The math behind perfect 5th tuning

[Robert Scott]

Richard Moody asks:

>Anyhow.... To this mind (I know that's a matter of opinion), the problem
>seems to be that 12 notes have be placed in an octave. You start with one
>note  say   n and end up with 2n.  You have to multiply that n by the same
>ratio    r      12 times to end up with 2n.  What is    r    ?
>
>Does this work??      n times a ratio twelve times equals two times n.
>
>Or symbolically   n(r^12) = 2n             hmmmmm   the two  n's   cancel
>
>	r^12 = 2      or r =  the twelth root of 2
>
>Is this right ???


    Yes, that's where it comes from.  Then for perfect 5th tuning I came up
with r ^ 7 = 1.5 by the same reasoning.  A perfect 5th is a frequency ratio of
3 : 2 or 1.5 and there are 7 half-steps.  In case you're wondering how to use
a hand calculator to solve this for r, just take the logarithm of both sides:

   log( r ^ 7 )  =  7 * log(r)  =  log(1.5)    so that
  log(r) = log(1.5) / 7     then use the inverse log button to get r.

Once this value of r is found, the octave in this tuning is the 12-th power
of that r.  (Use logarithms again.  It's easier than multiplying 12 times.)
I came up with  r ^ 12 = 2.003875474  which I claimed was sharp
by 3.26 cents.  This comes from taking the ratio  2.003875474 / 2 =
1.001937737.  So the octave is sharp by .1937737 %.  To convert from
percentage to cents, divide by .05946 which is 1 cent expressed as
a percentage.  That gives you 3.26 cents.

  -Bob Scott
   Ann Arbor, Michigan