User:Hans G. Oberlack/FS CV

Shows the largest quarter circle within a circle.

Elements
Base is the circle of given radius $$r_0$$ around point $$D$$ Inscribed is the largest possible quarter circle.

In order to find radius $$r_1$$ of the quarter circle, the following reasoning is used: Since point $$D$$ is the center of the circle we have: $$|DA| = |DB| = |DC| = r_0$$ The points $$A$$,$$B$$ and $$C$$ form a rectangular, isosceles triangle with: $$|AB| =|AC| = r_1$$ and $$|BC|=2 \cdot r_0$$ Applying the Pythagorean theorem on $$\triangle ABC$$ gives: $$|AB|^2+|AC|^2=|CB|^2$$ $$\Rightarrow r_1^2+r_1^2=(2 \cdot r_0)^2$$ $$\Rightarrow 2 \cdot r_1^2=4 \cdot r_0^2$$ $$\Rightarrow r_1^2=2 \cdot r_0^2$$ $$\Rightarrow r_1=\sqrt {2 \cdot r_0^2}$$ $$\Rightarrow r_1=\sqrt 2 \cdot r_0$$

Segments in the general case
0) The radius of the base circle $$ = r_0$$ 1) Radius of the quarter circle $$r_1 = \sqrt 2 \cdot r_0$$

Perimeters in the general case
0) Perimeter of base circle $$P_0 = 2 \cdot \pi \cdot r_0$$ 1) Perimeter of the quarter circle $$P_1= 2 \cdot r_1 + \frac {\pi \cdot r_1} {2} = (2  + \frac {\pi} {2}) \cdot r_1 = (2  + \frac {\pi} {2}) \cdot \sqrt 2 \cdot r_0=\frac {4+\pi} {2} \cdot \sqrt 2 \cdot r_0=(4+\pi) \cdot \frac {\sqrt 2} {2} \cdot r_0=(4+\pi) \cdot \frac {1} {\sqrt 2} \cdot r_0$$ $$\Rightarrow P_1=\frac {4+\pi} {\sqrt 2} \cdot r_0$$

Areas in the general case
0) Area of the base circle $$ A_0= \pi \cdot r_0^2$$ 1) Area of the inscribed quarter circle $$A_1 = \frac { \pi \cdot r_1^2} {4} = \frac{\pi}{4} \cdot (\sqrt 2 \cdot r_0)^2 = \frac{\pi}{4} \cdot 2 \cdot r_0^2  = \frac{\pi}{2} \cdot r_0^2 =\frac{1}{2} \cdot A_0$$

Centroids in the general 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 quarter circle: $$x_1=\frac {4 \cdot r_1}{3 \cdot \pi} - \cdot r_0 \cdot cos(45)= \frac {4 \cdot \sqrt 2 \cdot r_0} {3 \cdot \pi} - \cdot r_0 \cdot \frac {1} {\sqrt 2}= \left( \frac {4 \cdot \sqrt 2 } {3 \cdot \pi} - \frac {1} {\sqrt 2} \right) \cdot r_0= \frac {4 \cdot 2 - 3 \cdot \pi} {3 \cdot \pi \cdot \sqrt 2} \cdot r_0= \frac {8 - 3 \pi} {3 \pi \sqrt 2} \cdot r_0$$

$$y_1= x_1 = \frac {8 - 3 \pi} {3 \pi \sqrt 2} \cdot r_0 $$

Normalised case
In the normalised case the area of the base is set to 1. $$A_0=1 \quad \Rightarrow \pi \cdot r_0^2=1 \quad \Rightarrow r_0^2=\frac {1} {\pi} \quad \Rightarrow r_0=\frac {1} {\sqrt \pi}$$

Segments in the normalised case
0) Radius of the base circle $$ r_0 =\frac {1} {\sqrt \pi} = 0.56... $$ 1) Radius of the inscribed quarter circle $$ r_1=\sqrt 2 \cdot r_0 =\sqrt 2 \cdot \frac {1} {\sqrt \pi} = \sqrt \frac {2} {\pi} = 0.79...$$

Perimeters in the normalised case
0) Perimeter of base square $$ P_0= 2 \cdot \pi \cdot r_0 = 2 \cdot \pi \cdot \frac {1} {\sqrt \pi} = 2 \cdot \sqrt \pi =3.5449077...$$ 1) Perimeter of the inscribed quarter circle $$P_1=\frac {4+\pi} {\sqrt 2} \cdot r_0=\frac {4+\pi} {\sqrt 2} \cdot \frac {1} {\sqrt \pi}= \frac {4+\pi} {\sqrt {2 \cdot \pi}} = 2.8490832... $$ S) Sum of perimeters $$P_s = P_0 + P_1 =6.3939909...$$

Areas in the normalised case
0) Area of the base square $$ A_0= 1$$ 1) Area of the inscribed quarter circle $$ A_1= \frac{\pi}{2} \cdot r_0^2 = \frac{\pi}{2} \cdot (\frac {1} {\sqrt \pi})^2 = \frac{\pi}{2} \cdot \frac {1} { \pi} = 0.5$$

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 quarter circle:$$x_1=\frac {8 - 3 \pi} {3 \pi \sqrt 2} \cdot r_0 =\frac {8 - 3 \pi} {3 \pi \sqrt 2} \cdot \frac{1}{\sqrt \pi} = -0.0603095...\quad   y_1=x_1= -0.0603095...$$

Distances of centroids
The distance between the centroid of the base element and the centroid of the quarter circle is: $$d =\sqrt {(x_0-x_1)^2+(y_0-y_1)^2} = 0.0852905...$$

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. $$decimalpart (6.3939910...+0.0852906...)=decimalpart(6.4792816...)=.4792816$$ So the identifying number is: $$1.4792816$$