The moment of intertia is    J = SUM{m*r²}   where m are various point masses that compose a rotating body, and r are their respective distances from the center of rotation.  The units are kilogram meters-squared (kg*m²) or ounces inch-squared (oz*in²) or whatever consistent units you like.

[ using g*mm² ]

Having found out what is this J then you can go backwards and simulate it with one single _equivalent_ mass as you see (feel) it at a specified distance L from the center of rotation.

Let us look at our three essential components (order of magnitude values in parentheses):

1. The hammer.  The easiest element; the shank is light and the butt is close to the rotation center so they give negligible contributions.  Therefore we assume it is only a point mass m1 (about 1/2 oz for a heavy hammer) [14.18 g] at a radius r1 (about 6 in) [152.4 mm].

hammer:  J₁=m₁*r₁²     [ m_{1= 0.5*28.35= 14.18 grams,} r_{1= 6*25.4 = 152.4 mm}  ]

(= 0.5*6² = 18 oz*in² inertia)   [ = 14.18*(152.4)² = 329,341 g*mm² ]

2. The wippen.  For this estimate assume the wippen to be a uniform bar of mass m2 (1/2 oz) [14.18 g] and length r2 (4 in) [101.6 mm] rotating about its _end_.
By integration I get

wippen:  J₂ = (1/3)*m₂*r₂²      [ m_{2= 0.5*28.35= 14.18 grams,} r_{2= 4*25.4 = 101.6 mm}  ]
(= (1/3)*0.5*4² = 2.7 oz*in² inertia)  [ = (1/3)*14.18*(101.6)² = 48,791 g*mm² ]

3. The key.
Assume it to be a uniform bar of mass m3 (2 oz) [56.7 g] and length r3 (12 in) [304.8 mm] rotating about its _center_.
By integration I get

key:  J₃ = (1/12)*m₃*r₃²    [ m₃= 2*28.35= 56.7 grams,  r₃  =  12*25.4 = 304.8 mm  ]
(= (1/12)*2*12²  = 24 oz*in² inertia).  [ = (1/12)*56.7*(304.8)²  = 438,967  g*mm² ]

Now we try to go backwards to find equivalent masses as felt by the pianist's finger:

3. The key is pressed at L3 (5 in) [127 mm] from its rotation center. Its equivalent translational mass (mt) at that particular point is then

  mt₃ = J₃/L₃²     [J₃ =  438,967 , L₃  = 127  mm]
(= 24/5² = 0.96 oz mass equivalent)        [= 438,967/(127)² = 27.2 grams mass equivalent]

Observe that this differs from the mass of the key.

2. When the key is depressed s3 (0.4 in) [10.16mm], then the wippen somewhere, at L2, moves the same, let us assume at its middle (L2=2 in) [50.8 mm] (inspired by k2=2). Its _equivalent translational mass_ (as felt by the pianist) is then

  mt₂ = J₂/L₂²     [J₂ =  48,791 , L₂  = 50.8 mm]
(= 2.7/2² = 0.68 oz mass equivalent).        [= 48,791/(50.8)² =  19.36 grams mass equivalent]

1. When the key is depressed s3 (0.4 in), then the hammer somewhere, at L1, moves the same, let us assume at 1/5 of its length (L1=1.2 in) [30.48 mm] (inspired by k1=5). Its _equivalent translational mass_ (as felt by the pianist) is then

  mt₁ = J₁/L₁²      [J₁ =  329,341, L₁  = 30.48 mm]
(= 18/1.2² = 12.5 oz mass equivalent).        [= 329,341/(30.48)² =  354.5  grams mass equivalent]

The total mass as felt by the pianist is thus

  mt_tot =  mt_{1 +} mt_{2 +} mt₃    
(=  0.96 + 0.68 + 12.5 = 14.1 oz mass equivalent)  [= 27.2 +  19.36 + 354.5 grams mass equivalent]

The total equivalent mass is very much dominated by the hammer mass, as should be. When we double the mass of the key from 0.96 to 1.92 the new total will be 15.1 oz, an increase of 6.8%.

Observe that this mass is the _inertial_ component that is felt by the pianist. This has absolutely nothing to do with any static force applied to the key (as measured for instance with the Hickman balance). The static forces are influenced by gravity and any added counterwights and springs possibly included in the mechanical transmission chain.

Sigh  :-O

Johan Liljencrants