JICalc

>
> >Is anyone here in charge of JI Calc? I have grown attatched to it, and
> >have just upgraded my computer to a Power Mac, only to find that JI Calc
> >doesn't work! Will there be a Powermac version available soon?
>
> Adam -
>
> The problem is the same with the JI Ear Trainer. Both rely on an external
> command written by K. Romana Machado (http://www.fqa.com/romana/) that
> effected polyphonic microtones by bypassing the Mac Sound Manager.
> Unfortunately, when the hardware changed on the PowerMac, it no longer
> worked. Robert Rich tells me Romana has no plans to write new code for the
> PowerMac. If someone out there would like to take the time to write such
> code, it would be a simple matter to drop it into JICalc and the JI Ear
> Trainer. Any takers?

We've done plenty of Sound Manager code here, but I'm not quite sure
what is needed. The above URL doesn't respond. I'm sure with the old code
it wouldn't be too hard to come up with whatever was needed.

Dave

Dave Madole
Technical Director, Center for Contemporary Music
Listserv Administrator

Mills College
Oakland, CA 94613
510-430-2336

madole@mills.edu
>
> Bill
>
> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> ^ Bill Alves email: alves@hmc.edu ^
> ^ Harvey Mudd College URL: http://www2.hmc.edu/~alves/ ^
> ^ 301 E. Twelfth St. (909)607-4170 (office) ^
> ^ Claremont CA 91711 USA (909)621-8360 (fax) ^
> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>
>
>
>
> ------------------------------
>
> End of TUNING Digest 839
> ************************
>


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Scott_Purman@nile.com (Wed, 18 Sep 96, Subject: duh?) said:

> [C]an someone explain to me what the ratios are ratios of?
> or maybe tell me the ratios of the 12 tones in a
> traditional ET scale (or 13 tones, from C to C for
> instance). I think I could pick it up from
> there....thanks.

In answer to Scott Purnham, and on the assumption that
others new to the list may have similar queries, I offer
here a brief explanation of ratios, JI and 12TET. Apologies
to the great majority of list subscribers for cluttering up
their mailboxes -- you can skip to the next message.


What are they? -- The ratios of the frequencies of two
pitches. 3/2 is a perfect fifth, 5/4 a just major third,
81/64 a Pythagorean major third, 81/80 the syntonic comma
(difference between 5/4 and 81/64). To decipher the ratios,
you must extract the prime factors in the numerator and
denominator: e.g.

5/4 = 2^-2 * 5^1

81/80 = 2^-2 * 3^4 * 5^-1

"*" is multiplication, "^" is an exponent superscript sign
(so 3^4 is three-to-the-power-of-four, i.e. 81). I don't
know just how rusty your maths might be, so I'll remind you
that the negative exponent x^-1 means 1/x, so 2^-2 is 1/4,
while 2^-2 * 5^1 is 5/4. The intervals in JI Ear Trainer are
all smaller than an octave, so the numerator of the ratios
will always be greater than the denominator, but less than
twice the denominator; i.e. the ratios fall between 1/1
(unison) and 2/1 (the octave).

Since you will be familiar with the harmonic series, this
might be the most convenient point of entry. We are not
concerned here with absolute frequency, expressed in Hz, but
only with frequency relations. Define the fundamental as 1,
so the second harmonic, an octave higher has twice the
frequency, thus 2/1; the third harmonic, an octave and a
fifth higher, has three times the frequency, thus 3/1; the
fourth harmonic, two octaves higher is 4/1, i.e. (2^2)/1;
the fifth harmonic, two octaves and a (just) major third
higher is 5/1.

If a perfect fifth plus an octave is 3/1, how do you express
a perfect fifth alone in ratio terms? Just as multiplying
the frequency by 2 raises a pitch by an octave, so division
by 2 lowers a pitch by an octave; thus 3/2 is 3/1 less an
octave -- a perfect fifth above the fundamental. How do you
express a (just) major third in ratio terms? Since 5/1 is a
major third plus two octaves, divide 5 by 4 -- that is, by
2^2 -- to bring it down to a major third above the
fundamental; 5/4 is therefore a major third. But I also
mentioned a Pythagorean major third above; how is this
derived? Multiplying a frequency by 3 is equivalent to
raising a pitch by an octave and a fifth; now, say your
fundamental is C, then 3 times the fundamental is a G, 3^2
or 9 times is a D, 3^3 or 27 times is an A, and 3^4 or 81
times is an E -- this is the 81st harmonic. In order to
bring this E within the octave above the fundamental, we'll
have to descend by six octaves, so we divide 81 by 2^6 or
64, hence the ratio 81/64 for the Pythagorean major third.

How is the 81/64 third related to the 5/4 third? To find
which is higher, express the two ratios in terms of a common
denominator: thus we have 81/64 to compare with 80/64 --
this means that the Pythagorean third is higher. To express
the difference in ratio terms, divide 81/64 by 5/4; this is
equivalent to 81/64 * 4/5, which equals 81/80, i.e. the
syntonic comma I mentioned above. You can now manipulate the
ratios accordingly, without reference to the harmonic
series. Such ratios were in any case discussed by music
theorists long before the harmonic series and the frequency
of sound-waves were understood: the Chinese took the
relative lengths of bamboo pipes as their conceptual
starting-point, Pythagoras the relative weights of hammers,
and thereafter the divisions of a string became the most
common means of conceptualising ratios (the monochord of
mediaeval and Renaissance musicians was such a device -- it
was never intended for performance).

So far I've left "Pythagorean" and "just" undefined, but we
now have the means to rectify this. A Pythagorean interval
is expressible in terms of a ratio that includes only 2 and
3 (and their powers) in the numerator and denominator. A
just interval will have a ratio that also includes 5 (and
its powers) in the numerator and denominator. "Extended just
intonation" includes prime numbers higher than 5. Partch's
terminology is the most lucid for such purposes: Pythagorean
intervals are 3-limit intervals, just intervals are 5-limit,
while Partch used an 11-limit system. Note that each higher
number limit includes the previous limit, so the 3-limit
intervals form a proper subset of the 5-limit set; each
sytem is, theoretically speaking, an infinite set of
intervals (this doesn't conflict with the fact that one can
be a subset of another). A quick test of the knowledge so
far gained: express the major seventh in ratio terms, within
the 5-limit system. Just as we can construct such an
interval by adding a perfect fifth and a major third (in
this case, just), so we multiply 3/2 and 5/4, to obtain
15/8.

Finally, a word on the intervals of12-note equal temperament
(12TET); these are irrational proportions, so we will have
to abandon ratio terminology. The tuning system is
constructed upon that proportion which, when multiplied by
itself 12 times, equals 2. Thus we have to find an interval
above 1/1, the starting pitch, twelve of which will make up
an octave, i.e. 2/1. That interval is the twelfth root of 2
above the 1/1 starting point, the starting frequency
multiplied by 2^(1/12) -- that is, two to the power of a
twelfth, or the twelfth root of two. Thus the equal tempered
semitone is the proportion (2^[1/12])/1. How is the 12TET
perfect fifth expressed in such terms? Since this fifth is
seven 12TET semitones above an initial pitch, we want the
seventh power of the twelfth root of 2, i.e. (2^[1/12])^7 =
2^(7/12). How can this be compared to the 3/2 perfect fifth?
Simply find the decimal value of each; you'll find the 3/2
is very slightly higher. The 12TET system was adopted as a
tuning convenience in the late 18th/early 19th centuries in
Europe because it provided a good approximation of the
common 3-limit intervals, and a tolerable -- or at least
recognizable -- approximation of the 5-limit intervals. An
equivalent method for constructing 12TET is to find the
difference between 12 fifths and 7 octaves (the Pythagorean
comma), and remove the twelfth root of this difference from
each fifth; from this the "circle of fifths" is artificially
derived. Each 12TET fifth will therefore be 3/2 divided by
the twelfth root of the Pythagorean comma; I'll leave you to
prove for yourself that this is equivalent to 2^(7/12).

--
Jonathan Walker
Queen's University Belfast
mailto:kollos@cavehill.dnet.co.uk
http://www.music.qub.ac.uk/~walker/

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