How do I count with metronome for 6/8 measure especially for this music?

enter image description here

Thanks, Nader

  • 1
    I wonder very much about this being labelled 6/8. Unless the tempo is rather fast I'd say this is better served being written as 3/4 with all of the note values doubled. That said, if it really is supposed to be 6/8 then you still just count it like you normally would.
    – Fugu
    Commented Aug 15, 2017 at 19:15
  • 2
    @Fugu 3/4 would lend it a triple feel (three beats per measure) instead of the duplex feel of 6/8. 6/4 could work, but not 3/4! (Note also that doubling the note values puts it in 6/4. It already fits in 3/4.)
    – Richard
    Commented Aug 15, 2017 at 20:29
  • @Richard I think if you have sixteenth note triplets in one voice and a quarter note-eighth note repeating pattern in the other then counting this as compound (double) is weird unless the tempo is fast. Look at the two answers - both suggest counting the subunits of three. That together with my own intuition that you're going to have to count this as three unless the tempo is brisk suggests to me that this is not compound time. 6/4 doesn't play into this at all and doesn't really have anything to do with my suggestion.
    – Fugu
    Commented Aug 15, 2017 at 20:42
  • 1
    @Fugu Of course the asker is not in a position to change the time signature, but even if they were, 6/8 is a march feel and 3/4 is a waltz feel, so they are not equivalent time signatures, regardless of any tuplets present. Commented Aug 15, 2017 at 22:59
  • 4
    Seems like much ado about nothing. It's quite obviously 6/8 with triplets, as @nath answered. Why in the world anyone is trying to make it into something else is beyond me! Commented Aug 16, 2017 at 13:18

3 Answers 3


Wow, that's a good one! Triplets on triplets! It's sort of compound compound time.

6/8 is 1 2 3 4 5 6, taking three counts from each of the two main emphasised beats in each bar. So 1** 2** becomes 1 2 3 4 5 6. but this subdivides again, with some of the counts becoming triplets for themselves.

What I'd do is re-divide each bar into a sort of 12/8 feel. so a gentle 1,2,3,4 count, which would then be sub-divided into triplets. Like 1 2 3 2 2 3 3 2 3 4 2 3 5 2 3 6 2 3 This then puts it all into a fair timing grid.

I think at this point, the metronome would need to be set so each click represented the bold count number, the whole thing could be counted and played more easily. As in each click is the first of each 'sub-triplet'.

  • 2
    To keep it clear, I've always counted 6/8 as one-and-uh two-and-uh because the feel should be a compound march and I want to keep my thinking in a "one two one two" mode. For the triplets, I would probably invent a syllable (end) and count one and-ee-uh end-ee-uh two and-ee-uh end-ee-uh, but instead I think I would rather just count one-and-uh two-and-uh and fit the triplets evenly in the count. Not every rhythmic element should have its own count - that's not what counting is for. Commented Aug 15, 2017 at 21:13

this is a really nice one, I would proceed as follows:

enter image description here

would you mind adding the title and maybe the Op.Nr?


Since it's a 6/8, try to convey the feel of a duplet rhythm.

1) To count a 6/8 is 1 + + 2 + +. Start slow with the metronome with 1 beat equivalent to 1 quaver note; to get the correct rhythm.

2) To count the rhythm of your piece is to replace '+' with e.g. banana or any 3-syllables words.

3) New counting will be: [1] ba-na-na ba-na-na [2] ba-na-na ba-na-na...etc.

  • Not any three syllable word. 'Banana' generally has its emphasis on the middle syllable. As such, it's not particularly useful as the word to use. That needs to be a three syllable word with emphasis on the first syllable. Coconut, for example.
    – Tim
    Commented May 20, 2019 at 6:57

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