<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html xmlns:v="urn:schemas-microsoft-com:vml" = xmlns:o="urn:schemas-microsoft-com:office:office" = xmlns:w="urn:schemas-microsoft-com:office:word" = xmlns="http://www.w3.org/TR/REC-html40"> <head> <meta http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <meta name=ProgId content=Word.Document> <meta name=Generator content="Microsoft Word 9"> <meta name=Originator content="Microsoft Word 9"> <link rel=File-List href="cid:filelist.xml@01C3C6F6.96A89170">  <style>  </style>  </head> <body bgcolor=white lang=FR link=blue vlink=blue = style='tab-interval:35.4pt'> <div class=Section1> Excuse my question, = but why is not momentum used, while the hammer felt have some hysteresys (very = slow indeed) and the string/mass hammer mass relation is an important = parameter – while we don’t know how to keep it consistent in the treble where = the hammers are too heavy vs/strings.<o:p></o:p> <![if = !supportEmptyParas]> <![endif]><o:p></o:p> Do you mean that = momentum is only useful for tone production ?<o:p></o:p> <![if = !supportEmptyParas]> <![endif]><o:p></o:p> Best = Regards.<o:p></o:p> <![if = !supportEmptyParas]> <![endif]><o:p></o:p> Isaac = OLEG<o:p></o:p> <![if = !supportEmptyParas]> <![endif]><o:p></o:p> <![if = !supportEmptyParas]> <![endif]><o:p></o:p> -----Message d'origine----- De : pianotech-bounces@ptg.org [mailto:pianotech-bounces@ptg.org]De la part de Don A. Gilmore Envoyé : samedi 20 = décembre 2003 00:36 À : Pianotech Objet : Re: Cockeyed = hammers / Don Gilmore <![if = !supportEmptyParas]> <![endif]><o:p></o:p> Hi guys:<o:p></o:p>  <o:p></o:p> Before you all get too carried away, here is some food for = thought.<o:p></o:p>  <o:p></o:p> First of all, forget about momentum.  Momentum (and, once again, we need = to think in terms of angular momentum) is moment of inertia x angular velocity and is in units of slug-ft^2/sec or kg-m^2/s.  It is really only useful in calculating elastic = collisions between objects (like billiard balls, for example) that exhibit "conservation of momentum", or impulse calculations.  = Impulse is only useful if we are worried about constant forces, etc.  You were = all doing just fine with kinetic energy.<o:p></o:p>=  <o:p></o:p> Since the hammer is free from any outside influence between the time it is released by the action and the time it strikes the string, = we are talking about two totally independent things: how the action gets it up = to speed and what happens when it strikes the string.<o:p>= </o:p>  <o:p></o:p> As I mentioned before, the kinetic energy of the hammer is dependent = upon its rotational speed only since its mass does not = change.  No matter what kind of fancy things the action is doing when it is = accelerating the hammer, it all comes down to how fast it is going at the = time of release.  The "die is cast" at that point and you get = what you get.  Relating this to key force is complex, but as far = as energy is concerned, it all boils down to the angular velocity (rpm, = basically) at release.  <o:p></o:p>  <o:p></o:p> Now, how much of this energy is transferred to the string is another = story.  This is all decided by the geometry of the hammer, or more specifically, the relative positions of the center of gravity, the pivot point and the = contact face of the hammer.  There are only two places that can absorb any = energy at impact: at the string and at the pivot.  Obviously if you can reduce forces at the pivot to zero, any transfer of energy = will be to the string(s).  As I mentioned before this would be the case if the strings contacted the hammer at its center of = percussion.<o:p></o:p>  <o:p></o:p> Locating the center of percussion requires determining the center of gravity and = the radius of gyration.  Measuring the c.o.g. is a piece of cake.  = You can just balance the hammer on the edge of a ruler, then scoot it around = a bit and balance it again.  The two lines made by the edge of the ruler = will intersect at the center of gravity.<o:p></o:p>=  <o:p></o:p> The radius of gyration is a little trickier.  The radius of gyration = (k) is an imaginary distance from the pivot point to where the entire hammer = can be considered to act if it were a point mass.  In other words, if all = of the mass of the hammer were concentrated at a single point, k inches from = the pivot, it would have the same rotational behavior.  It is calculated in a composite manner similar to figuring the moment of inertia...in fact it = can be directly converted from the m.o.i. by dividing by the mass and taking = the square root.<o:p></o:p>  <o:p></o:p> From these two things, the center of percussion can be calculated.  It's = just <o:p></o:p>  <o:p></o:p> q = k^2 / r<o:p></o:p>  <o:p></o:p> where q is the distance from the pivot to the c.o.p., k is the radius of = gyration and r is the distance from the pivot to the c.o.g.<o:p>= </o:p>  <o:p></o:p> Now that you know where the point is, you can play around with the = hammer geometry to move it to where you need it.  You can also change the location of the c.o.g. (and thus c.o.p.) by strategically placing = weights, but remember that this will also increase the overall mass of the = hammer.<o:p></o:p>  <o:p></o:p> Hope this helps!<o:p></o:p>  <o:p></o:p> Don A. Gilmore Mechanical Engineer Kansas City<o:p></o:p> <div style='border:none;border-left:solid black 1.5pt;padding:0cm 0cm = 0cm 4.0pt'> ----- Original Message ----- = <o:p></o:p> <p class=MsoNormal = style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto; margin-left:3.75pt;background:#E4E4E4;border:none;mso-border-left-alt:sol= id black 1.5pt; padding:0cm;mso-padding-alt:0cm 0cm 0cm 4.0pt'> <div style='font-color:black'>From: <a href="mailto:ANRPiano@aol.com" title="ANRPiano@aol.com">ANRPiano@aol.com</a></fon= t> </div> <o:p></o:p> To:</= b> = <a href="mailto:pianotech@ptg.org" = title="pianotech@ptg.org">pianotech@ptg.org</a></fo= nt> <o:p></o:p>= Sent:= Friday, December = 19, 2003 7:18 AM<o:p></o:p> Subject:</fo= nt> Re: Cockeyed = hammers / Don Gilmore<o:p></o:p> <![if = !supportEmptyParas]> <![endif]><o:p></o:p> So my fine physics gents, put this = in layman's terms.  Are you suggesting moving the hammer up or down = the shank, changing the bore distance, changing the center of gravity of the hammer? Or all three?<o:p></o:p>  <o:p>= </o:p> Except for the last one, all would = involve other significant changes to make the hammer work, so please elaborate, = I am very curious.<o:p></o:p>  <o:p>= </o:p>  <o:p>= </o:p>  <o:p>= </o:p> Andrew = Remillard ANRPiano.com ANR Piano Service 2417 Maple Ave Downers Grove, IL 60515 630-852-5058<o:p></o:p> </div> </div> </body> </html>