Key Inertia (was Re: adjusting wippen ...)

[Phillip Ford]

Let's do a little energy napkin sketch. =20

Energy of mass m moving at velocity v=20
is (m*v^2)/2.

So if the key front is moving at speed=20
1, then a key lead halfway out is=20
moving at .5, while the hammer=20
moves at approximately 6.

If the hammer mass is 10g and the=20
lead mass is 10g, then the lead=20
energy is=20

(10*.5*.5)/2 =3D 1.25

while the hammer energy is=20

(10*6*6)/2 =3D 180!

That's a ratio of 1:144.
So yes, any energy=20
remaining in the key when it=20
bottoms out is wasted, but by far
most of the energy is in the hammer.
And it would appear that adding=20
a key lead makes less than a 1% change in
the TOTAL inertia (hammer inertia=20
is effectively much larger because=20
of the leverage involved).

Oversimplified - should all be
rotational - and neglects
the inertia of the unleaded key,
but I think it makes the point
that the SW and SWR are far and
away the most important components
of the overall inertia.

For better repetition though, you=20
still want low key inertia.  That would
appear - at least based on this example -=20
to be the chief benefit of assist springs.


-Mark Davidson

---------

You make an interesting point about energy.  I can't agree with you 
about inertia though.  You seem to be implying that key inertia is 
relatively insignificant in the total inertia picture.  This is 
counter to my experience.  I know that I can easily lead up a key so 
that the action is unplayable.  I also know that if I take an action 
that has a ton of lead in the keys, and make some changes to it, 
without reducing SW, so that the amount of lead in the keys has been 
reduced, that it will make a marked difference in the way the action 
plays.  On the other hand I can increase SW rather significantly, and 
while noticeable,  it won't render the action unplayable.

To do some different math let's assume:

SW = 10 g

Hammer CG is 13 cm from its center

WW = 18 g

Wippen CG is 7 cm from its center

Assume SWR of 5 - For a massless key this would imply 50 g of lead 
(or whatever) at the measuring point, which we'll assume is 23 cm 
from the key balance point.

Inertia of hammer about its center  = 10 x 13 x 13 = 1690 g cm^2

Inertia of wippen about its center = 18 x 7 x 7 = 882 g cm^2

Inertia of key lead about its center = 50 x 23 x 23 = 26450 g cm^2

The key doesn't look so insignificant to me.  Admittedly this is a 
severe example - the key itself has no distributed mass and the lead 
is all concentrated out at the end.  But still, if the key inertia is 
insignificant then these should be quibbles.  The hammer assumes high 
velocity from the key input because of all the leverage.  I don't see 
that all the leverage is affecting the hammer inertia though.  To 
accelerate the hammer, a torque has to be applied to the hammer 
shank.  This torque is being provided by jack force at the knuckle. 
This force is being reacted back through the mechanism to the 
capstan.  If you want more acceleration then more force has to be 
applied by the capstan.  The magnitude of the force is going to be 
dependent on the action geometry.  But changing this geometry 
(changing the amount of leverage) isn't changing the inertia of the 
hammer and shank, but changing the force at the capstan that's a 
manifestation of its inertia.  Its inertia is still small compared to 
that of the key.

Phil Ford


-- 
Phillip Ford
Piano Service and Restoration
1777 Yosemite Ave - 130
San Francisco, CA  94124