Neutral and interordinal intervals in MOS scales: Difference between revisions

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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:
: ''This page assumes that the reader is familiar with [[TAMNAMS]] MOS interval and step ratio names.''
# 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 is instead 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 and the chroma, respectively:
Given a tuning of a primitive (i.e. single-period) [[MOS]] pattern aLbs{{angbr|E}} with arbitrary [[equave]] E in a specific tuning (i.e. with a specific [[hardness]] value for L/s), we may define two types of notes "in the cracks of" interval categories defined by aLbs{{angbr|E}}:
* the hard equal tuning ((3a + b)-ed) contains neutrals of that mos but not interordinals.
# Given 1 ≤ ''k'' ≤ a + b &minus; 1, the '''neutral''' ''k''-mosstep or ''k''-step (abbrev. n''k''ms, 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. Depending on whether the imperfect generator is augmented or diminished, the corresponding semiperfect generator may be called '''semiaugmented''' or '''semidiminished'''.
* the soft equal tuning (3a + 2b)-ed) contains interordinals but not neutrals.
#: We use '''semichroma''' for the quantity c/2 where c = L &minus; s is the [[chroma]] of the MOS. The semichroma represents the difference between ordinary MOS intervals and their neutralized counterparts, in the sense that the following holds for a given interval class of ''k''-steps between 1-steps and (a + b &minus; 1)-steps, inclusive:
* 2(2a + b)-ed, twice the basic equal tuning, contains both types of intervals.
#: neutral ''k''-step = smaller ''k''-step + c/2 = larger ''k''-step &minus; c/2
# Given 0 ≤ ''k'' ≤ a + b &minus; 1, and assuming that the larger ''k''-step < the smaller (''k'' + 1)-step, the '''interordinal''' between ''k''-steps and (''k'' + 1)-steps, denoted ''k''×(''k'' + 1)(m)s (read "''k'' cross (''k'' + 1) (mos)step" or "''k'' inter (''k'' + 1) (mos)step"), is the interval exactly halfway between the larger ''k''-step and the smaller (''k'' + 1)-step.
#: If the smaller (''k'' + 1)-step is ''strictly larger'' than the larger ''k''-step in ''basic'' aLbs, ''k''×(''k'' + 1) is called a '''proper interordinal'''; otherwise, it is called an '''improper interordinal'''. If a > b, then aLbs{{angbr|E}} has a + 1 proper interordinals, including 0×1ms and (a+b&minus;1)×(a+b)ms.
#: We call s/2 (or 0×1ms) the '''interizer'''{{idiosyncratic}}. The interizer is of note since the following holds for any proper interordinal interval ''k''-inter-(''k'' + 1)-step:
#: ''k''-inter-(''k'' + 1)-step = larger ''k''-step + s/2 = smaller (''k'' + 1)-step &minus; s/2.
Neutral ''k''-steps generalize neutral interval categories based on the diatonic MOS, which are:
* neutral 1-diastep = neutral 2nd (A-(Bd)
* neutral 2-diastep = neutral 3rd (A-Ct)
* semiperfect 3-diastep = semiperfect 4th (A-Dt)
* semiperfect 4-diastep = semiperfect 5th (A-Ed)
* neutral 5-diastep = neutral 6th (A-Ft)
* neutral 6-diastep = neutral 7th (A-Gt)
 
Though the term ''interordinal'' is intended to be JI-agnostic and generalizable to non-diatonic mosses, the term comes from the fact that ''k''-steps in the diatonic MOS are conventionally called "(''k'' + 1)ths". Proper interordinals in other mosses generalize [[interordinal]] categories that are novel with respect to diatonic (aka "interseptimals"), which are:
* 0-inter-1-diastep = "unison-inter-2nd" = s/2
* 1-inter-2-diastep = "2nd-inter-3rd" = semifourth = chthonic = ultramajor 2nd
* 2-inter-3-diastep = "3rd-inter-4th" = semisixth = naiadic = ultramajor 3rd
* 4-inter-5-diastep = "5th-inter-6th" = semitenth = cocytic = inframinor 6th
* 5-inter-6-diastep = "6th-inter-7th" = semitwelfth = ouranic = inframinor 7th
* 6-inter-7-diastep = "7th-inter-octave" = octave &minus; s/2
Improper interordinals, in contrast, represent intervals that are technically between ordinal categories but occur within the MOS scale unlike proper interordinals which are wholly outside the interval categories defined by the MOS. The diatonic example of this is the tritone, which is interordinal but falls within diatonic interval categories as the [[12edo|basic tuning]] of diatonic tunes both the augmented 3-diastep and the diminished 4-diastep to 600 cents.
 
Given a primitive MOS aLbs with a > b, one can observe the following properties of the simplest equal tunings for the MOS, due to the way they divide the small step (s) and the chroma (c = L &minus; s). Note that s separates adjacent ordinal categories (i.e. [[interval class]]es) while c separates larger and smaller intervals in the same ordinal category.
* The basic equal tuning (2a + b)-edE contains neither neutrals nor interordinals, since both s and c are one edo step. (For the diatonic MOS 5L2s, this tuning is [[12edo]].)
* The monohard equal tuning (3a + b)-edE contains neutrals of that MOS but not interordinals, since c is two edo steps but s is one edo step. (For diatonic, this tuning is [[17edo]].)
* The monosoft equal tuning (3a + 2b)-edE contains interordinals but not neutrals, since c is one edo step and s is two edo steps. (For diatonic, this tuning is [[19edo]].)
* 2(2a + b)-edE, twice the basic equal tuning, contains both types of intervals, since both c and s are two edo steps. (For diatonic, this tuning is [[24edo]].)


== Examples ==
== Examples ==
These charts show where neutrals and interordinals corresponding to example 2/1-equivalent mosses are located in 2(2a + b)-edo (twice the basic edo tuning). Intervals larger than 600c have been omitted, as the structure is symmetrical about 600c.
=== Diatonic (5L2s) ===
=== Diatonic (5L2s) ===
<pre>
Basic 5L2s (diatonic, dia-): 12edo
Parent MOS: soft 2L3s (pentic, pt-)
1\24 0×1dias  (1st×2nd)
2\24 m1dias    (m2nd)
3\24 n1dias    (n2nd)
4\24 M1dias    (M2nd)      == m1pts
5\24 1×2dias  (2nd×3rd)  == n1pts
6\24 m2dias    (m3rd)      == M1pts
7\24 n2dias    (n3rd)      == 1×2pts
8\24 M2dias    (M3rd)      == d2pts
9\24 2×3dias  (3rd×4th)  == sP2pts
10\24 P3dias    (P4th)      == P2pts
11\24 sP3dias  (sP4th)
12\24 A3/d4dias (A4th/d5th) == 2×3pts
</pre>
=== m-chromatic (7L5s) ===
=== m-chromatic (7L5s) ===
<pre>
Basic 7L5s (m-chromatic, mchr-): 19edo
Parent MOS: soft 5L2s (diatonic, dia-)
1\38 0×1s
2\38 m1s
3\38 n1s
4\38 M1s      == m1dias (m2nd)
5\38 1×2s    == n1dias (n2nd)
6\38 m2s      == M1dias (M2nd)
7\38 n2s
8\38 M2s/m3s  == 1×2dias (2nd×3rd)
9\38 n3s
10\38 M3s      == m2dias (m3rd)
11\38 3×4s    == n2dias (n3rd)
12\38 m4s      == M2dias (M3rd)
13\38 n4s
14\38 M4s/d5s  == 2×3dias (3rd×4th)
15\38 sPs
16\38 P5s      == P3dias (P4th)
17\38 5×6s    == sP3dias (sP4th)
18\38 m6s      == A3dias (A4th)
19\38 n6s      == 3×4dias (4th×5th)
</pre>
=== Manual (4L1s) ===
<pre>
Basic 2/1-equivalent 4L1s (manual, man-): 9edo
Parent MOS: soft 1L3s (antetric, att-)
1\18 0×1mans
2\18 d1mans
3\18 sP1mans
4\18 P1mans  == P1atts
5\18 1×2mans == sP1atts
6\18 m2mans  == A1atts
7\18 n2mans  == 1×2atts
8\18 M2mans  == m2atts
9\18 2×3mans == n2atts
</pre>
=== Oneirotonic (5L3s) ===
=== Oneirotonic (5L3s) ===
=== Armotonic (7L2s) ===
<pre>
Basic 5L3s (oneirotonic, onei-): 13edo
Parent MOS: soft 3L2s (anpentic, apt-)
1\26 0×1oneis
2\26 m1oneis
3\26 n1oneis
4\26 M1oneis    == m1apts
5\26 1×2oneis    == n1apts
6\26 m2oneis    == M1apts
7\26 n2oneis
8\26 M2/d3oneis  == 1×2apts
9\26 sP3oneis
10\26 P3oneis    == P2apts
11\26 3×4oneis    == sP2apts
12\26 m4oneis    == A2apts
13\26 n4oneis    == 2×3apts
</pre>
 
== Interordinal-Neutral Theorem ==
The Interordinal-Neutral 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, generalizing an observation by [[User:Godtone]] relating neutrals and interordinals of [[5L 2s]] to those of the parent MOS [[2L 3s]].
=== Statement ===
Suppose a > b and gcd(a, b) = 1.
# Every proper interordinal of basic aLbs{{angbr|E}} save for 0×1ms and (a+b&minus;1)×(a+b)ms is a neutral or semiperfect interval of the parent MOS bL(a &minus; b)s{{angbr|E}}. The interizer of aLbs{{angbr|E}}, 0x1ms = s/2, is the semichroma of the parent MOS.
# Every interordinal interval of the parent MOS bL(a &minus; b)s{{angbr|E}} of basic aLbs{{angbr|E}} excluding 0×1ms and (a+b&minus;1)×(a+b)ms is a neutral or semiperfect interval of basic aLbs{{angbr|E}}.
# Except the neutral/semiperfect 1-step and the neutral/semiperfect (a + b &minus; 1)-step, every neutral or semiperfect interval of basic aLbs{{angbr|E}} is a proper interordinal of bL(a &minus; b)s{{angbr|E}}. The number (b &minus; 1) counts the places in 2(2a + b)edE (twice the basic MOS tuning for aLbs{{angbr|E}}) where the parent's interordinal is improper, being two steps away, instead of one step away, from each of the adjacent ordinal categories.
# aLbs{{angbr|E}} has a + 1 proper interordinals and b &minus; 1 improper interordinals.


== The Interordinal Theorem ==
=== Proof ===
Consider a primitive mos aLbs. Recall that (b - 1) satisfies:
Below we assume that the equave is 2/1, but the proof generalizes to any equave.


(b - 1) = |(brightest mode of basic aLbs, ignoring equaves) ∩ (darkest mode of basic aLbs, ignoring equaves)|
Consider a primitive MOS aLbs. Recall that (b &minus; 1) satisfies:
= #{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:
(b &minus; 1) = |(brightest mode of basic aLbs, ignoring equaves) ∩ (darkest mode of basic aLbs, ignoring equaves)|
* a > b.
= #{k : 0 < k < a + b and larger k-step of basic aLbs = smaller (k + 1)-step of basic aLbs} = # of potential improprieties,
* 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.


=== Lemma 1 ===
where ''potential improprieties'' are pairs of adjacent interval classes that witness the impropriety of a hard-of-basic tuning of the MOS. Part (4) immediately follows.
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).
==== Proof ====
floor((n+k)x) - floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) >= ceil((n+k)x - nx) - 1 = ceil(kx) - 1 = floor(kx).


=== Discretizing Lemma ===
Also recall that the following are equivalent for a MOS aLbs:
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.
* a > b.
==== Proof ====
* The parent MOS, which is bL(a&minus;b)s, has steps L + s and L. In this context, since aLbs is assumed to have hardness 2/1, bL(a&minus;b)s has hardness 3/2 thus is strictly proper.
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.


=== Statement of the theorem ===
Finally, recall that:
Suppose a > b and gcd(a, b) = 1.
# 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, but a neutral or semiperfect interval in basic aLbs is not always an interordinal interval in the parent mos.
# (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.
==== Proof ====
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:
* In basic aLbs, s = 1\n = 2\2n.  
* In basic aLbs, s = 1\n = 2\2n.  
* A concrete mos tuning is improper iff its hardness is > 2/1 and the number of s steps it has is > 1.
* A concrete MOS tuning is improper if and only if its hardness is > 2/1 and the number of s steps it has is > 1.
 
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 k×(k + 1)ms. For parts (1) and (2):
* (Note that we are assuming the basic tuning of the MOS) smaller (k + 1)-step of aLbs minus larger k-step of aLbs ≥ 0, with equality at improprieties (essentially by definition). At the values of k and k+1 that are proper, this equals s. To see why, observe that the number of generators represented by the difference must be b mod (a+b), since L is obtained by stacking b bright generators. The difference is at most &minus;1 generators, since the larger interval of the difference between the larger k-step and the larger k+1 step, a chroma sharper, does occur (in the brightest mode of the MOS), and cannot be less than &minus;a generators, lest the gap be nonpositive in the basic MOS, a contradiction since kx(k+1) is a proper interordinal. Hence the difference must be &minus;a generators, corresponding to s.
* As s is the chroma of bL(a &minus; b)s, it ''would'' be the difference between major and minor intervals in the parent MOS, assuming these interval sizes (smaller (k + 1)-step, larger k-step) occur in the parent; so k×(k + 1) would become neutral or semiperfect.
* To show that these actually occur in bL(a &minus; b)s, consider smaller and larger j-steps (1 ≤ j ≤ a &minus; 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 &minus; 1. These j's correspond to values of k such that larger k-step < smaller (k + 1)-step. Note that we are considering “junctures” between k-steps and (k + 1)-steps in aLbs, excluding k = 0 and k = a + b &minus; 1, so the total number of “junctures” to consider is finite, namely a + b &minus; 2. This proves parts (1) and (2).


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.
Part (3) is also immediate now: when larger k-step = smaller (k + 1)-step, larger (k + 1)-step &minus; smaller k-step = 2(L &minus; s) = 2s = L. The step L is 4 steps in 2n-edo. {{qed}}


Part (1) takes some step size arithmetic:
[[Category:MOS scale]]
* 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.
[[Category:Pages with proofs]]
* 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.
** To see why, suppose the difference is L (here k >= i >= 1, 0 < k < k+1 < a+b):
*** 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>Y=L^A sL...LsL...LsL...LsL...LsL^B</pre>
*** <pre>X=L^CsL...LsL...LsL^D</pre>
** Let r be the number of complete chunks (maximal contiguous strings of L's) in X, and let μ = n/(a+b). Note that the spacings between the s steps of a mos form a maximally even mos, and that since a > b, 3/2 < μ < 2. By the Discretizing Lemma, we have (|·| denotes length):
*** 1+A+B+floor((r+2)μ) <= |Y| <= 1+A+B+ceil((r+2)μ)
*** 1+C+D+floor(rμ) <= |X| <= 1+C+D+ceil(rμ)
*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor((r+2)μ) - ceil(rμ)
*** = (A+B)-(C+D)-1 + floor((r+2)μ) - floor(rμ)
*** (Lemma 1) >= (A+B)-(C+D)-1 + floor(2μ)
*** Hence, (C+D)-(A+B) >= floor(2μ).
*** Also, (C+D)-(A+B) <= 2*ceil(μ)-2 = 2(ceil(μ)-1) = 2*floor(μ). This is a contradiction: as 3/2 < μ < 2, we have floor(2μ) - 2*floor(μ) = 1.
* As s is the chroma of bL(a-b)s, it *would* be the difference between major and minor intervals in the parent mos, assuming these interval sizes (smaller k+1-step, larger k-step) occur in the parent; so kX(k+1) would become neutral or semiperfect.
* 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 (2).

Latest revision as of 12:30, 24 March 2024

This page assumes that the reader is familiar with TAMNAMS MOS interval and step ratio names.

Given a tuning of a primitive (i.e. single-period) MOS pattern aLbsE with arbitrary equave E in a specific tuning (i.e. with a specific hardness value for L/s), we may define two types of notes "in the cracks of" interval categories defined by aLbsE:

  1. Given 1 ≤ k ≤ a + b − 1, the neutral k-mosstep or k-step (abbrev. nkms, nks) 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. sPks), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) k-step. Depending on whether the imperfect generator is augmented or diminished, the corresponding semiperfect generator may be called semiaugmented or semidiminished.
    We use semichroma for the quantity c/2 where c = L − s is the chroma of the MOS. The semichroma represents the difference between ordinary MOS intervals and their neutralized counterparts, in the sense that the following holds for a given interval class of k-steps between 1-steps and (a + b − 1)-steps, inclusive:
    neutral k-step = smaller k-step + c/2 = larger k-step − c/2
  2. Given 0 ≤ k ≤ a + b − 1, and assuming that the larger k-step < the smaller (k + 1)-step, the interordinal between k-steps and (k + 1)-steps, denoted k×(k + 1)(m)s (read "k cross (k + 1) (mos)step" or "k inter (k + 1) (mos)step"), is the interval exactly halfway between the larger k-step and the smaller (k + 1)-step.
    If the smaller (k + 1)-step is strictly larger than the larger k-step in basic aLbs, k×(k + 1) is called a proper interordinal; otherwise, it is called an improper interordinal. If a > b, then aLbsE has a + 1 proper interordinals, including 0×1ms and (a+b−1)×(a+b)ms.
    We call s/2 (or 0×1ms) the interizer[idiosyncratic term]. The interizer is of note since the following holds for any proper interordinal interval k-inter-(k + 1)-step:
    k-inter-(k + 1)-step = larger k-step + s/2 = smaller (k + 1)-step − s/2.

Neutral k-steps generalize neutral interval categories based on the diatonic MOS, which are:

  • neutral 1-diastep = neutral 2nd (A-(Bd)
  • neutral 2-diastep = neutral 3rd (A-Ct)
  • semiperfect 3-diastep = semiperfect 4th (A-Dt)
  • semiperfect 4-diastep = semiperfect 5th (A-Ed)
  • neutral 5-diastep = neutral 6th (A-Ft)
  • neutral 6-diastep = neutral 7th (A-Gt)

Though the term interordinal is intended to be JI-agnostic and generalizable to non-diatonic mosses, the term comes from the fact that k-steps in the diatonic MOS are conventionally called "(k + 1)ths". Proper interordinals in other mosses generalize interordinal categories that are novel with respect to diatonic (aka "interseptimals"), which are:

  • 0-inter-1-diastep = "unison-inter-2nd" = s/2
  • 1-inter-2-diastep = "2nd-inter-3rd" = semifourth = chthonic = ultramajor 2nd
  • 2-inter-3-diastep = "3rd-inter-4th" = semisixth = naiadic = ultramajor 3rd
  • 4-inter-5-diastep = "5th-inter-6th" = semitenth = cocytic = inframinor 6th
  • 5-inter-6-diastep = "6th-inter-7th" = semitwelfth = ouranic = inframinor 7th
  • 6-inter-7-diastep = "7th-inter-octave" = octave − s/2

Improper interordinals, in contrast, represent intervals that are technically between ordinal categories but occur within the MOS scale unlike proper interordinals which are wholly outside the interval categories defined by the MOS. The diatonic example of this is the tritone, which is interordinal but falls within diatonic interval categories as the basic tuning of diatonic tunes both the augmented 3-diastep and the diminished 4-diastep to 600 cents.

Given a primitive MOS aLbs with a > b, one can observe the following properties of the simplest equal tunings for the MOS, due to the way they divide the small step (s) and the chroma (c = L − s). Note that s separates adjacent ordinal categories (i.e. interval classes) while c separates larger and smaller intervals in the same ordinal category.

  • The basic equal tuning (2a + b)-edE contains neither neutrals nor interordinals, since both s and c are one edo step. (For the diatonic MOS 5L2s, this tuning is 12edo.)
  • The monohard equal tuning (3a + b)-edE contains neutrals of that MOS but not interordinals, since c is two edo steps but s is one edo step. (For diatonic, this tuning is 17edo.)
  • The monosoft equal tuning (3a + 2b)-edE contains interordinals but not neutrals, since c is one edo step and s is two edo steps. (For diatonic, this tuning is 19edo.)
  • 2(2a + b)-edE, twice the basic equal tuning, contains both types of intervals, since both c and s are two edo steps. (For diatonic, this tuning is 24edo.)

Examples

These charts show where neutrals and interordinals corresponding to example 2/1-equivalent mosses are located in 2(2a + b)-edo (twice the basic edo tuning). Intervals larger than 600c have been omitted, as the structure is symmetrical about 600c.

Diatonic (5L2s)

 Basic 5L2s (diatonic, dia-): 12edo
 Parent MOS: soft 2L3s (pentic, pt-)
 1\24 0×1dias   (1st×2nd)
 2\24 m1dias    (m2nd)
 3\24 n1dias    (n2nd)
 4\24 M1dias    (M2nd)      == m1pts
 5\24 1×2dias   (2nd×3rd)   == n1pts
 6\24 m2dias    (m3rd)      == M1pts
 7\24 n2dias    (n3rd)      == 1×2pts
 8\24 M2dias    (M3rd)      == d2pts
 9\24 2×3dias   (3rd×4th)   == sP2pts
10\24 P3dias    (P4th)      == P2pts
11\24 sP3dias   (sP4th)
12\24 A3/d4dias (A4th/d5th) == 2×3pts

m-chromatic (7L5s)

Basic 7L5s (m-chromatic, mchr-): 19edo
Parent MOS: soft 5L2s (diatonic, dia-)
 1\38 0×1s
 2\38 m1s
 3\38 n1s
 4\38 M1s      == m1dias (m2nd)
 5\38 1×2s     == n1dias (n2nd)
 6\38 m2s      == M1dias (M2nd)
 7\38 n2s
 8\38 M2s/m3s  == 1×2dias (2nd×3rd)
 9\38 n3s
10\38 M3s      == m2dias (m3rd)
11\38 3×4s     == n2dias (n3rd)
12\38 m4s      == M2dias (M3rd)
13\38 n4s
14\38 M4s/d5s  == 2×3dias (3rd×4th)
15\38 sPs
16\38 P5s      == P3dias (P4th)
17\38 5×6s     == sP3dias (sP4th)
18\38 m6s      == A3dias (A4th)
19\38 n6s      == 3×4dias (4th×5th)

Manual (4L1s)

 Basic 2/1-equivalent 4L1s (manual, man-): 9edo
 Parent MOS: soft 1L3s (antetric, att-)
 1\18 0×1mans
 2\18 d1mans
 3\18 sP1mans
 4\18 P1mans  == P1atts
 5\18 1×2mans == sP1atts
 6\18 m2mans  == A1atts
 7\18 n2mans  == 1×2atts
 8\18 M2mans  == m2atts
 9\18 2×3mans == n2atts

Oneirotonic (5L3s)

 Basic 5L3s (oneirotonic, onei-): 13edo
 Parent MOS: soft 3L2s (anpentic, apt-)
 1\26 0×1oneis
 2\26 m1oneis
 3\26 n1oneis
 4\26 M1oneis     == m1apts
 5\26 1×2oneis    == n1apts
 6\26 m2oneis     == M1apts
 7\26 n2oneis
 8\26 M2/d3oneis  == 1×2apts
 9\26 sP3oneis
10\26 P3oneis     == P2apts
11\26 3×4oneis    == sP2apts
12\26 m4oneis     == A2apts
13\26 n4oneis     == 2×3apts

Interordinal-Neutral Theorem

The Interordinal-Neutral Theorem relates the neutral (resp. interordinal) intervals of aLbs (with a > b) with the interordinal (resp. neutral) intervals of its parent MOS, bL(a − b)s, generalizing an observation by User:Godtone relating neutrals and interordinals of 5L 2s to those of the parent MOS 2L 3s.

Statement

Suppose a > b and gcd(a, b) = 1.

  1. Every proper interordinal of basic aLbsE save for 0×1ms and (a+b−1)×(a+b)ms is a neutral or semiperfect interval of the parent MOS bL(a − b)sE. The interizer of aLbsE, 0x1ms = s/2, is the semichroma of the parent MOS.
  2. Every interordinal interval of the parent MOS bL(a − b)sE of basic aLbsE excluding 0×1ms and (a+b−1)×(a+b)ms is a neutral or semiperfect interval of basic aLbsE.
  3. Except the neutral/semiperfect 1-step and the neutral/semiperfect (a + b − 1)-step, every neutral or semiperfect interval of basic aLbsE is a proper interordinal of bL(a − b)sE. The number (b − 1) counts the places in 2(2a + b)edE (twice the basic MOS tuning for aLbsE) where the parent's interordinal is improper, being two steps away, instead of one step away, from each of the adjacent ordinal categories.
  4. aLbsE has a + 1 proper interordinals and b − 1 improper interordinals.

Proof

Below we assume that the equave is 2/1, but the proof generalizes to any equave.

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} = # of potential improprieties,

where potential improprieties are pairs of adjacent interval classes that witness the impropriety of a hard-of-basic tuning of the MOS. Part (4) immediately follows.

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

  • a > b.
  • 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.

Finally, recall that:

  • In basic aLbs, s = 1\n = 2\2n.
  • A concrete MOS tuning is improper if and only if its hardness is > 2/1 and the number of s steps it has is > 1.

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 k×(k + 1)ms. For parts (1) and (2):

  • (Note that we are assuming the basic tuning of the MOS) smaller (k + 1)-step of aLbs minus larger k-step of aLbs ≥ 0, with equality at improprieties (essentially by definition). At the values of k and k+1 that are proper, this equals s. To see why, observe that the number of generators represented by the difference must be b mod (a+b), since L is obtained by stacking b bright generators. The difference is at most −1 generators, since the larger interval of the difference between the larger k-step and the larger k+1 step, a chroma sharper, does occur (in the brightest mode of the MOS), and cannot be less than −a generators, lest the gap be nonpositive in the basic MOS, a contradiction since kx(k+1) is a proper interordinal. Hence the difference must be −a generators, corresponding to s.
  • As s is the chroma of bL(a − b)s, it would be the difference between major and minor intervals in the parent MOS, assuming these interval sizes (smaller (k + 1)-step, larger k-step) occur in the parent; so k×(k + 1) would become neutral or semiperfect.
  • 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. These j's correspond to values of k such that larger k-step < smaller (k + 1)-step. 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 finite, namely a + b − 2. This proves parts (1) and (2).

Part (3) is also immediate now: when larger k-step = smaller (k + 1)-step, larger (k + 1)-step − smaller k-step = 2(L − s) = 2s = L. The step L is 4 steps in 2n-edo. [math]\displaystyle{ \square }[/math]