Metallic MOS: Difference between revisions

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We are reasoning about MOS concepts in the abstract here. These truths about large and small steps are true whether they are 100¢ or 4516.8¢, and all we really care about are their ratios. So if we treat our small steps' size as 1 then we can treat our large steps' size as equal to the ''L'':''s'' ratio.

So the ''L'':''s'' ratio decreases by 1 because if an ''s''-sized chunk has been sliced off ~~<math>~~L~~</math>~~, and ~~<math>~~s~~</math>~~'s size is 1, then 1 should be subtracted from ~~<math>~~L~~</math>~~.

So the ''L'':''s'' ratio decreases by 1 because if an ''s''-sized chunk has been sliced off ''L'', and ''s''{{`s}} size is 1, then 1 should be subtracted from ''L''.

<math>

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</math>

But we're only getting started. This situation has recursive potential. We can now substitute <math>L + s</math> in for ~~<math>~~L~~</math>~~ as long as we also substitute in ''L'' for ''s'', and we'll still get a ratio that equals φ:

But we're only getting started. This situation has recursive potential. We can now substitute <math>L + s</math> in for ''L'' as long as we also substitute in ''L'' for ''s'', and we'll still get a ratio that equals φ:

<math>

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</math>

Now find any ''L'' in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ~~<math>~~L~~</math>~~ will be replaced with an ''L'' and an ''s''. After two iterations, the original ''L'' interval is now represented by two ''L''{{'s}} and an ''s''. And so forth.

Now find any ''L'' in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ''L'' will be replaced with an ''L'' and an ''s''. After two iterations, the original ''L'' interval is now represented by two ''L''{{'s}} and an ''s''. And so forth.

The same will hold for the right side of the interval pattern, for ''s'':

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</math>

Find any ~~<math>~~s~~</math>~~ in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ''s'' will be replaced with ''L''. After two iterations, the original ''s'' interval is now represented by an ''L'' and an ''s''. And so forth.

Find any ''s'' in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ''s'' will be replaced with ''L''. After two iterations, the original ''s'' interval is now represented by an ''L'' and an ''s''. And so forth.

Every MOS scale contains every scale earlier in its scale sequence. In other words, any interval that existed in an earlier scale will remain in all later scales. These earlier L's and s's that remain—only now spanning many ''L''{{'s}} and ''s''{{`s}} each—are precisely the larger intervals in the scale that also exhibit the φ ratio to each other.

Every MOS scale contains every scale earlier in its scale sequence. In other words, any interval that existed in an earlier scale will remain in all later scales. These earlier ''L''{{'s}} and s's that remain—only now spanning many ''L''{{'s}} and ''s''{{`s}} each—are precisely the larger intervals in the scale that also exhibit the φ ratio to each other.

=== Beyond golden cases ===

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</math>

We'll repeat the technique we used for the golden case: find any ~~<math>~~L~~</math>~~ in the horogram and observe how it gets split up as we iterate through the scale sequence. However, the complexity that silver introduces is that we don't look to the next iteration to see the next entry in the interval pattern; we have to skip an iteration. So if we just look at all the odd rings, ring 1, 3, 5, 7, etc. then we'll see the pattern. The same is true of ''s''.

We'll repeat the technique we used for the golden case: find any ''L'' in the horogram and observe how it gets split up as we iterate through the scale sequence. However, the complexity that silver introduces is that we don't look to the next iteration to see the next entry in the interval pattern; we have to skip an iteration. So if we just look at all the odd rings, ring 1, 3, 5, 7, etc. then we'll see the pattern. The same is true of ''s''.

And if we want to understand the interval pattern for {{nowrap|δ''s'' − 1}}, we'll look at the right and left sides separately:

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The reason the other cases exhibited such overlapping is that the small step size of the next ratio in the equivalence pattern became an ''L'', which is the same as the ''L'' size of the preceding ratio. However, for the silver mean's first isotope here, no such link exists, since ''s'' is substituted not for ''L'', but {{nowrap|''L'' + ''s''}}.

Another way of looking at this is: for δ''s'' and φ, it was the case that both ~~<math>~~s~~</math>~~ and ~~<math>~~L~~</math>~~'s interval sequences were the same, just offset from each other by a step. Whereas for {{nowrap|δ''s'' − 1}},''s'' and L's interval sequences are completely different.

Another way of looking at this is: for δ''s'' and φ, it was the case that both ''s'' and ''L''{{'s}} interval sequences were the same, just offset from each other by a step. Whereas for {{nowrap|δ''s'' − 1}}, ''s'' and ''L''{{'s}} interval sequences are completely different.

Finally, for the bronze ratio,

@@ Line 418: / Line 418: @@
 We are reasoning about MOS concepts in the abstract here. These truths about large and small steps are true whether they are 100¢ or 4516.8¢, and all we really care about are their ratios. So if we treat our small steps' size as 1 then we can treat our large steps' size as equal to the ''L'':''s'' ratio.
-So the ''L'':''s'' ratio decreases by 1 because if an ''s''-sized chunk has been sliced off <math>L</math>, and <math>s</math>'s size is 1, then 1 should be subtracted from <math>L</math>.
+So the ''L'':''s'' ratio decreases by 1 because if an ''s''-sized chunk has been sliced off ''L'', and ''s''{{`s}} size is 1, then 1 should be subtracted from ''L''.
 <math>
@@ Line 704: / Line 704: @@
 </math>
-But we're only getting started. This situation has recursive potential. We can now substitute <math>L + s</math> in for <math>L</math> as long as we also substitute in ''L'' for ''s'', and we'll still get a ratio that equals φ:
+But we're only getting started. This situation has recursive potential. We can now substitute <math>L + s</math> in for ''L'' as long as we also substitute in ''L'' for ''s'', and we'll still get a ratio that equals φ:
 <math>
@@ Line 743: / Line 743: @@
 </math>
-Now find any ''L'' in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, <math>L</math> will be replaced with an ''L'' and an ''s''. After two iterations, the original ''L'' interval is now represented by two ''L''{{'s}} and an ''s''. And so forth.
+Now find any ''L'' in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ''L'' will be replaced with an ''L'' and an ''s''. After two iterations, the original ''L'' interval is now represented by two ''L''{{'s}} and an ''s''. And so forth.
 The same will hold for the right side of the interval pattern, for ''s'':
@@ Line 756: / Line 756: @@
 </math>
-Find any <math>s</math> in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ''s'' will be replaced with ''L''. After two iterations, the original ''s'' interval is now represented by an ''L'' and an ''s''. And so forth.
+Find any ''s'' in the horogram and observe how it gets split up as we iterate through the scale sequence. In the next iteration, ''s'' will be replaced with ''L''. After two iterations, the original ''s'' interval is now represented by an ''L'' and an ''s''. And so forth.
-Every MOS scale contains every scale earlier in its scale sequence. In other words, any interval that existed in an earlier scale will remain in all later scales. These earlier L's and s's that remain—only now spanning many ''L''{{'s}} and ''s''{{`s}} each—are precisely the larger intervals in the scale that also exhibit the φ ratio to each other.
+Every MOS scale contains every scale earlier in its scale sequence. In other words, any interval that existed in an earlier scale will remain in all later scales. These earlier ''L''{{'s}} and s's that remain—only now spanning many ''L''{{'s}} and ''s''{{`s}} each—are precisely the larger intervals in the scale that also exhibit the φ ratio to each other.
 === Beyond golden cases ===
@@ Line 828: / Line 828: @@
 </math>
-We'll repeat the technique we used for the golden case: find any <math>L</math> in the horogram and observe how it gets split up as we iterate through the scale sequence. However, the complexity that silver introduces is that we don't look to the next iteration to see the next entry in the interval pattern; we have to skip an iteration. So if we just look at all the odd rings, ring 1, 3, 5, 7, etc. then we'll see the pattern. The same is true of ''s''.
+We'll repeat the technique we used for the golden case: find any ''L'' in the horogram and observe how it gets split up as we iterate through the scale sequence. However, the complexity that silver introduces is that we don't look to the next iteration to see the next entry in the interval pattern; we have to skip an iteration. So if we just look at all the odd rings, ring 1, 3, 5, 7, etc. then we'll see the pattern. The same is true of ''s''.
 And if we want to understand the interval pattern for {{nowrap|δ<sub>''s''</sub> − 1}}, we'll look at the right and left sides separately:
@@ Line 866: / Line 866: @@
 The reason the other cases exhibited such overlapping is that the small step size of the next ratio in the equivalence pattern became an ''L'', which is the same as the ''L'' size of the preceding ratio. However, for the silver mean's first isotope here, no such link exists, since ''s'' is substituted not for ''L'', but {{nowrap|''L'' + ''s''}}.
-Another way of looking at this is: for δ<sub>''s''</sub> and φ, it was the case that both <math>s</math> and <math>L</math>'s interval sequences were the same, just offset from each other by a step. Whereas for {{nowrap|δ<sub>''s''</sub> − 1}},''s'' and L's interval sequences are completely different.
+Another way of looking at this is: for δ<sub>''s''</sub> and φ, it was the case that both ''s'' and ''L''{{'s}} interval sequences were the same, just offset from each other by a step. Whereas for {{nowrap|δ<sub>''s''</sub> − 1}}, ''s'' and ''L''{{'s}} interval sequences are completely different.
 Finally, for the bronze ratio,