Re: [tuning] Re: J.S. Bach: Badinerie from Suite in B Minor (golden meantone)

From: John A. deLaubenfels

| [Danny:]
| >17-tone, 22-tone and 53-tone work differently by the way.
|
| Yeah, for starters they aren't meantone. Are they involved in a similar
| game?

They are but I don't know how. It's another Fibonacci-type sequence. The flats
and sharps cross each other, unlike 19/31/50/81-tET, so C# is *higher* than Db,
etc. This groups of ET's relies more on the circle of fifths.

Lemme see... 12 + 5 = 17, 17 + 5 = 22, what comes next?

~DaW~

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--- In tuning@y..., "Danny Wier" <dawier@y...> wrote:

> Lemme see... 12 + 5 = 17, 17 + 5 = 22, what comes next?

There's actually more than one golden convergence that this can refer
to . . . again, the archives from about two years ago are full of
this stuff.

--- In tuning@y..., "Danny Wier" <dawier@y...> wrote:
> From: John A. deLaubenfels

> | [Danny:]
> | >17-tone, 22-tone and 53-tone work differently by the way.
> |
> | Yeah, for starters they aren't meantone. Are they involved in a
similar
> | game?

> They are but I don't know how. It's another Fibonacci-type
sequence. The flats
> and sharps cross each other, unlike 19/31/50/81-tET, so C# is
*higher* than Db,
> etc. This groups of ET's relies more on the circle of fifths.

> Lemme see... 12 + 5 = 17, 17 + 5 = 22, what comes next?

Evidently 53 doesn't come at all--the generators are 3/5, 7/12, 10/17,
17/29, 27/46, 44/75, 71/121 ..., converging to w = (67+sqrt(5))/118;
which means an approximation of a fifth by 2^w, 2.14 cents sharp. The
meantone systems of 4/7, 7/12,11/19, 18/31, 29/50 ... on the other
hand converge to x = (15 - sqrt(5))/22, so that the "Golden Meantone"
I presume must be 2^x, 5.74 cents flat; about 4/15-comma meantone.

If you want something with 22 and 53 in it, the most plausible choice
I think is 5/22, 7/31, 12/53, 19/84 converging to a Golden Orwell,
but since I am writing a piece at the moment in orwell perhaps I am
prejudiced.

--- In tuning@y..., genewardsmith@j... wrote:
> --- In tuning@y..., "Danny Wier" <dawier@y...> wrote:
> > From: John A. deLaubenfels
>
> > | [Danny:]
> > | >17-tone, 22-tone and 53-tone work differently by the way.
> > |
> > | Yeah, for starters they aren't meantone. Are they involved in
a
> similar
> > | game?
>
> > They are but I don't know how. It's another Fibonacci-type
> sequence. The flats
> > and sharps cross each other, unlike 19/31/50/81-tET, so C# is
> *higher* than Db,
> > etc. This groups of ET's relies more on the circle of fifths.
>
> > Lemme see... 12 + 5 = 17, 17 + 5 = 22, what comes next?
>
> Evidently 53 doesn't come at all--the generators are 3/5, 7/12,
10/17,
> 17/29,

What happened to 22 ??

The
> meantone systems of 4/7, 7/12,11/19, 18/31, 29/50 ... on the other
> hand converge to x = (15 - sqrt(5))/22, so that the "Golden
Meantone"
> I presume must be 2^x, 5.74 cents flat; about 4/15-comma meantone.

Yup, Kornerup's Golden fifth is about 696.21 cents.

> If you want something with 22 and 53 in it, the most plausible
choice
> I think is 5/22, 7/31, 12/53, 19/84 converging to a Golden Orwell,
> but since I am writing a piece at the moment in orwell perhaps I am
> prejudiced.

So I take it you define the Orwell system as the system with a
generator of about 271.14 cents, interval of repetition of the
octave, and which approximates the primes as follows:

3:1 is approximated by 7 generators
5:1 " " " -3 generators
7:1 " " " 8 generators
11:1 " " " 2 generators

and with maximum error in the 11-limit of over 9.3 cents. Yes? A 22-
tone MOS would have 5 otonal and 5 utonal hexads . . . personally, I
like the number of complete chords to be well over half the number of
notes, but to each his own . . . looking forward to the music!

--- In tuning@y..., "Paul Erlich" <paul@s...> wrote:

> > Evidently 53 doesn't come at all--the generators are 3/5, 7/12,
> 10/17,
> > 17/29,

> What happened to 22 ??

I presume 22 was a mistake, and 29 was intended. Having used the 5 in
5+12=17, you can't reuse it by 5+17=22 if you are using the two-term
recurrence a_{n+1) = a_n + a_{n-1}, which is what we were talking
about. Since evidently the generator is some kind of fifth, 12+17=29
makes sense, and gives us a recurrence converging to a reasonable
fifth.

> > If you want something with 22 and 53 in it, the most plausible
> choice
> > I think is 5/22, 7/31, 12/53, 19/84 converging to a Golden
Orwell,
> > but since I am writing a piece at the moment in orwell perhaps I
am
> > prejudiced.

> So I take it you define the Orwell system as the system with a
> generator of about 271.14 cents, interval of repetition of the
> octave, and which approximates the primes as follows:

> 3:1 is approximated by 7 generators
> 5:1 " " " -3 generators
> 7:1 " " " 8 generators
> 11:1 " " " 2 generators

> and with maximum error in the 11-limit of over 9.3 cents. Yes?

That's it--an interesting system, to my mind.

A 22-
> tone MOS would have 5 otonal and 5 utonal hexads . . . personally,
I
> like the number of complete chords to be well over half the number
of
> notes, but to each his own . . . looking forward to the music!

You won't get it, at least if you insist on a high degree of
consonance--I'm using the 9-note MOS as a scale.

--- In tuning@y..., genewardsmith@j... wrote:
> --- In tuning@y..., "Paul Erlich" <paul@s...> wrote:
>
> > > Evidently 53 doesn't come at all--the generators are 3/5, 7/12,
> > 10/17,
> > > 17/29,
>
> > What happened to 22 ??
>
> I presume 22 was a mistake, and 29 was intended.

Why presume that? Perhaps a fifth around 710 cents was the "goal",
with the following mapping from primes to generators:

3 -> 1 gen.
5 -> 9 gen.
7 -> -2 gen.
and perhaps
11 -> -6 gen.

> Having used the 5 in
> 5+12=17, you can't reuse it by 5+17=22 if you are using the two-
term
> recurrence a_{n+1) = a_n + a_{n-1}, which is what we were talking
> about.

That may be what you are thinking about, but it wasn't stated.
Instead of assuming that Danny made a mistake, I went with what he
was saying, and noted that the convergence to a particular noble
generator may sometimes move by two steps in the same direction
before settling on the characteristic golden "zigzag".

> Since evidently the generator is some kind of fifth, 12+17=29
> makes sense, and gives us a recurrence converging to a reasonable
> fifth.

Only one of many.
>
> > > If you want something with 22 and 53 in it, the most plausible
> > choice
> > > I think is 5/22, 7/31, 12/53, 19/84 converging to a Golden
> Orwell,
> > > but since I am writing a piece at the moment in orwell perhaps
I
> am
> > > prejudiced.
>
> > So I take it you define the Orwell system as the system with a
> > generator of about 271.14 cents, interval of repetition of the
> > octave, and which approximates the primes as follows:
>
> > 3:1 is approximated by 7 generators
> > 5:1 " " " -3 generators
> > 7:1 " " " 8 generators
> > 11:1 " " " 2 generators
>
> > and with maximum error in the 11-limit of over 9.3 cents. Yes?
>
> That's it--an interesting system, to my mind.
>
> A 22-
> > tone MOS would have 5 otonal and 5 utonal hexads . . .
personally,
> I
> > like the number of complete chords to be well over half the
number
> of
> > notes, but to each his own . . . looking forward to the music!
>
> You won't get it, at least if you insist on a high degree of
> consonance--I'm using the 9-note MOS as a scale.

Well, you'll have four 8:10:11 chords, and not much else in the way
of "consonance". If not aiming for consonance, why use this
elaborately JI-related system?

--- In tuning@y..., "Paul Erlich" <paul@s...> wrote:
> --- In tuning@y..., genewardsmith@j... wrote:

> Why presume that? Perhaps a fifth around 710 cents was the "goal",
> with the following mapping from primes to generators:

> 3 -> 1 gen.
> 5 -> 9 gen.
> 7 -> -2 gen.
> and perhaps
> 11 -> -6 gen.

Is that what you get from the 7/12, 3/5, 10/17, 13/22, 23/39 ...
recurrence?

> > Having used the 5 in
> > 5+12=17, you can't reuse it by 5+17=22 if you are using the two-
> term
> > recurrence a_{n+1) = a_n + a_{n-1}, which is what we were talking
> > about.

> That may be what you are thinking about, but it wasn't stated.

"Golden" requires the Fibonacci recurrence to my mind; otherwise,
where's the gold?

> Instead of assuming that Danny made a mistake, I went with what he
> was saying, and noted that the convergence to a particular noble
> generator may sometimes move by two steps in the same direction
> before settling on the characteristic golden "zigzag".

I'm not sure what you are saying, but 12, 5, 17 goes down before it
goes up.

> > Since evidently the generator is some kind of fifth, 12+17=29
> > makes sense, and gives us a recurrence converging to a reasonable
> > fifth.
>
> Only one of many.

It's a much less cheesy fifth than yours, but the cheese discussion
did not lead to the conclusion that cheese was bad, I recall.

> Well, you'll have four 8:10:11 chords, and not much else in the way
> of "consonance". If not aiming for consonance, why use this
> elaborately JI-related system?

Whatever gives you that idea? I have a 1-9/7-3/2-12/7 chord, many
major thirds, and even more subminor thirds, not to mention a little
3/2 and 7/4, 8:11:14, etc. Plus, it has a nice quality melodically.

--- In tuning@y..., genewardsmith@j... wrote:
> --- In tuning@y..., "Paul Erlich" <paul@s...> wrote:
> > --- In tuning@y..., genewardsmith@j... wrote:
>
> > Why presume that? Perhaps a fifth around 710 cents was
the "goal",
> > with the following mapping from primes to generators:
>
> > 3 -> 1 gen.
> > 5 -> 9 gen.
> > 7 -> -2 gen.
> > and perhaps
> > 11 -> -6 gen.
>
> Is that what you get from the 7/12, 3/5, 10/17, 13/22, 23/39 ...
> recurrence?

That's not how I got it, no . . .
>
> > > Having used the 5 in
> > > 5+12=17, you can't reuse it by 5+17=22 if you are using the two-
> > term
> > > recurrence a_{n+1) = a_n + a_{n-1}, which is what we were
talking
> > > about.
>
> > That may be what you are thinking about, but it wasn't stated.
>
> "Golden" requires the Fibonacci recurrence to my mind; otherwise,
> where's the gold?

The gold is in the noble ratio that occurs in the limit of the
sequence, which can always be expressed in the form (A + B*phi)/(C +
D*phi) . . .

> > Instead of assuming that Danny made a mistake, I went with what
he
> > was saying, and noted that the convergence to a particular noble
> > generator may sometimes move by two steps in the same direction
> > before settling on the characteristic golden "zigzag".
>
> I'm not sure what you are saying, but 12, 5, 17 goes down before it
> goes up.

By same direction, I meant size of the generator. The number of notes
always goes up as you move down the scale tree.

To better understand what I am saying, look at
<http://www.anaphoria.com/sctree.PDF>. Note that to get to any one of
the particular vertical dotted lines, which represent noble generator
sizes, you may have to take more than one step in a given direction
at first, though eventually, you'll settle into the infinite zigzag
pattern. Even the Kornerup golden meantone that forms our prototype
does this: it goes:

1/2, 2/3, 3/5, 4/7, 5/12, 11/19, 18/31, 29/50, 47/81 . . .

and notice that the strict zigzag pattern, and the Fibonacci-like
structure, only begins at the 4/7 term.
>
> > > Since evidently the generator is some kind of fifth, 12+17=29
> > > makes sense, and gives us a recurrence converging to a
reasonable
> > > fifth.
> >
> > Only one of many.
>
> It's a much less cheesy fifth than yours, but the cheese discussion
> did not lead to the conclusion that cheese was bad, I recall.

Well, I was just assuming Danny didn't make a mistake, and as you
should now agree from my arguments here, there was absolutely no
reason to assume he did.
>
> > Well, you'll have four 8:10:11 chords, and not much else in the
way
> > of "consonance". If not aiming for consonance, why use this
> > elaborately JI-related system?
>
> Whatever gives you that idea?

You just said I won't like your piece if I'm hoping to hear
consonance.

>I have a 1-9/7-3/2-12/7 chord,

Ah yes -- the infamous "ASS" (see
<http://www.cix.co.uk/~gbreed/erlichs.htm>) -- the whole "Graham
complexity" thing made me forget about those -- though Margo's posts
today should have reminded me.

> many
> major thirds, and even more subminor thirds, not to mention a
little
> 3/2 and 7/4,

I was thinking of 3-part harmony or higher, but of course you're
right.

>8:11:14

Do you find that "consonant" in the register/timbre combination
you're using?

> etc. Plus, it has a nice quality melodically.

I thought you didn't believe MOS scales were especially good
melodically. Are you reconsidering that opinion?

--- In tuning@y..., "Paul Erlich" <paul@s...> wrote:

> > "Golden" requires the Fibonacci recurrence to my mind; otherwise,
> > where's the gold?

> The gold is in the noble ratio that occurs in the limit of the
> sequence, which can always be expressed in the form (A + B*phi)/(C
+
> D*phi) . . .

You won't get this with any old linear recurrence; the characteristic
polynomial must factor completley over Q(sqrt(5))--as x^2-x-1 does,
for instance.

> 1/2, 2/3, 3/5, 4/7, 5/12, 11/19, 18/31, 29/50, 47/81 . . .

This just looks wrong to me--the mediant of 2/3 and 3/5 is 5/8, etc.
Why not 1/2, 2/5, 3/7, 5/12, 8/19, 13/31, 21/50...? It has Fibonacci
numbers for numerators, which is sort of nifty, and it converges to a
meantone fourth--your version can't seem to decide if it wants to be
a fourth or a fifth, and is not generated by a recurrence where each
term is the mediant of the two previous terms.

> and notice that the strict zigzag pattern, and the Fibonacci-like
> structure, only begins at the 4/7 term.

It seems to me it begans at the 11/19 term, and I don't like it.
> > Whatever gives you that idea?

> You just said I won't like your piece if I'm hoping to hear
> consonance.

I said you wouldn't like it if you were hoping to hear complete
tetrads.

> >8:11:14

> Do you find that "consonant" in the register/timbre combination
> you're using?

Not as a place of rest.

> > etc. Plus, it has a nice quality melodically.

> I thought you didn't believe MOS scales were especially good
> melodically. Are you reconsidering that opinion?

What I really doubted was that Myhill's property was what was
required for melodic goodness; I think MOS are interesting for other
reasons. I havn't any theories of my own about melodic goodness,
though in my salad days of JI scale contruction I thought about
properness, which seems as if it ought to be important.

In-Reply-To: <9rae6l+rcpk@eGroups.com>
Paul wrote:

> So I take it you define the Orwell system as the system with a
> generator of about 271.14 cents, interval of repetition of the
> octave, and which approximates the primes as follows:
>
> 3:1 is approximated by 7 generators
> 5:1 " " " -3 generators
> 7:1 " " " 8 generators
> 11:1 " " " 2 generators

I've added this to the catalog at
<http://x31eq.com/catalog.htm>. Or rather, altered the
anonymous entry DK suggested. I also put the Shrutar temperament in
there. Interested parties in both cases my like to check the
attributions, because I'm not sure about them.

Graham

--- In tuning@y..., genewardsmith@j... wrote:

> I havn't any theories of my own about melodic goodness,
> though in my salad days of JI scale contruction I thought about
> properness, which seems as if it ought to be important.

I don't find the Pythagorean diatonic, which is improper, or the 22-
tET version of that, which is highly improper, to be disturbing in
the least. However, the fact that proper MOSs appear at the
convergents, and improper ones at the semiconvergents, is interesting
at least.

One generator consistent with Danny's spec would be

(7*phi + 2)/(17*phi + 5)

It's the limit of the Fibonacci-like sequence (the denominators give
the cardinalities of _all_ the proper MOSs of the generator):

2/5, 7/17, 9/22, 16/39, 25/61, 41/100 . . .

> > and notice that the strict zigzag pattern, and the Fibonacci-like
> > structure, only begins at the 4/7 term.
>
> It seems to me it begans at the 11/19 term, and I don't like it.

I don't know what you mean:

1/2 - left
2/3 - right
(3/5 - right)
4/7 - left
7/12 - right
11/19 - left
18/31 - right
29/50 - left
47/81 - right
etc.

How can you say it only begins at the 11/19 term?

--- In tuning@y..., graham@m... wrote:

> I've added this to the catalog at
> <http://x31eq.com/catalog.htm>. Or rather, altered the
> anonymous entry DK suggested. I also put the Shrutar temperament
in
> there. Interested parties in both cases my like to check the
> attributions, because I'm not sure about them.

Robert Valentine, I think, mentioned the Orwell 9- and/or 13-tone
MOSs in 31-tET some time ago. The Shrutar scale resulted from close
collaboration between myself and Dave Keenan.

--- In tuning@y..., "Paul Erlich" <paul@s...> wrote:

> 1/2 - left
> 2/3 - right
> (3/5 - right)
> 4/7 - left
> 7/12 - right
> 11/19 - left
> 18/31 - right
> 29/50 - left
> 47/81 - right

> How can you say it only begins at the 11/19 term?

This is Fibonacci-like, but it isn't what you gave before.

Here's a tuning challenge for you: find the rule of the sequence

3/7, 5/12, 5/12, 8/19, 10/24, 13/31, 18/43, 23/55, 31/74 ... and
determine what meantone fourth it converges to.

--- In tuning@y..., genewardsmith@j... wrote:
> --- In tuning@y..., "Paul Erlich" <paul@s...> wrote:
>
> > 1/2 - left
> > 2/3 - right
> > (3/5 - right)
> > 4/7 - left
> > 7/12 - right
> > 11/19 - left
> > 18/31 - right
> > 29/50 - left
> > 47/81 - right
>
> > How can you say it only begins at the 11/19 term?
>
> This is Fibonacci-like, but it isn't what you gave before.

What are you talking about? It's what both you and I gave for the
original, Kornerup golden meantone. Take another look!

> Here's a tuning challenge for you: find the rule of the sequence
>
> 3/7, 5/12, 5/12, 8/19, 10/24, 13/31, 18/43, 23/55, 31/74 ... and
> determine what meantone fourth it converges to.

Looks like your rule is

a(k+3) = a(k) + a(k+1)

This is not "golden" but rather one of the other "metallic"
generators we discussed some time ago. Let me think about it.