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RE: interference fit
On Sat, 24 Jan 1998 07:21:32 -0500, glen powell wrote:
>>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,
>Is not piston acceleration zero *at* TDC and *at* BDC as the piston is at a
>at that instant in time? Perhaps de/acceleration is highest immediately
It does kinda seem funny to say acceleration is max at TDC (or BDC) but
it's true, at least in the highest rpm range where the inertial forces dominate
over the frictional and cylinder pressure forces and where this kind of
discussion (rod breakage) applies.
In that domain, velocity is zero at TDC/BDC but remember the acceleration
profile is 90 degrees out of phase with velocity, at least in a freely
Starting at about 90 degrees BTDC, the crank is putting the brakes on the
trying to keep it from smashing through the head. Rod stretch time. At TDC, the
crank has succeeded in preventing disaster and now is jerking the piston
back toward the crank case. The rod is at its maximum stretched length at this
moment. The force producing that stretch is the peaking downward acceleration.
Since this same thing happens at BDC and since cylinder pressure would _add_
to the force stretching the rods at the bottom end, instead of opposing it, I'm
wondering if it would be a good idea to design a cam which got those exhaust
valves open very early in a turbo engine.
DeWitt Harrison firstname.lastname@example.org