Neutral and interordinal intervals in MOS scales
- 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 aLbs〈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〈E〉:
- 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. The following always 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, where c = L − s is the chroma of the mos.
- Given 1 ≤ k ≤ a + b − 2, 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. If a > b, then aLbs〈E〉 has b − 1 proper interordinals. 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 diatonic interordinals (aka "interseptimals"), which are:
- 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
Given a primitive mos aLbs with a > b, it's easy to 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 2L5s (pentic, pt-) 1\24 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 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 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 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 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 − b)s.
Statement
Suppose a > b and gcd(a, b) = 1.
- Every interordinal of basic aLbs〈E〉 is a neutral or semiperfect interval of the parent mos bL(a-b)s〈E〉.
- Every interordinal interval of the parent mos bL(a-b)s〈E〉 of basic aLbs〈E〉 is a neutral or semiperfect interval of basic aLbs〈E〉.
- Except the neutral/semiperfect 1-step and the neutral/semiperfect (a+b-1)-step, every neutral or semiperfect interval of basic aLbs〈E〉 is a proper interordinal of bL(a-b)s〈E〉. The number (b − 1) counts the places in 2(2a+b)edE (twice the basic mos tuning for aLbs〈E〉) where the parent's interordinal is improper, being two steps away, instead of one step away, from each of the adjacent ordinal categories.
Preliminaries for the 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 "improprieties".
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.
To prove the theorem, we need a couple lemmas.
Lemma 1
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
Consider an m-note 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.
Proof
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.
Proof of Theorem
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. Recall that:
- 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.
Part (1) and (2) take some step size arithmetic:
- 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, 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:
Y=L^A sL...LsL...LsL...LsL...LsL^B
X=L^CsL...LsL...LsL^D
- 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.
- To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
- 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.
Corollary
If a > b, then aLbs〈E〉 has b − 1 proper interordinals.