User:Hans G. Oberlack/FS H

The semicircle as base element.

General case

Segments in the general case

The radius of the semicircle $=r_{0}$

Perimeters in the general case

Perimeter of base semicircle$P_{0}=2\cdot r_{0}+\pi \cdot r_{0}=(2+\pi )\cdot r_{0}$

Areas in the general case

Area of the base semicircle $A_{0}={\frac {\pi \cdot r_{0}^{2}}{2}}$

Centroids in the general case

By definition the centroid points of a base shape are x_0=0 and y_0=0. Relatively is the lower left point of the of the base of the semicircle at
$\quad x_{L}=-r_{0}\quad y_{L}=-{\frac {4\cdot r_{0}}{3\cdot \pi }}$

Normalised case

In the normalised case the area of the base semicircle is set to 1.
So $A_{0}={\frac {\pi \cdot r_{0}^{2}}{2}}=1\quad \Rightarrow r_{0}^{2}={\frac {2}{\pi }}\quad \Rightarrow r_{0}={\sqrt {\frac {2}{\pi }}}$

Segments in the normalised case

Segment of the base semicircle $r_{0}={\sqrt {\frac {2}{\pi }}}=0.79788...$

Perimeter in the normalised case

Perimeter of base semicircle: $P_{0}=(2+\pi )\cdot r_{0}=(2+\pi )\cdot {\sqrt {\frac {2}{\pi }}}=4.102397...$

Area in the normalised case

Area of the base semicircle is by definition $A_{0}=1$

Centroids in the normalised case

Positions of the lower left point of the semicircle: $x_{L}=-r_{0}=-{\sqrt {\frac {2}{\pi }}}=-0.79788...\quad y_{L}=-{\frac {4\cdot r_{0}}{3\cdot \pi }}=-{\frac {4}{3\cdot \pi }}\cdot {\sqrt {\frac {2}{\pi }}}=-{\sqrt {\frac {32}{9\cdot \pi ^{3}}}}=-0.33863...$

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.

$decimalpart(4.1023974...+0)=decimalpart(4.1023974...)=.1023974...$

So the identifying number is: $0.1023974$