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:

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} \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{alignat} }

From the quadratic equation:

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} 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{alignat} }

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

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} 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{alignat} }

And so:

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

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

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