<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <META content="MSHTML 6.00.2900.2668" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY> <DIV><FONT face=Arial size=2>Hi David,</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Here it is, step by step...  But = then after all this mess, I managed to simplify it further, so read on down a bit....</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>First, convert that 15mm spread to = inches, since your tool is in inches:</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>15 mm / 25.4 mm/in = .59 = in</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Next, plug in:</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>A=downbearing angle<BR>  = 1 = deg</FONT></DIV> <DIV><FONT face=Arial size=2>d=deflection (gauge = reading)</FONT></DIV> <DIV><FONT face=Arial size=2>  = unknown (to be = determined)</FONT></DIV> <DIV><FONT face=Arial size=2>b=bridge width from front pin to back = pin</FONT></DIV> <DIV><FONT face=Arial size=2>  = .59 in<BR></FONT></DIV> <DIV><FONT face=Arial size=2>A=2 Arcsin(d/(1-b/2)),<BR>1=2 Arcsin(d/(1-.59/2))      (substitute)</FONT></DIV> <DIV><FONT face=Arial size=2>1 = 2 = Arcsin(d/(1-.295))     (simplify)</FONT></DIV> <DIV><FONT face=Arial size=2>1 = 2 Arcsin(d/.705)          (simplify)</FONT></DIV> <DIV><FONT face=Arial size=2>1/2 = Arcsin(d/.705)          = (divide both sides by two)</FONT></DIV> <DIV><FONT face=Arial size=2>.5 = Arcsin(d/.705)         &nbsp= ;  (simplify)</FONT></DIV> <DIV><FONT face=Arial size=2>sin(.5) = d/.705           &= nbsp;     (take the sine of both sides)</FONT></DIV> <DIV><FONT face=Arial size=2>.00873 = d/.705           &= nbsp;    (simplify)</FONT></DIV> <DIV><FONT face=Arial size=2>.00873*.705 = d            =    (multiply both sides by .705  (Mark, I think you divided by = accident!  I made the same mistake the first time through but caught the error.)</FONT></DIV> <DIV><FONT face=Arial size=2>d = .00615 in           &nbsp= ;        (simplify)</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>This is a bit less than the required = deflection of .009" in the case of a bridge of zero width.  If you think = about it, this makes sense.  The wider the bridge is, the greater an angle = you will create with a given amount of deflection.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Extrapolating from here, you can safely = say that a deflection of .01230 means 2 deg, a deflection of .01845 means 3 deg, = etc. -- again, up until a point, where the sine is no longer approximately equal = to the arc.  This relationship of course changes with a bridge of a = different width.  Then you'll need to pull out the calculator again and do = what I did above.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Of course I've overly complicated this = thing, as usual.  Using this same assumption, we can simplify the whole mess = even further, removing all apparent traces of trigonometry.  Try this:</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>A = angle of deflection</FONT></DIV> <DIV><FONT face=Arial size=2>d = deflection</FONT></DIV> <DIV><FONT face=Arial size=2>b = bridge width (in = inches)</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial color=#ff0000 size=2><STRONG>A = d / = (.00873 * (1-b/2))</STRONG></FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>or if you know b and A and want to = calculate d...</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial color=#ff0000 size=2><STRONG>d = A * = .00873 * (1-b/2)</STRONG></FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Using this form, I get...</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV> <DIV> <DIV><FONT face=Arial size=2>d = 1 * .00873 * = (1-.59/2)</FONT></DIV></DIV> <DIV><FONT face=Arial size=2>d = .00873 * .705</FONT></DIV> <DIV><FONT face=Arial size=2>d = .00615    (same answer)</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV></DIV> <DIV><FONT face=Arial size=2>Mark stated an assumption that the = front and rear angles had to be symmetrical, but we're already assuming that the sine = equals the arc.  Without getting into a trigometric discussion, basically = it doesn't matter any more whether the angles are symmetrical than it = matters that the sine doesn't really equal the arc.  Close enough.  I = wouldn't worry about the symmetry of the angles.<BR></FONT></DIV> <DIV><FONT face=Arial size=2>Hope this helps!  :-)</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Peace,</FONT></DIV> <DIV><FONT face=Arial size=2>Sarah</FONT></DIV> <DIV><FONT face=Arial size=2><IMG alt="" hspace=0 src="cid:009001c581e8$d51d9a50$0202a8c0@SarahDell4600" = align=baseline border=0>  </FONT></DIV> <DIV><FONT face=Arial size=2><grin></FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2> </DIV></FONT> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>----- Original Message ----- </FONT> <DIV><FONT face=Arial size=2>From: "David Love" <</FONT><A href="mailto:davidlovepianos@comcast.net"><FONT face=Arial size=2>davidlovepianos@comcast.net</FONT></A><FONT face=Arial size=2>></FONT></DIV> <DIV><FONT face=Arial size=2>To: "'Pianotech'" <</FONT><A href="mailto:pianotech@ptg.org"><FONT face=Arial size=2>pianotech@ptg.org</FONT></A><FONT face=Arial = size=2>></FONT></DIV> <DIV><FONT face=Arial size=2>Sent: Tuesday, July 05, 2005 8:37 = PM</FONT></DIV> <DIV><FONT face=Arial size=2>Subject: RE: Reading a dial gauge</FONT></DIV></DIV> <DIV><FONT face=Arial><BR><FONT size=2></FONT></FONT></DIV><FONT = face=Arial size=2>> Let's simplify a bit more.  Take the scenario below = and let's say there's a<BR>> 15mm spread between the front and rear bridge = pins.  What would the reading<BR>> need to be to indicate a total bearing = (front plus rear) of 1 degree.  I<BR>> just need a baseline.  I = use this for checking bearing on a strung piano at<BR>> tension.  A = couple mm difference in the bridge pin spacing isn't that<BR>> critical for = this operation.  <BR>> <BR>> David Love<BR>> </FONT><A href="mailto:davidlovepianos@comcast.net"><FONT face=Arial size=2>davidlovepianos@comcast.net</FONT></A><FONT face=Arial = size=2> <BR>> <BR>> -----Original Message-----<BR>> From: </FONT><A href="mailto:pianotech-bounces@ptg.org"><FONT face=Arial size=2>pianotech-bounces@ptg.org</FONT></A><FONT face=Arial = size=2> [mailto:pianotech-bounces@ptg.org] On Behalf<BR>> Of David = Love<BR>> Sent: Tuesday, July 05, 2005 5:33 PM<BR>> To: 'Pianotech'<BR>> Subject: = RE: Reading a dial gauge<BR>> <BR>> Let's simplify.  I zero the = three prongs on a level surface.  I place the<BR>> center plunger on = top of the string in the center of the bridge.  The two<BR>> outer = prongs are resting on either side of the bridge.  The dial reads .018.<BR>> = Do I have 1 degree of overall bearing?  And if it reads .036, I assume = I<BR>> have 2 degrees.  Do I have that right?<BR>> <BR>> David = Love<BR>> </FONT><A href="mailto:davidlovepianos@comcast.net"><FONT face=Arial = size=2>davidlovepianos@comcast.net</FONT></A><FONT face=Arial = size=2> <BR>> <BR>> -----Original Message-----<BR>> From: </FONT><A href="mailto:pianotech-bounces@ptg.org"><FONT face=Arial size=2>pianotech-bounces@ptg.org</FONT></A><FONT face=Arial = size=2> [mailto:pianotech-bounces@ptg.org] On Behalf<BR>> Of Sarah = Fox<BR>> Sent: Tuesday, July 05, 2005 4:22 PM<BR>> To: Pianotech<BR>> Subject: = Re: Reading a dial gauge<BR>> <BR>> Hi David,<BR>> <BR>> That's = a great question for a math nerd.  ;-)<BR>> <BR>> What angle are you = trying to measure?  Backscale segment to speaking <BR>> segment?  Backscale to bridge and speaking to bridge?  I assume you're = <BR>> putting the plunger at the center of the bridge.  Yes??<BR>> = <BR>> The sine of 1 deg is roughly .018.  That reading would correspond = to the <BR>> angle from "horizontal."  Then double the angle to go the = other way.  In <BR>> other words, the total angle between speaking and = backscale would be 2 deg. <BR>> A 1 deg angle would give you a = reading of .009.  This of course assumes that<BR>> <BR>> your bridge has = no width to it, like a violin's.  That's a very bad <BR>> = assumption, of course.<BR>> <BR>> The distance between the front and read bridge = pins is an important factor. <BR>> If you measure across the bridge, with the = plunger at the center of the <BR>> bridge (which is how you do it??), then = you would estimate the total <BR>> downbearing angle (between speaking and = backscale segments as...<BR>> <BR>> A=2 Arcsin(d/(1-b/2)),<BR>> = <BR>> where<BR>> <BR>> A=downbearing angle<BR>> d=deflection = (gauge reading)<BR>> b=bridge width from front pin to back pin<BR>> = <BR>> Downbearing pressure would be estimated as...<BR>> <BR>> P=2Td/(1-b/2),<BR>> <BR>> where<BR>> <BR>> P=downbearing = pressure<BR>> T=string tension<BR>> <BR>> Of course these = formulas assume an angle near zero and are only <BR>> approximately correct, = but they work very well for the first several <BR>> degrees.  They also = assume a sharp angle in the wire, with no arching across<BR>> <BR>> the bridge.<BR>> <BR>> Peace,<BR>> Sarah<BR>> <BR>> <BR>> = ----- Original Message ----- <BR>> From: "David Love" <</FONT><A href="mailto:davidlovepianos@comcast.net"><FONT face=Arial size=2>davidlovepianos@comcast.net</FONT></A><FONT face=Arial size=2>><BR>> To: "'Pianotech'" <</FONT><A href="mailto:pianotech@ptg.org"><FONT face=Arial size=2>pianotech@ptg.org</FONT></A><FONT face=Arial = size=2>><BR>> Sent: Tuesday, July 05, 2005 6:03 PM<BR>> Subject: Reading a dial = gauge<BR>> <BR>> <BR>> <BR>> So I picked up this dial gauge from Schaff = and I'm not really sure how it<BR>> works.  The center prong is the = plunger and the two outer prongs are 1" from<BR>> the center.  The smallest increments that can be read are .001".  So if I<BR>> put the = plunger on the center of the bridge with the two outer prongs<BR>> resting on = the front and back segments respectively, does a reading of .018"<BR>> equal 1 degree?  Or would that be 1/2 a degree since the total reading = must<BR>> be split between the two prongs at a total of 2" apart?<BR>> <BR>> = Yeah, I shoulda paid more attention in trig class.<BR>> <BR>> David = Love<BR>> </FONT><A href="mailto:davidlovepianos@comcast.net"><FONT face=Arial = size=2>davidlovepianos@comcast.net</FONT></A><BR><FONT face=Arial = size=2>> <BR>> <BR>> <BR>> 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