In geometry, a truncated cone is a body that is formed by rotating a rectangular trapezoid around that side of it that is perpendicular to the base. How to calculate volume of a truncated cone, everyone knows from a school geometry course, and in practice this knowledge is often used by designers of various machines and mechanisms, developers of some consumer goods, as well as architects.
Calculation of the volume of a truncated cone
Formula for calculating the volume of a truncated cone
The volume of a truncated cone is calculated by the formula:
| V | πh (R 2 + R × r + r 2) |
h- cone height
r- radius of the upper base
R- radius of the lower base
V- volume of a truncated cone
π - 3,14
With such geometric bodies, How truncated cones, V Everyday life everyone collides quite often, if not constantly. They are shaped in a wide variety of containers that are widely used in everyday life: buckets, glasses, some cups. It goes without saying that the designers who developed them probably used the formula by which it is calculated volume of a truncated cone, since this quantity has in this case very great importance, because it is this that determines such an important characteristic as the capacity of the product.
Engineering structures that represent truncated cones, can often be seen at large industrial enterprises, as well as thermal and nuclear power plants. This is exactly the shape of cooling towers - devices designed to cool large volumes of water by forcing a counter flow atmospheric air. Most often, these designs are used in cases where it is required to short time significantly reduce the temperature large quantity liquids. The developers of these structures must determine volume of a truncated cone the formula for calculating which is quite simple and known to all those who once studied well in high school.
Parts having this geometric shape, are quite often found in the design of various technical devices. For example, gear drives used in systems where it is necessary to change the direction of kinetic transmission are most often implemented using bevel gears. These parts are an integral part of a wide variety of gearboxes, as well as automatic and manual gearboxes used in modern cars.
Some cutting tools widely used in production, such as milling cutters, have a truncated cone shape. With their help, you can process inclined surfaces at a certain angle. To sharpen the cutters of metalworking and woodworking equipment, abrasive wheels, which are also truncated cones, are often used. Besides, volume of a truncated cone It is necessary for designers of turning and milling machines to determine which involve fastening cutting tools equipped with conical shanks (drills, reamers, etc.).
The development of the surface of a cone is a flat figure obtained by combining the side surface and base of the cone with a certain plane.
Options for constructing a sweep:
Development of a right circular cone
The development of the lateral surface of a right circular cone is a circular sector, the radius of which is equal to length generatrix of the conical surface l, and the central angle φ is determined by the formula φ=360*R/l, where R is the radius of the circle of the base of the cone.
In a number of tasks descriptive geometry The preferred solution is to approximate (replace) the cone with a pyramid inscribed in it and construct an approximate development, on which it is convenient to draw lines lying on the conical surface.
Construction algorithm
- We fit a polygonal pyramid into a conical surface. The more lateral faces an inscribed pyramid has, the more accurate the correspondence between the actual and approximate development.
- We construct the development of the lateral surface of the pyramid using the triangle method. We connect the points belonging to the base of the cone with a smooth curve.
Example
In the figure below in a straight line circular cone a regular hexagonal pyramid SABCDEF is inscribed, and the approximate development of its lateral surface consists of six isosceles triangles- sides of the pyramid.
Consider the 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 generatrix 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 in the drawing, lay off the segment S 0 A 0 =l, after which from points S 0 and A 0 we draw circles with radius S 0 B 0 =l and A 0 B 0 = A'B' respectively. We connect the intersection point of circles B 0 with points A 0 and S 0.
We construct 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 pyramid SABCDEF similarly to the triangle S 0 A 0 B 0 .
Points A, B, C, D, E and F, lying at the base of the cone, are connected by a smooth curve - an arc of a circle, the radius of which is equal to l.
Inclined cone development
Let's consider the procedure for constructing the development of the lateral surface inclined cone by approximation method.
Algorithm
- We inscribe the hexagon 123456 into the circle of the base of the cone. We connect points 1, 2, 3, 4, 5 and 6 with the vertex S. The pyramid S123456, constructed in this way, with a certain degree of approximation is a replacement for the conical surface and is used as such in further constructions.
- We determine the natural values of 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 projection plane and passing through the vertex S.
Thus, as a result of the rotation of edge S5, its new horizontal projection S’5’ 1 takes a position in which it is parallel to the frontal plane π 2. Accordingly, S’’5’’ 1 is the actual size of S5. - We construct a scan of the lateral 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 . The construction of each triangle is carried out on three sides. For example, △S 0 1 0 6 0 has length S 0 1 0 =S’’1’’ 0 , S 0 6 0 =S’’6’’ 1 , 1 0 6 0 =1’6’.
The degree to which the approximate development corresponds to the actual one depends on the number of faces of the inscribed pyramid. The number of faces is chosen based on the ease of reading the drawing, the requirements for its accuracy, the presence of characteristic points and lines that need to be transferred to the development.
Transferring a line from the surface of a cone to a development
Line n lying on the surface of the cone is formed as a result of its intersection with a certain plane (figure below). Let's consider the algorithm for constructing line n on a scan.
Algorithm
- We find the projections of points A, B and C at which line n intersects the edges of the pyramid S123456 inscribed in the cone.
- We determine the natural size of the segments SA, SB, SC by rotating around the projecting straight line. In the example under consideration, SA=S’’A’’, SB=S’’B’’ 1 , SC=S’’C’’ 1 .
- We find the position of points A 0 , B 0 , C 0 on the corresponding edges of the pyramid, plotting on the scan the segments S 0 A 0 =S''A'', S 0 B 0 =S''B'' 1, S 0 C 0 =S''C'' 1 .
- We connect points A 0 , B 0 , C 0 with a smooth line.
Development of a truncated cone
The method described below for constructing the development of a right circular truncated cone is based on the principle of similarity.
Definition of a truncated cone
A truncated cone can be obtained from a regular cone by intersecting 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 an ordinary cone) will be called a truncated cone.
He has two bases, which for a circular cone are circles, and one of them is larger than the other. Also, a truncated cone has height- a segment connecting two bases and perpendicular to each of them.
Online calculator
A 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 right circular cone. The volume of a given figure can be calculated in several ways.
Formula for the volume of a truncated cone using the radii of the bases and the distance between them
If we are given a circular truncated cone, then we can find its volume using the formula:
Volume of a truncated coneV = 1 3 ⋅ π ⋅ h ⋅ (r 1 2 + r 1 ⋅ r 2 + r 2 2) V=\frac(1)(3)\cdot\pi\cdot h\cdot(r_1^2+r_1\ cdot r_2+r_2^2)V=3 1 ⋅ π ⋅ h⋅(r 1 2 + r 1 ⋅ r 2 + r 2 2 )
R 1, r 2 r_1, r_2 r 1
,
r 2
- radii of the bases of the cone;
h h h- the distance between these bases (the height of the truncated cone).
Let's look at an example.
Problem 1Find the volume of a truncated cone if it is known that the area of the small base is equal to 64 π cm 2 64\pi\text( cm)^26 4 π cm2 , big - 169 π cm 2 169\pi\text( cm)^21 6 9 π cm2 , and its height is equal to 14 cm 14\text( cm) 1 4 cm.
Solution
S 1 = 64 π S_1=64\pi S 1
=
6 4 π
S 2 = 169 π S_2=169\pi S 2
=
1 6 9 π
h = 14 h=14 h =1
4
Let's find the radius of the small base:
S 1 = π ⋅ r 1 2 S_1=\pi\cdot r_1^2S 1 = π ⋅ r 1 2
64 π = π ⋅ r 1 2 64\pi=\pi\cdot r_1^26 4 π =π ⋅ r 1 2
64 = r 1 2 64 = r_1^2 6 4 = r 1 2
R 1 = 8 r_1=8 r 1 = 8
Likewise, for a large base:
S 2 = π ⋅ r 2 2 S_2=\pi\cdot r_2^2S 2 = π ⋅ r 2 2
169 π = π ⋅ r 2 2 169\pi=\pi\cdot r_2^21 6 9 π =π ⋅ r 2 2
169 = r 2 2 169 = r_2^2 1 6 9 = r 2 2
R 2 = 13 r_2=13 r 2 = 1 3
Let's calculate the volume of the cone:
V = 1 3 ⋅ π ⋅ h ⋅ (r 1 2 + r 1 ⋅ r 2 + r 2 2) = 1 3 ⋅ π ⋅ 14 ⋅ (8 2 + 8 ⋅ 13 + 1 3 2) ≈ 4938 cm 3 V= \frac(1)(3)\cdot\pi\cdot h\cdot (r_1^2+r_1\cdot r_2+r_2^2)=\frac(1)(3)\cdot\pi\cdot14\cdot(8 ^2+8\cdot 13+13^2)\approx4938\text( cm)^3V=3 1 ⋅ π ⋅ h⋅(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 cm3
Answer
4938 cm3. 4938\text( cm)^3.4 9 3 8 cm3 .
Formula for the volume of a truncated cone using the areas of the bases and their distance to the vertex
Let us 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 a truncated cone can be found as the difference in the volumes of two cones with corresponding bases and their distance (height) to the top of the cone.
Volume of a truncated coneV = 1 3 ⋅ S ⋅ H − 1 3 ⋅ s ⋅ h = 1 3 ⋅ (S ⋅ H − s ⋅ h) V=\frac(1)(3)\cdot S\cdot H-\frac(1) (3)\cdot s\cdot h=\frac(1)(3)\cdot (S\cdot H-s\cdot h)V=3 1 ⋅ S⋅H −3 1 ⋅ s⋅h =3 1 ⋅ (S⋅H −s⋅h)
S S S- area of the base of the large cone;
H H H- the height of this (large) cone;
s s s- area of the base of the small cone;
h h h- the height of this (small) cone;
Determine the volume of a truncated cone if the height of the full cone is H H H equal to 10 cm 10\text( cm)
Solution
R=5 R=5
Find the area of both bases of the cone:
S = π ⋅ R 2 = π ⋅ 5 2 ≈ 78.5 S=\pi\cdot R^2=\pi\cdot 5^2\approx78.5
s = π ⋅ r 2 = π ⋅ 4 2 ≈ 50.24 s=\pi\cdot r^2=\pi\cdot 4^2\approx50.24
Find the height of the small cone h h
H − h = 8 H-h=8
h = H − 8 h=H-8
h = 10 − 8 h=10-8
h = 2 h=2
The volume is equal to the formula:
V = 1 3 ⋅ (S ⋅ H − s ⋅ h) ≈ 1 3 ⋅ (78.5 ⋅ 10 − 50.24 ⋅ 2) ≈ 228 cm 3 V=\frac(1)(3)\cdot (S\cdot H-s\cdot h)\approx\frac(1)(3)\cdot (78.5\cdot 10-50.24\cdot 2)\approx228\text( cm)^3
Answer
228 cm3. 228\text( cm)^3.
Sometimes a task arises - to make a protective umbrella for an exhaust or chimney, an exhaust deflector for ventilation, etc. But before you start manufacturing, you need to make a pattern (or development) 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 can calculate it faster using a calculator (on a computer) than searching, downloading and dealing with these programs.
Let's start with a simple option - the development of a simple cone. The easiest way to explain the principle of pattern calculation 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 absolutely clear that the blank will be a circle with a cut out segment. Two parameters are known - diameter and height. Using the Pythagorean theorem, we calculate the diameter of the workpiece circle (do not confuse it with the radius ready cone). Half the diameter (radius) and height form right triangle. That's why:
So now we know the radius of the workpiece and can cut a circle.
Let's calculate the angle of the sector that needs to be cut out of the circle. We reason as follows: The diameter of the workpiece is equal to 2R, which means that the circumference is equal to Pi * 2 * R - i.e. 6.28*R. Let's denote it L. The circle is complete, i.e. 360 degrees. And the circumference of the finished cone is equal to Pi*D. Let's denote it 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 of these lengths. Let's apply the ratio rule. If 360 degrees gives us the full circumference of the workpiece, then the angle we are looking for should give us the circumference of the finished cone.
From the ratio formula we obtain the size of the angle X. And the cut sector is found by subtracting 360 - X.
From a round blank with radius R, you need to cut a sector with an angle (360-X). Don't forget to leave a small strip of material for overlap (if the cone attachment will overlap). After connecting the sides of the cut sector, we obtain a cone of a given size.
For example: We need a cone for an exhaust pipe hood with a height (H) of 100 mm and a diameter (D) of 250 mm. Using the Pythagorean formula, we obtain the radius of the workpiece - 160 mm. And the circumference of the workpiece is correspondingly 160 x 6.28 = 1005 mm. At the same time, the circumference of the cone we need is 250 x 3.14 = 785 mm.
Then we find that the angle ratio will be: 785 / 1005 x 360 = 281 degrees. Accordingly, you need to cut out a sector of 360 – 281 = 79 degrees.
Calculation of the pattern blank 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 draft 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 cross section of the full cone). But we will resort to the same simple mathematical constructions based on the Pythagorean theorem and similarity.
In fact, 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. We find the total height (P) from this ratio.
(D – Dm)/ 2H = D/2P
Hence P = D x H / (D-Dm).
Now knowing the total height of the cone, we can reduce the solution to the previous problem. Calculate the development of the workpiece as if for a full cone, and then “subtract” from it the development of its upper, unnecessary part. And we can directly calculate the radii of the workpiece.
Using the Pythagorean theorem, we obtain a larger radius of the workpiece - Rz. This Square root from the sum of the squares of the heights P and D/2.
The smaller radius Rm is the square root of the sum of the 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 full angle in the workpiece is 360 degrees.
Those. X / 360 = 3.14 x D / 6.28 x Rz
Hence X = 180 x D / Rz (This is the angle that must be left to get the circumference of the base). And you need to cut accordingly 360 - X.
For example: We need to make a truncated cone with a height of 250 mm, a base diameter of 300 mm, and a top hole diameter of 200 mm.
Find the height of the full cone P: 300 x 250 / (300 – 200) = 600 mm
Using the Pythagorean point, we find the outer radius of the workpiece Rz: Square root of (300/2)^2 + 6002 = 618.5 mm
Using the same theorem, we find the smaller radius Rm: Square root of (600 – 250)^2 + (200/2)^2 = 364 mm.
We determine the sector angle of our workpiece: 180 x 300 / 618.5 = 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 angle of the arc can have approximately 90-100 degrees of opening. We draw radii with an opening angle of 87.3 degrees. Our preparation is ready. Don't forget to allow an allowance for joining the edges if they are overlapped.
