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 \lt 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.

L
sL
LLs (2L 1s)
sLsLL (3L 2s)
LLsLLsLs (5L 3s)
sLsLLsLsLLsLL (8L 5s)

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 L₁ into

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

No, [math]L_2 \lt 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 L₁ = period and divide L₁.

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)

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 \lt 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 \lt 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.

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)

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 \lt 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 \lt 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

	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 ¢