I'm currently studying piano music theory and I have a hard time figuring out how much triplets, duplets, quadruplets, quintuplets, sextuplets, and septuplets are worth in different time signatures (simple, compounds hybrid meters) enter image description here If anyone has any tricks when they want to know how many beats/notes an irregular grouping is worth, please feel free to respond!

Thank you!


Just looking at your examples, some thoughts that may help are as follows:

I’m assuming triplets don’t give you too much trouble. They’re pretty straightforward.

For the larger groupings, based on your example page, you could think through the following if you get lost:

  1. For simple (duple) meter, find the largest number of notes in the grouping divisible by 4. Whatever the duration of that grouping is is the duration of your tuplet.

EXAMPLE: For a grouping of 9 sixteenths, the largest number of sixteenths in that grouping divisible by 4 is 8. 8 sixteenths equals 2 quarter notes, so your grouping fits into the space of two quarter notes.

  1. For compound meter, do the above but use multiples of 3 instead of 4.

EXAMPLE: For an grouping of 8 sixteenths, the largest number of sixteenths in that grouping divisible by 3 is 6. 6 sixteenths equals a dotted quarter, so your grouping fits into the space of a dotted quarter.

Hopefully this helps!

  • Triplets always equal one beat no matter if they are eighth notes, sixteenth notes, or quarter notes?
    – iiRosie1
    Dec 8 '19 at 15:26
  • 2
    Triplets are always equivalent to two of whatever note is used. So a triplet of eighths equals two eights = one quarter. A triplet of quarters equals two quarters or a half note. A triplet of sixteenths equals two sixteenths = one eighth.
    – Kevin H
    Dec 8 '19 at 15:32
  • Thank you so much! I understand it much better than the book now!
    – iiRosie1
    Dec 8 '19 at 15:33
  • Glad to help! Those can be tricky.
    – Kevin H
    Dec 8 '19 at 15:34
  • How about irregular groupings during irregular times? (Such as 5/8 or 7/4) Dec 10 '19 at 16:52

Honestly, I don't really "agree" with those crotchet- and minum-beat septuplets. I'd personally add an extra beam to both of those, in part to avoid the issue you're having here where there's an inconsistency between simple and compound metres. It's true that this leaves the pentuplet inconsistency unresolved (and unresolvable, barring something very confusing like quintuplet dotted-quavers), but:

These higher-value tuplets are read contextually, not according to strict rules. If you're in 4/4, and septuplet semiquavers are beamed as a single unit, with three other beats in the bar; well, you know what the composer means.

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