Neutral and interordinal intervals in MOS scales: Difference between revisions
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=== Statement of the theorem === | === Statement of the theorem === | ||
Suppose a > b and gcd(a, b) = 1. | 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. | |||
# The impropriety number (b - 1) of the mos aLbs 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. | # The impropriety number (b - 1) of the mos aLbs 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 ==== | ==== 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: | 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: | ||
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* 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. | ||
Part ( | 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 ( | 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. | * 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. | * 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. | ||