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Message: 10426 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 18:52:18

Subject: Re: Hanzos

From: Carl Lumma

>>>> >y recollection is that Paul H.'s algorithm assigns a unique >>>> lattice route (and therefore hanzo) to each 9-limit interval. >>>
>>> So what? You still get an infinite number representing each >>> interval, since you can multiply by arbitary powers of the dummy >>> comma 9/3^2. >>
>> An infinite number from where? If you look at the algorithm, that >> dummy comma has zero length. >
>How do you get that???? >Given a Fokker-style interval vector (I1, I2, . . . In):
[-2 0 0 1]
>1. Go to the rightmost nonzero exponent; add its absolute value >to the total. T=1 >2. Use that exponent to cancel out as many exponents of the opposite >sign as possible, starting to its immediate left and working right; >discard anything remaining of that exponent.
[-1 0 0 0] T=1
>3. If any nonzero exponents remain, go back to step one, otherwise >stop.
T=... whoops, I forgot an absolute value here. The correct value is 2. -Carl
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Message: 10427 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 20:21:06

Subject: Re: Canonical generators for 7-limit planar temperaments

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul G Hjelmstad" 
<paul.hjelmstad@u...> wrote:

> Thanks. So for example - 81/80, has generators 4/3 and 9/7. How does > one project onto the plane to obtain 1/7 and 5/7 respectively?
You lost me. Where did 1/7 and 5/7 come from? Do you mean project to exact coordinates on the 3/2-9/7 plane, or simply give them in terms of 3/2 and 9/7?
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Message: 10428 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 12:25:59

Subject: Re: Harmonized melody in the 7-limit

From: Carl Lumma

>> >f I didn't know better I'd say you were trying to BS me. What >> is a lattice of note classes? >
>It's the kind of lattice I was talking about--for each octave >ewquivalence class, we have a lattice point. Hence there is a lattice >point representing 9,9/4,9/8... etc but only one.
If you take the 2s out of the hanzos, we have that. -Carl
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Message: 10429 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 19:01:22

Subject: Re: Hanzos

From: Carl Lumma

>>>> >o what? You still get an infinite number representing each >>>> interval, since you can multiply by arbitary powers of the dummy >>>> comma 9/3^2. >>>
>>> An infinite number from where? If you look at the algorithm, that >>> dummy comma has zero length. >>
>> If it has length zero then we are not talking about a lattice at all, >> though a quotient of it (modding out the dummy comma) might be. In a >> symmetrical lattice it necessarily has the same length as, for >> example, 11/3^2, which is of length sqrt(1^2+2^2-1*2)=sqrt(3).
I don't know where sqrt would be coming from. I thought everything would have to have whole number lengths.
>> It >> does *not* have the same length as 11/9, which is of length one, of >> course. >
>Yes, you're onto something here. In the unweighted lattice there is >a point for 9/3^2, which lies on the diameter-1 hull.
That was based on the rather simplistic idea that one steps out and two back leaves you one away from where you started. -Carl ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * [with cont.] <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service * [with cont.] (Wayb.)
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Message: 10431 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 04:11:50

Subject: Re: 9-limit stepwise

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...>
wrote:

>> 116 [49/45, 10/9, 9/8, 8/7] [1, 1, 2, 2] 1.568046 6 > Pajara[6]
It's six notes of pajara, but not Pajara[6]. The 49/45 and 10/9 are seven fifths up and four octaves down--the apotome, 49/45 and 10/9 all being the same in pajara. The 9/8 and 8/7 of course are the same, both two fifths up and an octave down. So in terms of generators, it looks like [2,2,7,2,2,7], which adds up to 22 fifths. In 22-equal, it would be 443443, which is what you get from a generator of a half-octave plus 8/22=4/11, which is a 9/7, and as a DE is associated to a different temperament, with commas 50/49 and 245/243, which has come up before but for which I don't have a name. So which temperament is it? In practical terms, you probably take both, ending up with 22-et, and a 6-note scale which is *not* Pajara[6] but which is related to both pajara and the 9/7-generator temperament.
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Message: 10432 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 20:58:06

Subject: Re: Harmonized melody in the 7-limit

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
>>> If I didn't know better I'd say you were trying to BS me. What >>> is a lattice of note classes? >>
>> It's the kind of lattice I was talking about--for each octave >> ewquivalence class, we have a lattice point. Hence there is a lattice >> point representing 9,9/4,9/8... etc but only one. >
> If you take the 2s out of the hanzos, we have that.
We do not. 9/3^2 has no 2s in it, but it isn't the same as 1.
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Message: 10433 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 01:26:20

Subject: Re: Hanzos

From: Carl Lumma

>> >ne of us is still misunderstanding Paul Hahn's 9-limit approach. >> In the unweighted version 3, 5, 7 and 9 are all the same length. >
>In this system you don't exactly, have 7-limit notes and intervals. >You do have "hanzos", with basis 2,3,5,7,9.
The usual point of odd-limit is to get octave equivalence, and therefore I'd say the 2s should be dropped from the basis.
>The hahnzo |0 -2 0 0 1> is a comma, 9/3^2, which obviously would >play a special role. Hahnzos map onto 7-limit intervals, but not >1-1. Are you happy with the idea that two scales could be >different, since they have steps and notes which are distinct as >hahnzos, even though they have exactly the same steps and notes >in the 7-limit?
I think the answer here is yes, though I'm at a loss for why you're mapping hanzos to the 7-limit.
>We've got three hahnzos corresponding to 81/80;
My recollection is that Paul H.'s algorithm assigns a unique lattice route (and therefore hanzo) to each 9-limit interval. Certainly it can be used to find the set of lattice points within distance <= 2 of a given point. -Carl
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Message: 10434 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:04:55

Subject: Re: TOP and Tenney space webpage

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: >
>> I have no idea how to reconcile this with >> >> Yahoo groups: /tuning-math/message/9797 * [with cont.] >> >> and with the fact that it's obvious that this linear operator, when >> acting on a monzo, is the only one returns its pitch (or interval >> size) in cents. >
> I'm being contrary; it seems to me "measure" isn't really how we want > to look at it, since just intonation is now being viewed as one of an > infinite set of possible tuning maps.
True. It only measures pitch (or interval size) in the untempered case.
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Message: 10435 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 01:18:50

Subject: Re: Harmonized melody in the 7-limit

From: Carl Lumma

>> >ne of us is still misunderstanding Paul Hahn's 9-limit approach. >
>What in the world makes you think this has anything to do with me or >anything I've said? ""
The 9-limit would be different, for sure. The simple symmetrical lattice criterion wouldn't work, but it would be easy enough to find what does. If you call something which makes 3 half as large as 5 or 7 "symmetrical", it does. "" -Carl
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Message: 10436 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:05:58

Subject: Re: non-1200: Tenney/heuristic meantone temperament

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: >
>> Amazingly, Gene's page would have you believe you need to search > even
>> in the codimension 1 case! >
> You want I should derive the codimension 1 formula instead? Why not?
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Message: 10437 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 11:34:22

Subject: Stepwise 7-limit chord progressions

From: Gene Ward Smith

15/14

{[{1, 3/2, 5/4, 7/4}, {3/2, 9/7, 15/14, 12/7}], 
[{1, 3/2, 5/4, 7/4}, {25/14, 5/4, 15/14, 10/7}], 
[{1, 3/2, 5/4, 7/4}, {15/8, 3/2, 5/4, 15/14}], 
[{1,3/2, 6/5, 12/7}, {3/2, 9/7, 15/14, 12/7}], 
[{1, 5/3, 5/4, 10/7}, {25/14, 5/4,15/14, 10/7}], 
[{1, 3/2, 6/5, 12/7}, {15/8, 3/2, 5/4, 15/14}], 
[{1, 5/3, 5/4,10/7}, {15/8, 3/2, 5/4, 15/14}], 
[{1, 8/7, 10/7, 12/7}, {3/2, 9/7, 15/14, 12/7}], 
[{1, 8/7, 10/7, 12/7}, {25/14, 5/4, 15/14, 10/7}], 
[{1, 3/2, 6/5, 12/7}, {60/49, 15/14, 10/7, 12/7}], 
[{1, 5/3, 5/4, 10/7}, {60/49, 15/14, 10/7, 12/7}],
[{1, 8/7, 10/7, 12/7}, {60/49, 15/14, 10/7, 12/7}]}

16/15
{[{1, 7/5, 6/5, 8/5}, {64/35, 16/15, 8/5, 32/25}], 
[{1, 8/7, 4/3, 8/5},{64/35, 16/15, 8/5, 32/25}], 
[{1, 7/5, 6/5, 8/5}, {28/15, 16/15, 4/3, 8/5}], 
[{1, 8/7, 4/3, 8/5}, {28/15, 16/15, 4/3, 8/5}], 
[{1, 8/7, 4/3, 8/5}, {16/9, 32/21, 16/15, 4/3}], 
[{1, 5/3, 7/6, 4/3}, {28/15, 16/15, 4/3, 8/5}], 
[{1, 5/3, 7/6, 4/3}, {16/9, 32/21, 16/15, 4/3}]}

21/20
{[{1, 3/2, 5/4, 7/4}, {9/5, 21/20, 3/2, 6/5}], 
[{1, 7/5, 6/5, 8/5}, {21/20, 7/5, 7/4, 49/40}], 
[{1, 3/2, 5/4, 7/4}, {21/20, 21/16, 3/2, 7/4}], 
[{1, 3/2, 6/5, 12/7}, {42/25, 21/20, 7/5, 6/5}], 
[{1, 3/2, 6/5, 12/7}, {9/5, 21/20,3/2, 6/5}], 
[{1, 7/5, 6/5, 8/5}, {42/25, 21/20, 7/5, 6/5}], 
[{1, 3/2, 5/4, 7/4}, {21/20, 7/5, 7/4, 49/40}], 
[{1, 7/5, 6/5, 8/5}, {9/5, 21/20, 3/2, 6/5}], 
[{1, 3/2, 6/5, 12/7}, {21/20, 21/16, 3/2, 7/4}], 
[{1, 7/5, 7/6, 7/4}, {21/20, 7/5, 7/4, 49/40}], 
[{1, 7/5, 7/6, 7/4}, {42/25, 21/20, 7/5, 6/5}], 
[{1, 7/5, 7/6, 7/4}, {21/20, 21/16, 3/2, 7/4}]}

25/24
{[{1, 5/3, 5/4, 10/7}, {25/14, 25/24, 5/4, 25/16}], 
[{1, 3/2, 5/4, 7/4}, {5/3, 25/24, 5/4, 35/24}], 
[{1, 3/2, 5/4, 7/4}, {25/14, 25/24, 5/4, 25/16}],
[{1, 5/3, 7/6, 4/3}, {5/3, 25/24, 5/4, 35/24}], 
[{1, 5/3, 7/6, 4/3}, {5/3, 25/24, 25/21, 25/18}], 
[{1, 5/3, 5/4, 10/7}, {5/3, 25/24, 25/21, 25/18}], 
[{1, 5/3, 5/4, 10/7}, {5/3, 25/24, 5/4, 35/24}]}

36/35
{[{1, 7/5, 6/5, 8/5}, {36/35, 6/5, 12/7, 48/35}], 
[{1, 7/5, 6/5, 8/5},{9/5, 36/35, 6/5, 36/25}], 
[{1, 8/7, 10/7, 12/7}, {36/35, 6/5, 12/7, 48/35}],
[{1, 8/7, 10/7, 12/7}, {36/35, 9/7, 12/7, 72/49}], 
[{1, 3/2, 6/5, 12/7}, {9/5,36/35, 6/5, 36/25}], 
[{1, 3/2, 6/5, 12/7}, {36/35, 6/5, 12/7, 48/35}], 
[{1, 3/2, 6/5, 12/7}, {36/35, 9/7, 12/7, 72/49}]}

49/48
{[{1, 3/2, 5/4, 7/4}, {49/48, 7/6, 7/4, 35/24}], 
[{1, 3/2, 5/4, 7/4}, {49/32, 49/48, 7/4, 49/40}], 
[{1, 5/3, 7/6, 4/3}, {49/48, 7/6, 7/4, 35/24}], 
[{1, 5/3, 7/6, 4/3}, {49/30, 49/48, 7/6, 49/36}], 
[{1, 7/5, 7/6, 7/4}, {49/48, 7/6, 7/4, 35/24}], 
[{1, 7/5, 7/6, 7/4}, {49/30, 49/48, 7/6, 49/36}], 
[{1, 7/5,7/6, 7/4}, {49/32, 49/48, 7/4, 49/40}]}

50/49
{[{1, 8/7, 10/7, 12/7}, {25/14, 50/49, 10/7, 25/21}], 
[{1, 8/7, 10/7, 12/7}, {60/49, 50/49, 10/7, 80/49}], 
[{1, 5/3, 5/4, 10/7}, {25/14, 50/49, 10/7, 25/21}], 
[{1, 5/3, 5/4, 10/7}, {60/49, 50/49, 10/7, 80/49}]}


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Message: 10438 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:11:57

Subject: Re: Harmonized melody in the 7-limit

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: >
>> Also, could we screen based on which of the above >> combinations, and which orderings of those, produce the >> most low-numbered ratios in the scale? Or does such >> an approach fail on the grounds that it ignores temperament >> (aka TOLERANCE)? >
> One approach would be to use tempering to simplify the problem. If we > pick linear temperaments which reduce the four sizes of scale step to > two, we also automatically enforce Myhill's property.
That would seem to depend on the ordering, and if the period isn't an octave would seem to be impossible.
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Message: 10439 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 11:37:26

Subject: Re: Hanzos

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> My recollection is that Paul H.'s algorithm assigns a unique > lattice route (and therefore hanzo) to each 9-limit interval.
So what? You still get an infinite number representing each interval, since you can multiply by arbitary powers of the dummy comma 9/3^2.
> Certainly it can be used to find the set of lattice points > within distance <= 2 of a given point.
Hahn's alogorithm can, or hahnzos can, or what?
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Message: 10440 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:18:08

Subject: Re: DE scales with the stepwise harmonization property

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
>> Augmented[9] >> [28/25, 35/32, 15/14, 16/15] [1, 2, 3, 3] >> >> (28/25)/(35/32) = 128/125 >> (15/14)/(16/15) = 225/224 >
> Augmented[9], eh? How far is the 7-limit TOP version > from... > > ! > TOP 5-limit Augmented[9]. > 9 > ! > 93.15 > 306.77 > 399.92 > 493.07 > 706.69 > 799.84 > 892.99 > 1106.61 > 1199.76 > ! > > ...? > > -Carl
Just look at the horagram, Carl! 107.31 292.68 399.99 507.3 692.67 799.98 907.29 1092.66 1199.97
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Message: 10441 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 11:38:34

Subject: Re: Harmonized melody in the 7-limit

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
>>> One of us is still misunderstanding Paul Hahn's 9-limit approach. >>
>> What in the world makes you think this has anything to do with me or >> anything I've said? > > ""
> The 9-limit would be different, for sure. The simple symmetrical > lattice criterion wouldn't work, but it would be easy enough to > find what does. > > If you call something which makes 3 half as large as 5 or 7 > "symmetrical", it does. > ""
Can you point out where in the above quote you found the words "Paul Hahn?"
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Message: 10442 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:19:43

Subject: Re: Harmonized melody in the 7-limit

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:

>> One approach would be to use tempering to simplify the problem. If > we
>> pick linear temperaments which reduce the four sizes of scale step > to
>> two, we also automatically enforce Myhill's property. >
> That would seem to depend on the ordering, and if the period isn't an > octave would seem to be impossible.
It does depend on the ordering, of course. Mostly it seems to work; you can check if it is going to by adding up the multiplicities times the generator steps, and seeing if you get zero. In the case of pajara, a non-octave-period temperament, I showed how the generators add up to 22, so when you get 22 pajara generators, clearly what you've got to do is turn it into 22-equal.
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Message: 10443 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 04:07:13

Subject: Re: Harmonized melody in the 7-limit

From: Carl Lumma

>>>> >ne of us is still misunderstanding Paul Hahn's 9-limit approach. >>>
>>> What in the world makes you think this has anything to do with me or >>> anything I've said? >> >> ""
>> The 9-limit would be different, for sure. The simple symmetrical >> lattice criterion wouldn't work, but it would be easy enough to >> find what does. >> >> If you call something which makes 3 half as large as 5 or 7 >> "symmetrical", it does. >> "" >
>Can you point out where in the above quote you found the words "Paul >Hahn?"
What you said was that symmetrical lattice distance won't work. I asked why, and said Paul Hahn's version works. -Carl
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Message: 10444 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:20:44

Subject: Re: 9-limit stepwise

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> Here are some 9-limit stepwise harmonizable scales, with the same > bound on size of steps--8/7 is the largest. In order to keep the > numbers down, I also enforced that the size of the largest step (in > cents) is less than four times that of the smallest step--the > logarithmic ratio is the fourth number listed. > > In order I give scale type number on the list, scale steps, > multiplicities, largest/smallest, and number of steps in the scale. > > As you can see, the largest scale listed has 41 steps, which is > getting up there. (64/63)/(81/80)=5120/5103 and (49/48)/(50/49) = > 2401/2400; putting these together gives us hemififths, and hence > Hemififths[41] as a DE for this. Hemififths is into the > microtemperament range by most standards; it has octave-generator with > TOP values [1199.700, 351.365] and a mapping of > [<1 1 -5 -1|, <0 2 25 13|]. I've never tried to use it, and so far as > I know neither has anyone else, but this certainly gives a motivation. > Ets for hemififths are 41, 58, 99 and 140. We also get Schismic[29] > out of scale 4, Diaschismic[22] out of scale 5,
I thought Diaschismic had no unique definition in the 7-limit.
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Message: 10445 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 12:15:25

Subject: Re: Harmonized melody in the 7-limit

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> What you said was that symmetrical lattice distance won't work. It doesn't. > I asked why, and said Paul Hahn's version works.
That's a symmetrical lattice, but it isn't a lattice of note-classes. ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * [with cont.] <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service * [with cont.] (Wayb.)
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Message: 10446 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:22:48

Subject: Re: 9-limit stepwise

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote: > > On the other end of the size scale we have these. Paul, have you ever > considered Pajara[6] as a possible melody scale?
Seems awfully improper, but descending it resembles a famous Stravisky theme.
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Message: 10447 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 15:23:55

Subject: Re: DE scales with the stepwise harmonization property

From: Carl Lumma

>> >ugmented[9], eh? How far is the 7-limit TOP version >> from... >> >> ! >> TOP 5-limit Augmented[9]. >> 9 >> ! >> 93.15 >> 306.77 >> 399.92 >> 493.07 >> 706.69 >> 799.84 >> 892.99 >> 1106.61 >> 1199.76 >> ! >> >> ...? >> >> -Carl >
>Just look at the horagram, Carl! > >107.31 >292.68 >399.99 >507.3 >692.67 >799.98 >907.29 >1092.66 >1199.97
Oh! Where are the 7-limit horagrams? -C.
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Message: 10448 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:23:55

Subject: Re: 9-limit stepwise

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote: > > On the other end of the size scale we have these. Paul, have you ever > considered Pajara[6] as a possible melody scale?
I'm confused -- I thought the largest step was supposed to be less than 200 cents?
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Message: 10449 - Contents - Hide Contents

Date: Mon, 01 Mar 2004 23:27:24

Subject: Re: 9-limit stepwise

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote:
>> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> >> wrote: >> >> On the other end of the size scale we have these. Paul, have you > ever
>> considered Pajara[6] as a possible melody scale? >
> Seems awfully improper, but descending it resembles a famous > Stravisky theme.
What I should have asked was if you've tried 443443 as a melody scale in 22-et.
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