View source for User:Dummy index/Bimetallic MOS

See [[Metallic MOS]]. The article is unelegant with the definition of the split operation, so l write it in my own way.

== Golden case ==
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>.

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 = [1; 1, 1, 1, 1, ...]</math>.

Loop.

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

<pre>L
sL
LLs (2L 1s)
sLsLL (3L 2s)
LLsLLsLs (5L 3s)
sLsLLsLsLLsLL (8L 5s)
</pre>
== Silver case ==
First, Imagine <math>L = δ_s</math> and <math>s = 1</math> and divide L into

<math>\qquad L_1:s_1 = δ_s-1:1 = [1; 2, 2, 2, 2, ...]</math>.

OK, now <math>L_1 = δ_s-1</math> and <math>s_1 = s = 1</math>. Next, divide L<sub>1</sub> into

<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°)

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.

In this case, we actually can choose from two entry points:
* Let L = period and divide L, or
* Let L<sub>1</sub> = period and divide L<sub>1</sub>.
<pre>L (first entry point)
L s (left: second entry point)
sL L
ssL sL (from 1st: 2L 3s, from 2nd: 1L 2s)
LLLs LLs (from 1st: 5L 2s, from 2nd: 3L 1s)
LsLsLss LsLss (from 1st: 5L 7s, from 2nd: 3L 4s)
</pre>
== Bimetallic cases ==
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:s = \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>

More intense pattern. First, Imagine <math>L = 2\sqrt{2}+2</math> and <math>s = 1</math> and divide L into

<math>\qquad L_1:s_1 = 2\sqrt{2}+1:1 = [3; 1, 4, 1, 4, ...]</math>.

OK, now <math>L_1 = 2\sqrt{2}+1</math> and <math>s_1 = s = 1</math>. Next, divide L<sub>1</sub> into

<math>\qquad L_2:s_2 = 2\sqrt{2}:1 = [2; 1, 4, 1, 4, ...]</math>.

OK, now <math>L_2 = 2\sqrt{2}</math> and <math>s_2 = s = 1</math>. Next, divide L<sub>2</sub> into

<math>\qquad L_3:s_3 = 2\sqrt{2}-1:1 = [1; 1, 4, 1, 4, ...]</math>.

OK, now <math>L_3 = 2\sqrt{2}-1</math> and <math>s_3 = s = 1</math>. Next, divide L<sub>3</sub> into

<math>\qquad L_4:s_4 = 2\sqrt{2}-2:1 = [0; 1, 4, 1, 4, ...]</math>.

No, <math>L_4 < s_4</math>. Let <math>s_5 := L_4</math> and <math>L_5 := s_4</math>. (Rotate the rectangle 90°)

OK, now <math>L_5 = 1</math> and <math>s_5 = 2\sqrt{2}-2</math> and <math>L_5:s_5 = (\sqrt{2}+1) / 2 = [1; 4, 1, 4, 1, ...]</math>. Next, divide L<sub>5</sub> into

<math>\qquad L_6:s_6 = (\sqrt{2}-1)/2:1 = [0; 4, 1, 4, 1, ...]</math>.

No, <math>L_6 < s_6</math>. Let new <math>s := L_6</math> and new <math>L := s_6</math>. (Rotate the rectangle 90°)

OK, now <math>L = 2\sqrt{2}-2</math> and <math>s = 3-2\sqrt{2}</math> and <math>L:s = 2\sqrt{2}+2 = [4; 1, 4, 1, 4, ...]</math>.

Loop.

In this case, we actually can choose from five entry points.

== Generator size ==
{| class="wikitable"
!  !! Dividing ratio !! Generator size !! Remarks
|-
| Golden || <math>L_1:s_1 = φ-1:1</math><br /><small>i.e. <math>(s:L = 1:φ)</math></small> || g = 458.36 ¢, p = 1200 ¢ || logarithmic phi
|-
| Silver 1st isotope || <math>L_1:s_1 = δ_s-1:1</math> || g = 702.94 ¢, p = 1200 ¢ || argent fifth
|-
| Silver || <math>L_2:s_2 = δ_s-2:1</math><br /><small>i.e. <math>(s:L = 1:δ_s)</math></small> || g = 351.47 ¢, p = 1200 ¢|| Imaginary, argent neutral third
|-
| Bimetallic (unnamed A-1) || <math>L_1:s_1 = \sqrt{3}:1</math> || g = 760.77 ¢, p = 1200 ¢ || 
|-
| Bimetallic (unnamed A-2) || <math>L_2:s_2 = \sqrt{3}-1:1</math><br /><small>i.e. <math>(s_3:L_3 = 1:(\sqrt{3}+1)/2)</math></small> || g = 507.18 ¢, p = 1200 ¢ || Flattone
|-
| Bimetallic (unnamed A-3) || <math>L_4:s_4 = (\sqrt{3}-1)/2:1</math><br /><small>i.e. <math>(s:L = 1:\sqrt{3}+1)</math></small> || g = 321.54 ¢, p = 1200 ¢ || Superkleismic
|-
| Bimetallic (unnamed B-1) || <math>L_1:s_1 = 2\sqrt{2}+1:1</math> || g = 951.47 ¢, p = 1200 ¢ || Semaphore
|-
| Bimetallic (unnamed B-2) || <math>L_2:s_2 = 2\sqrt{2}:1</math> || g = 886.56 ¢, p = 1200 ¢ || Hanson
|-
| Bimetallic (unnamed B-3) || <math>L_3:s_3 = 2\sqrt{2}-1:1</math> || g = 775.74 ¢, p = 1200 ¢ || Squares
|-
| Bimetallic (unnamed B-4) || <math>L_4:s_4 = 2\sqrt{2}-2:1</math><br /><small>i.e. <math>(s_5:L_5 = 1:(\sqrt{2}+1)/2)</math></small> || g = 543.70 ¢, p = 1200 ¢ || 
|-
| Bimetallic (unnamed B-5) || <math>L_6:s_6 = (\sqrt{2}-1)/2:1</math><br /><small>i.e. <math>(s:L = 1:2\sqrt{2}+2)</math></small> || g = 205.89 ¢, p = 1200 ¢ || 
|}