Three triplet sixteen notes are worth one eighth note. The first F is worth two triplet sixteenth notes and the D after it is worth one triplet sixteenth note. Add them together and you get three triplet sixteenth notes which lasts one eighth note. It is the same for the next group of triplets notes. The last two notes are just normal sixteenth notes.My explanation might be unclear, so here is a visual representation of the notes and the counting which may or may not help:C A G F D C F F C1 + + 2 + + 3 + + 4 +Sorry if anything was unclear.
Just curious Are these obscure rhythm problems for a theory class or in a book of rhythm exercises? It is quite strange to see four in quick succession.
Three triplet eighth notes are worth one quarter note. The first D-flat is worth two triplet eighth notes and the last three notes (not including the grace note but including the second tied Db) are worth one triplet eighth note. That's three triplet eighth notes in total which is one quarter note.Db Db A G1 + + You need to make sure that the last three notes are worth one triplet eighth note and that the last G is twice as long as the note preceding it.
Okay I think I understand. So would it be correct to say that a triplet eighth note can essentially be divided up into 4 equal pieces/durations. Each piece being a triplet 32nd note. So the second Db and A each last for two triplet 32nds and the G lasts for a total of two triplet 32nds? Equivalently, the G lasts for 1 triplet sixteenth note. The second Db starts exactly on what would be the 3rd note of an eighth note triplet grouping.I understand a standard eighth note triplet grouping which is just 3 notes played over 1 quarter note beat unit. So when referring to a sixteenth note triplet grouping I believe that it is just each note of an eighth note triplet grouping divided into 2 pieces?When referring to a 32nd note triplet grouping I believe that it is just each note of an eighth note triplet group divided into 4 pieces?Is all of the above correct?