Results
| Direction | Ideal | Available (with losses) | Safe working |
|---|---|---|---|
| Extend (push) | 37.407 kN | 33.760 kN | 22.507 kN |
| Retract (pull) | 33.637 kN | 30.357 kN | 20.238 kN |
| Area | Value (mm²) | Value (m²) |
|---|---|---|
| Bore area | 3,117.25 | 3.117245e-3 |
| Rod area | 314.16 | 3.141593e-4 |
| Extend effective | 3,117.25 | 3.117245e-3 |
| Retract effective | 2,803.09 | 2.803086e-3 |
Example data
These sample rows are generated using the same formulas as the calculator.
| Scenario | Type | Pressure | Bore | Rod | Safe Extend | Safe Retract | Required P (Ext) |
|---|---|---|---|---|---|---|---|
| Hydraulic press push | Single-rod | 160.00 bar | 80.00 mm | 28.00 mm | 46.872 kN | 41.130 kN | — |
| Pneumatic actuator pull | Single-rod | 90.00 psi | 2.50 in | 1.00 in | 0.801 kN | 0.673 kN | — |
| Double-rod positioning | Double-rod | 6.00 bar | 50.00 mm | 16.00 mm | 1.543 kN | 1.543 kN | — |
| Rodless slide | Rodless | 500.00 kPa | 40.00 mm | 0.00 mm | 0.370 kN | 0.370 kN | — |
| Required pressure example | Single-rod | — | 63.00 mm | 20.00 mm | — | — | 63.98 bar |
Formula used
The ideal cylinder force is based on pressure times effective piston area: F = P × A.
- Bore area: Abore = π (D/2)²
- Rod area: Arod = π (d/2)²
- Single-rod retract area: Aret = Abore − Arod
- Double-rod effective area: A = Abore − Arod (both directions)
Losses are applied as a combined factor: LossFactor = Efficiency × (1 − Friction%) × cos(θ) × CylCount.
Safe working force is: Fsafe = (Fideal × LossFactor) ÷ SafetyFactor.
How to use this calculator
- Select Force from pressure or Required pressure.
- Choose the cylinder type: single-rod, double-rod, or rodless.
- Enter bore and rod diameters, then pick the length unit.
- Add efficiency, friction loss, cylinder count, angle, and safety factor.
- Click Calculate, then export CSV or PDF if needed.
Cylinder force fundamentals
Cylinder force is created when fluid pressure acts on piston area. The ideal relationship is F = P × A, where pressure is in Pa and area is in m². Example: 6 bar (600,000 Pa) on a 100 mm bore gives A ≈ 0.00785 m² and ideal extension force ≈ 4.71 kN.
Bore area, rod area, and effective area
The bore diameter sets piston area. Rod diameter reduces area on the retract side. With a 100 mm bore and 25 mm rod, piston area is ≈ 7,854 mm² and rod area ≈ 491 mm², so retract area is ≈ 7,363 mm². Smaller area means lower force at the same pressure.
Extension vs retraction force differences
Extension uses the full piston area on single-rod cylinders, so it is usually stronger. Retraction uses the annular area (piston minus rod), often 5–15% lower depending on rod size. Double-rod cylinders tend to have similar forces in both directions.
Pressure, efficiency, and friction losses
Real systems lose force due to seal friction, valve pressure drop, hose losses, and misalignment. Apply efficiency (such as 0.85–0.95) and any extra friction loss. If ideal force is 4.71 kN and total loss factor is 0.90, usable force becomes about 4.24 kN.
Multiple cylinders and angled loads
When cylinders share a load, total usable force is the sum, assuming equal pressure and similar geometry. For angled pushing, only the aligned component counts: Falong = F × cos(θ). At 30°, multiply by 0.866, so 4.24 kN becomes about 3.67 kN.
Safety factor and operating limits
Safety factor covers unknowns like sticking loads, temperature changes, and pressure fluctuations. Values of 1.25–2.0 are common depending on risk and duty cycle. Always compare required pressure with regulator, pump, valve, and cylinder ratings for continuous operation.
Choosing a bore size from results
If required pressure is near your supply limit, increase bore diameter, reduce losses, or use more cylinders. Force scales with area, so a modest bore increase can raise force noticeably. For pneumatic systems, remember compressibility can slow motion; sizing for flow is separate from sizing for force. Verify mounting, column strength, and the real load path before final selection.
FAQs
Why is retraction force usually smaller?
On a single-rod cylinder, the rod occupies part of the piston face during retraction. That reduces effective area, so force drops at the same pressure. Larger rods cause a larger reduction.
What efficiency value should I use?
Use 0.90 as a practical starting point. Choose 0.85 for worn seals, long hoses, or restrictive valves. Choose 0.95 for short runs with good components. Confirm with measurements if the force is critical.
How does cylinder angle change the force?
Only the component aligned with the load is effective. The calculator applies cos(θ) to estimate aligned force. At 45°, you keep about 70.7% of the cylinder force; at 60°, you keep 50%.
Can I calculate total force for multiple cylinders?
Yes. Enter the cylinder count and the calculator multiplies usable force, assuming equal pressure and equal load sharing. If cylinders differ or the geometry is uneven, calculate each cylinder separately and add the aligned forces.
Why does required pressure change with rod diameter?
Required pressure is F ÷ A. On retraction, effective area is piston area minus rod area, so a thicker rod reduces A. To produce the same force, the pressure must rise.
What safety factor should I choose?
For steady loads, 1.25–1.5 is common. For shock loads, sticking mechanisms, or uncertain friction, 1.75–2.0 is safer. Ensure the resulting required pressure stays within all component ratings.
Units are converted internally; double-check field assumptions for your hardware.