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From: | Wim van Dommelen |
Subject: | Re: subdividing sixteenth-note triplet beam groups |
Date: | Sat, 20 Oct 2012 18:10:19 +0200 |
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! |
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