<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; = charset=us-ascii"> <meta name=Generator content="Microsoft Word 10 (filtered)"> <style>  </style> </head> <body lang=EN-US link=blue vlink=blue> <div class=Section1> Very thorough—and much more = than I needed.  Thanks.  I wish I had my Lowell back.  = Anyway,  that does help and perhaps this dial gauge will turn out to be a much = easier way to tweak the bearing as you can set it on top of the bridge while = you turn nose bolts up and down and/or change the height of the set screws on the modified perimeter mountings.    Much appreciated.  = <div> David Love davidlovepianos@comcast.net<= /font> = </div> -----Original = Message----- From: = pianotech-bounces@ptg.org [mailto:pianotech-bounces@ptg.org] On Behalf Of Sarah Fox Sent: Tuesday, July 05, = 2005 10:09 PM To: Pianotech<fon= t size=2 face=Tahoma> Subject: Re: Reading a = dial gauge   <div> Hi David, </div> <div>   </div> <div> Here it is, step by = step...  But then after all this mess, I managed to simplify it further, so read = on down a bit.... </div> <div>   </div> <div> First, convert that 15mm = spread to inches, since your tool is in inches: </div> <div>   </div> <div> 15 mm / 25.4 mm/in = .59 = in </div> <div>   </div> <div> Next, plug = in: </div> <div>   </div> <div> A=downbearing angle   = 1 deg </div> <div> d=deflection (gauge = reading) </div> <div>   = unknown (to be = determined) </div> <div> b=bridge width from front = pin to back pin </div> <div>   = .59 = in </div> <div> A=2 = Arcsin(d/(1-b/2)), 1=2 Arcsin(d/(1-.59/2))      = (substitute) </div> <div> 1 = 2 Arcsin(d/(1-.295))     (simplify) </div> <div> 1 = 2 = Arcsin(d/.705)          (simplify) </div> <div> 1/2 = Arcsin(d/.705)          = (divide both sides by two) </div> <div> .5 = Arcsin(d/.705)         &nbsp= ;  (simplify) </div> <div> sin(.5) = d/.705           &= nbsp;     (take the sine of both sides) </div> <div> .00873 = d/.705           &= nbsp;    (simplify) </div> <div> .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.) </div> <div> d = .00615 in           &nbsp= ;        (simplify) </div> <div>   </div> <div> 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. </div> <div>   </div> <div> 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. </div> <div>   </div> <div> 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: </div> <div>   </div> <div> A = angle of = deflection </div> <div> d = = deflection </div> <div> b = bridge width (in = inches) </div> <div>   </div> <div> A = d / (.00873 * (1-b/2)) </div> <div>   </div> <div> or if you know b and A and = want to calculate d... </div> <div>   </div> <div> d = A * .00873 * (1-b/2) </div> <div>   </div> <div> Using this form, I = get... </div> <div>   </div> <div> <div> <div> d = 1 * .00873 * = (1-.59/2) </div> </div> <div> d = .00873 * = .705 </div> <div> d = = .00615    (same answer) </div> <div>   </div> </div> <div> 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. </div> <div> Hope this helps!  = :-) </div> <div>   </div> <div> Peace, </div> <div> Sarah </div> <div> <img width=110 = height=53 src="cid:image001.jpg@01C581B5.0A179AF0" align=baseline = border=0>  </div> <div> <grin></= p> </div> <div>   </div> <div>   </div> <div>   </div> <div> ----- Original Message = ----- <div> From: "David = Love" <<a href="mailto:davidlovepianos@comcast.net">davidlovepianos@comcast.net<= /span></a>> </div> <div> To: "'Pianotech'" = <<a href="mailto:pianotech@ptg.org">pianotech@ptg.org</a>> </div> <div> Sent: Tuesday, July 05, = 2005 8:37 PM </div> <div> Subject: RE: Reading a dial = gauge </div> </div> <div>   </div> > Let's simplify a bit more.  Take the scenario below and let's say there's a > 15mm spread between the front and rear bridge pins.  What = would the reading > need to be to indicate a total bearing (front plus rear) of 1 degree.  I > just need a baseline.  I use this for checking bearing on a = strung piano at > tension.  A couple mm difference in the bridge pin spacing = isn't that > critical for this operation.  > > David Love > <a href="mailto:davidlovepianos@comcast.net">davidlovepianos@comcast.net<= /span></a> > > -----Original Message----- > From: <a = href="mailto:pianotech-bounces@ptg.org">pianotech-bounces@ptg.org</s= pan></a> [mailto:pianotech-bounces@ptg.org] On Behalf > Of David Love > Sent: Tuesday, July 05, 2005 5:33 PM > To: 'Pianotech' > Subject: RE: Reading a dial gauge > > Let's simplify.  I zero the three prongs on a level = surface.  I place the > center plunger on top of the string in the center of the = bridge.  The two > outer prongs are resting on either side of the bridge.  The = dial reads .018. > Do I have 1 degree of overall bearing?  And if it reads .036, = I assume I > have 2 degrees.  Do I have that right? > > David Love > <a href="mailto:davidlovepianos@comcast.net">davidlovepianos@comcast.net<= /span></a> > > -----Original Message----- > From: <a = href="mailto:pianotech-bounces@ptg.org">pianotech-bounces@ptg.org</s= pan></a> [mailto:pianotech-bounces@ptg.org] On Behalf > Of Sarah Fox > Sent: Tuesday, July 05, 2005 4:22 PM > To: Pianotech > Subject: Re: Reading a dial gauge > > Hi David, > > That's a great question for a math nerd.  ;-) > > What angle are you trying to measure?  Backscale segment to = speaking > segment?  Backscale to bridge and speaking to bridge?  I = assume you're > putting the plunger at the center of the bridge.  Yes?? > > The sine of 1 deg is roughly .018.  That reading would = correspond to the > angle from "horizontal."  Then double the angle to = go the other way.  In > other words, the total angle between speaking and backscale would = be 2 deg. > A 1 deg angle would give you a reading of .009.  This of = course assumes that > > your bridge has no width to it, like a violin's.  That's a = very bad > assumption, of course. > > The distance between the front and read bridge pins is an important factor. > If you measure across the bridge, with the plunger at the center of = the > bridge (which is how you do it??), then you would estimate the = total > downbearing angle (between speaking and backscale segments = as... > > A=2 Arcsin(d/(1-b/2)), > > where > > A=downbearing angle > d=deflection (gauge reading) > b=bridge width from front pin to back pin > > Downbearing pressure would be estimated as... > > P=2Td/(1-b/2), > > where > > P=downbearing pressure > T=string tension > > Of course these formulas assume an angle near zero and are only = > approximately correct, but they work very well for the first = several > degrees.  They also assume a sharp angle in the wire, with no = arching across > > the bridge. > > Peace, > Sarah > > > ----- Original Message ----- > From: "David Love" <<a href="mailto:davidlovepianos@comcast.net">davidlovepianos@comcast.net<= /span></a>> > To: "'Pianotech'" <<a href="mailto:pianotech@ptg.org">pianotech@ptg.org</a>> > Sent: Tuesday, July 05, 2005 6:03 PM > Subject: Reading a dial gauge > > > > So I picked up this dial gauge from Schaff and I'm not really sure = how it > works.  The center prong is the plunger and the two outer = prongs are 1" from > the center.  The smallest increments that can be read are .001".  So if I > put the plunger on the center of the bridge with the two outer = prongs > resting on the front and back segments respectively, does a reading = of .018" > equal 1 degree?  Or would that be 1/2 a degree since the total reading must > be split between the two prongs at a total of 2" apart? > > Yeah, I shoulda paid more attention in trig class. > > David Love > <a href="mailto:davidlovepianos@comcast.net">davidlovepianos@comcast.net<= /span></a> > > > > = -------------------------------------------------------------------------= --- > ---- > > >> _______________________________________________ >> pianotech list info: <a href="https://www.moypiano.com/resources/#archives">http://www.ptg.org/mailman/l= istinfo/pianotech</a> >> > > > _______________________________________________ > pianotech list info: <a href="https://www.moypiano.com/resources/#archives">http://www.ptg.org/mailman/l= istinfo/pianotech</a> > > > _______________________________________________ > pianotech list info: <a href="https://www.moypiano.com/resources/#archives">http://www.ptg.org/mailman/l= istinfo/pianotech</a> > > > _______________________________________________ > pianotech list info: <a href="https://www.moypiano.com/resources/#archives">http://www.ptg.org/mailman/l= istinfo/pianotech</a> > > </div> </body> </html>