>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 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
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