>>> 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. rk is the distance Mg is from the fulcrum , not m1g. M is the mass of the key and action for the 'real key' case, as opposed to ye massless beam case, as Stephen refers to it, in which m1 is the action mass, since the key is massless. >> I am unsure of exactly what Ik is... Inertia of that point mass >> ?...Inertia at that same position ?? That should be I subscript k, but I don't know how to make those characters on my e-mail program. That would make things a little less confusing. Anyway, Ik is the moment of inertia of the key assembly (the key and the action components). As I understand it, Stephen has idealized the distributed mass of the key and action parts as a point mass M,located at a point rk, which will give the same moment of inertia as the distributed mass of the key and action parts. So yes, Ik is the inertia of that point mass. To quote from the paper: "Consider a real piano key and action components with distributed mass. This combination will be called the (unleaded) reference key and represented by a defined distributed mass beam with: (i)centre of mass (first moment about the fulcrum) and (ii) moment of inertia (second moment)." It seems a bit ambiguous to me, but as I read it, M is an idealized mass that represents the distributed mass of the key and action and it is located at center of mass rk. Ik is the moment of inertia of this point mass, so it should be Ik=Mrk^2. >>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" C doesn't seem to have an explicit definition in the paper, but it seems to be the center of mass of the key and action assembly. It's what I'm calling CG (center of gravity). Ik affects the slope of the acceleration line and so affects the location of the break point. >>> 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 Sorry about that. I threw CG in without definining it. Center of Gravity as stated above. Phil F
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