[Harp-L] re:harmonica math

[Joe Mahan <joe.mahan@xxxxxxxxxxx>]

Dave Payne wrote:

> Vern weote:
> "The formula is:  C = 1200 / log(2) * ( log(F1) - log(Fo) )"
>
>
> That's exactly what I was looking for, I think. Wow. That's some fancy math. I have no idea what that is, I think it has something to do with math, lol. I figured I'd have to study up on some math to figure it out. Thank you. Thanks to everyone else who chimed in as well. Lots of good info.
>
> Dave
>
>  
The math is not so bad, and there are other ways to get close enough.  
If you understand that every octave is a doubling of frequency, that's 
the factor of two. There are twelve (chromatic) semitones per octave, so 
to get from one semitone to the next higher semitone, you multiply the 
lower note by the twelfth root of two. The twelfth root of two is about 
1.05946, it's a constant value up and down the scale. What changes is 
how may Hz occupy each semitone.

So, to get from A=440 to A#, it's 440 times 1.05946, or 466.1624 Hz. 
Let's round it to 466 Hz to keep it simple.

A cent is 1/100 of a semitone. So, looking at cents between A = 440 and 
it's neighbor A# = 466, just subtract 440 from 466, and that's the semitone.
(26 Hz for this pair of notes)
Divide by 100, and that's a cent. (.26 Hz)


Another way to look at it? If you multiply 1.05946 by itself twelve 
times, the answer is 2, an octave.
(Well, very close to 2 due to rounding. And actually you do the 
multiplication 12 minus1 times(11), but I'm hoping to clarify not 
confuse, and I'm not sure I succeeded...)

Vern's method avoids the rounding errors.

Joe