I was playing some Rachmaninoff the other day, and I've found more rhythm fun !
The problem is on Op. 16 n°4, the Presto Moment Musical.
Here is the sheet : http://imslp.info/files/imglnks/usimg/5 ... ical_4.pdf
The question : Is the first b on the right hand simultaneous with the last note of the bar on the left hand ?
At first sight, it is not (sextuplets / triolets on the right hand are mentionned later in the piece), but if you look further on the sheet, the same melodic pattern b then e + g is repeated, this time being together with the left hand (see bar 2 versus bar 11 for instance). Are they played differently ? This is even worst than the Liszt case in my opinion. I'm not even talking of bar 13 where the last sight counts differently depending on the voice.
For a simple illustration on how that all doesn't make sense, just count bar 2 on the right hand :
Half rest (2/4) + Quarter rest (1/4) + Eighth rest (1/8 ) + semiquaver b (1/16) ; where is the missing 1/16 ? It does NOT make sense (it would if the eighth rest was dotted, but it is not).
Let's count bar 3 :
Dotted x 3 half note (2/4 + 1/4 + 1/8 + 1/16) + Semiquaver b (1/16) = 4/4 ; so the sixteenth b is played a little before the last left hand b.
And bar 5 :
Dotted x 2 quarter note (1/4 + 1/8 + 1/16) + Sixteenth note (1/16) + Quarter note (1/4) + Triolet (1/4) = 4/4 ; so the sixteenth f+a is played a little before the last note of the left hand sextuplet.
The main point of conflict I guess is bar 9 to 10
: the final b in this bar is both in the right hand accompagniement voice and in the melodic voice ; though there is only one written note, can there be two different notes ?
Let's count bar 9 :
Right hand upper voice : 4 x sextuplets (4 x 6 x 1/24 = 4/4)
Left hand : 4 x sextuplets (4 x 6 x 1/24 = 4/4)
The problematic right hand lower voice : the last "b" is CLEARLY part of the sextuplet, so it should be worth 1/24 (sextuplet sixteenth). But if you've followed me until here, you can see the same b in earlier bars is worth 1/16 (normal sixteenth). So, am I to deduce that while it is written in a sextuplet, it should also be played as a normal sixteenth ?
The fun bar 13 :
Upper accompagniement voice :
Quarter rest (1/4) + 2 sextuplets (2 x 6 x 1/24) + half a sextuplet (3 x 1/24) + eighth rest (1/8 ) = 1/4 + 2/4 + 1/8 + 1/8 = 4/4
Lower melodic voice :
Dotted x 2 half note (1/2 + 1/4 + 1/8 ) + semiquaver (1/16) = missing a 1/16.
Now, I could do a magic trick to make sense out of it :
If the rest is to be counter in the melodic voice (1/8 ), you would have 1/8+1/16 when adding the last semiquaver. But you could then say both the rest and the semiquaver are IN FACT part of the sextuplet preceding it : that would make 1/12 + 1/24 = 1/8, which would give the bar 4/4. But then - our last melodic note is part of a sextuplet, so it lasts 1/24 and not 1/16 like the other final melodic notes in the first bars. Plus you are counting a rest for two voices (maybe it's normal, that's still weird.), and even worse, counting that same rest differently depending on the voice, once 1/8 and once 1/12.
I've been playing this piece for a long time, always done it *almost* simultaneously in all instances with some rubato to mark the accent. Am I doing it wrong ? I think Rachmaninoff often uses this "almost together" bass and melody (the G minor Etude-tableau n°8 has some lovely passages like that), so that's why I played it like I did, but this bar 9 story is puzzling. The rest seemed obvious until then - because I assume, and recordings seem to confirm this, that it is played the same both in bar 2 and bar 9 ; that is to say b right hand then b left hand, then e+g right hand, though I can't really hear the b on the left hand because of my poor ear and recordings' quality.