User:Hans G. Oberlack/FS H



The semicircle as base element.

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: $$0.1023974$$