 The Clarinet BBoard
|
Author: PeterinToronto ★2017
Date: 2016-01-17 10:31
Hi again,
I was teaching this yesterday, and noticed something perhaps for the first time (or I'd forgotten it): in mvt.2 there is a tree of rhythmic equations at the start, 8th=8th, 16th=16th, 3 16ths=8th. And then as the music gets going Stravinsky states 16th=16th in a few places where there are groups of 2 and 3 16ths side-by-side.
I've always played this as the 16ths stay constant, so that you don't slow down to play the groups of 2 notes. But looking at this again, wouldn't this conflict with the first set of equations? If 8th=8th, and three 16ths=an 8th?
Or I am missing something (or many somethings!)?
Thanks a lot,
Peter
|
|
Reply To Message
|
|
Author: maxopf
Date: 2016-01-17 20:41
Attachment: Stravinsky1.png (62k)
Attachment: Stravinsky2.png (75k)
I have another question along the same lines. If 8th = 3 16ths but 16th=16th, would the attached images be subdivided properly? (The first image is the figure in the piece, the second image is my subdivision. The second grace note is supposed to be a 16th note, my bad.)
In other words, because 8th = 3 16ths but 16th=16th, does 8th + two 16ths = 5 16ths instead of 4?
Post Edited (2016-01-17 20:46)
|
|
Reply To Message
|
|
Author: PeterinToronto ★2017
Date: 2016-01-17 21:25
Thanks for the confirmation of my intuition! Max I think that Stravinsky means to treat the groups of 3 16ths as triplets that fit within the 8th pulse. So the spot you highlight can be counted easily in 8ths, with 2 8ths to the quarter (the grace-note before the beat) and the next bar all to an 8th count.
Hope this helps,
Peter
|
|
Reply To Message
|
|
Author: maxopf
Date: 2016-01-17 21:54
Attachment: Stravinsky1.png (64k)
That's what I thought initially. I highlighted the figure that's giving me trouble. The 8th = three 16ths, and the 16th rest + the 16th note = two 16ths. So the figure adds up to five 16th notes, which means it can't just be counted in 8th notes, correct?
Post Edited (2016-01-17 21:59)
|
|
Reply To Message
|
|
Author: Liquorice
Date: 2016-01-18 03:20
maxopf:
The C-flat is an appoggiatura (not a 16th). So you're highlighted figure adds up to four 16th notes.
|
|
Reply To Message
|
|
Author: maxopf
Date: 2016-01-18 03:53
I'm not counting the appoggiatura. The eighth note is equal to three 16ths because Stravinsky specified 8th = 3 16ths.
|
|
Reply To Message
|
|
Author: brycon
Date: 2016-01-18 05:50
Liquorice is correct, of course: that figure is four sixteenth-notes.
|
|
Reply To Message
|
|
Author: maxopf
Date: 2016-01-18 06:28
Attachment: Tempo.png (95k)
Of course I understand that it would normally be four sixteenth notes. What I'm asking is: because Stravinsky specified at the beginning of this movement that 8th = three 16ths and that 16th=16th, would that figure not add up to five sixteenth notes in the context of this movement? If not, why not?
Post Edited (2016-01-18 06:32)
|
|
Reply To Message
|
|
Author: Tony Pay ★2017
Date: 2016-01-18 11:33
maxopf wrote:
>> ...because Stravinsky specified at the beginning of this movement that 8th = three 16ths... >>
You have to read the legend from left to right. What Stravinsky wrote was that when you see three semiquavers BEAMED TOGETHER, they add up to the duration of a quaver.
Other semiquavers behave normally, so that a quaver has the duration of two semiquavers that aren't so beamed.
I suppose he was trying, without barlines, to avoid writing figure threes above triplets all over the place.
Tony
|
|
Reply To Message
|
|
The Clarinet Pages
|
 |
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
