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: $1.1415927$