User:Hans G. Oberlack/FS V

The Quartercircle as base element.

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: $$x_c=-\frac {4 \cdot r} {3 \cdot \pi} \quad   y_c=- \frac {4 \cdot r} {3 \cdot \pi}$$

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: $$0.0292122$$