<html xmlns:v="urn:schemas-microsoft-com:vml" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:w="urn:schemas-microsoft-com:office:word" xmlns:st1="urn:schemas-microsoft-com:office:smarttags" xmlns="http://www.w3.org/TR/REC-html40"> <head> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=us-ascii"> <meta name=Generator content="Microsoft Word 11 (filtered medium)"> <o:SmartTagType namespaceuri="urn:schemas-microsoft-com:office:smarttags" name="PersonName"/>  <style>  </style>  </head> <body bgcolor=white lang=EN-US link=blue vlink=blue> <div class=Section1> ROTFLOL !!!<o:p></o:p> <o:p> </o:p> JF<o:p></o:p> <o:p> </o:p> <div> <div class=MsoNormal align=center style='text-align:center'> <hr size=2 width="100%" align=center tabindex=-1> </div> From: pianotech-bounces@ptg.org [mailto:pianotech-bounces@ptg.org] On Behalf Of Amy Zilk Sent: Saturday, June 17, 2006 10:00 AM To: <st1:PersonName w:st="on">Pianotech List</st1:PersonName> Subject: Re: Doppler Cents Puzzler #2<o:p></o:p> </div> <o:p> </o:p> LOL!!!  You would want to shove the piano directly at the conductor at 3.4 mph.  Would that sufficiently flatten the conductor?  Amy <a href="mailto:ed440@mindspring.com">ed440@mindspring.com</a> wrote: <o:p></o:p> <div> Amy-<o:p></o:p> </div> <div> Excellent reply.<o:p></o:p> </div> <div> Now, please, can you tell us: If the conductor wants the piano at 442hz and I don't want to raise pitch at the pin block, how fast do I have to push the piano across the stage?<o:p></o:p> </div> <div> Mathematically challenged folks want to know.<o:p></o:p> </div> <div> Ed Sutton <o:p></o:p> </div> <blockquote style='border:none;border-left:solid blue 1.5pt;padding:0in 0in 0in 4.0pt; margin-left:0in;margin-top:5.0pt;margin-bottom:5.0pt'> -----Original Message----- From: Amy Zilk <AMY@ZILKNET.NET>Sent: Jun 17, 2006 9:18 AM To: <st1:PersonName w:st="on">schecter@pacbell.net</st1:PersonName>, <st1:PersonName w:st="on">Pianotech List</st1:PersonName> <PIANOTECH@PTG.ORG>Subject: Re: Doppler Cents Puzzler <ZZZHTML><ZZZHEAD><ZZZMETA content="text/html;charset=ISO-8859-1" http-equiv="Content-Type"></ZZZMETA><ZZZBODY text="#000000" bgcolor="#ffffff">Vladen's answer is the one I get. <o:p></o:p> </ZZZBODY></ZZZHEAD></ZZZHTML></PIANOTECH@PTG.ORG></AMY@ZILKNET.NET> f is the frequency of the source fo is the frequency heard by the observer c is the difference in cents (10) Vs is speed of sound in air (1100 ft/s) V is speed of observer (speed of the guy on bike -- the answer)<o:p></o:p> Doppler shift equation for stationary sound source; observer moving directly to or from the source:<o:p></o:p> <u1:p> </u1:p>    f/fo – 1 = V/Vs<o:p></o:p> <u1:p></u1:p>You have to get f/fo from the 10 cents difference.<o:p></o:p> <u1:p> </u1:p>f/fo = 2^(c/1200) = 1.005793<o:p></o:p> <u1:p></u1:p>Plug it in and get<o:p></o:p> <u1:p></u1:p>1.005793 – 1 = V/Vs<o:p></o:p> <u1:p> </u1:p>0.005793 = V/Vs<o:p></o:p> <u1:p> </u1:p>0.005793 * Vs = V<o:p></o:p> <u1:p> </u1:p>V = 0.005793 * (1100 ft/s) =  6.3723 ft/s  <o:p></o:p> 6.3723 ft/s (3600s/hr) / (5280 ft/mi ) = 4.3 mph<o:p></o:p> az<o:p></o:p> <div> <img border=0 width=32 height=32 id="_x0000_i1025" src="file:///C:%5CDocuments%20and%20Settings%5Cazilk%5CMy%20Documents%5CPiano%20Tech%5CAZP-email-sig.gif" moz-do-not-send=true><o:p></o:p> </div> Mark Schecter wrote: <o:p></o:p> Hi, Vladan. Well, your number and mine don't agree, and I'm not at all sure of mine. So I'm going to show how I got to my result, and if I'm wrong, I'd be delighted to know how. So here goes. The fact that the tone goes flat 10 cents when going away merely confirms that the difference between the stopped truck and the moving cycle produces a 10 cent differential in pitch. So I considered the pitch coming from the stopped truck to be 1, and the sound to be travelling at 1100 feet per second. In order to reach a pitch of 2, the cycle would have to be moving at the speed of sound toward the truck, to achieve a total of 2200 feet per second closing speed. With that thought in mind, I just calculated that a 10 cent increase in pitch equalled 10/1200 of the speed of sound, so: 10 cents higher than nominal pitch = 10/1200 * (speed of sound in air) or 1/120 * (1100 ft/sec) = 9.1666 ft/sec (speed of bicycle) 9.166 ft/sec * 3600 secs/hour = 33,000 ft/hour 33,000 / 5280 (ft/mi) = 6.25 mph However, you arrived at 2 meters/second, which equals 7200 meters/hour, which translates to 4.47 miles per hour. So would you tell me how you got there? Thanks! -Mark Schecter V T wrote: <o:p></o:p> 2 meters/second; I would have stopped for some ice cream. Vladan ===================== I was out riding my bicycle this calm quiet evening when I happened upon an ice cream truck playing music to attract customers.  The truck had stopped to dispense ice cream, but the music continued.  Since I always carry my ETD when I ride my bike, I quickly measured the pitch of a recurring note in the music and found it to be 10 cents sharp as I was riding straight towards the truck.  Then after I passed the truck, I measured the pitch again and found it to be 10 cents flat as I was riding directly away from it. How fast was I riding my bicycle? Robert Scott Ypsilanti, MI __________________________________________________ Do You Yahoo!? Tired of spam?  Yahoo! Mail has the best spam protection around <a href="http://mail.yahoo.com">http://mail.yahoo.com</a> <o:p></o:p> <o:p> </o:p> </blockquote> </div> </body> </html>