Hydraulic cylinder force formula

A hydraulic cylinder is a pressure-times-area actuator. The force formula is simple, but the area changes by stroke direction: extend uses the full bore area, while retract uses the smaller annular area left after the rod is subtracted. That rod-area difference is why a double-acting cylinder can push harder than it pulls at the same pressure.

The base formula

Start with Pascal's law in the form used for cylinder sizing:

F = P · A

where F is force, P is pressure and A is the effective piston area. In imperial units, psi × in² = lbf. In metric units, MPa × mm² = N because 1 MPa = 1 N/mm².

Push force on extension

On the extend stroke, pressure acts on the full circular piston face. If D is the bore diameter:

A_bore = (π/4) · D²

F_push = P · (π/4) · D²

This is the largest theoretical force a single-rod cylinder can make at a given pressure. A bigger bore increases force with the square of diameter, but it also slows the cylinder at a fixed pump flow because speed is v = Q / A.

Pull force on retraction

On the retract stroke, pressure is applied to the rod side of the piston. The rod blocks part of the area, so the working area is an annulus:

A_rod = (π/4) · d²

A_annulus = A_bore − A_rod = (π/4) · (D² − d²)

F_pull = P · (π/4) · (D² − d²)

The missing force is the rod-area differential:

F_loss = P · A_rod = P · (π/4) · d²

This is the number that often explains a retract problem. The same smaller annulus also means retract speed is faster than extend speed for the same flow.

Unit shortcuts

Inputs	Force shortcut	Use when
psi and in²	F_lbf = P_psi × A_in²	Imperial cylinder sizing
MPa and mm²	F_N = P_MPa × A_mm²	Clean metric math
bar and mm²	F_kN = P_bar × A_mm² / 10000	Common hydraulic pressure ratings
bar and cm²	F_kN = P_bar × A_cm² / 100	Catalog area tables

Worked example

Use a 50 mm bore, 22 mm rod and 160 bar working pressure. Convert pressure first: 160 bar = 16 MPa.

The bore area is:

A_bore = (π/4) · 50² = 1963.5 mm²

Therefore the extend force is:

F_push = 16 · 1963.5 = 31,416 N = 31.4 kN

The rod area is:

A_rod = (π/4) · 22² = 380.1 mm²

So the annular area is 1963.5 − 380.1 = 1583.4 mm², and the retract force is:

F_pull = 16 · 1583.4 = 25,334 N = 25.3 kN

The rod costs 31.4 − 25.3 = 6.1 kN on the pull stroke. That matches P · A_rod: 16 × 380.1 = 6082 N.

Pressure needed for a target force

If you already know the force you need, rearrange the same formula:

P = F / A

Use A_bore for an extend load and A_annulus for a retract load. If the required pressure is too close to the relief setting, step up the bore or reduce the load. The hydraulic pressure calculator runs this inverse calculation directly.

Practical checks after the formula

• Efficiency. Seal friction and rod-side back-pressure reduce real force. A first-pass derate of 5% to 15% is common, but measure or use the manufacturer data for final sizing.
• Rod buckling. A long extended rod pushing in compression is a column. Use the column buckling calculator when stroke length is long relative to rod diameter.
• Flow and speed. Larger area gives more force but less speed at fixed flow. Use the hydraulic cylinder calculator when you need force and stroke speed together.
• Side load. Cylinder formulas assume axial loading. Side load damages bearings, seals and rods; guide the load mechanically instead of asking the cylinder to act as a slide.

References

• Engineering ToolBox: Hydraulic Force vs. Pressure and Cylinder Size
• Engineering LibreTexts: Force Output of a Retracting Cylinder

Frequently asked questions

What is the hydraulic cylinder force formula?

The basic formula is F = P × A. For extension, use the full bore area: F_push = P × πD²/4. For retraction, subtract the rod area: F_pull = P × π(D²-d²)/4.

Why is retract force lower than extend force?

The piston rod occupies part of the rod-side piston face. Pressure acts only on the remaining annular area, so pull force is lower by P × πd²/4.

How do I calculate cylinder force in tons?

Calculate force first, then convert. In imperial units, lbf = psi × in² and US short tons are lbf / 2000. In metric, calculate newtons and divide by 1000 for kN or by 9.80665 for kgf.

Does the formula include friction?

No. The pressure-times-area result is theoretical. For a practical first pass, multiply by an efficiency factor such as 0.85 to 0.95, then account for back-pressure, side load and mounting losses separately.

Last reviewed: 2026-05-30.