Yahoo Tuning Groups Ultimate Backup tuning lattice wormholes

{Paul Erlich, TD 141.8:]
> what I called a "wormhole" (because it's a "warp" in the
> Euclidean fabric of the lattice) refers to the fact that
> certain composite-based ratios (such as 9:5) are in fact simpler
> than distances on a prime-based lattice would indicate. At least
> if you're talking about octave-equivalent lattices.

Can you explain what is the criteria for determining which
of the 'certain' composite ratios is 'simpler than distances
on a prime-based lattice would indicate'?

Don't mistake, by the 'lack-of-tone-in-speech email bug'
as Carl called it, that I'm being argumentative here.

I am sincerely interested in understanding exactly what
you mean. In my lattice formula the prime-factoring
itself makes composite factors such as 9 and 15 shorter
distances than their odd/integer values would indicate,
and to me this is the whole point of using prime instead
of odd factoring.

What would give 'certain' ratios, like that 9:5, an
even shorter distance? Forgive me if I missed something
in the weighting/complexity threads that would give the
answer - I'm still going thru those postings to try and
follow the whole line of reasoning.

-monz

Joseph L. Monzo....................monz@juno.com
http://www.ixpres.com/interval/monzo/homepage.html
|"...I had broken thru the lattice barrier..."|
| - Erv Wilson |
--------------------------------------------------

___________________________________________________________________
You don't need to buy Internet access to use free Internet e-mail.
Get completely free e-mail from Juno at http://www.juno.com/getjuno.html
or call Juno at (800) 654-JUNO [654-5866]

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

I wrote,

>> what I called a "wormhole" (because it's a "warp" in the
>> Euclidean fabric of the lattice) refers to the fact that
>> certain composite-based ratios (such as 9:5) are in fact simpler
>> than distances on a prime-based lattice would indicate. At least
>> if you're talking about octave-equivalent lattices.

Joe Monzo wrote,

>Can you explain what is the criteria for determining which
>of the 'certain' composite ratios is 'simpler than distances
>on a prime-based lattice would indicate'?

Well, in the type of octave-equivalent lattice that I outlined some time ago
and Paul Hahn more recently was involved in discussing, the construction is
as follows:

1st axis: steps of 3:2 have length log(3)

2nd axis: steps of 5:4 have length log(5)
the angle between the first and second axes is determined by the condition
that
steps of 5:3 have length log(5)

3rd axis: steps of 7:4 have length log(7)
the angle between the the first and third axes is determined by the
condition that
steps of 7:5 have length log (7)
the angle between the second and third axes is determined by the condition
that
steps of 7:6 have length log (7)

Obviously now we're in three dimensions. Now a prime-based lattice means:

4th axis: steps of 11:8 have length log(11)
so do steps of 11:5, 11:6, 11:7, and 11:9.

with the angles in 4-d determined by these lengths all being log(11), in
analogy with the lower-dimensional cases.

and so on, but since distances are measured using a triangular city-block
metric, and 9:5 is a step of 6:5 and a step of 3:2, the distance
corresponding to 9:5 is log(5)+log(3)=log(15). However, 9:5 is more
consonant than 11:5, which unfortunately has a shorter length of log(11). So
if we wish to keep the construction prime-based (giving a unique location to
each pitch-class), we must posit a "wormhole" making a 9:5 connection with
length log(9). Actually, I don't know if one actually needs to go into
non-Euclidean space to do this; can anyone calculate the Euclidean length of
a 9:5 in the lattice constructed above? If less than log(9), we can simply
draw a curved line to make the 9:5 connection with length log(9). If the
Euclidean distance is greater than log(9), we need a non-Euclidean "warp" in
space to make this connection, and that is what inspired the term
"wormhole".

>In my lattice formula the prime-factoring
>itself makes composite factors such as 9 and 15 shorter
>distances than their odd/integer values would indicate,
>and to me this is the whole point of using prime instead
>of odd factoring.

What exactly is your lattice formula that gives distances as output?

🔗monz@xxxx.xxx

[Paul Erlich, TD 156.14:]
> What exactly is your lattice formula that gives distances as
> output?

Well, we were discussing composites, and my formula doesn't
give those as outputs directly, but rather as the sum
of their composite prime vectors.

I suppose that for a simple composite of two prime factors,
the hypotenuse can be calculated from the the two prime
vectors by the Pythagorean theorem.

For more complicated ratios, I don't know how to do it
- I would think it's a trigonometry formula of some kind.

First I calculate the angle of the vector for each
prime as a function of its Semitone value, assuming
12 o'clock as 0 Semitones and travelling clockwise for
increasing Semitones:

PrimeAngle = ((3-Semitones)*PI)/6

Then I simply calculate x and y as the product of the
prime and the number of exponents it has, for each prime
factor in the ratio:

x = COS(PrimeAngle) * xOrigin * PrimeExponent * prime
y = SIN(PrimeAngle) * yOrigin * PrimeExponent * prime

> the angle between the first and second axes is determined by
> the condition that steps of 5:3 have length log(5)

Is there a formula that does this?

-monz

🔗Paul Hahn <Paul-Hahn@xxxxxxx.xxxxx.xxxx>

On Tue, 27 Apr 1999 monz@juno.com wrote:
[Paul Erlich:]
>> the angle between the first and second axes is determined by
>> the condition that steps of 5:3 have length log(5)
>
> Is there a formula that does this?

The actual, Euclidean distances in our proportional-distance
triangulated lattice are:

Between 5/4 and 9/8: sqrt((lg5)^2 + 2 * (lg3)^2) = 3.227315...
Between 7/4 and 9/8: sqrt((lg7)^2 + 2 * (lg3)^2) = 3.592416...

Since lg9 is about 3.169925, the distances are greater than desired, not
lesser, so Paul E.'s curved-line idea doesn't work.* Instead, a slight
"spacewarp" is necessary.

(*As stated, anyway. We could make the 5-limit and 7-limit lines curved
and the 9-limit lines straight instead; that would work.)

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "Hey--do you think I need to lose some weight?"
-\-\-- o
NOTE: dehyphenate node to remove spamblock. <*>

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

I didn't get the latest digest, Joe, so if this doesn't get posted, will you
do that for me?

I wrote,

>> What exactly is your lattice formula that gives distances as
>> output?

Joe Monzo wrote,

>Well, we were discussing composites, and my formula doesn't
>give those as outputs directly, but rather as the sum
>of their composite prime vectors.

>I suppose that for a simple composite of two prime factors,
>the hypotenuse can be calculated from the the two prime
>vectors by the Pythagorean theorem.

That's not true; your prime vectors are not at right angles. But basically
you're proposing a metric -- we've had disagreements over whether it should
be called Euclidean, Cartesian, or what -- which connects two notes with a
straight line and uses the distance of the line as a measure of complexity.
As we discussed before, in your lattice, that gives 15:8 considerably lower
complexity than 6:5, which doesn't seem right to me.

You were saying that your lattice makes composites appear simpler than their
odd or prime represenatations would suggest. That doesn't seem to be true.

>For more complicated ratios, I don't know how to do it
>- I would think it's a trigonometry formula of some kind.

You could either use the x and y values directly, or use the prime vectors
in conjunction with the law of cosines.

>> the angle between the first and second axes is determined by
>> the condition that steps of 5:3 have length log(5)

>Is there a formula that does this?

Since the triad forms an isosceles triangle with the 3:2 at the base, we can
bisect it into two right triangles. In each right triangle, the side along
the 3-axis has length log(3)/2, and the hypotenuse has length log(5). The
cosine of angle between the axes is therefore log(3)/(2*log(5)), so the
angle is 1.2225 radians or 70.04 degrees.

🔗monz@xxxx.xxx

[Paul Erlich, TD 158.9]
> As we discussed before, in your lattice, that gives 15:8
> considerably lower complexity than 6:5, which doesn't seem
> right to me.

???

On my lattice 15:8 and 6:5 have exactly the same complexity,
in terms of a direct vector lattice metric: both have
one rung on each of the 3- and 5- axes.

This doesn't change your argument, because your point is that
15:8 should have greater complexity than 6:5, neither lower *nor*
equal. I can agree with that.

[Erlich]
> You were saying that your lattice makes composites appear
> simpler than their odd or prime represenatations would suggest.
> That doesn't seem to be true.

Hold on - don't get it twisted.

This is what I originally said:

[me, monz, TD 155?]
> In my lattice formula the prime-factoring
> itself makes composite factors such as 9 and 15 shorter
> distances than their odd/integer values would indicate,
> and to me this is the whole point of using prime instead
> of odd factoring.

My lattice *are* a prime representation, so there's no way
that composites could appear simpler than their prime
representations, unless I have wormholes too.

I have more to say on this, but it will have to wait until
after I've done some calculations and formulated a more detailed
question.

-monz

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

>>> the angle between the first and second axes is determined by
>>> the condition that steps of 5:3 have length log(5)
>
>> Is there a formula that does this?

Paul Hahn wrote,

>The actual, Euclidean distances in our proportional-distance
>triangulated lattice are:

>Between 5/4 and 9/8: sqrt((lg5)^2 + 2 * (lg3)^2) = 3.227315...
>Between 7/4 and 9/8: sqrt((lg7)^2 + 2 * (lg3)^2) = 3.592416...

>Since lg9 is about 3.169925, the distances are greater than desired, not
>lesser, so Paul E.'s curved-line idea doesn't work.

I'm glad it doesn't -- that means my original "wormhole" intuition was
correct.

>* Instead, a slight
>"spacewarp" is necessary.

Yup, wormhole=spacewarp.

>(*As stated, anyway. We could make the 5-limit and 7-limit lines curved
>and the 9-limit lines straight instead; that would work.)

Well, if you started adding more and more dimensions, you end up having to
curve the prime-axis lines more and more, so I prefer fixing those as
straight (who wants a spaghetti lattice, anyway?) and having special
spacewarp/wormhole connections for the ratios of odd composites. Or, one
could envision the lines as tubes and use thinner tubes for the odd
composites, so that the volume of the tube-connections represents the
complexity of the ratio.

🔗perlich@xxxxxxxxxxxxx.xxx

I wrote,

>> As we discussed before, in your lattice, that gives 15:8
>> considerably lower complexity than 6:5, which doesn't seem
>> right to me.

Joe Monzo wrote,

>???

>On my lattice 15:8 and 6:5 have exactly the same complexity,
>in terms of a direct vector lattice metric:� both have
>one rung on each of the 3- and 5- axes.

Yes but you were just advocating not this city-block metric but a Euclidean metric (in which you would have to use the Pythagorean theorem or its non-right-triangle variant, the law of cosines). Look at your latest posts -- you're a slippery fellow!

>> You were saying that your lattice makes composites appear
>> simpler than their odd or prime represenatations would suggest.
>> That doesn't seem to be true.

>Hold on - don't get it twisted.

>This is what I originally said:

>[me, monz, TD 155?]
>> In my lattice formula the prime-factoring
>> itself makes composite factors such as 9 and 15 shorter
>> distances than their odd/integer values would indicate,
>> and to me this is the whole point of using prime instead
>> of odd factoring.

>My lattice *are* a prime representation, so there's no way
>that composites could appear simpler than their prime
>representations, unless I have wormholes too.

What about the odd part? That is present in your original statement.