What matters most?

[Mark Davidson]

>The issue is really one of the "moment of inertia" which is a measure of
>the resistance of an object to changes in its rotational motion, or torque.
>It is not quite the same as Newton's second law f=ma.
>
True.  You need to use the rotational version:

torque = inertia * angular_acceleration.  They are analagous though.

>If I recall my basic physics (and there's no guarantee of that), you are
>calculating the torque as a function of the angular acceleration, the
>distance from the axis of rotation and the mass.

Torque is analogous to force.  Measured as force * radius from center
of rotation (balance rail and center pins).

>In the key, the axis of
>rotation is clearly at the balance rail.  In the hammer shank assembly, I
>don't recall whether you calculate the axis of rotation from the flange
>center pin, or from the knuckle mounting,

Hammer center would be the center of rotation, torque would be knuckle
radius (e.g. 17mm) X force on knuckle.

>but I believe it is from the
>knuckle mounting since that is where the force is being applied in the
>second class lever.

>Similarly the wippen lever must be taken into account.
>The formula goes something like  t = m*r^2*a.

>Torque equals mass x distance from the axis of rotation
>x angular acceleration.  But I'm not
>sure about this.

Just force X distance.

>The torque in a system of compound levers, I believe, is
>a simple sum.

Take input torque, multiply by product of all the levers.  I think this
is effectively input_torque/SWR.

>The question would be, then, how  coordinated changes in the
>set of levers such that the overall action ratio remains equal effect the
>moment of inertia in each system.  In more practical language: what happens
>to the torque when the knuckle is at 16mm and the key ratio is at .48
>versus when the knuckle is at 17mm and the key ratio is at .52; all other
>things being equal.  The issue become complex because each lever has mass,
>distance from the axis of rotation and acceleration which are distintly
>different from the other (the key's angular acceleration is much different
>from that of the hammer shank).  What will be changing in this experiment
>is only the r (or does the angular acceleration of each component change,
>though the relationship between them remains the same???  Not sure.)
>Anyway, since the input "r" is squared, it's hard to imagine that changes
>in three unequal levers that keep the overall action ratio the same will
>not yield differences in the moment of inertia when all components are
>added together.
>
>I'll take my answer on the air--and good luck.

Still working on it... but let's start with this example:

1st put 50g on key to overcome BW and friction.  Any additional
force will move key.

Take

hammer ratio = 130 / 17
wippen ratio = 92 / 60
key ratio = 116 / 223

If we put 100 g at key front, then torque will be 100*223=22300 g mm.
Force at capstan will be 22300/116 = 192 g.  (same torque, smaller
radius, higher force).

Now put 192 g force on wippen to get 60*192=11520 g mm torque on wippen.
Output force at jack will be 11520/92=125 g.

Torque on shank is 125 * 17 = 2125g mm.  Effective force on shank at
radius of hammer is 2125/130=16 g.


-Mark