How to size a hydraulic cylinder

Sizing a hydraulic cylinder comes down to three coupled choices: the bore (which sets force), the rod diameter (which sets pull force, buckling resistance and regen behaviour), and how those interact with your pump’s flow (which sets speed). Get the bore right for the load, then check the rod and the speed. Below is the order I work in, with the formulas and a worked example.

1. Force comes from pressure times effective area

A cylinder is a pressure-times-area machine. The force it makes is the system pressure acting on whichever piston area is exposed:

F = P · A

On the extend (push) stroke, pressure acts on the full piston face, so the effective area is the full bore:

F_push = P · (π/4) · D²

On the retract (pull) stroke, the rod takes up part of the piston, so pressure acts only on the annular (ring-shaped) area that’s left:

F_pull = P · (π/4) · (D² − d²)

where D is the bore diameter and d is the rod diameter. The rod removes area, so pull force is always lower than push force for the same pressure. This asymmetry is the single most important fact in cylinder sizing: a cylinder that’s strong enough pushing can be too weak pulling.

2. Mind the units

The arithmetic is trivial; the unit slips are what bite. Work in N, mm and MPa and the numbers fall out cleanly, because 1 N/mm² = 1 MPa:

• 1 bar = 0.1 MPa = 0.1 N/mm²
• force (N) = pressure (MPa or N/mm²) × area (mm²)

So a pressure quoted in bar should be divided by 10 to get MPa before you multiply by an area in mm². Skip that step and your force comes out 10× too big. For a compact formula reference with imperial and metric shortcuts, see the hydraulic cylinder force formula.

3. Size the bore for your target force

Start from the worst-case force the cylinder must deliver — usually the push stroke — and the pressure your system can reliably supply. Rearranging the push equation for the bore:

D = √( 4F / (π · P) )

Compute the ideal bore, then round up to the next standard ISO bore (40, 50, 63, 80, 100, 125, 160 mm…). Rounding up gives you force margin; rounding down leaves you short. Don’t size to your relief-valve setting either — leave headroom so pressure spikes don’t put the cylinder at its limit on every cycle.

4. Size the rod for buckling, pull and regen

The rod is chosen for three reasons, and the governing one depends on the application:

• Buckling. A long, extended rod under a compressive load is a column, and columns buckle (Euler). On long strokes the rod diameter is set by the Euler buckling check long before stress matters, which is why long-stroke cylinders often have surprisingly fat rods. Mounting style matters here — a pinned-pinned cylinder buckles at a lower load than a rigidly fixed one.
• Pull force. The rod sets the annular area (π/4)(D²−d²), so it directly limits how hard the cylinder can pull. If the pull stroke does real work, check it explicitly.
• Regen ratio. In a regeneration circuit the bore-to-annulus area ratio sets the speed and force during regen. A common 2:1 cylinder (annular area half the bore area) gives equal extend/retract speed in regen.

5. Speed comes from flow divided by area

Once the areas are fixed, the pump’s flow rate sets the stroke speed:

v = Q / A

Because the extend and retract areas differ, extend and retract speeds differ too for the same flow — retract is faster (smaller annular area), extend is slower (full bore). The trade-off is fundamental: a bigger bore gives more force but moves slower at a given flow. If a cycle is too slow, you either raise flow (a bigger pump — see the pump flow & HP calculator) or accept a smaller bore and higher pressure.

Worked example: 63 mm bore at 160 bar

Take a D = 63 mm bore running at 160 bar. First convert the pressure: 160 bar = 16 MPa = 16 N/mm².

Push force. The full bore area is (π/4)·63² = 3117.2 mm², so:

F_push = 16 × 3117.2 = 49 875 N ≈ 49.9 kN

Pull force with a d = 36 mm rod. The rod removes (π/4)·36² = 1017.9 mm², leaving an annular area of 3117.2 − 1017.9 = 2099.3 mm²:

F_pull = 16 × 2099.3 = 33 588 N ≈ 33.6 kN

So this cylinder pushes with about 49.9 kN but pulls with only 33.6 kN — the 36 mm rod costs roughly a third of the force on the return stroke. That’s exactly the kind of asymmetry you must design around, and it’s what the calculator reports for these inputs.

Putting it together

The full sizing loop is: (1) find the worst-case force; (2) solve D from that force and your working pressure, then round up to a standard ISO bore; (3) pick a rod for buckling on the stroke length plus the pull force you need; (4) check extend and retract speeds against your pump flow and adjust the bore, pressure or pump if the cycle time is off. If you’re also choosing engine-style bore and stroke for displacement, the bore & stroke calculator handles that side. To run the force numbers for any bore, rod and pressure,

Frequently asked questions

Why is a cylinder’s pull force lower than its push force?

On the rod side the rod occupies part of the piston, so the pressure acts on the annular area (π/4)(D²−d²) instead of the full bore (π/4)D². Less effective area at the same pressure means less force — and, for a fixed flow, a faster stroke.

How do I pick the bore for a target force?

Rearrange F = P·(π/4)·D² to D = √(4F/(π·P)), using consistent units — force in N, pressure in MPa (N/mm²), area in mm². Then round up to the next standard ISO bore so you meet the force with margin.

What sets the rod diameter?

Three things: the pull force you need on the annular side, resistance to buckling on long strokes (an Euler column check), and the regeneration ratio if you run a regen circuit. On long-stroke or pushing-only cylinders, buckling usually governs.

Last reviewed: 2026-05-29.