Pianotech

Blaine,

Both you and Jon are correct in assuming the squareness of the strike angle and hammer core to shank angle to deliver the best energy transfer, but both miss the most important issue – center of oscillation.

The loss of power transfer to the string and the strain on the flange is experienced when the tip of the hammer (string contact point) is not inertially balanced during contact. We all have experienced the "sting" in our hands when we hit a baseball that is too far out or too close to the "sweet spot". This sweet spot is the point on the bat (or hammer shank assembly) where the mass x speed is balanced on each side of the contact point – that is approximately 2/3 on a uniform rod. The logic is this; the outer regions of a bat are traveling faster than the inner (closer to the rotation point) portions of the bat. To balance the inertia, more mass is needed inward, toward the rotational point – because it is traveling slower. Less mass is needed toward the outer portions of contact because it is traveling faster. Physics 101

Blaine,

Both you and Jon are correct in assuming the squareness of the strike angle and hammer core to shank angle to deliver the best energy transfer, but both miss the most important issue – center of oscillation.

The loss of power transfer to the string and the strain on the flange is experienced when the tip of the hammer (string contact point) is not inertially balanced during contact. We all have experienced the "sting" in our hands when we hit a baseball that is too far out or too close to the "sweet spot". This sweet spot is the point on the bat (or hammer shank assembly) where the mass x speed is balanced on each side of the contact point – that is approximately 2/3 on a uniform rod. The logic is this; the outer regions of a bat are traveling faster than the inner (closer to the rotation point) portions of the bat. To balance the inertia, more mass is needed inward, toward the rotational point – because it is traveling slower. Less mass is needed toward the outer portions of contact because it is traveling faster. Physics 101

Blaine,

Both you and Jon are correct in assuming the squareness of the strike angle and hammer core to shank angle to deliver the best energy transfer, but both miss the most important issue – center of oscillation.

The loss of power transfer to the string and the strain on the flange is experienced when the tip of the hammer (string contact point) is not inertially balanced during the hammer travel. We all have experienced the "sting" in our hands when we hit a baseball that is too far out or too close to the "sweet spot". This sweet spot is the point on the bat (or hammer shank assembly) where the mass x speed is balanced on each side of the contact point – that is approximately 2/3 on a uniform rod. The logic is this; the outer regions of a bat are traveling faster than the inner (closer to the rotation point) portions of the bat. To balance the inertia, more mass is needed inward, toward the rotational point – because it is traveling slower. Less mass is need toward the outer point of contact because it is traveling faster. Physics 101

Cobrun,

No. the added mass of the shank near the flange has nothing to do with the "balance inertia", as I defined it – although it exasperates it.  The added mass is simply a strength (stiffness) issue. If a shank is uniformly constructed, as evident on the hammer end, it would be too weak to sustain the necessary acceleration applied at the knuckle.

Also, my post was not comprehensive in explaining "center of oscillation" (aka, center of percussion). I would first state that the hammer/shank assemblies of both the upright and grand piano are far from obtaining the balanced inertia because of the "stuff" the hammer/shank assembly must negotiate around – in grands, the pinblock, in uprights the damper assembly.  Certainly, the construction of a typical "clapper" of a bell is evident of balanced inertia. Notice that some well-balanced clappers have a small added mass located beyond the spherical mass that strikes the bell.  This added mass is needed to balance the inertia on both sides of the contact point of the clapper. If one were to calculate the mass from the lower half (hanging bell) from the upper half (including the shaft) of the clapper contact point you would realize that there would be an inertia imbalance without that smaller added weight. Remember, the shaft is part of the balanced inertia equation.

Skipping over added minutia; to affect the ideal balanced inertia of a piano hammer/shank assembly, it would require a certain added weight on the outer portion of the hammer core center line and below (grand hammer description) the shank. The current typical hammer/shank designs cause the center of inertia to be traveling, not at 90°, which produces the maximum transfer of power, but approximately 5° off perpendicular. Additionally, that balanced offset sends a shock wave back down the shank where it is reflected back and forth between the hammer and flange. That is the "sting" I described with the baseball bat.  This is why high-speed photography shows the hammer hitting the string several times when struck.

P.S. Wessel Nickel & Gross hammer shanks greatly reduce this off balance phenonium.

Blaine,

Both you and Jon are correct in assuming the squareness of the strike angle and hammer core to shank angle to deliver the best energy transfer, but both miss the most important issue – center of oscillation.

The loss of power transfer to the string and the strain on the flange is experienced when the tip of the hammer (string contact point) is not inertially balanced during contact. We all have experienced the "sting" in our hands when we hit a baseball that is too far out or too close to the "sweet spot". This sweet spot is the point on the bat (or hammer shank assembly) where the mass x speed is balanced on each side of the contact point – that is approximately 2/3 on a uniform rod. The logic is this; the outer regions of a bat are traveling faster than the inner (closer to the rotation point) portions of the bat. To balance the inertia, more mass is needed inward, toward the rotational point – because it is traveling slower. Less mass is needed toward the outer portions of contact because it is traveling faster. Physics 101