User:Michael Retriever/Ellipse perimeters

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This is a comparative test that attempts to verify the efficiency of various formulas (infinite binomial expansions) for the perimeter of an ellipse; to see whether they need more or less terms within their infinite expansions to reach a satisfactory result. The charts have been done with MS Excel.


Contents

[edit] Formulas

1st
perimeter = 2 \pi az;\,

z=(1+f')^{-1}\left[1+\left({1\over 2}\right)^2f'^2+\left({1\over 3^2}\right)\left({1\cdot 3\over 2\cdot 4}\right)^2f'^4+\left({1\over 5^2}\right)\left({1\cdot 3\cdot 5\over 2\cdot 4\cdot 6}\right)^2f'^6+\dots\right]\,
Used within the charts with the form:
perimeter = 2 \pi az;\,

z=\left[{1\over 1+f'}+\left({1\over 2}\right)^2{f'^2\over 1+f'}+\left({1\over 3^2}\right)\left({1\cdot 3\over 2\cdot 4}\right)^2{f'^4\over 1+f'}+\left({1\over 5^2}\right)\left({1\cdot 3\cdot 5\over 2\cdot 4\cdot 6}\right)^2{f'^6\over 1+f'}+\dots\right]\,
2nd
perimeter = 2 \pi az;\,

z=\left[1 - \left({1\over 2}\right)^2e^2 - \left({1\over 3}\right)\left({1\cdot 3\over 2\cdot 4}\right)^2e^4 - \left({1\over 5}\right)\left({1\cdot 3\cdot 5\over 2\cdot 4\cdot 6}\right)^2e^6 - \dots\right]\,
3rd (website link)
perimeter = 2 \pi bz;\,

z=\left[1 + \left({1 \over 2}\right)^2 n - \left({1 \over 2\cdot 4}\right)^2 3n^2 + \left({1 \cdot 3 \over 2\cdot 4\cdot 6}\right)^2 5n^3 - \left({1 \cdot 3 \cdot 5\over 2\cdot 4\cdot 6\cdot 8}\right)^2 7n^4 + \dots\right]\,
Used within the charts with the form:
perimeter = 2 \pi az;\,

z=\left[\left(\frac{b}{a}\right) + \left({1 \over 2}\right)^2 \left(\frac{b}{a}\right) n- \left({1\over 3}\right)\left({1\cdot 3\over 2\cdot 4}\right)^2 \left(\frac{b}{a}\right) n^2 + \left({1\over 5}\right)\left({1\cdot 3\cdot 5\over 2\cdot 4\cdot 6}\right)^2 \left(\frac{b}{a}\right) n^3 - \dots\right]\,
Equations
b/a=\cos\left(o\!\varepsilon\right);\, o\!\varepsilon=\arccos\left(\frac{b}{a}\right);\,

f'=\frac{a-b}{a+b}=\tan^2\left(\frac{o\!\varepsilon}{2}\right);\, \left(1+f'\right)^{-1} = \cos^2 \left(\frac{o\!\varepsilon}{2}\right);\,

e=\sqrt{1-\frac{b^2}{a^2}}=\sin\left(o\!\varepsilon\right);\, n=\frac{a^2}{b^2}-1=\tan^2\left(o\!\varepsilon\right)\,
The various elements of the formulas can be expressed through elementary equations or through trigonometric equations. For more information you can visit eccentricity and angular eccentricity.

[edit] Polygonal paths

Check explanation for the polygonal path. The polygonal path 1 measures the distance between two points in the ellipse's quarter of a perimeter every 0.01 degrees. The polygonal path 2 measures the distance between two points in the ellipse's quarter of a perimeter every 0.001 degrees.

[edit] Comparative test

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[edit] Conclusion

As we can see, the first formula is the most efficient one.