Re: Broken beams' slopes

[David Kastrup]

Carl Sorensen <address@hidden> writes:

> 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.

per 3 eighth, or 1.67 ss per eighth.

> If you choose the slope of 1.2 for both sides,

1.67 ss

> 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,

Yes.

> and the c stem would be longer than it would be without the beam on
> the other side of the break.

Yes.

> If you force the b and c stems to be the same length, the a and f
> beams would be too short.

Why would one force them to be the same length?  If we have an
equalizing infinite force applying, it would force the b stem to be
1.67ss shorter than the c stem.  Of course, the usual beam scoring would
apply in order to make sure that no stem gets too short.

I don't get your point.

-- 
David Kastrup