Cylinder Force Calculator
Find the force a pneumatic or hydraulic cylinder produces from its bore diameter, rod diameter, and operating pressure. Enter dimensions in inches or mm and pressure in psi, bar, kPa, or MPa, and pick extend or retract — the tool returns force in pounds-force and newtons.
Example: with Bore diameter 2 · Diameter unit inches · Rod diameter (for retract) 1 · Operating pressure 100 · Pressure unit psi → Cylinder force: 314 lbf.
- In newtons1,397 N (1.40 kN)
- Effective piston area3.142 in² effective
Computed by the calculator below using its default values. Change any input to see your own numbers.
Cylinder force is pressure times piston area, F = P·A. On the retract stroke the rod blocks part of the piston, so the same pressure pulls with less force than it pushes.
Push, pull, and the rod
A fluid cylinder makes force the simplest way possible: pressure acting on the piston face, F = P·A. The piston area comes from the bore diameter, area = π/4·bore², so a 2-inch bore at 100 psi pushes with about 314 pounds. Double the pressure and you double the force; go up one bore size and force climbs with the square of the diameter. That is the whole design lever for sizing an actuator to a job.
The retract stroke is weaker, and the reason is the rod. On the rod side of the piston, the rod itself occupies part of the area, so the pressure has less surface to act on: effective area is π/4·(bore² − rod²). A cylinder that pushes 314 pounds might pull only 236 with the same pressure. Designers exploit this asymmetry, or avoid it with double-rod cylinders when equal force both ways matters.
How it’s calculated
Extend force = P · (π/4·bore²). Retract force = P · (π/4·(bore² − rod²)). Diameters convert to inches (mm ÷25.4); pressure to psi (bar ×14.50377, kPa ×0.1450377, MPa ×145.0377). Force in lbf = psi × area (in²); newtons = lbf × 4.4482216.
Gives the theoretical static force at the stated pressure. Real output is lower from seal friction (often a few percent) and any back-pressure on the opposite port; dynamic loads and buckling of long rods are separate checks.
Extend force at 100 psi by bore
| Bore | Piston area | Force (lbf) |
|---|---|---|
| 1 in | 0.785 in² | 79 |
| 1.5 in | 1.767 in² | 177 |
| 2 in | 3.142 in² | 314 |
| 3 in | 7.069 in² | 707 |
| 4 in | 12.566 in² | 1,257 |
Computed with F = P·π/4·bore² at 100 psi; rounded.
Common mistakes
- Using the rod-side (retract) area for a push stroke, or the full bore for a pull — they differ by the rod area.
- Mixing pressure units; 100 psi and 100 bar differ by about 14.5 times.
- Ignoring seal friction and back-pressure, which make real force a few percent below the ideal.
- Forgetting that force scales with the square of the bore, not linearly with diameter.
Frequently asked questions
What is the cylinder force formula?
Force = pressure × piston area. Extending, the area is π/4 × bore²; retracting, it is π/4 × (bore² − rod²) because the rod blocks part of the piston.
Why is retract force lower than extend force?
On the rod side, the rod occupies part of the piston area, so the pressure acts on less surface. The same pressure therefore pulls with less force than it pushes.
Does this work for both pneumatic and hydraulic cylinders?
Yes. The F = P·A relationship is identical; hydraulics simply run at much higher pressures, so they make far more force from the same bore.
Is the calculated force what I actually get?
It is the ideal static force. Subtract a few percent for seal friction and any back-pressure on the other port, and check long rods separately for buckling.