File:FS HS dia.png

Summary[edit]

The semicircle as base element. Inscribed is the largest regular hexagon.

General case[edit]

Segments in the general case[edit]

0) The radius of the semicircle: ${\displaystyle r_{0}}$
1) The side length of the inscribed hexagon: ${\displaystyle a_{1}={\frac {2}{\sqrt {13}}}\cdot r_{0}\quad \approx 0.555\cdot r_{0}}$, 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 hexagon: ${\displaystyle P_{1}=6\cdot {\frac {2}{\sqrt {13}}}\cdot r_{0}\quad \approx 3.328\cdot r_{0}}$

Areas in the general case[edit]

0) Area of the base semicircle ${\displaystyle A_{0}={\frac {\pi \cdot r_{0}^{2}}{2}}\approx 1.571\cdot r_{0}^{2}}$
1) Area of the inscribed hexagon ${\displaystyle A_{1}={\frac {3{\sqrt {3}}}{2}}\cdot a_{1}^{2}={\frac {3{\sqrt {3}}}{2}}\cdot {\frac {4}{13}}r_{0}^{2}={\frac {6{\sqrt {3}}}{13}}\cdot r_{0}^{2}\quad \approx 0.799r_{0}^{2}}$

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 hexagon relative to the base centroid is: ${\displaystyle S_{1}=\left(0+\left({\frac {\sqrt {3}}{\sqrt {13}}}-{\frac {4}{3\cdot \pi }}\right)\cdot i\right)\cdot r_{0}\quad \approx (0+0.056i)\cdot r_{0}}$, 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 the inscribed hexagon: ${\displaystyle a_{1}={\frac {2}{\sqrt {13}}}\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 0.443}$

Perimeter in the normalised case[edit]

0) Perimeter of base semicircle: ${\displaystyle P_{0}=(2+\pi )\cdot r_{0}=(2+\pi )\cdot {\sqrt {\frac {2}{\pi }}}=4.1023974}$
1) Perimeter of inscribed hexagon: ${\displaystyle P_{1}=6\cdot {\frac {2}{\sqrt {13}}}\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 2.6555203}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}=6.7579177}$

Area in the normalised case[edit]

0) Area of the base semicircle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed hexagon ${\displaystyle A_{1}={\frac {6{\sqrt {3}}}{13}}\cdot {\frac {2}{\pi }}\approx 0.509}$

Centroids in the normalised case[edit]

0) Centroid of the base shape: ${\displaystyle S_{0}=0+0i}$
1) Centroid of the inscribed hexagon: ${\displaystyle S_{1}=\left(0+\left({\frac {\sqrt {3}}{\sqrt {13}}}-{\frac {4}{3\cdot \pi }}\right)\cdot i\right)\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 0+0.045i}$

Distances of centroids[edit]

The distance between the centroid of the base semicircle and the centroid of the circle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}\quad \approx 0.0446586}$
Sum of distances: ${\displaystyle d_{s}=d_{01}\quad \approx 0.0446586}$

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.

${\displaystyle decimalpart(6.7579177+0.0446586)=decimalpart(6.8025764)=.8025764}$

So the identifying number is: ${\displaystyle 1.8025764}$

Calculations[edit]

Given elements[edit]

(1) ${\displaystyle |AM|=|BM|=|MD|=r_{0}}$
(2) ${\displaystyle |DC|=a_{1}}$
(3) ${\displaystyle |EM|={\frac {1}{2}}\cdot a_{1}}$
(4) ${\displaystyle |ED|={\sqrt {3}}\cdot a_{1}}$, diagonal of 2 of hexagon

Calculation 1[edit]

${\displaystyle |ED|^{2}+|EM|^{2}=|MD|^{2}\quad }$, applying Pythagorean theorem to the rectangular triangle ${\displaystyle \triangle {EMD}}$
${\displaystyle \Leftrightarrow |ED|^{2}+|EM|^{2}=r_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow |ED|^{2}+\left({\frac {1}{2}}\cdot a_{1}\right)^{2}=r_{0}^{2}\quad }$, applying equation (3)
${\displaystyle \Leftrightarrow \left({\sqrt {3}}\cdot a_{1}\right)^{2}+\left({\frac {1}{2}}\cdot a_{1}\right)^{2}=r_{0}^{2}\quad }$, applying equation (4)
${\displaystyle \Leftrightarrow 3\cdot a_{1}^{2}+{\frac {1}{4}}\cdot a_{1}^{2}=r_{0}^{2}\quad }$, multiplying
${\displaystyle \Leftrightarrow {\frac {3\cdot 4+1}{4}}\cdot a_{1}^{2}=r_{0}^{2}\quad }$, rearranging
${\displaystyle \Leftrightarrow {\frac {13}{4}}\cdot a_{1}^{2}=r_{0}^{2}\quad }$, rearranging
${\displaystyle \Leftrightarrow a_{1}^{2}={\frac {4}{13}}\cdot r_{0}^{2}\quad }$, rearranging
${\displaystyle \Leftrightarrow a_{1}={\frac {2}{\sqrt {13}}}\cdot r_{0}\quad \approx 0.555\cdot r_{0}}$, extracting the root

Calculation 2[edit]

${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}M}}+{\overrightarrow {MS_{1}}}}$
${\displaystyle \Leftrightarrow S_{1}=(0+0i)-|S_{0}M|i+|MS_{1}|i}$
${\displaystyle \Leftrightarrow S_{1}=-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i+|MS_{1}|i\quad }$, definition of centroid of semicircle
${\displaystyle \Leftrightarrow S_{1}=-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i+{\frac {1}{2}}\cdot |ED|i\quad }$, since the centroid of a hexagon is in the middle of the diagonals
${\displaystyle \Leftrightarrow S_{1}=-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i+{\frac {1}{2}}\cdot {\sqrt {3}}\cdot a_{1}\cdot i\quad }$, applying equation (4)
${\displaystyle \Leftrightarrow S_{1}=-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i+{\frac {1}{2}}\cdot {\sqrt {3}}\cdot {\frac {2}{\sqrt {13}}}\cdot r_{0}\cdot i\quad }$, applying result of calculation 1
${\displaystyle \Leftrightarrow S_{1}=-\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i+{\frac {\sqrt {3}}{\sqrt {13}}}\cdot r_{0}\cdot i\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{1}=-\left({\frac {4}{3\cdot \pi }}\right)\cdot r_{0}\cdot i+{\frac {\sqrt {3}}{\sqrt {13}}}\cdot r_{0}\cdot i\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{1}=\left({\frac {\sqrt {3}}{\sqrt {13}}}-{\frac {4}{3\cdot \pi }}\right)\cdot r_{0}\cdot i\quad \approx (0+0.056i)\cdot r_{0}}$, rearranging

Licensing[edit]

I, the copyright holder of this work, hereby publish it under the following license:

This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.

You are free:

• to share – to copy, distribute and transmit the work
• to remix – to adapt the work

Under the following conditions:

• 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.

https://creativecommons.org/licenses/by-sa/4.0CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 truetrue