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MachineCalcs

Hydraulic Cylinder Force Calculator

Push and pull force from bore, rod diameter and system pressure — with the rod-area differential shown explicitly. Standard ISO bore sizes. Metric and imperial.

Inputs

mm
mm
bar
L/min

Results

Output force(F)
31 420N

31.42 kN · 7 063 lbf · 3.2 t

Push stroke (full bore).

Push force (full bore)31 420N

7 063 lbf

Pull force (annulus)25 330N

5 695 lbf

Rod area subtracted.

Rod-area differential6 082N

1 367 lbf

Force the rod removes on the pull stroke (P × rod area).

Piston area1 963mm²

Rod area380.1mm²

Annulus area1 583mm²

Stroke speed(v)169.8mm/s

From flow ÷ working area.

  • Theoretical force from pressure × area; excludes seal friction, back-pressure and dynamic losses (typical mechanical efficiency 0.85–0.95).
  • Pull (retract) force is lower than push by exactly the rod-area differential, because the rod removes area on the rod side.

How it works

A hydraulic cylinder’s force is simply pressure acting over an area. On the push (extend) stroke, pressure acts on the full piston face: F_push = P · (π/4) · bore² On the pull (retract) stroke the rod takes up part of that face, so pressure acts only on the annulus — the bore area minus the rod area: F_pull = P · (π/4) · (bore² − rod²)

The difference between the two is exactly the rod-area differential, P · (π/4) · rod² — the force the rod removes on the pull stroke. This calculator shows that line explicitly, because it is the number people miss when a retracting cylinder won’t pull what they expected. The same rod area makes the cylinder retract faster than it extends for a given flow, since speed = flow ÷ area.

Worked example

A 50 mm bore cylinder with a 22 mm rod at 160 bar (16 MPa). The piston area is ≈ 19.6 cm², so the push force is ≈ 31.4 kN (3.2 tonne, 7,060 lbf). The rod area is ≈ 3.8 cm², so on retract pressure acts on the ≈ 15.8 cm² annulus and the pull force is ≈ 25.3 kN. The shortfall — the rod-area differential — is ≈ 6.08 kN, exactly P × rod area. At 20 L/min the cylinder extends at ≈ 170 mm/s. Load this page and the calculator shows these numbers.

Reference data

Standard metric cylinder bore and rod sizes (ISO 6020-2), with the piston and annulus areas and the force each develops at 100 bar. Force scales linearly with pressure — at 200 bar, double these; at 250 bar, multiply by 2.5.

Standard ISO 6020-2 bore/rod combinations. MM1 = smaller rod (more pull force); MM2 = larger rod (stiffer).
Bore (mm) Rod (mm) Series Piston area (cm²) Annulus area (cm²) Push @100 bar (kN) Pull @100 bar (kN)
25 12 MM1 4.91 3.78 4.91 3.78
25 18 MM2 4.91 2.36 4.91 2.36
32 14 MM1 8.04 6.5 8.04 6.5
32 22 MM2 8.04 4.24 8.04 4.24
40 18 MM1 12.6 10 12.6 10
40 28 MM2 12.6 6.41 12.6 6.41
50 22 MM1 19.6 15.8 19.6 15.8
50 36 MM2 19.6 9.46 19.6 9.46
63 28 MM1 31.2 25 31.2 25
63 45 MM2 31.2 15.3 31.2 15.3
80 36 MM1 50.3 40.1 50.3 40.1
80 56 MM2 50.3 25.6 50.3 25.6
100 45 MM1 78.5 62.6 78.5 62.6
100 70 MM2 78.5 40.1 78.5 40.1
125 56 MM1 123 98.1 123 98.1
125 90 MM2 123 59.1 123 59.1
160 70 MM1 201 163 201 163
160 110 MM2 201 106 201 106
200 90 MM1 314 251 314 251
200 140 MM2 314 160 314 160

Source: ISO 6020-2 mounting & rod-diameter series. Verify against the cylinder manufacturer's catalogue for the exact model.

Frequently asked questions

How do I calculate hydraulic cylinder force?
Force = pressure × area. The push (extend) force uses the full bore area, F = P·(π/4)·bore². The pull (retract) force uses the annulus — the bore area minus the rod area: F = P·(π/4)·(bore²−rod²). Enter bore, rod and pressure above.
Why is the pull force less than the push force?
On the retract stroke the rod occupies part of the piston face, so pressure acts on the annulus (bore area minus rod area) instead of the full bore. The shortfall equals pressure × rod area — shown on this page as the rod-area differential.
What pressure should I enter?
Use the actual working pressure (relief-valve setting) the cylinder sees, not the pump’s maximum rating. Typical industrial and mobile systems run 100–250 bar (1,450–3,600 psi).
Does this include friction and efficiency losses?
No — this is the theoretical force from pressure × area. Real output is roughly 85–95% of it after seal friction and back-pressure on the rod side.
How is the stroke speed found?
Speed = flow ÷ working area. Extend uses the piston area; retract uses the smaller annulus area, so for the same flow the cylinder retracts faster than it extends.
Does this work in metric and imperial?
Yes — toggle SI/Imperial in the header. Pressure switches between bar and psi, and force is shown in N, kN, lbf and tonne at once.

Method & assumptions

  • Theoretical force from pressure × area — excludes seal friction, back-pressure and dynamic losses. Real mechanical efficiency is typically 0.85–0.95.
  • Pressure is the actual working pressure at the cylinder, not the pump rating.
  • Rod buckling (column loading) is not checked here — long rods in compression need a separate stroke/rod-diameter check.

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