<!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 6.00.2800.1276" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY style="FONT-SIZE: 14pt; FONT-FAMILY: Arial; BACKGROUND-COLOR: = #ffffff" bgColor=#ffffff> <DIV><FONT size=2>Hi guys:</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV> <DIV><FONT face=Arial size=2>Before you all get too carried away, = here is some food for thought.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>First of all, forget about = momentum.  Momentum (and, once again, we need to think in terms of <EM>angular</EM> = 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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>As I mentioned before, the kinetic = energy of the hammer is dependent upon its <EM>rotational speed only </EM>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.  </FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>From these two things, the center of = percussion can be calculated.  It's just </FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>q = k^2 / r</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>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.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Hope this helps!</FONT></DIV> <DIV><FONT size=2></FONT> </DIV> <DIV><FONT size=2>Don A. Gilmore<BR>Mechanical Engineer<BR>Kansas City</FONT></DIV></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=ANRPiano@aol.com = href="mailto:ANRPiano@aol.com">ANRPiano@aol.com</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> Friday, December 19, 2003 = 7:18 AM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Cockeyed hammers / = Don Gilmore</DIV> <DIV><BR></DIV> <DIV><FONT size=4>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?</FONT></DIV> <DIV><FONT size=4></FONT> </DIV> <DIV><FONT size=4>Except for the last one, all would involve other = significant changes to make the hammer work, so please elaborate, I am very curious.</FONT></DIV> <DIV><FONT size=4></FONT> </DIV> <DIV> </DIV> <DIV> </DIV> <DIV><FONT lang=0 face=Arial size=4 PTSIZE="14" = FAMILY="SANSSERIF">Andrew Remillard<BR>ANRPiano.com<BR>ANR Piano Service<BR>2417 Maple Ave = <BR>Downers Grove, IL = 60515<BR>630-852-5058</FONT></DIV></BLOCKQUOTE></BODY></HTML>