---------------------- multipart/alternative attachment Phillip Ford wrote: > > 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 wrote that (m1g) wrong (looking at his first figure), and understood it wrong to begin with. I'd thought for a bit that he was saying more or less that rk was the distance of the << capstan >> from the fulcrum. I found that was wrong last nite after writing. But I am unsure of what you are saying here as well.. isnt Mg a constant (both in quantity and position) and rk more or less the effective balance point at any given time for any given counterleading situation ? > > >> I am unsure of exactly what Ik is... Inertia of that point mass > >> ?...Inertia at that same position ?? > > .. 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)." > Yes... that also became clear last nite as well... grin ... helps to read the formulas and definitions on the last page. > 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. Hmm... I'd rather thought the M was an idealized mass as you said, but located at the << capstan >> and the center of mass was a seperate quantity. This would mean that the mass at the center of mass would be the product of these two. If so the values I plugged in last nite were more or less ok. Sort of a different way of describing the leverage ?? ... I mean in a fully balanced key rk is zero so Mrk and for that matter Mrk2 become zero as well, as they should in that case as it would take zipp diddly force to get the key moving then :) . This would also mean that Ik would be zero there..... :)...yet M itself never becomes zero. Hmmmmm... I feel another one of these headaches comming on ... hehe. At anyrate... this seems to be the point where I am least certain of what Stephen means. And I'd like to clear it up for certain as then I would know for sure how to plug what values into the equations > >>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. Isnt the captilized G the symbol for the gravitational constant ?.. is that what you mean to use here ? And Ik ....as I understand it, Stephen says that the slope of the acceleration line has nothing to do with the breakpoint location. The breakpoint location is where all different slopes intersect... so I dont quite get what you mean by this sentence. Cheers RicB -- 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 ---------------------- multipart/alternative attachment An HTML attachment was scrubbed... URL: https://www.moypiano.com/ptg/pianotech.php/attachments/23/39/b7/a7/attachment.htm ---------------------- multipart/alternative attachment--
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