<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=windows-1252"> <META content="MSHTML 5.50.4522.1800" name=GENERATOR></HEAD> <BODY bgColor=#ffffff> <DIV>Well, OK Paul and Mike and others. I don't really understand it, = but I'll assume you are right with this "concentration factor". But I guess as = you pointed out Paul, that if your pins break at the becket, it kinda makes = all this moot. Your pin with the 1/0 top will have the same shear strength as any = other 1/0 pin - because the weak link is the area of the becket. Your pins = sound intriguing. Where are they available?</DIV> <DIV> </DIV> <DIV>Terry Farrell</DIV> <DIV> <SPAN id=__#Ath#SignaturePos__></SPAN> </DIV> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style="FONT: 10pt arial">----- Original Message ----- </DIV> <DIV style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: = black"><B>From:</B> <A title=mjbkspal@execpc.com = href="mailto:mjbkspal@execpc.com">Mike and Jane Spalding</A> </DIV> <DIV style="FONT: 10pt arial"><B>To:</B> <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">pianotech@ptg.org</A> </DIV> <DIV style="FONT: 10pt arial"><B>Sent:</B> Sunday, January 27, 2002 = 2:23 PM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Tuning Pin = Size</DIV> <DIV><BR></DIV> <DIV><FONT size=2>Paul and Terry,</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Whenever there is an abrupt change in size/shape, = there will be a stress concentration.  This is what you've got at the inside = corner where the 1/0 upper portion of your pin meets the 2/0 or larger lower portion.</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Picture a pin of uniform diameter, with a = straight line drawn along it.  Now twist the pin, look at the line: it's a = uniform spiral, like a barber pole, or a candy cane.</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Now take the stepped pin, draw the straight line = along the lower portion, in towards the  center along the step, up along = the upper portion.  Twist the pin, look at the line.  Still generally = a spiral, but:  The spiral on the upper (smaller diameter) section = is faster than the spiral on the lower portion.  Where the upper = spiral meets the step, there's a distortion in the spiral which is your = stress concentration.  </FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Stress concentrations can be minimized by = radiusing the inside corner, by optimizing feeds and speeds in the lathe, and = polishing the radius after machining.  </FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Best case, I would guess Paul's pins still have a = 20% to 30% stress concentration factor at the step.</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Hope this helps</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Mike Spalding</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style="FONT: 10pt arial">----- Original Message ----- </DIV> <DIV style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: = black"><B>From:</B> <A title=larudee@pacbell.net href="mailto:larudee@pacbell.net">larudee@pacbell.net</A> </DIV> <DIV style="FONT: 10pt arial"><B>To:</B> <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">pianotech@ptg.org</A> </DIV> <DIV style="FONT: 10pt arial"><B>Sent:</B> Sunday, January 27, = 2002 12:19 PM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Tuning Pin = Size</DIV> <DIV><BR></DIV>Terry, <P>I am not an engineer, but I consulted one while researching for = my patent, and that's how it was explaiened to me.  The problem is = that when you twist or flex the top portion, the bottom portion doesn't = twist or flex as much.  If you twist or flex the pin that is 2" in = diameter in the base and .276" in the top portion to the breaking point, it's = going to break at the transition point every time.  If your argument = were correct, it would break randomly at any point along the top = portion.  Are there any engineers who would care to elaborate?  Carl? <P>Paul <P>Farrell wrote: <BLOCKQUOTE TYPE="CITE"> <STYLE></STYLE> I'm trying to understand this. Let's say we have a 0.276 = -in.  dia. tuning pin that is 2-in. long. Let's say it has a shear strength = of 300 inch-pounds. Meaning of course if you install the pin in a new = Baldwin, put a tuning hammer on it (or a torque wrench) and try to turn it, = when you get to a shear force of 300 inch-pounds, it will shear into = two pieces - leaving one piece in your tuning lever tip and the other in the block. Now take a similar pin, but make it 6 inches long. Do = the same things, and it should shear at 300 inch-pounds of torque. Length = should not matter (you will of course get more twist with the longer pin = before it shears). Now take a 0.286-in.  dia. tuning pin that = is 2-in. long. Let's say it has a shear strength of 350 inch-pounds. Do the = same things to it and it will shear at a torque of 350 = inch-pounds. Now take a pin with a bottom of 0.286-in. dia. and a top of 0.276-in. = dia. Put it in that same nasty Baldwin block - or a strong vice - or = whatever - just so it doesn't move - at it will shear at a torque of 300 inch-pounds. Now take a pin with a bottom of 2-in. dia. and a = top of 0.276-in. dia. Put it in that same nasty Baldwin block - or a = strong vice - or whatever - just so it doesn't move - at it will shear at a = torque of 300 inch-pounds. The larger base would act just like the pinblock = with the constant diameter 0.276-in pin in it. They would both shear at 300 = inch-pounds. Or so it would seem to me. Concentrating = shear forces? How does it do that? Terry Farrell <SPAN id=__#Ath#SignaturePos__></SPAN> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style="FONT: 10pt arial">----- Original Message = -----</DIV> <DIV style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: = black"><B>From:</B> <A title=larudee@pacbell.net = href="mailto:larudee@pacbell.net">larudee@pacbell.net</A></DIV> <DIV style="FONT: 10pt arial"><B>To:</B> <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">pianotech@ptg.org</A></DIV> <DIV style="FONT: 10pt arial"><B>Sent:</B> Sunday, January 27, = 2002 11:27 AM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Tuning Pin Size</DIV> Terry, <P>All of what you mention affects shearing, but the bottom = portion also affects it by concentrating the shear forces at the point where = the diameter changes.  In other words, the greater the = difference, the more torsion and flex will end at that point and the less those = forces will be distributed thoughout the pin. <P>Paul <P>Farrell wrote: <BLOCKQUOTE TYPE="CITE"> "The larger the size difference = between the two portions, the greater the risk." Why would that be? I = should think the point at which a pin would shear would depend = entirely on the metal composition (let's assume this is constant), its = diameter, and the tightness of the pin/block fit (torque). As you make = any pin size fit tighter in the block, it will get closer to its shear = point. As you make any pin smaller in diameter, you will move toward = a lower shear point. Diameter and torque - I think that is all. Why = would the diameter contrast between the top and bottom portion affect = its shear strength? Is there something about the machining process? Or = do you mean (by the above quote): 'The smaller the diameter of the = top portion of the pin, the greater the risk of shearing' = (because, of course, the smaller diameter pin will have a lower shear = strength, and will shear at a lower pin torque). How would the diameter of = the bottom portion of the pin affect the shear strength? I am = assuming that the rebuilder will drill/ream/whatever the hole to a = proper diameter for the diameter of the pin bottom portion. Terry Farrell <SPAN = id=__#Ath#SignaturePos__></SPAN></BLOCKQUOTE></BLOCKQUOTE></BLOCKQUOTE>= </BLOCKQUOTE></BLOCKQUOTE></BODY></HTML>