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# 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 )

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### Rod Force F 2 diagram

- when pressure act on opposite side of rod.

#### 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.

## Related Topics

### • Hydraulics and Pneumatics

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