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Message: 10500 - Contents - Hide Contents Date: Wed, 03 Mar 2004 21:47:12 Subject: Re: Hanzos 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 "Paul Erlich" <perlich@a...> > wrote: >>>>> Why would anyone ever use it on non-lattice points? >>>> Constructing scales, for starters. >> Which scales involve non-lattice points?Scales which have centers in holes--the octahedra and tetahedra--instead of lattice points, for starters.

Message: 10501 - Contents - Hide Contents Date: Wed, 03 Mar 2004 21:57:31 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>>> Hardly arbitrary if you accept the above 48 transformations as >>> interesting, since they are orthogonal. >>>> I don't get it. How does this list of transformations depend on your >> metric? >> It defines a metric.That seems impossible to me. Please convince me.>>> I'm pointing out that Euclidean gives us more information. >>>> But all kinds of other shapes, besides a sphere, would also imply >> these same distinct shells. >> They imply distict shells, but not the *same* distinct shells.But what's special about the shells you get from the Euclidean metric as opposed to the shells you get from using, instead of a sphere, some other rounded shape with all the symmetries of the rhombic dodecahedron?>>>>> Is it in >>>>> fact so obvious that 49/48<-->50/49 is a very different >> propostion>>>>> than 15/14<-->21/20, and if so, why? >>>>>>>> It's immediately evident from looking at the lattice. >>>>>> It's immediately obvious from looking at the *Euclidean* lattice. >>> Aren't you proving my point for me? >>>> Absolutely not. For instance, for 50:49 or any other "para" dyad with >> Hahn distance 2, there's only one note consonant with both pitches in >> the dyad. So it's straightforward to take the "lattice" (or whatever >> you want to call it) and transform 50:49 to any dyad which has this >> property (such as 9:8, 25:18, 25:16, etc.) while it would take some >> severe gymnastics to transform it to one that doesn't have this >> property, >> Which is clear from the Euclidean lattice, which says that 9/8, 25/18, > 25/16, 50/49 are all located at a distance of 2 from the unison, and > (this is harder; you can use the invariant stuff I mentioned as one > method) that they are all in the same geometric relationship to the > lattice. >It is hardly clear using Hahn distance,I repeat: Hahn distance is not intended to give configuration information anyway. But configuration, not metrics, is what I was referring to above.> which says all of the > above are at a distance of 2, but also says 10/9, 16/15, 25/24, 36/35 > etc for which the Euclidean distance is sqrt(3) are at a distance of > 2, and even 15/14, 21/20 etc for which the Euclidean distance is > sqrt(2) are at a distance of 2. The Hahn distance is making far less > refined distictions, and is not providing the help in sorting these > questions out that the Euclidean distance immediately gives.We are going around in circles with this conversation.>>>>> It's tough if I am going to be >>>>> dumped on half the time for obscurity, and the other half for >>>>> condescendingly explaining the blindingly obvious. >>>>>>>> It's possible to explain the blindingly obvious without being >>>> condescending. Don't take it personally, but do try to improve. >>>>>> First I need to learn what is blindingly obvious and what isn't. >>>> No you don't -- just explain it without being condescending, and even >> if people already know it, they won't be offended. >> Your claim is that I was being condescending. On what basis are you > making it? What, exactly, was condescending about my remark?I don't care to dwell on this any longer, but for the record, here it was: "Yes, of course they are meaningful. If you want to try to linearly transform 50/49 to 49/48, be my guest, but don't be surprised . . ."

Message: 10502 - Contents - Hide Contents Date: Wed, 03 Mar 2004 00:00:15 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:>> ***[-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c] >> >> I now get a positive semidefinite quadratic form by substituting > this>> in Q; if I normalize by clearing denominators, I get >> >> Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc >> **Afraid I am stuck again. How is this fed back into Q? Sorry to be > dense, I've stared at this too long...It's the result of feeding *** into Q.

Message: 10504 - Contents - Hide Contents Date: Wed, 03 Mar 2004 01:01:18 Subject: Re: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:> You created the Hanzos thread, you tell me what it's about.You told me for whatever reason that I didn't understand distinct 3 and 9 generators, so I gave how I thought about it.

Message: 10505 - Contents - Hide Contents Date: Wed, 03 Mar 2004 03:33:27 Subject: Stepwise scales From: Gene Ward Smith If we put together the tetrads joining 1 to 15/14, 16/15, 21/20, 25/24, 36/35, 49/48 or 50/49 stepwise, we get 30 tetrads. If we add to these the complements under inversion (which in terms of the lattice coordinates is x-->[-1,-1,-1]-x) we get 50 tetrads. These tetrads all have the property that at least one of 1, 15/14, 14/15, 16/15, 15/16...is a chord element. Their notes comprise 55 notes to the octave, giving the following scale: [1, 50/49, 49/48, 36/35, 25/24, 21/20, 16/15, 15/14, 35/32, 10/9, 28/25, 9/8, 8/7, 7/6, 25/21, 6/5, 60/49, 49/40, 5/4, 32/25, 9/7, 64/49, 21/16, 4/3, 49/36, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 72/49, 3/2, 32/21, 49/32, 14/9, 25/16, 8/5, 80/49, 49/30, 5/3, 42/25, 12/7, 7/4, 16/9, 25/14, 9/5, 64/35, 28/15, 15/8, 40/21, 48/25, 35/18, 96/49, 49/25] If we look at the intervals this gives us, the three smallest are 2401/2400, 225/224, and 1029/1024, making it a natural candidate for miracle tempering. Tempering by miracle gives a 43 note scale, as follows: [-26, -21, -20, -19, -18, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7,-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,16, 18, 19, 20, 21, 26] ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ 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.)

Message: 10506 - Contents - Hide Contents Date: Thu, 04 Mar 2004 06:31:19 Subject: Re: Hanzos 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 "Paul Erlich" <perlich@a...> > wrote: >>>>>> Hardly arbitrary if you accept the above 48 transformations as >>>> interesting, since they are orthogonal. >>>>>> I don't get it. How does this list of transformations depend on > your >>> metric? >>>> It defines a metric. >> That seems impossible to me. Please convince me.It's easiest to see in fcc coordinates, where the group consists of sign changes and permutations of [x, y, z]. What homogenous polynomial of degree two in x, y and z is invariant under the operations of the group? It is easy to check that x^2+y^2+z^2 fits the bill, and not difficult to show it is the *only* degree two polynomial, up to a constant factor, which does. That, however tells us that Euclidean distance from the unison is preserved by the operations of the group, or in other words, the group is a group of orthogonal transformations. The geometry comes right out of the group. One way to look at it is to put the origin at the center of a hexany instead, and what you are looking at are rotations and reflections--Euclidean self-congruences--of an octahedron.> But what's special about the shells you get from the Euclidean metric > as opposed to the shells you get from using, instead of a sphere, > some other rounded shape with all the symmetries of the rhombic > dodecahedron?Obviously the sphere is far easier to manage, just for starters. You might try to get a polynomial of even degree in x, y, z which is not a power of x^2+y^2+z^2 but is invariant under the group operations and which has the same value on the twelve points [+-1, +1, 0] and its permutations, and use this to define a norm. There are independent invariants of degree four and six; the simplest possibility would be to use [(x^2+y^2+z^2)^2 + K (x+y+z)(-x+y+z)(x-y+z)(x+y-z)]^(1/4) as the norm, where K is a constant. What values of K make this a norm? Hell if I know. :)> I repeat: Hahn distance is not intended to give configuration > information anyway. But configuration, not metrics, is what I was > referring to above.Then you should *love* Euclidean metrics, which have angles and all the rest of it, and are ideal for analyzing configurations.> I don't care to dwell on this any longer, but for the record, here it > was: > > "Yes, of course they are meaningful. If you want to > try to linearly transform 50/49 to 49/48, be my guest, but don't be > surprised . . ."That is offensive mostly because you choose to be offended by it; and I would guess you choose to be offended by it because of ongoing--perhaps chronic--irritation with me. If you are upset because Dave left, he gave his reasons, and I wasn't in them, by the way. I was irritated when Carl seemed to suggest I didn't know beans about symmetric lattices, but he meant nothing offensive by what he said, so my reaction was really my problem. ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ 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.)

Message: 10507 - Contents - Hide Contents Date: Thu, 04 Mar 2004 16:34:17 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>>> Scales which have centers in holes--the octahedra and >>> tetahedra--instead of lattice points, for starters. >>>> OK. But why would a *sphere* be the most useful delimiting shape for >> constructing such scales? >> I didn't claim it would--in fact, you were the one expressing > skepticism about the value of the Hahn metric for non-lattice points.I retract that skepticism. But all your "shell" posts used the euclidean metric, and this question was aimed at that.> I feel it wouldn't, since the resulting>> scales can actually have 'concavities' of a mild sort, indicating >> suboptimality in terms of consonances-per-note. >> Concavities? You are claiming it is possible for the convex hull to > contain lattice points not inside the sphere?No -- that's why I put 'concavities' in quotes. But the smallest polyhedron large enough to contain all the edges in the scale will have concavities in general if you use a sphere (i.e., euclidean metric), which is indicative of suboptimality in terms of consonances- per-note.> In any case, with spheres the shells are less populous, so you get > more scale possibilities--for good or ill, I don't know.I think, beyond a certain shell, you begin to miss out on the most desirable scale possibilities entirely: those where the smallest polyhedron large enough to contain all the edges in the scale is convex.

Message: 10508 - Contents - Hide Contents Date: Thu, 04 Mar 2004 16:39:55 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- 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 "Paul Erlich" <perlich@a...> >> wrote: >>>>>>>> Hardly arbitrary if you accept the above 48 transformations as >>>>> interesting, since they are orthogonal. >>>>>>>> I don't get it. How does this list of transformations depend on >> your >>>> metric? >>>>>> It defines a metric. >>>> That seems impossible to me. Please convince me. >> It's easiest to see in fcc coordinates, where the group consists of > sign changes and permutations of [x, y, z]. What homogenous polynomial > of degree two in x, y and z is invariant under the operations of the > group? It is easy to check that x^2+y^2+z^2 fits the bill, and not > difficult to show it is the *only* degree two polynomial, up to a > constant factor, which does.But not all metrics are degree two polynomials.> That, however tells us that Euclidean > distance from the unison is preserved by the operations of the group, > or in other words, the group is a group of orthogonal transformations. > The geometry comes right out of the group. One way to look at it is to > put the origin at the center of a hexany instead, and what you are > looking at are rotations and reflections--Euclidean > self-congruences--of an octahedron.Convince me that Hahn's metric doesn't yield the exact same list of 48 orthogonal transformations.>>> But what's special about the shells you get from the Euclidean metric >> as opposed to the shells you get from using, instead of a sphere, >> some other rounded shape with all the symmetries of the rhombic >> dodecahedron? >> Obviously the sphere is far easier to manage, just for starters. You > might try to get a polynomial of even degree in x, y, z which is not a > power of x^2+y^2+z^2 but is invariant under the group operations and > which has the same value on the twelve points [+-1, +1, 0] and its > permutations, and use this to define a norm. There are independent > invariants of degree four and six; the simplest possibility would be > to use > > [(x^2+y^2+z^2)^2 + K (x+y+z)(-x+y+z)(x-y+z)(x+y-z)]^(1/4) > > as the norm, where K is a constant. What values of K make this a norm? > Hell if I know. :)But you seem to be retracting your claim about that the list of orthogonal transformations defines a metric -- yes?

Message: 10509 - Contents - Hide Contents Date: Thu, 04 Mar 2004 18:04:26 Subject: Re: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> No -- that's why I put 'concavities' in quotes. But the smallest > polyhedron large enough to contain all the edges in the scale will > have concavities in general if you use a sphere (i.e., euclidean > metric), which is indicative of suboptimality in terms of consonances- > per-note.Do you have an example of this?

Message: 10510 - Contents - Hide Contents Date: Thu, 04 Mar 2004 18:17:32 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>> No -- that's why I put 'concavities' in quotes. But the smallest >> polyhedron large enough to contain all the edges in the scale will >> have concavities in general if you use a sphere (i.e., euclidean >> metric), which is indicative of suboptimality in terms of consonances- >> per-note. >> Do you have an example of this?I'm sure you've posted some already. Basically, if the convex hull is not a rhombic dodecahedron but has the same symmetry as one, you're suboptimal in terms of consonances-per-note -- you can improve the ratio by either adding or taking away notes until you do have a rhombic dodecahedron.

Message: 10511 - Contents - Hide Contents Date: Thu, 04 Mar 2004 18:21:39 Subject: Re: Hanzos From: Paul Erlich --- 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 "Paul Erlich" <perlich@a...> > wrote: >>>>> No -- that's why I put 'concavities' in quotes. But the smallest >>> polyhedron large enough to contain all the edges in the scale > will>>> have concavities in general if you use a sphere (i.e., euclidean >>> metric), which is indicative of suboptimality in terms of > consonances- >>> per-note. >>>> Do you have an example of this? >> I'm sure you've posted some already. Basically, if the convex hull is > not a rhombic dodecahedron but has the same symmetry as one, you're > suboptimal in terms of consonances-per-note -- you can improve the > ratio by either adding or taking away notes until you do have a > rhombic dodecahedron.Or cuboctahedron (there may be other possibilities). If you can create a depiction of the scales you obtained from spherical shells, you'll see the 'mild concavities' I'm talking about. This would be a lot easier to depict in ASCII if we were talking about 5-limit instead of 7-limit . . .

Message: 10512 - Contents - Hide Contents Date: Thu, 04 Mar 2004 18:52:45 Subject: Re: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> But not all metrics are degree two polynomials.The point is that not all finite groups of matricies gives you such a thing, and that in this case we are dealing with a group which has the orthogonality property, which means the group reprsentation itself has a Eucildean character.> Convince me that Hahn's metric doesn't yield the exact same list of > 48 orthogonal transformations.When I talked about 49/48->50/49, you became upset with me, and now you ask me this! Let's see--a Hahn isometry would require that you send 3,5,7 to three independent 7-limit consonances, which gives you 12 choose 3 = 220 possibilities, clearly a lot more than 48. If we have something in Hahn shell n, does it necessarily stay in Hahn shell n? Either way, we are counting the number consonance steps it takes to get to the note-class, so it seems it should. I suppose the thing to do is to work out an example, but it certainly isn't clear how Hahn distinquishes the 48 orthogonal transformations among the rest.> But you seem to be retracting your claim about that the list of > orthogonal transformations defines a metric -- yes?No, it clearly defines a metric--singling out the Euclidean metric because of the uniqueness of the degree two invariant; you seem to take that to mean a unique metric. I have invariants of degrees four and six to play with, and any homogenous polynomial I cook up from them which leads to convex unit balls (from setting the polynomial equal to one) will define a metric which is designed to have the specific property that the 48 transformations are isometries of this metric.

Message: 10513 - Contents - Hide Contents Date: Thu, 04 Mar 2004 19:05:11 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>> But not all metrics are degree two polynomials. >> The point is that not all finite groups of matricies gives you such a > thing, and that in this case we are dealing with a group which has the > orthogonality property, which means the group reprsentation itself has > a Eucildean character. >>> Convince me that Hahn's metric doesn't yield the exact same list of >> 48 orthogonal transformations. >> When I talked about 49/48->50/49, you became upset with me, and now > you ask me this! Let's see--a Hahn isometry would require that you > send 3,5,7 to three independent 7-limit consonances, which gives you > 12 choose 3 = 220 possibilities, clearly a lot more than 48.Gene, you are wrong. A Hahn isometry would require that you send 3,5,7 to three independent 7-limit consonances *which are consonant with one another*! Only 48 possibilities.> but it certainly isn't clear how Hahn > distinquishes the 48 orthogonal transformations among the rest.As long as we're following the definition of orthogonal transformation that you gave, you're transforming the entire lattice, so this seems perfectly clear to me.>> But you seem to be retracting your claim about that the list of >> orthogonal transformations defines a metric -- yes? >> No, it clearly defines a metric--singling out the Euclidean metric > because of the uniqueness of the degree two invariant; you seem to > take that to mean a unique metric.What else could "defines a metric" be taken to mean, if not "yields a unique choice of metric"?

Message: 10514 - Contents - Hide Contents Date: Thu, 04 Mar 2004 19:12:47 Subject: Re: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> Gene, you are wrong. A Hahn isometry would require that you send > 3,5,7 to three independent 7-limit consonances *which are consonant > with one another*! Only 48 possibilities.Yeah, that occurred to me after posting.> What else could "defines a metric" be taken to mean, if not "yields a > unique choice of metric"?I don't think it's possible for a finite group to do that; what *is* possible is that you get a unique positive-definite quadratic form out of it.

Message: 10515 - Contents - Hide Contents Date: Thu, 04 Mar 2004 19:19:52 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>> Gene, you are wrong. A Hahn isometry would require that you send >> 3,5,7 to three independent 7-limit consonances *which are consonant >> with one another*! Only 48 possibilities. >> Yeah, that occurred to me after posting. >>> What else could "defines a metric" be taken to mean, ifnot "yields a>> unique choice of metric"? >> I don't think it's possible for a finite group to do that;So you shouldn't say it "defines a metric", unless I'm missing something.

Message: 10516 - Contents - Hide Contents Date: Thu, 04 Mar 2004 21:48:23 Subject: Hahn norm formula From: Gene Ward Smith For a note class represented by 3^a 5^b 7^c, I get that it should be ||(a,b,c)||_Hahn = max(|a|,|b|,|c|,|b+c|,|a+c|,|a+b|,|a+b+c|) This is a pretty simple formula, and applicable to non-lattice points. I'll take a look at scales which arise from it.

Message: 10517 - Contents - Hide Contents Date: Thu, 04 Mar 2004 22:47:32 Subject: Re: Hahn norm formula From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> For a note class represented by 3^a 5^b 7^c, I get that it should be > > ||(a,b,c)||_Hahn = max(|a|,|b|,|c|,|b+c|,|a+c|,|a+b|,|a+b+c|) > > This is a pretty simple formula, and applicable to non-lattice points. > I'll take a look at scales which arise from it.The number of notes in Hahn shell n turns out to be 10n^2+2; about this the handbook of integer seqences says "Points on surface of cuboctahedron (or icosahedron): a(0) = 1, for n> 0, a(n) = 10n^2 + 2 (coordination sequence for f.c.c. lattice)." So we seem to be doing things right here. If we take the union of the Hahn shells, we get lattice-point-centered Hahn scales. About the integer sequence for this the handbook says "Centered icosahedral (or cuboctahedral) numbers, also crystal ball sequence for f.c.c. lattice. Comments: Called 'magic numbers' in some chemical contexts." The latter stikes me as nifty. Crystal ball scales? Magic number scales? I like the first especially. Here are the first few Hahn shells: Shell 0 [1] Shell 1 [8/7, 7/6, 6/5, 5/4, 4/3, 7/5, 10/7, 3/2, 8/5, 5/3, 12/7, 7/4] Shell 2 [50/49, 49/48, 36/35, 25/24, 21/20, 16/15, 15/14, 35/32, 10/9, 28/25, 9/8, 25/21, 60/49, 49/40, 32/25, 9/7, 64/49, 21/16, 49/36, 48/35, 25/18, 36/25,35/24, 72/49, 32/21, 49/32, 14/9, 25/16, 80/49, 49/30, 42/25, 16/9, 25/14, 9/5, 64/35, 28/15, 15/8, 40/21, 48/25, 35/18, 96/49, 49/25] Shell 3 [126/125, 64/63, 128/125, 28/27, 256/245, 360/343, 343/320, 27/25, 160/147, 49/45, 192/175, 54/49, 125/112, 384/343, 245/216, 343/300, 147/128, 144/125, 125/108, 400/343, 75/64, 288/245, 147/125, 32/27, 343/288, 175/144, 128/105, 216/175, 56/45, 432/343, 63/50, 80/63, 125/98, 245/192, 35/27, 125/96, 98/75, 75/56, 343/256, 168/125, 27/20, 200/147, 175/128, 343/250, 480/343, 45/32, 64/45, 343/240, 500/343, 256/175, 147/100, 40/27, 125/84, 512/343, 112/75, 75/49, 192/125, 54/35, 384/245, 196/125, 63/40, 100/63, 343/216, 45/28, 175/108, 105/64, 288/175, 576/343, 27/16, 250/147, 245/144, 128/75, 343/200, 216/125, 125/72, 256/147, 600/343, 432/245, 343/192, 224/125, 49/27, 175/96, 90/49, 147/80, 50/27, 640/343, 343/180, 245/128, 27/14, 125/64, 63/32, 125/63] Here are some crystal ball scales: Ball 1 13 notes [1, 8/7, 7/6, 6/5, 5/4, 4/3, 7/5, 10/7, 3/2, 8/5, 5/3, 12/7, 7/4] Ball 2 55 notes [1, 50/49, 49/48, 36/35, 25/24, 21/20, 16/15, 15/14, 35/32, 10/9, 28/25,9/8, 8/7, 7/6, 25/21, 6/5, 60/49, 49/40, 5/4, 32/25, 9/7, 64/49, 21/16, 4/3, 49/36, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 72/49, 3/2, 32/21, 49/32, 14/9,25/16, 8/5, 80/49, 49/30, 5/3, 42/25, 12/7, 7/4, 16/9, 25/14, 9/5, 64/35, 28/15, 15/8, 40/21, 48/25, 35/18, 96/49, 49/25] Ball 3 147 notes [1, 126/125, 64/63, 50/49, 49/48, 128/125, 36/35, 28/27, 25/24, 256/245,360/343, 21/20, 16/15, 15/14, 343/320, 27/25, 160/147, 49/45, 35/32, 192/175,54/49, 10/9, 125/112, 384/343, 28/25, 9/8, 245/216, 8/7, 343/300, 147/128, 144/125, 125/108, 400/343, 7/6, 75/64, 288/245, 147/125, 32/27, 25/21, 343/288, 6/5, 175/144, 128/105, 60/49, 49/40, 216/175, 56/45, 5/4, 432/343, 63/50, 80/63, 125/98, 245/192, 32/25, 9/7, 35/27, 125/96, 64/49, 98/75, 21/16, 4/3, 75/56,343/256, 168/125, 27/20, 200/147, 49/36, 175/128, 48/35, 343/250, 25/18, 480/343, 7/5, 45/32, 64/45, 10/7, 343/240, 36/25, 500/343, 35/24, 256/175, 72/49, 147/100, 40/27, 125/84, 512/343, 112/75, 3/2, 32/21, 75/49, 49/32, 192/125, 54/35, 14/9, 25/16, 384/245, 196/125, 63/40, 100/63, 343/216, 8/5, 45/28, 175/108, 80/49, 49/30, 105/64, 288/175, 5/3, 576/343, 42/25, 27/16, 250/147, 245/144, 128/75, 12/7, 343/200, 216/125, 125/72, 256/147, 600/343, 7/4, 432/245, 16/9, 25/14, 343/192, 224/125, 9/5, 49/27, 175/96, 64/35, 90/49, 147/80, 50/27, 640/343, 28/15, 15/8, 40/21, 343/180, 245/128, 48/25, 27/14, 35/18, 125/64,96/49, 49/25, 63/32, 125/63]

Message: 10518 - Contents - Hide Contents Date: Thu, 04 Mar 2004 23:33:09 Subject: Re: Hahn norm formula From: Gene Ward Smith Here are shells around a deep hole, the octadron around the hole is the first one listed. I then take the union to get deep hole centered scales. The number of inhabitants of shell n is 6n^2, and the total number in a ball is n(n+1)(2n+1); this is six times the number of stacked canon balls in a square pyramid--ie, 1^2+2^2+3^2+ ... + n^2 = n(n+1)(2n+1)/6. Shell 1 (hexany) [1, 21/20, 6/5, 7/5, 3/2, 7/4] Shell 2 [49/48, 36/35, 15/14, 35/32, 28/25, 9/8, 8/7, 63/50, 9/7, 4/3, 48/35, 10/7, 36/25, 35/24, 147/100, 49/32, 63/40, 49/30, 5/3, 147/80, 28/15, 15/8, 48/25, 49/25] Shell 3 [50/49, 1029/1000, 49/45, 192/175, 54/49, 441/400, 10/9, 343/300, 144/125, 75/64, 288/245, 189/160, 25/21, 343/288, 175/144, 216/175, 56/45, 245/192, 1029/800, 64/49, 98/75, 75/56, 343/256, 175/128, 343/250, 441/320, 25/18, 45/32, 112/75, 189/125, 32/21, 75/49, 192/125, 384/245, 196/125, 45/28, 1029/640, 80/49, 288/175, 27/16, 245/144, 216/125, 432/245, 441/250, 16/9, 343/192, 224/125, 175/96, 90/49, 189/100, 40/21, 343/180, 245/128, 27/14] Ball 1 [1, 21/20, 6/5, 7/5, 3/2, 7/4] Ball 2 30 notes [1, 49/48, 36/35, 21/20, 15/14, 35/32, 28/25, 9/8, 8/7, 6/5, 63/50, 9/7,4/3, 48/35, 7/5, 10/7, 36/25, 35/24, 147/100, 3/2, 49/32, 63/40, 49/30, 5/3, 7/4, 147/80, 28/15, 15/8, 48/25, 49/25] Ball 3 84 notes [1, 50/49, 49/48, 36/35, 1029/1000, 21/20, 15/14, 49/45, 35/32, 192/175,54/49, 441/400, 10/9, 28/25, 9/8, 8/7, 343/300, 144/125, 75/64, 288/245, 189/160, 25/21, 343/288, 6/5, 175/144, 216/175, 56/45, 63/50, 245/192, 9/7, 1029/800, 64/49, 98/75, 4/3, 75/56, 343/256, 175/128, 48/35, 343/250, 441/320, 25/18, 7/5, 45/32, 10/7, 36/25, 35/24, 147/100, 112/75, 3/2, 189/125, 32/21, 75/49, 49/32, 192/125, 384/245, 196/125, 63/40, 45/28, 1029/640, 80/49, 49/30, 288/175, 5/3, 27/16, 245/144, 216/125, 7/4, 432/245, 441/250, 16/9, 343/192, 224/125, 175/96, 90/49, 147/80, 28/15, 15/8, 189/100, 40/21, 343/180, 245/128,48/25, 27/14, 49/25]

Message: 10519 - Contents - Hide Contents Date: Thu, 04 Mar 2004 01:34:56 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- 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 "Paul Erlich" <perlich@a...> >> wrote: >>>>>>> Why would anyone ever use it on non-lattice points? >>>>>> Constructing scales, for starters. >>>> Which scales involve non-lattice points? >> Scales which have centers in holes--the octahedra and > tetahedra--instead of lattice points, for starters.OK. But why would a *sphere* be the most useful delimiting shape for constructing such scales? I feel it wouldn't, since the resulting scales can actually have 'concavities' of a mild sort, indicating suboptimality in terms of consonances-per-note.

Message: 10520 - Contents - Hide Contents Date: Thu, 04 Mar 2004 05:47:12 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:> Message 9767, you have 1/13, 9/13 as the squares of the projected > generators. I can see where the denominator comes from, but where > do 1 and 9 come from?From a mistake somewhere.>> Q81 is semidefinite, since Q81(81/80)=0. We can get a positive >> definite form, and hence a lattice, by going to two dimensions with >> basis 3/2 and 9/7: >> >> Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2 >> Once again, I don't see how (3/2) and (9/7) replace 3 and 5Do you see that you can get every 7-limit interval of the 81/80-planar system from 2, 3/2 and 9/7? We have 5~(3/2)^4, and 7 ~ 4 (3/2)^2 (7/9)

Message: 10521 - Contents - Hide Contents Date: Thu, 04 Mar 2004 05:52:44 Subject: Re: Hanzos From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:>> Scales which have centers in holes--the octahedra and >> tetahedra--instead of lattice points, for starters. >> OK. But why would a *sphere* be the most useful delimiting shape for > constructing such scales?I didn't claim it would--in fact, you were the one expressing skepticism about the value of the Hahn metric for non-lattice points. I feel it wouldn't, since the resulting> scales can actually have 'concavities' of a mild sort, indicating > suboptimality in terms of consonances-per-note.Concavities? You are claiming it is possible for the convex hull to contain lattice points not inside the sphere? I think for any norm, including Euclidean and Hahn, that is ruled out by the triangle inequality. In any case, with spheres the shells are less populous, so you get more scale possibilities--for good or ill, I don't know. ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ 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.)

Message: 10522 - Contents - Hide Contents Date: Fri, 05 Mar 2004 19:29:26 Subject: Re: Hahn norm formula 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: >>> For a note class represented by 3^a 5^b 7^c, I get that it should be >> >> ||(a,b,c)||_Hahn = max(|a|,|b|,|c|,|b+c|,|a+c|,|a+b|,|a+b+c|) >> >> This is a pretty simple formula, and applicable to non-lattice points. >> I'll take a look at scales which arise from it. >> The number of notes in Hahn shell n turns out to be 10n^2+2; about > this the handbook of integer seqences says "Points on surface of > cuboctahedron (or icosahedron): a(0) = 1, for n> 0, a(n) = 10n^2 + 2 > (coordination sequence for f.c.c. lattice)." So we seem to be doing > things right here. If we take the union of the Hahn shells, we get > lattice-point-centered Hahn scales. About the integer sequence for > this the handbook says "Centered icosahedral (or cuboctahedral) > numbers, also crystal ball sequence for f.c.c. lattice. > Comments: Called 'magic numbers' in some chemical contexts." > > The latter stikes me as nifty. Crystal ball scales? Magic number > scales? I like the first especially. > > > Here are the first few Hahn shells: > > Shell 0 > [1] > > Shell 1 > [8/7, 7/6, 6/5, 5/4, 4/3, 7/5, 10/7, 3/2, 8/5, 5/3, 12/7, 7/4]Union of shell 0 and shell 1 is known as the 7-limit Tonality Diamond. Thanks for your great work here, Gene.

Message: 10523 - Contents - Hide Contents Date: Fri, 05 Mar 2004 20:43:00 Subject: Re: Hahn norm formula From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> Union of shell 0 and shell 1 is known as the 7-limit Tonality Diamond. > > Thanks for your great work here, Gene.Thanks! Not only has Crystal Ball One been seen often before, Crystal Ball Two has already appeared--it is exactly the 55 note scale which I posted in message 9922, in connection with stepwise harmonizing scales. Despite its apparent theoretical interest, Crystal Ball Two should suffice as a name, as I at least don't believe in scales with steps of size 2401/2400. Tempering it out leads to 51 notes, which is how many it has in hemiwuerschmidt and ennealimmal also.

Message: 10524 - Contents - Hide Contents Date: Fri, 05 Mar 2004 20:52:42 Subject: Re: Hahn norm formula From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: >>> Union of shell 0 and shell 1 is known as the 7-limit Tonality Diamond. >> >> Thanks for your great work here, Gene. >> Thanks! Not only has Crystal Ball One been seen often before, Crystal > Ball Two has already appeared--it is exactly the 55 note scale which I > posted in message 9922, in connection with stepwise harmonizing scales. > Despite its apparent theoretical interest, Crystal Ball Two should > suffice as a name,Paul Hahn might have called it the 7-limit radius 2 scale or perhaps more likely, the Level 2 7-limit Diamond. Now what do you get if you center around a major tetrad (shallow hole)?

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