Hydraulic cylinder calculator: push and pull force, speed and oil volume
This hydraulic cylinder calculator gives the theoretical force of a double-acting cylinder from bore, rod diameter and pressure – pushing on the full piston area, pulling on the smaller ring area. Add the pump flow and the stroke to get speed, oil volume and time. A 4 in bore with a 1.5 in rod at 2,000 psi pushes 25,133 lbf and pulls 21,598 lbf.
Example: Units: imperial (in, psi, gpm) · Bore (piston diameter) 4 · Rod diameter 1.5 · Pressure 2,000 · Pump flow (optional) 10 · Stroke (optional) 12 → Cylinder force: Push 25,133 lbf · pull 21,598 lbf. Source: Bosch Rexroth RE 17332 (edition 2021-07), hydraulic cylinder series CDH1/CGH1/CSH1, p. 14 “Diameters, areas, forces, flow”: areas A1/A2/A3, area ratio φ, forces F1/F2/F3 as theoretical static force without efficiency, flow at 0.1 m/s (read 09.10.2026). Updated: .
How it is calculated
Force = pressure × area. Extending, the oil presses on the whole piston area A1 = π × D² ÷ 4. Retracting, it only reaches the ring area A3 = A1 − A2, because the rod covers A2 = π × d² ÷ 4. Bosch Rexroth lists exactly these three areas and forces in its cylinder data sheet and calls them the “theoretical static cylinder force (without consideration of the efficiency)”.
Examples at 2,000 and 2,500 psi
| Bore × rod | Pressure | Push F1 | Pull F3 |
|---|---|---|---|
| 2 × 1 in | 2,000 psi | 6,283 lbf | 4,712 lbf |
| 3 × 1.5 in | 2,000 psi | 14,137 lbf | 10,603 lbf |
| 4 × 1.5 in | 2,000 psi | 25,133 lbf | 21,598 lbf |
| 4 × 2 in | 2,500 psi | 31,416 lbf | 23,562 lbf |
| 5 × 2.5 in | 2,500 psi | 49,087 lbf | 36,816 lbf |
In metric units the same rule reads F [kN] = p [bar] × A [cm²] ÷ 100. Rexroth’s example: a 100 mm piston at 250 bar pushes 196.35 kN; with a 56 mm rod the area ratio is 1.46.
Speed, oil volume and cycle time
Speed = flow ÷ area. With 10 gpm the 4 × 1.5 in cylinder extends at 3.06 in/s and retracts faster, at 3.57 in/s, because the ring area is smaller. A 12 in stroke takes 0.65 gal and 3.9 s out, 0.56 gal and 3.4 s back. Rexroth gives the reverse view: at 0.1 m/s the 100 mm piston needs 47.1 l/min. To convert pressure use bar to psi; for the physics behind F = p × A see the force calculator and for the barrel volume the cylinder calculator.
Frequently asked questions
How do you calculate hydraulic cylinder force?
Multiply pressure by area. A 4 in bore has 12.57 in²; at 2,000 psi that is 25,133 lbf of push force.
Why is the pull force lower than the push force?
When retracting, the rod covers part of the piston, so only the ring area A1 − A2 counts. With a 1.5 in rod that is 10.80 in², or 21,598 lbf at 2,000 psi.
How do I calculate cylinder speed?
Divide flow by area. 10 gpm (2,310 in³/min) on 12.57 in² gives 3.06 in/s extending.
What is the area ratio φ?
Piston area divided by ring area, A1 ÷ A3. Rexroth lists 1.46 for a 100 mm piston with a 56 mm rod: the cylinder retracts 1.46 times faster than it extends.
What is the regenerative (differential) force?
In a regenerative circuit both sides are under pressure and only the rod area A2 pushes: less force (3,534 lbf for the 1.5 in rod at 2,000 psi) but a much faster extend speed.
Is the calculated force what the cylinder really delivers?
No. It is the theoretical static force; seal friction, back pressure and efficiency reduce it. The rated pressure and rod buckling must be checked separately.
Sources and legal basis
- Bosch Rexroth RE 17332 (edition 2021-07), hydraulic cylinder series CDH1/CGH1/CSH1, p. 14 “Diameters, areas, forces, flow”: areas A1/A2/A3, area ratio φ, forces F1/F2/F3 as theoretical static force without efficiency, flow at 0.1 m/s (read 09.10.2026)
- NIST SP 811, Appendix B.8 – conversion factors: 1 bar = 10⁵ Pa, 1 psi = 6 894.757 Pa, 1 lbf = 4.448 222 N, 1 in = 2.54 cm, 1 gal (U.S.) = 3.785 412 L, 1 kgf = 9.806 65 N (read 09.10.2026)
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