Hydraulic Force vs. Pressure and Cylinder Size
Calculate hydraulic cylinder force.
The force produced on the rod side (1) of a double acting hydraulic piston  can be expressed as
F_{1} = P_{1 }(π (d_{2}^{2}  d_{1}^{2}) / 4) (1)
where
F_{1} = rod force (lb, N)
d_{1} = rod diameter (in, mm)
d_{2} = piston diameter (in, mm)
P_{1} = pressure in the cylinder on the rod side (psi, N/mm^{2})
 1 bar = 105 N/m^{2} = 0.1 Nm/mm^{2}
The force produced on the opposite of rod side (2)  can be expressed as
F_{2} = P_{2} (π d_{2}^{2 }/ 4) (2)
where
F_{2} = rod force (lb, N)
P_{2} = pressure in the cylinder (opposite rod) (psi, N/mm^{2})
Hydraulic Force Calculator
Imperial Units
Pressure acting on rod side
Pressure acting on opposite of rod side
Piston diameter  d_{2} (in)
Cylinder pressure  d_{2 }(psi)
Metric Units
Pressure acting on rod side
Piston diameter  d_{2} (mm)
Rod diameter  d_{1} (mm)
Cylinder pressure  P_{1 }(bar)
Pressure acting on opposite of rod side
Piston diameter  d_{2} (mm)
Cylinder pressure  d_{2 }(bar)
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Rod Force F_{2} diagram
 when pressure act on opposite side of rod.
Imperial Units
Metric Units
 1 psi (lb/in^{2}) = 144 psf (lb_{f}/ft^{2}) = 6,894.8 Pa (N/m^{2}) = 6.895x10^{3} N/mm^{2} = 6.895x10^{2} bar
 1 N/m^{2} = 1 Pa = 1.4504x10^{4} lb/in^{2} = 1x10^{5} bar = 4.03x10^{3} in water = 0.336x10^{3} ft water = 0.1024 mm water = 0.295x10^{3} in mercury = 7.55x10^{3} mm mercury = 0.1024 kg/m^{2} = 0.993x10^{5} atm
 1 lb_{f} (Pound force) = 4.44822 N = 0.4536 kp
 1 N (Newton) = 0.1020 kp = 7.233 pdl = 7.233/32.174 lb_{f} = 0.2248 lb_{f} = 1 (kg m)/s^{2} = 10^{5} dyne = 1/9.80665 kg_{f}
 1 in (inch) = 25.4 mm
 1 m (meter) = 39.37 in = 100 cm = 1000 mm
Rod Force F_{1} diagram
 when pressure act on the same side of the rod.
Imperial Units
Related Topics

Hydraulics and Pneumatics
Hydraulic and pneumatic systems  fluids, forces, pumps and pistons.
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