Re: [Tex-music] trios on 6/8 mesure, how many beams ?
I learned from my young age that a group of notes could be superabundant (italian: sovrabbondanti) or insufficient (it: insufficienti). In your sample the group in the second bar is superabundat (4 instead of 3 quarter note), whilst the group in the thrid bar is insufficient (4 instead of 6 eight note). I can't undertand why one indication should be correct and the other wrong. Luigi At 07.22 05/02/04, you wrote: I have here a discussion with (real) musiciens, who asked me to type some music. I followed the default shape of Don's pmx but they critize the number of beams on trios. They say the 3th mesure on the example (commented with "wrong?" must have a beam. I known that Don (always) is right but i put the question here to collect arguments in the heavy discussion we have. So comments are welcome = 1 1 6 8 6 8 0 0 1 220 .0 t ./ Ap AT %1 b44 b b / %2 b24dx4 D"right?" a g fs / %3 [m1 b24dx4n6 D"wrong?" a g fs ] / %4 b4dx2 b bd / %5 b4x3 b b b2 / %6 b4dx4 b b b bd / % m4848 b24x3 b b / b84 a g fs / b8 b b4 / b4x3 b b b4 / b8x3 b b b4d / b4x4 b b b b / === thanks Andre ___ Tex-music mailing list [EMAIL PROTECTED] http://icking-music-archive.org/mailman/listinfo/tex-music
RE: [Tex-music] trios on 6/8 mesure, how many beams ?
Andre wrote > I known > that Don (always) is right ... I appreciate the thought, but I disagree with it. But if I'm not always right, maybe I'm wrong about this, and maybe therefore I really am always right. And I know a barber who shaves everyone in town except himself. Here's one argument why 4 notes in 3 quarters should have no flags: If you had three notes in three quarters, like the first bar in Andre's example, each one has length 1.0 quarters and 0 flags. If you had 3 notes in 2 quarters, like the 7th bar in Andre's example, each one has length 0.667 quarters and 0 flags. With 4 notes in three quarters, each note has length 0.75 quarters. That's BETWEEN the lengths of notes in two cases where no one (except maybe Andre's friends) would argue that there should be no flags. In case anyone is interested, the formula that PMX uses to compute the number of flags in an xtuplet is INT(13.429+(0.952*LN(nnotes)-LN(32*wholes))/0.69315)-10 where "wholes" is the fractional number of whole notes duration for the entire xtuplet, and nnotes is the number of notes. I derived this formula myself. If anyone wants to know why it doesn't depend directly on wholes/nnotes (the length of each individual note), I'll have to get back to you because I don't remember exactly, but I think I have a record somewhere. --Don ___ Tex-music mailing list [EMAIL PROTECTED] http://icking-music-archive.org/mailman/listinfo/tex-music
RE: [Tex-music] trios on 6/8 mesure, how many beams ?
> > I have here a discussion with (real) musiciens, who asked > > me to type some> > music. I followed the default shape of > > Don's pmx but they criticise the number of beams on trios. > > They say the 3th> > mesure on the example (commented with > > "wrong?" must have a > beam. I known> > that Don (always) > > is right but i put the question here to > collect arguments > > in the heavy discussion we have. > > So comments are welcome > The behaviour of PMX is right: the second measure is correct, > the third one is false. Technically true - the logic is that the printed notes in x-tuplets should sound shorter than the same notes ``normally''. However I have seen both notations in good quality printed music, and in fact feel that the ``wrong'' notation is more common. As long as the result is clear, one would hope that it shouldn't matter too much! Okay, back to lurking now (With gratitude as always to everyone responsible for a system I use frequently, with enjoyment.) Andrew. ___ Tex-music mailing list [EMAIL PROTECTED] http://icking-music-archive.org/mailman/listinfo/tex-music
[Tex-music] trios on 6/8 mesure, how many beams ?
On Thu, 5 Feb 2004 07:22:14 +0100 "Andre Van Ryckeghem" <[EMAIL PROTECTED]> wrote: > I have here a discussion with (real) musiciens, who asked me to type some > music. I followed the default shape of > Don's pmx but they critize the number of beams on trios. They say the 3th > mesure on the example (commented with "wrong?" must have a beam. I known > that Don (always) is right but i put the question here to collect arguments > in the heavy discussion we have. > > So comments are welcome The behaviour of PMX is right: the second measure is correct, the third one is false. In the last measure of your example, the 4 under the beam is redundant and superfluous. Olivier ___ Tex-music mailing list [EMAIL PROTECTED] http://icking-music-archive.org/mailman/listinfo/tex-music
[Tex-music] trios on 6/8 mesure, how many beams ?
I have here a discussion with (real) musiciens, who asked me to type some music. I followed the default shape of Don's pmx but they critize the number of beams on trios. They say the 3th mesure on the example (commented with "wrong?" must have a beam. I known that Don (always) is right but i put the question here to collect arguments in the heavy discussion we have. So comments are welcome = 1 1 6 8 6 8 0 0 1 220 .0 t ./ Ap AT %1 b44 b b / %2 b24dx4 D"right?" a g fs / %3 [m1 b24dx4n6 D"wrong?" a g fs ] / %4 b4dx2 b bd / %5 b4x3 b b b2 / %6 b4dx4 b b b bd / % m4848 b24x3 b b / b84 a g fs / b8 b b4 / b4x3 b b b4 / b8x3 b b b4d / b4x4 b b b b / === thanks Andre ___ Tex-music mailing list [EMAIL PROTECTED] http://icking-music-archive.org/mailman/listinfo/tex-music
