User:Dummy index/Bimetallic MOS: Difference between revisions

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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 ==

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>.

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.

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

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<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.

Line 40:

* Let L1 = period and divide L1.

<pre>L (first entry point)

L s (second entry point)

L s (left: second entry point)

sL L

ssL sL (from 1st: 2L 3s, from 2nd: 1L 2s)

Line 46:

LsLsLss LsLss (from 1st: 5L 7s, from 2nd: 3L 4s)

</pre>

== ~~Bimetal~~ case ==

== 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 L1 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 L3 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 L1 = period and divide L1, or

* Let L3 = period and divide L3.

<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 L1 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 L2 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 L3 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 L5 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> i.e. <math>(s:L = 1:φ)</math> || 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> i.e. <math>(s:L = 1:δ_s)</math> || 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> i.e. <math>(s_3:L_3 = 1:(\sqrt{3}+1)/2)</math> || g = 507.18 ¢, p = 1200 ¢ || Flattone

|-

| Bimetallic (unnamed A-3) || <math>L_4:s_4 = (\sqrt{3}-1)/2:1</math> i.e. <math>(s:L = 1:\sqrt{3}+1)</math> || 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> i.e. <math>(s_5:L_5 = 1:(\sqrt{2}+1)/2)</math> || g = 543.70 ¢, p = 1200 ¢ ||

|-

| Bimetallic (unnamed B-5) || <math>L_6:s_6 = (\sqrt{2}-1)/2:1</math> i.e. <math>(s:L = 1:2\sqrt{2}+2)</math> || g = 205.89 ¢, p = 1200 ¢ ||

|}

@@ Line 1: / Line 1: @@
-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 ==
-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>.
-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.
-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
@@ Line 30: / Line 30: @@
 <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.
@@ Line 40: / Line 40: @@
 * Let L<sub>1</sub> = period and divide L<sub>1</sub>.
 <pre>L (first entry point)
-L s (second entry point)
+L s (left: second entry point)
 sL L
 ssL sL (from 1st: 2L 3s, from 2nd: 1L 2s)
@@ Line 46: / Line 46: @@
 LsLsLss LsLss (from 1st: 5L 7s, from 2nd: 3L 4s)
 </pre>
-== Bimetal case ==
+== 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 ¢ ||
+|}