Yahoo Tuning Groups Ultimate Backup tuning Margo Schulter's JI 'virtual 12-eq' Artusi scale

Hmmm... I missed out on a few really interesting things
while I was computer-challenged last summer.

I found an interesting scale proposed and explained by
Margo Schulter in the old Mills TD 1511.2, from 1998-Aug-21.

I haven't yet read in depth either Margo's posting or the
numerous responses to it, only scanned them. But I didn't
see any prime-factor version of the ratios, which surprised
me, given how gigantic some of them were.

So I did it, and was intrigued by the periodicities I noticed
in the progression of the exponents in this 5-limit system.
I've reproduced Margo's original table, with the prime-factor
formula for each ratio added under the cents:

--------------------------------------------------------------------
Note Interval Ratio/
(Cents)
--------------------------------------------------------------------
c Unison 1:1
(0.000000)
2^((13*0)+0)*3^-(7*0)*5^-0

g 5 16384:10935
(700.001280)
2^((13*1)+1)*3^-(7*1)*5^-1

d M2 134217728:
119574225
(200.002560)
2^((13*2)+1)*3^-(7*2)*5^-2

a M6 2199023255552:
1307544150375
(900.003840)
2^((13*3)+2)*3^-(7*3)*5^-3

e M3 18014398509481984:
14297995284350625
(400.005120)
2^((13*4)+2)*3^-(7*4)*5^-4

b M7 295147905179352825856:
156348578434374084375
(1100.006400)
2^((13*5)+3)*3^-(7*5)*5^-5

f# A4/d5 2417851639229258349412352:
1709671705179880612640625
(600.007680)
2^((13*6)+3)*3^-(7*6)*5^-6

c# A1/m2 19807040628566084398385987584:
18695260096141994499225234375
(100.008961)
2^((13*7)+3)*3^-(7*7)*5^-7

g# A5/m6 324518553658426726783156020576256:
204432669151312709849027937890625
(800.010241)
2^((13*8)+4)*3^-(7*8)*5^-8

d# A2/m3 2658455991569831745807614120560689152:
2235471237169604482199120500833984375
(300.011521)
2^((13*9)+4)*3^-(7*9)*5^-9

a# A6/m7 43556142965880123323311949751266331066368:
24444877978449625012847382676619619140625
(1000.012801)
2^((13*10)+5)*3^-(7*10)*5^-10

e# A3/4 356811923176489970264571492362373784095686656:
267304740694346649515486129568835535302734375
(500.014081)
2^((13*11)+5)*3^-(7*11)*5^-11

b# 8 5846006549323611672814739330865132078623730171904:
2922977339492680612451840826835216578535400390625
(1200.015361)
2^((13*12)+6)*3^-(7*12)*5^-12

Of course, the 1:1 can be simply described as n^0, but
I expanded it here to make it fit the formula.

And to make it easier to see how the formula works, here are
the prime-factor notations by themselves:

2^((13* 0)+0) * 3^-(7* 0) * 5^- 0
2^((13* 1)+1) * 3^-(7* 1) * 5^- 1
2^((13* 2)+1) * 3^-(7* 2) * 5^- 2
2^((13* 3)+2) * 3^-(7* 3) * 5^- 3
2^((13* 4)+2) * 3^-(7* 4) * 5^- 4
2^((13* 5)+3) * 3^-(7* 5) * 5^- 5
2^((13* 6)+3) * 3^-(7* 6) * 5^- 6
2^((13* 7)+3) * 3^-(7* 7) * 5^- 7
2^((13* 8)+4) * 3^-(7* 8) * 5^- 8
2^((13* 9)+4) * 3^-(7* 9) * 5^- 9
2^((13*10)+5) * 3^-(7*10) * 5^-10
2^((13*11)+5) * 3^-(7*11) * 5^-11
2^((13*12)+6) * 3^-(7*12) * 5^-12

The formula itself is:
2^((13*degree)+number) * 3^-(7*degree) * 5^-degree

where 'number' is the part I haven't figured out.

The exponents of prime-factor 2 could also be calculated
as (14*degree)-number.

-monz

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

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🔗monz@xxxx.xxx

[me, monz, TD 171.2]

> 2^((13* 0)+0) * 3^-(7* 0) * 5^- 0
> 2^((13* 1)+1) * 3^-(7* 1) * 5^- 1
> 2^((13* 2)+1) * 3^-(7* 2) * 5^- 2
> 2^((13* 3)+2) * 3^-(7* 3) * 5^- 3
> 2^((13* 4)+2) * 3^-(7* 4) * 5^- 4
> 2^((13* 5)+3) * 3^-(7* 5) * 5^- 5
> 2^((13* 6)+3) * 3^-(7* 6) * 5^- 6
> 2^((13* 7)+3) * 3^-(7* 7) * 5^- 7
> 2^((13* 8)+4) * 3^-(7* 8) * 5^- 8
> 2^((13* 9)+4) * 3^-(7* 9) * 5^- 9
> 2^((13*10)+5) * 3^-(7*10) * 5^-10
> 2^((13*11)+5) * 3^-(7*11) * 5^-11
> 2^((13*12)+6) * 3^-(7*12) * 5^-12
>
> The formula itself is:
> 2^((13*degree)+number) * 3^-(7*degree) * 5^-degree
>
> where 'number' is the part I haven't figured out.

Well, I *do* know that 'number' is approximately 'degree/2'.

Perhaps one of you mathematicians can figure out
exactly what works for this variable.

-monz

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

🔗monz@xxxx.xxx

[me, monz, TD 171.2]
> The formula itself is:
> 2^((13*degree)+number) * 3^-(7*degree) * 5^-degree

Sorry - bad nomenclature here:

'degree' really should be called 'fifth', or perhaps
'Quint' as used by Paul Erlich and many German theorists.

This would make 'number' approximately 'Quint/2',
and I would appreciate a more exact quantification of 'number'
from anyone who can figure it out.

-monz

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

🔗monz@xxxx.xxx

[me, monz, TD 171.2]
>
> 2^((13* 0)+0) * 3^-(7* 0) * 5^- 0
> 2^((13* 1)+1) * 3^-(7* 1) * 5^- 1
> 2^((13* 2)+1) * 3^-(7* 2) * 5^- 2
> 2^((13* 3)+2) * 3^-(7* 3) * 5^- 3
> 2^((13* 4)+2) * 3^-(7* 4) * 5^- 4
> 2^((13* 5)+3) * 3^-(7* 5) * 5^- 5
> 2^((13* 6)+3) * 3^-(7* 6) * 5^- 6
> 2^((13* 7)+3) * 3^-(7* 7) * 5^- 7
> 2^((13* 8)+4) * 3^-(7* 8) * 5^- 8
> 2^((13* 9)+4) * 3^-(7* 9) * 5^- 9
> 2^((13*10)+5) * 3^-(7*10) * 5^-10
> 2^((13*11)+5) * 3^-(7*11) * 5^-11
> 2^((13*12)+6) * 3^-(7*12) * 5^-12

Hmmm... even tho the ratios in this scale are contained
within the 5-limit (odd or prime), factoring them the
way I have shows that with a different interpretation of
'limit' this could be construed as a 13-prime-limit system,
where 7 and 13 are quite prominent.

Interesting...

Can a mathematician explain if there's any significance in
this idea?

-monz

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

🔗Brett Barbaro <barbaro@xxxxxxxxx.xxxx>

> [me, monz, TD 171.2]
>>
>> 2^((13* 0)+0) * 3^-(7* 0) * 5^- 0
>> 2^((13* 1)+1) * 3^-(7* 1) * 5^- 1
>> 2^((13* 2)+1) * 3^-(7* 2) * 5^- 2
>> 2^((13* 3)+2) * 3^-(7* 3) * 5^- 3
>> 2^((13* 4)+2) * 3^-(7* 4) * 5^- 4
>> 2^((13* 5)+3) * 3^-(7* 5) * 5^- 5
>> 2^((13* 6)+3) * 3^-(7* 6) * 5^- 6
>> 2^((13* 7)+3) * 3^-(7* 7) * 5^- 7
>> 2^((13* 8)+4) * 3^-(7* 8) * 5^- 8
>> 2^((13* 9)+4) * 3^-(7* 9) * 5^- 9
>> 2^((13*10)+5) * 3^-(7*10) * 5^-10
>> 2^((13*11)+5) * 3^-(7*11) * 5^-11
>> 2^((13*12)+6) * 3^-(7*12) * 5^-12
>
>
> Hmmm... even tho the ratios in this scale are contained
> within the 5-limit (odd or prime),

Hi, Joe. They are contained in the 5-prime-limit but not the 5-odd-limit.

> factoring them the
> way I have shows that with a different interpretation of
> 'limit' this could be construed as a 13-prime-limit system,
> where 7 and 13 are quite prominent.
> Interesting...
>
>
> Can a mathematician explain if there's any significance in
> this idea?

The exponents on the primes in the first interval are 13, 7, and 1. All the
other intervals are created by superimposing multiple copies of that first
interval. Therefore all exponents will be multiples of 13, 7, and 1. The
"number" in the power of two is simply the accumulated number of octave
displacements requires to keep the tuning within one octave.