Pneumatics

Force — Double-Acting Cylinder

Calculate push (cap-end) and pull (rod-end) force of a double-acting pneumatic cylinder, including friction correction and pressure drop.

Units:
in LinkedIn
💡Pull force is always less than push force because the rod reduces the effective piston area. Add minimum 25% safety margin for dynamic applications. Relation: 1 kg ≈ 9.81 N.
F [N] = A [m²] × (P − ΔP_correction) [Pa] × (1 − friction%/100)
P P → F Ø bore
Input
Cylinder bore iInternal bore diameter of the cylinder in millimetres.
mm
Rod diameter iPiston rod diameter in mm. Standard pairings: Ø50→Ø20, Ø63→Ø25, Ø80→Ø32, Ø100→Ø40 mm.
mm
Working pressure iAvailable supply pressure at the cylinder port in bar gauge.
bar
Pressure correction iDeduction for pressure loss in hoses, fittings and valve. Standard allowance: 0.3–0.5 bar.
bar
Seal friction iInternal friction of piston and rod seals as a percentage of theoretical force. Typically 5–10%.
%
Result
Cap-end area (push) [cm²]
Rod-end area (pull) [cm²]
Push force (cap-end) [N]
Push force [kg]
Pull force (rod-end) [N]
Pull force [kg]
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