<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)">  <title>Re: [CAUT]  Re. Link to Young Paper</title> <o:SmartTagType namespaceuri="urn:schemas-microsoft-com:office:smarttags" name="PlaceName"/> <o:SmartTagType namespaceuri="urn:schemas-microsoft-com:office:smarttags" name="PlaceType"/> <o:SmartTagType namespaceuri="urn:schemas-microsoft-com:office:smarttags" name="State"/> <o:SmartTagType namespaceuri="urn:schemas-microsoft-com:office:smarttags" name="place"/>  <style>  </style> </head> <body lang=EN-US link=blue vlink=purple> <div class=Section1> Hi Fred,<o:p></o:p> <o:p> </o:p> At a convention someone mentioned this as “the” way one should take the FAC measurements. Wasn’t this in the journal once too? I’m still unclear as to the “step by step” instructions. (1, 2, 3…) For the gadget impaired, (me) would you give those instructions???<o:p></o:p> <o:p> </o:p> Thanks,<o:p></o:p> Jim Busby<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: caut-bounces@ptg.org [mailto:caut-bounces@ptg.org] On Behalf Of Fred Sturm Sent: Friday, January 19, 2007 4:47 PM To: caut Subject: Re: [CAUT] Re. Link to Young Paper<o:p></o:p> </div> <o:p> </o:p> On 1/19/07 3:29 PM, "David Brown" <David.C.Brown.2@asu.edu> wrote:<o:p></o:p> Dear Fred- I wonder if you could expand on this a bit more. I attempted to calculate FAC this way after reading your post but I must have done something wrong. The difference between the first and fourth partial of A4 seemed to yield a wider octave ( larger A number, to be expected I suppose) than the normal reading,  but the difference between the first and fourth partial of C 6 yielded numbers well into the 20’s or 30’s! I tried the the difference between the first and second partial of C7 as well and still no real usable numbers for me. There must be a way to enlighten me! Hope all is well in <st1:State w:st="on"><st1:place w:st="on">New Mexico</st1:place></st1:State>. Regards- David<o:p></o:p> Hi David,     The A4/A6 difference needs to be multiplied by 0.8, to factor out the difference between 1st and 2nd partial and leave the difference between 2nd and 4th partial. The C6/C8 difference needs to be multiplied by 0.2 to factor out the difference between 2nd and 4th partial and leave the difference between 1st and 2nd partial. The difference between 1st and second partial is 1/4 the difference between 2nd and 4th partial (it’s a logarithmic scale, based on a square of the difference between partial numbers, kind of).     So, if you read A4/A6 as 10 (zero A4 and read its 4th partial at A6 as plus 10 cents), you enter 8 as the A number. When you have calculated the tuning, the number for A4 (the tuning offset, tuning it’s 4th partial) will be 10. For C6/C8, if the number is 35 (C6 zeroed, its 4th partial at C8 read as plus 35 cents), you enter 7. (Obviously when you enter the number, you have to go to the right note name and octave, and scroll cents up or down).     Is this clearer?     (BTW, David is referring, I believe, to some posts I sent to pianotech around ten years ago, when I was coming to terms with my SAT). Regards, Fred Sturm <st1:place w:st="on"><st1:PlaceType w:st="on">University</st1:PlaceType> of <st1:PlaceName w:st="on">New Mexico</st1:PlaceName></st1:place>   <o:p></o:p> </div> </body> </html>