Inertia

Ed Sutton ed440@mindspring.com
Fri, 26 Dec 2003 22:35:05 -0500


Jim-

This is pretty convincing to me (who is not an engineer) that good enough is good
enough.

Assuming 4 large leads at 12 gr. each, somehow fitted in a tiny space, at front
they would represent about 1/7 the total force of acceleration, at 1/2 about 1/26,
at 1/4 about 1/101.  Maintaining frontweight by doubling the lead weights, would
give 1/13 total force at 1/2 distance.  (And I doubt anyone's going to be using 8
large leads on even the worst actiopn.)

Good enough in this case would mean "stay within Stanwood's front weight ceilings
and keep the leads toward the back, but don't go crazy over it."

Does this seem to be the understanding you intended?

Thanks.

Ed Sutton
----- Original Message -----
From: "James Ellis" <claviers@nxs.net>
To: <caut@ptg.org>
Sent: Thursday, December 25, 2003 2:45 PM
Subject: Inertia


> >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
>
>
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