Hydraulic Cylinder Calculator
Computes the output force produced by a hydraulic cylinder given its bore diameter, working pressure, and mechanical efficiency. Used by hydraulic system designers to size cylinders for presses, lifts, and mobile equipment.
About this calculator
A hydraulic cylinder converts fluid pressure into linear mechanical force. The theoretical push force on the extend stroke is: F = (π / 4) × D² × P, where D is the bore diameter (inches) and P is the working pressure (PSI). This yields force in pounds. Accounting for mechanical efficiency η (%), the actual output force becomes: F = (π × (D/2)² × P × η) / 100. On the retract stroke, the effective area is reduced by the rod cross-section: A_retract = π/4 × (D² − d²), where d is the rod diameter. Efficiency losses (typically 90–98% for well-maintained cylinders) arise from seal friction, internal leakage, and fluid viscosity. PSI × area in square inches gives force directly in pound-force (lbf).
How to use
Suppose a cylinder has a bore diameter of 4 inches, working pressure of 2,000 PSI, and 95% mechanical efficiency. Step 1 — compute bore area: A = π × (4/2)² = π × 4 = 12.566 in². Step 2 — apply pressure: F_theoretical = 12.566 × 2,000 = 25,133 lbf. Step 3 — apply efficiency: F_actual = 25,133 × 0.95 ≈ 23,876 lbf (about 10.8 tonnes). This cylinder can therefore push roughly 23,876 lbf under these conditions, which you would use to confirm it meets your load requirement with an appropriate safety margin.
Frequently asked questions
How does rod diameter affect the retract force of a hydraulic cylinder?
On the retract stroke, fluid acts on the annular area between the bore and the rod, which is smaller than the full bore area. The retract force is F = π/4 × (D² − d²) × P × η/100. A larger rod diameter reduces the effective area significantly — for example, a 4-inch bore with a 2.5-inch rod has an annular area of 7.67 in² versus 12.57 in² for the extend stroke, producing about 39% less retract force at the same pressure. This asymmetry must be accounted for in applications like clamping or pulling where the retract force is the working stroke.
What working pressure is typical for industrial hydraulic cylinders?
Industrial hydraulic systems commonly operate between 1,500 and 3,000 PSI, with heavy-duty systems reaching 5,000 PSI or more. Mobile equipment such as excavators often runs at 3,000–5,000 PSI to keep cylinder sizes compact. Higher pressure allows smaller, lighter cylinders to produce the same force, but demands higher-quality seals, harder chrome-plated rods, and more robust hoses and fittings. Always include a pressure relief valve set 10–20% above the working pressure to protect the circuit from shock loads.
Why is mechanical efficiency important when sizing a hydraulic cylinder?
Mechanical efficiency accounts for the energy lost to seal friction, rod packing drag, and internal leakage within the cylinder. A new, well-lubricated cylinder may achieve 97–99% efficiency, but worn seals or high-viscosity fluid at cold temperatures can reduce it to 90% or lower. Neglecting efficiency leads to under-sized cylinders that cannot produce the required force under real operating conditions. It is good engineering practice to design for at least 90% efficiency and to build in a safety factor of 1.25–2× over the calculated required force.