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Created page with "See Metallic MOS. The article is uncomfortable with the definition of the split operation, so I'll write it in my own way. == Golden case == You know a process cutting th..."
 
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See [[Metallic MOS]]. The article is uncomfortable with the definition of the split operation, so I'll write it in my own way.
See [[Metallic MOS]]. The article is unelegant with the definition of the split operation, so l write it in my own way.


== Golden case ==
== Golden case ==
You know a process cutting the square from golden rectangle. Imagine <math>L = φ</math> and <math>s = 1</math> and divide L into
You know a process cutting off a square from a golden rectangle. Imagine <math>L = φ</math> and <math>s = 1</math> and divide L into


<math>\qquad L_1:s_1 = φ-1:1 = [0; 1, 1, 1, 1, ...]</math>.
<math>\qquad L_1:s_1 = φ-1:1 = [0; 1, 1, 1, 1, ...]</math>.


No, <math>L_1 < s_1</math>. Let new <math>s := L_1</math> and new <math>L = s_1</math>. (Rotate the rectangle 90°)
No, <math>L_1 < s_1</math>. Let new <math>s := L_1</math> and new <math>L := s_1</math>. (Rotate the rectangle 90°)


OK, now <math>L = 1</math> and <math>s = φ-1</math> and <math>L:s = φ</math>.
OK, now <math>L = 1</math> and <math>s = φ-1</math> and <math>L:s = φ = [1; 1, 1, 1, 1, ...]</math>.


Loop.
Loop.


In thinking MOS pattern, direction of division is altered when every rotating.
In thinking MOS pattern, direction of division is altered when every rotation.


<pre>L
<pre>L
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<math>\qquad L_2:s_2 = δ_s-2:1 = [0; 2, 2, 2, 2, ...]</math>.
<math>\qquad L_2:s_2 = δ_s-2:1 = [0; 2, 2, 2, 2, ...]</math>.


No, <math>L_2 < s_2</math>. Let new <math>s := L_2</math> and new <math>L = s_2</math>. (Rotate the rectangle 90°)
No, <math>L_2 < s_2</math>. Let new <math>s := L_2</math> and new <math>L := s_2</math>. (Rotate the rectangle 90°)


OK, now <math>L = 1</math> and <math>s = δ_s-2</math> and <math>L:s = δ_s</math>.
OK, now <math>L = 1</math> and <math>s = δ_s-2</math> and <math>L:s = δ_s = [2; 2, 2, 2, 2, ...]</math>.


Loop.
Loop.
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* Let L<sub>1</sub> = period and divide L<sub>1</sub>.
* Let L<sub>1</sub> = period and divide L<sub>1</sub>.
<pre>L (first entry point)
<pre>L (first entry point)
L s (second entry point)
L s (left: second entry point)
sL L
sL L
ssL sL (from 1st: 2L 3s, from 2nd: 1L 2s)
ssL sL (from 1st: 2L 3s, from 2nd: 1L 2s)
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LsLsLss LsLss (from 1st: 5L 7s, from 2nd: 3L 4s)
LsLsLss LsLss (from 1st: 5L 7s, from 2nd: 3L 4s)
</pre>
</pre>
== Bimetal case ==
== Bimetallic case ==
First, Imagine <math>L = \sqrt{3}+1</math> and <math>s = 1</math> and divide L into
 
<math>\qquad L_1:s_1 = \sqrt{3}:1 = [1; 1, 2, 1, 2, ...]</math>.
 
OK, now <math>L_1 = \sqrt{3}</math> and <math>s_1 = s = 1</math>. Next, divide L<sub>1</sub> into
 
<math>\qquad L_2:s_2 = \sqrt{3}-1:1 = [0; 1, 2, 1, 2, ...]</math>.
 
No, <math>L_2 < s_2</math>. Let <math>s_3 := L_2</math> and <math>L_3 := s_2</math>. (Rotate the rectangle 90°)
 
OK, now <math>L_3 = 1</math> and <math>s_3 = \sqrt{3}-1</math> and <math>L_3:s_3 = (\sqrt{3}+1) / 2 = [1; 2, 1, 2, 1, ...]</math>. Next, divide L<sub>3</sub> into
 
<math>\qquad L_4:s_4 = (\sqrt{3}-1)/2:1 = [0; 2, 1, 2, 1, ...]</math>.
 
No, <math>L_4 < s_4</math>. Let new <math>s := L_4</math> and new <math>L := s_4</math>. (Rotate the rectangle 90°)
 
OK, now <math>L = \sqrt{3}-1</math> and <math>s = 2-\sqrt{3}</math> and <math>L_3:s_3 = \sqrt{3}+1 = [2; 1, 2, 1, 2, ...]</math>.
 
Loop.
 
In this case, we actually can choose from three entry points:
* Let L = period and divide L, or
* Let L<sub>1</sub> = period and divide L<sub>1</sub>, or
* Let L<sub>3</sub> = period and divide L<sub>3</sub>.
<pre>L (first entry point)
L s (left: second entry point)
sL L (right: third entry point)
LLs Ls (from 1st: 3L 2s, from 2nd: 2L 1s)
LsLss Lss (from 1st: 3L 5s, from 2nd: 2L 3s, from 3rd: 1L 2s)
sLLsLLL sLLL (from 1st: 8L 3s, from 2nd: 5L 2s, from 3rd: 3L 1s)
LLsLsLLsLsLs LLsLsLs (from 2nd: 7L 5s, from 3rd: 4L 3s)
</pre>
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