Hydraulic Cylinder Calculator
Enter bore, rod diameter and system pressure to get extend and retract forces — the rod side is weaker, and this shows exactly by how much.
Last updated: 2026-09-15
How the calculation works
- Force is pressure times area. Extending uses the full bore area; retracting uses the annulus (bore minus rod).
- The retract figure shows the force penalty of the rod — typically 30–50% less than extend at the same pressure.
Formula
F_extend = P × π/4 × D_bore² F_retract = P × π/4 × (D_bore² − D_rod²)
| Symbol | Meaning | Unit |
|---|---|---|
F | Force | N |
P | Hydraulic pressure | MPa |
A | Effective piston area | m² |
Worked example
A 63 mm bore, 28 mm rod at 160 bar: extend area = 31.2 cm² → 49.9 kN. Retract area = 25.0 cm² → 40.0 kN — 20% less on the return stroke, which matters when the cylinder pulls a load back.
Interpreting the result
The extend/retract asymmetry is a design lever: rodless cylinders push and pull equally; big rods trade retract force for buckling stiffness. Real output is 5–10% below calculated (seals, line losses), and long strokes in compression need rod buckling checks before force checks.
Assumptions
- Standard double-acting cylinder with round rod.
- Neglects seal friction and back-pressure on the exhaust side.
Limitations
- Regenerative circuits (rod and cap sides connected) behave differently.
- Telescopic and plunger cylinders have distinct areas.
Frequently asked questions
How much force does a 4-inch cylinder make at 2000 psi?
F = 2000 × π × 2² = 25,133 lbf ≈ 12.5 tons extending. Retracting with a 2-inch rod: 18,850 lbf.
Why is retract force lower?
The rod occupies area on the retract side, so the same pressure acts on less piston area — the annulus.