<!doctype html public "-//w3c//dtd html 4.0 transitional//en"> <html> Couldnt sleep... :) Wanted to try something out here to try and clear up where I am confused about things. Lets take a very shot in the dark type example keyset. Given the following :     <img SRC="cid:part1.3EDE675E.95514862@grieg.uib.no" height=146 width=404> Where r =10 cm; rb = 5 cm; rk = 2.5 cm;  m = 30 grams and M = 140 grams. (just accept these as working values) whats going to be our breakpoint acceleration, and what will be the force needed to get there ? Before trying to work this out... I took M of 140 grams because thats a fairly typical weight that could be expected to be resting on the capstan of a key in the middle / low part of a piano. I'm unsure as to whether this is right  to insert this here...... so correct if wrong. In anycase.... g is the acceleration due to gravity right ?... ie. circa 9.8 ms^2 ? That gives if so (r/rb)g =  19.6ms^2 which should then be the breakpoint acceleration for this example. Fbr then is (140 * 2.5 + Ik/5 )( 9.8/10), and if Ik can be understood to be Mrk^2 then thats 140 * 6.25 = 875 which ends up yielding a force of 514.5 grams. Which if even remotely correct would mean we are confined to pretty light play levels for this soft zone. Ok .... this should suffice to show exactly where I am confused about what to plug in where in the formulas at the appendix. So... please straighten me out on this one :) RicB   -- Richard Brekne RPT, N.P.T.F. UiB, Bergen, Norway <A HREF="mailto:rbrekne@broadpark.no">mailto:rbrekne@broadpark.no</A> <A HREF="http://home.broadpark.no/~rbrekne/ricmain.html">http://home.broadpark.no/~rbrekne/ricmain.html</A> <A HREF="http://www.hf.uib.no/grieg/personer/cv_RB.html">http://www.hf.uib.no/grieg/personer/cv_RB.html</A>  </html>