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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