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:
$|AB|^{2}+|AC|^{2}=|CB|^{2}$
$\Rightarrow r_{1}^{2}+r_{1}^{2}=(2\cdot r_{0})^{2}$
$\Rightarrow 2\cdot r_{1}^{2}=4\cdot r_{0}^{2}$
$\Rightarrow r_{1}^{2}=2\cdot r_{0}^{2}$
$\Rightarrow r_{1}={\sqrt {2\cdot r_{0}^{2}}}$
$\Rightarrow r_{1}={\sqrt {2}}\cdot r_{0}$

General case

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

Black-and-White version

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: $1.4792816$