Circumference

The circumference is the distance around a closed curve. Circumference is a kind of perimeter.

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

Or, substituting the diameter for the radius:

where r is the radius and d is the diameter of the circle, and Ï€ (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 Ï€ 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 $\sqrt{(dx)^2+(f'(x)dx)^2} = \left( \sqrt{1+f'^2(x)} \right) dx.$

Thus the circle circumference can be calculated as dara:)

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

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

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

Pi (Ï€) is the ratio of the circumference of a circle to its diameter.

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,

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

$\begin{align}\mbox{E2}\left[0,90^\circ\right]&= \mbox{Integral}'s\mbox{ divided difference};\\ Pr&=a\times\mbox{E2}\left[0,90^\circ\right] \quad(\mbox{perimetric radius});\\ c&=2\pi\times Pr.\end{align}\,\!$

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

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

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

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

In graph theory the circumference of a graph refers to the longest cycle contained in that graph.

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Last updated on Monday September 24, 2007 at 05:23:47 PDT (GMT -0700)
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