<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><div>on July 15, 2008, David Love wrote:</div><div><br><br><blockquote type="cite"><span class="Apple-style-span" style="border-collapse: separate; color: rgb(0, 0, 0); font-family: Helvetica; font-size: 12px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-border-horizontal-spacing: 0px; -webkit-border-vertical-spacing: 0px; -webkit-text-decorations-in-effect: none; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0; "><div bgcolor="white" lang="EN-US" link="blue" vlink="purple" style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><div style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; "><span class="Apple-style-span" style="color: rgb(20, 79, 174); -webkit-text-stroke-width: -1; ">An examination of the formula for frequency of a string as a function of tension (or BP%), diameter, length (BP%) and looking at the differences in the rate of change depending on tension levels should yield more clarity.   </span></div><div class="Section1"><p class="MsoNormal" style="margin-top: 0in; margin-right: 0in; margin-left: 0in; margin-bottom: 0.0001pt; font-size: 12pt; font-family: 'Times New Roman'; "><font size="2" color="navy" face="Arial"><span style="font-size: 10pt; font-family: Arial; color: navy; "> </span></font></p><div style="margin-top: 0in; margin-right: 0in; margin-left: 0in; margin-bottom: 0.0001pt; font-size: 12pt; font-family: 'Times New Roman'; "><font size="2" color="navy" face="Arial"><span style="font-size: 10pt; font-family: Arial; color: navy; ">For example, take two strings of equal length producing equal frequency (the dependent variable) but with different diameters (gauges—they will have different amounts of tension and they will also have different BP%) and then change the length equal amounts and you should see a difference in the change in frequency between the two.  </span></font></div></div></div></span></blockquote></div><br><div>Although I am far from an expert in rescaling or the physics of piano strings, I don't believe that the part of this about BP% is accurate.  My understanding is that two strings of equal length producing equal frequency will both be at exactly the same percentage of their respective breaking points.  For instance:</div><div><br></div><div>Note 40 (C4), speaking length, 715mm , diameter of 0.040",  tension 199.957lbs, %breakpoint <b>49.436</b>%</div><div>Note 40 (C4), speaking length, 715mm , diameter of 0.038",  tension 180.461lbs, %breakpoint <b>49.436</b>%</div><div><br></div><div>(figures from Pscale)</div><div><br></div><div>In other words, though the second string is smaller in diameter, and will be at a lower tension when producing the note "middle C," since its diameter is in fact smaller, its breaking point will be lower, and it will be at exactly the same <b>percentage</b> of its breaking point.  As I understand it, this is because, as its diameter changes, its breaking point changes proportionally.  See pages 33-35 in John Travis' "A Guide to Restringing" (in the section by James Hayes) for a simple experiment you can do to demonstrate this empirically.</div><div><br></div><div>The aspect of the string's behavior that does change in the example above is <b>inharmonicity</b>:</div><div><br></div><div>Example above with 0.040 diameter, inharmonicity = 0.192</div><div>Example above with 0.037 diameter, inharmonicity = 0.173</div><div><br></div><div>If I am wrong in my conceptual understanding, someone please correct me.  I want to be sure that I understand this correctly.</div><div><br></div><div>I will also point out that, from my archive reading and past posts, there are at least two formulae out there for calculating breakpoint and breakpoint percentage, so some of you might plug my numbers into your formula and get a different value for %breakpoint. However, that value should be identical for the two wire diameters given above.</div><div><br></div><div>Joe DeFazio</div><div>Pittsburgh</div></body></html>