The force produced by a double acting hydraulic piston on the rod side can be expressed as
F1 = π / 4 (d22 - d12) P1 (1)
where
F1 = rod pull force (lb)
d1 = rod diameter (inches)
d2 = piston diameter (inches)
P1 = pressure in the cylinder (rod side) (lff/in)
The force produced opposite the rod can be expressed as
F2 = π / 4 d22 P2 (2)
where
F2 = rod push force (lb)
P2 = pressure in the cylinder (opposite rod) (lff / in)
The diagram below indicates the rod pushing force for cylinders with different diameters and pressures.
where
F1 = rod pull force (lb)
d1 = rod diameter (inches)
d2 = piston diameter (inches)
P1 = pressure in the cylinder (rod side) (lff/in)
The force produced opposite the rod can be expressed as
F2 = π / 4 d22 P2 (2)
where
F2 = rod push force (lb)
P2 = pressure in the cylinder (opposite rod) (lff / in)
The diagram below indicates the rod pushing force for cylinders with different diameters and pressures.
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