A hydraulic cylinder turns fluid pressure into a straight push or pull. The physics is one line — force equals pressure times area — but two areas matter. On the extend stroke oil pushes on the full face of the piston, so you use the bore area. On the retract stroke the rod occupies part of that face, leaving only the ring-shaped annulus for oil to act on, which is why a cylinder always pulls with less force than it pushes. The same two areas, divided into the pump flow, give you the extend and retract speeds.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: Pascal's law and standard fluid-power sizing practice.
The cylinder equations
In metric, pressure in bar times area in mm² times 0.1 gives force in newtons (1 bar = 0.1 N/mm²). In imperial, pressure in psi times area in square inches gives force directly in pounds. The annulus area is always the bore area minus the rod area, so a fat rod buys you faster retraction at the cost of pull force — a deliberate trade-off in press and clamp design.
Worked example — a 50 mm press cylinder
Scenario: A 50 mm bore, 28 mm rod cylinder runs at 150 bar with 20 L/min of flow.
So this cylinder pushes with ~3 tonnes, pulls with ~2.1 tonnes, and extends at about 170 mm/s. Because the retract annulus is smaller, the same 20 L/min retracts it faster — around 247 mm/s. If you needed more push, raising pressure to 210 bar lifts the push force to ~41 kN without changing the cylinder.
Frequently Asked Questions
F = P × A. On extend, the full bore area pushes: F = P × π/4 × D². A 50 mm bore at 150 bar gives ~29.5 kN.
The rod covers part of the piston, so retract oil acts only on the annulus: F = P × π/4 × (D²−d²). A bigger rod means less pull force.
v = Q/A. Extend uses the bore area; retract uses the smaller annulus, so it retracts faster. Keep units consistent.
Industrial systems usually run 100–250 bar (1,500–3,600 psi); mobile gear can reach 350 bar+. Keep working pressure below the rated max.
Divide newtons by 9,810 for tonnes-force: 29,500 N ÷ 9,810 ≈ 3.0 t. In imperial, 1 ton-force = 2,000 lbf.