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<DIV><FONT face="Bookman Old Style" color=#0000ff>Comments below:</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face="Bookman Old Style" size=2>> William,<BR>> <BR>> Vladan
is correct. The mass vs force concepts are, unfortunately,
not<BR>> intuitive. It's not made any easier by the fact that we use
the same units<BR>> for both mass and force. (real physicists use
different units) The<BR>> readings you refer to, taken at the front of
the key, are force (or weight,<BR>> if you will). A given value of key
front weight could be the result of an<BR>> infinite number of combinations
of mass and distance from the fulcrum. <BR>> Might be a large mass close to
the fulcrum, might be a smaller mass further<BR>> from the fulcrum. So
measuring front weight does not tell us the mass. To<BR>> determine the
rotational moment of inertia, it is necessary to know where<BR>> the mass is,
and how big it is. <BR></DIV></FONT>
<DIV><FONT face="Bookman Old Style" size=2><FONT color=#0000ff size=3>Right, I
agree. My question is more directed at the concept of whether more mass
closer to the fulcrum yields a different MOI than less mass farther from the
fulcrum. Is the fact that some of the weight of the key/lead combination
is supported at the fulcrum at all relevant in our real world example where the
pianist plays at the front of the key? That is, if we change the point at
which we apply the force on the key, will that alter our calculations of
inertia?</FONT></DIV>
<DIV></FONT><FONT face="Bookman Old Style" size=2></FONT> </DIV>
<DIV><FONT face="Bookman Old Style" size=2></FONT> </DIV>
<DIV><FONT face="Bookman Old Style"><FONT color=#0000ff>Also, to Vladan (anyone)
again, in your example of the two differing leads at different places, my brief
experimenting seems to support this concept. That is, I just weighed off a
keystick and tared the scale to zero with no weights added. Adding a 10g
weight at 5cm from the fulcrum gave me a FW of 3.5g. To get the same 3.5g
FW with a 5g weight, that weight needed to be not 10cm, but 11.9cm from the
fulcrum.</FONT></FONT></DIV>
<DIV><FONT face="Bookman Old Style" color=#0000ff></FONT> </DIV>
<DIV><FONT face="Bookman Old Style" color=#0000ff></FONT><FONT
face="Bookman Old Style"><FONT color=#0000ff>So, following your example
calculations, I have:</FONT></FONT></DIV>
<DIV><FONT face="Bookman Old Style" color=#0000ff>10g*5cm*5cm =
250gcm^2</FONT></DIV>
<DIV><FONT face="Bookman Old Style" color=#0000ff>vs.</FONT></DIV>
<DIV><FONT face="Bookman Old Style" color=#0000ff>5g*11.9cm*11.9cm =
708.1gcm^2</FONT></DIV>
<DIV><FONT face="Bookman Old Style"><FONT
color=#0000ff></FONT></FONT> </DIV>
<DIV><FONT face="Bookman Old Style" color=#0000ff>Egads, it's worse than I
thought. A much greater inertia with less weight further from the fulcrum,
as you predicted, with same FW. Now, I only question if this is the
appropriate formula for calculating this relationship. I'm not suggesting
you are wrong, I simply don't have as clear an understanding of all that we
should consider for this problem. As I said, I'll go read Stephen's paper
now, and see what I can glean from that.</FONT></DIV>
<DIV><FONT face="Bookman Old Style"><FONT
color=#0000ff></FONT></FONT> </DIV>
<DIV><FONT face="Bookman Old Style"><FONT
color=#0000ff>Thanks,</FONT></FONT></DIV>
<DIV><FONT face="Bookman Old Style"><FONT color=#0000ff>This is
fun,</FONT></FONT></DIV>
<DIV><FONT face="Bookman Old Style"><FONT color=#0000ff>William R.
Monroe</FONT></FONT></DIV>
<DIV><FONT face="Bookman Old Style"><FONT color=#0000ff></FONT> </DIV>
<DIV><FONT size=2></FONT><FONT size=2></FONT><FONT size=2></FONT><FONT
size=2></FONT><FONT size=2></FONT><FONT size=2></FONT><FONT
size=2></FONT><BR></FONT><FONT face="Bookman Old Style" size=2>> > >
I=m*r^2<BR>> > ><BR>> > > I=moment of inertia<BR>> >
> m=mass of the lead<BR>> > > r=distance from the balance rail hole
to the lead<BR>> > ><BR>> > > Note that the formula takes the
square of "r".<BR>> > ><BR>> > > Now, let's look at our
options for lead placement.<BR>> > > Suppose that we can use a 12 gram
weight placed 10 cm<BR>> > > away from the balance rail hole to get the
desired<BR>> > > balance weight. As an alternative, we could also
use<BR>> > > a 24 gram weight placed 5 cm away from the balance<BR>>
> > rail hole and still have the same balance weight. Is<BR>>
> > there a difference between the two?<BR>> > ><BR>> >
> Yes, there is. Looking at the formula above, the<BR>> > >
inertia increases with "r" squared. In the first<BR>> > > case,
the inertia will be 12*10*10=1200 gcm. In the<BR>> > > second
case, the inertia will be 24*5*5=600 gcm.<BR>> > > Placing the 24g
weight 5cm away from the balance rail<BR>> > > hole reduced the moment
of inertia by 50%. That's a<BR>> > > lot. You can minimize
the inertia by placing the<BR>> > > weights closer to the balance rail
hole and increasing<BR>> > > the amount of lead accordingly so that the
balance<BR>> > > weight meets the target.<BR>> > ><BR>>
> > There is a fundamental trade-off between balance<BR>> > >
weight and key stick inertia. If we add a lot of lead<BR>> > > to
make the balance weight low, the action will feel<BR>> > > very light,
but only when you move the key slowly. If<BR>> > > you try to
play a loud note, all that lead will have<BR>> > > to be accelerated
and the key will feel heavy.<BR>> > ><BR>> > > If you put very
little (or no) lead into the key and<BR>> > > accept a high balance
weight, the action will be heavy<BR>> > > when you measure it with your
weights, but it will<BR>> > > feel light when you play
fast/hard/loud.<BR>> > ><BR>> > > The question is: What is
better? My own preference is<BR>> > > for a heavier balance
weight with less inertia.<BR>> > ><BR>> > > Vladan<BR>>
><BR>> <BR>> <BR>> <BR>> </FONT></DIV></BODY></HTML>