Hydraulic cylinders lift diggers, presses and car lifts with immense forces from modest pressures. The force is simply pressure times the piston area – smaller on the retract stroke because the rod takes up part of the area.
How to use the Piston Force Calculator
- Bore (piston) diameter D
- Rod diameter d
- Pressure P
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the cylinder bore, rod diameter and pressure.
Inputs explained
| Field | What to enter |
|---|---|
| Bore (piston) diameter D | Units: mm, cm, m, in, µm |
| Rod diameter d | Units: mm, cm, m, in, µm |
| Pressure P | Units: bar, MPa, psi, kPa |
What each result means
- Piston area = pi×[Bore (piston) diameter D]^2÷4
- Push (extend) force = [Pressure P]×pi×[Bore (piston) diameter D]^2÷4
- Pull (retract) force = [Pressure P]×pi×([Bore (piston) diameter D]^2 – [Rod diameter d]^2)÷4
Example values
How the main result changes when Bore changes and everything else stays at the example values above:
| Bore (piston) diameter D (mm) | Push (extend) force (kN) |
|---|---|
| 31.5 | 7.7931 |
| 47.25 | 17.5345 |
| 63 | 31.1725 |
| 78.75 | 48.707 |
| 94.5 | 70.138 |
| 126 | 124.6898 |
Formula
F_push = P·π·D²/4, F_pull = P·π·(D² − d²)/4
Worked example
Inputs — Bore (piston) diameter D: 63 mm; Rod diameter d: 36 mm; Pressure P: 100 bar.
Results — Piston area = 31.1725 cm²; Push (extend) force = 31.17245 kN; Pull (retract) force = 20.99369 kN.
Key terms
- Annulus area
- Piston area minus rod area.
Frequently asked questions
Why is the pull force smaller?
The rod reduces the effective area on the retract side.
What pressure do hydraulic systems use?
Typically 100–350 bar; pneumatic systems 6–10 bar.
Do I need a safety factor?
Yes – account for friction and pressure losses (often 10–25%).
How does area scale with diameter?
With its square – doubling the bore quadruples the force.