Lengths used to calculate various action ratios:
(L1) Length of the front part of the key from the pivot to the pianist's finger
(L2) Length of the back portion of the key from pivot to capstan
(L3) Length from wippen pivot to capstan contact point.
(L4) Length from wippen pivot to the jack/knuckle contact point.
(L5) Length from hammer-shank pivot to jack/knuckle contact point.
(L6) Length from hammer-shank pivot to hammer/string contact point.
Density of various action materials:
Density of spruce
Density of maple
Density of lead
For these calculations I will assume that the keystick has a constant cross section with no features other than the lead counter weights installed in it. The actul key tapers toward its back end, but also includes a back check that more or less offsets the mass and inertia lost by the tapering. As the intertia calculations will show, keystick dimensions do not have to be estimated with a high degree of accuracy.
Total Length of key
Width of key
Height of key
Length of key from its front edge to its pivot point
Length of key from its back edge to its pivot point
Length from key pivot to first lead weight
Length from key pivot to second lead weight
The example key has two lead weights installed in it. Each one is a cylinder of about 0.49" diameter.
Volume of each lead weight
Mass of each lead weight
Mass of back portion of key
Mass of front portion of key less lead weights
The force, expressed as mass, exerted by the key on the pianist's finger:
To calculate the moment of inertia of the keystick I use the standard formula for the moment of inertia of a thin rod.
moment of inertia of a thin rod
For these calculations I assume that the hammer shank has a constant cross section. The hammer is considered to be a point mass.
Diameter of hammer shank
Mass of hammer/shank assembly with the assembly supported at the hammer-shank pivot point, the hammer shaft horizontal, the hammer pointing downward, with the hammer resting on the weight-measuring scale.
Mass of hammer shank (volume times density).
To compute the mass of the hammer, we must subtract the weight of the shank from WThamp. We can consider the mass of the shank to be a point source located at the shank's center of mass, which because the shank is considered to be uniform, is at its physical center. Because the shank was effectively surrported at its end, and its center of mass is at its center point, the measured weight of the shank is one half of its total mass.
The force, expressed as mass, required to support the hammer/shank assembly at the knuckle is the sum of the weights of the hammer and hammer shank times their lever ratio.
Weight at knuckle
Moment of inertia of hammer/shank assembly
The wippen consists of quite a number of pieces, but to simplify calculations, its mass distribution is assumed to be uniform along its length. Although this assumption may seem dubious, calculations will show that the moment of inertia of the wippen as felt by the pianist is of so little consequence that the assumption is more than adequate.
Total mass of the wippen
Force, expressed as a mass, required to support the wippen at the capstan contact point, with the wippen's pivot simply supported
Moment of inertia of the wippen using the formula for a thin rod
Following are the various lever ratios in the action.
Strike ratios vary somewhat, but a value between 5 and 6 is typical.
Static force supplied by the pianist's finger to support the hammer
The total downweight is the sum of the static force due to gravity plus the action friction. Friction is typically in the 8 to 12 gram range, and total downweight is typically 50 to 55 grams, so the value computed above is about what is expected.
Inertia at pianist's finger
The inertia experienced by the pianist is the sum of the interias of the hammer assembly, the wippen, and the key. As previously derived, the inertia is reflected through the lever system as the square of the lever ratios.
Contribution of hammer assy
Contribution of wippen
Contribution of key (previously calculated)
Force due to inertia
We can now calculate the force due to inertia required to actuate the key. For simplicity, let's assume that the pianist applies a constant force to the key. We will start by calculating the torque.
Let's say it takes 0.1 second to depress the key
Torque equals the inertia times the angular acceleration
From the standard formula,
we can calculate the angular acceleration
The key travel is typically about 10 mm for most pianos
From the standard formula
Torque supplied by the pianist to overcome inertia
Torque divided by the lever arm calculates the force supplied by the pianist
How does this force compare to the static downweight of the action? To compare we must convert the weight, which, due to convention, we have been expressing as mass, to force. To do so we multiply by the acceleration of gravity.
Static downweight--assume friction plus static force equals 50 gm, which we must convert to force in order to compare with the force due to inertia.