[pianotech] Force equivalents in different actions

[David Love]

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