Quintuplet, a correct one?

[OCTO]

I have completely got blank in my head.

Is this the correct one?

ScreenShot 2026-06-15 at 20.06.47.png (17.29 KiB) Viewed 3382 times

[David Ward]

I think that according to the strictest traditional practice it ought to be five quavers (eighths) in the time of three, on the principle that a tuplet is for more in the time of fewer. Thus I was taught long ago; but is that still considered best? Or indeed, was what I was taught correct? Now, I too feel uncertain…

For clarity you might keep what you have and specify 5:6.

[John Ruggero]

I would go with the five 1/8 notes. That way the notes appear to be slower than those in the first measure, not faster.

[OCTO]

I would go with the five 1/8 notes. That way the notes appear to be slower than those in the first measure, not faster.

That is a valid statement indeed.

Hmm, I was thinking about that "rule" regarding tuplets in simple VS compound meter.

If I remember correctly, in simple meter, tuplets are counted as increasing toward the regular division. In compound meter, tuplets are counted as decreasing toward the regular division.

In my case, one voice contains regular sixteens, while two other voices contain a quintuplet and a quadruplet.

The first is 5:6 and another is 2:3+2:3=4

ScreenShot 2026-06-16 at 01.33.40.png (25.17 KiB) Viewed 3328 times

Interestingly, I think that this is not of a concern:

ScreenShot 2026-06-16 at 01.42.51.png (7.23 KiB) Viewed 3328 times

[LSL2026]

Yeah, here’s my takeaway:

The way tuplets first evolved in compound meter was by “borrowing” the division from the parallel simple meter since that is what simple meters already did.

Taking 2/4 and 6/8 as an example, a beat borrowed from 6/8 was a triplet in 2/4.

So sensibly, a beat from 2/4 borrowed into 6/8 was a duplet.

But yes, this messed up the scheme making it go “the wrong way”.

Simple meters keep the same value/beam until you reach the next subdivision.

4, 5, 6, and 7 all carry 2 beams, and 8+ carries 3 beams until you get to 16 notes per beat.

But 6/8 already used a single beam for BEFORE (under) 3. So it “should have been” a quarter note duplet, but instead it’s an 8th note duplet.

Meaning the 4 per beat would be like 16ths in simple meters, so the beam “immediately” switches after each division.

So 2 and 3 per beat are 1 beam, then 4 5 and 6 are 2, and then 7+ are 3, up to 12.

Gould addresses this, and it seems the preferred solution is to use a ratio with note value instead.

That said, it’s pretty clear that this group of 5 notes is one measure!!!

Still, I think the ratios would be better here, because there’s a mixture of measure long tuplets, and beat long tuplets.

Otherwise, assuming 6/16, if 6 notes per measure are 2 beams, then anything up to and including that immediately after 3 notes is going to be two beams too: meaning the quadruplet and quintuplet “should have” 2 beams too.

After all the quadruplet is actually 2 16th note tuplets, right

[David Ward]

FWIW this is how I notated a quintuplet and quasi quadruplets in 6/8 in a violin piece written in 1992 (which I'm now tinkering with in Dorico).

[Magnus Johansson]

Hey! I agree with David and John and would notate it like in the following example. It is extra important in the score, I think.

Screenshot_20260616_105022.png (6.85 KiB) Viewed 3294 times

[MalteM]

@OCTO IMHO this is definitely incorrect: The quartuplet in the first part and the quartuplet in the third part don’t match. One of them must be a quartuplet.

[Magnus Johansson]

@OCTO IMHO this is definitely incorrect: The quartuplet in the first part and the quartuplet in the third part don’t match. One of them must be a quartuplet.

Yes, the 32nds should be 16ths instead:

Screenshot_20260616_122340.png (2.65 KiB) Viewed 3277 times

[OCTO]

My brain will explode.

I need a mathematician, preferably a Nobel-prize recipient.

@OCTO IMHO this is definitely incorrect: The quartuplet in the first part and the quartuplet in the third part don’t match. One of them must be a quartuplet.

Yes, the 32nds should be 16ths instead:

Yes, I see it.  And I understand it.

But in the lower quadruplet, we have four notes in the time of two notes, not three, and each of those notes is dotted.

In the upper quadruplet, on the other hand, we have four notes in the time of three notes, not two.

I might be very mistaken but I can't get it.

Here. In Sibelius it is not allowed to create any 5:6. So here is "workaround" (i.e. without a plugin). That gives us 5 :

Screenshot 2026-06-16 at 15.19.42.png (34.5 KiB) Viewed 3256 times