# RE: Piston acceleration (was interferance etc)

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This is good stuff, but could you please splain it is layman's terms?

Conceptually, I find it difficult to understand how acceleration is greatest
at TDC, as the piston stops moving up, is momentarily motionless, and
the starts to accelerate downward. How can acceleration be greatest at the
instant in time that upward motion has ceased, the piston is motionless
and then downward acceleration commences? It seems to me that acceleration
must do to zero at some point as the piston approaches its point of zero motion.

How about some plots, one showing piston velocity and another showing piston
acceleration, superimposed and showing the complete cycle from TDC to TDC,
he aksd.

Guessing here again, and mebbe this is where my logic falls apart, I would think that
max piston velocity would occur at mid-stroke and that acceleration would be zero
at mid-stroke?

Thanks!

-glen

Well, I pulled some references...

According to the Bosch handbook, the inertia force is proportional to
N^2 * (cos(alpha) + lambda * cos(2*alpha))
where N is RPM and alpha is crank angle (TDC is 0),
lambda is stroke/(2*rod length).

cos(x) is at a maximum (1) for x = 0 and minimum (-1) for x = 180

This gives acceleration proportional to

N^2 * (1 + lambda) at TDC
N^2 * (1 - lambda) at BDC

The magnitude is greatest at TDC.

This is also covered in Corky Bell's book in the first few pages.

"Internal Combustion Engine Fndamentals" by John B Heywood
has a formula for velocity and differentiating it for acceleration,
I got:

2 * PI^2 * N^2 * stroke (1 +/- lambda)

Seems consistent.

It's notable that the force on the con rods increases with RPM squared,
so think twice before removing those rev limiters...

Orin.

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