Moment of Inertia of grand action parts.

[Mark Davidson]

Ron N wrote:

>Except it doesn't work that way in a real action. The peak velocity of a
>hammer in a real action is limited by the compliance of the parts, not the
>input impulse. That is primarily, but not entirely, key flex.

Not disagreeing with this, but I don't think we've necessarily been talking
about "peak" velocity, just velocity in general.  Somewhere between ppp,
where BW and friction are the main issues, and fff, where action saturation
occurs, is a whole range of other dynamics that are very much affected by
these discussions of inertia.  And that's the range where people do most of
their playing.

>Gravity only counts in static balance measurements.

Gravity is a constant force, whether you're moving or not.  And the force
between two objects does change when there is vertical acceleration, like
standing in an elevator when it starts to go down, or starts going up.  If
the elevator were to fall at the rate of gravitational acceleration you will
cease to exert a force on the floor of the elevator.  If it falls faster
than the rate of gravitational acceleration due to some other force on the
elevator, then the top of the elevator will start to push on you.

Now key leads are in a rotational system, which isn't as simple as an
elevator.  There are horizontal and rotational forces as well as vertical.
However there is still some rate of angular acceleration of the key at which
the same finger force on the key will produce the same angular acceleration.

We can solve for that force as follows.

Using the formula I recently posted, we have

(1) KFA = (FF-DW) * L^2 / I

KFA = key front acceleration
FF = finger force
DW = downweight
L = front key radius
I = moment of inertia

If we add a lead with mass M at radius L this changes to

(2) KFA = (FF-DW+M) * L^2 / (I + M*L^2)

We want the same key front acceleration, (KFA same), so set (1) equal to
(2):

(FF-DW) * L^2 / I = (FF-DW+M) * L^2 / (I + M*L^2)

With a bit of algebra, you can solve for FF, the force that gives the same
acceleration with and without keylead of mass M at radius L.

FF = DW + I / L^2

Not sure how useful that is, but maybe we can put the question to bed.

Conceptually, adding a lead decreases BW but increases inertia.  So it makes
sense that you can get the key moving with less force, but at some point you
will have to exert more force for the same acceleration.  The crossover
point is what I just solved for.

-Mark