[CM] cmn factorize

Timothy Ernest Johnson tejohnso@uiuc.edu
Fri, 7 Apr 2006 08:47:29 -0700


hi. i'm new to the list -- just a comment about the notation
issue:

yes, on p. 212, example 11-24 there is a nested tuplet figure
with 3 triplet 16ths inside of one triplet 8th. i think the
reason for only two beams is that each tuplet level only
relates to the previous level -- so that the triplet 16ths
only relate to the triplet 8th notes as if they were the
prevailing meter (they know nothing of other levels)

however if we had 9 16ths as a single group this would be
relating to the 16ths at the 'bottom' level and therefore
would clearly be 9:8 and would require three beams.

---- Original message ----
>Date: Fri, 7 Apr 2006 07:13:13 -0700
>From: "Bill Schottstaedt" <bil@ccrma.Stanford.EDU>  
>Subject: Re: [CM] cmn factorize  
>To: Dave Phillips <dlphillips@woh.rr.com>
>Cc: Ren Bastian <rbastian@free.fr>, cmdist@ccrma.Stanford.EDU
>
>But look at page 212 -- he gives exactly the case we were
talking about
>and uses 2 beams for 1/9.  The earlier case is made simple by
leaving
>the discussion in terms of fancy meters with simple quarter
note divisions,
>so there 9 is clearly 9:2 in our thinking (and not problematic). 
>Also it irks me that he thinks he can smirk at Ives -- the
bastard!
>
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Timothy E. Johnson
visiting lecturer
School of Music
University of Illinois/Urbana-Champaign