Key Inertia

[Don A. Gilmore]

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Gentlemen:

I've been following this thread for a while and the physics is getting a =
little fast and loose, though quite interesting!

First of all, the kinetic energy of the hammer will not 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.

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.

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 =
not half the mass multiplied by the square of the velocity.  It would be =
half the moment of inertia (mass moment of inertia) of the hammer and =
shank multiplied by the square of the angular 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. =20

Thus the energy added to the vibrating string(s) is only reliant upon =
the hammer's angular velocity...period.

Don A. Gilmore
Mechanical Engineer
  ----Original Message -----=20
  From: Mark Davidson=20
  To: Pianotech=20
  Sent: Wednesday, December 17, 2003 5:52 PM
  Subject: Key Inertia


  John Hartmann wrote:

  >The points are:

  >1) The only force that gets transferred to the string is that portion =
of=20
  >the force applied at the key that works to accelerate the hammer and =
shank.

  Agree.  But note that not all of the hammer's energy is transferred to =
the string. The hammer bounces off and is still moving, so it still has =
energy, which is also wasted.  But I think you mean that none of the key =
or wippen energy makes it to the string, with which I agree.

  >2) The portion of the force applied to the key used to accelerate the =

  >key and wippen do not go into into the string but are wasted.

  Agree.

  >3) Removing mass from the key and wippen will make the action more=20
  >efficient.

  Hmm.  Starts to get fuzzy here.  If you keep the same speed, so less =
energy is wasted, then I agree. But if you remove mass and play the key =
with the same force, you will have more speed, not the same speed.

  >4) Given our ideal action with no bending and friction losses =
changing=20
  >the hammer mass does not effect the efficiency of the action.

  Same problem as 3.  If you don't change the mass of hammer and wippen, =
and keep the same key speed, you will waste the same amount of energy.  =
But again, if you add mass to the hammer and keep the same force on the =
key, you will have less acceleration, less speed, less key/wippen energy =
wasted.


  So I guess I agree with 1 and 2, but for 3 and 4 we need to define =
"efficiency" before we can talk about whether it stays the same or not.

  -Mark
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