Key Leads and Inertia Experiment

[Richard Brekne]

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Phillip Ford wrote:

> >Where r =10 cm; rb = 5 cm; rk = 2.5 cm;  m = 30 grams and M = 140 grams.
> >(just accept these as working values)
> >

>
> Ric,
>
> This all looks OK except that you need to be consistent in your units. For your distance measurements you're using cm, but for g you are using m/s^2.  You need to be using 980 cm/s^2 or change your distance measurements to meters.
>
> Phil F
>

Grin.. I DID that didnt I ?? so what... this changes things to.... :)


     That gives if so (r/rb)g =  19.6 ms^2 which should then be the breakpoint
     acceleration for this example.

     Fbr then is (140 * .025 + Ik / .05 )( 9.8 / .1), and if Ik can be understood to
     be Mrk2 then thats 140 * .0252 = which gives .0875

     (140 * .025 + .0875 / .05) * 98 = 514.5 grams of force needed to reach the breakpoint acceleration.

btw... droped a 500 gram weight on a few keys today to sort of assess the blow strength. I let the 500 grams start about 1 cm above the key... and this yeilded what I would call a mp - m level of play.

So... assuming the above is more or less correct... our breakpoints are limited to fairly soft levels of play.


--
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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