On Jul 7, 2005, at 1:49 PM, [EMAIL PROTECTED] wrote:
Putting the mechanics aside for a moment, could someone please explain
what you can do with 12/12 that you CANNOT do using standard meters,
or combinations thereof ?
Not so much 12/12, but say 5/12.
Let's say you were honking along happily in 4/4, mixing eighths,
sixteenths, and eighth-note-triplets freely, as those young kids today
are wont to do. Then suddenly, you just want 3 eighth notes in a bar.
Great, a bar of 3/8 (or 1/Q. ) and there you go. A standard solution
exists that everyone easily understands.
But then later, you are playing some triplets which work out perfectly,
but you ONLY NEED FIVE OF THEM, not six. If you needed 6, then a bar
of 2/4 with triplets marked normally would be great. But if you want a
new downbeat after you've only played FIVE eighth-note-triplets, then
you're out of luck in standard metre systems. Then you would need a bar
indicating 5 (or really 3+2) over whatever eighth-note triplets are in
relation to a quarter note. Hey, we do the math, and you get 12
triplets in a whole, which makes them 1/12th notes.
So you mark 5/12, and put in three eighths beamed together followed by
2 eighths beamed together, and I would put a bracketed 3 tuplet over
the first group, and the same over the second group (even though there
are only TWO notes in it) for clarity.
There must be a good cause to write something that most accomplished
musicians may have difficulty sight reading because of some obscure
meter.
Yes. One would only use it if it clarified the musical gesture. If I
could accomplish it with an ordinary metric modulation instead, I would
do it.
But let's say again, in the same happily honking 4/4, that you are
constantly doing this odd-triplet thing, but at one point actually have
4 pulses worth of triplets. Rather than switch back to 4/4 with tuplets
for one measure, I might be tempted to make that measure 12/12. "Might
be" is the operative word. 12/12 is not really in my vocabulary (12/8
barely is!) and I would do my darndest to find a conventional solution
first.
But that's how it would work.
Christopher
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