Difference between revisions of "PrA"

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(Created page with "'''PrA''' is a simple ad-hoc parameter to define the "non-circularity" or [https://en.wikipedia.org/wiki/Ellipse eccentricity] of a 2D object. This quantity is simply: :<math>...")
 
 
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\end{alignat}
 
\end{alignat}
 
</math>
 
</math>
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And thus any non-circular object will have a larger PrA.
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And thus any non-circular object will have a larger PrA. An infinitely eccentric object would have <math>\mathrm{PRA} \to \infty</math>.
 +
===Ellipse===
 +
If the object is an [https://en.wikipedia.org/wiki/Ellipse ellipse], with equation:
 +
: <math>\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1</math>
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Then the width is <math>2a</math> and height <math>2b</math> (we assume <math>a \ge b</math>), the foci are <math>(\pm c, 0)</math> for <math display="inline">c = \sqrt{a^2-b^2}</math>. The eccentricity is:
 +
: <math>e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}</math>
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A circle has <math>e=0</math>, while increasingly squashed ellipses have values of <math>e</math> closer and closer to <math>1</math>. The area of an ellipse is:
 +
: <math>A = \pi a b</math>
 +
The perimeter is not analytic but can be approximated very roughly by:
 +
: <math>P \approx \pi (a +b)</math>
 +
Which yields:
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:<math>
 +
\begin{alignat}{2}
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\mathrm{PRA} & = \frac{Pr}{A} \\
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  & = \frac{P \left( \sqrt{\frac{A}{\pi}} \right) }{A} \\
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  & \approx \frac{\pi(a+b) }{\pi a b} \left( \sqrt{\frac{\pi a b}{\pi}} \right) \\
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  & \approx \frac{(a+b) }{ a b} \sqrt{a b} \\
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  & \approx \frac{(a+b) }{ \sqrt{a b} } \\
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\end{alignat}
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</math>
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One can establish a relationship between eccentricity and PrA by setting <math>b=1</math> and considering <math>a \in [1, \infty]</math>:
 +
:<math>
 +
\begin{alignat}{2}
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\mathrm{PRA} & \approx \frac{(a+1) }{ \sqrt{a} } \\
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e & = \sqrt{1 - \frac{1}{a^2}}
 +
\end{alignat}
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</math>
 +
In particular:
 +
:<math>
 +
\begin{alignat}{2}
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\frac{(a+1) }{ \sqrt{a} } & = \mathrm{PRA}  \\
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(a+1)^2 & = \mathrm{PRA}^2 a  \\
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a^2+2a+1- \mathrm{PRA}^2 a & = 0  \\
 +
(1)a^2+(2-\mathrm{PRA}^2 )a+(1) & = 0  \\
 +
\end{alignat}
 +
</math>
 +
From the quadratic equation:
 +
:<math>
 +
\begin{alignat}{2}
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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}
 +
</math>
 +
Since <math>a \to \infty</math> as <math>P \to \infty</math>, we select the positive branch.
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:<math>
 +
\begin{alignat}{2}
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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}
 +
</math>
 +
And so:
 +
:<math>
 +
\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}
 +
</math>
 +
We can convert into a width:height ratio (<math>a/b</math>) as:
 +
:<math>
 +
\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}
 +
</math>

Latest revision as of 10:55, 31 May 2022

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

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} r = \sqrt{\frac{A}{\pi}} \end{alignat} }

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

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} \\ & = \frac{(2 \pi r)(r)}{\pi r^2} \\ & = 2 \end{alignat} }

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

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} \geq 2 \end{alignat} }

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

Ellipse

If the object is an ellipse, with 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 \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1}

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

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

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

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 = \pi a b}

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

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 P \approx \pi (a +b)}

Which yields:

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} \\ & = \frac{P \left( \sqrt{\frac{A}{\pi}} \right) }{A} \\ & \approx \frac{\pi(a+b) }{\pi a b} \left( \sqrt{\frac{\pi a b}{\pi}} \right) \\ & \approx \frac{(a+b) }{ a b} \sqrt{a b} \\ & \approx \frac{(a+b) }{ \sqrt{a b} } \\ \end{alignat} }

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

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} & \approx \frac{(a+1) }{ \sqrt{a} } \\ e & = \sqrt{1 - \frac{1}{a^2}} \end{alignat} }

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 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 \to \infty} 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 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:

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 (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} }
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