Keylead inertia and leverage (was Re: Ideal leading pattern:)

[Richard Brekne]

Nice explanation, leaving all friction issues aside for the moment I'd like to
continue on this and raise this little head scratcher I have been wondering
about that doesnt seem to add up.

Take an action model of a grand and remove all the weights from the key.
Otherwise set the model up with standard regulation, and Strikeweights. Now do
the following. First figure out what front weight you need to get a ball park
Balance Weight figure. Then find the two extremes for placement of this weight
along the length of the key. By that I mean find out how much weight you need as
close in to the balance rail as possible to get that FW, and then how much you
need out towards the end of the key.

When you have that figured out then do the standard old fashioned dw /uw
measurements with the weights in each configuration. Now the only thing I want
you to look at is how the actuall "measurement process" behaves. I have found
that with the weight placed out at then end of the key the measurements
typically stop up half way through the key stroke, but with the weights in close
to  the center the measurment typically goes smoothly all the way down to
letoff, or all the way up to hammer rest.

Since the only difference is placement of key leads this doesnt seem like a
friction issue from the top action, rather an inertia /momentum issue. But
intuitively at least it would seem like the "smooth" reaction to the measurment
should happen under circumstances of greatest inertial, yet the opposite seems
to be the case if we take the inertia formula...

So how do you explain this ?

I am not sure how much it matters as everything else also points to leads being
best placed towards the center...unless there is more to the factor of inertia
relating to the basic weight of the lever then the formular discussed accounts
for.... In anycase it seems curious to me and I wouldnt mind an explaination
that makes sense.


Mike and Jane Spalding wrote:

>
> Let's take the key weight example, and we'll round off the numbers to make
> it easier, if not necessarily correct for your specific piano:  (You may
> find it makes more sense if you sketch this as you read through)
>
> The key measures 8" from front rail pin to balance rail pin.  You want to
> increase the keyweight by 4 grams, and are considering putting the weight at
> 4" (4 grams times 8" divided by 4" = 8 grams required) or at 2" (4 grams
> times 8" divided by 2" = 16 grams required).
>
> Let's let A equal the acceleration of the key, at the front rail pin, for a
> mezzo-forte blow.  The acceleration at 2" would be A times 2" divided by 8",
> or A/4.  The force at 2" would be M * A/4, or 16*A/4, or 4*A.
>
> The acceleration at 4" would be A times 4" divided by 8", or A/2.  The force
> at 4" would be M * A/2, or 8*A/2, or 4*A.
>
> So, in both cases, the force AT THE WEIGHT is equal to 4*A.  But the force
> felt by the pianist, AT THE FRONT RAIL, is reduced by the leverage:  For the
> weight at 4", the force is reduced by 4" divided by 8", so the pianist feels
> 2*A.  For the weight at 2", the force is reduced by 2" divided by 8", so the
> pianist feels A.  Bottom line, the closer the weight can be placed to the
> balance rail, the less inertial resistance the pianist will feel.
>
> Mike

--
Richard Brekne
RPT, N.P.T.F.
Bergen, Norway
mailto:Richard.Brekne@grieg.uib.no