Determine of cents of a note in 440hz will translate

[ElectroNewby]

Solved!

 

[MrAl]

Hi,

[Note: The original OP post was a question about how to calculate the frequency of notes in between semi tones]


According to Wikipedia, one cent is the ratio between the two frequencies that is equal to 2^(1/1200).

Thus, if A=440Hz then A plus one cent will be:
440*(2^(1/1200)) Hertz

A plus two cents will be:
440*(2^(1/1200))^2 Hertz

A plus three cents will be:
440*(2^(1/1200))^3 Hertz

A plus N cents will be:
440*(2^(1/1200))^N Hertz

A note with frequency F plus N cents will be:
Fnew=F*(2^(1/1200))^N Hertz

Also,

Fnew/F=(2^(1/1200))^N
or
Fnew/F=2^(N/1200)
or
log(Fnew/F)=log(2)*N/1200
or
1200*log(Fnew/F)/log(2)=N

So if you know Fnew and F and Fnew>F you can find N and you might have to round the answer.

If however Fnew<F then use:
N=1200*log(F/Fnew)/log(2)

or use the first formula and interpret a negative N as 'down' in cents.

 

[ElectroNewby]

Thank you sir! Logaritms, there are so mean

I found out a chard that has exactly what I was looking for here:

**broken link removed**

 

[MrAl]

Hello,

Ok but you erased your original post, why did you do that? Other people might want to know the same thing at a later date.