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Topic: When "regular" quintuplet and so on? (Read 4833 times) previous topic - next topic

When "regular" quintuplet and so on?

I read of various methods for obtaining quintuplet and similar irregular groups.
But WHEN noteworthy composer will have native tools about these musical signs? Perhaps Noteworthy 3.0 ?
I think this software is now mature enough for a complete control of the score, isn't it?




 

Re: When "regular" quintuplet and so on?

Reply #1 –
The request for "n-tuplets" has come up many times.  Most often it is for doublets (n=2) and for quintuplets (n=5).  I'm curious how one would define what these mean:

  • For a triplet, the duration of the note is always multiplied by 2/3 I believe, regardless of time signature.
  • For a doublet in 6/8 time, is the duration of the note always multiplied by 3/2?  Does a doublet make sense in every possible time signature?  Can a doublet note duration always simply be multiplied by 3/2?
  • For a quintuplet in 4/4 time, is the duration of the note always multiplied by 4/5?  What about 6/8 time, where it should perhaps be multiplied by 3/5?

If n-tuplets were to be done, could there be simple rules as to change in duration?  Or would the rules have to depend on the time signature?  Or should the "n:x" notation always be forced, where n notes would occupy the same time as x notes would, and the duration of each would be multiplied by x/n?  Thoughts anyone?

Re: When "regular" quintuplet and so on?

Reply #2 –
I thought about replying, but thought biavianco had already checked out the workarounds such as these, but regretted they were necessary.

Putting 5 eighth notes into the space 3 may take some thought.

P.S. As in the other tuplets, the main problem is what to do with ordinary notes on other staves.  For a dotted quarter, tie it to a quarter on the played staff and add a hidden quarter rest on the displayed staff.  For two dotted eighths, the solution is similar.

For three eighth notes on another staff, the 12 64th notes expand to 20.  Since 20 is not divisible by 3, each note would have to expand to either 7 or 6 64th notes, but that would be an approximation, such as midi triplets.  The solution might be to tripletize those beats and the number of 64th notes would expand to 30 and each regular beat would require 10 64th notes.  The 8th note already takes up 8 of those, so they would be padded with a 32nd (tied note or hidden rest).

P.P.S.  Can't see the forest for the trees.  Find the lowest common multiple of 5 and 3 and expand those three notes into 15 beats (or 30 for the whole measure if necessary in 6/8).
Since 1998

Re: When "regular" quintuplet and so on?

Reply #3 –
Quote from: Randy Williams –
I'm curious how one would define what these mean:
... Thoughts anyone?
Some answers here: http://en.wikipedia.org/wiki/Tuplet
controversy here: http://en.wikipedia.org/wiki/Talk:Tuplet#Sextuplet_interpretation

Just to show how complex this can become:
<Image Link>
Discussion of the above image here: https://ccrma.stanford.edu/software/cmn/cmn/cmn.html#beat-subdivision
Registered user since 1996

Re: When "regular" quintuplet and so on?

Reply #4 –
Quote
Some answers here: http://en.wikipedia.org/wiki/Tuplet

Actually, that's where I got most of my info from above.  :-)

I'm just wondering how the folks who've requested duplets/n-lets over the years wanted them to behave, if they were ever to be implemented native to NWC.  The chances of them getting into NWC are directly proportional to how complicated they'd need to be to satisfy those that have asked for it.  So where are all the folks who've been complaining about this?