User:Hans G. Oberlack/QV 1.6776922

Shows the largest quarter circle within a square.

Elements
Base is the square $$[ABCD]$$ of side length s. Inscribed is the largest possible quarter circle of radius r.

Segments in the general case
0) The side length of the base square $$ = s$$ 1) Radius of the quarter circle $$r = s$$

Perimeters in the general case
0) Perimeter of base square$$ = 4 \cdot s$$ 1) Perimeter of the quarter circle$$= 2 \cdot r + \frac {\pi \cdot r} {2} = (2 + \frac {\pi} {2}) \cdot r = (2  + \frac {\pi} {2}) \cdot s$$

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

Centroids in the general case
Centroid positions are measured from the lower left point of the square. 0) Centroid positions of the base square: $$x_c=\frac {s}{2} \quad   y_c=\frac {s}{2}$$ 1) Centroid positions of the inscribed quarter circle: $$x_c=\frac {4 \cdot r}{3 \cdot \pi}=\frac {4 \cdot s}{3 \cdot \pi} \quad   y_c=\frac {4 \cdot r}{3 \cdot \pi}=\frac {4 \cdot s}{3 \cdot \pi}$$

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 quarter circle $$ r = s = 1$$

Perimeters in the normalised case
0) Perimeter of base square $$ P_0= 4 $$ 1) Perimeter of the inscribed quarter circle $$P_1 = (2 + \frac {\pi} {2}) \cdot 1 =3.570796...$$ S) Sum of perimeters $$P_s = (6 + \frac {\pi} {2})  =7.570796...$$

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

Centroids in the normalised case
Centroid positions are measured from the lower left point of the square. 0) Centroid positions of the base square: $$x_0=\frac {1}{2} \quad   y_0=\frac {1}{2}$$ 1) Centroid positions of the inscribed quarter circle:$$x_1=\frac {4 \cdot 1}{3 \cdot \pi}= 0.42441318... \quad  y_1=\frac {4 \cdot 1}{3 \cdot \pi}= 0.42441318...$$

Distances of centroids
The distance between the centroid of the base element and the centroid of the quarter circle is: $$d =\sqrt {(x_0-x_1)^2+(y_0-y_1)^2} = 0,1068959...$$

Identifying number
Apart of the base element there is only one 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. $$decimalpart (7.5707963...+0.1068959...)=decimalpart(7.6776922...)=.6776922$$ So the identifying number is: $$1.6776922$$