A hydraulic cylinder's force is pressure times piston area, F = P × A, with A = π/4 × bore². The push stroke uses the full bore; the pull stroke uses the smaller annular area (bore minus rod). This calculator gives both forces and the areas, in metric or imperial.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: F = P·A and area formulas, recomputed in code.
The formula
In metric, multiply pressure in bar by 0.1 to get N/mm², then by the area in mm² for newtons. In imperial, multiply psi directly by area in square inches for pounds-force. The pull (rod-side) force is always less than the push force because the rod occupies part of the piston face on the return stroke. These are ideal forces — seal friction and back-pressure reduce the real output slightly.
Worked examples
50 mm bore, 100 bar, 25 mm rod:
2 in bore, 1500 psi:
4 in bore, 2000 psi:
Doubling the bore quadruples the area and the force, since area grows with the square of the diameter. The rod reduces pull force noticeably — here a 25 mm rod cuts the 50 mm cylinder's return force by about a quarter.
Frequently Asked Questions
F = P × A, A = π/4 × bore². 50 mm at 100 bar ≈ 19.6 kN.
Push uses full bore area; pull uses bore − rod (annular) area, so it's smaller.
N = bar × 0.1 × mm². 100 bar on 1963 mm² ≈ 19,635 N.
A = π/4 × D². 2 in bore ≈ 3.14 in²; 50 mm ≈ 1963 mm².
Theoretical — real is a few % lower from friction and back-pressure.