RE: interference fit

[glen powell <gpowell@acacianet.com>]

>> <snip>
>> Quite simply, the acceleration of the piston is at its greatest at TDC
>> and BDT.  To accelerate anything, you have to supply a force.  At TDC,
>> <snip>

>> Is not piston acceleration zero *at* TDC and *at* BDC as the piston is at a dead stop
>> at that instant in time? Perhaps de/acceleration is highest immediately before/after
>> TDC/BDC?

>Speed is 0, acceleration is rate of change of speed and definitely isn't 0.
>The simplest example of speed being instantaneously 0 while under acceleration
>is throwing something in the air.  As it comes to a halt before falling,
>its speed is 0.  Acceleration is a constant 1g though.

>You have to calculate the speed the piston reaches as it rises in the
>bore and the acceleration required to bring it to a halt at TDC.
>The frictional forces (and weight if the bore is close to vertical)
>are orders of magnitude _less_ than that required to decelerate the
>piston.  Therefore the force must be supplied by the rod.

>I'll check the references tonight.

>Orin.

OK, I see your point but I'm still not clear on it. I think of it (prolly wrong) this way:

The piston is moving up toward TDC.
At some instant in time it will be at rest at TDC.
For it to achieve the state of rest or 'no motion' the piston must decelerate to 0.
At this instant in time it has no motion and no acceleration.
The piston then must accelerate from 0 to proceed to move back down.
So, in my primitive and prolly wrong analysis, the piston decelerates to 0
as it approaches TDC, @ TDC it has no motion and no acceleration, it then
reverses direction and accelerates downward, so at some instant in time
when upward deceleration is zero and the piston is motionless in space,
downward acceleration is also zero, or it could not be motionless.
There must be an instant in time between upward motion and downward motion
where both speed and acceleration are both zero. (?)

Mebbe I am wrong because it may be possible for the piston to go from upward
deceleration to downward deceleration without reaching zero acceleration at TDC
while it is instantaneously at rest at TDC and before downward deceleration commences?

-glen