[linux-audio-user] various questions (long)

Tim Hall tim at glastonburymusic.org.uk
Tue Mar 4 07:51:00 EST 2003

Semi OT side issue warning. Absolutely fascinating. This explains much about 
why I often fail to understand explanations.

On Tuesday 04 March 2003 09:38, Maarten de Boer wrote:
> this is really easy. i would not even call this math ;-)
> with samplerate R, 1 sec = R samples
> at x BPM, 1 min = 60 sec = x beats. so the duration of 1 beat = 60/x sec
> combined: 1 beat = R*60/x samples.
> so at 120 BPM, 44100 Hz, 1 beat == 22050 samples. but you probably want
> your loops to be 4 beats, so 88200 samples.

With all due respect Maarten, You _have_ managed to make it look like rather 
complex math. Either that, or I'm dyslexic. I would explain it thus:

each beat lasts 60/bpm seconds, multiply by the number of beats in the bar. 
That will give you the length in seconds, you an convert that to length in Hz 
as above if you need to. 4 beats to the bar _is_ the default setting, if you 
like. but best to check otherwise you may end up with a loop twice or half 
the length.

It's really ok to have odd numbers of beats to the bar if you want, but 
that's just personal bias. 


tim hall

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