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
Metric Units
Pressure acting on rod side
Pressure acting on opposite of rod side
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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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