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Hydraulic Cylinder Force Calculator

Calculate hydraulic cylinder force from bore diameter and system pressure. Enter values for instant results with step-by-step formulas.

Reviewed by Daniel Agrici, Founder & Lead Developer

Reviewed by Daniel Agrici, Founder & Lead Developer

Formula

Force = Pressure x Area | Push Force = P x (Pi x (Bore/2)^2) | Pull Force = P x (Pi x (Bore/2)^2 - Pi x (Rod/2)^2)

Where P = system pressure in PSI, Bore = bore diameter, Rod = rod diameter. Push force uses the full bore area, while pull force uses the annulus area (bore area minus rod area).

Worked Examples

Example 1: Industrial Press Cylinder

Problem:A hydraulic press uses a cylinder with a 6-inch bore, 3.5-inch rod, and operates at 3000 PSI. Calculate push and pull forces.

Solution:Bore area = Pi x (6/2)^2 = Pi x 9 = 28.274 sq in\nRod area = Pi x (3.5/2)^2 = Pi x 3.0625 = 9.621 sq in\nAnnulus area = 28.274 - 9.621 = 18.653 sq in\nPush force = 28.274 x 3000 = 84,823 lbs = 42.41 tons\nPull force = 18.653 x 3000 = 55,959 lbs = 27.98 tons

Result:Push Force: 84,823 lbs (42.41 tons) | Pull Force: 55,959 lbs (27.98 tons)

Example 2: Mobile Equipment Lift Cylinder

Problem:An excavator arm cylinder has a 4-inch bore, 2.5-inch rod, and operates at 4500 PSI. Find the forces and oil volume for a 24-inch stroke.

Solution:Bore area = Pi x (4/2)^2 = 12.566 sq in\nRod area = Pi x (2.5/2)^2 = 4.909 sq in\nAnnulus area = 12.566 - 4.909 = 7.657 sq in\nPush force = 12.566 x 4500 = 56,549 lbs\nPull force = 7.657 x 4500 = 34,458 lbs\nExtend volume = 12.566 x 24 = 301.6 cu in = 1.305 gal\nRetract volume = 7.657 x 24 = 183.8 cu in = 0.796 gal

Result:Push: 56,549 lbs | Pull: 34,458 lbs | Extend Vol: 1.305 gal | Retract Vol: 0.796 gal

Frequently Asked Questions

How is hydraulic cylinder force calculated?

Hydraulic cylinder force is calculated using the fundamental relationship between pressure and area: Force equals pressure multiplied by the effective piston area. For the push (extend) stroke, the full bore area is used, which is the cross-sectional area of the piston calculated as Pi times the bore radius squared. For the pull (retract) stroke, the rod area must be subtracted from the bore area because the rod occupies space on the rod side of the piston. This net area is called the annulus area. Higher system pressure or a larger bore diameter directly increases the available force output.

Why is the pull force less than the push force?

The pull force is always less than the push force because the piston rod reduces the effective area on the retract side of the cylinder. When the cylinder extends, hydraulic fluid acts on the full bore area of the piston. When it retracts, the rod passes through the piston seal, effectively reducing the area that fluid pressure can act upon. This reduced area is called the annulus area, and it equals the bore area minus the rod cross-sectional area. The larger the rod diameter relative to the bore, the greater the difference between push and pull forces, which is an important consideration in cylinder sizing.

References

Reviewed by Daniel Agrici, Founder & Lead Developer · Editorial policy