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  How do you convert Cents to Hertz

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Author Topic:   How do you convert Cents to Hertz
George Kimery

From: Limestone, TN, USA

posted 15 July 2006 05:02 AM     profile     
I want to try Billy Cooper's tuning, which is in Cents. My Tu-112H tuner is much better for HZ because it is the large, top scale. Looking at an old Korg tuner I have, it looks like minus 10 HZ would be minus 40 Cents.So, do I just divide the Cents by 4 to convert to HZ? i.e. minus 10 cents would be minus 2.5 HZ? minus 7 cents would be minus 1.75 HZ. etc. etc.
Ken Fox

From: Ray City, GA USA

posted 15 July 2006 05:54 AM     profile

That should help!

[This message was edited by Ken Fox on 15 July 2006 at 05:56 AM.]

George Kimery

From: Limestone, TN, USA

posted 15 July 2006 06:08 AM     profile     
Thanks, Ken, but if I understood the links correctly, they more or less convert HZ to cents. The HZ is what I need, I already have the cents. Also, I am looking for a simpler way of doing it with a calculator, not having to put in numbers in a computer program or link.
Charlie C Harrison

From: Decatur, Mississippi, USA

posted 15 July 2006 06:40 AM     profile

This may help!

Jon Light

From: Brooklyn, NY

posted 15 July 2006 06:54 AM     profile     
George--your first statement is approximately correct:

One Hertz = 4 Cents.

1 Cent = 1/4 Hertz

If this is less than astronomically accurate, it seems to be close enough for jazz. Or country music. This looks about right on my old Korg DT-1 and it is consistent with my Jeff Newman chart.

[This message was edited by Jon Light on 15 July 2006 at 06:54 AM.]

Jack Stoner

From: Inverness, Florida

posted 15 July 2006 07:27 AM     profile     
Jon is right on. The 4 cents to 1 Hz is what I use.
Dave Mudgett

From: Central Pennsylvania, USA

posted 15 July 2006 10:14 AM     profile     
The approximate conversion factor 4 is based on differences from a center frequency of about 440 Hz. It's different for different center frequencies. One can compute Cents change in going from frequency f0 to frequency f1 using logarithms in the following formula:

Cents = K*(log(f1)-log(f0))

where 1200 = K*log(2)

If one uses common logarithms (base 10), this constant comes out to

K = 1200/log10(2) or approximately 3986.3

So, using 440 as the center frequency, f0, then the number of Cents change going to 441 Hz is

Cents(440-to-441) = 3986.3*(log10(441)-log10(440))

or about 3.93 Cents.

Here's a table of Cents difference for some frequencies around 440 Hz:

435 Hz: -19.78 Cents
436 Hz: -15.81 Cents
437 Hz: -11.84 Cents
438 Hz: -7.89 Cents
439 Hz: -3.94 Cents
440 Hz: 0 Cents
441 Hz: +3.93 Cents
442 Hz: +7.85 Cents
443 Hz: +11.76 Cents
444 Hz: +15.67 Cents
445 Hz: +19.56 Cents

So, the conversion factor 4 Cents/Hz is just fine for our purposes of tuning around 440 Hz.

But note that this conversion factor is not constant - for example, for a center frequency one octave lower, 220 Hz, the factor doubles:

3986.3*(log10(221)-log10(220)) or about 7.85

Similarly, for a center frequency one octave higher, the factor is cut in half:

3986.3*(log10(881)-log10(880)) or about 1.97

This is because of the basic definition of a Cent as a percentage of the tonal change. The ratio 442/440 is the same as 221/220, hence the same number of Cents change are identical.

Even these figures are not astronomically accurate.

Joseph Meditz

From: San Diego, California USA

posted 15 July 2006 03:36 PM     profile     
To do this on a (scientific) calculator, all you need to remember is that there are 1200 cents to an octave. The formula is:

a*2^(c/1200) = b.

For example, using the information from Dave's table, we can convert 19.56 cents to Hz for a 440.
Here a = 440, c = 19.56 and b is what we want to know.

440*2^(19.56/1200) = 444.9994 Hz.


I had to edit this since I typed a "2" instead of a "c" in the forumula. Sorry about that! And by way of answering another question, you can, for example, find out the frequency of a B that is off by 19.56 cents by noting that each half step is 100 cents. So, for the note B, two half steps above A440, the frequency would be

440*2^(219.56/1200) = 499.495 Hz.
while a perfectly tuned ET B would be
440*2^(200/1200) = 493.8833 Hz


[This message was edited by Joseph Meditz on 16 July 2006 at 05:03 PM.]

John Bechtel

From: Nashville, Tennessee,U.S.A.

posted 15 July 2006 09:10 PM     profile     
I gave up on Hz vs. cents about (3)-weeks ago, during the last controversy over tuning E9 to 440Hz. or 442Hz.! hee~hee!

“Big John”
a.k.a. {Keoni Nui}
Current Equipment

John McGann

From: Boston, Massachusetts, USA

posted 16 July 2006 04:24 AM     profile     
So 100 cents to the next half step in ET unless it's JI or something else where you keep the change?

50 cents would be a quarter tone?

Larry Jamieson

From: Walton, NY USA

posted 16 July 2006 06:41 AM     profile     
John, Last I knew, 50 cents was half a dollar...
George Kimery

From: Limestone, TN, USA

posted 16 July 2006 12:27 PM     profile     
Thanks a lot guys. All of the input was interesting, some above my head in the math, but Charlie's link with the C chart gave me exactly what I was looking for. Also, the divide by 4 approximation rule is quite good enough for my tin ears. The chart just saves me having to calculate. The forum is wonderful. Thanks again to all those who took the time to respond.
Jim Bates

From: Alvin, Texas, USA

posted 16 July 2006 07:27 PM     profile     
Answer to John: Yes, 50 cents is a quarter tone.


John McGann

From: Boston, Massachusetts, USA

posted 17 July 2006 10:25 AM     profile     
Thanks Jim!!!
chas smith

From: Encino, CA, USA

posted 17 July 2006 12:07 PM     profile     
Um, you don't convert cents to frequency, because cents is a measure of distance (interval) between frequencies. For instance, 1200 cents is the distance between octaves. So, between the octave A 220 and A440, there are 220 integer frequencies, that are covered by 1200 cents. Between the octave A1760 and A3520, there are 1760 integer frequencies covered by 1200 cents.

What I use to get the cents between 2 frequencies is:
Log( freq1) - Log(freq2) / Log( 1200th root of 2)

Morton Kellas

From: Chazy, NY, USA 12921

posted 18 July 2006 09:15 AM     profile     
That seems to make cents to me.
William Steward

From: Grand Cayman, Cayman Islands

posted 18 July 2006 08:19 PM     profile     
Morton that hertz!

From: United Kingdom

posted 19 July 2006 02:14 AM     profile     
Frequently !!
Earnest Bovine

From: Los Angeles CA USA

posted 19 July 2006 07:57 AM     profile     
I don't know about cents, but I did find help with converting from cps to Hertz. This info is from an article that I remember from my dad's ham radio magazines around 1960.

Frequency Conversion: The unit of frequency measurement used to be Cycles Per Second (cps). Alternatively, some people used Kilocycles per Second or Megacycles per Second. Around 1960, the world converted to frequency measurements in Hertz. Abbreviated, it is Hz. The change was to honor Heinrich Hertz.
If you still have magazines and books describing frequency in Cycles per Second or Megacycles, you will want to convert those numbers to Hz (Hertz) or MHz (Megahertz). To do that, use this chart:

The graph uses a semi-logarithmic scale to simplify the conversion. However, it does not cover the complete range of frequencies that you may need. For out-of-range values, you can extrapolate the curve by hand, or use the following mathematical method to convert cps to Hz: Multiply the frequency in cycles per second by cycles per second and then take the square root of the result.

Jim Bates

From: Alvin, Texas, USA

posted 19 July 2006 08:16 PM     profile     
It is interesting to note that while you can calculate (or look up in a table) the freqency versus cents to an accuracy of hundreths or thousanths of a cent, you are not going to be able to set a strobe or needle tuner with such accuracy.

In my other life I am a piano tuner, and when I am requested to tune a piano (raise the pitch) to A=442, I simply set the strobe for 440+10 cents and the piano will be very, very close.

Why? Because when all of the 235 or so piano strings are tightened (raised 10 cents from the A=440, the extra force on the bridges and soundboard typically cause a 1 - 1.5% drop in pitch. (If you raise 10 cents, you will only net a 8.5 or 9 cent gain overall for the piano.) This works for most cases. Also, I carry an A=442 tuning fork to check the results.

I wouldn't worry too much about a cent or two off. Most of us can't hold the bar over the frets to that accuracy. Spend less time fine tuning and more time playing!


Dave Mudgett

From: Central Pennsylvania, USA

posted 19 July 2006 08:48 PM     profile     
How accurately one needs to calculate things like this depends on the application. The danger of being too approximate is that, when one considers a lot of multiple intervals, errors in approximation can accumulate, leading to large errors in the final result.

The correct way to do calculations is to keep the accuracy of intermediate results to a large number of significant figures, and then round off to the accuracy you desire when you're done.

That's a good one, EB. Are you sure that's not from a social sciences journal? Sorry, ducking for cover - it was a joke. I musta read one too many articles by Yale mathematician Serge Lang.

Of course, the reason for using logs in the Hz - Cents conversion is that the function of frequency vs. #semitones of tonal change is not linear, but exponential - frequency doubles for each octave of tonal change.

Jim Bates

From: Alvin, Texas, USA

posted 20 July 2006 01:37 PM     profile     
A little correction: I meant to say that a "10-15% drop (1 to 1.5 cents) for a pitch raise of 10 cents is within normal".


Mike Wheeler

From: Columbus, Ohio, USA

posted 25 July 2006 06:29 AM     profile     
Earnest, my friend, that chart is simply a log chart showing CPS to HERTZ in a 1:1 relationship. In fact, the only thing that changed regarding frequency, was the name. There is no conversion to be done at all between CPS and Hertz. Thought you'd like to know. :-)
Robert Leaman

From: Murphy, North Carolina, USA

posted 25 July 2006 07:47 AM     profile     
Mike Wheeler is absolutely correct. Sixty (60) cycles per second equals sixty Hertz (60Hz) or 38.19718634+ radians.
Bobby Sparks

From: Williamston, North Carolina, USA

posted 25 July 2006 09:03 AM     profile     
Earnest...I think ya got'em!!


Donny Hinson

From: Balto., Md. U.S.A.

posted 25 July 2006 02:20 PM     profile     
Thank you Mike...for stating the obvious. Being from the old "cycle/kilocycle/megacycle era" (and with all due respect to Heinrich), I always thought that the replacement of the term "cycle" with "Hertz" was one of the dumbest ideas ever to come along.


Yeah, yeah, so does 120.

Mike Wheeler

From: Columbus, Ohio, USA

posted 25 July 2006 05:10 PM     profile     
Well, Donny, you know as well as I do that the electronic super-gurus like to keep things cryptic so that not just any bozo can understand this stuff. lol!

I have been told I have a talent for stating the obvious. ha, ha.

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