Re: Conrod force/acceleration

[Audinamx <Audinamx@aol.com>]

>From my point of view it seems that the terms "acceleration" and
"deceleration" are confusing folks.  Acceleration and deceleration are exactly
the same thing (changes in velocity) only in different directions.  In that
context "acceleration" is force in the direction of motion, causing velocity
to increase, while deceleration is force against the direction of motion,
causing velocity to decrease.  The nature of having a piston attached to an
offset pin on a rotating crankshaft nominally running at a constant rate of
rotation is that the velocity of the piston will have a sinusoidal
characteristic.  It is true that in theory as the piston is precisely at TDC
(and BDC for that matter) the bottom end of the rod is moving perpendicular to
the direction of the cylinder, so the piston is stationary ... but it turns
out that the acceleration that the piston is experiencing is the *greatest* at
this point.  This is why the rod experiences the greatest stretching at TDC.
The other characteristic of simple harmonic motion is that it is when the
piston is at the center of its travel and the velocity is at its highest that
the piston experiences a brief moment of no acceleration!  What this means is
that just prior to that midpoint the crank is pulling on the conrod (speeding
it up, "accelerating" it), and right after it is pushing the rod (slowing it
down, "decelerating" it).

I think Wolff's example was the best illustration that just because you are
instantaneously at zero velocity does not mean there is zero acceleration.
After you let go of that ball the only thing acting on it is the force of
gravity (32 ft/sec^2).  If you throw the ball straight up it will come to a
stop before beginning to fall back to the ground ... all done under constant
_acceleration_ ...

I hope that this helps to explain the physics of the situation ...

Steve Buchholz
San Jose, CA (USA)