Moment of Inertia of grand action parts.

[Don A. Gilmore]

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