Science & Engineering
Hydraulic cylinder force.
Push force F = P × bore area and pull force F = P × (bore area − rod area), from bore, rod diameter and gauge pressure, with optional efficiency and the pressure for a target force. Runs in your browser.
Cylinder
Result
Fill the form and press Calculate.
Worked example
A 100 mm bore cylinder with a 70 mm rod at 200 bar (20 MPa): bore area = π × 0.1² ÷ 4 = 7,854 mm², so pushing F = 20 MPa × 7,854 mm² = 157.1 kN. Pulling, the rod area (3,848 mm²) is lost: annulus = 4,006 mm², F = 80.1 kN, about 51% of the push force.
How it works
Pascal's law says pressure in an enclosed fluid acts equally on every surface, so force equals pressure times area, F = P × A. On the push stroke the fluid acts on the whole piston face, A = π D² ÷ 4. On the pull stroke the rod occupies part of the piston face, so the fluid acts only on the ring left over, A = π (D² − d²) ÷ 4, and the pull force is always lower than the push force at the same pressure.
Efficiency, if you enter it, simply scales both forces. The target-force option divides the force you want by the area (and efficiency) to give the pressure required. The same relations hold for a pneumatic (air) cylinder; use the air gauge pressure. Pressure must be gauge pressure, not absolute.
Not for safety-critical design
Planning and learning aid only. It is not for structural, lifting, pressure-system or other safety-critical design, and it is not professional engineering advice. Have a qualified engineer check anything that carries people or could fail dangerously.
Sources and basis: calculate.co.nz: Hydraulic cylinder force calculator; Wikipedia: Pascal's law. Checked 2026-10-16. Planning aid only; see the note above.
FAQ
How do you calculate hydraulic cylinder force?
Force equals pressure times area, F = P × A. Pushing, use the full bore area π D² ÷ 4. Pulling, use the bore area minus the rod area, π (D² − d²) ÷ 4.
Why is the pull force lower than the push force?
On the pull stroke the rod takes up part of the piston face, so the oil acts on a smaller ring-shaped area at the same pressure.
Is this force the real force I will get?
It is the theoretical force. Seal friction makes actual output slightly lower, and back-pressure on the return side reduces it further. Use the efficiency box to allow for it.
Does it work for pneumatic cylinders?
Yes, the same F = P × A applies with air gauge pressure. Air is compressible, so speed and force during motion can differ from a hydraulic cylinder.
Which pressure do I enter?
Gauge pressure, as read on a pressure gauge, not absolute pressure. Never exceed the cylinder's rated working pressure.