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Try dividing any two frequencies that differ by 10 cents. It's a
constant. Ten cents difference anywhere in the spectrum is a constant
frequency ratio of ~1.0579. The doppler shift is found in terms of
this ratio and you can convert the 10 cents difference into the ratio
of frequencies. It's counterintuitive for me (but not for Mark), that
it doesn't matter which pitch you are measuring but that's the way it
works.<br>
az<br>
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John Delmore wrote:
<blockquote cite="mid200606211723.k5LHNltp014059@ptg.org" type="cite">
<pre wrap="">Hi Mark:
You're not all wet --perhaps just a little damp!!
It's not your speed that changes if you measure different pitches--it's the
size of the cents around that pitch. So, if you're working with a
difference in cents from a certain frequency, you'll get a different result.
If the question had been phrased in Hz, you wouldn't have this problem.
For example:
A440+10 cents = 442.5489 Hz, a difference of 2.5489 Hz
A440-10 cents = 437.4658 Hz, a difference of 2.5342 Hz
A220+10 cents = 221.2744 Hz, a difference of 1.2744 Hz
And A220-10 cents = 218.7329 Hz, a difference of 1.2671 Hz.
(I hope I got these right, I'm using the "2^1/1200:1" definition that I
learned thanks to this discussion)
I'm working on the general formula to determine "cents difference" from a
pitch for any frequency, but that's a little more math than I have time for
right now...it involves the dreaded logarithmus naturalis, I think.
</pre>
</blockquote>
Log base two, not natural logs (base e).<br>
az<br>
<blockquote cite="mid200606211723.k5LHNltp014059@ptg.org" type="cite">
<pre wrap="">
I hope these minutia aren't too far afield...they're a lot more "on-topic"
than "that other thread"...
John
</pre>
</blockquote>
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