User:Hans G. Oberlack/FS V

The Quartercircle as base element.

General case

Segments in the general case

The radius of the quarter circle $=r$

Perimeter in the general case

Perimeter of base quarter circle $P_{0}=2\cdot r+{\frac {\pi \cdot r}{2}}=(2+{\frac {\pi }{2}})\cdot r$

Areas in the general case

Area of the base quarter circle $A_{0}={\frac {\pi \cdot r^{2}}{4}}$

Centroids in the general case

By definition the centroid points of a base shape are $x_{0}=0\quad y_{0}=0$. Relatively is the lower left point of the base of the quarter circle at:
$x_{c}=-{\frac {4\cdot r}{3\cdot \pi }}\quad y_{c}=-{\frac {4\cdot r}{3\cdot \pi }}$

Normalised case

Area in the normalised case

In the normalised case the area of the base quarter circle is set $A_{0}=1$

Segments in the normalised case

Since $A_{0}={\frac {\pi \cdot r^{2}}{4}}=1\quad \Rightarrow r^{2}={\frac {4}{\pi }}\quad \Rightarrow r={\sqrt {\frac {4}{\pi }}}={\frac {2}{\sqrt {\pi }}}=1.128379...$

Perimeter in the normalised case

Perimeter of base quarter circle $P_{0}=(2+{\frac {\pi }{2}})\cdot r=(2+{\frac {\pi }{2}})\cdot {\frac {2}{\sqrt {\pi }}}=4.029212...$

Centroids in the normalised case

Positions of the lower left corner of base quarter circle: $x_{L}=-{\frac {4\cdot r}{3\cdot \pi }}=-{\frac {4}{3\cdot \pi }}\cdot {\sqrt {\frac {4}{\pi }}}=-{\sqrt {\frac {64}{9\cdot \pi ^{3}}}}=-0.479...\quad y_{L}=x_{L}=-0.479...$

Identifying number

Apart of the base element there is no other shape allocated. Therefore the integer part of the identifying number is 0.
The decimal part of the identifying number is the decimal part of the sum of the perimeters and the distances of the centroids in the normalised case.

$decimalpart(4.0292122...+0)=decimalpart(4.0292122...)=.0292122...$

So the identifying number is: $0.0292122$