sweep.

Algorithm

1. Find the projections of points A, B and C at which line n intersects the edges of the pyramid S123456 inscribed in the cone.
2. Determine the actual size of the segments SA, SB, SC by rotating around the projecting line. In this example, SA \u003d S''A''''SB \u003d S''B'''1, SC \u003d S''C''''1.
3. We find the position of the points A 0, B 0, C 0 on the corresponding edges of the pyramid, postponing the segments S 0 A 0 \u003d S `` A '', S 0 B 0 \u003d S `` B '' 1, S 0 C 0 \u003d S``C '' 1.
4. Connect the points A 0, B 0, C 0 with a smooth line.

Truncated cone flat pattern

The method of constructing a sweep of a straight circular truncated cone described below is based on the principle of similarity.

It is necessary to build a flat pattern of surfaces and transfer the line of intersection of surfaces to the flat pattern. This problem is based on surfaces ( cone and cylinder) with their line of intersection given in previous problem 8.

To solve such problems in descriptive geometry, you need to know:

- the procedure and methods for constructing unfolded surfaces;

- mutual correspondence between the surface and its development;

- special cases of building sweeps.

Decision proceduresproblems

1. Note that a sweep is a figure obtained in
as a result of cutting the surface along some generatrix and gradually unbending it until it is completely aligned with the plane. Hence the sweep of a straight circular cone - a sector with a radius equal to the length of the generatrix and a base equal to the circumference of the base of the cone. All sweeps are built only from natural values.

Figure 9.1

- the circumference of the base of the cone, expressed in natural size, we divide by a number of shares: in our case - 10, the accuracy of building the sweep depends on the number of shares ( figure 9.1.a);

- we postpone the received shares, replacing them with chords, on the length
arc drawn with a radius equal to the length of the generatrix of the cone l \u003d | Sb |. We connect the beginning and end of the counting of the shares to the top of the sector - this will be the sweep of the lateral surface of the cone.

Second way:

- we build a sector with a radius equal to the length of the generatrix of the cone.
Note that both in the first and in the second case the extreme right or left generators of the cone l \u003d | Sb | are taken as the radius, since they are expressed in natural size;

- at the top of the sector, we postpone the angle a, determined by the formula:

Figure 9.2

where r - the value of the radius of the base of the cone;

l - the length of the generatrix of the cone;

360 - constant value converted to degrees.

To the unfolded sector, we build the base of the cone of radius r.

2. According to the conditions of the problem, it is required to move the intersection line
surfaces of the cone and cylinder for a scan. To do this, we use the properties of one-to-one between the surface and its flat pattern, in particular, note that every point on the surface corresponds to a point on the flat pattern and each line on the surface corresponds to a line on the flat pattern.

Hence follows the sequence of transferring points and lines
from the surface to the sweep.

Figure 9.3

For sweeping the cone. Let us agree that the surface of the cone is cut along the generatrix S’ a’ ... Then the points 1, 2, 3,…6
will lie on circles (arcs on the sweep) with radii respectively equal to the distances taken along the generatrix S’ A’ from the top S’ to the corresponding secant plane with points 1’ , 2’, 3’…6’ -| S1|, | S2|, | S3|….| S6 | (Figure 9.1.b).

The position of the points on these arcs is determined by the distance taken from the horizontal projection from the generatrix Sa, along the chord to the corresponding point, for example, to the point c, ac \u003d 35 mm ( figure 9.1.a). If the distance along the chord and the arc is very different, then to reduce the error, you can divide a larger number of fractions and put them on the corresponding sweep arcs. In this way, any points are transferred from the surface to its flat pattern. The resulting points will be connected by a smooth curve along the pattern ( figure 9.3).

For cylinder sweep.

A cylinder sweep is a rectangle with a height equal to the generatrix height and a length equal to the circumference of the cylinder base. Thus, to construct a sweep of a straight circular cylinder, it is necessary to construct a rectangle with a height equal to the height of the cylinder, in our case 100mm, and a length equal to the circumference of the base of the cylinder, determined by the known formulas: C=2 R\u003d 220mm, or by dividing the circumference of the base into a series of shares, as indicated above. We attach the base of the cylinder to the upper and lower parts of the resulting scan.

Let us agree that the cut is made along the generatrix AA 1 (A’ A’ 1 ; AA1) ... Note that the cut should be made along the characteristic (control) points for more convenient construction. Considering that the length of the sweep is the circumference of the base of the cylinder C, from point A’= A’ 1 section of the frontal projection, we take the distance along the chord (if the distance is large, then it must be divided into shares) to the point B’ (in our example - 17mm) and put it on the sweep (along the length of the base of the cylinder) from point A. From the obtained point B, draw a perpendicular (generatrix of the cylinder). Dot 1 should be on this perpendicular) at a distance from the base taken from the horizontal projection to the point. In our case, the point 1 lies on the axis of symmetry of the sweep at a distance 100/2 \u003d 50mm (fig. 9.4).

Figure 9.4

And we do this to find all other points on the sweep.

We emphasize that the distance along the length of the sweep to determine the position of the points is taken from the frontal projection, and the distance along the height - from the horizontal one, which corresponds to their natural values. We connect the resulting points with a smooth curve along the pattern ( figure 9.4).

In the variants of the problems, when the intersection line splits into several branches, which corresponds to the complete intersection of surfaces, the methods of constructing (transferring) the intersection line to the flat pattern are similar to those described above.

Section: Descriptive Geometry /

Sometimes the task arises - to make a protective umbrella for a chimney or chimney, an exhaust deflector for ventilation, etc. But before you start manufacturing, you need to make a pattern (or scan) for the material. There are all sorts of programs on the Internet for calculating such sweeps. However, the problem is so easy to solve that you will quickly calculate it using a calculator (in your computer) than you will search, download and deal with these programs.

Let's start with a simple option - a simple cone sweep. The easiest way to explain the principle of calculating a pattern is with an example.

Let's say we need to make a cone with a diameter of D cm and a height of H centimeters. It is quite clear that a circle with a cut out segment will act as a blank. Two parameters are known - diameter and height. By the Pythagorean theorem, we calculate the diameter of the workpiece circle (do not confuse with the radius finished cone). Half the diameter (radius) and the height form a right-angled triangle. Therefore:

So now we know the radius of the workpiece and can cut the circle.

Let's calculate the angle of the sector to be cut from the circle. We argue as follows: The diameter of the workpiece is 2R, which means that the circumference is Pi * 2 * R - i.e. 6.28 * R. Let us denote it by L. The circle is complete, i.e. 360 degrees. And the circumference of the finished cone is Pi * D. We denote it by Lm. It is naturally less than the circumference of the workpiece. We need to cut a segment with an arc length equal to the difference between these lengths. Let's apply the ratio rule. If 360 degrees gives us the complete circumference of the workpiece, then the desired angle should give the circumference of the finished cone.

From the ratio formula we obtain the size of the angle X. And the cut-out sector is found by subtracting 360 - X.

A sector with an angle (360-X) must be cut out of a round blank with a radius R. Remember to leave a small strip of overlap material (if the cone mounts will overlap). After connecting the sides of the cut sector, we get a cone of a given size.

For example: We need a cone for an umbrella chimney with a height (H) of 100 mm and a diameter (D) of 250 mm. According to the Pythagoras formula, we obtain the workpiece radius - 160 mm. And the circumference of the workpiece, respectively, is 160 x 6.28 \u003d 1005 mm. At the same time, the circumference of the cone we need is 250 x 3.14 \u003d 785 mm.

Then we get that the ratio of the angles will be: 785/1005 x 360 \u003d 281 degrees. Accordingly, it is necessary to cut the sector 360 - 281 \u003d 79 degrees.

Calculation of the blank pattern for a truncated cone.

Such a part is sometimes needed in the manufacture of adapters from one diameter to another or for Volpert-Grigorovich or Khanzhenkov deflectors. They are used to improve traction in a chimney or ventilation pipe.

The task is slightly complicated by the fact that we do not know the height of the entire cone, but only its truncated part. In general, there are three initial numbers: the height of the truncated cone H, the diameter of the lower hole (base) D, and the diameter of the upper hole Dm (at the section of the full cone). But we will resort to the same simple mathematical constructions based on the Pythagorean theorem and similarity.

Indeed, it is obvious that the value (D-Dm) / 2 (half the difference in diameters) will relate to the height of the truncated cone H in the same way as the base radius to the height of the entire cone, as if it were not truncated. Find the total height (P) from this ratio.

(D - Dm) / 2H \u003d D / 2P

Hence P \u003d D x H / (D-Dm).

Now, knowing the total height of the cone, we can reduce the solution to the previous problem. Calculate the sweep of the workpiece as if for a full cone, and then "subtract" from it the sweep of its upper, unnecessary part to us. And we can calculate directly the radii of the workpiece.

We obtain, according to the Pythagorean theorem, a larger radius of the workpiece - Rz. It is the square root of the sum of the squares of the heights P and D / 2.

The smaller radius Rm is the square root of the sum of squares (P-H) and Dm / 2.

The circumference of our workpiece is 2 x Pi x Rz, or 6.28 x Rz. And the circumference of the base of the cone is Pi x D, or 3.14 x D. The ratio of their lengths will give the ratio of the angles of the sectors, if we assume that the total angle in the workpiece is 360 degrees.

Those. X / 360 \u003d 3.14 x D / 6.28 x Rz

Hence X \u003d 180 x D / Rz (This is the angle that must be left in order to get the circumference of the base). And you need to cut out, respectively, 360 - X.

For example: We need to make a truncated cone 250 mm high, base diameter 300 mm, top hole diameter 200 mm.

We find the height of the full cone P: 300 x 250 / (300 - 200) \u003d 600 mm

According to t. Pythagoras we find the outer radius of the workpiece Rz: Square root of (300/2) ^ 2 + 6002 \u003d 618.5 mm

Using the same theorem, we find the smaller radius Rm: The square root of (600 - 250) ^ 2 + (200/2) ^ 2 \u003d 364 mm.

Determine the angle of the sector of our workpiece: 180 x 300 / 618.5 \u003d 87.3 degrees.

On the material we draw an arc with a radius of 618.5 mm, then from the same center - an arc with a radius of 364 mm. The arc angle can have approximately 90-100 degrees of opening. Draw radii with an opening angle of 87.3 degrees. Our blank is ready. Do not forget to give an allowance for joining the edges if they overlap.

we take the perpendiculars to each segment, on them we lay down the real values \u200b\u200bof the generatrices of the cylinder, taken from the frontal projection. Connecting the obtained points together, we get a curve.

To obtain a full sweep, we add a circle (base) and the actual size of the section (ellipse) to the sweep of the side surface, built along its major and minor axes or along points.

5.3.4. Creation of a flattened cone flat pattern

IN in a particular case, the sweep of a cone is a flat figure consisting of a circular sector and a circle (the base of the cone).

IN in the general case, the unfolding of the surface is carried out according to the principle of unfolding a polyhedral pyramid (that is, by the method of triangles) inscribed in a conical surface. The greater the number of faces of the pyramid inscribed in the conical surface, the smaller the difference between the actual and approximate sweeps of the conical surface will be.

The construction of the sweep of the cone begins with drawing from the point S 0 an arc of a circle with a radius equal to the length of the generatrix of the cone. On this arc, 12 parts of the circumference of the base of the cone are laid and the resulting points are connected to the top. An example of an image of a full scan of a truncated cone is shown in Fig. 5.7.

Lecture 6 (beginning)

MUTUAL CROSSING OF SURFACES. METHODS FOR CONSTRUCTION OF MUTUAL CROSSING OF SURFACES.

METHOD OF AUXILIARY SECTIONAL PLANES AND SPECIAL CASES

6.1. Mutual intersection of surfaces

Intersecting with each other, the surfaces of the bodies form various broken or curved lines, which are called lines of mutual intersection.

To construct intersection lines of two surfaces, you need to find points that simultaneously belong to two specified surfaces.

When one of the surfaces completely penetrates the other, 2 separate intersection lines are obtained, called branches. In the case of a cut-in, when one surface partially enters another, the line of intersection of the surfaces will be one.

6.2. Intersection of faceted surfaces

The intersection line of two polyhedrons is a closed spatial polyline. Its links are the lines of intersection of the faces of one polyhedron with the faces of another, and the vertices are the points of intersection of the edges of one polyhedron with the faces of another. Thus, to build a line of intersection of two polyhedra, you need to solve the problem either on the intersection of two planes (facet method), or on the intersection of a straight line with a plane (edge \u200b\u200bmethod). In practice, both methods are usually used in combination.

Intersection of a pyramid with a prism. Consider the case of intersection

pyramid with a prism, the lateral surface of which is projected by π3 onto the outline bases (quadrangle). We begin construction with a profile projection. When drawing points, we use the edge method, that is, when the edges of the vertical pyramid intersect the edges of the horizontal prism (Fig. 6.1).

Analysis of the problem statement shows that the line of intersection of the pyramid and the prism splits into 2 branches, one of the branches is a flat polygon, points 1, 2, 3, 4 (points of intersection of the edges of the pyramid with the face of the prism). Their horizontal, frontal and profile projections are located on the projections of the corresponding edges and are determined by the communication lines. Similarly, points 5, 6, 7 and 8 can be found belonging to another branch. Points 9, 10, 11, 12 are determined from the condition that the upper and lower edges of the prism are parallel to each other, that is, 1 "2" is parallel to 5 "10", etc.

You can use the construction clipping planes method. The construction plane intersects both surfaces along the broken lines. The mutual intersection of these lines gives us the points belonging to the desired intersection line. We select α "" "and β" "" as auxiliary planes. Using the plane α "" "

we find projections of points 1 ", 2", 3 ", 4", and planes β "" "- points 5", 6 ", 9", 10 ", 11", 12 ". Points 7 and 8 are determined as in the previous method ...

6.3. Intersection of faceted surfaces

from surfaces of revolution

Most of the technical parts and objects are composed of a combination of various geometric bodies. Intersecting with each other,

the surfaces of these bodies form various straight or curved lines, which are called lines of mutual intersection.

To build a line of intersection of two surfaces, you need to find points that would simultaneously belong to two surfaces.

When a polyhedron intersects with a surface of revolution, a spatial curved intersection line is formed.

If there is a complete intersection (penetration), then two closed curved lines are formed, and if an incomplete intersection, then one closed spatial intersection line.

To construct the line of mutual intersection of the polyhedron with the surface of revolution, the method of auxiliary cutting planes is used. The construction plane intersects both surfaces along curved lines and along broken lines. The mutual intersection of these lines gives us the points belonging to the desired intersection line.

Let it be required to construct the projection of the line of intersection of the surfaces of the cylinder and the triangular prism. As seen from Fig. 6.2, all three faces of the prism participate in the intersection. Two of them are directed at a certain angle to the axis of rotation of the cylinder, therefore, they intersect the surface of the cylinder in ellipses, one face is perpendicular to the axis of the cylinder, i.e., intersects it in a circle.

Solution plan:

1) find the points of intersection of the edges with the surface of the cylinder;

2) find the lines of intersection of the faces with the surface of the cylinder. As seen from Fig. 6.2, the lateral surface of the cylinder is horizontal

tally-projecting, that is, perpendicular to the horizontal plane of the projections. The lateral surface of the prism is profile-projection, that is, each of its facets is perpendicular to the profile plane of the projections. Consequently, the horizontal projection of the line of intersection of the bodies coincides with the horizontal projection of the cylinder, and the profile projection - with the profile projection of the prism. Thus, in the drawing, you only need to build a frontal projection of the intersection line.

We begin construction by drawing characteristic points, that is, points that can be found without additional construction. These are points 1, 2 and 3. They are located at the intersection of the outline generatrices of the frontal projections of the cylinder with the frontal projection of the corresponding edge of the prism using communication lines.

Thus, the intersection points of the prism edges with the cylinder surface are plotted.

In order to find intermediate points (there are four such points in total, but let us designate one of them as A) of the intersection lines of the cylinder with the prism faces, we intersect both surfaces with a projection plane or a level plane. Take, for example, the horizontal plane α. The α plane intersects the prism faces along two straight lines, and the cylinder intersects in a circle. These lines intersect at point A "(one point is signed, and the rest are not), which belongs to both the surface of the cylinder (lies on the circle that belongs to the cylinder) and the surface of the prism (lies on straight lines that belong to the faces of the prism).

The straight lines, along which the faces of the prism intersect with the plane α, were found first on the profile projection of the polyhedron (where they were projected to point A "" "and a symmetric point), and then, using communication lines, they were constructed on the horizontal projection of the prism. Point A and symmetric points were obtained at the intersection of the horizontal projection of the intersection lines (plane α with the prism) with the circle and using the communication lines are found on the frontal projection.

16.1. Drawings of unfolded surfaces of prisms and cylinders.

For the manufacture of machine tool fences, ventilation pipes and some other products, their sweeps are cut out of sheet material.

A scan of the surfaces of any straight prism is a flat figure made up of side faces - rectangles and two bases - polygons.

For example, in the unfolding of the surfaces of a hexagonal prism (Fig. 139, b), all the faces are equal rectangles with a width a and a height h, and the bases are regular hexagons with a side equal to a.

Figure: 139. Construction of the drawing of the unfolding of the surfaces of the prism: a - two views; b - unfolding surfaces

Thus, you can build a drawing of the unfolded surfaces of any prism.

The unfolding of the cylinder surfaces consists of a rectangle and two circles (Fig. 140, b). One side of the rectangle is equal to the height of the cylinder, the other is the circumference of the base. In the drawing of the sweep, two circles are attached to the rectangle, the diameter of which is equal to the diameter of the bases of the cylinder.

Figure: 140. Construction of the drawing of the unfolded surfaces of the cylinder: a - two views; b - unfolding surfaces

16.2. Drawings of unfolded surfaces of a cone and pyramid.

The sweep of the surfaces of the cone is a flat figure, consisting of a sector - a sweep of the lateral surface and a circle - the base of the cone (Fig. 141, 6).

Figure: 141. Construction of a drawing of a sweep of the surfaces of a cone: a - two views; b - unfolding surfaces

The constructions are done like this:

1. Draw an axial line and from the point s "on it describe a radius equal to the length s" a "of the generatrix of the cone, an arc of a circle. The circumference of the base of the cone is laid on it.

   Point s "is connected to the end points of the arc.

2. A circle is attached to the resulting figure - a sector. The diameter of this circle is equal to the diameter of the base of the cone.

The length of a circle when constructing a sector can be determined by the formula C \u003d 3.14xD.

The angle a is calculated by the formula a \u003d 360 ° xD / 2L, where D is the diameter of the circumference of the base, L is the length of the generatrix of the cone, it can be calculated by the Pythagorean theorem.

Figure: 142. Construction of the drawing of the unfolded surfaces of the pyramid: a - two views; b - unfolded surfaces

A drawing of a sweep of the surfaces of the pyramid is built as follows (Fig. 142, b):
From an arbitrary point O describe an arc of radius L equal to the length of the side edge of the pyramid. On this arc, four segments are laid, equal to the side of the base. The extreme points are connected by straight lines with the point O. Then a square is attached equal to the base of the pyramid.

Pay attention to how the drawings of the sweeps are drawn up. A special sign is placed above the image. From the fold lines, which are dash-dotted with two points, draw leader lines and write on the shelf "Fold Lines".

1. How to build a drawing of the unfolded surfaces of a cylinder?
2. What inscriptions are applied on the drawings of unfolded surfaces of objects?