Formulas
Bore area = π × bore² ÷ 4
Annulus area = bore area − π × rod² ÷ 4
Push force = pressure × bore area
Pull force = pressure × annulus area
Speed (in/s) = gpm × 231 ÷ 60 ÷ area
231 is the number of cubic inches in a US gallon.
Worked example
A 3 in bore, 1.5 in rod cylinder at 2,000 psi pushes 14,137 lbf (7.1 tons) and pulls 10,603 lbf. With a 10 gpm pump it extends at 5.4 in/s and retracts faster, at 7.3 in/s, because the rod side holds less oil.
Practical notes
- Size the cylinder for 75–80% of the required force at your relief setting to leave margin for friction and pressure losses.
- Long strokes in compression need a rod buckling check. Rod manufacturers publish stop-tube and buckling charts.
- Retract flow out of the cap end is larger than pump flow by the area ratio. Size return lines and filters for it.
Frequently asked questions
- How much force does a 2 inch hydraulic cylinder have?
- At 2,000 psi, a 2 in bore pushes 3.14 in² × 2,000 = 6,283 lbf (about 3.1 tons). Pull force depends on the rod size.
- Why does my cylinder retract faster than it extends?
- The rod takes up volume on the rod side, so the same flow fills it faster. The speed difference equals the bore-to-annulus area ratio.
Results are estimates for planning and education. Verify against the current code edition your jurisdiction has adopted, manufacturer data and a licensed professional before you build or install.