Re: subdividing sixteenth-note triplet beam groups

[Wim van Dommelen]

Hi Steve,

Thanks for pointing out, I was indeed too quick in replying, looks "logical" and simpler, but didn't test that specific line of code, stupid. I went back to where I did this once upon a time and saw there is also another possibility I used. However that also needs some modifications:

\version "2.16.0"

\relative c' {

    \times 2/3 { c16 c \set stemRightBeamCount = #1 c }

    \times 2/3 { \set stemLeftBeamCount = #1 c c c }

}

With the beam-count setting you remove the connecting beam, is slightly easier to remember when you use also other types of beam connections (or parts of it). It is discussed in the NR, search for paragraph "Manual Beams".

Regards,

Wim.

On 20 Oct 2012, at 17:36 , Steve Yegge wrote:

Nick's works -- thanks Nick!

Wim, your solution doesn't work; not sure why.  It would certainly be simpler if it did.

-steve

On Sat, Oct 20, 2012 at 5:55 AM, Wim van Dommelen <address@hidden> wrote:

And why not very simply:

    \times 2/3 { c16[ c c c c c] }

Regards,

Wim.

On 20 Oct 2012, at 00:15 , Nick Payne wrote:

On 20/10/12 08:58, Steve Yegge wrote:

I would like this:

  \times 2/3 { c16 c c c c c }

to render with two beamed groups of three notes each, with the two groups connected by a single beam, like so:

c c c  c c c

| | |  | | |

=====  =====

============

I have tuplet numbers and brackets set to transparent.

\relative c' {
    \set tupletSpannerDuration = #(ly:make-moment 1 4)
    \set baseMoment = #(ly:make-moment 1 8)
    \set subdivideBeams = ##t
    \override TupletNumber #'stencil = ##f
    \times 2/3 { \repeat unfold 24 { c16 } } |
}

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