This is a multi-part message in MIME format. ---------------------- multipart/related attachment ------=_NextPart_001_0092_01C581C7.4E0E4440 Hi David, Here it is, step by step... But then after all this mess, I managed to = simplify it further, so read on down a bit.... First, convert that 15mm spread to inches, since your tool is in inches: 15 mm / 25.4 mm/in =3D .59 in Next, plug in: A=3Ddownbearing angle =3D 1 deg d=3Ddeflection (gauge reading) =3D unknown (to be determined) b=3Dbridge width from front pin to back pin =3D .59 in A=3D2 Arcsin(d/(1-b/2)), 1=3D2 Arcsin(d/(1-.59/2)) (substitute) 1 =3D 2 Arcsin(d/(1-.295)) (simplify) 1 =3D 2 Arcsin(d/.705) (simplify) 1/2 =3D Arcsin(d/.705) (divide both sides by two) .5 =3D Arcsin(d/.705) (simplify) sin(.5) =3D d/.705 (take the sine of both sides) .00873 =3D d/.705 (simplify) .00873*.705 =3D 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.) d =3D .00615 in (simplify) 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. 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. 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: A =3D angle of deflection d =3D deflection b =3D bridge width (in inches) A =3D d / (.00873 * (1-b/2)) or if you know b and A and want to calculate d... d =3D A * .00873 * (1-b/2) Using this form, I get... d =3D 1 * .00873 * (1-.59/2) d =3D .00873 * .705 d =3D .00615 (same answer) 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. Hope this helps! :-) Peace, Sarah =20 <grin> ----- Original Message -----=20 From: "David Love" <davidlovepianos@comcast.net> To: "'Pianotech'" <pianotech@ptg.org> Sent: Tuesday, July 05, 2005 8:37 PM Subject: RE: Reading a dial gauge > 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. =20 >=20 > David Love > davidlovepianos@comcast.net=20 >=20 > -----Original Message----- > From: pianotech-bounces@ptg.org [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 >=20 > 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? >=20 > David Love > davidlovepianos@comcast.net=20 >=20 > -----Original Message----- > From: pianotech-bounces@ptg.org [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 >=20 > Hi David, >=20 > That's a great question for a math nerd. ;-) >=20 > What angle are you trying to measure? Backscale segment to speaking=20 > segment? Backscale to bridge and speaking to bridge? I assume you're = > putting the plunger at the center of the bridge. Yes?? >=20 > The sine of 1 deg is roughly .018. That reading would correspond to = the=20 > angle from "horizontal." Then double the angle to go the other way. = In=20 > other words, the total angle between speaking and backscale would be 2 = deg.=20 > A 1 deg angle would give you a reading of .009. This of course = assumes that >=20 > your bridge has no width to it, like a violin's. That's a very bad=20 > assumption, of course. >=20 > The distance between the front and read bridge pins is an important = factor.=20 > If you measure across the bridge, with the plunger at the center of = the=20 > bridge (which is how you do it??), then you would estimate the total=20 > downbearing angle (between speaking and backscale segments as... >=20 > A=3D2 Arcsin(d/(1-b/2)), >=20 > where >=20 > A=3Ddownbearing angle > d=3Ddeflection (gauge reading) > b=3Dbridge width from front pin to back pin >=20 > Downbearing pressure would be estimated as... >=20 > P=3D2Td/(1-b/2), >=20 > where >=20 > P=3Ddownbearing pressure > T=3Dstring tension >=20 > Of course these formulas assume an angle near zero and are only=20 > approximately correct, but they work very well for the first several=20 > degrees. They also assume a sharp angle in the wire, with no arching = across >=20 > the bridge. >=20 > Peace, > Sarah >=20 >=20 > ----- Original Message -----=20 > From: "David Love" <davidlovepianos@comcast.net> > To: "'Pianotech'" <pianotech@ptg.org> > Sent: Tuesday, July 05, 2005 6:03 PM > Subject: Reading a dial gauge >=20 >=20 >=20 > 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? >=20 > Yeah, I shoulda paid more attention in trig class. >=20 > David Love > davidlovepianos@comcast.net >=20 >=20 >=20 > = -------------------------------------------------------------------------= --- > ---- >=20 >=20 >> _______________________________________________ >> pianotech list info: https://www.moypiano.com/resources/#archives >>=20 >=20 >=20 > _______________________________________________ > pianotech list info: https://www.moypiano.com/resources/#archives >=20 >=20 > _______________________________________________ > pianotech list info: https://www.moypiano.com/resources/#archives >=20 >=20 > _______________________________________________ > pianotech list info: https://www.moypiano.com/resources/#archives >=20 > ------=_NextPart_001_0092_01C581C7.4E0E4440 An HTML attachment was scrubbed... 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