Lengths of the ellipse axes. What is an ellipse: formula for the circumference of an ellipse

Circumference is a closed plane curve, all points of which are equidistant from a given point (the center of the circle). The distance from any point of the circle \(P\left((x,y) \right)\) to its center is called radius. The center of the circle and the circle itself lie in the same plane. Equation of a circle of radius \(R\) with center at the origin ( canonical equation circle ) has the form
\((x^2) + (y^2) = (R^2)\).

Equation of a circle radius \(R\) with center at an arbitrary point \(A\left((a,b) \right)\) is written as
\((\left((x - a) \right)^2) + (\left((y - b) \right)^2) = (R^2)\).



Equation of a circle passing through three points , written in the form: \(\left| (\begin(array)(*(20)(c)) ((x^2) + (y^2)) & x & y & 1\\ (x_1^2 + y_1^2) & ((x_1)) & ((y_1)) & 1\\ (x_2^2 + y_2^2) & ((x_2)) & ((y_2)) & 1\\ (x_3^2 + y_3^2) & ((x_3)) & ((y_3)) & 1 \end(array)) \right| = 0.\\\)
Here \(A\left(((x_1),(y_1)) \right)\), \(B\left(((x_2),(y_2)) \right)\), \(C\left(( (x_3),(y_3)) \right)\) are three points lying on the circle.



Equation of a circle in parametric form
\(\left\( \begin(aligned) x &= R \cos t \\ y &= R\sin t \end(aligned) \right., \;\;0 \le t \le 2\pi\ ),
where \(x\), \(y\) are the coordinates of the points of the circle, \(R\) is the radius of the circle, \(t\) is the parameter.

General equation of a circle
\(A(x^2) + A(y^2) + Dx + Ey + F = 0\)
subject to \(A \ne 0\), \(D^2 + E^2 > 4AF\).
The center of the circle is located at the point with coordinates \(\left((a,b) \right)\), where
\(a = - \large\frac(D)((2A))\normalsize,\;\;b = - \large\frac(E)((2A))\normalsize.\)
The radius of the circle is
\(R = \sqrt (\large\frac(((D^2) + (E^2) - 4AF))((2\left| A \right|))\normalsize) \)

Ellipse is a plane curve for each point of which the sum of the distances to two given points ( ellipse foci ) is constant. The distance between the foci is called focal length and is denoted by \(2c\). The middle of the segment connecting the foci is called the center of the ellipse . An ellipse has two axes of symmetry: the first or focal axis, passing through the foci, and the second axis perpendicular to it. The points of intersection of these axes with the ellipse are called peaks. The segment connecting the center of the ellipse with the vertex is called semi-axis of the ellipse . The semi-major axis is denoted by \(a\), the semi-minor axis by \(b\). An ellipse whose center is at the origin and whose semi-axes lie on coordinate lines is described by the following canonical equation :
\(\large\frac(((x^2)))(((a^2)))\normalsize + \large\frac(((y^2)))(((b^2)))\ normalsize = 1.\)



The sum of the distances from any point of the ellipse to its foci constant:
\((r_1) + (r_2) = 2a\),
where \((r_1)\), \((r_2)\) are the distances from an arbitrary point \(P\left((x,y) \right)\) to the foci \((F_1)\) and \(( F_2)\), \(a\) is the semimajor axis of the ellipse.



The relationship between the semi-axes of the ellipse and the focal length
\((a^2) = (b^2) + (c^2)\),
where \(a\) is the semi-major axis of the ellipse, \(b\) is the semi-minor axis, \(c\) is half the focal length.

Ellipse eccentricity
\(e = \large\frac(c)(a)\normalsize

Equations of ellipse directrixes
The directrix of an ellipse is a straight line perpendicular to its focal axis and intersecting it at a distance \(\large\frac(a)(e)\normalsize\) from the center. The ellipse has two directrixes located on opposite sides of the center. The directrix equations are written in the form
\(x = \pm \large\frac(a)(e)\normalsize = \pm \large\frac(((a^2)))(c)\normalsize.\)

Equation of an ellipse in parametric form
\(\left\( \begin(aligned) x &= a\cos t \\ y &= b\sin t \end(aligned) \right., \;\;0 \le t \le 2\pi\ ),
where \(a\), \(b\) are the semi-axes of the ellipse, \(t\) is the parameter.

General equation of ellipse
\(A(x^2) + Bxy + C(y^2) + Dx + Ey + F = 0\),
where \((B^2) - 4AC

General equation of an ellipse whose semi-axes are parallel to the coordinate axes
\(A(x^2) + C(y^2) + Dx + Ey + F = 0\),
where \(AC > 0\).

Ellipse perimeter
\(L = 4aE\left(e \right)\),
where \(a\) is the semimajor axis of the ellipse, \(e\) is the eccentricity, \(E\) is complete elliptic integral of the second kind.

Approximate formulas for the perimeter of an ellipse
\(L \approx \pi \left[ (\large\frac(3)(2)\normalsize\left((a + b) \right) - \sqrt (ab) ) \right],\;\;L \approx \pi \sqrt (2\left(((a^2) + (b^2)) \right)),\)
where \(a\), \(b\) are the semi-axes of the ellipse.

Area of ​​the ellipse
\(S = \pi ab\)

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Oval is a closed box curve that has two axes of symmetry and consists of two support circles of the same diameter, internally conjugate by arcs (Fig. 13.45). An oval is characterized by three parameters: length, width and radius of the oval. Sometimes only the length and width of the oval are specified, without defining its radii, then the problem of constructing an oval has a large variety of solutions (see Fig. 13.45, a... d).

Methods for constructing ovals based on two identical reference circles that touch (Fig. 13.46, a), intersect (Fig. 13.46, b) or do not intersect (Fig. 13.46, c) are also used. In this case, two parameters are actually specified: the length of the oval and one of its radii. This problem has many solutions. It's obvious that R > OA has no upper bound. In particular R = O 1 O 2(see Fig. 13.46.a, and Fig. 13.46.c), and the centers O 3 And O 4 are determined as the points of intersection of the base circles (see Fig. 13.46, b). According to the general point theory, mates are determined on a straight line connecting the centers of arcs of osculating circles.

Constructing an oval with touching support circles(the problem has many solutions) ( rice. 3.44). From the centers of the reference circles ABOUT And 0 1 with a radius equal, for example, to the distance between their centers, draw arcs of circles until they intersect at points ABOUT 2 and O 3.

Figure 3.44

If from points ABOUT 2 and O 3 draw straight lines through the centers ABOUT And O 1, then at the intersection with the support circles we obtain the connecting points WITH, C 1, D And D 1. From points ABOUT 2 and O 3 as from centers of radius R 2 draw arcs of conjugation.

Constructing an oval with intersecting reference circles(the problem also has many solutions) (Fig. 3.45). From the intersection points of the reference circles C 2 And O 3 draw straight lines, for example, through centers ABOUT And O 1 until they intersect with the reference circles at the junction points C, C 1 D And D 1, and radii R2, equal to the diameter of the reference circle - the conjugation arc.

Figure 3.45 Figure 3.46

Constructing an oval along two specified axes AB and CD(Fig. 3.46). Below is one of many possible solutions. A segment is plotted on the vertical axis OE, equal to half the major axis AB. From point WITH how to draw an arc with a radius from the center SE to the intersection with the line segment AC at the point E 1. Towards the middle of the segment AE 1 restore the perpendicular and mark the points of its intersection with the axes of the oval O 1 And 0 2 . Build points O 3 And 0 4 , symmetrical to the points O 1 And 0 2 relative to the axes CD And AB. Points O 1 And 0 3 will be the centers of reference circles of radius R1, equal to the segment About 1 A, and the points O2 And 0 4 - centers of conjugation arcs of radius R2, equal to the segment O 2 C. Straight lines connecting centers O 1 And 0 3 With O2 And 0 4 At the intersection with the oval, the connecting points will be determined.

In AutoCAD, an oval is constructed using two reference circles of the same radius, which:

1. have a point of contact;

2. intersect;

3. do not intersect.

Let's consider the first case. Construct the segment OO 1 =2R, axis parallel X, at its ends (points O and O 1) the centers of two supporting circles of radius R and the centers of two auxiliary circles of radius R 1 = 2R are placed. From the intersection points of the auxiliary circles O 2 and O 3, arcs CD and C 1 D 1 are built, respectively. The auxiliary circles are removed, then the inner parts of the support circles are cut off relative to the arcs CD and C 1 D 1. In Figure ъъ the resulting oval is highlighted with a thick line.

Figure Constructing an oval with touching support circles of the same radius

In astronomy, when considering the movement of cosmic bodies in orbits, the concept of “ellipse” is often used, since their trajectories are characterized by precisely this curve. In the article we will consider the question of what the marked figure represents, and also give the formula for the length of the ellipse.

What is an ellipse?

According to the mathematical definition, an ellipse is a closed curve for which the sum of the distances from any of its points to two other specific points lying on the main axis, called foci, is a constant value. Below is a figure that explains this definition.

In the figure, the sum of the distances PF" and PF is equal to 2 * a, that is, PF" + PF = 2 * a, where F" and F are the foci of the ellipse, "a" is the length of its semi-major axis. The segment BB" is called the semi-minor axis, and distance CB = CB" = b - length of the semi-minor axis. Here point C determines the center of the figure.

The picture above also shows a simple rope and two nails method that is widely used for drawing elliptic curves. Another way to get this figure is to perform it at any angle to its axis, which is not equal to 90 o.

If the ellipse is rotated along one of its two axes, then it forms a three-dimensional figure, which is called a spheroid.

Formula for the circumference of an ellipse

Although the figure in question is quite simple, the length of its circumference can be accurately determined by calculating the so-called elliptic integrals of the second kind. However, the self-taught Indian mathematician Ramanujan, at the beginning of the 20th century, proposed a fairly simple formula for the length of an ellipse, which approaches the result of the marked integrals from below. That is, the value of the value in question calculated from it will be slightly less than the actual length. This formula looks like: P ≈ pi *, where pi = 3.14 is the number pi.

For example, let the lengths of the two semi-axes of the ellipse be equal to a = 10 cm and b = 8 cm, then its length P = 56.7 cm.

Everyone can check that if a = b = R, that is, an ordinary circle is considered, then Ramanujan’s formula reduces to the form P = 2 * pi * R.

Note that in school textbooks another formula is often given: P = pi * (a + b). It is simpler, but also less accurate. So, if we apply it to the case considered, we obtain the value P = 56.5 cm.

When we are dealing with round tubs, everything is quite simple. Indeed, there are diameters - upper and lower, there is the height of the rivets, it is not difficult to calculate the perimeter... All that remains is to make a template and plan it yourself, gaining the required total width of the rivets. But what if our product is oval? How many templates are needed to make it, and what kind? How is this smooth line formed, moving from small radii at the ends of the product to large sides with a relatively slight bend?

To understand this issue, let's start with the method described by G. Ya. Fedotov in the book “Secrets of Cooperation.” This is what the author offers us in the chapter “Ankerok”, dedicated to the manufacture of this portable flat barrel, which has an oval cross-section.

Geometric method for calculating oval parameters according to Fedotov

As you know, an oval consists of four mating arcs - two large and two small. The frame seems to be assembled from the rivets of a large and small barrel. In fact, this is how it is. Only, of course, the master makes two types of rivets specially - some, as it were, for a small barrel, others - for a large one. Then, arranging them in a certain order, he tightens them with hoops, obtaining a frame with pressed sides and an oval cross-section.

In order to accurately determine what kind of rivets of one and another type should be, how many of them should be included in the frame set, it is necessary to perform some calculations. First of all, on a life-size sheet of paper, draw an oval cross-section of the frame in its widest part. Using a compass, draw an auxiliary circle, the diameter of which should be equal to the height of the barrel. ( Under the height in this case G.Ya. Fedotov implies the major axis of the oval - this can be seen from the figure). Its center is marked by two mutually perpendicular axial lines. The vertical axis is divided into five equal parts. Two small circles are drawn around points 1 and 4, tangent to the large auxiliary circle. Straight lines are drawn through the intersection points of the horizontal center line with the auxiliary circle and the centers of small circles. At the intersection of these lines with the arcs of small circles there will be so-called connecting points. They are connected using a compass with large arcs. The centers of these arcs will be at the intersection of the horizontal centerline and the major arc of the auxiliary circle.

Guided by the oval drawn on paper, two templates are made. The contours of one of them should correspond to the small arc of the oval, and the other - to the large one.

In order to determine exactly how many rivets are required to assemble the frame of the barrel, it is necessary to determine its perimeter. It will be equal to the sum of the lengths of the major and minor arcs. The length of each arc is found as follows. First, determine the perimeter of complete circles, part of which are the arcs that make up the oval. The perimeters are set according to the formula 2πR, where π=3.14. Then, dividing the perimeter of the small circle into 3 parts, the length of the small arc is obtained. In turn, the perimeter of the great circle is divided into six parts and the length of the great arc is determined. The total length of the two arcs is doubled and the perimeter of the oval is obtained.

Isn't it all simple? This method really works, and works flawlessly.

But what if our oval product is a bathtub with a volume of 500 liters?

Drawing it in full size is not the easiest task. But you need two such drawings - for the upper and lower oval.

Scaling? It is fraught with inaccuracies...

From the geometry of construction given by G. Ya. Fedotov, it is not difficult to derive formulas with the help of which the same quantities can be obtained without drawing anything on paper.

Algebraic method for calculating oval parameters according to Fedotov

Despite the fact that Gennady Yakovlevich does not give these formulas in the book, we will still call the method by his name, since it is correct only for the drawing given above, and, in fact, simply replaces it.

So, let L be the length of the oval, l its width, r the radius of the small circle, R the radius of the large circle.

1) Find the radius of the small circle:

r=L/5

2) Find the auxiliary quantity h - the distance between the point of intersection of the axial lines and the center of the small circle A 1:

h=1.5r

3) Find the auxiliary quantity c - the distance between two parallel lines B 2 A 1 and A 2 B 1:

c= √ [(L/2) 2 +h 2 ]

4) Find the radius of the major arc R:

R=c+r

5) Find the auxiliary quantity q - the distance between point B 1 (B 2) and the point of intersection of the large arc of the oval and the horizontal center line:

q=L-R

6) Find the width of the oval l:

l=L-2q

7) Multiply the radii R and r by 2 to find the parameters D and d. These are our diameters - those needed for making templates.

8) Find the length of the minor arc m:

m=πd/3

9) Find the length of the major arc M:

M=πD/6

10) And finally, find the perimeter of the oval p:

p=2(M+m)

This calculation will have to be repeated to find the parameters of the second oval (the bottom or top of our bathtub).

When calculating an oval according to Fedotov, you need to keep in mind some features.

Firstly, the master can only set the length of the oval L. Its width l is already calculated, that is, it turns out to be strictly tied to a certain length value. In other words, if we need to change the width, we will have to change the length. It is not comfortable.

Secondly, when calculating using this method, it turns out that the large and small arcs of our product have different tapers. So, for a 500 liter bath,
which is calculated in exactly this way, the diameters of the large arcs at the top and bottom are 204 and 234 cm, respectively, and the diameters of the small ones are 52 and 60. Thus, with a riveting height of 85 cm, the taper coefficient for the small arc is 0.094, and for the large one - 0.353. For such an oval, the patterns described in the article “Taper of a cooper’s product” do not work, and the reliability of fixing wooden hoops at a certain height has to be determined experimentally.

Universal formulas for calculating oval parameters

However, it turns out that the vertical axis of the oval in our drawing does not have to be divided exactly into five parts. It can be made into four parts, or into three, or into six. Moreover, it is generally not necessary to divide it into equal parts. The angle formed by the horizontal axial line and the AB lines can generally be anything (within the limits of the drawing, of course).

Let us denote this angle by the symbol γ. And let the axes of the oval (its length and width, respectively) be equal to a and b.

Then the universal formulas for calculating the oval parameters will look like this:

R=[(b/2*(sin(γ)-1)+(a/2*cos γ)] /

r=[(b/2*cos (γ/2)) - (a/2*sin (γ/2))] / [(cos (γ/2)-sin(γ/2)]

Do they look scary? Hmm, perhaps that's true. But, using these formulas, we can freely set three parameters: the length of the oval, its width and the auxiliary angle γ. This means that we can calculate an oval with any given overall dimensions a and b, and more than one. With the same values ​​of a and b, we can get as many different ovals as we can come up with different values ​​of the auxiliary angle γ that fit into the drawing.

Let's explain with an example. Let us need to calculate an oval whose axes are 150 and 84 cm, respectively (the parameters of the large oval of our 500 liter bathtub). The table shows how the diameters D and d, the lengths of the major and minor arcs M and m, as well as the perimeter of the oval p will change depending on the change in the angle γ.

All these ovals will have slightly different contours, but the same overall dimensions - 150x84 cm.

At the same time, by setting values ​​for the large and small ovals of our product, we can freely set the same taper for the large and small arcs, which will make our ovals seem to fit evenly into one another when viewed from above. For such products, the difference between large and small diameters will be the same, and, therefore, the taper coefficient will be the same. An example of such a product is our ottoman,
having the following parameters: diameters of large arcs - 96 and 90 cm, diameters of small arcs - 36 and 30 cm, lengths of large and small ovals - 66 and 60 cm, and their widths - 44 and 38 cm. As you can see, the difference is in diameters, so in overall dimensions everywhere equal to 6 cm. The taper coefficient for a riveting height of 45 cm is 0.133. Wooden hoops are tensioned equally over the entire surface of the product and securely fixed at a given height.

In order to avoid the need to carry out complex calculations each time, it is enough to enter the above formulas once into some computing program. Below you can download an Excel document in which you enter only the values ​​of a and b (you need to enter the same values ​​in all rows), after which the program will automatically generate all the necessary parameters of such ovals for a wide range of angles γ. Just be sure not to enter anything by hand in the other columns so as not to replace the formulas with numeric values.