Re: A China-Sumer connection

From: Comm (tjsrno_at_spampost.com)
Date: 03/02/05 

Date: Wed, 02 Mar 2005 23:27:53 GMT

"Antonio Marques" <m.ap@sapo.pt> wrote in message
news:d04p4v$rhp$1@worf.visyn.net...
> As if it were a big thing:
>
> Given any frequency, you can know the note by
>
> . Picking the A below (for instance, since it was 23, let's pick 55/4; A
> can be obtained by dividing 440 by 2 any number of times)
> . Calculating m = 12 log2 (frequency/A)
> . Checking the nearest value in the table:
>
> 0 a
> 1 a#
> 2 b
> 3 c
> 4 c#
> 5 d
> 6 d#
> 7 e
> 8 f
> 9 f#
> 10 g
> 11 g#
>
> In this case, let's use Smalltalk:
>
> f := 23.
> A := 55/4.
> m := 12 * ((f/A) ln / 2 ln).
>
> And m is revealed as 8.90643, which in fact is close to the 9 for f#.
>
23 cycles per second gives you the note F# (F sharp) that is 4 octaves below
the F# that's just below 440 A. No, it's not a big deal - but figuring
that out without a calculator for someone who wanted to know and who can't
do that kind of calculation? It was a favor. It took some time. I don't
have a calculator or even know of one capable of taking the 12th root of a
number. I took the cube root of 2 - the square of that answer and the
square of that again. That gives the 12th root of 2. You can divide A 440
in half - IF YOU KNOW A is 440. The person asking me did not know that at
all. Point is, he couldn't do the math on it anyway. But, you can get A440
down to 27.5 for a very low A, 4 octaves below 440A. Then divide each 1/2
note by that 12th root of 2 number. F#. No matter how you look at it, it's
a lot of paperwork. (More than doing Income Tax).
>

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