I am looking for sliver polygons (term used in geographic information systems for long thin 2D surfaces) and am using the following formula to identify which polygons have a smaller area to circumference ratio (a.k.a. the thinness ratio): $$ \frac{4 \cdot \pi \cdot \text{area}}{\text{perimeter}^2} $$ That much I understand. But what is not fully clear, is the $4 \cdot \pi$ bit and why the length has to be squared. Can someone explain this in simple terms?
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- The length are squared so that the ratio is invariant under scaling. 2) The $4\pi$ is a normalization factor to set the ratio to $1$ for a circle. In fact, by isoperimetric inequality, all shapes has ratio $\le 1$ and $= 1$ when and only when the shape is a circle.
– achille hui Jun 23 '15 at 14:23 -
This has been cross-posted to http://gis.stackexchange.com/q/151939 – PolyGeo Jun 24 '15 at 09:51
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The reasoning behind the Thinness ratio is that we compare the measured area with the theoretical area of a circle with the same perimeter (circumference). If they are similar their ratio will approach to one.
$$ Th = \frac{A_{measured}}{A_{theoretical}}\tag{1}$$
So the first question we have to ask is, what's the theoretical area of a circle with the same perimeter? The circle area is: $$A_{theoretical}=\pi\times r^{2} \tag{2}$$
But we don't have the radius however we know that the circumference (aka perimeter) is equal to: $$ (p)erimeter = 2\pi\times r$$ Hence: $$ r = \frac{p}{2\pi}$$ Substututing in equation 2: $$A_{theoretical}=\pi\times (\frac{p}{2\pi})^{2} = \frac{p^{2}}{4 \pi}$$ Finnaly substituting in 1: $$ Th = \frac{4\pi \times A_{measured}}{p^{2}}$$
G M
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