Re: [proposal] easy triplets and tuplets - Draft 3

[Reinhold Kainhofer]

On 2012-10-08 00:21, Joseph Rushton Wakeling wrote:
> On 10/07/2012 11:52 PM, Reinhold Kainhofer wrote:
> > There is, however, no check whether the fraction with the durations makes
> > sense and matches the real tuplet (in most cases, itwill not).
>
> Yes, that's what I mean.  I'd like to see something where the fractions
> and durations are all derived from the tuplet syntax.

Actually, thinking of it, it would actually be quite simple to calculate 
the displayed fraction with durations from the given durations and the 
tuplet fraction (except that there is no way to distinguish 3:2 and 4:6).

(m*dur1):(n*dur2) = tuplet fraction

can be easily solved as
m/n = (tuplet fraction)*dur2/dur1

and then reduced to minimal m and n. This could then be displayed as
  m{dur1}:n{dur2}

Cheers,
Reinhold


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Reinhold Kainhofer, address@hidden, http://www.kainhofer.com
 * Financial & Actuarial Math., Vienna Univ. of Technology, Austria
 * http://www.fam.tuwien.ac.at/, DVR: 0005886
 * Edition Kainhofer, Music Publisher, http://www.edition-kainhofer.com