Middle C in Hertz
Adrian A Osborne
What frequency is middle C in hertz on a piano keyboard.
Ta!
Adrian O.
Frank Weston
In equal temperament for a piano tuned with no stretch, middle C would
be 261.6255653 Hz. You will rarely find such a piano.
Frank Weston
Denman Maroney
275 = 440/2 x 5/4
"Adrian A Osborne" <adr...@cytanet.com.cy> wrote:
>What frequency is middle C in hertz on a piano keyboard.
>
>Ta!
>
>Adrian O.
Robert Scott
No, 440/2 x 5/4 is not the frequency of middle C on a modern piano.
That would be true if the minor third (A-C) were tuned beatless,
which is pretty far from an equal tempered minor third.
Bob Scott
Ann Arbor, Michigan
Gil G Silberman
>On Mon, 14 Apr 1997 17:52:41 GMT, den...@pipeline.com (Denman Maroney)
>wrote:
>>275 = 440/2 x 5/4
In article <3352c515...@news.mindspring.com>,
Larry Craven <l...@mindspring.com> wrote:
>Close, but no cigar! If A=440, then middle-C is exactly 261.6 Hz,
>harmonics notwithstanding. That is 1/2 the Hz rate of the standard C
>tuning fork, which at 523.2 is actually an octave above mid-C.
I don't know about tuning forks, but if A is 440 and you are
in equal (even) temperment, then C should be at 440/(2^(9/12)),
or 261.6256 hertz.
The 2 is there because every octave down halves the frequency.
The 9/12 is there because you have gone down 9 semitones from
middle A in a scale that has 12 semitones per octave.
-- gil, coming to you at 133 X 2^20 hertz (or is that 133,000,000?)
http://www.hooked.net/~bigbug
Larry Craven
>275 = 440/2 x 5/4
>
>"Adrian A Osborne" <adr...@cytanet.com.cy> wrote:
>
>>What frequency is middle C in hertz on a piano keyboard.
Close, but no cigar! If A=440, then middle-C is exactly 261.6 Hz,
harmonics notwithstanding. That is 1/2 the Hz rate of the standard C
tuning fork, which at 523.2 is actually an octave above mid-C.
Larry
Albert Lord
>>What frequency is middle C in hertz on a piano keyboard.
Correcting my notation error:
C4=(2^1/12)^3 * 440/2 or C4=2^3/12 * 440/2
=261.6 Hz
Albert Lord.
--
jeanette
On 16 Apr 1997 03:17:45 GMT, Albert Lord <72624...@CompuServe.COM>
wrote:
>>>What frequency is middle C in hertz on a piano keyboard.
>
>>C4=(2^-12)^3 * 440/2
>
>Correcting my notation error:
>C4=(2^1/12)^3 * 440/2 or C4=2^3/12 * 440/2
> =261.6 Hz
How do you find out the Hz of other notes?
Jeanette
Donna and Ian
The frequency of a note a semitone above a given note
is the given note frequency multiplied by the twelfth
root of 2 when the scale is tuned to the equal tempered
scale. It is easier to continue with symbols.
Let the frequency of a given note be n Hz.
So f = n Hz.
Then a semitone above, f = n * (2 ^ (1/12)).
It follows that a whole tone above n is
f = n * (2 ^ (1/12)) * (2 ^ (1/12))
= n * ((2 ^ (1/12)) ^ 2)
Generally, the frequency m semitones above f = n,
f = n * ((2 ^ (1/12)) ^ m)
= n * (2 ^ (m/12))
Another way of calculating middle C from A440 is
middle C = 440 * (2 ^ (-9/12)) since middle C
is 9 semitones below A440 ( m = -9 ).
Ian
--
To reply by email, please remove the NS from the address supplied.
Ragtimbill
The frequency of any note can be determined by using a multiplying factor
and a starting frequency (usually A440). The multiplying factor is the
12th root of 2, namely 1.0594631. Thus, to find the Bb above A, multiply
A by the factor:
440 X 1.0594631 = 466.16. Then, for the next note, multiply the Bb by the
factor:
466.16 X 1.0594631 = 493.88. C would then equal 523.25. Middle C would
therefore be equal to half of that, or 216.63. To go down, use the factor
to divide by: G# would be A/facror, or: 440 / 1.0594631 = 415.3.
Hope this helps you.
Bill Rowland
(retired piano technician, now a computer programmer)
Bill Rowland
2100 N. 26th
Broken Arrow, OK 74014
ragti...@aol.com
Samuel Marin
Hi !
This is all true, technically speaking. However, our ears don't really hear
this. There are some modifications perceptible in the lowest and higher
pitches (you'll need a professional or a book to help you on this). Over
that, there are true problems in selecting the tunig inside the scales. The
tempered keyboard is a reference most of us use nowadays, but it's
unfortunately not the mathematical rule. Check out a professional or a
book's advice on this, again, if you need be precise. (and want to have
things sound good)
--
samue...@infonie.fr
Ragtimbill <ragti...@aol.com> a écrit dans l'article
<19970501031...@ladder01.news.aol.com>...
wcri...@telerama.lm.com
Ragtimbill (ragti...@aol.com) wrote:
> The frequency of any note can be determined by using a multiplying factor
> and a starting frequency (usually A440). The multiplying factor is the
> therefore be equal to half of that, or 216.63. To go down, use the factor
> to divide by: G# would be A/facror, or: 440 / 1.0594631 = 415.3.
Hey Ragtime, How can middle C be below A below middle C? Of course,
slight anomaly in dividing by two. (g) Anotherbill
--
---------------
wcri...@lm.com
Ragtimbill
Ragtimbill (ragti...@aol.com) wrote:
> The frequency of any note can be determined by using a multiplying
factor
> and a starting frequency (usually A440). The multiplying factor is the
> 466.16 X 1.0594631 = 493.88. C would then equal 523.25. Middle C would
> therefore be equal to half of that, or 216.63. To go down, use the
factor
>>Hey Ragtime, How can middle C be below A below middle C? Of course,
slight anomaly in dividing by two. (g) Anotherbill <<
--
A440 is the A above middle C. Three semitones above that is the C one
octave above Middle C. Every note's frequency is exactly twice that of
the octave below it. Therefore, to derive Middle C, simply divide the
frequency of the octave above it by two. Sorry I did not make that clear.