How this instrument works
A pneumatic cylinder turns air pressure into straight-line push using nothing more exotic than Pascal's principle: pressure applied to a confined fluid acts equally over every surface it touches. Multiply that pressure by the area of the piston face it's pushing against and the result is force — F = P × A. A 6 bar supply pushing on a 20 cm² piston face delivers 1,200 N, whether that force is doing useful work against a clamp, a stamping die, or nothing at all; the formula has no idea what's bolted to the other end of the rod.
The two readings this instrument gives, extend force and retract force, differ because a double-acting piston does not present the same working area in both directions. Extending, the full bore face A_bore takes the push. Retracting, the piston rod already occupies part of that same face on the return side, so air only acts on the annulus left over, A_bore minus A_rod. A fatter rod, chosen for buckling resistance on a long stroke, quietly eats into retract force even though the bore and supply pressure haven't changed at all.
What the formula leaves out is everything that happens once the piston starts moving: seal friction, exhaust-port restriction, and end-of-stroke cushioning each subtract a real slice, typically 3 to 10 percent, from this theoretical number. A cylinder's datasheet lists this calculation's output as the rated force and expects whoever specifies the cylinder to build in a working margin on top, rather than sizing a job to land exactly on the figure this page returns.
- Enter Air supply pressure — the regulated line pressure feeding the cylinder, commonly around 6 bar for shop air.
- Enter Bore area — the full piston face's cross-sectional area, taken from the cylinder's bore diameter.
- Enter Rod cross-sectional area — the piston rod's own cross-sectional area, smaller than the bore area.
- Read Extend-stroke force (full bore area) — the push delivered as the rod extends out.
- Read Retract-stroke force (bore minus rod area) — the pull delivered as the rod retracts, always the lower of the two figures.
Worked example — a 20 cm² bore cylinder at 6 bar
A machine builder is checking whether a double-acting cylinder will clamp hard enough before it gets welded into a fixture. The cylinder has a 20 cm² bore and a 3 cm² rod, running on shop air regulated to 6 bar. Converted to the SI units the formula runs on, that is a bore area of 0.002 m², a rod area of 0.0003 m², and a supply pressure of 600,000 Pa. Enter those three figures and Extend-stroke force returns 1200.0 N: F = 600,000 × 0.002 = 1,200 N, the full bore pushing at line pressure.
Retract-stroke force comes back lower, 1020.0 N: F = 600,000 × (0.002 − 0.0003) = 600,000 × 0.0017 = 1,020 N. The rod's own 3 cm² is simply gone from the working face on the way back in, so the fixture releases with about 15 percent less force than it clamped shut with. That 180 N gap is real geometry, not a rounding error, which is exactly why a cylinder's datasheet lists push and pull as two separate numbers.
Questions
Why is the retract force lower than the extend force?
Because the piston rod subtracts area from the working face on the way back. Air fills the full bore on the extend stroke, A_bore, but on retract it acts only on the annulus left after the rod's own cross-section is removed, A_bore − A_rod. With a 20 cm² bore and a 3 cm² rod at 6 bar, that is 1,200 N pushing against 1,020 N pulling — a real, geometric asymmetry, not a pressure loss.
Does this include friction from the seals?
No — this is theoretical force at zero piston velocity, F = P × A with no losses included. Real cylinders lose roughly 3 to 10 percent of that figure to seal drag, cushioning, and exhaust back-pressure once the piston is actually moving, so a sizing engineer typically keeps 20 to 30 percent of margin above the calculated force rather than specifying a cylinder right at the limit.
What pressure unit should I use?
Enter Air supply pressure in the unit your gauge reads; the field's unit menu converts bar or psi to the pascals the formula runs on internally. Shop air commonly sits at 6 bar, about 600,000 Pa or 87 psi, which is the pressure used in the worked example on this page.
Why do single-acting cylinders only need an extend force?
A single-acting cylinder applies compressed air on one side only and relies on a spring, or gravity, to return the piston, so there is no retract-stroke air pressure to compute — only F_extend = P × A_bore applies. The retract figure on this page assumes a double-acting cylinder, where air is supplied to both ports and pushes the rod back in as well as out.
How much more force does a bigger bore actually buy?
Force scales with the square of the bore diameter, since area itself is proportional to diameter squared. Doubling the bore diameter roughly quadruples A_bore, and therefore roughly quadruples extend force at the same pressure — which is why a modest step from a 32 mm bore to a 63 mm bore, not quite double the diameter, delivers nearly four times the push.
Does this formula also work for hydraulic cylinders?
Yes — F = P × A is the same relation for any pressurized fluid, air or oil; only the typical pressure range differs. Hydraulic systems commonly run at 100 to 350 bar rather than pneumatic's 5 to 10 bar, so an identical bore delivers dramatically more force, which is exactly why hydraulics gets chosen over pneumatics when very high force has to come from a compact cylinder.