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*To*: "'Orin Eman'" <orin@wolfenet.com>, Quattro List <quattro@coimbra.ans.net>*Subject*: RE: Piston acceleration (was interferance etc)*From*: glen powell <gpowell@acacianet.com>*Date*: Sat, 24 Jan 1998 22:47:56 -0500*Sender*: owner-quattro@coimbra.ans.net

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.

**Follow-Ups**:**Re: Piston acceleration (was interferance etc)***From:*Arun Rao <rao@pixar.com>

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