Yahoo Tuning Groups Ultimate Backup tuning Average Complexity/Diatonic Stuff

On Sat, Mar 5, 2011 at 8:44 PM, Carl Lumma <carl@...> wrote:
>
> Mike wrote:
> >I talked about this all before. It's a rough concept and I'm
> >interested in testing it, and I'm not sure exactly what's going to
> >work the best. Mean and std. dev are two different metrics I'd like to
> >test out.
>
> Those are easy enough to compute. Not sure what you're computing
> the mean of though. Also we should move this to tuning.

I'm computing the mean complexity of all of the chords in the scale.
Ideally, I'd be computing the mean concordance by HE of all of the
chords in the scale, or with the HE transform I've hypothesized, but
those don't exist yet, so we'll have to settle for this for now. I
guess you'd have to normalize it such that 4:5:6 and 4:5:6:7 have
comparable complexity, so (a;b:c:...)^(1/N) is probably the way to go,
where N is the number of notes in the scale.

> >I have no idea how to run these tests, though, because I have no idea
> >how to open MATLAB and figure out what the least complex meantone
> >mapping for C C# D is.
>
> I don't know what you mean with this notation.

That is standard meantone notation. If you're looking at meantone[12],
and you're trying to get the average complexity of every chord in the
scale, you'll first have to get the complexity of every chord in the
scale. C C# D is the triad formed by 0-1-2 in meantone[12].

There are an infinite number of intervals that C C# D maps to in
meantone - they could be 1/1 25/24 9/8, or it could be 81/80 135/128
10/9. Since I'm using Tenney height here as a replacement for HE, it's
probably a safe assumption to make that when there are an infinite
series of intervals that a tempered interval could be, that the
simplest one is how it will be perceived. That is, C-D in quarter
comma meantone is more likely to be heard as 9/8 than 10/9, since 9/8
has a larger field of attraction than 10/9.

What I don't understand is how to do this for chords. C-D-E in quarter
comma meantone could be mapped as 64:72:81 or 81:90:100, but the
simplest possible mapping for it is 8:9:10, and that's probably how
also it will be heard. So a good way to start doing this since we
don't have HE available is to find the simplest possible mapping for
whatever chord you're looking at, get the complexity for that mapping,
and then do all of the statistical analysis based off of the
complexities of the simplest possible mappings for every chord in the
scale.

> >How do I come up with the complexity for that
> >chord and make some kind of histogram when I don't even know what the
> >mapping for the chord is?
>
> Don't you just take all the chords in the scale and score their
> concordance?

OK, but I don't want to do this by hand, I want to write a MATLAB
program that does it for me if I dial in a mapping and a number of
notes. Given the information 81/80, 7, how do I figure out the
complexity of the chord 0-1-2? 81/80 and 7 specifies the meantone
diatonic scale, and the concordance for 0-1-2 should be
(8*9*10)^(1/3). How do I figure out that 8:9:10 is the mapping to use
for the complexity calculation, when it could also be 64:72:81 or
81:90:100 or 72:80:81?

For porcupine[7], assuming all generators are going up, if I want to
find the complexity of 7-0-1-2-3, how on earth do I work out that
that's an octave-equivalent 8:9:10:11:12? My linear algebra mojo is
rising, but it isn't THAT good.

> >Precisely, what I want to do is this:
> >1) Come up with a one-sided PMF for a scale that represents the
> >concordance (measured right now in complexity) of a random chord in
> >that scale.
>
> PMF?

Probability mass function.

> >2) I predict that the meantone pentatonic scale will have a higher
> >mean concordance than the meantone diatonic scale, and the meantone
> >diatonic scale will have a higher mean concordance than the meantone
> >chromatic scale, for obvious reasons.
> >3) I predict that the maximum value of concordance in the PMF will be
> >higher in meantone[12] then meantone[7] and likewise with meantone[5].
> >4) Thus predictable changes in both the mean and standard deviation
> >will occur with this PMF if more notes are added, and experiment will
> >show which of the two, if any, turn out to be more relevant. There are
> >advantages to both.
>
> If you want to use this on larger scales you'll have to normalize
> it to step size. But since you only want small scales to come
> out good, you might not normalize it.

Do you mean normalize it to average step size, meaning to normalize it
to the number of notes in the scale? The normalized version would be
interesting as well, although not what I was aiming for.

> >>> Until I have a program that can actually
> >>>compute the TH for every chord in a scale and then find out the mean
> >>>and stdev of the resulting distribution, I won't know.
>
> Yeah I read that. Have no idea what it means.

TH stands for Tenney Height, sorry.

-Mike

🔗Mike Battaglia <battaglia01@...>

🔗Carl Lumma <carl@...>

Mike Battaglia <battaglia01@...> wrote:

Geez Mike, slow down.

> I'm computing the mean complexity of all of the chords in the
> scale. Ideally, I'd be computing the mean concordance by HE of
> all of the chords in the scale, or with the HE transform I've
> hypothesized, but those don't exist yet, so we'll have to settle
> for this for now.

Triadic harmonic entropy exists, and it's about to get slightly
better. Since you mostly have been talking about triads with
this business, why not start with them? To my knowledge it's
never been done. In a later message you say you think the
n-adic entropy will resemble the average m-adic entropy where
m < n. I agree. We know there are many chords where all the
dyads are good that aren't themselves good. But I think triads
are already muuch stronger. So basically just do what Paul did
with his entropy minimzier but for triads.

> That is standard meantone notation. If you're looking at
> meantone[12], and you're trying to get the average complexity
> of every chord in the scale, you'll first have to get the
> complexity of every chord in the scale. C C# D is the triad
> formed by 0-1-2 in meantone[12].

Please let me again suggest you not use "standard notation".

> What I don't understand is how to do this for chords.

That's because there isn't a reasonable way to do it for
arbitrary chords.

> So a good way to start doing this since we don't have
> HE available is to find the simplest possible mapping for
> whatever chord you're looking at, get the complexity for that
> mapping, and then do all of the statistical analysis based
> off of the complexities of the simplest possible mappings
> for every chord in the scale.

Howabout this reasoning: If you mash n notes on a keyboard,
you're most likely to get the simplest chord of n notes, just
because they outnumber more complex n-note chords. So that's
just Graham complexity relative to the scale size. And since
large chords tend to be discordant anyway, we can focus on
the simplest triad or tetrad generated by the mapping group
or subgroup. I think Gene may have said earlier to just use
Graham complexity. It's closely related to chord coverage,
which tells us it ought to be good for mashability too.

> > If you want to use this on larger scales you'll have to
> > normalize it to step size. But since you only want small
> > scales to come out good, you might not normalize it.
>
> Do you mean normalize it to average step size, meaning to
> normalize it to the number of notes in the scale?

They're not quite the same but yes.

-Carl

🔗Mike Battaglia <battaglia01@...>

On Sun, Mar 6, 2011 at 6:21 AM, Carl Lumma <carl@...> wrote:
>
> Mike Battaglia <battaglia01@...> wrote:
>
> Geez Mike, slow down.

Can't slow down!

> > I'm computing the mean complexity of all of the chords in the
> > scale. Ideally, I'd be computing the mean concordance by HE of
> > all of the chords in the scale, or with the HE transform I've
> > hypothesized, but those don't exist yet, so we'll have to settle
> > for this for now.
>
> Triadic harmonic entropy exists, and it's about to get slightly
> better. Since you mostly have been talking about triads with
> this business, why not start with them? To my knowledge it's
> never been done. In a later message you say you think the
> n-adic entropy will resemble the average m-adic entropy where
> m < n. I agree. We know there are many chords where all the
> dyads are good that aren't themselves good. But I think triads
> are already muuch stronger. So basically just do what Paul did
> with his entropy minimzier but for triads.

OK, let's do that. Do we have tabulated data for it somewhere?

I should say that I think for the full results to come through on
this, we'd have to do a tetradic minimizer for tetrads of the form
0.5a:a:b:c, which means we're minimizing the entropy of triads that
have the root note doubled down an octave. Diminished chords will rank
as more discordant this way, but I think they'll be less concordant
than minor chords the other way.

> > That is standard meantone notation. If you're looking at
> > meantone[12], and you're trying to get the average complexity
> > of every chord in the scale, you'll first have to get the
> > complexity of every chord in the scale. C C# D is the triad
> > formed by 0-1-2 in meantone[12].
>
> Please let me again suggest you not use "standard notation".

My example was a meantone example, and meantone notation is a way to
identify chords in meantone. What's the problem?

> > What I don't understand is how to do this for chords.
>
> That's because there isn't a reasonable way to do it for
> arbitrary chords.

There's no reasonable way to find the simplest possible mapping for an
arbitrary chord in a temperament?

> > So a good way to start doing this since we don't have
> > HE available is to find the simplest possible mapping for
> > whatever chord you're looking at, get the complexity for that
> > mapping, and then do all of the statistical analysis based
> > off of the complexities of the simplest possible mappings
> > for every chord in the scale.
>
> Howabout this reasoning: If you mash n notes on a keyboard,
> you're most likely to get the simplest chord of n notes, just
> because they outnumber more complex n-note chords. So that's
> just Graham complexity relative to the scale size.

When you say "simplest," you mean in terms of Graham complexity now? I
was talking about complexity as in how it refers to Tenney Height.

> And since large chords tend to be discordant anyway, we can focus on
> the simplest triad or tetrad generated by the mapping group
> or subgroup. I think Gene may have said earlier to just use
> Graham complexity. It's closely related to chord coverage,
> which tells us it ought to be good for mashability too.

That might be an even simpler way to do it. We should compare the
results for all of these.

-Mike

🔗Carl Lumma <carl@...>

Mike Battaglia <battaglia01@...> wrote:

>> Geez Mike, slow down.
>
> Can't slow down!

You should. I think it would be a win.

>> Triadic harmonic entropy exists, and it's about to get slightly
>> better. Since you mostly have been talking about triads with
>> this business, why not start with them? To my knowledge it's
>> never been done. In a later message you say you think the
>> n-adic entropy will resemble the average m-adic entropy where
>> m < n. I agree. We know there are many chords where all the
>> dyads are good that aren't themselves good. But I think triads
>> are already muuch stronger. So basically just do what Paul did
>> with his entropy minimzier but for triads.
>
> OK, let's do that. Do we have tabulated data for it somewhere?

Yes, in Steve's folder here or mine, as you know.

> I should say that I think for the full results to come through
> on this, we'd have to do a tetradic minimizer for tetrads of
> the form 0.5a:a:b:c, which means we're minimizing the entropy
> of triads that have the root note doubled down an octave.
> Diminished chords will rank as more discordant this way, but
> I think they'll be less concordant than minor chords the
> other way.

And I think triads are sufficient, at least for now. One of
the things you'll need to get precise about is how to handle
octave equivalence. We've got two octaves of 3HE data, so
perhaps just extend the scale under test to two octaves and
pull triads out of that. If we're going up to 12/oct, that's
3|24 or about 2000 chords to score, which is no problem.

>>> C C# D is the triad formed by 0-1-2 in meantone[12].
>>
>> Please let me again suggest you not use "standard notation".
>
> My example was a meantone example, and meantone notation is a
> way to identify chords in meantone. What's the problem?

Standard notation is usually understood in 12-ET. If it's
meantone you'll also need to say where the wolf is. Once you
do that you should realize that there are many more chords
than C C# D formed by 0-1-2 in the scale, unless C is zero,
which you haven't specified.

> > there isn't a reasonable way to do it for arbitrary chords.
>
> There's no reasonable way to find the simplest possible
> mapping for an arbitrary chord in a temperament?

There are, but for most chords the mapping is irrelevant,
in that it leads to JI chords with geomean(a*b*c...) > 10.

>> Howabout this reasoning: If you mash n notes on a keyboard,
>> you're most likely to get the simplest chord of n notes, just
>> because they outnumber more complex n-note chords. So that's
>> just Graham complexity relative to the scale size.
>
> When you say "simplest," you mean in terms of Graham complexity
> now? I was talking about complexity as in how it refers to
> Tenney Height.

Yes. The latter is sometimes also called complexity, but
calling it discordance in this thread may reduce confusion.

> That might be an even simpler way to do it. We should compare
> the results for all of these.

The first idea (looking up the 3HE of all triads) is a
straightforward problem. 3-combinations, lookup, mean or
std., done.

The latter idea is something very close to what Igs was doing.
Find the simplest triad or tetrad within an octave generated
by the mapping group. Subtract from the scale size then divide
that by the scale size. That gives you the chords per note.
Multiply by the chord size (3 or 4 as appropriate) and you get
the chord coverage per note. Scala reports chord coverage in
its show locations output by the way.

-Carl

🔗Mike Battaglia <battaglia01@...>

On Sun, Mar 6, 2011 at 4:47 PM, Carl Lumma <carl@...> wrote:
>
> Mike Battaglia <battaglia01@...> wrote:
>
> >> Geez Mike, slow down.
> >
> > Can't slow down!
>
> You should. I think it would be a win.

I'll try, but the problem is that I'm already onto taking the mellin
transform of the magnitude spectrum of a signal and trying to see if
that yields useful periodicity information, because I stayed up to 4
AM last night working some stuff out that I didn't bother to post in
an uninhibited orgy of spam to tuning-math. My brain works in quick
spurts and then it's done.

> > I should say that I think for the full results to come through
> > on this, we'd have to do a tetradic minimizer for tetrads of
> > the form 0.5a:a:b:c, which means we're minimizing the entropy
> > of triads that have the root note doubled down an octave.
> > Diminished chords will rank as more discordant this way, but
> > I think they'll be less concordant than minor chords the
> > other way.
>
> And I think triads are sufficient, at least for now. One of
> the things you'll need to get precise about is how to handle
> octave equivalence. We've got two octaves of 3HE data, so
> perhaps just extend the scale under test to two octaves and
> pull triads out of that. If we're going up to 12/oct, that's
> 3|24 or about 2000 chords to score, which is no problem.

I was going to use reduced-octave chords for all of this. Maybe a
better approach is to go with simplest octave-equivalent mapping
chords within 2 octaves, meaning 8:9:11:14 turns into 4:7:9:11.

> >>> C C# D is the triad formed by 0-1-2 in meantone[12].
> >>
> >> Please let me again suggest you not use "standard notation".
> >
> > My example was a meantone example, and meantone notation is a
> > way to identify chords in meantone. What's the problem?
>
> Standard notation is usually understood in 12-ET. If it's
> meantone you'll also need to say where the wolf is. Once you
> do that you should realize that there are many more chords
> than C C# D formed by 0-1-2 in the scale, unless C is zero,
> which you haven't specified.

I meant C C# D as in contrast to C Db D. I guess C is zero. I left out
the C as being zero on purpose, because it doesn't matter for this
experiment - if we're going to hit every single triad in the scale,
then it doesn't matter which mode of the (let's say 31-tet) meantone
chromatic scale you're using.

> > There's no reasonable way to find the simplest possible
> > mapping for an arbitrary chord in a temperament?
>
> There are, but for most chords the mapping is irrelevant,
> in that it leads to JI chords with geomean(a*b*c...) > 10.

Yes, but it's just supposed to be a rough metric for evaluating
consonance for more than triads. I guess if you want you could do
something like min(geomean(a,b,c,...),10), which traps the discordance
at a certain cutoff and avoids skewing the results.

> The first idea (looking up the 3HE of all triads) is a
> straightforward problem. 3-combinations, lookup, mean or
> std., done.
>
> The latter idea is something very close to what Igs was doing.
> Find the simplest triad or tetrad within an octave generated
> by the mapping group. Subtract from the scale size then divide
> that by the scale size. That gives you the chords per note.

What do you mean by chords per note? So for meantone[12], we're
looking at 4:5:6. 12 - 3 is 9, and 9/12 is 3/4, and that's the amount
of chords per note. What does 3/4 chords per note mean...?

> Multiply by the chord size (3 or 4 as appropriate) and you get
> the chord coverage per note. Scala reports chord coverage in
> its show locations output by the way.

So you get 9/4 coverage of the scale per note? I'm completely lost now.

The three things we're proposing to test, and what we need to do them, are:

1) Mean/standard deviation/etc of discordance of all of the chords in
the scale. We can place a strict limit on Tenney height discordance at
10 if you think that'll be better. Either way, that to automate this I
need to figure out what every random chord is mapped to, in simplest
terms, or else I can't get its Tenney Height.
2) Triadic HE minimizer. To do this, we need to work out the octave
equivalent-stuff. I recommend going with the most concordant,
simplest, etc way to voice the triad that fits into two octaves (e.g.
8:9:10 becomes 4:5:9, 8:9:14 becomes 4:7:9, etc).
3) Find the simplest chord of n notes and work out what its
concordance is, and then work out its chord coverage. I don't really
understand this one.

The three things that would be ideal to do in the future are:
1) Replace every instance of Tenney height above with HE.
2) Work out the HE signal processing transform.
3) Test the concordance of the scale as a whole, and see if it's
quicker and gives similar results.

Yes?

-Mike

🔗Carl Lumma <carl@...>

Mike Battaglia <battaglia01@...> wrote:

> > There are, but for most chords the mapping is irrelevant,
> > in that it leads to JI chords with geomean(a*b*c...) > 10.
>
> Yes, but it's just supposed to be a rough metric for evaluating
> consonance for more than triads. I guess if you want you could
> do something like min(geomean(a,b,c,...),10), which traps the
> discordance at a certain cutoff and avoids skewing the results.

geomean(a*b*c...) > 10 doesn't mean the discordance is high,
it means Tenney Height is useless.

> > The first idea (looking up the 3HE of all triads) is a
> > straightforward problem. 3-combinations, lookup, mean or
> > std., done.
> >
> > The latter idea is something very close to what Igs was doing.
> > Find the simplest triad or tetrad within an octave generated
> > by the mapping group. Subtract from the scale size then divide
> > that by the scale size. That gives you the chords per note.
>
> What do you mean by chords per note? So for meantone[12],
> we're looking at 4:5:6.

Maybe.

> 12 - 3 is 9, and 9/12 is 3/4, and that's the amount
> of chords per note.

Right.

> What does 3/4 chords per note mean...?

It means you're more likely to get a concordant chord when
you mash the keyboard.

> > Multiply by the chord size (3 or 4 as appropriate) and you get
> > the chord coverage per note. Scala reports chord coverage in
> > its show locations output by the way.
>
> So you get 9/4 coverage of the scale per note? I'm completely
> lost now.

This just normalizes for larger chords. What you get when you
mash depends on the number of ways a given note can participate
in a concordance. That's chords/note, except if it's triads
it's representing fewer relationships to the other notes in the
scale than if it's tetrads. So now do a show locations 4:5:6
on the diatonic scale. It says all 7 notes are covered, and
two are covered more than once (the C major mode both C and G
participate in two major triads).

> The three things we're proposing to test, and what we need to
> do them, are:
> 1) Mean/standard deviation/etc of discordance of all of the
> chords in the scale. We can place a strict limit on Tenney
> height discordance at 10 if you think that'll be better.

I think 3HE on all the triads is a better bet.

> Either way, that to automate this I need to figure out what
> every random chord is mapped to, in simplest terms, or else
> I can't get its Tenney Height.

If you want to do it this way it's a lattice search problem.
For a scale like meantone[12] which is a MOS of a rank 2
temperament, you get the linear val for the scale. Then you
tune the JI lattice out to some radius using this val. Then
you draw circles around the notes of the incoming chord on
your labeled lattice and finally, pick the smallest circle.

> 2) Triadic HE minimizer. To do this, we need to work out

Running the minimizer is harder than just scoring given
scales. The minimizer requires Monte Carlo search. Scoring
is just combinations and lookup, as I said.

> the octave
> equivalent-stuff. I recommend going with the most concordant,
> simplest, etc way to voice the triad that fits into two
> octaves (e.g. 8:9:10 becomes 4:5:9, 8:9:14 becomes 4:7:9, etc).

You're thinking about starting with JI chords. I'm saying
just take all 3-combinations of two octaves of the scale and
look up their 3HE. You should be able to do it an hour.

> 3) Find the simplest chord of n notes and work out what its
> concordance is, and then work out its chord coverage. I don't
> really understand this one.

That's basically just the Graham complexity.

(scale size - G.C.) * chord size) / scale size

should be the average chord coverage. Just use the one-octave
voicing of the mapping group. 4:5:6:7 for 7-limit etc.

> The three things that would be ideal to do in the future are:
> 1) Replace every instance of Tenney height above with HE.
> 2) Work out the HE signal processing transform.
> 3) Test the concordance of the scale as a whole, and see if it's
> quicker and gives similar results.
>
> Yes?

Yes but the above two methods are no slouch and should be
easy enough. If you run them both for a bunch of scales we
can see if I'm right that they're basically the same.

-Carl

🔗Mike Battaglia <battaglia01@...>

Carl,

Lots of stuff here, I think I'm on the same page now. Just a few things:

On Sun, Mar 6, 2011 at 6:45 PM, Carl Lumma <carl@...> wrote:
>
> > Either way, that to automate this I need to figure out what
> > every random chord is mapped to, in simplest terms, or else
> > I can't get its Tenney Height.
>
> If you want to do it this way it's a lattice search problem.
> For a scale like meantone[12] which is a MOS of a rank 2
> temperament, you get the linear val for the scale. Then you
> tune the JI lattice out to some radius using this val. Then
> you draw circles around the notes of the incoming chord on
> your labeled lattice and finally, pick the smallest circle.

What do you mean pick the smallest circle? You just said to "draw
circles," what determines the size of these circles to begin with so I
can pick the smallest one?

> > 2) Triadic HE minimizer. To do this, we need to work out
>
> Running the minimizer is harder than just scoring given
> scales. The minimizer requires Monte Carlo search. Scoring
> is just combinations and lookup, as I said.

I thought scoring was minimized. OK, so just scoring.

> > the octave
> > equivalent-stuff. I recommend going with the most concordant,
> > simplest, etc way to voice the triad that fits into two
> > octaves (e.g. 8:9:10 becomes 4:5:9, 8:9:14 becomes 4:7:9, etc).
>
> You're thinking about starting with JI chords. I'm saying
> just take all 3-combinations of two octaves of the scale and
> look up their 3HE. You should be able to do it an hour.

This might have to be next week.

> > 3) Find the simplest chord of n notes and work out what its
> > concordance is, and then work out its chord coverage. I don't
> > really understand this one.
>
> That's basically just the Graham complexity.
>
> (scale size - G.C.) * chord size) / scale size
>
> should be the average chord coverage. Just use the one-octave
> voicing of the mapping group. 4:5:6:7 for 7-limit etc.

You mean the average chord coverage for that chord, in the sense that
you defined before? This is the same thing as the first type of chord
coverage test you proposed?

> > Yes?
>
> Yes but the above two methods are no slouch and should be
> easy enough. If you run them both for a bunch of scales we
> can see if I'm right that they're basically the same.

Feel free to start trying to hack away at it but I'm going to be out
of action for a bit. If someone can work out some kind of way to at
least work out the Tenney height for an arbitrary chord in some
temperament, I can load it up and at least do that test. Otherwise the
rest will have to wait.

-Mike

🔗Carl Lumma <carl@...>

Mike Battaglia <battaglia01@...> wrote:

> > If you want to do it this way it's a lattice search problem.
> > For a scale like meantone[12] which is a MOS of a rank 2
> > temperament, you get the linear val for the scale. Then you
> > tune the JI lattice out to some radius using this val. Then
> > you draw circles around the notes of the incoming chord on
> > your labeled lattice and finally, pick the smallest circle.
>
> What do you mean pick the smallest circle? You just said to
> "draw circles," what determines the size of these circles to
> begin with so I can pick the smallest one?

What you actually do is find the volume of the smallest convex
lattice region which includes all the points of the chord
once each.

> > > 2) Triadic HE minimizer. To do this, we need to work out
> >
> > Running the minimizer is harder than just scoring given
> > scales. The minimizer requires Monte Carlo search. Scoring
> > is just combinations and lookup, as I said.
>
> I thought scoring was minimized. OK, so just scoring.

Scoring is just looking up the triads on Steve's list and
then taking the mean or stdev or whatever you like. That
gives you the mashability number for a given scale.

Minimizing is *finding* the most mashable scale directly.
Paul's dyadic minimizer did, FWIW, find the meantone
pentatonic scale. The triadic minimizer is going to be a
lot harder because of the additional interrelationships.

> > > 3) Find the simplest chord of n notes and work out what its
> > > concordance is, and then work out its chord coverage. I don't
> > > really understand this one.
> >
> > That's basically just the Graham complexity.
> >
> > (scale size - G.C.) * chord size) / scale size
> >
> > should be the average chord coverage. Just use the one-octave
> > voicing of the mapping group. 4:5:6:7 for 7-limit etc.
>
> You mean the average chord coverage for that chord, in the
> sense that you defined before? This is the same thing as
> the first type of chord coverage test you proposed?

Yes, it's the same formula I mentioned before.

> Feel free to start trying to hack away at it but I'm going to
> be out of action for a bit.

I'm swamped. It's your idea, and it's a good one. You should
see it through to completion! If that's next week that's no
problem.

> If someone can work out some kind of way to at
> least work out the Tenney height for an arbitrary chord

I've told you to forget Tenney height about as many times
as I can. You don't have to listen to me but then let's
stop discussing it. Anyway I told you how I'd do it (first
paragraph here).

> I can load it up and at least do that test.

If you can do that you can do the 3HE thing! -- it's WAAAY
easier. -Carl