PrA

PrA is a simple ad-hoc parameter to define the "non-circularity" or eccentricity of a 2D object. This quantity is simply:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \mathrm{PRA} = \frac{Pr}{A} \end{alignat} }

Where $P$ is the object's perimeter, $A$ is its surface area, and $r$ is an effective size (radius), computed based on the corresponding circle of the same area:

${\begin{alignedat}{2}r={\sqrt {\frac {A}{\pi }}}\end{alignedat}}$

This definition of PrA is convenient, since it provides a simple measure of eccentricity. In particular, for a circle one expects:

${\begin{alignedat}{2}\mathrm {PRA} &={\frac {Pr}{A}}\\&={\frac {(2\pi r)(r)}{\pi r^{2}}}\\&=2\end{alignedat}}$

Since a circle has the minimal perimeter (for a given area), this is a limiting value of PrA:

${\begin{alignedat}{2}\mathrm {PRA} \geq 2\end{alignedat}}$

And thus any non-circular object will have a larger PrA. An infinitely eccentric object would have $\mathrm {PRA} \to \infty$.

Ellipse

If the object is an ellipse, with equation:

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1$

Then the width is $2a$ and height $2b$ (we assume $a\geq b$), the foci are $(\pm c,0)$ for ${\textstyle c={\sqrt {a^{2}-b^{2}}}}$. The eccentricity is:

$e={\frac {c}{a}}={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}$

A circle has $e=0$, while increasingly squashed ellipses have values of $e$ closer and closer to $1$. The area of an ellipse is:

$A=\pi ab$

The perimeter is not analytic but can be approximated very roughly by:

$P\approx \pi (a+b)$

Which yields:

${\begin{alignedat}{2}\mathrm {PRA} &={\frac {Pr}{A}}\\&={\frac {P\left({\sqrt {\frac {A}{\pi }}}\right)}{A}}\\&\approx {\frac {\pi (a+b)}{\pi ab}}\left({\sqrt {\frac {\pi ab}{\pi }}}\right)\\&\approx {\frac {(a+b)}{ab}}{\sqrt {ab}}\\&\approx {\frac {(a+b)}{\sqrt {ab}}}\\\end{alignedat}}$

One can establish a relationship between eccentricity and PrA by setting $b=1$ and considering $a\in [1,\infty ]$:

${\begin{alignedat}{2}\mathrm {PRA} &\approx {\frac {(a+1)}{\sqrt {a}}}\\e&={\sqrt {1-{\frac {1}{a^{2}}}}}\end{alignedat}}$

In particular:

${\begin{alignedat}{2}{\frac {(a+1)}{\sqrt {a}}}&=\mathrm {PRA} \\(a+1)^{2}&=\mathrm {PRA} ^{2}a\\a^{2}+2a+1-\mathrm {PRA} ^{2}a&=0\\(1)a^{2}+(2-\mathrm {PRA} ^{2})a+(1)&=0\\\end{alignedat}}$

From the quadratic equation:

${\begin{alignedat}{2}a&={\frac {-(2-\mathrm {PRA} ^{2})\pm {\sqrt {(2-\mathrm {PRA} ^{2})^{2}-4(1)(1)}}}{2(1)}}\\&={\frac {1}{2}}\left(-2+\mathrm {PRA} ^{2}\pm {\sqrt {(2-\mathrm {PRA} ^{2})^{2}-4}}\right)\\&={\frac {1}{2}}\left(\mathrm {PRA} ^{2}-2\pm {\sqrt {(2-\mathrm {PRA} ^{2})^{2}-4}}\right)\\&={\frac {1}{2}}\left(\mathrm {PRA} ^{2}-2\pm {\sqrt {4-4\mathrm {PRA} ^{2}+\mathrm {PRA} ^{4}-4}}\right)\\&={\frac {1}{2}}\left(\mathrm {PRA} ^{2}-2\pm {\sqrt {\mathrm {PRA} ^{4}-4\mathrm {PRA} ^{2}}}\right)\\&={\frac {1}{2}}\left(\mathrm {PRA} ^{2}-2\pm \mathrm {PRA} {\sqrt {\mathrm {PRA} ^{2}-4}}\right)\\\end{alignedat}}$

Since $a\to \infty$ as $P\to \infty$, we select the positive branch.

${\begin{alignedat}{2}a&={\frac {1}{2}}\left(\mathrm {PRA} ^{2}-2+\mathrm {PRA} {\sqrt {\mathrm {PRA} ^{2}-4}}\right)\\a^{2}&={\frac {1}{4}}\left(\mathrm {PRA} ^{2}-2+\mathrm {PRA} {\sqrt {\mathrm {PRA} ^{2}-4}}\right)^{2}\\\end{alignedat}}$

And so:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} e & = \sqrt{1 - \frac{1}{a^2}} \\ & = \sqrt{1 - \frac{4}{\left( \mathrm{PRA}^2-2 + \mathrm{PRA} \sqrt{\mathrm{PRA}^2 - 4} \right)^2}} \\ \end{alignat} }

We can convert into a width:height ratio ($a/b$) as:

${\begin{alignedat}{2}{\frac {a}{b}}&={\sqrt {1-e^{2}}}\\&={\frac {1}{2}}\left(\mathrm {PRA} ^{2}-2+\mathrm {PRA} {\sqrt {\mathrm {PRA} ^{2}-4}}\right)\\\end{alignedat}}$