Hydraulic Cylinder Force Calculator
Push and pull force from hydraulic pressure, bore diameter, and rod diameter in metric or imperial units.
What it does
A hydraulic cylinder turns fluid pressure into linear force. The oil pushes on a piston, the piston pushes the rod, and the rod moves the load. The size of the force depends on two things: the pressure in the circuit and the area of the piston that the pressure acts on.
Extension and retraction are not symmetric. When the rod extends, oil fills the full bore, so the whole circle of the piston does the work. When the rod retracts, the rod takes up part of that circle, and oil only pushes on the ring shaped area that remains. A thick rod removes more area and lowers the pull force.
The formula is simple but the units need care. Pressure in bar is 100 kilopascals, or 0.1 newtons per square millimeter. A 100 mm bore has an area of 7,853.98 square millimeters, so 200 bar, which is 20 N/mm2, produces 157,079 newtons, or 157.1 kN.
Imperial units work the same way. Pressure in psi multiplied by area in square inches gives pounds of force directly. A 4 inch bore at 3,000 psi has 12.566 square inches of area and produces about 37,699 pounds of force before losses.
The result is a design starting point, not a rating. Real cylinders lose force to seal friction, rod gland drag, and the back pressure of oil returning to the tank. Rod buckling, mounting strength, and the relief valve setting also limit what the cylinder can safely do. A margin of at least 25% over the calculated load is common practice.
How to use
- Choose metric for bar and millimeters or imperial for psi and inches.
- Enter the pressure the cylinder sees, not the pump's maximum rating.
- Enter the bore diameter from the cylinder data sheet.
- Enter the rod diameter, or zero for a plunger style cylinder with no rod.
- Read the extension force first, then the lower retraction force.
- Compare both forces with the actual load and add a margin for friction and back pressure.
Formula
- F
- Cylinder force in newtons or pounds force.
- p
- Pressure at the cylinder, in N/mm2 (MPa) or psi.
- D
- Bore diameter in millimeters or inches.
- d
- Rod diameter in the same unit as the bore.
- A_extend
- Full piston area used when the rod extends.
- A_retract
- Annulus area left around the rod when it retracts.
One bar equals 0.1 N/mm2, so a pressure in bar can be multiplied by the area in mm2 and divided by 10 to get newtons. Metric outputs are shown in kilonewtons and square centimeters. Imperial outputs stay in pounds force and square inches. The bore must be larger than the rod, and a zero rod diameter is allowed for a ram.
Worked example
Example: 200 bar on a 100 mm bore with a 50 mm rod
Metric defaults: 200 bar, 100 mm bore, 50 mm rod.
- Bore area: pi x 100^2 / 4 = 7,853.98 mm2, or 78.54 cm2.
- Pressure in N/mm2: 200 bar x 0.1 = 20 N/mm2.
- Extension force: 20 x 7,853.98 = 157,079.6 N, or 157.1 kN.
- Rod area: pi x 50^2 / 4 = 1,963.50 mm2.
- Annulus area: 7,853.98 - 1,963.50 = 5,890.49 mm2, or 58.90 cm2.
- Retraction force: 20 x 5,890.49 = 117,809.7 N, or 117.8 kN.
Reference table
| Bore | Bore area | Extension force | Load at 25% margin |
|---|---|---|---|
| 50 mm | 19.63 cm2 | 39.3 kN | 31.4 kN |
| 80 mm | 50.27 cm2 | 100.5 kN | 80.4 kN |
| 100 mm | 78.54 cm2 | 157.1 kN | 125.7 kN |
| 125 mm | 122.72 cm2 | 245.4 kN | 196.3 kN |
Limits
- Gives theoretical force at the cylinder and ignores seal friction, rod gland drag, and oil viscosity.
- Does not subtract back pressure on the return side, which can cut effective force notably on a fast circuit.
- Does not check rod buckling, pin strength, mounting bolts, or cylinder wall thickness.
- Assumes steady pressure; pressure spikes and relief valve behavior are outside the model.
Common mistakes
- Using the bore radius instead of the diameter in the area formula.
- Using the full bore area for retraction force and overstating the pull.
- Mixing bar and psi in the same calculation.
- Sizing the cylinder at the relief valve setting and leaving no margin for friction.
- Forgetting that a long rod can buckle before it reaches the calculated force.
Try it yourself
Quick checks. Answers sit tucked under each question.
250 bar, 80 mm bore, 40 mm rod?
Extension 125.7 kN, retraction 94.2 kN, bore area 50.27 cm2, annulus 37.70 cm2.
2,500 psi, 3 inch bore, 1.5 inch rod?
Extension 17,671 lbf, retraction 13,254 lbf, bore area 7.07 in2, annulus 5.30 in2.
300 bar, 50 mm bore, 25 mm rod?
Extension 58.9 kN, retraction 44.2 kN, bore area 19.63 cm2, annulus 14.73 cm2.
Questions
How do I calculate hydraulic cylinder force?
Multiply the pressure by the piston area. For extension use the full bore area; for retraction use the bore area minus the rod area.
Why is retraction force lower?
The rod occupies part of the piston, so less area is exposed to oil on the retract side. A larger rod reduces the pull force further.
What pressure should I enter?
The pressure actually available at the cylinder while it moves. Pump rating, hose losses, and the relief valve setting all affect it.
How much safety margin do I need?
Many designs use 1.25 to 1.5 times the calculated load to cover friction, back pressure, and unexpected shock loads.
Does oil flow change the force?
Flow changes speed, not force at a given pressure. A larger pump makes the cylinder move faster but does not raise the force.
Can I use this for a pneumatic cylinder?
The same area and pressure math applies, but air compressibility, much lower pressures, and different seal friction make the results less precise.
Sources
- NASA Glenn: Pascal's principle
- OpenStax University Physics: Pascal's principle and hydraulics
- Embry-Riddle Aeronautical University: Fluid statics