Ideal, static hydraulics. Assumes an incompressible fluid, no friction or losses, and full-bore (extend) area. Real cylinders lose force to seals, back-pressure and the rod side.
Engineering · hydraulics

Hydraulic Force

Pressure times area is force — and how a small push lifts a big load.
F = P · A · Pascal
100 bar · Ø50 mm

Cylinder

metric
bar
mm
N

The cylinder

fluid under pressure force Pascal: pressure is equal everywhere in the fluid
Result
Detail
Field notes

Pressure, area, force

How it works

Why hydraulics multiply force

Hydraulics rest on one line: F = P · A. Pressure pushing on an area produces a force. And by Pascal's principle, pressure applied to a confined fluid is the same everywhere in it — so a small piston and a large piston connected by fluid feel the same pressure but produce different forces. The bigger piston wins, in proportion to its area. That's a hydraulic press, a jack, a digger arm.

Worked example

A pump makes 100 bar (10 MPa). Push it against a 50 mm bore piston — area 19.6 cm² — and you get 19.6 kN of force, about 2 tonnes. Feed the same pressure to a 150 mm piston (9× the area) and it pushes harder: ~177 kN.

The catch is stroke: the big piston moves 9× less for the same fluid. Force up, distance down — energy is conserved, exactly like a lever.

Why does area go up with the square of diameter?

A circle's area is π/4·d². Double the bore and you quadruple the area — and the force. It's why a small increase in cylinder diameter buys a big jump in capacity.

bar, psi, MPa — how do they relate?

1 bar ≈ 14.5 psi ≈ 0.1 MPa ≈ 100 kPa. Hydraulic systems often run 100–350 bar (roughly 1,500–5,000 psi). The tool converts as you switch units.

Do I get force for free?

No. The output piston moves less in exact proportion to the force gain, so the work (force × distance) is the same on both sides, minus real-world losses. Hydraulics trade distance for force, like any machine.

What about the rod side?

On a cylinder's retract stroke, the rod takes up part of the piston, so the effective area — and the force — are smaller. This tool models the full-bore (extend) area; subtract the rod area for retract.