Re: How should tupletSpannerDuration actually work?

[Hans Aberg]

On 13 Jan 2013, at 14:13, David Kastrup <address@hidden> wrote:

> Hans Åberg <address@hidden> writes:
> 
>> On 12 Jan 2013, at 12:53, David Kastrup <address@hidden> wrote:
>> 
>>> "Phil Holmes" <address@hidden> writes:
>>> 
>>>>> I have a hard time considering the output of
>>>> 
>>>>> \version "2.16.0"
>>>>> 
>>>>> \relative c' {
>>>>> \set tupletSpannerDuration = #(ly:make-moment 1 2)
>>>>> \times 2/3 { c8 d e f g a g f e d c d }
>>>>> \set tupletSpannerDuration = #(ly:make-moment 1 4)
>>>>> \times 2/3 { c4 d e f g a g f e d c d }
>>>>> }
>>>>> 
>>>>> 
>>>>> useful:
>>>> 
>>>> I think that looks OK?
>>> 
>>> No.  In the first case, the beaming should be 2+2+2 rather than 3+3: the
>>> whole point is that there is supposed to be a subdivision into three,
>>> and 3+3 defeats that.
>> 
>> Hindemith, "Elementary training", p. 116, gives the rule that triplets
>> are tripartite and sextuplet bipartite.
> 
> But the tuplet brace extends the tripartition to six notes, yielding
> three groups of two, contrary to the beaming.

Right. On p. 116a, he gives a example of 3 on six 16th notes. It may be beamed 
with all notes beamed together, that is, "in one", but they can be beamed with 
one long beam and the second beam as three groups of two.

Possibly there is problem if one one writes triple and not as 3:2.

Hans