Hi, David: What we're dealing with is momentum. If you change the mass or the velocity in equal but opposite percentages, you'll get the same momentum. It is linear, not exponential. Formulas that have a factor raised to a power other than one (like 5 squared equals 25) will be non linear. Replacing a set of hammers that are heavier will require more force to accelerate them at the same rate as the originals. That will give them more momentum. An interesting question would be to find out whether the sound produced is the same if a heavier hammer with less velocity is compared to a lighter hammer with more velocity. The laws of physics would say it would be the same either way, as the same energy is imparted to the string. My question is, does it sound the same given the same input? Using a weight dropped on a key from a certain height, you could make the experiment. Or use the record feature of a player system. The pianist's perception and the instantaneous acceleration of the key is a difficult thing to measure. I'm sure the people at PianoDisc, Yamaha, QRS, and other player manufacturers are swimming in this stuff. How to make a mechanism to impart the energy to a string in exactly the same way as it was initially played by the artist is a very complicated affair. Given all the differences of action ratios, hammer weights, etc., between pianos, making a machine to replicate the original performance isn't easy. But the same principles are at work with mass, velocity, momentum and time. There are differing technologies for energizing the solenoids amongst the different player manufacturers. They don't just turn on and off, but are energized at various levels according to where the plungers are in the stroke. Similarly, the pianist is energizing the key at different points in the keystroke. My question is, if you change the hammer weight and compensate for it, will the pianist notice the change? Is the system now more efficient (and hence seem to have more power, whether real or perceived)? The laws of physics doesn't change, but the question is, how does the performer perceive the changes you make? Ex-engineering student, Paul McCloud San Diego -----Original Message----- From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On Behalf Of David Love Sent: Tuesday, March 23, 2010 7:47 PM To: pianotech at ptg.org Subject: Re: [pianotech] Force equivalents in different actions Forgive the awkward formulation of the question. Let's put it a different way. All other things being equal, it is my observation that a higher action leverage will produce more power than a lower leverage action. I presume this is because for the same velocity that the key travels downward, the hammer speed is greater in the higher leverage action. For example, if I press a key in an action whose action ratio is 6:1 at a velocity of, say, 10 mph, then the hammer moves at a velocity of 60 mph because, after all, it must cover 6 times the distance. If I press a key in an action whose ratio is 5:1 at the same 10 mph velocity, the hammer moves at 50 mph (let's put aside shank flexing and such for the moment). Since the velocity of the 6:1 hammer is greater it imparts more energy to the string, it has more power. Thus, if I change the action from a 6:1 to a 5:1 ratio and want the same power output then I stands to reason that I would have to change the mass of the hammer. By how much would I have to increase the mass in that situation in order to achieve the same level of power. More to the point, if my concern were in exceeding the original power output, where should I draw the line in terms of increased mass. Is it a straight relationship? If I reduce the hammer speed by 10% will an increase in the mass of the hammer by an equal percentage create the same power output? Or is the relationship non linear. This difference, btw, is very well illustrated by the Stanwood adjustable leverage action where you can easily hear (and feel) the difference. David Love www.davidlovepianos.com -----Original Message----- From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On Behalf Of John Delacour Sent: Tuesday, March 23, 2010 5:04 PM To: pianotech at ptg.org Subject: Re: [pianotech] Force equivalents in different actions At 16:24 -0700 23/3/10, David Love wrote: >... The answer I've suggested seems too simple on the surface (of >course we're assuming hammers of equal density and not consider for >the moment the increase in blow distance required for a lower AR >which will increase the force * distance part of the equation.) ... It's rather difficult to understand the question because of the loose way you use the special terms, but it seems you are seeking to attain equal momentum in both set-ups, and momentum = mass x velocity so the matter is pretty simple. The following article might flesh things out a bit, if you need that: <http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html> JD No virus found in this incoming message. Checked by AVG - www.avg.com Version: 8.5.437 / Virus Database: 271.1.1/2736 - Release Date: 03/23/10 19:33:00
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