Re: Optimum diatonic meantones

[Me, Dave Keenan]
>>Can you explain how 3x^3+4x=16 is derived from >equal-beating
>>considerations? I guess that x here represents >the fifth-size as a
>>frequency ratio rather than its logarithmic error >in commas.

[Paul Erlich]
>Yes. According to Jorgenson, Smith wanted a fifth and a major sixth to
>beat at the same rate when their lower note was the same.

I still don't see how to get 3x^3+4x=16, where x is the frequency ratio of
the fifth. But I agree that 5/18 comma is the answer. I get it from 3x =
5(1-3x) where x is the flatness of the fifth in commas.

Certainly 3x^3+4x=16, gives very close to 5/18 comma but how does it come
from the requirement you describe. I understand the powers but not the
multipliers or the constant or the addition. I can explain my formula very
easily as follows.

For a fifth and sixth on the same lower note, the beats in the fifth (2:3)
are due to the 3rd partial of the lower note while those in the sixth (3:5)
are due to its 5th partial. If the error in the fifth is x commas, then the
error in the sixth (in meantone) is 1-3x commas (ignoring the sign of the
error). Hence equal beats occur (to a very good approximation due to the
small logarithmic deviations) when 3x = 5(1-3x). Actually there are two
solutions (by changing the sign of one side) but the above gives mimimum
beats.

-- Dave Keenan
http://dkeenan.com

1. can anyone shed any light into the confusion below (at the bottom
of this message) about robert smith's 1749 meantone tuning system,
maybe 5/18-comma meantone or maybe a few thousanths of a cent
different from that, maybe defined by setting the beat rates of the
fifth and major sixth (built from a common lower pitch) equal to one
another?

2. it seems that if one is concerned with having rhythmically "in-
time" beat rates, not just for root-position triads, but for all
possible chords, and if one is justified in ignoring the piano's
inharmonicity, one can simply quantize the whole tuning to a single
harmonic series. for example, if the whole tuning is quantized to the
nearest Hz, then every possible combination of notes will produce a
waveform, beats and all, that repeats itself every second. i think a
great test of the whole concept would be if someone (gene?) found an
excellent integer-relation for one complete octave-span of 12-equal,
and then one transposed the solution to fill out the rest of the
octaves. if the whole concept really has merit, than the resulting
tuning system should be audibly superior to 12-equal for the kinds of
music where 12-equal is supposedly ideal (i.e., atonal music or music
with plenty of dissonant chords and little "centricity" about one
particular region in the circle of fifths).

3. the natives here get testy when stuff gets too mathematical, thus
the tuning-math list was established some time ago. gene is already
posting more about this subject there than here, so at least robert
should look over there, and maybe this whole discussion should be
moved there . . .

--- In tuning@yahoogroups.com, Dave Keenan <d.keenan@xx.xxx.xxx wrote:
> [Me, Dave Keenan]
> >>Can you explain how 3x^3+4x=16 is derived from >equal-beating
> >>considerations? I guess that x here represents >the fifth-size as
a
> >>frequency ratio rather than its logarithmic error >in commas.
>
> [Paul Erlich]
> >Yes. According to Jorgenson, Smith wanted a fifth and a major
sixth to
> >beat at the same rate when their lower note was the same.
>
> I still don't see how to get 3x^3+4x=16, where x is the frequency
ratio of
> the fifth. But I agree that 5/18 comma is the answer. I get it from
3x =
> 5(1-3x) where x is the flatness of the fifth in commas.
>
> Certainly 3x^3+4x=16, gives very close to 5/18 comma but how does
it come
> from the requirement you describe. I understand the powers but not
the
> multipliers or the constant or the addition. I can explain my
formula very
> easily as follows.
>
> For a fifth and sixth on the same lower note, the beats in the
fifth (2:3)
> are due to the 3rd partial of the lower note while those in the
sixth (3:5)
> are due to its 5th partial. If the error in the fifth is x commas,
then the
> error in the sixth (in meantone) is 1-3x commas (ignoring the sign
of the
> error). Hence equal beats occur (to a very good approximation due
to the
> small logarithmic deviations) when 3x = 5(1-3x). Actually there are
two
> solutions (by changing the sign of one side) but the above gives
mimimum
> beats.
>
> -- Dave Keenan
> http://dkeenan.com

--- In tuning@yahoogroups.com, "wallyesterpaulrus
<wallyesterpaulrus@y...>" <wallyesterpaulrus@y...> wrote:

> 1. can anyone shed any light into the confusion below (at the
bottom
> of this message) about robert smith's 1749 meantone tuning system,
> maybe 5/18-comma meantone or maybe a few thousanths of a cent
> different from that, maybe defined by setting the beat rates of the
> fifth and major sixth (built from a common lower pitch) equal to
one
> another?

Where in the world did this posting come from? Yahoo seems screwed
up, in that I haven't seen it yet.

If you want the fifth and major sixth to beat the same, then if f is
the fifth, in a meantone system f^3/2 will be the major sixth. Hence
you want (3(f^3/2)-5)/(2f-3) to be either +1 or -1; this leads to two
polynomial equations for f, f^3-4f-4=0 and 3f^3+4f-16=0. The second
has a real root very close to 5/18-comma meantone.

> 2. it seems that if one is concerned with having rhythmically "in-
> time" beat rates, not just for root-position triads, but for all
> possible chords, and if one is justified in ignoring the piano's
> inharmonicity, one can simply quantize the whole tuning to a single
> harmonic series.

It's a thought. Maybe someone should try it and see how it sounds;
you could call it JI if you wanted to.

--- In tuning@yahoogroups.com, "Gene Ward Smith <gwsmith@s...>"
<gwsmith@s...> wrote:
> --- In tuning@yahoogroups.com, "wallyesterpaulrus
> <wallyesterpaulrus@y...>" <wallyesterpaulrus@y...> wrote:
>
> > 1. can anyone shed any light into the confusion below (at the
> bottom
> > of this message) about robert smith's 1749 meantone tuning
system,
> > maybe 5/18-comma meantone or maybe a few thousanths of a cent
> > different from that, maybe defined by setting the beat rates of
the
> > fifth and major sixth (built from a common lower pitch) equal to
> one
> > another?
>
> Where in the world did this posting come from?

the year 1999.

> Yahoo seems screwed
> up, in that I haven't seen it yet.

it was posted long before you joined this list.

> If you want the fifth and major sixth to beat the same, then if f
is
> the fifth, in a meantone system f^3/2 will be the major sixth. Hence
> you want (3(f^3/2)-5)/(2f-3) to be either +1 or -1; this leads to
two
> polynomial equations for f, f^3-4f-4=0 and 3f^3+4f-16=0. The second
> has a real root very close to 5/18-comma meantone.

just a coincidence, or is this "very close"ness a general feature of
these sorts of beasts? by the way, in a previous message, you
mentioned "algebraic integers" and i had no idea what you were
talking about, or why. please explain on the tuning-math list.

> > 2. it seems that if one is concerned with having rhythmically "in-
> > time" beat rates, not just for root-position triads, but for all
> > possible chords, and if one is justified in ignoring the piano's
> > inharmonicity, one can simply quantize the whole tuning to a
single
> > harmonic series.
>
> It's a thought. Maybe someone should try it and see how it sounds;
> you could call it JI if you wanted to.

yup -- we need someone to come up with at least one example of "it"
first. can you be that someone?

--- In tuning@yahoogroups.com, "wallyesterpaulrus
<wallyesterpaulrus@y...>" <wallyesterpaulrus@y...> wrote:

> > If you want the fifth and major sixth to beat the same, then if f
> is
> > the fifth, in a meantone system f^3/2 will be the major sixth.
Hence
> > you want (3(f^3/2)-5)/(2f-3) to be either +1 or -1; this leads to
> two
> > polynomial equations for f, f^3-4f-4=0 and 3f^3+4f-16=0. The
second
> > has a real root very close to 5/18-comma meantone.

> just a coincidence, or is this "very close"ness a general feature
of
> these sorts of beasts?

It's due to the linearization Robert mentioned. If s is the major
sixths approximation, and f is fifths, then for them to beat the same
you want (2f-3)/(3s-5) = +-1, which means y = (f-3/2)/(s-5/3) = +-3/2.
By the same argument as before, the ratio of log errors in any log
base (eg, base 81/80) will be approximated by
(2/3)(f-3/2) / (3/5) (s-5/3), or 10/9 y. If the error of the fifth in
terms of a comma is x, the error of the sixth is 3x-1, and this means
(9/10) x/(3x-1) = +-3/2, approximately, or x=5/12 or 5/18.