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Hydraulic Cylinder Calculator

Push and pull force, rod speed, oil volume and cycle time for any cylinder

Updated · Free, no signup

Inches or millimetres.

Inches or millimetres.

psi or bar.

US gpm or L/min.

Inches or millimetres.

Push / pull force

25,133 lbf push · 18,850 lbf pull

Extend (push) force

25,133 lbf

Retract (pull) force

18,850 lbf

Push force in US tons

12.57 tons

Extend speed

3.06

in/s (imperial) or mm/s (metric).

Retract speed

4.08

in/s (imperial) or mm/s (metric).

Oil volume to extend

1.306

US gallons or litres.

Oil volume to retract

0.979

US gallons or litres.

Extend time

7.83 s

Full cycle time

13.71 s

  • Push force is about 12.6 US tons (11.4 tonnes-force).
  • The rod takes up 25% of the piston area, so the cylinder retracts 1.33× faster than it extends but pulls with less force.

Extend vs retract force (lbf or kN)

About the Hydraulic Cylinder Calculator

This hydraulic cylinder calculator sizes a double-acting cylinder from five numbers: bore diameter, rod diameter, system pressure, pump flow and stroke length. It returns the push (extend) force on the full piston area, the pull (retract) force on the smaller annulus area, the extend and retract speeds, the oil needed for each stroke, and the time for a full cycle.

Use it to pick a cylinder for a log splitter, press, loader or lift, to check whether a pump is big enough for the speed you want, or to work out what pressure you need for a given force. The force calculation is the same for pneumatic cylinders — just enter the air pressure (speed depends on valve and air flow, so ignore those results for air). Choose imperial (inches, psi, gpm) or metric (mm, bar, L/min).

Results are theoretical: real cylinders lose a few percent of force to seal friction and back-pressure on the return side, and real speeds are slightly lower because of pump slip and relief valve leakage. Allow a margin of 10–20% when selecting components.

How to use the hydraulic cylinder calculator

  1. 1Choose imperial or metric units.
  2. 2Enter the bore and rod diameters from the cylinder data sheet.
  3. 3Enter the working (relief) pressure and the pump flow.
  4. 4Enter the stroke length.
  5. 5Read the push and pull force, speeds, oil volume and cycle time.

Formula and method

F = P × A · A_ext = π·D²/4 · A_ret = π(D² − d²)/4 · v = Q / A · V = A × stroke · t = stroke / v

Pressure acting on an area produces force, so the push force is pressure times the full piston area and the pull force is pressure times the annulus area (piston area minus rod area). With pressure in psi and area in square inches the force is in pounds; with bar and mm the calculator converts to pascals and square metres and reports kilonewtons.

Rod speed is the pump flow divided by the area being filled, so the cylinder retracts faster than it extends because the annulus is smaller. The oil volume per stroke is the area times the stroke (231 cubic inches per US gallon), and the stroke time is the stroke divided by the speed. All values are ideal — no friction, leakage or back-pressure.

F
Cylinder force
P
Working pressure
D, d
Bore (piston) diameter and rod diameter
Q
Pump flow rate
v
Rod speed
V
Oil volume per stroke

Worked examples

4" bore × 2" rod at 2,000 psi, 10 gpm, 24" stroke

The piston area is π × 4² ÷ 4 = 12.57 in², so push force = 2,000 × 12.57 = 25,133 lbf (12.6 tons). The annulus is 12.57 − 3.14 = 9.42 in², giving 18,850 lbf pull. 10 gpm is 38.5 in³/s, so the rod extends at 3.06 in/s, takes 7.8 s for 24 inches, and the full cycle takes about 13.7 s.

Metric: 63 mm bore × 36 mm rod, 160 bar, 20 L/min, 300 mm

Piston area = 3,117 mm², so 16 MPa × 3,117 mm² = 49.9 kN push; the annulus of 2,099 mm² gives 33.6 kN pull. 20 L/min is 333,333 mm³/s, so the rod extends at about 107 mm/s and covers 300 mm in 2.8 s, using 0.94 L of oil.

Log splitter: 3" bore × 1.5" rod at 1,500 psi, 5 gpm, 12"

A 3-inch bore has 7.07 in² of area, so 1,500 psi gives about 10,600 lbf, or 5.3 tons of splitting force. At 5 gpm (19.25 in³/s) the ram moves at 2.7 in/s and needs about 4.4 s to extend 12 inches.

Frequently asked questions

How do you calculate hydraulic cylinder force?+

Multiply the pressure by the piston area. For a 4-inch bore at 2,000 psi, area = π × 4² ÷ 4 = 12.57 in² and force = 2,000 × 12.57 ≈ 25,133 lbf. For the retract force, subtract the rod area from the piston area first.

Why is retract force lower than extend force?+

On the rod side the oil pushes on the annulus — the piston area minus the rod area — which is smaller than the full piston face. Less area at the same pressure means less force. The same smaller area also makes the cylinder retract faster at a given flow.

How do I calculate cylinder speed from gpm?+

Convert gpm to cubic inches per second (gpm × 231 ÷ 60) and divide by the piston area in square inches. A 10 gpm pump feeding a 4-inch bore (12.57 in²) gives 38.5 ÷ 12.57 ≈ 3.06 inches per second.

How many tons is my hydraulic cylinder?+

Divide the push force in pounds by 2,000 to get US tons. A 4-inch bore at 2,500 psi produces 31,416 lbf, or about 15.7 tons. Log splitter ratings are often quoted this way at the relief pressure.

Does this work for pneumatic cylinders?+

Yes for force: a pneumatic cylinder also produces pressure × area, so a 2-inch bore at 90 psi gives about 283 lbf push. Speed is different because air is compressible and depends on valve size and air supply, so ignore the speed and time results for air.

How much oil does a hydraulic cylinder hold?+

Multiply piston area by stroke for the extend volume and annulus area by stroke for the retract volume; divide cubic inches by 231 to get US gallons. Make sure the reservoir can supply the difference between the two when all cylinders extend.

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