Re: subdividing sixteenth-note triplet beam groups

2012-10-20 Thread 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 modificat

Re: subdividing sixteenth-note triplet beam groups

2012-10-20 Thread Steve Yegge
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 wrote: > And why not very simply: > > \times 2/3 { c16[ c c c c c] } > > Regards, > Wim. > > On 20 Oct 2012, at 00:

Re: subdividing sixteenth-note triplet beam groups

2012-10-20 Thread Wim van Dommelen
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 connect

Re: subdividing sixteenth-note triplet beam groups

2012-10-19 Thread Nick Payne
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

subdividing sixteenth-note triplet beam groups

2012-10-19 Thread Steve Yegge
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. I tried #set subdivideBeam