Key Leads and Inertia

[Richard Brekne]

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Phillip Ford wrote:

> >If we start off with these very rough figures, and figure an "action
> >weight" of 100 grams on a the back of the key, then the soft zone is
> >going to pretty much be a pianisimo thing. Certainly by the time you hit
> >a blow of 500 grams we would be in the so called hard zone... no matter
> >which way you turn it if these assumptions are anything close to correct.
>
> Now I'm getting interested in the ballpark in which this breakpoint plays for 'normally leaded' actions.  If it's down in the ppp range then all of this talk may be for nothing.  Shifting it around by changing lead locations might still not move the breakpoint out into the dynamic range that the pianist is using most of the time.

Which would mean of course that we are stuck with the same choices as before.... more or less. We'd have only the opportunity to provide uniform intertia... in some sense or another.


> >One other point I am a bit unclear on. The article seems to say that the
> >position of the breakpoint is independant of any key leverage ratio. The
> >location of the breakpoint is given as a = (r/rb)g. rk is given to denote
> >the distance from the fulcrum the <<action weight>> is applied to the
> >back of the key. In each case this yields then also the amount of input
> >force by the finger that is needed to achieve this breakpoint
> >accelleration... also independant of the keys actual leverage...??
> >Perhaps I misunderstand this ?
> >
> >RicB
>
> The acceleration at the breakpoint is determined by the relative location of the lead to the finger force.

That much we talked about last time.. easy enough to see... but...

> However the force at the breakpoint is dependent on key parameters.  You need both to locate the breakpoint.

This is where I am unclear. When he says....

     "The dynamic breakpoint lies on the
     acceleration vs force graph for the reference key at the point where acceleration is a = (r/rb)g.
     This point defines the acceleration and corresponding applied force for which the reference and all
     leaded keys respond in exactly the same way."

and...

     "Alternatively, if the location of the lead in the key is known, the
     breakpoint acceleration can be calculated as given by a = (r/rb)g and the breakpoint can be found
     on the graph corresponding to the reference key."

this seems to say that the breakpoint location, and the force neccessary to achieve that acceleration is a function the front key parameters and the weight of the action.


>  I would think that Ik and rk are going to be dependent on the key ratio.  Changing rk changes the key ratio.

Grin... since rk is per definition the distance m1g is from the fulcrum... I guess so. I am unsure of exactly what Ik is... Inertia of that point mass ?...Inertia at that same position ??   And how does any of this then affect the position or occurance of the break point ? I'd also like a word put to the term "C"

>  Changing rk also moves mass M changing the CG of the key and consequently Ik.

Yes, I buy this much... but all that seems cancelled out by the time he reduces to a = (r/rb)g... and he uses the term W1... ie weight... which doesnt change just because you move it. Perhaps I am just looking sideways at things here :)... but I would think the weight of a thing, and its effect at different points on a lever were two different things.  Anyways... this is where I am getting stuck for the
moment.

Just what exactly is this "C"...  and did you mean Cg instead of CG... if not where did a G come in ? just so I dont assume something I shouldnt :)


>
> Phil F
>
> __

--
Richard Brekne
RPT, N.P.T.F.
UiB, Bergen, Norway
mailto:rbrekne@broadpark.no
http://home.broadpark.no/~rbrekne/ricmain.html
http://www.hf.uib.no/grieg/personer/cv_RB.html


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