Hydraulic cylinder force
Size a hydraulic cylinder from its bore, rod and working pressure: push force on the full piston area, pull force on the annulus, and the stroke speed the pump flow produces in each direction. Efficiency is applied to force, so seal friction is not quietly ignored.
| Quantity | Value | Basis |
|---|---|---|
| Bore area | 7854.0 mm² | π/4 × 100² |
| Annulus area | 5890.5 mm² | π/4 × (100² − 50²) |
| Push force | 119.38 kN | 16.0 MPa × bore area × 95% |
| Pull force | 89.54 kN | pressure × annulus area × efficiency |
| Push force (tonne-force) | 12.17 tf | kN ÷ 9.80665 |
| Extend speed | 42.4 mm/s | 20.0 L/min ÷ bore area |
| Retract speed | 56.6 mm/s | flow ÷ annulus area |
| Hydraulic power | 5.33 kW | pressure × flow |
| Pull / push ratio | 0.750 | annulus area ÷ bore area |
Efficiency covers seal friction and internal leakage. Rod-side force is always lower because the rod removes area from the annulus.
A_b = π d_b² / 4 · full piston face
A_a = π (d_b² − d_r²) / 4 · rod side, smaller
F = p A η · η = seal and leakage efficiency
v = Q / A · same flow, smaller area, faster stroke
P = p Q · hydraulic power into the cylinder
A clear path from inputs to results.
A hydraulic cylinder produces force from pressure and speed from flow, and the two are independent: raising the pressure does not make the rod move faster, and opening the valve does not make it push harder. The rod also removes area from the piston, so every cylinder pulls with less force and retracts faster than it extends. This calculator gives you both directions from bore, rod, pressure and pump flow.
How to use
- Enter the bore diameter — the inside diameter of the barrel.
- Enter the rod diameter; the rod side has less area and therefore less force.
- Set the working pressure in bar.
- Enter the pump flow reaching this cylinder in litres per minute.
- Adjust efficiency to account for seal friction, typically 90–95%.
- Read push and pull force, both stroke speeds, and hydraulic power.
The formula
Reference: Pascal's principle and continuity; ISO 4413 hydraulic fluid power — general rules and safety requirements.
Worked example
63 mm bore, 36 mm rod, 210 bar, 12 L/min, 92% efficiency
- Bore 63 mm, rod 36 mm.
- Working pressure 210 bar.
- Pump flow 12 L/min.
- Mechanical efficiency 92%.
- 1
Pressure: 210 bar = 21 N/mm².
- 2
Bore area: π/4 × 63² = 3,117.25 mm².
- 3
Push force: 21 × 3,117.25 × 0.92 = 60,225 N = 60.23 kN.
- 4
Annulus area: π/4 × (63² − 36²) = 2,099.37 mm², so pull force is 21 × 2,099.37 × 0.92 = 40.56 kN.
- 5
Flow: 12 L/min = 200,000 mm³/s. Extend speed 200,000 / 3,117.25 = 64.2 mm/s; retract 200,000 / 2,099.37 = 95.3 mm/s.
- 6
Hydraulic power: 21 MPa × 12 L/min = 4.20 kW.
Push 60.23 kN · pull 40.56 kN · extends 64.2 mm/s · retracts 95.3 mm/s · 4.20 kW.
Interpreting the result
The cylinder pulls with only 67% of its push force, because the rod occupies a third of the piston area. That ratio is fixed by geometry and never changes with pressure. Speed behaves the other way: the smaller annulus retracts faster for the same flow, which is why a cylinder often needs a flow control on the retract side to avoid a hard return.
Assumptions & limitations
- Steady-state force and speed. Acceleration, oil compressibility and line losses are not modelled.
- Efficiency is applied to force only, as a single factor covering seal friction and internal leakage.
- Pressure is the working pressure at the cylinder, not the pump setting; relief and directional-valve losses reduce it.
- The rod diameter is clamped below the bore, so an impossible rod is treated as a nearly solid piston.
- Rod buckling, cylinder wall thickness and port sizing are separate checks and can govern before oil pressure does.
Frequently asked questions
Why is pull force lower than push force?
On the retract stroke the oil acts on the annulus — the piston area minus the rod cross-section. A 100 mm bore with a 50 mm rod has 7,854 mm² of push area but only 5,890 mm² of pull area, so it pulls with 75% of the force.
How do I pick a bore size?
Start from the required force at your working pressure and add margin: A = F / (p × η). Round up to the next standard bore — 40, 50, 63, 80, 100, 125, 160, 200, 250 mm — then check that the resulting speed at your pump flow is acceptable.
Can a long stroke buckle the rod?
Yes, and it is often the real limit. A slender rod in compression behaves as a column: check the Euler load for the stroke length and the rod's second moment of area. A stroke much longer than the bore needs rod guidance or a larger rod.
Does higher pressure make the cylinder faster?
No. Speed is set by flow divided by area. More pressure raises force; more flow raises speed. In practice a pump has a pressure-compensated limit, so force and speed trade against each other at the pump's power ceiling.
What efficiency should I use?
90–95% for a healthy cylinder with good seals. Older or high-cycle cylinders run lower. If the cylinder is doing precise force work, use 0.9 and treat the difference as margin rather than pretending the seals are perfect.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.