Re: Subdivide beams in triplets

[Pierre Perol-Schneider]

Hi Nick,

Here you go:

\version "2.18.2"

%% Solution 1
\relative c''{
  \set subdivideBeams = ##t
  \set baseMoment = #(ly:make-moment 1 8)
  \tweak text #"9"
  \tuplet 3/2 %% or 6/4
  { a16[ bes a bes c bes c d c] }
}

%% Solution 2
\relative c''{
  \tuplet 9/4 {
    a16 bes
    \set stemLeftBeamCount = #2
    \set stemRightBeamCount = #1
    a
    \set stemLeftBeamCount = #1
    \set stemRightBeamCount = #2
    bes c
    \set stemLeftBeamCount = #2
    \set stemRightBeamCount = #1
    bes
    \set stemLeftBeamCount = #1
    \set stemRightBeamCount = #2   
    c d c
  }
}

Cheers,

Pierre

2015-09-22 7:00 GMT+02:00 Mark Stephen Mrotek <address@hidden>:

Nick,

Try changing the 9/4 to 3/2. Does that work?

Mark

-----Original Message-----
From: lilypond-user-bounces+carsonmark=address@hidden
[mailto:lilypond-user-bounces+carsonmark=address@hidden] On Behalf Of
Nick McManus
Sent: Monday, September 21, 2015 9:53 PM
To: address@hidden
Subject: Subdivide beams in triplets

Hi all,

I'm having some trouble achieving subdivided beams in this particular
example:

\version "2.18.2"

\relative c''{
  \set subdivideBeams = ##t
  \set baseMoment = #(ly:make-moment 1 8)
  \tuplet 9/4 {a16 bes a bes c bes c d c} }

I would like to achieve a set of 9 semiquavers (16th notes) with the beam
subdivided every three notes.  How do I go about this?

Thanks,

Nick McManus


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