MOInertia

[Ed Sutton]

Richard-

Take another look at Jim Ellis' post of a few days ago, and see if it doesn't
answer your concern.

It seems to me he is talking about the relative amount of work that must get done
to make the hammer bonk the string at any velocity.  This amount doesn't change as
key ratios change, the only thing that changes is how far and fast (relatively)
the key must move, and that appears to be covered in his analysis.  I believe the
analysis would be the same if the hammer were at the end of a 5/1 lever with no
intermediate parts.

I've copied Jim's post below.

Happy New Year!

Ed Sutton

>
> I've asked this point on two occasions already and we havn't gotten
> really into the point. Its one thing to point out the individual MOI of
> the parts as John, and others have done. But how these turn out after
> leverage is figured would be another thing. If it came anywhere close to
> the roughly 5:1 ratio figure... then suddenly the Hammer and Shank MOI
> become reduced by a factor of 5 and we are talking a whole different
> degree of significance for the key MOI.
>

Jim wrote:

>From reading the latest posts, it's obvious that any further discussion of
"inertia" is hopeless.  So, I'll change tracks here, and try something else.

Suppose we have a total action ratio of 1:5 from key front to hammer head.
Suppose we have a 12 gram hammer head.
Suppose we have a 12 gram key lead.

We put the key lead half way between the key front and the balance hole.
The lead weight will now move 1/10 as far as the hammer head.
The lead weight will move 1/10 as fast as the hammer head.

Kinetic energy = 0.5 mass x velocity squared.

We can't calculate the actual kinetic energy of the hammer head or the
lead, because we don't know what the velocity is.  But what we do know is
that the hammer head has the same mass as the lead weight, and it will be
moving 10 times as fast.  Since KE = 1/2 mV sq, and 1x1=1, and 10x10=100,
it's obvious that the V sq of the hammer head is 100 times that of the lead
weight.  Since the masses are the same, it's also obvious that the kinetic
energy of the hammer head will be 100 times that of the lead weight.

Work done = force x distance.  Since the distance the key front travels is
the same, it's also obvious that it will take only 1/100 of the force at
the key front to accelerate the lead weight as it will to accelerate the
hammer head.

If we put the lead weight out at the front of the key (where we don't put
lead weights) it will have 1/25 the kinetic energy of the hammer head.  If
we put it 1/4 the way out in front of the balance rail, it will have 1/400
the kinetic energy of the hammer head.

The force required to accelerate or decelerate the lead weight over a given
distance will be proportional to the kinetic energy given to, or given up
by, the lead weight.  The same principle is true of the hammer head.
Therefore, if the lead weight is half way between the balance hole and the
key front, it will require 100 times as much force at the key front to
accelerate the hammer head as it will to accelerate the lead weight.

The static contidions will be very different.  The force exerted by the
lead weight will be directly proportional the the lever ratio - whether
it's a simple lever, or a compound lever.  The same will be true of the
hammer head.

I hope this helps.

Jim Ellis