Harmonic Series coincidence

Michael,

you said:"You also said 16/15 (1.06666) is the most dissonant interval to your ear...but then (ironically) you use 15/14 and other intervals incredibly close to it as intervals in your scale."

For me, in melody, anything goes, so the 15/14 is fine in melody. As I said before, using sine waves the 5/6 is the narrowest "allowable" interval, whether a dyad or as part of a six note chord. The 15/14 (in this case a note, not an interval) however can be used in plenty of other chords that don't contain 1/1.(e.g 15/14, 3/2, 15/7, 3/1.)

>>"Pardon my impatience (though I'm sure I could find this out by reading your book for a long enough period of time...but what exactly would makes 1/1 and 2/1 have the "same strength"?"

In 12TET 1/1 and 2/1 *have* the same strength but if you are choosing 12 perfectly just (relative to each other) notes for a scale you will always have to choose between 1/1 or 2/1 as the strongest tonic, unless you use the square root of 2 for the tritone but this wouldn't be 'just'. This is explained in chapter 7 of my book (have you read it yet?)

Here's a funny coincidence I came across today. The "ideal" harmonic series has frequencies: x, 2x, 3x, 4x etc. and the amplitudes are y, y/2, y/3, y/4 etc.

If you treat the harmonic series as a chord then perhaps the 'overall' strength of each harmonic is greater than the initial strength (e.g. the third harmonic on its own has an 'initial' strength of 0.3333). But when you pair it with all the other harmonics (say the first sixteen), using my formula, then the 'overall' value of the third harmonic should be greater than 0.3333.

First Harmonic... each pairning is (1/x + 1/y - diss(x/y))/2f
(x<y and the ratios are *not* simplified). 'f' is the number of the higher harmonic and is a 'weight' (i.e. the strength of each pairing is only as strong as the quieter element of the pair).
If x/y<=0.9375 then diss(x/y) = x/y. Else diss=(1 - x/y)*15.

1 + (1/1 + 1/2 - 1/2)/2f +
(1/1 + 1/3 - 1/3)/2f +
(1/1 + 1/4 - 1/4)/2f + etc to infinity

Third Harmonic...

0.3333 + (1/1 + 1/3 - 1/3)/2f +
(1/2 + 1/3 - 2/3)/2f +
(1/3 + 1/4 - 3/4)/2f + etc to infinity

If you work out the overall value of the first harmonic then it has no limit, it (slowly) approaches infinity. I suspect they all do. However the strength value of how the, say, first harmonic relates to the, say, third harmonic tends toward a finite limit. In other words if the first harmonic has an overall strength of 1, then (after a zillion iterations) the second harmonic approaches 0.5, the third approaches 0.3333, the fourth approached 0.25 etc. This is probably a tautology but it says a lot for my formula:
(1/x + 1/y - diss(x/y))

So the next step is to write a program (using, say, the first sixteen harmonics of each note in an interval) that works out the consonance values of "ideal" intervals, using my formulas and 'weighting' each calculation according to the strength value of the weaker 'element' in each pair. If this approach is right then we'll have a 'rough guide' to consonance for instruments whose timbres are close to the 'ideal'.

Again, take a look at chapters 4, 6, 7, 10 and 11 of my book.

John.

PS Doty possibly meant to say 'bowed' strings instead of 'plucked' strings.