# 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 / 4 (d_{2}^{2}- d_{1}^{2}) P_{1}(1)

where

F_{1}= rod pull force (lb)

d_{1}= rod diameter (inches)

d_{2}= piston diameter (inches)

P_{1}= pressure in the cylinder (rod side) (lf_{f}/in^{2})

The force produced opposite the rod can be expressed as

F_{2}= π / 4 d_{2}^{2}P_{2}(2)

where

F_{2}= rod push force (lb)

P_{2}= pressure in the cylinder (opposite rod) (lf_{f}/in^{2})

##### 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:

*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 lb*_{f}(Pound force) = 4.44822 N = 0.4536 kp*1 in (inch) = 25.4 mm*

### Pull Diagram

Rod pulling force for hydraulic cylinders are indicated below:

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