Ellipses are weird.

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[dyld]

Love how the ellipse perfectly matches his head

[themibo]

another way of finding the area of an ellipse is by applying the transformation made by M, a matrix of entries a, 0, 0, b. The area of the unit circle is scaled by the determinant of this new matrix which is precisely equal to ab. Thus the ellipse has area of pi ab

[torgeirl]

The arc length can be calculated using a string and wrapping it around the ellipse: source my engineering degree

[davidbrisbane]

Srinivasa Ramanujan came up this good approximation for the perimeter of an ellipse.

p ≅ π[3(a + b) - √[(3a + b)(a + 3b)]].

An even better approximation he came up with is

p ≅ π(a + b)[1 + 3h/[10 - √(4 - 3h )]], where
h = (a - b)²/(a + b)².

[alejrandom]

You can also look at the elipse as a diagonal matrix acting on a unit circle. Determinant is ab, so area is πab.

[chaosredefined]

A while back, Matt Parker of Standup Math made a video on this. He went through a lot of approximations. As a result of that video, I went and played around and made an approximation of my own. First, ensure that a > b, and define d = a + b. Then the perimeter is within 1% of (3 b^2 + 4 d^2) / (b + d). For those who don't like the d = a+b term, instead, you get (4a^2 + 8ab + 7b^2) / (2a + b).

As I said, this is accurate within 1% of the true value. There are more accurate formula, but they get a lot more complex. I can also increase the accuracy to 0.25%, by making the numbers more clunky.

[andrewhone]

I wouldn't call elliptic integrals "impossible" - trigonometric functions, or exponentials, are just as impossible to calculate without a calculator, a computer, or a book of tables. For instance, try calculating sin(1), or exp(2.5)....

[skalas]

I'm not at this level of mathematics yet, but I think it is because:
Pi is a constant for circles(e = 1(one value only)), while ellipses can have 0 <= e < 1. I know that there are series that sum up to Pi so I think there must be some series that can compute "Pi" for a given e

[davidbrisbane]

Of course, using the appropriate expanded form of the "unintegratable" integral, we can calculate the circumference of an ellipse to any degree of precision required.

p = 2πa[1 - (1/2)²e² - (1*3/2*4)²e⁴/3
- (1*3*5/2*4*6)²e⁶/5 - ... ].

[PrimeVibeGaming]

Again Sir Srinivasa Ramanujan supermacy, two most accurate relation to find elliptic circumference

[AdrianBoyko]

Did you know that, in spherical geometry, the branches of a hyperbola are each an ellipse?

[robshaw]

Matt Parker has a deep dive into this topic ("Stand Up Maths" channel)

[ralvarezb]

The wonderful world of elliptic integrals and Arithmetic Geometric Mean. See "Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity" (Borwein)

[mclark]

Eliptic functions were used to prove Fermat

[nbooth]

Arclengths in general are a problem, not just for ellipses.

[brandon.m]

The formula is no more complicated or impossible than the area of the ellipse, the area of a circle, or the circumference of a circle. While the integral is not elementary, pi appears in all of these formulas. Can we calculate pi exactly? I think not and it is approximated the same way as you would the integral, so I’m not sure what about the ellipse’s circumference is so different that is “impossible” while the others are “easy”.

[jwstolk]

The circumference of an ellipse requires an infinite sum, just like the circumference of a circle. (pi is an infinite sum)

[davidbrisbane]

If you used Excel, you can write a function as follows to calculate the perimeter of an ellipse.

Here's an example of how you can calculate the perimeter of an ellipse using Excel:

1. Open Microsoft Excel and create a new worksheet.
2. In cell A1, enter the heading "Semi-Major Axis (a)".
3. In cell A2, enter the value for the semi-major axis of the ellipse (e.g., `7.2`).
4. In cell B1, enter the heading "Semi-Minor Axis (b)".
5. In cell B2, enter the value for the semi-minor axis of the ellipse (e.g., `4.5`).
6. In cell C1, enter the heading "Decimal Places (d)".
7. In cell C2, enter the desired number of decimal places for accuracy (e.g., `3`).

8. In cell D1, enter the heading "Perimeter".
9. In cell D2, enter the following formula to calculate the perimeter:

```
=4 * A2 * ELLIPKE(1 - (B2/A2)^2) / SQRT(1 - (B2/A2)^2)
```

10. In cell D2, format the cell to display the desired number of decimal places by selecting the cell, right-clicking, and choosing "Format Cells." In the "Number" tab, select "Number" as the category and set the decimal places to the value in cell C2.

After following these steps, the cell D2 will display the computed perimeter of the ellipse based on the given values of semi-major axis (`a`), semi-minor axis (`b`), and the desired number of decimal places (`d`).

The output would be ... 29.707

[gibson]

I m still waiting for the perimeter. We need Ramanujan, since we all suck, really

[MichaelRothwell]

Nice! Can you give the link to the videos on elliptic integrals you mentioned?