# Hydraulic Force

## Hydraulic force - area formulas and calculator

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The force produced by a double acting hydraulic piston on the rod side can be expressed as

F_{1}= (π(d/ 4) P_{2}^{2}- d_{1}^{2})_{1}(1)

where

F_{1}= rod pull force (lb, N)

d_{1}= rod diameter (in, m)

d_{2}= piston diameter (in, m)

P_{1}= pressure in the cylinder (rod side) (lf_{f}/in^{2 }(psi), N/m^{2}(Pa))

The force produced opposite the rod can be expressed as

F_{2}= (π d_{2}^{2 }/ 4)P_{2}(2)

where

F_{2}= rod push force (lb, N)

P_{2}= pressure in the cylinder (opposite rod) (lf_{f}/in^{2 }(psi),N/m^{2}(Pa))

##### Calculate Pull Force - with pressure acting on rod side

##### Calculate Push Force - with pressure acting opposite rod side

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### Push Diagram

Rod pushing force for hydraulic cylinders are indicated below:

*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*

### Pull Diagram

Rod pulling force for hydraulic cylinders are indicated below:

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