Re: automatic subdivision of beams

[Carl Sorensen]

On 1/5/12 7:07 AM, "Xavier Scheuer" <address@hidden> wrote:

>On 5 January 2012 12:38, Stefan Thomas <address@hidden>
>wrote:
>> Dear community,
>> is it possible to get automatic subdivision of beams, e.g. for
>> 32seconds in a 4/4 bar?
>
>It is not currently possible.  It would be really nice, though.
>In 'scm/time-signature-settings.scm' one can find
>
>  ;;;   (subdivide . grouping-rules)  (not yet implemented, reserved
>for future use)
>
>I asked Carl but he "doesn't have a schedule for automatic subdivision
>on anything other than the beat."


As I've studied more about this, I don't think it can even be done in the
general case and still be consistent with conventional music notation.

The rule of subdivision according to Ross is that the beams carrying over
the subdivision should match the value of the beat that is being
subdivided.

So subdividing 5/16 as (2 . 3), how many beams should we have between the
group of 2 and the group of 3?  The group of 2 should expect 1 beam
(because the group is 1/8), but what about the group of 3.

Now if we subdivide 5/16 as (3 . 2), how many beams between the groups?
What is the justification for having a single beam extend beyond the group
of 3?  This group is an 8. group, and there is no 8. beaming.

If we follow the rules expressed by Ross, we should only subdivide on
beats, and the number of beams between subdivisions should represent the
length of the beat.

I'd welcome any discussion on how this ought to behave differently, but
unless I get a good vision of the right way to do this, I won't be doing
so.

Thanks,

Carl