<!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 bgColor=#ffffff background=""> <DIV><FONT face=Arial size=2>Gentlemen:</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>I've been following this thread for a = while and the physics is getting a little fast and loose, though quite interesting!</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>First of all, the kinetic energy of the = hammer will <EM>not</EM> all be transmitted to the string.  In fact only a = small portion will be.  When you bounce a ball on the ground all of the = balls energy is not transmitted to the ground.  The hammer will bounce = back, retaining much of its energy, which is subsequently dissipated into the = back check.  For it to have transmitted all of its energy it would have = to come to a dead stop on the string and fall back naturally with gravity only, = which is not the case.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Secondly, all that matters is the = angular velocity of the hammer when it is released by the mechanism to travel = freely.  Albeit this is only for a small distance, it still is no longer under = the influence of any outside force by the time it reaches the = string(s).  So its kinetic energy (or, what is more useful in this case, its momentum) = is preordained by the time it strikes the string.  No matter how its = angular velocity is arrived at through the complex motions of the action, it is = all that matters once the hammer is free.  Think of it like the muzzle = velocity of a gun: though very complex things go on in the barrel of the gun, the = whole thing becomes a relatively simple ballistics problem once it leaves = the muzzle.  Angular velocity is everything here.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Thirdly, the dynamic motion of the = hammer has been described herein as a linear problem, which it is not.  The = kinetic energy of the hammer is <EM>not </EM>half the mass multiplied by the = square of the velocity.  It would be half the moment of inertia = (<EM>mass</EM> moment of inertia) of the hammer and shank multiplied by the square of the <EM>angular</EM> velocity (in radians per second).  The difference = between this quantity right before the string is contacted minus the quantity = after it leaves the string (neglecting gravity) is the total loss of = energy.  Some goes into vibrating the string and some goes into deforming the = felt, etc. (mostly dissipated as heat).  Ostensibly this is proportionate to = the energy of the hammer before impact.  </FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Thus the energy added to the vibrating = string(s) is only reliant upon the hammer's angular velocity...period.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Don A. Gilmore</FONT></DIV> <DIV><FONT face=Arial size=2>Mechanical Engineer</FONT></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=mark.davidson@mindspring.com href="mailto:mark.davidson@mindspring.com">Mark Davidson</A> </DIV> <DIV style="FONT: 10pt arial"><B>To:</B> <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">Pianotech</A> </DIV> <DIV style="FONT: 10pt arial"><B>Sent:</B> Wednesday, December 17, = 2003 5:52 PM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Key Inertia</DIV> <DIV><BR></DIV> <DIV><FONT face="Courier New" size=2>John Hartmann = wrote:</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>>The points = are:<BR><BR>>1) The only force that gets transferred to the string is that portion of = <BR>>the force applied at the key that works to accelerate the hammer and shank.<BR></FONT></DIV> <DIV><FONT face="Courier New" size=2>Agree.  But note that = not all of the hammer's energy is transferred to the string. </FONT><FONT = face="Courier New" size=2>The hammer bounces off and is still moving, so it still has = energy, which is </FONT><FONT face="Courier New" size=2>also wasted.  = But I think you mean that none of the key or wippen energy makes it to = </FONT><FONT face="Courier New" size=2>the string, with which I = agree.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT><BR><FONT = face="Courier New" size=2>>2) The portion of the force applied to the key used to = accelerate the <BR>>key and wippen do not go into into the string but are wasted.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>Agree.<BR><BR>>3) Removing = mass from the key and wippen will make the action more = <BR>>efficient.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>Hmm.  Starts to get = fuzzy here.  If you keep the same speed,</FONT><FONT face="Courier = New" size=2> so less energy is wasted, then I agree. </FONT><FONT face="Courier = New" size=2>But if you remove mass and play the key with the same force, = you will have </FONT><FONT face="Courier New" size=2>more speed, not the = same speed.<BR><BR>>4) Given our ideal action with no bending and = friction losses changing <BR>>the hammer mass does not effect the efficiency = of the action.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>Same problem as 3.  If = you don't change the mass of hammer and wippen, and </FONT><FONT face="Courier = New" size=2>keep the same key speed, you will waste the same amount of energy.  But again, </FONT><FONT face="Courier New" size=2>if = you add mass to the hammer and keep the same force on the key, </FONT><FONT face="Courier New" size=2>you will have less acceleration, less = speed, less key/wippen energy wasted.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>So I guess I agree with 1 and = 2, but for 3 and 4 we need to define "efficiency" before </FONT><FONT = face="Courier New" size=2>we can talk about whether it stays the same or = not.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>-Mark</FONT></DIV></BLOCKQUOTE></BODY></HTML>