How to use it
- Choose metric (mm, bar, L/min) or US units (in, psi, gpm).
- Enter the bore and rod diameters.
- Enter the working pressure, the pump flow and the stroke.
- Read the forces, speeds, times and oil volumes in both directions.
Formulas
- Piston area: A = π D² / 4; annulus area: A_r = π (D² − d²) / 4.
- Force: F_push = p × A, F_pull = p × A_r.
- Speed: v_ext = Q / A, v_ret = Q / A_r.
- Oil volume: V = A × stroke (extend), A_r × stroke (retract).
- Time: t = V / Q.
Unit steps used: 1 bar = 0.1 N/mm²; 1 L/min = 10⁶ / 60 mm³/s; 1 US gallon = 231 in³.
Worked example
A 63 mm bore with a 36 mm rod at 100 bar, 20 L/min and 500 mm stroke:
- A = π × 63² ÷ 4 = 3,117.25 mm², A_r = π × (63² − 36²) ÷ 4 = 2,099.37 mm²;
- push 10 × 3,117.25 = 31.17 kN, pull 20.99 kN;
- 20 L/min = 333,333 mm³/s, so it extends at 106.93 mm/s and retracts at 158.78 mm/s;
- extending takes 1.559 L and 4.68 s, retracting 1.05 L and 3.15 s.
In US units, a 2 in bore with a 1 in rod at 1,000 psi pushes 3,141.59 lbf and pulls 2,356.19 lbf; 5 gpm (19.25 in³/s) moves it out at 6.13 in/s.
Tips
- Check the rod for buckling on long strokes under push load; the force here does not include it.
- Regenerative circuits route rod-side oil back to the cap side for faster extension with less force.
- Pump size: flow × pressure gives the hydraulic power; 20 L/min at 100 bar is about 3.3 kW before losses.