File:FS CRCC2(2) dia.png

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English

Circle containing the largest right isosceles triangle, the next largest circle, the second next largest circle and the second next largest circle

Summary edit

DescriptionFS CRCC2(2) dia.png	English: Circle containing the largest right isosceles triangle, the next largest circle, the second next largest circle and the second next largest circle Deutsch: Kreis, der das größte gleichschenklige rechte Dreieck, den nächstgrößten Kreis, den zweitgrößten Kries und den folgenden zweitgrößten Kreis enthält
Date	15 December 2023
Source	Own work
Author	Hans G. Oberlack

Elements edit

Base is the circle of given radius $r_{0}$ around point $S_{0}$
Inscribed is the largest possible right triangle with side length $a_{1}$ .
Added is the largest circle $r_{2}$ around point $S_{2}$
Added is the second largest circle $r_{3}$ around point $S_{3}$
Added is the second largest circle $r_{4}$ around point $S_{4}$

General case edit

Segments in the general case edit

0) The radius of the base circle $=r_{0}$
1) Side of the triangle $a_{1}={\sqrt {2}}\cdot r_{0}\quad$ , see CRC dia
2) The radius of the inscribed circle $r_{2}={\frac {1}{2}}\cdot r_{0}$
3) The radius of the second inscribed circle $r_{3}={\frac {1}{4}}\cdot r_{0}\quad$ ,see Calculation 1
4) The radius of the next second inscribed circle $r_{4}=r_{3}={\frac {1}{4}}\cdot r_{0}\quad$ , for symmetry reasons

Perimeters in the general case edit

0) Perimeter of base circle $P_{0}=2\cdot \pi \cdot r_{0}$
1) Perimeter of the triangle $P_{1}=({\sqrt {2}}+1)\cdot 2r_{0}$
2) Perimeter of inscribed circle $P_{2}=2\cdot \pi \cdot r_{2}=2\cdot \pi \cdot {\frac {1}{2}}\cdot r_{0}=\pi \cdot r_{0}$
3) Perimeter of second inscribed circle $P_{3}=2\cdot \pi \cdot r_{3}=2\cdot \pi \cdot {\frac {1}{4}}\cdot r_{0}={\frac {\pi }{2}}\cdot r_{0}$
4) Perimeter of next second inscribed circle $P_{4}=P_{3}=2\cdot \pi \cdot r_{3}=2\cdot \pi \cdot {\frac {1}{4}}\cdot r_{0}={\frac {\pi }{2}}\cdot r_{0}$

Areas in the general case edit

0) Area of the base circle $A_{0}=\pi \cdot r_{0}^{2}$
1) Area of the inscribed triangle $A_{1}={\frac {a_{1}^{2}}{2}}={\frac {({\sqrt {2}}r_{0})^{2}}{2}}={\frac {2r_{0}^{2}}{2}}=r_{0}^{2}={\frac {A_{0}}{\pi }}$
2) Area of the inscribed circle $A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot ({\frac {1}{2}}\cdot r_{0})^{2}={\frac {\pi }{4}}\cdot r_{0}^{2}={\frac {A_{0}}{4}}$
3) Area of the second inscribed circle $A_{3}=\pi \cdot r_{3}^{2}=\pi \cdot ({\frac {1}{4}}\cdot r_{0})^{2}={\frac {\pi }{16}}\cdot r_{0}^{2}={\frac {A_{0}}{16}}$
4) Area of the next second largest inscribed circle $A_{4}=A_{3}=\pi \cdot r_{3}^{2}=\pi \cdot ({\frac {1}{4}}\cdot r_{0})^{2}={\frac {\pi }{16}}\cdot r_{0}^{2}={\frac {A_{0}}{16}}$

Covered surface of the base shape $C_{b}={\frac {A_{1}+A_{2}+A_{3}+A_{4}}{A_{0}}}={\frac {{\frac {A_{0}}{\pi }}+{\frac {A_{0}}{4}}+{\frac {A_{0}}{16}}+{\frac {A_{0}}{16}}}{A_{0}}}={\frac {16+6\pi }{16\pi }}\quad \approx 69.3\%$

Centroids in the general case edit

Centroids as graphically displayed and orientated edit

The centroid positions of the following shapes will be expressed orientated so that the first shape n with $S_{n}\neq S_{0}$ will be of type $S_{n}=0-ki$ with $k>0$ . This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: $S_{0}=0+0i$
1) Orientated centroid position of the inscribed triangle: $S_{1}=\left(0-{\frac {1}{3}}\cdot i\right)\cdot r_{0}\quad \approx (0-0.333i)\cdot r_{0}\quad$
2) Orientated centroid position of the inscribed circle: $S_{2}=r_{2}\cdot (0+i)=\left(0+{\frac {1}{2}}\cdot i\right)\cdot r_{0}=(0+0.500i)\cdot r_{0}$
3) Orientated centroid position of the inscribed circle: $S_{3}=\left({\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot r_{0}\quad \approx (0.707+0.25i)\cdot r_{0}$
4) Orientated centroid position of the next inscribed circle: $S_{4}=\left(-{\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot r_{0}\quad \approx (-0.707+0.25i)\cdot r_{0}$

Normalised case edit

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 edit

0) Radius of the base circle $r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564$
1) Side length of the inscribed triangle $a_{1}={\sqrt {2}}\cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\frac {2}{\pi }}}\quad \approx 0.798$
2) Radius of the inscribed circle $r_{2}={\frac {1}{2{\sqrt {\pi }}}}\quad \approx 0.282$
3) Radius of the second inscribed circle $r_{3}={\frac {1}{4{\sqrt {\pi }}}}\quad \approx 0.141$
4) Radius of the next inscribed circle $r_{4}=r_{3}={\frac {1}{4{\sqrt {\pi }}}}\quad \approx 0.141$

Perimeters in the normalised case edit

0) Perimeter of base circle $P_{0}=2\cdot \pi \cdot r_{0}=2\cdot \pi \cdot {\frac {1}{\sqrt {\pi }}}=2\cdot {\sqrt {\pi }}\quad \approx 3.545$
1) Perimeter of the inscribed triangle $P_{1}=({\sqrt {2}}+1)\cdot 2\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 2.724$
2) Perimeter of inscribed circle: $P_{2}=\pi \cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\pi }}\quad \approx 1.772$
3) Perimeter of second inscribed circle $P_{3}={\frac {\sqrt {\pi }}{2}}\quad \approx 0.886$
4) Perimeter of next inscribed circle $P_{4}=P_{3}={\frac {\sqrt {\pi }}{2}}\quad \approx 0.886$

Areas in the normalised case edit

0) Area of the base circle $A_{0}=1$
1) Area of the inscribed triangle $A_{1}=r_{0}^{2}=({\frac {1}{\sqrt {\pi }}})^{2}={\frac {1}{\pi }}\quad \approx 0.318$
2) Area of the inscribed circle $A_{2}={\frac {A_{0}}{4}}=0.250$
3) Area of the second inscribed circle $A_{3}={\frac {\pi }{16}}\cdot r_{0}^{2}={\frac {\pi }{16}}\cdot ({\frac {1}{\sqrt {\pi }}})^{2}={\frac {\pi }{16}}\cdot {\frac {1}{\pi }}={\frac {1}{16}}=0.0625$
3) Area of the next inscribed circle $A_{4}=A_{3}={\frac {\pi }{16}}\cdot r_{0}^{2}={\frac {\pi }{16}}\cdot ({\frac {1}{\sqrt {\pi }}})^{2}={\frac {\pi }{16}}\cdot {\frac {1}{\pi }}={\frac {1}{16}}=0.0625$

Centroids in the normalised case edit

Centroids as graphically displayed and orientated edit

The centroid positions of the following shapes will be expressed orientated so that the first shape n with $S_{n}\neq S_{0}$ will be of type $S_{n}=0-ki$ with $k>0$ . This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: $S_{0}=0+0i$
1) Orientated centroid position of the inscribed triangle: $S_{1}=\left(0-{\frac {1}{3}}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left(0-{\frac {1}{3{\sqrt {\pi }}}}\cdot i\right)\quad \approx (0-0.188i)\quad$
2) Orientated centroid position of the inscribed circle: $S_{2}=r_{2}\cdot (0+i)=\left(0+{\frac {1}{2}}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left(0+{\frac {1}{2{\sqrt {\pi }}}}\cdot i\right)\quad \approx (0+0.282i)$
3) Orientated centroid position of the inscribed circle: $S_{3}=\left({\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left({\frac {1}{\sqrt {2\pi }}}+{\frac {1}{4{\sqrt {\pi }}}}\cdot i\right)\quad \approx (0.399+0.141i)$
4) Orientated centroid position of the inscribed circle: $S_{4}=\left(-{\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left(-{\frac {1}{\sqrt {2\pi }}}+{\frac {1}{4{\sqrt {\pi }}}}\cdot i\right)\quad \approx (-0.399+0.141i)$

Calculations edit

Calculation 1 edit

(1) $\quad {\overline {S_{0}DS_{2}}}$ is perpendicular on ${\overline {BC}}$ since the circle around $S_{2}$ is tangent to ${\overline {BC}}$
and $\quad {\overline {ES_{3}}}$ is perpendicular on ${\overline {BC}}$ since the circle around $S_{3}$ is tangent to ${\overline {BC}}$
$\quad \rightarrow |ES_{0}|=|DS_{3}|\quad$ and $|ES_{3}|=|DS_{0}|=r_{3}$
(2) $\quad {\overline {S_{0}S_{3}F}}$ is perpendicular in $F$ since the circle around $S_{3}$ is tangent to the circle around $S_{0}$ with $\quad |FS_{0}|=r_{0}\quad$ and $\quad |FS_{3}|=r_{3}$
$\quad \rightarrow |S_{0}S_{3}|=r_{0}-r_{3}\quad$
(3) $\quad {\overline {S_{2}GS_{3}}}$ is perpendicular in $G$ since the circle around $S_{3}$ is tangent to the circle around $S_{2}$ with $\quad |GS_{2}|=r_{2}\quad$ and $\quad |GS_{3}|=r_{3}$
$\quad \rightarrow |S_{2}S_{3}|=r_{2}+r_{3}\quad$
(4) $\quad |S_{0}DS_{2}|=r_{2}$
$\quad \Leftrightarrow \quad |DS_{0}|+|DS_{2}|=r_{2}\quad$
$\quad \Leftrightarrow \quad r_{3}+|DS_{2}|=r_{2}\quad$ , applying eqaution (1)
$\quad \Leftrightarrow \quad |DS_{2}|=r_{2}-r_{3}\quad$
(5) $\quad |ES_{0}|^{2}+|ES_{3}|^{2}=|S_{0}S_{3}|^{2}$ , applying the Pythagorean theorem
$\quad \Leftrightarrow |ES_{0}|^{2}+r_{3}^{2}=|S_{0}S_{3}|^{2}\quad$ , applying equation (1)
$\quad \Leftrightarrow |ES_{0}|^{2}+r_{3}^{2}=(r_{0}-r_{3})^{2}\quad$ , applying equation (2)
$\quad \Leftrightarrow |ES_{0}|^{2}+r_{3}^{2}=r_{0}^{2}+r_{3}^{2}-2r_{0}r_{3}\quad$ , applying binomial theorem
$\quad \Leftrightarrow |ES_{0}|^{2}=r_{0}^{2}-2r_{0}r_{3}\quad$

(6) $\quad |DS_{3}|^{2}+|DS_{2}|^{2}=|S_{2}S_{3}|^{2}$ , applying the Pythagorean theorem
$\quad \Leftrightarrow |DS_{3}|^{2}+|DS_{2}|^{2}=|S_{2}S_{3}|^{2}\quad$
$\quad \Leftrightarrow |ES_{0}|^{2}+|DS_{2}|^{2}=|S_{2}S_{3}|^{2}\quad$ , applying equation (1)
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}+|DS_{2}|^{2}=|S_{2}S_{3}|^{2}\quad$ , applying equation (5)
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}+(r_{2}-r_{3})^{2}=|S_{2}S_{3}|^{2}\quad$ , applying equation (4)
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}+(r_{2}-r_{3})^{2}=(r_{2}+r_{3})^{2}\quad$ , applying equation (3)
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}+(r_{2}^{2}+r_{3}^{2}-2r_{2}r_{3})=(r_{2}+r_{3})^{2}\quad$ , applying binomial theorem
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}+r_{2}^{2}+r_{3}^{2}-2r_{2}r_{3}=r_{2}^{2}+r_{3}^{2}+2r_{2}r_{3}\quad$ , applying binomial theorem
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}-2r_{2}r_{3}=2r_{2}r_{3}\quad$ , deducting $r_{2}^{2}+r_{3}^{2}$
$\quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}=4r_{2}r_{3}\quad$ , adding $2r_{2}r_{3}$
$\quad \Leftrightarrow r_{0}^{2}=4r_{2}r_{3}+2r_{0}r_{3}\quad$ , adding $2r_{0}r_{3}$
$\quad \Leftrightarrow r_{0}^{2}=(4r_{2}+2r_{0})\cdot r_{3}\quad$
$\quad \Leftrightarrow r_{3}={\frac {r_{0}^{2}}{4r_{2}+2r_{0}}}\quad$
$\quad \Leftrightarrow r_{3}={\frac {r_{0}^{2}}{4{\frac {1}{2}}r_{0}+2r_{0}}}\quad$ , since $r_{2}={\frac {1}{2}}r_{0}$
$\quad \Leftrightarrow r_{3}={\frac {r_{0}^{2}}{2r_{0}+2r_{0}}}\quad$
$\quad \Leftrightarrow r_{3}={\frac {r_{0}^{2}}{4r_{0}}}\quad$
$\quad \Leftrightarrow r_{3}={\frac {1}{4}}\cdot r_{0}\quad$

Calculation 2 edit

The centroid $S_{3}$ can be written as $S_{3}=x_{3}+y_{3}i$
$\quad \Leftrightarrow S_{3}-S_{0}=x_{3}+y_{3}i-(x_{0}+y_{0}i)\quad$ , since $S_{0}=x_{0}+y_{0}i$
$\quad \Leftrightarrow S_{3}-S_{0}=x_{3}-x_{0}+(y_{3}-y_{0})i\quad$
$\quad \Leftrightarrow {\overrightarrow {S_{0}S_{3}}}={\overrightarrow {x_{0}x_{3}}}+{\overrightarrow {y_{0}y_{3}}}\quad$ expressed as vectors
$\quad \Leftrightarrow {\overrightarrow {S_{0}S_{3}}}={\overrightarrow {S_{0}D}}+{\overrightarrow {DS_{3}}}\quad$ using vectors of known length
$\quad \Leftrightarrow x_{3}-x_{0}+(y_{3}-y_{0})i={\overrightarrow {S_{0}D}}+{\overrightarrow {DS_{3}}}\quad$
$\quad \Leftrightarrow x_{3}-x_{0}+(y_{3}-y_{0})i={\frac {|S_{0}D|(1+i)}{\sqrt {2}}}+{\overrightarrow {DS_{3}}}\quad$ since ${\overrightarrow {S_{0}D}}$ is on the 45-degree axis
$\quad \Leftrightarrow x_{3}-x_{0}+(y_{3}-y_{0})i={\frac {|S_{0}D|(1+i)}{\sqrt {2}}}+{\frac {|DS_{3}|(1+i)}{\sqrt {2}}}\cdot (-i)\quad$ since ${\overrightarrow {DS_{3}}}$ is perpendicular to ${\overrightarrow {S_{0}D}}$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {|S_{0}D|(1+i)}{\sqrt {2}}}+{\frac {|DS_{3}|(1+i)}{\sqrt {2}}}\cdot (-i)\quad$ since $x_{0}=0$ and $y_{0}=0$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{3}(1+i)}{\sqrt {2}}}+{\frac {|DS_{3}|(1+i)}{\sqrt {2}}}\cdot (-i)\quad$ since $|S_{0}D|=r_{3}$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{3}(1+i)}{\sqrt {2}}}+{\frac {|ES_{0}|(1+i)}{\sqrt {2}}}\cdot (-i)\quad$ since $|DS_{3}|=|ES_{0}|$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{3}(1+i)}{\sqrt {2}}}+{\frac {|ES_{0}|(-i-i^{2})}{\sqrt {2}}}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{3}(1+i)}{\sqrt {2}}}+{\frac {|ES_{0}|(1-i)}{\sqrt {2}}}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {|ES_{0}|(1-i)}{\sqrt {2}}}\quad$ since $r_{3}={\frac {r_{0}}{4}}$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {{\sqrt {r_{0}^{2}-2r_{0}r_{3}}}\cdot (1-i)}{\sqrt {2}}}\quad$ since $|ES_{0}|^{2}=r_{0}^{2}-2r_{0}r_{3}$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {{\sqrt {r_{0}^{2}-2r_{0}{\frac {r_{0}}{4}}}}\cdot (1-i)}{\sqrt {2}}}\quad$ since $r_{3}={\frac {r_{0}}{4}}$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {{\sqrt {r_{0}^{2}-r_{0}^{2}{\frac {2}{4}}}}\cdot (1-i)}{\sqrt {2}}}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {{\sqrt {r_{0}^{2}(1-{\frac {1}{2}})}}\cdot (1-i)}{\sqrt {2}}}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {r_{0}{\sqrt {\frac {1}{2}}}\cdot (1-i)}{\sqrt {2}}}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i={\frac {r_{0}(1+i)}{4{\sqrt {2}}}}+{\frac {r_{0}\cdot (1-i)}{{\sqrt {2}}{\sqrt {2}}}}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i=\left({\frac {(1+i)}{4{\sqrt {2}}}}+{\frac {(1-i)}{2}}\right)\cdot r_{0}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i=\left({\frac {(1+i)}{4{\sqrt {2}}}}+{\frac {2{\sqrt {2}}(1-i)}{4{\sqrt {2}}}}\right)\cdot r_{0}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i=\left({\frac {1+i+2{\sqrt {2}}-2{\sqrt {2}}i}{4{\sqrt {2}}}}\right)\cdot r_{0}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i=\left({\frac {1+i+2{\sqrt {2}}-2{\sqrt {2}}i}{4{\sqrt {2}}}}\right){\frac {\sqrt {2}}{\sqrt {2}}}\cdot r_{0}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i=\left({\frac {{\sqrt {2}}+i{\sqrt {2}}+4-4i}{4\cdot 2}}\right)\cdot r_{0}\quad$
$\quad \Leftrightarrow x_{3}+y_{3}i=\left({\frac {({\sqrt {2}}+4)+({\sqrt {2}}-4)i}{8}}\right)\cdot r_{0}\quad \approx (0.677-0.323i)\cdot r_{0}$

Calculation 3 edit

$\quad S_{w}={\frac {A_{0}\cdot S_{0}+A_{1}\cdot S_{1}+A_{2}\cdot S_{2}+A_{3}\cdot S_{3}}{A_{0}+A_{1}+A_{2}+A_{3}}}$
$\quad \Leftrightarrow S_{w}={\frac {A_{0}\cdot S_{0}+A_{1}\cdot S_{1}+A_{2}\cdot S_{2}+A_{3}\cdot S_{3}}{A_{0}+{\frac {A_{0}}{\pi }}+{\frac {A_{0}}{4}}+{\frac {A_{0}}{16}}}}$
$\quad \Leftrightarrow S_{w}={\frac {A_{0}\cdot S_{0}+A_{1}\cdot S_{1}+A_{2}\cdot S_{2}+A_{3}\cdot S_{3}}{\frac {(21\pi +16)\cdot A_{0}}{16\pi }}}$
$\quad \Leftrightarrow S_{w}=\left(A_{0}\cdot S_{0}+A_{1}\cdot S_{1}+A_{2}\cdot S_{2}+A_{3}\cdot S_{3}\right)\cdot {\frac {16\pi }{(21\pi +16)\cdot A_{0}}}$
$\quad \Leftrightarrow S_{w}=\left(A_{0}\cdot S_{0}+{\frac {A_{0}}{\pi }}\cdot S_{1}+{\frac {A_{0}}{4}}\cdot S_{2}+{\frac {A_{0}}{16}}\cdot S_{3}\right)\cdot {\frac {16\pi }{(21\pi +16)\cdot A_{0}}}$
$\quad \Leftrightarrow S_{w}=\left(S_{0}+{\frac {1}{\pi }}\cdot S_{1}+{\frac {1}{4}}\cdot S_{2}+{\frac {1}{16}}\cdot S_{3}\right)\cdot {\frac {16\pi \cdot A_{0}}{(21\pi +16)\cdot A_{0}}}$
$\quad \Leftrightarrow S_{w}=\left(0+{\frac {1}{\pi }}\cdot S_{1}+{\frac {1}{4}}\cdot S_{2}+{\frac {1}{16}}\cdot S_{3}\right)\cdot {\frac {16\pi }{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left({\frac {16}{16\pi }}\cdot S_{1}+{\frac {4\pi }{16\pi }}\cdot S_{2}+{\frac {\pi }{16\pi }}\cdot S_{3}\right)\cdot {\frac {16\pi }{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left(16\cdot S_{1}+4\pi \cdot S_{2}+\pi \cdot S_{3}\right)\cdot {\frac {1}{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left(16\cdot (-{\frac {1}{3{\sqrt {2}}}})\cdot (1+i)\cdot r_{0}+4\pi \cdot S_{2}+\pi \cdot S_{3}\right)\cdot {\frac {1}{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left(16\cdot (-{\frac {1}{3{\sqrt {2}}}})\cdot (1+i)\cdot r_{0}+4\pi \cdot ({\frac {1}{2{\sqrt {2}}}})\cdot (1+i)\cdot r_{0}+\pi \cdot \left({\frac {({\sqrt {2}}+4)+({\sqrt {2}}-4)i}{8}}\right)\cdot r_{0}\right)\cdot {\frac {1}{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left((-{\frac {16}{3{\sqrt {2}}}})\cdot (1+i)+({\frac {4\pi }{2{\sqrt {2}}}})\cdot (1+i)+\pi \cdot \left({\frac {({\sqrt {2}}+4)+({\sqrt {2}}-4)i}{8}}\right)\right)\cdot {\frac {r_{0}}{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left((-{\frac {16\cdot 2\cdot 8}{48{\sqrt {2}}}})\cdot (1+i)+({\frac {4\pi \cdot 3\cdot 8}{48{\sqrt {2}}}})\cdot (1+i)+\pi \cdot 3\cdot 2\cdot {\sqrt {2}}\left({\frac {({\sqrt {2}}+4)+({\sqrt {2}}-4)i}{48{\sqrt {2}}}}\right)\right)\cdot {\frac {r_{0}}{21\pi +16}}$
$\quad \Leftrightarrow S_{w}=\left((-256\cdot (1+i)+96\pi \cdot (1+i)+\pi \cdot 6{\sqrt {2}}\left(({\sqrt {2}}+4)+({\sqrt {2}}-4)i\right)\right)\cdot {\frac {r_{0}}{(21\pi +16)(48{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left((-128\cdot (1+i)+48\pi \cdot (1+i)+\pi \cdot 3{\sqrt {2}}\left(({\sqrt {2}}+4)+({\sqrt {2}}-4)i\right)\right)\cdot {\frac {2r_{0}}{(21\pi +16)(48{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left(-128-128i+48\pi +48\pi i+3\pi {\sqrt {2}}{\sqrt {2}}+4\cdot 3\pi {\sqrt {2}}+3i\pi {\sqrt {2}}{\sqrt {2}}-4\cdot 3\pi {\sqrt {2}}i\right)\cdot {\frac {r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left(-128-128i+48\pi +48\pi i+3\pi \cdot 2+12\pi {\sqrt {2}}+6i\pi -12i\pi {\sqrt {2}}\right)\cdot {\frac {r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left(-128-128i+48\pi +48\pi i+6\pi +12\pi {\sqrt {2}}+6i\pi -12i\pi {\sqrt {2}}\right)\cdot {\frac {r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left(-128-128i+54\pi +54\pi i+12\pi {\sqrt {2}}-12i\pi {\sqrt {2}}\right)\cdot {\frac {r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left(-128+54\pi +12\pi {\sqrt {2}}-128i+54\pi i-12i\pi {\sqrt {2}}\right)\cdot {\frac {r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left(54\pi +12\pi {\sqrt {2}}-128+54\pi i-12i\pi {\sqrt {2}}-128i\right)\cdot {\frac {r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}=\left((27\pi +6\pi {\sqrt {2}}-64)+(27\pi -6\pi {\sqrt {2}}-64)\cdot i\right)\cdot {\frac {2r_{0}}{(21\pi +16)(24{\sqrt {2}})}}$
$\quad \Leftrightarrow S_{w}={\frac {\left((27\pi +6\pi {\sqrt {2}}-64)+(27\pi -6\pi {\sqrt {2}}-64)\cdot i\right)}{(21\pi +16)(12{\sqrt {2}})}}\cdot r_{0}\quad \approx (0.034-0.004i)\cdot r_{0}$
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Calculation 4 edit

Calculating $S_{3}$ in the orientated position:
$S_{3}=S_{0}+|S_{0}E|+|ES_{3}|$
$\quad \Leftrightarrow S_{3}=0+|S_{0}E|+|ES_{3}|i$
$\quad \Leftrightarrow S_{3}=|S_{0}E|+r_{3}i\quad$ , see calculation 1
$\quad \Leftrightarrow S_{3}={\sqrt {|S_{0}S_{3}|^{2}-|ES_{3}|^{2}}}+r_{3}i\quad$ , applying Pythagorean theoreme
$\quad \Leftrightarrow S_{3}={\sqrt {|S_{0}S_{3}|^{2}-r_{3}^{2}}}+r_{3}i\quad$ , applying calculation 1
$\quad \Leftrightarrow S_{3}={\sqrt {(r_{0}-r_{3})^{2}-r_{3}^{2}}}+r_{3}i\quad$ , applying calculation 1
$\quad \Leftrightarrow S_{3}={\sqrt {r_{0}^{2}-2r_{0}r_{3}+r_{3}^{2}-r_{3}^{2}}}+r_{3}i\quad$ , applying binomial formula
$\quad \Leftrightarrow S_{3}={\sqrt {r_{0}^{2}-2r_{0}r_{3}}}+r_{3}i\quad$ , eliminating
$\quad \Leftrightarrow S_{3}={\sqrt {r_{0}^{2}-2r_{0}{\frac {r_{0}}{4}}}}+{\frac {r_{0}}{4}}i\quad$ , since $r_{3}={\frac {r_{0}}{4}}$
$\quad \Leftrightarrow S_{3}={\sqrt {r_{0}^{2}-{\frac {1}{2}}r_{0}^{2}}}+{\frac {r_{0}}{4}}i\quad$ , rearranging
$\quad \Leftrightarrow S_{3}={\sqrt {r_{0}^{2}-{\frac {1}{2}}r_{0}^{2}}}+{\frac {r_{0}}{4}}i\quad$ , rearranging
$\quad \Leftrightarrow S_{3}={\sqrt {\frac {1}{2}}}\cdot r_{0}+{\frac {1}{4}}\cdot r_{0}\cdot i\quad$ , rearranging
$\quad \Leftrightarrow S_{3}=\left({\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot r_{0}\quad \approx (0.707+0.25i)\cdot r_{0}$ , rearranging

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to remix – to adapt the work

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attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original.

File history

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	Date/Time	Thumbnail	Dimensions	User	Comment
current	10:32, 15 December 2023		1,503 × 1,428 (98 KB)	Hans G. Oberlack (talk \| contribs)	Uploaded own work with UploadWizard

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Horizontal resolution	129.92 dpc
Vertical resolution	129.92 dpc