Thread: Dodecaphonics
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Old 2006-11-18, 17:05
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My previous post did not make sense. Then let me say it this way (take some time to understand this logic. Try it out seven hundrets of times and then say I am right.):

If you make a string 2/3 of its length, it will become a perfect fifth. This is one of the axioms of music and intervals.
If you make a string of its length, it will become a perfect octave. Also one of the axioms of music and intervals.
So if the root of a string is F, and we shorten the length of the string by 2/3, then it becomes a C.
C => G
G => D
D => A
A => E
E => B
B => F#
F# => C#
C# => G#
G# => D#
D# => A#
A# => E#

If we take a perfect octave from F, we get an F
After seven times taking a perfect octave, we get an F.
This F is the half of the half of the half of the half of the half of the half of the half of the length of the F we were starting at. So that is to the seventh power of a string's length.

On piano, we are at the F button with both the ways of taking intervals.

According to the twelve times we take a perfect fifth from the root, we get the wave-length that is 2/3 to the twelfth power of the root. This is 4096 divided by 531441, which is nearby 0.0077073.
This is less than the wave-length of the seventh octave, which is to the seventh power, which is 0.0078125.
The frequence of the E# is therefore higher than the frequence of the F. So the E# is higher than the F. This is very logic and easily traceable.

If we take even more perfect fifths and octaves from the F (in total these are 84 perfect fifths and 49 octaves), we get an F with twelve sharps and a custom F. We will see that the F with twelve sharps is definitely not thesame as the F:
The F with twelve sharps is 2/3 to the 84th power which is nearby 1.6156 * 10^-15.
The custom F is now 49 octaves higher than the original F, thus to the 49th power, which is nearby 1.7764 * 10^-15.
These wave-lengths are not thesame.

This proof is simple, clear and perfect. So johnmansley, your theorem is down.


The fifths on a piano are therefore not perfect. There has been lot of discussing about this subject ever before they made the piano and even before the harpsichord was found, and they made a compromis so that the error would be minimal. So on piano, indeed, there are twelve.
For a violin or for other strings, and also for brass instruments and percussion, nevertheless this matters, and you are not about to have the opinion that those are no part of western music.

That passage in which I sayd that a C# is 15/16 of the string of the root, that was indeed not true.


In dodecaphonics, most of the time we do not use a C with twelve sharps or flats. We most of the time use only one sharp or flat at one note. But nevertheless, the number of possible dodecaphonic melodies is INFINITE. We just keep it simple (on piano with 12!, so only 479,001,600 possibilities for a dodecaphonic order - for melody is not really the good word for this).

I am a student of math and music, so this was a subject of some masterclass lessons at a university.
Quote:
Originally Posted by johnmansley
forget about butchering music theory
How can I?
Quote:
Originally Posted by johnmansley
just count them

For once more: how can I count an infinite collection, except for making a calculation?
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