cents to cps - the formula

[Richard Moody]

> Actually, this formula is for going from cps to cents.  To go
> the other way, use:


Boy these math "majors" catch every little detail.  Actually I wanted
to find the cents from a ratio of frequencies, or cents FROM cps.  
So how come I put as the subject cents TO cps I dunno unless  rigor
mortis of a brain cell.  
	I was glad to get Scott's formula but couldn't quite figure how to
put it into a spread sheet.  However I noticed a formula I had and I
think it is the same so I leave that to the math experts. 
	f*2^(cents/1200)= F  
	It looks like getting a new F by multiplying the old f by the twelth
root of 2   or 2^(1/12).  Of course that give the F a semi tone up. 
Now if you wanted the F of a third up, that would be f*2^(4/12)  
right?? But here we have the octave divided into not 12 parts, but
1200, and the third ratio is then 400/1200.  So f*2^(400/1200) gives
F of the third above f.   
	For those who want to know how much one cent really is take a freq
say 220 and enter it into your favorite spread sheet like this,
220*2^(1/1200) and you get........ 220.1271137. or a little over a
tenth of a cycle per second, or one beat in 10 seconds.  But suppose
we are tuning a third up from 220 and the partials involved are 220 *
5 -C# 4* 4.   So one cent for 1100 is 1100*2^(1/1200) or.....
1100.635568, or 6 beats in ten seconds or 3 beats in five seconds. So
if you err by a cent you will be off the amount of a tempered fifth. 
Or if you judging beats, if you err flat by that amount the third
will be slightly slower that the the G#--C third is supposed to be. 
So because of the high freq of the partials used and the nature of
cents, it is fair to say the ear can be accurate to half a cent.  It
also might be fair to say the machine should have greater accuracy on
the lower notes than on the higher notes, while the opposite is true
for the ear.  

	This formula is a little simpler than the one first posted.  It came
out of the Encyclopedia Britannica, the article on
"Temperament"(1967)


1200/LOG(2)*LOG(F/f) compared to 1200*(log(F/f)/log(2)) (which could
be simplified algebraically to the same) 

Richard Moody 
----------
> From: Robert Scott <rscott@wwnet.net>
> To: pianotech@ptg.org
> Subject: Re: cents to cps - the formula
> Date: Tuesday, March 10, 1998 4:32 PM
> 
> Richard Moody wrote:
> 
> >ps   Anyone come up with something simpler? 1200*(log(F/f)/log(2))
> 
> Actually, this formula is for going from cps to cents.  To go
> the other way, use:
> 
>  F = f * exp( c * log(2) / 1200 )
> 
> where f is a starting pitch in cps, c is the number of cents to
> change f, and F is the resulting new pitch in cps.  exp() is the
> inverse of the log() function.  In all these formulas it doesn't
> matter whether you use base 10 logarithms or natural (base "e") 
> logarithms, so long as you are consistent.
> 
> Bob Scott
> Ann Arbor, Michigan