Re: Broken beams' slopes

[Carl Sorensen]

On 8/27/11 7:44 AM, "David Kastrup" <address@hidden> wrote:

> Carl Sorensen <address@hidden> writes:
> 
>> On 8/27/11 7:21 AM, "David Kastrup" <address@hidden> wrote:
>> 
>>> Janek Warchoł <address@hidden> writes:
>>> 
>>>> I wonder if this solution would yield good results: keep beam slope
>>>> before and after break identical (except for some beam quanting,
>>>> perhaps, but that's less than 0.3 ss), but modify stem lengths: make
>>>> them as long as they would be if there were no beam on the other side
>>>> of the break.
>>> 
>>> I would expect this to yield mostly reasonably results.  I'd also keep
>>> beam orientation.  But it might make sense to dole out a bit of spring
>>> force (just decidedly less than infinite) for making the vertical beam
>>> positions at the break match.
>> 
>> 
>> It would seem that this algorithm would fail for  a simple broken beam
>> 
>> a8[ b \break c f]
> 
> Care to elaborate?

The a to b beam would have a slope of 1 ss per eighth note.

The c to f beam  would have a slope of 3 ss per eighth note.

the a to f beam would have a slope of 5 ss per  4 eighth notes, or 1.2 ss
per eighth note.

If you choose the slope of 1.2 for both sides, then it seems to me that the
b stem will be longer than it would be without the beam on the other side of
the break, and the c stem would be longer than it would be without the beam
on the other side of the break.  If you force the b and c stems to be the
same length, the a and f beams would be too short.

Thanks,

Carl