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Re: Broken beams' slopes

From: Janek Warchoł
Subject: Re: Broken beams' slopes
Date: Sun, 28 Aug 2011 11:52:45 +0200

W dniu 27 sierpnia 2011 15:51 użytkownik Carl Sorensen
<address@hidden> napisał:
> 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.

Sorry, Carl, but i don't get it at all.  (btw, in which octave is your example?)
Why "c to f beam  would have a slope of 3 ss per eighth note."?  f
notehead is only 1.5 ss higher than c, and beams are usually damped,
so the beam slope in c[ f] is less than 1.5 ss.
Perhaps i didn't explain my suggestion clear enough.  Please take a
look at the attachment - that's how i imagine beam breaking could
- first, imagine an unbroken beam.
- break the beam while retaining the slope.
- adjust them a bit vertically: in the lower octave beam (left side)
the notes before the break have a bit long stems, but they couldn't be
shorter because beam must stop at middle line.  Notes after break have
too short stems - these can be adjusted by moving the beam up about
0.5 ss. On the right side, stems before break are quite ok, and stems
after the break can be shortened a bit by moving the beam up.

I don't see how this could fail or produce bad output - ?


Attachment: broken beams.png
Description: PNG image

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