File:FS HR dia.png

Summary[edit]

The semicircle as base element with given radius ${\displaystyle r_{0}}$
. Inscribed is the largest right triangle.

General case[edit]

Segments in the general case[edit]

0) Radius of the semicircle: ${\displaystyle r_{0}}$
1) Side length of the right triangle: ${\displaystyle a_{1}={\sqrt {2}}\cdot r_{0}}$, see Calculation 1

Perimeters in the general case[edit]

0) Perimeter of base semicircle: ${\displaystyle P_{0}=2\cdot r_{0}+\pi \cdot r_{0}=(2+\pi )\cdot r_{0}\quad \approx 5.142\cdot r_{0}}$
1) Perimeter of inscribed right triangle: ${\displaystyle P_{1}=2\cdot a_{1}+2\cdot r_{0}=2\cdot {\sqrt {2}}\cdot r_{0}+2\cdot r_{0}=2\cdot r_{0}\cdot ({\sqrt {2}}+1)\quad \approx 4.828\cdot r_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=(4+\pi +2{\sqrt {2}})\cdot r_{0}\quad \approx 9.97\cdot r_{0}}$

Areas in the general case[edit]

0) Area of the base semicircle ${\displaystyle A_{0}={\frac {\pi }{2}}\cdot r_{0}^{2}\quad \approx 1.57\cdot r_{0}^{2}}$
1) Area of the inscribed right triangle ${\displaystyle A_{1}={\frac {a_{1}^{2}}{2}}={\frac {({\sqrt {2}}\cdot r_{0})^{2}}{2}}=r_{0}^{2}\quad }$

Centroids in the general case[edit]

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid of the inscribed right triangle relative to the base centroid is: ${\displaystyle S_{1}=\left(0+{\frac {\pi -4}{3\cdot \pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.091i)\cdot r_{0}\quad }$, see Calculation 2

Normalised case[edit]

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

Segments in the normalised case[edit]

0) Radius of the base semicircle ${\displaystyle r_{0}={\sqrt {\frac {2}{\pi }}}\quad \approx 0.798}$
1) Side length of inscribed right triangle ${\displaystyle a_{1}={\sqrt {2}}\cdot {\sqrt {\frac {2}{\pi }}}={\frac {2}{\sqrt {\pi }}}\quad \approx 1.128}$

Perimeter in the normalised case[edit]

0) Perimeter of base semicircle: ${\displaystyle P_{0}=(2+\pi )\cdot r_{0}=(2+\pi )\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 4.1023974}$
1) Perimeter of inscribed right triangle: ${\displaystyle P_{1}=2\cdot ({\sqrt {2}}+1)\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 3.8525275}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 7.9549249}$

Areas in the normalised case[edit]

0) Area of the base semicircle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed right triangle ${\displaystyle A_{1}=r_{0}^{2}=\left({\sqrt {\frac {2}{\pi }}}\right)^{2}={\frac {2}{\pi }}\quad \approx 0.637}$

Centroids in the normalised case[edit]

0) ${\displaystyle S_{0}=0+0i}$
1) ${\displaystyle S_{1}=\left(0+{\frac {\pi -4}{3\cdot \pi }}\cdot i\right)\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 0-0.073i}$

Distances of centroids[edit]

The distance between the centroid of the base semicircle and the centroid of the circle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}\quad \approx 0.0726712}$
Sum of distances: ${\displaystyle d_{s}=d_{01}\quad \approx 0.0726712}$

Identifying number[edit]

Apart of the base element there is only one 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.

${\displaystyle decimalpart(7.9549249+0.0726712)=decimalpart(8.0275961)=.0275961}$

So the identifying number is: ${\displaystyle 1.0275961}$

Calculations[edit]

Known elements[edit]

(0) Given is the radius of the base semicircle ${\displaystyle r_{0}}$
(1) ${\displaystyle \quad |AB|=|BC|=|BD|=r_{0}}$
(2) ${\displaystyle \quad |AD|=|CD|=a_{1}}$
(3) ${\displaystyle \quad |AC|=2\cdot r_{0}}$

Calculation 1[edit]

${\displaystyle \quad |CD|^{2}=|BC|^{2}+|BD|^{2}\quad }$, applying the pythagorean theoreme on the rectangular triangle ${\displaystyle \triangle BCD}$
${\displaystyle \quad \Leftrightarrow |CD|^{2}=r_{0}^{2}+|BD|^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |CD|^{2}=r_{0}^{2}+r_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |CD|^{2}=2\cdot r_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow a_{1}^{2}=2\cdot r_{0}^{2}\quad }$, applying equation (2)

Calculation 2[edit]

${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}B}}+{\overrightarrow {BS_{1}}}}$
${\displaystyle \Leftrightarrow S_{1}=(0+0i)-\left(0+\left({\frac {4\cdot |BD|}{3\cdot \pi }}\right)\cdot i\right)+(0+{\frac {1}{3}}\cdot |BD|\cdot i)\quad }$, applying the formulas for centroids of semicircles and triangles
${\displaystyle \Leftrightarrow S_{1}=(0+0i)-\left(0+\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i\right)+(0+{\frac {1}{3}}\cdot r_{0}\cdot i)\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow S_{1}=(0+0+0)+\left(0i-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i+{\frac {1}{3}}\cdot r_{0}\cdot i\right)\quad }$, summing the real and the imaginary terms
${\displaystyle \Leftrightarrow S_{1}=0+\left({\frac {1}{3}}\cdot r_{0}\cdot i-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i\right)\quad }$
${\displaystyle \Leftrightarrow S_{1}=0\cdot r_{0}+\left({\frac {1}{3}}\cdot i\cdot r_{0}-{\frac {4}{3\cdot \pi }}\cdot i\cdot r_{0}\right)\quad }$
${\displaystyle \Leftrightarrow S_{1}=0\cdot r_{0}+\left({\frac {1}{3}}-{\frac {4}{3\cdot \pi }}\right)\cdot i\cdot r_{0}\quad }$
${\displaystyle \Leftrightarrow S_{1}=0\cdot r_{0}+\left({\frac {\pi -4}{3\cdot \pi }}\right)\cdot i\cdot r_{0}\quad }$
${\displaystyle \Leftrightarrow S_{1}=\left(0+{\frac {\pi -4}{3\cdot \pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.091i)\cdot r_{0}}$

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