Relationship between Bohlen–Pierce and octave-ful temperaments: Difference between revisions

== Relationship of rank-2 "Lambda" temperament to octave-ful temperaments ==

The rank-2 temperament under discussion here is the 3.5.7 temperament that tempers out only 245/243 (not any other commas such as 3125/3087). Its mapping matrix is {{val|1 1 2}}, {{val|0 2 -1}} and the two generators are ~1902 cents and ~440 cents. (The generator represents 9/7 and two of them make a 5/3 because of 245/243 vanishes.) In 3/1-equivalence world, its MOS sequence goes 4, 5, 9, 13..., and the 9-note MOS is what's known as the BP "Lambda" scale.

What if we want a rank-2 system? In that case we must find a suitable interval in the system to represent 2/1. In theory any of the many 7-limit temperaments tempering out 245/243 (the [[sensamagic family]]) could be considered a relative of Bohlen-Pierce. But most of them divide either 3/1 or the ~440 cent generator into some number of equal parts, and so don't have the same generators/lattice as rank-2 BP itself. (In math terms the "index" is greater than 1.) For example, [[magic]] divides 3/1 into 5 equal parts. [[Superpyth]], on the other hand, leaves 3/1 unsplit but makes 9/7 no longer the generator - instead if has complexity 6.

To find a rank-2 temperament more closely analogous to Lambda temperament but containing the prime 2, 2/1 must be mapped to some already existing note of Lambda temperament rather than a new note obtained by splitting up the 3/1 period or ~440 cent generator. There are two obvious candidates for this mapping of 2/1: +7 generators and -6 generators. (Note 7+6=13 so in 13ed3 these notes are "enharmonically equivalent" to each other.)

Consider the first possibility, that 2/1 is mapped to +7 generators in the BP lambda scale. Seven generators up from C is H#, so this means H# represents 2/1 above C. In equal-tempered BP H# is only 1170 cents, so in order for this to accurately represent 2/1 the generator needs to be widened by a few cents, from ~440 cents up to ~443 cents.

Now consider what happens if 2/1 is mapped to -6 generators instead. Six generators down from C (in the Lambda scale notation) is Jb, so now Jb represents 2/1 above C. In this case the generator must be narrowed from ~440 to ~436 cents in order for Jb to end up at a reasonable 2/1.

Furthermore, since 2/1 is being mapped to (2, -6) (up two ~1902 cent periods and down six ~436 cent generators), and both 2 and -6 are even numbers, that means 2/1 splits into two equal parts in this system. (This is in contrast to sensi in which 2/1 is intact and doesn't split into any equal parts.)