4000/3993: Difference between revisions
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'''4000/3993''', the '''wizardharry comma''' or '''pine comma''' is an [[unnoticeable comma|unnoticeable]] [[11-limit]] [[comma]] with a size of roughly 3.03 [[cents]]. It is the amount by which a stack of three [[11/10]] submajor seconds falls short of the [[4/3]] perfect fourth | '''4000/3993''', the '''wizardharry comma''' or '''pine comma''' is an [[unnoticeable comma|unnoticeable]] [[11-limit]] (specifically [[2.3.5.11 subgroup|2.3.5.11-subgroup]]) [[comma]] with a size of roughly 3.03 [[cents]]. It is the amount by which a stack of three [[11/10]] submajor seconds falls short of the [[4/3]] perfect fourth; in other words it is equal to (4/3)/(11/10)<sup>3</sup>. | ||
In terms of commas, it is the difference between [[100/99]] ( | In terms of commas, it is the difference between [[100/99]] ({{S|10}}) and [[121/120]] ({{S|11}}), itself having an [[S-expression]] of S10/S11. It is also the difference between [[540/539]] ([[S-expression|S12/S14]]) and [[9801/9800]] ({{S|99}}). It factors into 13-limit commas as ([[1575/1573]])⋅([[2080/2079]]) or ([[625/624]])⋅([[6656/6655]]). | ||
== Temperaments == | == Temperaments == | ||
[[Tempering out]] this comma means the fourth is divided into an equal stepped tetrachord, the step of which is a ''trienfourth'' (from 1/3 of a fourth) or ''submajor second'' and because it is an [[ultraparticular]] it makes [[12/11]] and [[10/9]] equidistant from 11/10. | [[Tempering out]] this comma means the fourth is divided into an equal stepped tetrachord, the step of which is a ''trienfourth'' (from 1/3 of a fourth) or ''submajor second'' and because it is an [[ultraparticular]] it makes [[12/11]] and [[10/9]] equidistant from 11/10. | ||
On the low-accuracy end, this may be reminiscent of how the [[250/243|porcupine comma]] splits 4/3 into three 10/9's instead. In fact, 2.3.5.11-subgroup [[porcupine]] equates 10/9 with 11/10 and 12/11 so that the equally-split fourth represents 9:10:11:12, making | On the low-accuracy end, this may be reminiscent of how the [[250/243|porcupine comma]] splits 4/3 into three 10/9's instead. In fact, 2.3.5.11-subgroup [[porcupine]] equates 10/9 with 11/10 and 12/11 so that the equally-split fourth represents 9:10:11:12, making porcupine as efficient and elegant as it can reasonably be. (As a [[rank-3]] detemperament of porcupine, one way to extend it to include prime 7 is by tempering out [[385/384]], which septimal porcupine also tempers out.) | ||
Tempering it out along with the [[schisma]] results in the rank-2 [[tertiaschis]] temperament. Tempering it out with the trimitone comma [[8019/8000]] ([[S-expression|S9/S10]], so that three [[10/9]]'s are also an [[11/8]]) implies also tempering out the [[semiparticular]] [[243/242]] ([[S-expression|S9/S11]]) = ([[3/2]])/([[11/9]])<sup>2</sup> leading to [[larry]] in the [[gravity family]]. Tempering it out with both the | Tempering it out along with the [[schisma]] results in the rank-2 [[tertiaschis]] temperament. Tempering it out with the trimitone comma [[8019/8000]] ([[S-expression|S9/S10]], so that three [[10/9]]'s are also an [[11/8]]) implies also tempering out the [[semiparticular]] [[243/242]] ([[S-expression|S9/S11]]) = ([[3/2]])/([[11/9]])<sup>2</sup> leading to [[larry]] in the [[gravity family]]. Tempering it out with both the schisma and trimitone comma gives a description of [[65edo]] in the no-7's 11-limit, making it an excellent way to extend schismic to include prime 11. | ||
Another strategy is to take advantage of the size of | Another strategy is to take advantage of the size of 121/120 (S11) so as to equate it with [[144/143]] ({{S|12}}) = ([[16/13]])/(11/9), for those seeking to keep the undecimal and tridecimal neutral thirds distinct, thus tempering out the marveltwin comma, [[325/324]] ([[S-expression|S10/S12]]), a comma with various advantages. | ||
Observing that 4000/3993 = (540/539)/(9801/9800) also shows a natural path, if one is willing to split the octave in half, leading to [[hades]]. | Observing that 4000/3993 = (540/539)/(9801/9800) also shows a natural path, if one is willing to split the octave in half, leading to [[hades]]. | ||