How to make a flat pattern - a pattern for a cone or truncated cone of specified dimensions. Simple sweep calculation

In geometry, a truncated cone is a body that is formed by the rotation of a rectangular trapezoid about its lateral side, which is perpendicular to the base. How do they count truncated cone volume, everyone knows from the school geometry course, but in practice this knowledge is often used by designers of various machines and mechanisms, developers of some consumer goods, as well as architects.

Calculating the volume of a truncated cone

The formula for calculating the volume of a truncated cone

The volume of the truncated cone is calculated by the formula:

πh (R 2 + R × r + r 2)

h - cone height

r - radius of the upper base

R - radius of the bottom base

V - volume of the truncated cone

π - 3,14

With geometric bodies such as truncated cones, in everyday life, everyone collides quite often, if not constantly. A wide variety of containers, widely used in everyday life, have their shape: buckets, glasses, some cups. It goes without saying that the designers who developed them probably used the formula by which the truncated cone volume, since this value is very important in this case, because it is it that determines such an important characteristic as the capacity of the product.

Engineering structures, which are truncated cones, can often be seen at large industrial enterprises, as well as thermal and nuclear power plants. This is the shape of cooling towers - devices designed to cool large volumes of water by injecting a counter flow of atmospheric air. Most often, these designs are used in cases where it is required to significantly reduce the temperature of a large amount of liquid in a short time. The developers of these structures are required to determine truncated cone volume the formula for calculating which is quite simple and known to all those who at one time studied well in high school.

Parts with this geometric shape are often found in the design of various technical devices. For example, gears used in systems where it is required to change the direction of a kinetic gear are most often implemented using bevel gears. These parts are integral to a wide variety of gearboxes and automatic and manual transmissions used in modern vehicles.

The shape of a truncated cone has some cutting tools widely used in production, for example, milling cutters. With their help, it is possible to process inclined surfaces at a certain angle. For sharpening cutters of metalworking and woodworking equipment, abrasive wheels are often used, which are also truncated cones. Moreover, truncated cone volume it is required to determine the designers of lathes and milling machines, which involve fastening a cutting tool equipped with tapered shanks (drills, reamers, etc.).

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

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.
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 constructed 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

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.
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.
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

Find the projections of points A, B and C at which line n intersects the edges of the pyramid S123456 inscribed in the cone.
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.
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.
Connect the points A 0, B 0, C 0 with a smooth line.

Truncated cone flat pattern

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

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. Using 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. Be sure to leave a small strip of overlap material (if the cone mount 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 a little 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 radius of the base 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 a seam allowance if the edges overlap.

Defining a frustum

A truncated cone can be obtained from an ordinary cone by crossing such a cone with a plane parallel to the base. Then the figure that is located between two planes (this plane and the base of the usual cone) will be called a truncated cone.

He has two bases, which are circles for a circular cone, and one of them is larger than the other. Also the truncated cone has the height - a segment connecting two bases and perpendicular to each of them.

Online calculator

The truncated cone can be direct, then the center of one base is projected into the center of the second. If the cone inclined, then such projection does not take place.

Consider a straight circular cone. The volume of a given figure can be calculated in several ways.

The formula for the volume of a truncated cone in terms of the radii of the bases and the distance between them

If we are given a circular truncated cone, then we can find its volume by the formula:

Truncated cone volume

V \u003d 1 3 ⋅ π ⋅ h ⋅ (r 1 2 + r 1 ⋅ r 2 + r 2 2) V \u003d \\ frac (1) (3) \\ cdot \\ pi \\ cdot h \\ cdot (r_1 ^ 2 + r_1 \\ V \u003dh ⋅3 1 ⋅ π ⋅ r 1, r 2 r_1, r_2(r 1 2 + r 1 ⋅ r 2 + r 2 2 )

- radii of the bases of the cone; r 1 , r 2 h h
h - the distance between these bases (the height of the truncated cone). Let's look at an example.

Problem 1

Find the volume of the truncated cone if it is known that the area of \u200b\u200bthe small base is

64 π cm 2 64 \\ pi \\ text (cm) ^ 2 6 4 πcm , large -2 169 π cm 2 169 \\ pi \\ text (cm) ^ 2 1 6 9 π, and its height is , large -2 14 cm 14 \\ text (cm) Decision 1 4 , large -.

S 1 \u003d 64 π S_1 \u003d 64 \\ pi

S 6 4 π 1 = S 2 \u003d 169 π S_2 \u003d 169 \\ pi
1 6 9 π 6 4 π 2 = h \u003d 14 h \u003d 14
h \u003d find the radius of the small base:1 4

S 1 \u003d π ⋅ r 1 2 S_1 \u003d \\ pi \\ cdot r_1 ^ 2

64 π \u003d π ⋅ r 1 2 64 \\ pi \u003d \\ pi \\ cdot r_1 ^ 26 4 π 1 = π ⋅ r 1 2

6 4 π \u003d64 \u003d r 1 2 64 \u003d r_1 ^ 2π ⋅ r 1 2

R 1 \u003d 8 r_1 \u003d 8 6 4 = r 1 2

Likewise, for a large base: r 1 = 8

S 2 \u003d π ⋅ r 2 2 S_2 \u003d \\ pi \\ cdot r_2 ^ 2

169 π \u003d π ⋅ r 2 2 169 \\ pi \u003d \\ pi \\ cdot r_2 ^ 26 4 π 2 = π ⋅ r 2 2

1 6 9 π \u003d169 \u003d r 2 2 169 \u003d r_2 ^ 2π ⋅ r 2 2

{!LANG-cab958ebedc3558a32da2fc61856a942!} 1 6 9 = r 2 2

R 2 \u003d 13 r_2 \u003d 13 r 2 = 1 3

Let's calculate the volume of the cone:

V \u003d 1 3 ⋅ π ⋅ h ⋅ (r 1 2 + r 1 ⋅ r 2 + r 2 2) \u003d 1 3 ⋅ π ⋅ 14 ⋅ (8 2 + 8 ⋅ 13 + 1 3 2) ≈ 4938 cm 3 V \u003d \\ frac (1) (3) \\ cdot \\ pi \\ cdot h \\ cdot (r_1 ^ 2 + r_1 \\ cdot r_2 + r_2 ^ 2) \u003d \\ frac (1) (3) \\ cdot \\ pi \\ cdot14 \\ cdot (8 ^ 2 + 8 \\ cdot 13 + 13 ^ 2) \\ approx4938 \\ text (cm) ^ 3h ⋅3 1 ⋅ π ⋅ r 1, r 2 r_1, r_2(r 1 2 + r 1 ⋅ r 2 + r 2 2 ) = 3 1 ⋅ π ⋅ 1 4 ⋅ (8 2 + 8 ⋅ 1 3 + 1 3 2 ) ≈ 4 9 3 8 , large -3

Answer

4938 cm 3. 4938 \\ text (cm) ^ 3.4 9 3 8 , large -3 .

The formula for the volume of a truncated cone in terms of the areas of the bases and their distance to the top

Let's say we have a truncated cone. Let's mentally add the missing piece to it, thereby making it a “regular cone” with a top. Then the volume of the truncated cone can be found as the difference between the volumes of two cones with the corresponding bases and their distance (height) to the top of the cone.

Truncated cone volume

V \u003d 1 3 ⋅ S ⋅ H - 1 3 ⋅ s ⋅ h \u003d 1 3 ⋅ (S ⋅ H - s ⋅ h) V \u003d \\ frac (1) (3) \\ cdot S \\ cdot H- \\ frac (1) (3) \\ cdot s \\ cdot h \u003d \\ frac (1) (3) \\ cdot (S \\ cdot Hs \\ cdot h)h ⋅3 1 ⋅ S ⋅H -3 1 ⋅ s ⋅find the radius of the small base:3 1 ⋅ (S ⋅H -s ⋅h)

S S S - the area of \u200b\u200bthe base of the large cone;
H H H - the height of this (large) cone;
s s s - the area of \u200b\u200bthe base of the small cone;
h - the distance between these bases (the height of the truncated cone). - the height of this (small) cone;

Problem 2

Determine the volume of the truncated cone if the height of the full cone H H H equals 10 cm 10 \\ text (cm) $10 cm, the radius of the lower base R - 5 cm, top r - 4 cm, and the height of the truncated cone is 8 cm .$

S 1 \u003d 64 π S_1 \u003d 64 \\ pi

$\u003d 5 R \u003d 5$

Find the areas of both bases of the cone:

$\u003d π ⋅ R 2 \u003d π ⋅ 5 2 ≈ 78.5 S \u003d \\ pi \\ cdot R ^ 2 \u003d \\ pi \\ cdot 5 ^ 2 \\ approx78.5$

$\u003d π ⋅ r 2 \u003d π ⋅ 4 2 ≈ 50.24 s \u003d \\ pi \\ cdot r ^ 2 \u003d \\ pi \\ cdot 4 ^ 2 \\ approx50.24$

Find the height of the small cone $h :$

$\u003d 8 H-h \u003d 8$

$\u003d H - 8 h \u003d H-8$

$\u003d 10 - 8 h \u003d 10-8$

$\u003d 2 h \u003d 2$

The volume is equal to the formula:

$\u003d 1 3 ⋅ (S ⋅ H - s ⋅ h) ≈ 1 3 ⋅ (78.5 ⋅ 10 - 50.24 ⋅ 2) ≈ 228 cm 3 V \u003d \\ frac (1) (3) \\ cdot (S \\ cdot Hs \\ cdot h) \\ approx \\ frac (1) (3) \\ cdot (78.5 \\ cdot 10-50.24 \\ cdot 2) \\ approx228 \\ text (cm) ^ 3$