User:Hans G. Oberlack/QC 1.1415927

Shows the largest circle within a square.

General case
Base is the square $$[ABCD]$$ of side length s. The radius r of the inscribed circle has the length $$ \frac {s}{2}$$

Segments in the general case
0) The side length of the base square $$ = s$$ 1) Radius of the inscribed circle $$r_1 = \frac{s}{2}$$

Perimeters in the general case
0) Perimeter of base square $$P_0 = 4 \cdot s$$ 1) Perimeter of the circle $$P_1= 2 \cdot \pi \cdot r_1 = \pi \cdot s$$

Areas in the general case
0) Area of the base square $$A_0 = s^2$$ 1) Area of the inscribed circle $$A_1 = \pi \cdot r_1^2 = \pi \cdot (\frac{s}{2})^2= \pi \cdot \frac{s^2}{4} = \frac{\pi}{4} \cdot s^2 =\frac{\pi}{4} \cdot A_0 $$

Centroids in the general case
Centroid positions are measured from centroid point of the base shape 0) Centroid positions of the base square: $$x_0=0 \quad   y_0=0$$ 1) Centroid positions of the inscribed circle: $$x_1=0 \quad   y_1=0$$

Normalised case
In the normalised case the area of the base is set to 1. $$||ABCD||=1 \Rightarrow s^2=1 \Rightarrow s=1 $$

Segments in the normalised case
0) Segment of the base square $$ s = 1 $$ 1) Segment of the inscribed circle $$ r_1 = \frac{1}{2} = 0.5$$

Perimeters in the normalised case
0) Perimeter of base square $$ P_0= 4 $$ 1) Perimeter of the inscribed circle $$P_1= 2 \cdot \pi \cdot r_1 = 2 \cdot \pi \cdot \frac {1}{2} = \pi =3.1415926... $$

Areas in the normalised case
0) Area of the base square $$ A_0= 1$$ 1) Area of the inscribed circle $$ A_1= \pi \cdot r_1^2 = \pi \cdot (\frac {1}{2})^2 = \frac {\pi}{4} = 0.785398... $$

Centroids in the normalised case
Centroid positions are measured from the centroid point of the base shape 0) Centroid positions of the base square: $$x_0=0 \quad   y_0=0$$ 1) Centroid positions of the inscribed circle: $$x_1=0 \quad   y_1=0$$

Identifying number
Apart of the base element there is only one other shape allocated. Therefore the integer part of the identifying number is 1. 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. $$decimalpart (3.1415927...+0)=decimalpart(3.1415927...)=0.1415927...$$ So the identifying number is: $$1.1415927$$