Hello Mark and John,
I am not suggesting compensating for the distance to the vertical center of
the key, although for accuracy perhaps this is important. I think this should
be disregarded for a preliminary effort as already has been done with other
lesser factors. Many times, the distance from the front of the key to the
balance rail hole is longer in the front "half" than the back even on grands,
and, particularly so on uprights. What I suggest is to transfer the nominal
axis of rotation implicit in the use of this formula from an axis taken at the
center of gravity to the one at the real center of rotation which is at the
balance rail hole by using the Parallel axis theorem. This is done by the
simple expedient of adding another term which is md^2. The distance between
the parallel axes is indicated by d. The distance to the vertical center of
the body is, I think, as you suggest, insignificant.
I(g) = 1/12ml^2 is, by definition a formula for the value, excepting
complications as I said the other day, when an axis is taken at the center of
gravity. On a Knabe concert grand I measured these two "halves" the other day,
again ignoring the flaring of the key, and estimated the distance between the
balance point of the key and the actual balance rail hole to be 1'3/4".
This can be applied to the first part of John's proposed calculation
which is the case without the key leads installed and then the effect of the
weights added calculated subsequently. . However, I have done some work
recently on a new, simple but powerful method of dynamic action analysis which
I will offer here, although I intend to publish it elsewhere, that I believe
offers a better way to represent the actual effect of alterations in action
parts. In the case of the keystick, using
I(g1) = 1/12ml^2 + md^2, as expressed above,
which requires one to find the actual center of balance of the key and the
distance from this to the balance rail hole; and .
upon adding or removing weights the actual center of balance will be changed
with a corresponding change in d. Label these d(1) and d(2). Find the new
center of balance, recalculate using the parallel axis theorem and the effect
of the changes in the weighting is:
dW = (1/12ml^2 +md(2)^2) - (1/12ml^2 + md(1)^2).
(dW) is deltaW, not downweight.
Essentially this is I(g2) - I(g1) and takes into account the effect of the
changes on the center of balance induced by changes in key leading, or what is
better, anything added or taken away from the key, for example the capstan or
changes in the weight of the backcheck. This expression does this by measuring
within the limits implied by applying this particular formula to these
particular items (the keys), which is still somewhat questionable in my mind,
and may do a good job of estimating the effect of alterations can be had
relatively easily, I believe. Although one should remember that an actual,
rigid, eccentric three dimensional body that is anaxisymmetric such as a key,
implies a much more involved effort to arrive at an accurate I. Even so, the
effect which altering weight has on its MOI can still be demonstrated by the
changes in distance from the center of rotation to the in the center of
balance and these can be expressed using the Parallel axis theorem.
In my opinion, this is the correct way to calculate the dynamic effect of
alterations keyleads, or whatever kind in the keystick.
It also offers, as I will demonstrate at greater length in another post,
a way to measure the effect of changes in the dynamics of the action itself,
for example changes in weight, size, moment arms, some aspects of friction
etc. of the hammer, shanks, whippen and other parts. The simple principle
which underlies this is that one can do for changes in action parts, exactly
what has been demonstrated above, that is analyze changes in the moments of
inertia that are implicit by virtue of changes in the mass distribution of
rotating parts.
This is done, first, by observing changes induced in the balance point of
the part and analyzing the effect of changes on the part which are expressed as
different values of the total I found at the axial point, through a use of the
Parallel Axis theorem as used above. This is an important application, but
much more so is the application of the principle to the entire action.
Secondly, and more comprehensively, a similar analysis can be performed when
the capstan is loaded by the action train and, subsequently, found after
arbitrarily changing parts and then subjecting these to a similar treatment.
This is no less than an comprehensive expression of the touch of the action
itself. These changes, taken at the loaded key, are a proxy for the actual
summed changes in moment of inertia, angular momentum and angular acceleration
which take place when changes are made in the various components of the
action, and, again, represent an aggregate that is the touch itself.
Furthermore, perhaps even more importantly, it then becomes possible
through the use of this simple prinicple to impose an objective measure upon
the dyanamic characteristics of the action when in play possibly through the
application of an intergral or differential equation which I only sense vaguely
at the moment and have no real understanding of as yet.
Regards, Robin Hufford.
Mark Davidson wrote:
> Robin Hufford wrote:
>
> >The formula becomes(using I(est)) to indicate
> >rotation at the balance rail,
> >I(est) = 1/12ml^2 + md^2
>
> If I understand you correctly, d is basically half the height of the key.
> I.e., you're rotating the key not around its center top-to-bottom, but
> around its bottom, so the center is displaced by the distance of half the
> key height. If so, for a 20 inch key that is 1 inch high this represents
> only a 0.75% difference from using (1/12 ml^2) alone. In the land of rough
> estimates, I would probably not make too much of this.
>
> -Mark
>
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