So, we see in our example that forces due to inertia are significant.  The example used a nominal time of 0.1 seconds to depress a key, but what is a typical range of times?  Experiments on my own piano show that key travel time for a hard blow can be as little as 0.02 seconds, and for a soft blow, 0.25 seconds.  The analysis shows that the force due to inertia varies as the reciprocal of the time squared, so for the 0.02-second blow, force due to inertia becomes 10.75 newtons, which is over 20 times the static force! 

 

Obviously, the inertia of the action has a highly significant effect on action feel.  In fact, one wonders why down weight and friction are typically the only parameters given in discussing action feel.  My conjecture is that static forces are easily measured with only a few weights.  Direct measurement of inertia, however, requires more sophisticated equipment—equipment whose cost would not be overly burdensome using today’s sophisticated and inexpensive electronics, but which would have been quite expensive and bulky using older mechanical technology.  Also, adjusting the down weight requires only the addition or subtraction of lead weights in the keys, whereas consistent inertia requires accurately adjusting hammer mass and/or action ratios—both if consistent tone is desired.

 

An important observation is that the inertia, as reflected to the pianist’s finger, of the hammer assembly is almost 7 times that of the key, and about 60 times greater than that of the wippen with the tested parts.  Some may be surprised at this finding and wonder if it could possibly be true.  There is a simple way to get an intuitive grasp of the great differences in inertias.

 

Let’s think about the velocities of the various action parts when a key is depressed, because as the famous Newton’s law shows, it takes force to accelerate a mass to a velocity.  The pianist’s finger attains some velocity as the key is depressed.  The capstan attains about ½ of this velocity because it is about ½ the distance from the key’s pivot point as is the pianist’s finger.  The jack attains a velocity a bit less than the finger velocity because it is less than twice as far from the wippen pivot as is the point where the key pushes on the wippen.

 

Think how much smaller and lighter the wippen is as compared to the key.  So, it is not surprising then that the force required to accelerate the wippen is much less than that of the key.  Now, remember that the hammer velocity is usually 5 to 6 times that of the pianist’s finger.  Other things being equal, force goes up as the square of the velocity attained.  So, it is sensible that the hammer’s inertia dominates.

 

Hammer mass and the hammer-shank pivot to knuckle/jack contact distance are the most important parameters to control to provide consistent inertia forces.  This distance is described as Lk in the analysis.  The reasons for this become obvious when inspecting the formula for hammer inertia as felt by the pianist’s finger, Ihamf.  Ignoring inertia of both key and wippen, this formula shows that the inertia varies directly as hammer mass and by the square of the action ratios.  Given the expectation that the action ratios may vary from nominal because of some typical tolerance in the placement of the pivot points, the greatest ratio error will arise from the shortest action length.  Lk, is by far the shortest distance, and so is the principal problem in this regard.  Using the example action parameters, a change in Lk of only 0.5 mm (0.020”) causes Ihamf to change by about 6%. Of course, any error in Lk can be compensated by a change in hammer mass, but hammer mass affects tone.

 

It is inevitable that the question of what inertia is best should arise.  A definitive answer cannot be provided, but we can explore what the effects caused by low or high inertia are. To do so, we must take a short diversion to explore the relationship between inertia and kinetic energy in the hammer assembly using a simplified hammer assembly model.  In the following analysis key and wippen inertias are ignored because they are so much less than that of the hammer.

 

Simplified Hammer/Shank Model

 

The picture below is a model of a hammer supported by its shank, and pivoting about its normal pivot point.  The hammer is modeled simply as a mass mham, the shank is considered massless, and the length of the shank is Lh. 

 

 

 

 

 


To start, let's see how much torque we need to apply to impart a specify kinetic energy to the hammer.  We will use the following four formulas from the physics of rotational dynamics.

                                                                 Equation 1

                                                           Equation 2

                                                               Equation 3  

                                                                   Equation 4

where T=torque, q=angle (radians), w=angular velocity, I=moment of inertia, and t=time.

If Equations 3 and 4 are solved for t and then equated, the formula  is derived.  The terms of Equation 2 can be solved for w2 to yeild  .  Substituting these relations into Equation 1 yields.

                                                                     Equation 5

 

Let’s consider Equation 5 for a moment.  It could be considered the canonical equation for energy because it is explicitly in energy’s fundamental units, force and distance, or as it is in the rotational case, torque and angle.  It shows that as long as the torque applied to the hammer assembly and the distance it travels remains constant, then inertia has no effect on the energy imparted to the hammer.  How can this be, you might ask.  The answer lies in the variable not show in Equation 5—time.  We can substitute Equations 3 and 4 into Equation 2, and derive Equation 6, which describes energy in what is probably a more useful form.

                                                                      Equation 6

 

What this equation shows is that there is a tradeoff between moment of inertia, time, and hammer travel-distance for constant energy.  So, if the hammer assembly is lowered in moment of inertia, the key must be depressed more quickly, or the hammer must travel a longer distance in order for energy to remain constant.  Analogously, if hammer moment of inertia is increased, the key must be depressed more slowly or the hammer must travel a shorter distance for energy to remain constant.

 

Clearly, the time it takes to press a key must be held within some range for a piano to operate and feel normal.  For example, in a fast passage, the pianist has no choice but to press the key rapidly in order to be able to play subsequent notes.  So, high hammer inertia will force rapid passages to be played with much force—so much for pianissimo.  Low hammer inertia will result in a piano with less volume, even with hard key strikes.

 

One might wonder if, for a given amount of energy supplied to the hammer, the ratio of static down weight vs. dynamic force due to inertia can be varied.    Somewhat surprisingly, the answer is yes.  To do the analysis, we will consider only the forces caused by the hammer mass and neglect all other inertias.  Please refer to the analysis for variable definitions.  Let’s start by defining a new variable Rq­, which is the ratio of the hammer’s angular motion divided by the key’s angular motion.  From the basic equations, se know that

where Ik is the inertia as felt by the pianist’s finger

 

The energy imparted to the hammer is

                                                 Equation 7

 

Also from the action analysis we know that the force due to hammer inertia at the key is

                                                  Equation 8

where qk is the angular distance the key moves

 

The static force at the pianist’s finger is

                                                     Equation 9

 

 

If we solve Equation 7 for t2 and substitute it into Equation 8, we derive

                                                Equation 10

 

If we now solve Equation 7 for Rqqk and substitute that into Equation 10, we get

 

This result is interesting.  It says that for any given amount of energy imparted to the hammer, we can change the ratio between static and dynamic force by varying the hammer mass.  No other action parameter matters.  So, one can reduce the static force by reducing the hammer mass, or vice versa.  Of course, various action ratios and distances would have to be adjusted concomitantly.

 

In conclusion, we have seen that inertia plays a crucial role in the piano action, that ratios and hammer mass must be accurately held for inertia to be well controlled, that the wippen has virtually no effect on inertia, and the key has less than most people probably thought.  Finally we have learned that the ratio between static force and dynamic force can be adjusted by changing hammer mass. 

 

 

 

 

 

1The standard formulas presented here can be found in any number of physics books.  I got them from Physics, the excellent text by Halliday and Resnick.

 

Acknowledgements

Thanks to Christopher Brown of Concord Piano LLC for allowing me to weigh and measure action parts from his Steinway B.

Thanks to ADE Technologies for use of one of their non-contact distance measuring gauges to measure the range of times it takes to press a key.