Rod Push Force

Last modified by

on

Jul 24, 2020, 6:28:07 PM

Created by

on

Jul 12, 2014, 11:37:39 AM

F 2 = \frac{π}{4} \cdot (d_{2}^{2}) \cdot p_{2}

$F2 = pi/4*( d_2 ^2)* p_2$

piston diameter (inches)

$"piston diameter (inches)"$

pressure in the cylinder (opposite rod) (lff/in2)

$" pressure in the cylinder (opposite rod) (lff/in2)"$

•

π

$pi$ by MichaelBartmess

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RodPushForce-illustration.png

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