File:FS QR dia.png

Summary[edit]

General case[edit]

Segments in the general case[edit]

0) Base is the square ${\displaystyle [ABCD]}$ of side length ${\displaystyle a_{0}}$.
1) The side of the inscribed right triangle ${\displaystyle \triangle {\overline {ABD}}}$ has the length ${\displaystyle a_{1}=a_{0}}$

Perimeters in the general case[edit]

0) Perimeter of base square ${\displaystyle P_{0}=4\cdot a_{0}}$
1) Perimeter of the triangle ${\displaystyle P_{1}=a_{1}+{\sqrt {2}}\cdot a_{1}+a_{1}=(2+{\sqrt {2}})\cdot a_{1}=(2+{\sqrt {2}})\cdot a_{0}}$

Areas in the general case[edit]

0) Area of the base square ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed triangle ${\displaystyle A_{1}={\frac {1}{2}}\cdot a_{1}^{2}={\frac {1}{2}}\cdot a_{0}^{2}}$

Centroids in the general case[edit]

Centroid positions are measured from centroid point of the base shape
0) Centroid positions of the base square: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed triangle: ${\displaystyle x_{1}=y_{1}=-{\frac {1}{6}}\cdot a_{0}\quad }$ ,see Calculation 1

Normalised case[edit]

In the normalised case the area of the base is set to 1.
${\displaystyle ||ABCD||=1\Rightarrow a_{0}^{2}=1\Rightarrow a_{0}=1}$

Segments in the normalised case[edit]

0) Segment of the base square ${\displaystyle a_{0}=1}$
1) Segment of the inscribed triangle ${\displaystyle a_{1}=a_{0}=1}$

Perimeters in the normalised case[edit]

0) Perimeter of base square ${\displaystyle P_{0}=4}$
1) Perimeter of the inscribed triangle ${\displaystyle P_{1}=(2+{\sqrt {2}})\cdot a_{0}=2+{\sqrt {2}}\quad \approx 3.4142136}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 7.4142136}$

Areas in the normalised case[edit]

0) Area of the base square ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle ${\displaystyle A_{1}={\frac {1}{2}}\cdot a_{0}=0.5}$

Centroids in the normalised case[edit]

Centroid positions are measured from the centroid point of the base shape
0) Centroid positions of the base square: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed circle: ${\displaystyle x_{1}=y_{1}=-{\frac {1}{6}}\quad \approx 0.167}$

Distances of normalised centroids[edit]

The distance between the centroid of the base element and the centroid of the circle is:
${\displaystyle d={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}\approx 0.2357023}$

Identifying number[edit]

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.

${\displaystyle decimalpart(7.4142136+0.2357023)=decimalpart(7.6499158)=0.6499158}$

So the identifying number is: ${\displaystyle 1.6499158}$

Calculations[edit]

Calculation 1[edit]

${\displaystyle \quad x_{1}=|AX_{1}|-|AX_{0}|}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {a_{1}}{3}}-{\frac {a_{0}}{2}}}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {a_{0}}{3}}-{\frac {a_{0}}{2}}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {1}{3}}-{\frac {1}{2}}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {2}{6}}-{\frac {3}{6}}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=-{\frac {1}{6}}\cdot a_{0}}$

Licensing[edit]

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