>>FWIW. The number is 32.2 fps^2 (9.8 m/s^2) at the earth's
surface.
>>http://www.tcaep.co.uk/science/constant/detail/gravityaccelerationdueto.htm
>>Gravitationally,
>>Phil F
>Thanks Phil,
>
>Maybe I have this wrong could you help? The leads in a key lighten the
>touch when the key is pushed slowly. So for soft playing it looks like
>they assist in playing the key. At what point do the leads begin to
hinder
>the touch? I am making an assumption but it looks to me like they help up
>to the point they are accelerated past the speed they would fall. After
>that additional energy needs to be applied to move the lead faster. As a
>key is played faster and faster more of it begins to accelerate past the
>falling point. A lead out toward the end could be hindering the touch and
>a lead near the balance point would still be helping to depress the key.
>
>Please excuse me for making a few uneducated guesses I know it's not the
>usual high level math and science found on this list. I would be glad if
>someone could straighten me out on this.
>
>John Hartman RPT
John,
I'll do the best I can. Pardon me if I'm repeating something that you already know. For a system to be statically in balance the sum of the forces and the sum of the torques on the system must equal zero. If we isolate the key, then the hammer, shank, and whippen are applying a force at the capstan. The key's balance point, or axis of rotation, is at the balance rail (on top of the punching) where the pin passes through the key. The torque being applied to the key by the force on the capstan is this force times its arm (which is the distance from the capstan to the balance point). The balance rail is providing a reacting force on the key. Since this force is applied at the balance point its arm is zero, so it applies no torque to the key. The player's finger is applying a force downward at the front end of the key. The torque being applied by the finger is the finger force times its arm (the distance from the spot where the finger is touching to the key balance point). For the sake of simplicity assume that the center of gravity of the key is at the balance point (in other words if we place the key on a knife edge at the balance point it would sit there, not fall one way or the other). The weight of
the key is therefore applied at the balance point and exerts no
torque. So, for the key to be in balance the sum of the forces and torques given above must be zero. Let's take an example:
C = force applied at capstan = 120 grams
F = force applied by finger = ?
B = force on balance rail = ?
W = weight of key (force of key) = 100 grams
x = distance from capstan to balance point = 100 mm
y = distance from finger to balance point = 200 mm
Sum of forces = zero so C + F + B + W = 0
Sum of torques = zero so Cx + Fy = 0
If you solve the equations you'll get F = 60 grams. So the players finger has to apply a force of 60 grams to keep the key in balance (in other words the key has a 60 gram downweight). Let's say you don't like this and want to reduce it. So you want to add some weight in the front of the key. You can see that the torque being applied by the lead weight is in the same direction as the torque being applied by the finger and is counteracting the torque being applied by the capstan force. So, the more lead you add
the less force needs to be applied by the finger to balance the force applied at the capstan. Let's take another example (let's say you want a 50 gram downweight):
C = force applied at capstan = 120 grams
F = force applied by finger = 50 grams
B = force on balance rail = ?
W = weight of key = 100 grams
L = weight of lead (force of lead) = ?
x = distance from capstan to balance point = 100 mm
y = distance from finger to balance point = 200 mm
z = distance from balance point to lead weight = 50mm (case 1) or 100mm (case 2)
Sum of forces = zero so C + F + B + W + L = 0
Sum of torques = zero so Cx + Fy + Lz = 0
Let's consider two cases. In case 1 the lead is located 50 mm from the balance point. Solving the equations gives a lead weight of L = 40 grams. In case 2 the lead is located 100mm from the balance point. In this case solving the equations gives L = 20 grams. As you see, if you put the lead further out you need less of it to give the same static balance. Let's hold that thought a moment and move on -
The key rotates about a point, it does not move in a straight line. The equation of motion for a body rotating about an axis is:
T = I A
T = torque applied to body
I = mass moment of inertia of body
A = angular acceleration of body
Assuming that a given torque is being applied by the finger, then as the inertia of the key and leads go up the acceleration has to
decrease. Another way to say this is, if you want to achieve a certain amount of acceleration of the key, then as the inertia of the key and leads go up more torque has to be applied by the finger to the key to achieve this acceleration.
The moment of inertia is determined by the dimensions and mass of the body. For a point mass the moment of inertia is I = m R^2 where
m = mass of body
R = distance of body from axis of rotation
Because the leads are compact and their dimensions are small relative to the key, assuming a point mass seems a reasonable assumption.
In the examples above the key will have a given moment of inertia. Adding lead weights will increase that moment of inertia. For the two cases given:
Case 1
The increase in inertia is I = 40 grams x G x (50 mm)^2 note G =
acceleration of gravity (which started this whole thing) to convert the weight to a mass.
So increase in inertia is 100,000 G g mm^2
Case 2
The increase is I = 20 X G x (100)^2 = 200,000 G g mm^2.
As you see, the increase in inertia is twice as much for case 2 as for case 1 even though we added only half as much lead. For the same torque applied by the finger in both cases, the key in case 2 will accelerate less quickly. This is the principle behind the Steinway accelerated action. Concentrating more lead closer to the key balance point gives the desired downweight with less inertia than less lead further out on the key.
Inertia enters the picture whenever the key is moving. As the equation shows, the torque is proportional to the acceleration. So, the finger (the pianist) will notice the effects of inertia more as the acceleration that it is trying to achieve increases. When the key is still, or barely moving as in soft play, the torque resistance that the finger will experience will be small because the acceleration is small.
The inertia is always working against the finger whether the key is going down or up. This is somewhat counterintuitive. If you put a bunch of lead in the front of the key (say you had 10 gram downweight) you might think that the key would go down faster. Certainly if you took the action stack off, the key would drop like a rock. However, this key will not accelerate downward as fast as a key without all that lead (which might have, say, a
70 gram downweight) because its inertia is so much higher.
So, there is some balance to be struck between slow play and fast
play. You could set up an action that would require almost no effort to play when playing slowly or softly, but would play like a truck if you tried to play it fast. Or you could set up an action to be 'quicker' but it would feel heavier for slow or soft play.
Also, you can achieve a desired downweight (or frontweight if you prefer) with different key inertia depending on where you choose to locate the lead or leads. It would certainly be possible to have key 1 and key 88 have different front weights but the same inertia by proper location of the leads, if you thought that this was desirable.
I hope that this is helpful.
Phil F
Phillip Ford
Piano Service & Restoration
1777 Yosemite Ave - 130
San Francisco, CA 94124
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