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Message: 6925 - Contents - Hide Contents Date: Thu, 19 Jun 2003 08:36:33 Subject: Fwd: triads behave as i thoughtt From: wallyesterpaulrus a relatively recent post . . . --- In harmonic_entropy@xxxxxxxxxxx.xxxx wally paulrus <wallyesterpaulrus@y...> wrote: even with the primitive voronoi algorithm, Yahoo groups: /harmonic_entropy/files/trivoro.gif * [with cont.] the area associated with each triad a:b:c is, as i suspected, inversely proportional to the geometric mean of the terms, i.e., proportional to 1/cuberoot(a*b*c) -- as long as a*b*c is not too close to the limit used for all the triads in the computation: Yahoo groups: /harmonic_entropy/files/triadic.gif * [with cont.] this is in perfect agreement with the guess i posted in this list some time ago . . . --------------------------------- Do you Yahoo!? Yahoo! Web Hosting - Let the expert host your site --- End forwarded message ---

Message: 6927 - Contents - Hide Contents Date: Fri, 20 Jun 2003 20:35:08 Subject: Re: 5-comma list From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" <paul.hjelmstad@u...> wrote:> Gene, could you repost your nice 5-comma list (The one that shows > everything,comma name, period-generator, rms, complexity badness) I > searched the archives up and down and can't find it. Thanks . . . . . . . . . . . . . . . Is Yahoo groups: /tuning-math/message/5080 * [with cont.]what you want? There's also the following: Yahoo groups: /tuning-math/message/5920 * [with cont.] Yahoo groups: /tuning-math/message/5054 * [with cont.]

Message: 6929 - Contents - Hide Contents Date: Fri, 20 Jun 2003 21:59:38 Subject: Re: 5-comma list From: wallyesterpaulrus --- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" <paul.hjelmstad@u...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote:>> --- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" >>> >>> Gene, could you repost your nice 5-comma list (The one that shows >>> everything,comma name, period-generator, rms, complexity badness) > I>>> searched the archives up and down and can't find it. Thanks >> . >> . >> . >> . >> . >> . >> . >> . >> . >> . >> . >> . >> . >> . >> . >> >> Is Yahoo groups: /tuning-math/message/5080 * [with cont.]>> what you want? There's also the following: >> >> >> Yahoo groups: /tuning-math/message/5920 * [with cont.] >> >> Yahoo groups: /tuning-math/message/5054 * [with cont.]> Thanks. 5920 is best, but there was a "better" one. Luckily its > printed out and at home. Funny it's no longer in the archivewhy do you say it's no longer in the archive???????????????? if you printed it out, surely it has a message number or something on it? are you saying that message is gone? p.s. i put this table of 5-limit commas and associated temperaments together using contributions by gene, graham, dave, and others: Yahoo groups: /tuning/database? * [with cont.] method=reportRows&tbl=10&sortBy=3 (be sure to click on "next" repeatedly to see the whole table) this is meant to accompany the five graphs that you see by mousing- over the various zoom levels near the top of: Definitions of tuning terms: equal temperament... * [with cont.] (Wayb.)

Message: 6930 - Contents - Hide Contents Date: Sun, 22 Jun 2003 05:35:21 Subject: 12 equal to meantone conversion algorithm From: Gene Ward Smith It seems to me that a pretty good start on this could be done algorithmically, and that in any event the result would be interesting and listenable. The method I've been pondering would definately not solve any 128/125 drift problem; one may regard that as a bug or a feature. Given a set of octave reduced note classes, we can represent each element as a unit vector or equivalently, a complex number of absolute value one. Raise these to the seventh power to get a set of complex numbers organized around a circle of fifths, then add all of the complex numbers (sum_i z_i^7); if the result is zero, return zero; otherwise divide by the absolute value to get another number of absolute value unity. This is the 12-equal midpoint. We now take the previous chord's 12-equal midpoint and find which side of it the new midpoint it is closest to (with a tiebreaker if needed.) We add that on to the previous chord's meantone midpoint, which is a real number, not a number mod 1 or 12 or 2 pi or 360 (however you are measuring a circle) to get the meantone midpoint of the new chord. If the new chord has a 12-equal midpoint of zero, we add zero and simply use the old midpoint. The first chord has a meantone midpoint equal to its 12-equal midpoint, allowing us to start the recursion (if the first chord is an augmented triad or diminished seventh, we probably need to go to the first unambiguous chord and work backwards and forwards.) We now take a range of +-6 around the meantone midpoint, and use it to decide how to map the 12 note classes to meantone for this chord, using the previous midpoint as a tiebreaker if needed. I'm actually thinking of trying to code this in C, despite the fact that I've written only in Maple for the last 15 years, but of course if I could con someone into doing it for me that would be even better. I think it would be a wonderful addition to Scala and Ada would be just fine as a language to write it in; Python also springs to mind as a good language. :)

Message: 6931 - Contents - Hide Contents Date: Sun, 22 Jun 2003 12:59:35 Subject: Re: 12 equal to meantone conversion algorithm From: Carl Lumma Pretty simple, compared to this: * [with cont.] (Wayb.) -Carl

Message: 6932 - Contents - Hide Contents Date: Mon, 23 Jun 2003 00:47:35 Subject: Re: 12 equal to meantone conversion algorithm From: wallyesterpaulrus how does this compare with dave keenan's ideas: An algorithm for dynamically choosing the best... * [with cont.] (Wayb.) ? --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> It seems to me that a pretty good start on this could be done > algorithmically, and that in any event the result would be > interesting and listenable. The method I've been pondering would > definately not solve any 128/125 drift problem; one may regard that > as a bug or a feature. > > Given a set of octave reduced note classes, we can represent each > element as a unit vector or equivalently, a complex number of > absolute value one. Raise these to the seventh power to get a set of > complex numbers organized around a circle of fifths, then add all of > the complex numbers (sum_i z_i^7); if the result is zero, return > zero; otherwise divide by the absolute value to get another number of > absolute value unity. This is the 12-equal midpoint. > > We now take the previous chord's 12-equal midpoint and find which > side of it the new midpoint it is closest to (with a tiebreaker if > needed.) We add that on to the previous chord's meantone midpoint, > which is a real number, not a number mod 1 or 12 or 2 pi or 360 > (however you are measuring a circle) to get the meantone midpoint of > the new chord. If the new chord has a 12-equal midpoint of zero, we > add zero and simply use the old midpoint. The first chord has a > meantone midpoint equal to its 12-equal midpoint, allowing us to > start the recursion (if the first chord is an augmented triad or > diminished seventh, we probably need to go to the first unambiguous > chord and work backwards and forwards.) > > We now take a range of +-6 around the meantone midpoint, and use it > to decide how to map the 12 note classes to meantone for this chord, > using the previous midpoint as a tiebreaker if needed. > > I'm actually thinking of trying to code this in C, despite the fact > that I've written only in Maple for the last 15 years, but of course > if I could con someone into doing it for me that would be even > better. I think it would be a wonderful addition to Scala and Ada > would be just fine as a language to write it in; Python also springs > to mind as a good language. :)

Message: 6933 - Contents - Hide Contents Date: Mon, 23 Jun 2003 03:33:46 Subject: Re: 12 equal to meantone conversion algorithm From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "wallyesterpaulrus" <wallyesterpaulrus@y...> wrote:> how does this compare with dave keenan's ideas: > > An algorithm for dynamically choosing the best... * [with cont.] (Wayb.)Thanks for pointing this out. They are similar in both using a sliding window. Despite the fact that I use complex numbers, my algorithm is actually simpler and more straightforward. If we did the ascending chromatic scale example, using my method we simply get ascending diatonic semitones; hence we do not arrive back in the key of C, but we don't experience any irregularities.

Message: 6934 - Contents - Hide Contents Date: Mon, 23 Jun 2003 04:08:59 Subject: Re: 12 equal to meantone conversion algorithm From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> Thanks for pointing this out. They are similar in both using a > sliding window. Despite the fact that I use complex numbers, my > algorithm is actually simpler and more straightforward. If we did the > ascending chromatic scale example, using my method we simply get > ascending diatonic semitones; hence we do not arrive back in the key > of C, but we don't experience any irregularities.I should point out that this is only true if the chromatic scale is not harmonized. My algorithm, while very simple, does not work on the crude level of note-for-note, but is based on note-sets. If the chromatic passage was harmonized in a way which leads back to C, for instance by C-A-D-cm-C-F-D-G-fm-F-C7-G7-C, back to C we would come.

Message: 6935 - Contents - Hide Contents Date: Mon, 23 Jun 2003 06:13:12 Subject: Complex number meantone conversion argument function From: Gene Ward Smith In case anyone else wants to mess with this, here is Maple code for the "argument", -6 < arg <= 6, of a set of 12-equal note-classes. This is the basic function which gets the job done, and as you can see, it's not complicated. arg := proc (s) # meantone conversion argument function # "s" is set or list of octave-reduced notes local i, u, w; if nops(s) = 0 then RETURN(0) fi; u := 0; w := (3^(1/2) + I)/2; for i to nops(s) do u := u+w^(7*s[i]) od; u := simplify(u); if u = 0 then RETURN(0) fi; u:=evalf(6*Im(ln(u))/Pi); if abs(u - round(u)) < 0.00001 then u := round(u) fi; u end:

Message: 6936 - Contents - Hide Contents Date: Mon, 23 Jun 2003 08:15:40 Subject: Re: Complex number meantone conversion argument function From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> In case anyone else wants to mess with this, here is Maple code for > the "argument", -6 < arg <= 6, of a set of 12-equal note-classes. This > is the basic function which gets the job done, and as you can see, > it's not complicated.There's no need for messing with raising to the 7th power if we start with the right root of unity, and a symmetric chord should not return zero, which confuses it with C; so this returns "sym": arg := proc (s) # meantone conversion argument function # "s" is set or list of octave-reduced notes local i, u, w; if nops(s) = 0 then RETURN(sym) fi; u := 0; w := -(3^(1/2) + I)/2; for i to nops(s) do u := u+w^s[i] od; u := simplify(u); if u = 0 then RETURN(sym) fi; u:=evalf(6*Im(ln(u))/Pi); if abs(u - round(u)) < 0.00001 then u := round(u) fi; u end:

Message: 6937 - Contents - Hide Contents Date: Mon, 23 Jun 2003 10:27:09 Subject: Re: 12 equal to meantone conversion algorithm From: Graham Breed Gene Ward Smith wrote:> I should point out that this is only true if the chromatic scale is > not harmonized. My algorithm, while very simple, does not work on the > crude level of note-for-note, but is based on note-sets. If the > chromatic passage was harmonized in a way which leads back to C, for > instance by C-A-D-cm-C-F-D-G-fm-F-C7-G7-C, back to C we would come.I also have some code for 12-equal to meantone conversions, and I managed to write it without using complex numbers! It's at http://www.microtonal.co.uk/gesualdo.zip - Type Ok * [with cont.] (Wayb.) I don't know how out of date that is, but I do have a more recent version on my Revo. One difference is that I have found passages in Gesualdo that it doesn't convert correctly. There's a total of 3 wrong chords. From what I remember, these can be resolved by using a different gamut depending on whether he used a Bb in the key signature. The gamut restrictions could possible be removed by using a flexible key center. Take everything relative to the average number of steps on the spiral of fifths, calculated recursively as total(n) = k*total(n-1) + note number(n) = k*number(n-1) + 1 center(n) = total(n)/number(n) Then, k=0 means it calculates each new chord looking at only the previous chord, and k=1 makes it remember all previous chords equally, which will give very conservative results. I didn't implement this for the Gesualdo program, because it isn't a rule he was likely to follow, but it might work for an automated tuning program. Graham

Message: 6938 - Contents - Hide Contents Date: Mon, 23 Jun 2003 10:52:34 Subject: Re: 12 equal to meantone conversion algorithm From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> I don't know how out of date that is, but I do have a more recent > version on my Revo. One difference is that I have found passages in > Gesualdo that it doesn't convert correctly. There's a total of 3 wrong > chords. From what I remember, these can be resolved by using a > different gamut depending on whether he used a Bb in the key signature.You think you have problems. I just tried my algorithm on Schumann's Foreign Lands and Peoples. The fact that it didn't end up in the same key it started with wasn't as disturbing as the obvious clinkers. Of course, I imagine Renaissance music is easier to covert to meantone!

Message: 6939 - Contents - Hide Contents Date: Mon, 23 Jun 2003 13:49:59 Subject: Re: 12 equal to meantone conversion algorithm From: Manuel Op de Coul Have a look at this too: ÖFAI technical reports * [with cont.] (Wayb.) Manuel

Message: 6940 - Contents - Hide Contents Date: Tue, 24 Jun 2003 05:58:24 Subject: Re: 12 equal to meantone conversion algorithm From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> You think you have problems. I just tried my algorithm on Schumann's > Foreign Lands and Peoples. The fact that it didn't end up in the same > key it started with wasn't as disturbing as the obvious clinkers.I screwed up; the Schumann in fact presents zero difficulties for conversion to meantone so far as I can determine. You can find a 31-equal version and the midi I retuned here: . . . Yahoo groups: /tuning_files/files/31 equal/for... * [with cont.] Yahoo groups: /tuning_files/files/originals/sr... * [with cont.]

Message: 6941 - Contents - Hide Contents Date: Tue, 24 Jun 2003 06:07:59 Subject: Re: Interval Database Experiences From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Porres" <decuritiba@y...> wrote:> I just checked an explanation about continued fractions and it's > amazing, I see you're using convergents and semi-convergents on your > Excell table, it seemed at first kinda complicated, but I guess I'll > dig it... > > but hey, what does it have to do with that pdf file ( Self Similar > Pitch Structures - Clampitt ) ??? I downloaded that once and couldn't > figure it out to, let me dig the traditional harmony first, I'm still > studying Bach's counterpoint technique.Sorry to take so long to reply. It is the same mathematics put to a different purpose. In Clampitt's paper it is applied in the logarithmic pitch domain, whereas you're using it in the linear frequency domain. Clampitt is finding ratios representing fractions of an octave (or other interval of periodicity) (i.e. degrees of equal temperaments) where you are finding ratios representing frequency ratios (i.e. justly intoned intervals when the numbers are small). It's nice that the same mathematical tool has these two different applications to tuning. You might say Clampitt is applying it to melodic properties while you are applying it to harmonic ones.> ha ha ha, good job on your notation research, specially by keeping up > the good humor, since you're involved in Scala, would you know of a > complete table of name intervals? I guess that would be useful...I'm not really "involved in Scala", but the file 'intnam.par' (and its equivalents in other languages) that comes with Scala, is very useful. With a little work it can be imported into a spreadsheet and sorted by interval size or whatever. -- Dave Keenan

Message: 6943 - Contents - Hide Contents Date: Wed, 25 Jun 2003 20:20:46 Subject: (unknown) From: wallyesterpaulrus --- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" <paul.hjelmstad@u...> wrote:> I've been studying the "taxicab" diagrams huh?? > with the comma/temperment > names. I can convince myself as to why Augmented, Diminished and > Aristoxean fall along straight lines.oh, you must mean the big equal temperament graph with the linear temperament lines . . . is the meaning of the axes on that graph clear to you?> Could someone explain why the > meantone temperments, for example, fall along a straight line. > (I see why knowing 2 out of 3 of perfect fifth, minor third and major > third determine the third one, pretty obvious,)Could someone give me > a general mathematical formula/description that explains why all the > comma/temperments fall along straigt lines? Thanksthe comma vanishing implies a certain linear relation between any two of the consonances. let's say perfect fifth and major third, why don't we? then for any meantone, and only for meantones, 4*(perfect fifth) - 2400 cents = major third. is it clear to you that this is the equation of a straight line in perfect fifth - major third space?

Message: 6945 - Contents - Hide Contents Date: Wed, 25 Jun 2003 21:02:35 Subject: (unknown) From: wallyesterpaulrus --- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" <paul.hjelmstad@u...> wrote:> Thanks, that's kind of what I thought but I am still amazed by it.are you also amazed that the slope of each temperament-line is identical to the slope of the corresponding comma-vector on the hexagonal "small 5-limit intervals" graph below? (i presume we're both looking at The Proxomitron Reveals... * [with cont.] (Wayb.) arts.org/dict/eqtemp.htm)

Message: 6946 - Contents - Hide Contents Date: Wed, 25 Jun 2003 22:27:05 Subject: (unknown) From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" <paul.hjelmstad@u...> wrote:>> 4*(perfect fifth) - 2400 cents = major third. >> >> is it clear to you that this is the equation of a straight line in >> perfect fifth - major third space? Yes> Thanks, that's kind of what I thought but I am still amazed by it. Yahoo groups: /tuning-math/files/dualzoomk.gif * [with cont.] Yahoo groups: /tuning-math/files/dualzoomn.gif * [with cont.]If you want to get really boggled check out the dual zoomers above, and the rest of them in the files.

Message: 6948 - Contents - Hide Contents Date: Sun, 29 Jun 2003 10:14:29 Subject: The Cawapu Comma From: Gene Ward Smith The de facto pitch bend range is +-200 cents, leading to a de facto pitch resolution of a cawapu, which is 1/4096 of a cent (as opposed to the midipu, which does not seem important in practice but which would give you 1/4 cawapu resolution in case you thought you needed it.) This lead me to wonder what tuning properties, in theory, the 12 * 4096 = 49152 et would have. It turns out to have a 7-limit comma which stands out from the pack: (4375/4374)/(250047/250000) = (4375/4374)^2/(2401/2400) = 78125000/78121827. I think a good name for this comma would the the cawapu comma. There's another noteworthy comma, this one actually less than a cawapu in size, for the 11-limit version of the 49152 et. This one is (2401/2400)^3/(5632/5625) = 7^24 / 2^24 3 5^2 11.

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