Neutral and interordinal intervals in MOS scales: Difference between revisions
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5\24 1×2dias (2nd×3rd) == n1pts | 5\24 1×2dias (2nd×3rd) == n1pts | ||
6\24 m2dias (m3rd) == M1pts | 6\24 m2dias (m3rd) == M1pts | ||
7\24 n2dias (n3rd) == | 7\24 n2dias (n3rd) == 1×2pts | ||
8\24 M2dias (M3rd) == d2pts | 8\24 M2dias (M3rd) == d2pts | ||
9\24 2×3dias (3rd×4th) == sP2pts | 9\24 2×3dias (3rd×4th) == sP2pts | ||
10\24 P3dias (P4th) == P2pts | 10\24 P3dias (P4th) == P2pts | ||
11\24 sP3dias (sP4th) | 11\24 sP3dias (sP4th) | ||
12\24 A3/d4dias (A4th/d5th) == | 12\24 A3/d4dias (A4th/d5th) == 2×3pts | ||
</pre> | </pre> | ||
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6\38 m2s == M1dias (M2nd) | 6\38 m2s == M1dias (M2nd) | ||
7\38 n2s | 7\38 n2s | ||
8\38 M2s/m3s == | 8\38 M2s/m3s == 1×2dias (2nd×3rd) | ||
9\38 n3s | 9\38 n3s | ||
10\38 M3s == m2dias (m3rd) | 10\38 M3s == m2dias (m3rd) | ||
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12\38 m4s == M2dias (M3rd) | 12\38 m4s == M2dias (M3rd) | ||
13\38 n4s | 13\38 n4s | ||
14\38 M4s/d5s == | 14\38 M4s/d5s == 2×3dias (3rd×4th) | ||
15\38 sPs | 15\38 sPs | ||
16\38 P5s == P3dias (P4th) | 16\38 P5s == P3dias (P4th) | ||
17\38 5×6s == sP3dias (sP4th) | 17\38 5×6s == sP3dias (sP4th) | ||
18\38 m6s == A3dias (A4th) | 18\38 m6s == A3dias (A4th) | ||
19\38 n6s == | 19\38 n6s == 3×4dias (4th×5th) | ||
</pre> | </pre> | ||
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3\18 sP1mans | 3\18 sP1mans | ||
4\18 P1mans == P1atts | 4\18 P1mans == P1atts | ||
5\18 | 5\18 1×2mans == sP1atts | ||
6\18 m2mans == A1atts | 6\18 m2mans == A1atts | ||
7\18 n2mans == | 7\18 n2mans == 1×2atts | ||
8\18 M2mans == m2atts | 8\18 M2mans == m2atts | ||
9\18 | 9\18 2×3mans == n2atts | ||
</pre> | </pre> | ||
=== Oneirotonic (5L3s) === | === Oneirotonic (5L3s) === | ||
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6\26 m2oneis == M1apts | 6\26 m2oneis == M1apts | ||
7\26 n2oneis | 7\26 n2oneis | ||
8\26 M2/d3oneis == | 8\26 M2/d3oneis == 1×2apts | ||
9\26 sP3oneis | 9\26 sP3oneis | ||
10\26 P3oneis == P2apts | 10\26 P3oneis == P2apts | ||
11\26 3×4oneis == sP2apts | 11\26 3×4oneis == sP2apts | ||
12\26 m4oneis == A2apts | 12\26 m4oneis == A2apts | ||
13\26 n4oneis == | 13\26 n4oneis == 2×3apts | ||
</pre> | </pre> | ||
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=== Proof of Theorem === | === 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 | 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. | * 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 iff its hardness is > 2/1 and the number of s steps it has is > 1. | ||
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*** Hence, (C+D)-(A+B) ≥ 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. | *** 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 | * 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). | * 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). | ||