# Hydraulic Force

## Calculate hydraulic cylinders forces

The force produced on the rod side *(1)* of a double acting hydraulic piston - can be expressed as

F_{1}=P(π_{1 }(d/ 4) (1)_{2}^{2}- d_{1}^{2})

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, bar)

The force produced on the opposite of rod side *(2)* - can be expressed as

F_{2}=P(π d_{2}_{2}^{2 }/ 4)(2)

where

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

P_{2}= pressure in the cylinder (opposite rod) (psi,bar)

### Hydraulic Force Calculator

#### Imperial Units

##### Pressure acting on rod side

Piston diameter - d_{2}(in)

Rod diameter - d_{1}(in)

Cylinder pressure - P_{1 }(psi)

##### 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

_{2}

- 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

_{1}

- when pressure act on the same side of the rod.