Moment of Inertia of grand action parts.

[Don A. Gilmore]

Oops.  I just realized a major flaw in my reasoning.  Other components of
the action are also moving right before the key bottoms out.  They would add
to the total momentum of the system as well.  So much for simple.

Let me give this more than two seconds of thought and I'll see what I can
come up with.  Sorry for the false alarm.

Don A. Gilmore
Mechanical Engineer
Kansas City

----- Original Message ----- 
From: "Don A. Gilmore" <eromlignod@kc.rr.com>
To: "Pianotech" <pianotech@ptg.org>; "Mark Davidson"
<mark.davidson@mindspring.com>
Sent: Sunday, December 28, 2003 10:21 AM
Subject: Re: Moment of Inertia of grand action parts.


> Gentlemen:
>
> Up to now I have just been trying to straighten out the proper use of
> formulas.  Ironically, I wasn't even sure what the final goal was.  If you
> want ro relate the force applied to a key to what velocity will be
produced
> in the hammer when it is released, it may be easier than you think.
>
> If we first consider the simplified case of a constant force on the key
then
> I think the concept of impulse/momentum might be useful.  A constant force
> on the key would be the case if you laid a weight on it.  The force would
be
> constant throughout the stroke.  The "impulse" given to the mechanism
would
> be the weight (force) multiplied by the time in seconds it takes to strike
> the stop at the bottom (forgive me if I don't know all the terms you guys
> use for these things).  The impulse at the key is equal to the change in
> momentum of the system.
>
> The interesting thing is that momentum is conserved regardless of the
masses
> of individual mechanism parts or friction.  You could have any Rube
Goldberg
> contraption between the key and the hammer and the relation between
impulse
> and momentum would be the same.
>
> Now that I have everyone excited, how easy is this to do?  You need to be
> able to accurately time the fall of the weight.  You also have to have a
way
> to determine the velocity of the weight just before it hits bottom, though
> we might be able to get away with assuming that the acceleration is
constant
> and approximate a final velocity.  And you need to know the m.o.i. of the
> hammer, as usual.
>
> Here's how it might go.  We put a 5 lb weight on the key and time how long
> it takes to bottom out.  We determine the impulse by multiplying this time
> by the weight, W
>
> Implulse = Wt
>
> Then if we assume constant acceleration we can get the terminal velocity
by
>
> v = 2h / t
>
> where h is the vertical distance we travel.  From this we can calculate
the
> momentum of the weight.
>
> momentum = mv = Wv / g
>
> where g is the acceleration of gravity (we're just converting pounds to
> slugs).
>
> OK, we know that impulse is equal to the change in momentum and the
momentum
> was zero before we started, so whatever Ft comes out to be will be equal
to
> the momentum of the weight *and* the angular momentum of the hammer added
> together.  Now that we know the weight momentum we can subtract it from
the
> total (Ft) and what's left over should be the momentum of the hammer.
>
> Now we just need the velocity of the hammer.  Angular momentum is
>
> angular momentum = Iw
>
> or moment of inertia times angular velocity.  If we rearrange this
equation
> we have
>
> w = a.m. / I
>
> So if we divide the leftover momentum by the moment of inertia of the
> hammer, we should get its angular velocity at release.
>
> If you wanted to really get fancy, we could also do this if the force is
not
> constant.  If we had a way to graph the force against time (time along the
x
> axis and force on the y axis), the total impulse would be the area under
the
> curve.
>
> Don A. Gilmore
> Mechanical Engineer
> Kansas City
>
> ----- Original Message ----- 
[link redacted at request of site owner - Jul 25, 2015]
> To: "Mark Davidson" <mark.davidson@mindspring.com>; "Pianotech"
> <pianotech@ptg.org>
> Sent: Sunday, December 28, 2003 8:20 AM
> Subject: Re: Moment of Inertia of grand action parts.
>
>
> > Mark,
> >
> > Thanks for sharing this with me. Just yesterday I came to exactly the
> > same conclusion. I am doing a drawing to show the acceleration ratios of
> > the wip and shank in relation to the key. The fact that we both came up
> > with the same formula is encouraging but to be sure we need to have Don
> > go over it.
> >
> > Have you plugged in the MOIs? It looks like the shank and hammer
> > contribute about 12 times or more of the total I as felt at the key. If
> > the formula is right it shows how unimportant changes to the key MOI is
> > in relation to overall efficiency. Also, if there is any benefit to
> > pattern leading it is not to make the action feel even from note to
> > note. Adding lead to the key is not the big evil commonly thought unless
> > it has some effect on repetition.
> >
> > I think we are going to find that the biggest problem with increasing
> > the mass of the action parts is the losses due to bending and
compliance.
> >
> > I still need to complete the kinetic model of the action but I can see
> > ahead to the next step. Maybe you are already there. Is there a way to
> > convert the kinetic forces developed at different levels of play into
> > static loads. Then we can see how these loads bend the shank and key. It
> > would be great if this could lead to a formula for finding the terminal
> > velocity of the hammer.
> >
> > John Hartman RPT
>
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