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ellipse perimeter calculator online

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ellipse perimeter

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Ellipse perimeter calculator online

presented on our website is a subsection

online calculator of the perimeter of geometric shapes

... That is why you can not only

set the accuracy of calculations

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our

online calculator

, without super-efforts go to the calculation

perimeter

 any of the following geometric shapes: triangle, rectangle, square, parallelogram, rhombus, trapezoid, circle, sector of the circle, regular polygon.

Also, you can literally go to

online calculator of the area of ​​geometric shapes

and calculate

square

triangle

,

rectangle

,

square

,

parallelogram

,

diamond

,

trapezium

,

circle

,

ellipse

,

sectors of the circle

,

regular polygon

also in several ways

and with

detailed solution

.

Ellipse

is a closed curve on a plane, which can be obtained as the intersection of a plane and a circular

cylinder

, or as an orthographic projection

circles

on a plane.

Circle

is a special case

ellipse

... As well as

hyperbole

and

parabola

,

ellipse

is an

conical section

and

quadric

.

ellipse

is intersected by two parallel straight lines, then the segment connecting the midpoints of the segments formed at the intersection of the straight lines and

ellipse

, will always go through

center of ellipse

... This property makes it possible to construct using a compass and a ruler to obtain

center of ellipse

.

Evolute

ellipse

there is

asteroid

which is stretched along the short axis.

With this

You can do

calculating the perimeter of an ellipse

 in the following ways:

-

calculation of the perimeter of an ellipse through two semiaxes

;

-

calculating the perimeter of an ellipse through two axes

.

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calculating the perimeter of an ellipse

.

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In astronomy, when considering the motion of cosmic bodies in orbits, the concept of "ellipse" is often used, since their trajectories are characterized by this very curve. Consider in the article the question of what the marked figure is, and also give the formula for the length of an 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, and called foci, is a constant value. Below is a figure that explains this definition.

You will be interested in:

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 is the length of the minor semiaxis. Here point C defines the center of the figure.

The figure above also shows a simple rope and two studs method that is widely used to draw elliptical curves. Another way to get this figure is to cross-section the cone at any angle to its axis that is not 90o.

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

Ellipse circumference formula

Although the figure under consideration is quite simple, its circumference can be precisely determined by calculating the so-called elliptic integrals of the second kind. However, the Hindu self-taught mathematician Ramanujan, at the beginning of the 20th century, proposed a fairly simple formula for the length of an ellipse, which approximates the result of the above integrals from below. That is, the value of the considered value calculated from it will be slightly less than the real length. This formula has the form: P ≈ pi *, where pi = 3.14 is the number pi.

For example, let the lengths of the two semi-axes of the ellipse be 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 is reduced to the form P = 2 * pi * R.

Note that school textbooks often use a different formula: P = pi * (a + b). It is simpler, but also less accurate. So, if we apply it for the considered case, then we get the value P = 56.5 cm.

In astronomy, when considering the motion of cosmic bodies in orbits, the concept of "ellipse" is often used, since their trajectories are characterized by this very curve. Consider in the article the question of what the marked figure is, and also give the formula for the length of an 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, and 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 is the length of the minor semiaxis. Here point C defines the center of the figure.

The figure above also shows a simple rope and two studs method that is widely used to draw elliptical curves. Another way to get this figure is to perform 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 volumetric figure, which is called a spheroid.

Ellipse circumference formula

Although the figure under consideration is quite simple, its circumference can be precisely determined by calculating the so-called elliptic integrals of the second kind. However, the Hindu self-taught mathematician Ramanujan, at the beginning of the 20th century, proposed a fairly simple formula for the length of an ellipse, which approximates the result of the above integrals from below. That is, the value of the considered value calculated from it will be slightly less than the real length. This formula has the form: P ≈ pi *, where pi = 3.14 is the number pi.

For example, let the lengths of the two semi-axes of the ellipse be 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 is reduced to the form P = 2 * pi * R.

Note that school textbooks often use a different formula: P = pi * (a + b). It is simpler, but also less accurate. So, if we apply it for the considered case, then we get the value P = 56.5 cm.

    Circumference a closed plane curve is called, all points of which are equidistant from a given point (center of a 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. The equation of a circle of radius \ (R \) centered at the origin ( canonical circle equation ) has the form
    \ ((x ^ 2) + (y ^ 2) = (R ^ 2) \).

    Circle equation radius \ (R \) centered 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 as: \ (\ 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) \) - three points lying on a 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 \)
    provided \ (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 called a flat curve, for each point of which the sum of the distances to two given points ( foci of an ellipse ) 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 ... The 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 an ellipse ... The semi-major axis is denoted by \ (a \), the semi-minor axis is denoted by \ (b \). The ellipse, the center of which is at the origin, and the semiaxes lie on the coordinate lines, is described as follows 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) \) - distances from an arbitrary point \ (P \ left ((x, y) \ right) \) to focuses \ ((F_1) \) and \ (( F_2) \), \ (a \) - semi-major axis of the ellipse.

    Relationship between semi-axes of an ellipse and 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 the half of the focal length.

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

    Directrix Ellipse Equations
    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. Directrix equations are written as
    \ (x = \ pm \ large \ frac (a) (e) \ normalsize = \ pm \ large \ frac (((a ^ 2))) (c) \ normalsize. \)

    Ellipse equation 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 semiaxes of the ellipse, \ (t \) is a parameter.

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

    General equation of an ellipse whose semiaxes 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 semi-major axis of the ellipse, \ (e \) is the eccentricity, \ (E \) - 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 semiaxes of the ellipse.

    Ellipse area
    \ (S = \ pi ab \)

When we are dealing with round tubs, everything is pretty simple. Indeed, there are diameters - upper and lower, there is the height of the rivets, it is not difficult to calculate the perimeter ... It remains only to make a template and plan it for yourself, gaining the required total width of the rivets. But what if our product is oval? How many templates do you need to make it, and which ones? How is this smooth line formed, passing from small radii at the ends of the product to large, having a relatively slight bend, sides?

To understand this issue, let's start with the method described by G. Ya. Fedotov in the book "Secrets of the cooper's craft". 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 the parameters of the oval according to Fedotov

As you know, an oval consists of four mating arcs - two large and two small. The frame is as if assembled from rivets of a large and a small keg. In fact, this is how it is. Only, of course, the master makes two types of rivets specially - one as if for a small keg, others - for a large one. Then, arranging them in a certain order, it pulls them together with hoops, receiving a skeleton with pressed sides and an oval section.

In order to determine exactly what rivets of both types should be, how many of them should be included in the skeleton set, it is necessary to perform some calculations. First of all, an oval cross-section of the skeleton is drawn on a sheet of life-size paper in its widest part. An auxiliary circle is drawn with a compass, the diameter of which should be equal to the height of the barrel. ( Under the height in this case G.Ya. Fedotov means the major axis of the oval - this can be seen from the figure). Its center is marked with 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 points of intersection of the horizontal center line with the auxiliary circle and the centers of small circles. At the intersection of these lines with arcs of small circles, there will be the so-called conjugation points. They are connected with a compass in large arcs. The centers of these arcs will be at the intersection of the horizontal centerline and the major arc of the construction circle.

Guided by an 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 establish exactly how many rivets are required to assemble the barrel's frame, 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. Perimeters are set using the formula 2πR, where π = 3.14. Then, by dividing the perimeter of the minor circle into 3 parts, the length of the minor 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 - everything is simple? This method really works, and it works flawlessly.

But what if our oval product is a 500 liter bathtub?

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

Scaling? Fraught with inaccuracies ...

From the geometry of construction, given by G. Ya. Fedotov, it is easy to deduce formulas with the help of which the same values ​​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 - 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 value h - the distance between the point of intersection of the center lines and the center of the small circle A 1:

h = 1.5r

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

c = √ [(L / 2) 2 +h 2]

4) Find the radius of the large arc R:

R =c +r

5) Find the auxiliary value 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 and find the parameters D and d. These are our diameters - those that are needed for making templates.

8) Find the length of the small arc m:

m =πd / 3

9) Find the length of the large arc M:

M =πD / 6

10) And finally, we 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 (bottom or top of our bath).

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

First, the master can only set the length of the oval L. Its width l is already calculated, that is, it turns out to be rigidly 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 large and small arcs of our product have different taper. So, for a bath of 500 liters,
which is calculated in 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 rivet 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 regularities described in the article "Taper of a cooper product" do not work, and the reliability of fixing wooden hoops at a certain height has to be determined empirically.

Universal formulas for calculating the parameters of the oval

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 in four parts, or three, or six. Moreover, it is generally not necessary to divide it into equal parts. The angle formed by the horizontal centerline and AB lines in general can be anything (within 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 parameters of the oval 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 it is. But on the other hand, using these formulas, we can freely set three parameters: the length of the oval, its width and the auxiliary angle γ. And this means that we can calculate an oval with any given overall dimensions a and b, and even more than one. With the same values ​​of a and b, we can get as many different ovals as we can think of different values ​​of the auxiliary angle γ that fit into the drawing.

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

Oval length, a, cm

Oval width, b, cm

Large arc diameter, D, cm

Small arc diameter, d, cm

Large arc length, M, cm

Small arc length, m, cm

Oval perimeter, p, cm

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

At the same time, setting the 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 be evenly inscribed into one another, if we look at them from above. For such products, the difference between large and small diameters will be the same, and, therefore, the taper coefficient will also be the same. An example of such a product is our ottoman,
having the following parameters: the diameters of the large arcs are 96 and 90 cm, the diameters of the small arcs are 36 and 30 cm, the lengths of the large and small ovals are 66 and 60 cm, and their widths are 44 and 38 cm. As you can see, the difference is both in diameters, and in overall dimensions it is everywhere equal to 6 cm. The taper coefficient at a rivet height of 45 cm is 0.133. Wooden hoops over the entire surface are stretched on the product in the same way and securely fixed at a given height.

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