MOS diagrams

The moment-of-symmetry process of unfolding a scale takes, for most people, a conceptual leap or two. Below are visualizations of the process:

• From the Wilson Archives on Kraig Grady's [Anaphoria.com:
  • The first set of 32 horograms.
  • The Scale Tree is the basis of the horograms.
  • Moments of Symmetry, of Equal Divisions of the Octave.
• From David Finnamore's Elevenminstrel.com: 5- To 9-Tone, Octave-Repeating Scales From Wilson's Golden Horagrams of the Scale Tree.
• Charles Lucy describes a technique involving dis-continuous chains of fifths (i.e. skipping some).
• Joe Monzo's helixes could also be of use here...

L and s

The mechanics of scale generation are such that — when iterating from one scale to the next densest one — all large steps in the preceding scale become one large step and one small step in the new scale.

Another way to think about this is that a small-step-sized chunk has been split off of each of the former large steps. The remainder can be either larger or smaller than the small step

• If it is larger, then it stays the large step.
• If it is smaller, then it becomes the new small step, and everything that used to be a small step is now a large step.

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 [math]\displaystyle{ 1 }[/math] then we can treat our large steps’ size as equal to the [math]\displaystyle{ L{:}s }[/math] ratio.

So the [math]\displaystyle{ L{:}s }[/math] ratio decreases by [math]\displaystyle{ 1 }[/math] because if an [math]\displaystyle{ s }[/math]-sized chunk has been sliced off [math]\displaystyle{ L }[/math], and [math]\displaystyle{ s }[/math]’s size is [math]\displaystyle{ 1 }[/math], then [math]\displaystyle{ 1 }[/math] should be subtracted from [math]\displaystyle{ L }[/math].

When [math]\displaystyle{ L - s > s }[/math]:

[math]\displaystyle{ \begin{align} L’{:}s’ &= (L - s){:}s \\ &= (L - 1){:}1 \\ &= L - 1 \end{align} }[/math]

When [math]\displaystyle{ L - s < s }[/math], the result is simply reciprocated:

[math]\displaystyle{ \begin{align} L’{:}s’ &= s{:}(L - s) \\ &= 1{:}(L - 1) \\ &= \frac{1}{L - 1} \end{align} }[/math]