Neutral and interordinal intervals in MOS scales: Difference between revisions

Given a tuning of a primitive (single-period) [[mos]] pattern aLbs with a > b, we may define two types of notes "in the cracks of" interval categories defined by aLbs:

# Given 1 <= ''k'' <= a + b - 1, the '''neutral''' ''k''-step (abbrev. n''k''s) is the interval exactly halfway between the smaller ''k''-step and the larger ''k''-step of the mos. When the mos is generated by a (perfect) ''k''-step, this may instead be called the '''semiperfect''' ''k''-step (abbrev. sP''k''s), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) ''k''-step.

# Given 1 <= ''k'' <= a + b &minus; 2, and assuming that the larger ''k''-step < the smaller (''k'' + 1)-step, the '''interordinal''' between ''k''-steps and (''k'' + 1)-steps, denoted ''k''x(''k'' + 1)s or ''k''X(''k'' + 1)s (read "''k'' cross (''k'' + 1) step"), is the interval exactly halfway between the larger ''k''-step and the smaller (''k'' + 1)-step. The name comes from the fact that ''k''-steps in the diatonic mos are conventionally called "(''k'' + 1)ths".

Given such a mos, it's easy to notice the following properties of the simplest equal tunings for the mos, due to the way they divide the small step (s) and the chroma (L &minus; s):

* The basic equal tuning (2a + b)-ed contains neither neutrals nor interordinals.

* The monohard equal tuning (3a + b)-ed contains neutrals of that mos but not interordinals.

* The monosoft equal tuning (3a + 2b)-ed contains interordinals but not neutrals.

* 2(2a + b)-ed, twice the basic equal tuning, contains both types of intervals.

=== Armotonic (7L2s) ===

  1\18

  4\18 P1mans == P1atts

  5\18 1x2mans == sP1atts

  6\18 m2mans == A1atts

  7\18 n2mans == 1x2atts

  8\18 M2mans == m2atts

  9\18 2x3mans == n2atts

10\18 m3mans == M2atts

11\18 n3mans == 2x3atts

12\18 M3mans == d3atts

13\18 3x4mans == sP3atts

14\18 P4mans == P3atts

15\18 sP4mans

16\18 A4mans

17\18

== The Interordinal Theorem ==

The Interordinal Theorem relates the neutral (resp. interordinal) intervals of aLbs (with a > b) with the interordinal (resp. neutral) intervals of its parent mos, bL(a &minus; b)s.

# Every interordinal in basic aLbs (an interval that is exactly halfway between the larger k-step and the smaller (k+1)-step) is a neutral or semiperfect interval in the parent mos bL(a-b)s.

# Every interordinal interval in the parent mos bL(a-b)s of basic aLbs is a neutral or semiperfect interval in basic aLbs. The converse does not hold unless b = 1; (b - 1) counts the places in 2(2a+b)edo (twice the basic mos tuning for aLbs) where assigning interordinals to the parent mos of basic aLbs fails.

=== Preliminaries ===

Consider a primitive mos aLbs. Recall that (b - 1) satisfies:

(b - 1) = |(brightest mode of basic aLbs, ignoring equaves) ∩ (darkest mode of basic aLbs, ignoring equaves)|

= #{k : 0 < k < a+b and larger k-step of basic aLbs = smaller (k+1)-step of basic aLbs}.

Also recall that the following are equivalent for a mos aLbs:

* The parent mos, which is bL(a-b)s, has steps L + s and L. In this context, since aLbs is assumed to have hardness 2/1, bL(a-b)s has hardness 3/2 thus is strictly proper.

To prove the theorem, we need a couple lemmas.

Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n+k)x) - floor(nx) >= floor(kx).

floor((n+k)x) - floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) >= ceil((n+k)x - nx) - 1 = ceil(kx) - 1 = floor(kx).

Consider an m-note [[maximal evenness|maximally even]] mos of an n-equal division, and let 1 <= k <= m-1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n.

The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'<equave> or ceil(n'/m')\n'<equave>, implying the lemma.

Let n = 2a + b (the basic edo tuning of aLbs) and suppose that m\(2n) is an interordinal (where m must be odd) between k-steps and (k+1)-steps, denoted kX(k+1)ms. Recall that:

* A concrete mos tuning is improper iff its hardness is > 2/1 and the number of s steps it has is > 1.

Part (1) takes some step size arithmetic:

* Larger k+1-step of aLbs minus larger k-step of aLbs = Smaller k+1-step of aLbs minus smaller k-step of aLbs must be L, if 0 < k < k+1 < a+b. The reason is that larger 1-step = L, larger 2-step = LL, because there are more L’s than s’s.

* Smaller k+1-step of aLbs minus larger k-step of aLbs >= 0, with =0 at improprieties. At the values of k and k+1 that are proper, this equals s.

*** X = Larger (k+1)-step = (i+2)L + (k-i-1)s

*** Smaller (k+1)-step = (i+1)L + (k-i)s

*** Larger k-step = iL + (k-i)s

*** Y = Smaller k-step = (i-1)L + (k-i+1)s

** Since the smaller k-step has two more s steps than the larger (k+1)-step, Y has two more complete chunks of L's than X:

*** <pre>X=L^CsL...LsL...LsL^D</pre>

* To show that these actually occur in bL(a-b)s, consider smaller and larger j-steps ( 1 <= j <= a-1) in the parent mos. These intervals also occur in the mos aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a-1. Note that we are considering “junctures” between k-steps and k+1-steps in aLbs, excluding k = 0 and k = a+b-1, so the total number of “junctures” to consider is a+b-2. This proves part (1).

Part (2) is easier to see: where basic aLbs is improper, larger k-step = smaller k+1-step, and larger k+1-step - smaller k-step = 2(L-s) = 2s = L. But the step L is not two steps in 2n-edo.