Hidroteknik Calculators
Hydraulic Cylinder Force Calculation
The push and pull force a hydraulic cylinder produces depends on the system pressure and the effective piston area. Below you will find the formula, units and a worked example.
⚙ Open the Hydraulic Calculator →Core Formula
Push force (piston side)
D = cylinder bore (inner) diameter
Pull force (rod side)
d = rod diameter. The rod area reduces the effective area, so pull force is always lower than push force.
Units
| Quantity | Symbol | Unit |
|---|---|---|
| Pressure | P | bar (1 bar = 0.1 N/mm²) |
| Diameter | D, d | mm |
| Area | A | mm² (÷100 → cm²) |
| Force | F | N (÷9.81 → kgf) |
Worked Example
Given: Bore D = 80 mm, rod d = 45 mm, pressure P = 160 bar.
1) Piston area: A = π × 80² / 4 = 5026.5 mm²
2) Push force: F = 160 × 0.1 × 5026.5 = 80,424 N ≈ 80.4 kN ≈ 8.2 tonne
3) Annular area: A = π × (80² − 45²) / 4 = 3436 mm²
4) Pull force: F = 160 × 0.1 × 3436 = 54,976 N ≈ 55.0 kN ≈ 5.6 tonne
Frequently Asked Questions
What is the hydraulic cylinder force formula?
Force = Pressure × Area (F = P × A). For push, the area is the piston area (π×D²/4); for pull, it is the annular area (π×(D²−d²)/4).
Why are push and pull force different?
On the rod side the effective area is reduced by the rod cross-section, so at the same pressure the pull force is always lower than the push force.
How do I use pressure in bar?
1 bar = 0.1 N/mm². With area in mm², F(N) = P(bar) × 0.1 × A(mm²). Rule of thumb: F(kgf) ≈ P(bar) × A(cm²).