Elliptic paraboloid canonical equation. Ellipsoid

The proven property of the tangent to a parabola is very important, since it follows from it that the rays emanating from the focus of a concave parabolic mirror, i.e., a mirror whose surface is obtained from the rotation of the parabola around its axis, are reflected by a parallel beam, namely parallel mirror axes (Fig.).

This property of parabolic mirrors is used in the construction of searchlights, in the headlights of any car, as well as in reflecting telescopes. Moreover, in the latter case, inversely, the rays coming from the heavenly body; almost parallel, are concentrated near the focus of the telescope mirror, and since the rays coming from different points of the luminary are much non-parallel, they are concentrated near the focus in different points, so that near the focus an image of the luminary is obtained, the greater the greater the focal length of the parabola. This image is already viewed through a microscope (telescope eyepiece). Strictly speaking, only rays strictly parallel to the axis of the mirror are collected at one point (the focus), while parallel rays going at an angle to the axis of the mirror are collected only almost to one point, and the further this point is from the focus, the more image more blurry. This circumstance limits the “field of view of the telescope.”

Let its inner surface be a mirror surface; this parabolic mirror is illuminated by a beam of light rays parallel to the axis of the op-amp. All rays parallel to the op-amp axis, after reflection, will intersect at one point on the op-amp axis (focus F). The design of parabolic telescopes is based on this property. Rays from distant stars come to us in the form of a parallel beam. By making a parabolic telescope and placing a photographic plate at its focus, we get the opportunity to amplify the light signal coming from the star.

The same principle underlies the creation of a parabolic antenna, which allows amplification of radio signals. If you place a light source at the focus of a parabolic mirror, then after reflection from the surface of the mirror, the rays coming from this source will not be scattered, but will be collected into a narrow beam parallel to the axis of the mirror. This fact is used in the manufacture of spotlights and lanterns, various projectors, the mirrors of which are made in the shape of paraboloids.

The above-mentioned optical property of a parabolic mirror is used to create mirror telescopes, various solar heating installations, and also searchlights. By placing a powerful point source of light at the focus of a parabolic mirror, we obtain a dense stream of reflected rays parallel to the axis of the mirror.

When a parabola rotates around its axis, a figure is obtained that is called a paraboloid. If the inner surface of the paraboloid is made mirror and a beam of rays is directed at it, parallel to the axis symmetry of a parabola, then the reflected rays will converge at one point, which is called the focus. At the same time, if the light source is placed at the focus, then the rays reflected from the mirror surface of the paraboloid will be parallel and not scattered.

The first property makes it possible to obtain a high temperature at the focus of the paraboloid. According to legend, this property was used by the ancient Greek scientist Archimedes (287-212 BC). While defending Syracuse in the war against the Romans, he built a system of parabolic mirrors that allowed the reflected rays of the sun to be focused on the Roman ships. As a result, the temperature at the foci of the parabolic mirrors turned out to be so high that a fire broke out on the ships and they burned down.

The second property is used, for example, in the manufacture of spotlights and car headlights.

Hyperbola

4. The definition of a hyperbola gives us a simple way to construct it with a continuous movement: take two threads, the difference in lengths of which is 2a, and attach one end of these threads to points F" and F. If you hold the other two ends together with your hand and move along the threads with the point of a pencil, taking care that the threads are pressed to the paper, stretched and touching, starting from the drawing tip to the point where the ends meet, the tip will draw part of one of the branches of the hyperbola (the larger the longer the threads are taken) (Fig.).

Reversing the roles of points F" and F, we obtain part of another branch.

For example, On the topic “2nd order curves” you can consider the following problem:

Task. Two railway stations A and B are located at a distance of s km from one another. To any point M, cargo can be delivered from station A either by direct road transport (first route) or by railway to station B, and from there by car (second route). The railway tariff (price of transportation of 1 ton per 1 km) is m rubles, the road transport tariff is n rubles, n > m, the loading and unloading tariff is k rubles. Determine the area of ​​influence of railway station B, i.e., the area to which it is cheaper to deliver cargo from station A by mixed means - by rail, and then by road, i.e. determine the geometric location of points for which the second path is more profitable than the first.

Solution. Let us denote AM = r, BM = r, then the cost of delivery (transportation and loading-unloading) along the route AM is equal to nr + k, and the cost of delivery along the path ABM is equal to ms + 2k + ng. Then the points M, for which both values ​​are equal, satisfy the equation nr + k = ms+2k+nг, or

ms + k = nr - ng

r - r = = const > O,

therefore, the line delimiting the region is one of the branches of the hyperbola | r - r | = const. For all points of the plane lying on the same side as point A of this hyperbola, the first path is more advantageous, and for points lying on the other side - the second, therefore the branch of the hyperbola outlines the area of ​​influence of station B.

Variant of this problem.

Two railway stations A and B are located at a distance of l km from each other. To point M, cargo can be delivered from station A either by direct road transport, or by rail to station B, and from there by car (Fig. 49). In this case, the railway tariff (price of transporting 1 ton per 1 km) is m rubles, loading and unloading costs k rubles (per 1 ton) and the road transport tariff is n rubles (n > m). Let's determine the so-called zone of influence of railway station B, i.e., the zone to which it is cheaper to deliver cargo from A using a mixed route: by rail and then by road.

Solution. The cost of delivering 1 ton of cargo along the AM route is r n, where r = AM, and along the ABM route it will be equal to 1m + k + r n. We need to solve the double inequality r n 1m+ k+ r n and determine how the points on the plane (x, y) are distributed, to which it is cheaper to deliver the cargo either by the first or second route.

Let us find the equation of the line forming the boundary between these two zones, i.e. the locus of points for which both paths are “equally beneficial”:

r n = 1m+ k+ r n

From this condition we obtain r - r = = const.

Therefore, the dividing line is a hyperbola. For all external points of this hyperbola, the first path is more advantageous, and for internal points - the second. Therefore, the hyperbola will outline the zone of influence of station B. The second branch of the hyperbola will outline the zone of influence of station A (the cargo is delivered from station B). Let's find the parameters of our hyperbola. Its major axis is 2a = , and the distance between the foci (which are stations A and B) in this case is 2c = l.

Thus, the condition for the possibility of this problem, determined by the relation a< с, будет

This problem connects the abstract geometric concept of a hyperbola with a transport and economic problem.

The required locus of points is the set of points lying inside the right branch of the hyperbola containing point B.

6. I know " Agricultural machines» important performance characteristics of a tractor working on a slope, showing its stability are the longitudinal inclination angle and the lateral roll angle.

For simplicity we will consider wheeled tractor. The surface on which the tractor operates (at least a fairly small part of it) can be considered a plane (plane of movement). The longitudinal axis of the tractor is the projection of the straight line connecting the midpoints of the front and rear axles onto the plane of movement. The lateral roll angle is the angle formed with the horizontal plane of a straight line, perpendicular to the longitudinal axis and lying in the plane of movement.

When studying the topic “Lines and planes in space” in a mathematics course, we consider the following problems:

a) Find the angle of longitudinal inclination of a tractor moving along a slope if the angle of inclination of the slope and the angle of deviation of the tractor’s trajectory from the longitudinal direction are known.

b) The maximum lateral roll angle of the tractor is the maximum permissible angle of inclination of the slope across which the tractor can stand without tipping over. What tractor parameters are enough to know to determine the maximum lateral roll angle; how to find this one
corner?

7. The presence of rectilinear generatrices is used in construction equipment. The founder of the practical application of this fact is the famous Russian engineer Vladimir Grigorievich Shukhov (1853-1939). V. G. Shukhov carried out the design of masts, towers and supports made up of metal beams located along rectilinear generatrices single-sheet hyperboloid of revolution. The high strength of such structures, combined with lightness, low manufacturing cost and elegance, ensures their widespread use in modern construction.

8. LAWS OF MOTION OF A FREE RIGID BODY

For a free body, all types of motion are equally possible, but this does not mean that the motion of a free body is disordered and does not obey any laws; on the contrary, the translational motion of a rigid body, regardless of its external shape, is constrained by the law of the center of mass and is reduced to the movement of one point, and the rotational motion is by the so-called main axes of inertia or ellipsoid of inertia. Thus, a stick thrown into free space, or grain flying out of a sorter, etc., moves forward as one point (center of mass), and at the same time rotates around the center of mass. In general, during translational motion, any rigid body, regardless of its shape or complex machine can be replaced by one point (center of mass), and in case of rotation - by an ellipsoid of inertia , whose radius vectors are equal to --, where / is the moment of inertia of this body relative to the axes passing through the center of the ellipsoid.

If the moment of inertia of a body changes for some reason during rotation, then the speed of rotation will change accordingly. For example, during an overhead jump, acrobats compress into a ball, causing the moment of inertia of the body to decrease and the rotation speed to increase, which is what is needed for the success of the jump. In the same way, after slipping, people stretch their arms to the sides, which causes the moment of inertia to increase and the speed of rotation to decrease. In the same way, the moment of inertia of the harvest rake about the vertical axis is variable during its rotation about the horizontal axis.

There are two types of paraboloids: elliptic and hyperbolic.

Elliptical paraboloid is a surface that, in some system of Cartesian rectangular coordinates, is defined by the equation

An elliptical paraboloid has the shape of an infinite convex bowl. It has two mutually perpendicular planes of symmetry. The point with which the origin of coordinates is combined is called the vertex of the elliptic paraboloid; the numbers p and q are called its parameters.

A hyperbolic paraboloid is a surface defined by the equation

Hyperbolic paraboloid has the shape of a saddle. It has two mutually perpendicular planes of symmetry. The point with which the origin of coordinates is combined is called the vertex of a hyperbolic paraboloid; numbers R And q are called its parameters.

Exercise 8.4. Let us consider the construction of a hyperbolic paraboloid of the form

Let it be necessary to construct a part of a paraboloid lying in the ranges: xО[–3; 3], atО[–2; 2] with step D=0.5 for both variables.

Performance. First you need to solve the equation for the variable z. In the example

Let's enter the variable values X to column A. To do this, in the cell A1 enter a symbol X. To cell A2 the first value of the argument is entered - the left limit of the range (–3). To cell A3- the second value of the argument is the left limit of the range plus the construction step (–2,5). Then, selecting the block of cells A2:AZ, using autofill we get all the values ​​of the argument (we drag the lower right corner of the block to the cell A14).

Variable values at enter into the line 1 . To do this, in the cell IN 1 The first value of the variable is entered - the left limit of the range (–2). To cell C1- the second value of the variable - the left limit of the range plus the construction step (– 1,5). Then, selecting the block of cells B1:C1,by autofilling we get all the values ​​of the argument (we drag the lower right corner of the block to the cell J1).

Next, enter the variable values z. To do this, the table cursor must be placed in the cell AT 2 and enter the formula - = $A2^2/18 -B$1^2/8, then press the key Enter. In a cell AT 2 appears 0. Now you need to copy the function from the cell AT 2. To do this, use autofill (drawing to the right) to copy this formula first into the range B2:J2, after which (by pulling down) - into the range B2:J14.

As a result, in the range B2:J14 A table of hyperbolic paraboloid points will appear.

To plot a chart on the toolbar Standard you need to press a button Chart Wizard. In the dialog box that appears Chart Wizard (Step 1 of 4): Chart Type indicate the type of diagram - Surface, and view - Wire (transparent) surface(top right diagram in the right window). Then press the button Further in the dialog box.

In the dialog box that appears Chart Wizard (Step 2 of 4): Data Source charts you need to select the tab Range data and in the field Range use the mouse to indicate the data interval B2:J14.

Next, you need to indicate the rows or columns where the data rows are located. This will determine the orientation of the axes X And u. In the example, the switch Rows in Using the mouse pointer, set it to the position of the columns.

Select the Row tab and in the field X-axis labels indicate the range of signatures. To do this, activate this field by clicking the mouse pointer in it and enter the range of axis labels X -A2:A14.

Enter the values ​​of the axis labels u. To do this, in the work field Row select the first entry Row 1 and by activating the work field Name with the mouse pointer, enter the first value of the variable y: –2. Then into the field Row select the second entry Row 2 and into the work field Name enter the second value of the variable y: –1.5. Repeat this way until the last entry - Row 9.

After the required entries appear, click the button Further.

The third window requires you to enter a chart title and axis names. To do this, select the tab Headings by clicking on it with the mouse pointer. Then to the work field Chart title enter the name from the keyboard: Hyperbolic paraboloid. Then enter in the same way into the work fields X-axis (categories),Y axis (data series) And Z axis (values) corresponding names: x, y And z.

A hyperbolic paraboloid also belongs to second-order surfaces. This surface cannot be obtained using an algorithm that uses the rotation of a certain line relative to a fixed axis.

A special model is used to construct a hyperbolic paraboloid. This model includes two parabolas located in two mutually perpendicular planes.

Let parabola I be located in a plane and motionless. Parabola II makes a complex movement:

▫ its initial position coincides with the plane
, and the vertex of the parabola coincides with the origin of coordinates: =(0,0,0);

▫ then this parabola moves parallel transfer, and its top
makes a trajectory coinciding with parabola I;

▫ two different initial positions of parabola II are considered: one – the upward branches of the parabola, the second – the downward branches.

Let's write down the equations: for the first parabola I:
– invariably; for the second parabola II:
– initial position, equation of motion:
It's not hard to see that the point
has coordinates:
. Since it is necessary to display the law of motion of a point
: this point belongs to parabola I, then the following relations must always be satisfied: =
And
.

From the geometric features of the model it is easy to see that the movable parabola sweeps up some surface. In this case, the equation of the surface described by parabola II has the form:

or→
. (1)

The shape of the resulting surface depends on the distribution of the signs of the parameters
. There are two possible cases:

1). Signs of quantities p And q coincide: parabolas I and II are located on the same side of the plane OXY. Let's accept: p = a 2 And q = b 2 . Then we get the equation of the known surface:

→ elliptical paraboloid . (2)

2). Signs of quantities p And q are different: parabolas I and II are located along different sides from the plane OXY. Let p = a 2 And q = - b 2 . Now we get the surface equation:

→hyperbolic paraboloid . (3)

It is not difficult to imagine the geometric shape of the surface defined by equation (3) if we recall the kinematic model of the interaction of two parabolas involved in the movement.

In the figure, parabola I is conventionally shown in red. Only the neighborhood of the surface at the origin of coordinates is shown. Due to the fact that the shape of the surface expressively hints at a cavalry saddle, this neighborhood is often called - saddle .

In physics, when studying the stability of processes, types of equilibrium are introduced: stable - a hole, convex downward, unstable - a surface convex upward, and intermediate - a saddle. Equilibrium of the third type is also classified as a type of unstable equilibrium, and only on the red line (parabola I) is equilibrium possible.

§ 4. Cylindrical surfaces.

When considering surfaces of revolution, we identified the simplest cylindrical surface - a cylinder of revolution, that is, a circular cylinder.

In elementary geometry, a cylinder is defined by analogy with the general definition of a prism. It's quite complicated:

▫ let us have a flat polygon in space
– let’s denote it as , and the polygon coincides with it
– let’s denote it as
;

▫ applicable to polygon
movement parallel translation: points
move along trajectories parallel to a given direction ;

▫ if you stop polygon transfer
, then its plane
parallel to the plane ;

▫ the surface of a prism is called: a collection of polygons ,
– grounds prisms and parallelograms
,
,... – side surface prisms.

IN Let's use the elementary definition of a prism to construct a more general definition of a prism and its surface, namely, we will distinguish:

▫ an unbounded prism is a polyhedral body bounded by edges ,,... and the planes between these edges;

▫ a limited prism is a polyhedral body bounded by edges ,,... and parallelograms
,
,...; the lateral surface of this prism is a set of parallelograms
,
,...; prism bases – a set of polygons ,
.

Let us have an unlimited prism: ,,... Let's intersect this prism with an arbitrary plane . Let's intersect the same prism with another plane
. In cross-section we get a polygon
. In general, we assume that the plane
not parallel to the plane . This means that the prism was not built by parallel translation of the polygon .

The proposed construction of a prism includes not only straight and inclined prisms, but also any truncated ones.

In analytical geometry, we will understand cylindrical surfaces so generally that an unbounded cylinder includes an unbounded prism as a special case: one has only to assume that the polygon can be replaced by an arbitrary line, not necessarily closed - guide cylinder. Direction called generatrix cylinder.

From all that has been said, it follows: to define a cylindrical surface, it is necessary to specify a guide line and the direction of the generatrix.

Cylindrical surfaces are obtained on the basis of plane curves of the 2nd order, serving guides For forming .

At the initial stage of studying cylindrical surfaces, we will accept simplifying assumptions:

▫ let the guide of the cylindrical surface always be located in one of the coordinate planes;

▫ direction of generatrix coincides with one of the coordinate axes, that is, perpendicular to the plane in which the guide is defined.

The accepted restrictions do not lead to loss of generality, since it remains possible due to the choice of sections by planes And
build arbitrary geometric shapes: straight, inclined, truncated cylinders.

Elliptical cylinder .

Let us take an ellipse as the guide of the cylinder :
, located in the coordinate plane

: elliptical cylinder.

Hyperbolic cylinder .

:

, and the direction of the generatrix determines the axis
. In this case, the equation of the cylinder is the line itself : hyperbolic cylinder.

Parabolic cylinder .

Let us take a hyperbola as the guide of the cylinder :
, located in the coordinate plane
, and the direction of the generatrix determines the axis
. In this case, the equation of the cylinder is the line itself : parabolic cylinder.

Comment: considering general rules constructing equations of cylindrical surfaces, as well as the presented particular examples of elliptic, hyperbolic and parabolic cylinders, we note: constructing a cylinder for any other generatrix, for the accepted simplifying conditions, should not cause any difficulties!

Let us now consider more General terms constructing equations of cylindrical surfaces:

▫ the guide of the cylindrical surface is located in an arbitrary plane of space
;

▫ direction of generatrix in the adopted coordinate system is arbitrary.

We depict the accepted conditions in the figure.

▫ cylindrical surface guide located in an arbitrary plane space
;

▫ coordinate system
obtained from the coordinate system
parallel transfer;

▫ guide location in the plane the most preferable: for a 2nd order curve we will assume that the origin of coordinates coincides with center symmetry of the curve under consideration;

▫ direction of generatrix arbitrary (can be specified in any of the ways: vector, straight line, etc.).

In what follows we will assume that the coordinate systems
And
match up. This means that the 1st step of the general algorithm for constructing cylindrical surfaces, reflecting parallel translation:
→
, previously completed.

Let us recall how parallel transfer is taken into account in the general case by considering a simple example.

Example 6–13 : In the coordinate system
as:
=0. Write down the equation of this guide in the system
.

Solution:

1). Let us designate an arbitrary point
: in system
How
, and in the system
How
.

2). Let's write down the vector equality:
=
+
. In coordinate form this can be written as:
=
+
. Or in the form:
=
–
, or:
=.

3). Let us write the equation of the cylinder guide in the coordinate system
:

Answer: transformed guide equation: =0.

So, we will assume that the center of the curve representing the cylinder guide is always located at the origin of the system coordinates
in the plane .

Rice. IN . Basic drawing for building a cylinder.

Let's make one more assumption that simplifies the final steps of constructing a cylindrical surface. Since by using rotation of the coordinate system it is not difficult to align the direction of the axis
coordinate systems
with plane normal , and the directions of the axes
And
with guide symmetry axes , then we will assume that as the initial position of the guide we have a curve located in the plane
, and one of its symmetry axis coincides with the axis
, and the second with the axis
.

Comment: since the operations of parallel translation and rotation around a fixed axis are quite simple, the accepted assumptions do not limit the applicability of the developed algorithm for constructing a cylindrical surface in the most general case!

We have seen that when constructing a cylindrical surface in the case where the guide located in the plane
, and the generatrix is ​​parallel to the axis
, it is enough to determine only the guide .

Since a cylindrical surface can be uniquely determined by specifying any line obtained in the section of this surface by an arbitrary plane, we will accept the following general algorithm for solving the problem:

1 ▫ . Let the direction of the generatrix cylindrical surface given by vector . Let's design a guide , given by the equation:
=0, to a plane perpendicular to the direction of the generatrix , that is, onto the plane
. As a result, the cylindrical surface will be specified in the coordinate system
equation:
=0.

2 ▫
around the axis
at an angle
: meaning of angle
compatible with the system
, and the equation of the conical surface is transformed into the equation:
=0.

3 ▫ . Apply rotation of the coordinate system
around the axis
at an angle
: meaning of angle is quite clear from the figure. As a result of rotation, the coordinate system
compatible with the system
, and the equation of the conical surface is transformed into
=0. This is the equation of a cylindrical surface for which the guide was given and generator in the coordinate system
.

The example below illustrates the implementation of the written algorithm and the computational difficulties of such problems.

Example 6–14 : In the coordinate system
the cylinder guide equation is given as:
=9. Write an equation for a cylinder whose generators are parallel to the vector =(2,–3,4).

R
decision:

1). Let us project the cylinder guide onto a plane perpendicular to . It is known that such a transformation turns a given circle into an ellipse, the axes of which will be: large =9, and small =
.

This figure illustrates the design of a circle defined in a plane
to the coordinate plane
.

2). The result of designing a circle is an ellipse:
=1, or
. In our case it is:
, Where
==.

3
). So, the equation of a cylindrical surface in the coordinate system
received. Since according to the conditions of the problem we must have the equation of this cylinder in the coordinate system
, then it remains to apply a coordinate transformation that transforms the coordinate system
to the coordinate system
, at the same time the equation of the cylinder:
into an equation expressed in terms of variables
.

4). Let's take advantage basic drawing, and write down all the trigonometric values ​​necessary to solve the problem:

==,
==,
==.

5). Let us write down the formulas for transforming coordinates when moving from the system
to the system
:
(IN)

6). Let us write down the formulas for transforming coordinates when moving from the system
to the system
:
(WITH)

7). Substituting Variables
from system (B) to system (C), and also taking into account the values ​​of the trigonometric functions used, we write:

=
=
.

=
=
.

8). It remains to substitute the found values And into the cylinder guide equation :
in the coordinate system
. Having completed carefully all algebraic transformations, we obtain the equation of a conical surface in the coordinate system
: =0.

Answer: cone equation: =0.

Example 6–15 : In the coordinate system
the cylinder guide equation is given as:
=9, =1. Write an equation for a cylinder whose generators are parallel to the vector =(2,–3,4).

Solution:

1). It’s easy to see that this example differs from the previous one only in that the guide has been moved in parallel by 1 upward.

2). This means that in relations (B) one should accept: =-1. Taking into account the expressions of system (C), we will correct the entry for the variable :

=
.

3). The change is easily taken into account by adjusting the final equation for the cylinder from the previous example:

Answer: cone equation: =0.

Comment: it is easy to see that the main difficulty with multiple transformations of coordinate systems in problems with cylindrical surfaces is accuracy And endurance in algebra marathons: long live the education system adopted in our long-suffering country!

The height of a paraboloid can be determined by the formula

The volume of the paraboloid touching the bottom is equal to half the volume of a cylinder with base radius R and height H, the same volume occupies the space W’ under the paraboloid (Fig. 4.5a)

Fig.4.5. The ratio of volumes in a paraboloid touching the bottom.

Wп – volume of the paraboloid, W’ – volume under the paraboloid, Hп – height of the paraboloid

Fig.4.6. The ratio of volumes in a paraboloid touching the edges of the cylinder Hp is the height of the paraboloid., R is the radius of the vessel, Wl is the volume under the height of the liquid in the vessel before the start of rotation, z 0 is the position of the vertex of the paraboloid, H is the height of the liquid in the vessel before the start of rotation.

In Fig. 4.6a, the liquid level in the cylinder before the start of rotation is H. The volume of liquid Wl before and after rotation is maintained and is equal to the sum of the volume Wt of the cylinder with height z 0 plus the volume of liquid under the paraboloid, which is equal to the volume of the paraboloid Wp with height Hn

If the paraboloid touches the upper edge of the cylinder, the height of the liquid in the cylinder before the start of rotation H divides the height of the paraboloid Hn into two equal parts, the lowest point (vertex) of the paraboloid is located in relation to the base (Fig. 4.6c)

In addition, the height H divides the paraboloid into two parts (Fig. 4.6c), the volumes of which are equal to W 2 = W 1. From the equality of the volumes of the parabolic ring W 2 and the parabolic cup W 1, Fig. 4.6c

When the surface of the paraboloid intersects the bottom of the vessel (Fig. 4.7) W 1 =W 2 =0.5W ring

Fig. 4.7 Volumes and heights when the surface of a paraboloid intersects the bottom of the cylinder

Heights in Fig. 4.6

volumes in Fig. 4.6.

Location of the free surface in the vessel

Fig.4.8. Three cases of relative rest during rotation

1. If the vessel is open, Po = Ratm (Fig. 4.8a). During rotation, the top of the paraboloid falls below the initial level-H, and the edges rise above the initial level, the position of the top

2. If the vessel is completely filled, covered with a lid, has no free surface, is under excess pressure Po>Patm, before rotation the surface (PP) on which Po=Patm will be above the level of the lid at a height h 0i =M/ ρg, H 1 =H+ M/ρg.

3. If the vessel is completely filled, it is under vacuum Po<Ратм, до вращения поверхность П.П., на которой Ро=Ратм будет находиться под уровнем крышки на высоте h 0и =-V/ρg, Н 2 =Н-V/ρg ,

4.7. Rotation at high angular velocity (Fig. 4.9)

When a vessel containing liquid rotates at a high angular velocity, the force of gravity can be neglected compared to centrifugal forces. The law of pressure change in a liquid can be obtained from the formula

(4.22),

The surfaces of the level form cylinders with a common axis around which the vessel rotates. If the vessel is not completely filled before rotation begins, the pressure P 0 will act along the radius r = r 0 , instead of expression (4.22) we will have

in which we take g(z 0 - z) = 0,

Rice. 4.9 Location of surfaces of rotation in the absence of gravity.

Radius of the inner surface for known H and h

Around its axis, you can get an ordinary elliptical. It is a hollow isometric body whose sections are ellipses and parabolas. An elliptic paraboloid is given by:
x^2/a^2+y^2/b^2=2z
All principal sections of a paraboloid are parabolas. When cutting the XOZ and YOZ planes, only parabolas are obtained. If you draw a perpendicular section relative to the Xoy plane, you can get an ellipse. Moreover, the sections, which are parabolas, are specified by equations of the form:
x^2/a^2=2z; y^2/a^2=2z
Sections of the ellipse are given by other equations:
x^2 /a^2+y^2/b^2=2h
An elliptical paraboloid at a=b turns into a paraboloid of revolution. The construction of a paraboloid has a number of features that need to be taken into account. Start the operation by preparing the basis - a drawing of the graph of the function.

In order to start building a paraboloid, you must first build a parabola. Draw a parabola in the Oxz plane as shown in the figure. Give the future paraboloid a certain height. To do this, draw a straight line so that it touches the upper points of the parabola and is parallel to the Ox axis. Then draw a parabola in the Yoz plane and draw a straight line. You will get two paraboloid planes perpendicular to each other. After this, in the Xoy plane, construct a parallelogram that will help draw an ellipse. Inscribe an ellipse into this parallelogram so that it touches all its sides. After these transformations, erase the parallelogram, and what remains is a three-dimensional image of a paraboloid.

There is also a hyperbolic paraboloid, which has a more concave shape than an elliptical one. Its sections also have parabolas and, in some cases, hyperbolas. The main sections along Oxz and Oyz, like those of an elliptic paraboloid, are parabolas. They are given by equations of the form:
x^2/a^2=2z; y^2/a^2=-2z
If you draw a section relative to the Oxy axis, you can get a hyperbola. When constructing a hyperbolic paraboloid, use the following equation:
x^2/a^2-y^2/b^2=2z - equation of a hyperbolic paraboloid

Initially construct a fixed parabola in the Oxz plane. Draw a moving parabola in the Oyz plane. After this, set the height of the paraboloid h. To do this, mark two points on the fixed parabola, which will be the vertices of two more movable parabolas. Then draw another coordinate system O"x"y" to plot the hyperbolas. The center of this coordinate system should coincide with the height of the paraboloid. After all the constructions, draw those two movable parabolas mentioned above so that they touch the extreme points of the hyperbolas. In the result is a hyperbolic paraboloid.