# Difference between revisions of "PrA"

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

{\displaystyle {\begin{alignedat}{2}\mathrm {PRA} ={\frac {Pr}{A}}\end{alignedat}}}

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

{\displaystyle {\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:

{\displaystyle {\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:

{\displaystyle {\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 ${\displaystyle \mathrm {PRA} \to \infty }$.

### Ellipse

If the object is an ellipse, with equation:

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

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

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

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

${\displaystyle A=\pi ab}$

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

${\displaystyle P\approx \pi (a+b)}$

Which yields:

{\displaystyle {\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 ${\displaystyle b=1}$ and considering ${\displaystyle a\in [1,\infty ]}$:

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

In particular:

{\displaystyle {\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}}}

{\displaystyle {\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 ${\displaystyle a\to \infty }$ as ${\displaystyle P\to \infty }$, we select the positive branch.
{\displaystyle {\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}}}
{\displaystyle {\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}}}