# Hydraulic Force vs. Pressure and Cylinder Size

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

_{ 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.

#### Imperial Units

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Hydraulic and pneumatic systems - fluids, forces, pumps and pistons.

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