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English

Second largest circle adjacent to largest circle in a circle with the largest isosceles triangle

Summary

DescriptionFS CRCC2 dia.png	English: Second largest circle adjacent to largest circle in a circle with the largest isosceles triangle Deutsch: Zweitgrößter Kreis neben dem größten Kreis in einem Kreis mit dem größten gleichschenkligen Dreieck
Date	24 April 2022
Source	Own work
Author	Hans G. Oberlack

Elements

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 $S_{3}$

General case

Segments in the general case

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

Perimeters in the general case

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 $P_{3}=2\cdot \pi \cdot r_{3}=2\cdot \pi \cdot {\frac {1}{4}}\cdot r_{0}={\frac {\pi }{2}}\cdot r_{0}$

Areas in the general case

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 $A_{3}=\pi \cdot r_{3}^{2}=\pi \cdot ({\frac {1}{4}}\cdot r_{0})^{2}={\frac {\pi }{16}}\cdot r_{0}^{2}={\frac {A_{0}}{16}}$

Covered surface of the base shape $C_{b}={\frac {A_{1}+A_{2}+A_{3}}{A_{0}}}={\frac {{\frac {A_{0}}{\pi }}+{\frac {A_{0}}{4}}+{\frac {A_{0}}{16}}}{A_{0}}}={\frac {16+5\pi }{16\pi }}\quad \approx 63.1\%$

Centroids in the general case

1) Centroids as graphically displayed

0) Centroid position of the base circle: $S_{0}=0+0i$
1) Centroid positions of the inscribed triangle: $S_{1}=-{\frac {1}{3{\sqrt {2}}}}\cdot (1+i)\cdot r_{0}\quad \approx -0.236\cdot (1+i)\cdot r_{0}\quad$ , see CR dia
2) Centroid positions of the inscribed circle: $S_{2}={\frac {r_{2}}{\sqrt {2}}}\cdot (1+i)={\frac {r_{0}}{2{\sqrt {2}}}}\cdot (1+i)\quad \approx 0.354(1+i)\cdot r_{0}$ , see CRC dia
3) Centroid positions of the second inscribed circle: $S_{3}={\frac {({\sqrt {2}}+4)+({\sqrt {2}}-4)i}{8}}\cdot r_{0}\quad \approx (0.677-0.323i)\cdot r_{0}$ , see Calculation 2

W) Weighted average centroid: $S_{w}={\frac {\left((27\pi +6\pi {\sqrt {2}}-64)+(27\pi -6\pi {\sqrt {2}}-64)\cdot i\right)}{(21\pi +16)(12{\sqrt {2}})}}\cdot r_{0}\quad \approx (0.034-0.004i)\cdot r_{0}\quad$ , see Calculation 3

2) Orientated centroids

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: $S_{3}=\left({\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot r_{0}\quad \approx (0.707+0.25i)\cdot r_{0}$

Normalised case

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 $r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564$
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 $r_{3}={\frac {1}{4{\sqrt {\pi }}}}\quad \approx 0.181$

Perimeters in the normalised case

0) Perimeter of base circle $P_{0}=2\cdot \pi \cdot r_{0}=2\cdot \pi \cdot {\frac {1}{\sqrt {\pi }}}=2\cdot {\sqrt {\pi }}\quad \approx 3.5449077$
1) Perimeter of the inscribed triangle $P_{1}=({\sqrt {2}}+1)\cdot 2\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 2.7241483$
2) Perimeter of inscribed circle: $P_{2}=\pi \cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\pi }}\quad \approx 1.7724539$
3) Perimeter of second inscribed circle $P_{3}={\frac {\sqrt {\pi }}{2}}\quad \approx 0.8862269$

S) Sum of perimeters $P_{s}=P_{0}+P_{1}+P_{2}\quad \approx 8.9277368$

Areas in the normalised case

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 $A_{3}={\frac {\pi }{16}}\cdot r_{0}^{2}={\frac {\pi }{16}}\cdot ({\frac {1}{\sqrt {\pi }}})^{2}={\frac {\pi }{16}}\cdot {\frac {1}{\pi }}={\frac {1}{16}}=0.0625$

Centroids in the normalised case

1) Centroids as graphically displayed

Centroid positions are measured from the centroid point of the base shape.
0) Centroid position of the base circle: $S_{0}=0+0i$
1) Centroid position of the inscribed triangle: $S_{1}=x_{1}+y_{1}i=-{\frac {1}{3{\sqrt {2\pi }}}}(1+i)\quad \approx -0.133\cdot (1+i)$
2) Centroid position of the inscribed circle: $S_{2}=x_{2}+y_{2}i={\frac {1}{2{\sqrt {2}}}}\cdot {\frac {1}{\sqrt {\pi }}}\cdot (1+i)\quad \approx 0.199\cdot (1+i)$
3) Centroid position of the second inscribed circle: $S_{3}=x_{3}+y_{3}i={\frac {({\sqrt {2}}+4)+({\sqrt {2}}-4)i}{8}}\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 0.382-0.182i$

W) Weighted average centroid: $S_{w}={\frac {\left((27\pi +6\pi {\sqrt {2}}-64)+(27\pi -6\pi {\sqrt {2}}-64)\cdot i\right)}{(21\pi +16)(12{\sqrt {2}})}}\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 0.019-0.002i\quad$

2) Orientated centroids

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: $S_{3}=\left({\frac {1}{\sqrt {2}}}+{\frac {1}{4}}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left({\frac {1}{\sqrt {2\pi }}}+{\frac {1}{4{\sqrt {\pi }}}}\cdot i\right)\quad \approx (0.399+0.141i)$

Calculations

Calculation 1

(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

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

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

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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