>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 ? Here is an excerpt from the appendix of Stephen's paper: ". Moment of inertia of the reference key is Ik. . Centre of mass of the reference key is rk, assumed to be behind the fulcrum point. . Mass of the reference key is M." And: "For simplicity refer to the combined (lead free) key stick and action components as the reference key." I think this choice of words is confusing. He uses the word key to refer to the key stick and action components together. Normal piano tech usage of key is for the key stick. So when we start talking about the KEY, are we talking about the keystick or the 'reference' key? A better choice of words might be key assembly, or reference assembly, or unleaded assembly, or some such, for the combined unleaded key stick and action components. I read the above to say that M is the mass of the key and action components (whippen, shank, and hammer) without any balancing leads. I'm assuming that the action components are being idealized as a mass that would be located at the capstan point. The key stick itself has a mass and a center of mass. Combine the two and you get total mass for the 'reference key' M with its center of mass located at rk. So Mg is the weight of the unleaded key stick and action components together. rk is their center of mass. Both of these are constant for a given note of an action and are independent of the balance leading. I believe that he is treating the balance leading separately rather than changing the mass and center of mass of the reference key every time the leading is changed. >>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. My understanding was that M was the combined mass of the keystick and action components located at their combined center of mass. I think this might have been clearer if Stephen had called out a mass for the action components, showed where he located it (I assume at the capstan), called out a mass for the key stick, showed where he located it, and then done the arithmetic to arrive at the combined mass M and the center of mass C. The formula for moment of inertia = mass x radius^2 only applies to point masses. For key leads this is a pretty good assumption. For action components lumped at the capstan I suppose this is a good assumption (we probably can't easily assume anything else for a first cut). For a large or long item like a key this is a poor assumption. To idealize the key by lumping it's entire mass at one point and saying its inertia is mass x radius^2 is an oversimplification, perhaps too much so even for simple cases like the ones we're looking at. For example, if the key was a perfectly symmetrical rectangle in profile that was centered on the balance point, its center of mass would be at the balance point. Looking at mass x radius^2 would indicate that it had no inertia. This is certainly not the case. (For example, for a rectangle of length l and height h, it's moment of inertia would be lh/3(l^2 + h^2) if I remember correctly). A better assumption might be to say that Ik is the moment of inertia of the keystick, whatever it is by calculation, plus action mass x capstan location^2. >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... rk is not changing by the definition he gave. Balancing lead is not included in M and does not affect rk. Inertia does not go to zero for the reason I gave above, because the key is not a point mass. It has significant inertia and thus requires significant torque to get it moving, even if its center of mass is directly over the balance point. >> >>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 ? CG is an abbreviation for the words Center of Gravity. G is not g, the gravitational constant. Forget I brought it up, it's justing confusing things. Let's stick with center of mass. >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 The breakpoint is located on the acceleration line of the unbalanced key at an acceleration of (r/rb)g. The slope of that acceleration line is determined by Ik. If you change the inertia of the unbalanced key then you change the slope of the acceleration line, which in turn changes the force corresponding to the point (r/rb)g. For a given unbalanced key acceleration line, then the point where the acceleration lines of the keys with balancing lead intersect it is independent of the slopes of the balanced key lines. My point was that at a given acceleration you either measure the force applied to get that acceleration, or having measured some other force and acceleration points and established a line, you just read the force off the chart at the given acceleration. You either use the slope of the line to determine the force or you use the measured forces and accelerations to determine the slope of the line. Phil F Phillip Ford Piano Service & Restoration 1777 Yosemite Ave - 130 San Francisco, CA 94124
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