A fragment from Debussy's Ballade

This piece is in common time (4 4), but I'm having difficulty counting it.

I think the whole lower staff (all the note groups) should have the triplet mark but it doesn't (is this an error of the editor?).

And then for the upper one, I get the first voice on the top adds up very obviously to 4 beats but for the secondary voice, should I assume those are triplets as well? And then why would they not add up to 4 as well?

There are no rests so I'm not sure how that one works out. Thanks!

3 Answers 3


They're all triplets. Once you realise there's continual triplets throughout the bar, it falls into place easily. The engraver has taken a few shortcuts.


Every group of three sixteenth notes in this measure is a triplet. This could have been written much more clearly, but often the triplet marking is not continued once the pattern is established.


It could have been written more clearly. It's 8 lots of triplets, that alternate between hands. So, count 1&a2&a3&a4&a5&a6&a7&a8&a. Sounds complex! But the phrasing puts 1&a2&a as the first phrase, and the others follow suite.

The melody is shown with up stems, there being one concealed, played r.h. but written in bass clef - that's the F (#?) ninth dot from the end of the bar, coming in on the count of 6&a.

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