Re: Separate a tuplet from the rest under a common beam

[Archer Endrich]

Hello,

I often have varied groupings under one beam and have used the solution given by Michael Werner.  However, I have also created the following to simplify entering the commands and the look of the score file:

bcLnone = \set stemLeftBeamCount = #0
bcRone = \set stemRightBeamCount = #1
bcLone = \set stemLeftBeamCount = #1
bcRtwo = \set stemRightBeamCount = #2
bcLtwo = \set stemLeftBeamCount = #2

Such as in:

\times 2/3 {f'16([ b, \bcRone f' ~} \bcLone f16 b,] ~ b16[ \bcRone f' ~ \bcLone f32-- b,16.] f'32[ b, ~ b8 f'16] ~ \times 4/6 {f16 b, \bcRone f' ~ \bcLone f b, f'] ~} | \times 2/3 {f16[ b, \bcRone f' ~} \bcLone f32 b, f' b,] f'32[ b,32 ~ b8]) r16 fis'32([ b, ~ b8 fis'16] ~ \times 4/6 { fis16[ b, \bcRone fis' ~ \bcLone fis b, fis'])} |

Archer

On 14/12/2022 16:24, Michael Werner wrote:

Resending to include the list ... oops.

Something like this, perhaps?

\version "2.23.82"

\relative  {
  \clef treble
  \key c \minor
  \time 2/4
  \tuplet 3/2 { g'16[( aes \set stemRightBeamCount = #1 g) } \set stemLeftBeamCount = #1 fis g] a8 b |
}

On Wed, Dec 14, 2022 at 10:42 AM Volodymyr Prokopyuk <volodymyrprokopyuk@gmail.com> wrote:

Hello,

Problem

How can I separate a \tuplet from two sixteenths using a common eightingth beam?

Example code

\version "2.23.81"

\relative {
  \clef treble
  \key c \minor
  \time 2/4
  \tuplet 3/2 { g'='16[( aes g) } fis g] a8 b |
}

Desired result

I've tried to adjust the baseMoment, beatStructure, subdivideBeats and strictBeatBeaming but without success.

Thank you very much,

Vlad

beamcountexample.ly
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