F=ma applied to slider crank piston motion

Kalus · Feb 4, 2012

I'm struggling to apply F=ma to the motion of a slider crank, more specifically the piston. I want to find out how much force is nessesary to keep the piston in motion. Essentially, in the mechanism there is acceleration and deceleration, does that mean only at some points in the cycle there is force needed to move the piston?

Using the equations derived for the motion of a piston:

http://upload.wikimedia.org/wikipedia/en/math/1/6/8/1686ee2f8b1d67ce1eb1aa4fb4b0daac.png

Three lines of derivation for the derivative x' of a slider crank piston position: x' = dx/dA = -r sin A + ((1/2)(-2) r^2 sin A cos A)/sqrt(l^2 - r^2 sin^2 A), then simplified to -r sin A - r^2 sin A cos A / sqrt(l^2 - r^2 sin^2 A).

$Typeset equations giving the second derivative x'' of piston displacement with respect to crank angle A, written as -r cos A minus fractions involving sqrt(l^2 - r^2 sin^2 A) and r^2 sin A cos A terms.$

Gives you this:

Line graph of piston motion versus crank angle from -360 to 360 degrees, showing position, velocity and acceleration curves for three crank radii (R=1.80 green, 2.00 red, 2.20 blue) at L=6.00.

Kalus · Feb 4, 2012

Sorry, I should add why I'm asking this. I have the following diagram:

Graph of force per unit area in N/mm² against crank angle φ from 0 to 720 degrees, with curves labelled f_g, f_j and f, the resultant.

Where f_g is the combustion force, f_j is the inertial force and f is the resultant force. I'm trying to work out what's providing the force to overcome the acceleration around 650 to 90 degrees.

256bits · Feb 4, 2012

shaft flywheel

F=ma applied to slider crank piston motion

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