A hydraulic cylinder's rod speed is set by how fast oil is pumped in versus how much area it has to fill: v = Q/A. The bore area is π/4 × bore², so for a fixed flow a bigger bore moves slower — there's simply more volume to fill each stroke. That's the speed-versus-force trade: a large bore gives more push at a given pressure but extends more gently, while a small bore is quick but weaker. The same flow against the smaller rod-side area also makes a cylinder retract faster than it extends.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the continuity relation v = Q/A for an incompressible fluid.
The cylinder speed equations
Find the bore area from the diameter, then divide the flow by it for the speed. The unit-friendly form folds in the conversions (1 L = 10⁶ mm³, 1 m = 1000 mm) so you can work straight from L/min and mm. To solve the other way — the flow needed for a target speed — use Q = v × A; for the bore that gives a speed, A = Q/v then bore = √(4A/π). The rod-side (retract) area is the bore area minus the rod's cross-section.
Worked example — a press cylinder
Scenario: A cylinder with a 50 mm bore is fed 20 L/min of oil. How fast does the rod extend?
The rod extends at about 10.19 m/min, or 0.17 m/s. Halve the flow to 10 L/min and it slows to 5.09 m/min; double it to 40 L/min and it speeds to 20.37 m/min — speed tracks flow directly. Switch to a 100 mm bore at the original 20 L/min and the area quadruples, dropping the speed to about 2.5 m/min, but that bigger bore now pushes four times harder at the same pressure.
Frequently Asked Questions
v = Q/A, A = π/4·bore². 50 mm bore, 20 L/min → 20×1000/1963.5 = 10.19 m/min.
More area to fill per stroke. Double the bore → 4× area → ¼ speed for the same flow.
More flow or a smaller bore. v ∝ Q and v ∝ 1/A.
The rod reduces the rod-side area, so the same flow gives a higher speed (and less force).
Bore in mm, flow in L/min → m/min. v(m/min) = Q×1000/A(mm²). ÷60×1000 for mm/s.