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<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>No dice, Amy, you can’t use this dimensionless
ratio in the Doppler equation. <span style='mso-spacerun:yes'> </span>You
have to have a frequency.<span style='mso-spacerun:yes'> </span>It DOESN’T
matter what pitch (frequency) you measure. <span
style='mso-spacerun:yes'> </span>The problem comes when you try to use a
logarithmic scale (cents) in a linear equation (Hz in Doppler).<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>JD<o:p></o:p></span></font></p>
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<p class=MsoNormal><b><font size=2 color=black face=Tahoma><span
style='font-size:10.0pt;font-family:Tahoma;color:windowtext;font-weight:bold'>From:</span></font></b><font
size=2 color=black face=Tahoma><span style='font-size:10.0pt;font-family:Tahoma;
color:windowtext'> pianotech-bounces@ptg.org [mailto:pianotech-bounces@ptg.org]
<b><span style='font-weight:bold'>On Behalf Of </span></b>Amy Zilk<br>
<b><span style='font-weight:bold'>Sent:</span></b> Wednesday, June 21, 2006
5:46 PM<br>
<b><span style='font-weight:bold'>To:</span></b> <st1:PersonName w:st="on">Pianotech
List</st1:PersonName><br>
<b><span style='font-weight:bold'>Subject:</span></b> Re: Doppler puzzle</span></font><font
color=black><span style='color:windowtext'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=3 color=black face="Times New Roman"><span
style='font-size:12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 color=black face="Times New Roman"><span
style='font-size:12.0pt'>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<o:p></o:p></span></font></p>
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