<HTML>In a message dated 1/25/2002 8:58:08 AM Pacific Standard Time, A440A@AOL.COM writes: <BLOCKQUOTE TYPE=CITE style="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">Subj:pin comparisons Date:1/25/2002 8:58:08 AM Pacific Standard Time From:<A HREF="mailto:A440A@AOL.COM">A440A@AOL.COM</A> Reply-to:<A HREF="mailto:pianotech@ptg.org">pianotech@ptg.org</A> To:<A HREF="mailto:pianotech@ptg.org">pianotech@ptg.org</A> Sent from the Internet    Hi Ed</BLOCKQUOTE>    I like your common sense approach here. To my mind it doesn't really matter how   .    <BLOCKQUOTE TYPE=CITE style="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">big the pin is, within reason  a no.1 thru 4 pin as long as they render easily in the    block. know what I mean. If I pin a new block with number ones and they're to tight it 's still hard just as hard to tune as the too tight no.4               >>Dale Erwin</BLOCKQUOTE> <BLOCKQUOTE TYPE=CITE style="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px"> Greetings,    I have just finished Paul's article this month in the journal.  Interesting views can be found on all aspects of tuning pins, since we all spend so much time with them,  and I am wondering what the numbers can tell me. Specifically, the difference in tuning control between sizes of pins.  I don't think that the increased radius of the larger pin is a significant factor.   I find (admittedly, I am NO math whiz, so if I have missed a step, please disregard everything that follows), that by determining the circumferance of two different sizes of pins, and then relating that to degrees of movement, the differences begin to seem academic.     Here is how it looks to me:    A pin that is .272" in diameter has a cir. of .85408".  This equates to .00237" per degree of rotation.  A .286" pin, by the same calculation has .0025" per degree of rotation. This means that the larger pin will move the string approx.  .0001" more per degree of rotation.  If we consistantly move pins by increments of 6 degrees in fine tuning, then the difference in pin size accounts for maybe .0006" (that is 6 ten-thousandths!!) difference in string length being pulled around the pin.   ( I have omitted the 1/2 string diameter from the circumferance equations,since that is a variable on a per string basis, though increasing the diameters of the two calculations would further reduce the percentage difference between them).    Since difference as it relates to tuning is based on changing the tension per degree of rotation, and tension/pitch relationships are functions of the square,  I have to ask just how much difference can be found from  .0006" of string movement, at the pin?  I don't think it would be a discernable quantity. Others? Regards, Ed Foote </BLOCKQUOTE> </HTML>