Definitions

Circumference

Circumference

[ser-kuhm-fer-uhns]
The circumference is the distance around a closed curve. Circumference is a kind of perimeter.

Circumference of a circle

is the length around a circle The circumference of a circle can be calculated from its diameter using the formula:

c=picdot{d}.,!

Or, substituting the diameter for the radius:

c=2picdot{r}=picdot{2r},,!

where r is the radius and d is the diameter of the circle, and π (the Greek letter pi) is defined as the ratio of the circumference of the circle to its diameter (the numerical value of pi is 3.141 592 653 589 793...).

If desired, the above circumference formula can be derived without reference to the definition of π by using some integral calculus, as follows:

The upper half of a circle centered at the origin is the graph of the function f(x) = sqrt{r^2-x^2}, where x runs from -r to +r. The circumference (c) of the entire circle can be represented as twice the sum of the lengths of the infinitesimal arcs that make up this half circle. The length of a single infinitesimal part of the arc can be calculated using the Pythagorean formula for the length of the hypotenuse of a rectangular triangle with side lengths dx and f'(x)dx, which gives us sqrt{(dx)^2+(f'(x)dx)^2} = left(sqrt{1+f'(x)^2} right) dx.

Thus the circle circumference can be calculated as

c = 2 int_{-r}^r sqrt{1+f'(x)^2}dx = 2 int_{-r}^r sqrt{1+frac{x^2}{r^2-x^2}}dx = 2 int_{-r}^r sqrt{frac{1}{1-frac{{x}^2}{{r}^2}}}dx

The antiderivative needed to solve this definite integral is the arcsine function:

c = 2r big[arcsin(frac{x}{r}) big]_{-r}^{r} = 2r big[arcsin(1)-arcsin(-1) big] = 2r(tfrac{pi}{2}-(-tfrac{pi}{2})) = 2pi r.

Circumference of an ellipse

The circumference of an ellipse is more problematic, as the exact solution requires finding the complete elliptic integral of the second kind. This can be achieved either via numerical integration (the best type being Gaussian quadrature) or by one of many binomial series expansions.

Where a,b are the ellipse's semi-major and semi-minor axes, respectively, and o!varepsilon,! is the ellipse's angular eccentricity,

o!varepsilon=arccos!left(frac{b}{a}right)=2arctan!left(!sqrt{frac{a-b}{a+b}},right);,!

begin{align}mbox{E2}left[0,90^circright]&= mbox{Integral}'smbox{ divided difference}; Pr&=atimesmbox{E2}left[0,90^circright] quad(mbox{perimetric radius}); c&=2pitimes Pr.end{align},!

There are many different approximations for the mbox{E2}left[0,90^circright] divided difference, with varying degrees of sophistication and corresponding accuracy.

In comparing the different approximations, the tan!left(frac{o!varepsilon}{2}right)^2,! based series expansion is used to find the actual value:

begin{align}mbox{E2}left[0,90^circright] &=cos!left(frac{o!varepsilon}{2}right)^2 frac{1}{UT}sum_{TN=1}^{UT=infty}{.5choose{}TN}^2tan!left(frac{o!varepsilon}{2}right)^{4TN}, &=cos!left(frac{o!varepsilon}{2}right)^2Bigg(1+frac{1}{4}tan!left(frac{o!varepsilon}{2}right)^4 +frac{1}{64}tan!left(frac{o!varepsilon}{2}right)^8 &qquadqquadqquad;,+frac{1}{256}tan!left(frac{o!varepsilon}{2}right)^{12} +frac{25}{16384}tan!left(frac{o!varepsilon}{2}right)^{16} +...Bigg);end{align},!

Muir-1883

Probably the most accurate to its given simplicity is Thomas Muir's:
begin{align}Pr
&approxleft(frac{a^{1.5}+b^{1.6}}{2}right)^frac{1}{1.5}=aleft(frac{1+cos!left(o!varepsilonright)^{1.5}}{2}right)^frac{1}{1.5}, &quadapprox{a}timescos!left(frac{o!varepsilon}{2}right)^2left(1+frac{1}{4}tan!left(frac{o!varepsilon}{2}right)^4right);end{align},!

Ramanujan-1914 (#1,#2)

Srinivasa Ramanujan introduced two different approximations, both from 1914
begin{align}1.;Pr&approxpiBig(3(a+b)-sqrt{big(3a+bbig)big(a+3bbig)}Big),
&quad=pi{a}bigg(6cos!left(frac{o!varepsilon}{2}right)^2sqrt{big(3+cos!left(o!varepsilonright)big)big(1+3cos!left(o!varepsilonright)big)}bigg);end{align},!

begin{align}2.;Pr&approxfrac{1}{2}Big(a+bBig)Bigg(1+frac{3big(frac{a-b}{a+b}big)^2}{10+sqrt{4-3big(frac{a-b}{a+b}big)^2}}Bigg);
&quad=atimescos!left(frac{o!varepsilon}{2}right)^2Bigg(1+frac{3tan!big(frac{o!varepsilon}{2}big)^4}{10+sqrt{4-3tan!big(frac{o!varepsilon}{2}big)^4}}Bigg);end{align},!

The second equation is demonstratively by far the better of the two, and may be the most accurate approximation known.

Letting a = 10000 and b = a×cos{}, results with different ellipticities can be found and compared:

b Pr Ramanujan-#2 Ramanujan-#1 Muir
9975

 9987.50391 11393 

 9987.50391 11393 

 9987.50391 11393 

 9987.50391 11389
9966

 9983.00723 73047

 9983.00723 73047

 9983.00723 73047

 9983.00723 73034
9950

 9975.01566 41666

 9975.01566 41666

 9975.01566 41666

 9975.01566 41604
9900

 9950.06281 41695

 9950.06281 41695

 9950.06281 41695

 9950.06281 40704
9000

 9506.58008 71725

 9506.58008 71725

 9506.58008 67774

 9506.57894 84209
8000

 9027.79927 77219

 9027.79927 77219

 9027.79924 43886

 9027.77786 62561
7500

 8794.70009 24247

 8794.70009 24240

 8794.69994 52888

 8794.64324 65132
6667

 8417.02535 37669

 8417.02535 37460

 8417.02428 62059

 8416.81780 56370
5000

 7709.82212 59502

 7709.82212 24348

 7709.80054 22510

 7708.38853 77837
3333

 7090.18347 61693

 7090.18324 21686

 7089.94281 35586

 7083.80287 96714
2500

 6826.49114 72168

 6826.48944 11189

 6825.75998 22882

 6814.20222 31205
1000

 6468.01579 36089

 6467.94103 84016

 6462.57005 00576

 6431.72229 28418
 100

 6367.94576 97209

 6366.42397 74408

 6346.16560 81001

 6303.80428 66621
  10

 6366.22253 29150

 6363.81341 42880

 6340.31989 06242

 6299.73805 61141
   1

 6366.19804 50617

 6363.65301 06191

 6339.80266 34498

 6299.60944 92105
iota

 6366.19772 36758

 6363.63636 36364

 6339.74596 21556

 6299.60524 94744

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