Neutral and interordinal intervals in MOS scales: Difference between revisions
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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): | 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): | ||
* 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. | * 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. | * 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). | ||