File:FS CHQ dia.png

Summary[edit]

The circle as base element. Inscribed is the largest semicircle. Also inscribed is the largest square.

General case[edit]

Segments in the general case[edit]

0) Radius of the circle ${\displaystyle r_{0}}$
1) Radius of the inscribed semicircle: ${\displaystyle r_{1}=r_{0}\quad }$
2) Side length of the inscribed square: ${\displaystyle a_{2}={\frac {2}{\sqrt {5}}}\cdot r_{0}\quad \approx 0.894\cdot r_{0}\quad }$, see Calculation 1

Perimeters in the general case[edit]

0) Perimeter of base circle: ${\displaystyle P_{0}=2\pi \cdot r_{0}\quad \approx 6.238\cdot r_{0}}$
1) Perimeter of the inscribed semicircle: ${\displaystyle P_{1}=2\cdot r_{1}+\pi \cdot r_{1}=(2+\pi )\cdot r_{0}\quad \approx 5.142\cdot r_{0}}$
2) Perimeter of the inscribed square: ${\displaystyle P_{2}=4\cdot a_{2}=4\cdot {\frac {2}{\sqrt {5}}}\cdot r_{0}={\frac {8}{\sqrt {5}}}\cdot r_{0}\quad \approx 3.578\cdot r_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=2\pi \cdot r_{0}+(2+\pi )\cdot r_{0}+{\frac {8}{\sqrt {5}}}\cdot r_{0}=(3\pi +2+{\frac {8}{\sqrt {5}}})\cdot r_{0}\quad \approx 15.002\cdot r_{0}}$

Areas in the general case[edit]

0) Area of the base circle ${\displaystyle A_{0}=\pi \cdot r_{0}^{2}}$
1) Area of the inscribed semicircle: ${\displaystyle A_{1}={\frac {\pi \cdot r_{1}^{2}}{2}}={\frac {\pi }{2}}\cdot r_{0}^{2}={\frac {1}{2}}\cdot A_{0}}$
2) Area of the inscribed square: ${\displaystyle A_{2}=a_{2}^{2}={\frac {4}{5}}\cdot r_{0}^{2}=0.800\cdot r_{0}^{2}={\frac {4}{5\pi }}\cdot A_{0}}$

Covered surface of base shape: ${\displaystyle C_{b}={\frac {A_{1}+A_{2}}{A_{0}}}={\frac {{\frac {1}{2}}A_{0}+{\frac {4}{5\pi }}A_{0}}{A_{0}}}={\frac {1}{2}}+{\frac {4}{5\pi }}={\frac {5\pi +8}{10\pi }}\quad \approx 75.5\%}$

Centroids in the general case[edit]

1) Centroids as graphically displayed[edit]

0) Centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed semicircle: ${\displaystyle S_{1}=\left({\frac {4\cdot r_{1}}{3\pi }}\right){\frac {1}{\sqrt {2}}}\cdot (1+i)=\left({\frac {2{\sqrt {2}}}{3\pi }}\right)\cdot (1+i)\cdot r_{0}\quad \approx 0.3\cdot (1+i)\cdot r_{0}\quad }$
2) Centroid position of the inscribed square: ${\displaystyle S_{2}=-{\frac {1}{\sqrt {10}}}\cdot (1+i)\cdot r_{0}\quad \approx -0.316\cdot (1+i)\cdot r_{0}\quad }$, see Calculation 2
W) Weighted centroid: ${\displaystyle S_{w}={\frac {4}{3}}\cdot {\frac {5{\sqrt {5}}-6}{(15\pi +8)\cdot {\sqrt {10}}}}\cdot (1+i)\cdot r_{0}\quad \approx 0.04\cdot (1+i)\cdot r_{0}}$, see Calculation 3

2) Orientated centroids[edit]

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed semicircle: ${\displaystyle S_{1}=\left({\frac {4\cdot r_{1}}{3\pi }}\right)\cdot (0-i)=\left(0-{\frac {4}{3\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.3i)\cdot r_{0}\quad }$
2) Centroid position of the inscribed square: ${\displaystyle S_{2}=0+{\frac {a_{2}}{2}}\cdot i=(0+{\frac {1}{\sqrt {5}}}\cdot i)\cdot r_{0}\quad \approx (0+0.447i)\cdot r_{0}\quad }$

Normalised case[edit]

In the normalised case the area of the base circle is set to 1.
So ${\displaystyle A_{0}=\pi \cdot r_{0}^{2}=1\quad \Rightarrow r_{0}^{2}={\frac {1}{\pi }}\quad \Rightarrow r_{0}={\sqrt {\frac {1}{\pi }}}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564}$

Segments in the normalised case[edit]

0) Radius of the base circle ${\displaystyle r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564}$
1) Radius inscribed semicircle: ${\displaystyle r_{1}=r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564}$
2) Side length of the inscribed square: ${\displaystyle a_{2}={\frac {2}{\sqrt {5}}}\cdot {\frac {1}{\sqrt {\pi }}}={\frac {2}{\sqrt {5\pi }}}\quad \approx 0.505}$

Perimeters in the normalised case[edit]

0) Perimeter of base circle: ${\displaystyle P_{0}=2\pi \cdot r_{0}=2\pi \cdot {\frac {1}{\sqrt {\pi }}}=2{\sqrt {\pi }}\quad \approx 3.5449077}$
1) Perimeter of the inscribed semicircle:${\displaystyle P_{1}=(2+\pi ){\frac {1}{\sqrt {\pi }}}\quad \approx 2.9008330}$
2) Perimeter of the inscribed square: ${\displaystyle P_{2}=4\cdot a_{2}={\frac {8}{\sqrt {5}}}\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 2.0185060}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}=(3\pi +2+{\frac {8}{\sqrt {5}}})\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 8.4642467}$

Areas in the normalised case[edit]

0) Area of the base circle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed semicircle: ${\displaystyle A_{1}={\frac {\pi }{2}}\cdot \left({\frac {1}{\sqrt {\pi }}}\right)^{2}={\frac {\pi }{2}}\cdot {\frac {1}{\pi }}\quad =0.500}$
2) Area of the inscribed square: ${\displaystyle A_{2}=a_{2}^{2}={\frac {4}{5}}\cdot \left({\frac {1}{\sqrt {\pi }}}\right)^{2}={\frac {4}{5\pi }}\quad \approx 0.255}$

Centroids in the normalised case[edit]

1) Centroids as graphically displayed[edit]

0) Centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$

1) Centroid position of the inscribed semicircle: ${\displaystyle S_{1}=\left({\frac {2{\sqrt {2}}}{3\pi }}\right)\cdot (1+i)\cdot {\frac {1}{\sqrt {\pi }}}=\left({\frac {2{\sqrt {2}}}{3\pi {\sqrt {\pi }}}}\right)\cdot (1+i)\quad \approx 0.169\cdot (1+i)\quad }$

2) Centroid position of the inscribed square: ${\displaystyle S_{2}=-{\frac {1}{\sqrt {10}}}\cdot (1+i)\cdot {\frac {1}{\sqrt {\pi }}}=-{\frac {1}{\sqrt {10\pi }}}\cdot (1+i)\quad \approx -0.178\cdot (1+i)\quad }$

2) Orientated centroids[edit]

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the ma-thematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the inscribed semicircle: ${\displaystyle S_{1}=\left(0-{\frac {4}{3\pi }}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left(0-{\frac {4}{3\pi {\sqrt {\pi }}}}\cdot i\right)\quad \approx (0-0.239i)\quad }$
2) Centroid position of the inscribed square: ${\displaystyle S_{2}=(0+{\frac {1}{\sqrt {5}}}\cdot i)\cdot {\frac {1}{\sqrt {\pi }}}=(0+{\frac {1}{\sqrt {5\pi }}}\cdot i)\quad \approx (0+0.252i)\quad }$

Calculations[edit]

Given elements and equations[edit]

(1) ${\displaystyle |AS_{0}|=|BS_{0}|=|CS_{0}|=r_{0}\quad }$, since A,B and C are points on the perimeter of the circle
(2) ${\displaystyle |CD|=|DE|=|EF|=|FC|=a_{2}\quad }$, since CDEF is a square
(3) ${\displaystyle |DS_{0}|=|ES_{0}|=|CG|=|FG|={\frac {a_{2}}{2}}\quad }$, for symmetry reasons
(4) ${\displaystyle |GS_{0}|=a_{2}\quad }$,applying equations (2) and (3)

Calculation 1[edit]

${\displaystyle \quad |GS_{0}|^{2}+|CG|^{2}=|CS_{0}|^{2}\quad }$, applying Pythagorean theoreme
${\displaystyle \quad \Leftrightarrow a_{2}^{2}+|CG|^{2}=|CS_{0}|^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow a_{2}^{2}+\left({\frac {a_{2}}{2}}\right)^{2}=|CS_{0}|^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow a_{2}^{2}+\left({\frac {a_{2}}{2}}\right)^{2}=r_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow a_{2}^{2}+{\frac {a_{2}^{2}}{4}}=r_{0}^{2}\quad }$
${\displaystyle \quad \Leftrightarrow {\frac {5}{4}}\cdot a_{2}^{2}=r_{0}^{2}\quad }$
${\displaystyle \quad \Leftrightarrow a_{2}^{2}={\frac {4}{5}}\cdot r_{0}^{2}\quad }$
${\displaystyle \quad \Leftrightarrow a_{2}={\frac {2}{\sqrt {5}}}\cdot r_{0}\quad }$

Calculation 2[edit]

${\displaystyle S_{2}=S_{0}-{\frac {|GS_{0}|}{2}}{\frac {1}{\sqrt {2}}}\cdot (1+i)}$
${\displaystyle \Leftrightarrow S_{2}=S_{0}-{\frac {a_{2}}{2}}{\frac {1}{\sqrt {2}}}\cdot (1+i)\quad }$, applying equation (4)
${\displaystyle \Leftrightarrow S_{2}=S_{0}-{\frac {2}{2{\sqrt {5}}}}\cdot r_{0}{\frac {1}{\sqrt {2}}}\cdot (1+i)\quad }$
${\displaystyle \Leftrightarrow S_{2}=S_{0}-{\frac {1}{\sqrt {5}}}{\frac {1}{\sqrt {2}}}\cdot (1+i)\cdot r_{0}\quad }$
${\displaystyle \Leftrightarrow S_{2}=S_{0}-{\frac {1}{\sqrt {10}}}\cdot (1+i)\cdot r_{0}\quad }$
${\displaystyle \Leftrightarrow S_{2}=-{\frac {1}{\sqrt {10}}}\cdot (1+i)\cdot r_{0}\quad }$, since ${\displaystyle S_{0}=0\cdot (1+i)}$

Calculation 3[edit]

${\displaystyle \quad S_{w}={\frac {A_{0}\cdot S_{0}+A_{1}\cdot S_{1}+A_{2}\cdot S_{2}}{A_{0}+A_{1}+A_{2}}}={\frac {A_{1}\cdot S_{1}+A_{2}\cdot S_{2}}{A_{0}+A_{1}+A_{2}}}\quad }$ , since ${\displaystyle S_{0}=0+0i=0}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {A_{0}}{2}}\cdot S_{1}+A_{2}\cdot S_{2}}{A_{0}+{\frac {A_{0}}{2}}+A_{2}}}\quad }$ , since ${\displaystyle A_{1}={\frac {A_{0}}{2}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {A_{0}}{2}}\cdot S_{1}+{\frac {4\cdot A_{0}}{5\pi }}\cdot S_{2}}{A_{0}+{\frac {A_{0}}{2}}+{\frac {4\cdot A_{0}}{5\pi }}}}\quad }$ , since ${\displaystyle A_{2}={\frac {4\cdot A_{0}}{5\pi }}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {1}{2}}\cdot S_{1}+{\frac {4}{5\pi }}\cdot S_{2}}{1+{\frac {1}{2}}+{\frac {4}{5\pi }}}}\quad }$ , excluding ${\displaystyle A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {5\pi }{10\pi }}\cdot S_{1}+{\frac {8}{10\pi }}\cdot S_{2}}{{\frac {10\pi }{10\pi }}+{\frac {5\pi }{10\pi }}+{\frac {8}{10\pi }}}}\quad }$ ,rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {5\pi \cdot S_{1}+8\cdot S_{2}}{10\pi +5\pi +8}}\quad }$ ,rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {5\pi \cdot S_{1}+8\cdot S_{2}}{15\pi +8}}\quad }$ ,rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {5\pi \cdot \left({\frac {2{\sqrt {2}}}{3\pi }}\right)\cdot (1+i)\cdot r_{0}+8\cdot S_{2}}{15\pi +8}}\quad }$ ,since ${\displaystyle S_{1}=\left({\frac {2{\sqrt {2}}}{3\pi }}\right)\cdot (1+i)\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {5\pi \cdot \left({\frac {2{\sqrt {2}}}{3\pi }}\right)\cdot (1+i)\cdot r_{0}-8\cdot {\frac {1}{\sqrt {10}}}\cdot (1+i)\cdot r_{0}}{15\pi +8}}\quad }$ ,since ${\displaystyle S_{2}=-{\frac {1}{\sqrt {10}}}\cdot (1+i)\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {5\pi \cdot \left({\frac {2{\sqrt {2}}}{3\pi }}\right)-8\cdot {\frac {1}{\sqrt {10}}}}{15\pi +8}}\cdot (1+i)\cdot r_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {\left({\frac {10{\sqrt {2}}}{3}}\right)-{\frac {8}{\sqrt {10}}}}{15\pi +8}}\cdot (1+i)\cdot r_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {\left({\frac {10{\sqrt {20}}}{3{\sqrt {10}}}}\right)-{\frac {24}{3{\sqrt {10}}}}}{15\pi +8}}\cdot (1+i)\cdot r_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {10{\sqrt {20}}-24}{(15\pi +8)\cdot (3{\sqrt {10}})}}\cdot (1+i)\cdot r_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{w}=2\cdot {\frac {5{\sqrt {4\cdot 5}}-12}{(15\pi +8)\cdot (3{\sqrt {10}})}}\cdot (1+i)\cdot r_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{w}=2\cdot {\frac {5\cdot 2{\sqrt {5}}-12}{(15\pi +8)\cdot (3{\sqrt {10}})}}\cdot (1+i)\cdot r_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {4}{3}}\cdot {\frac {5{\sqrt {5}}-6}{(15\pi +8)\cdot {\sqrt {10}}}}\cdot (1+i)\cdot r_{0}\quad \approx 0.04\cdot (1+i)\cdot r_{0}}$ , rearranging

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