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The force produced by a double acting hydraulic piston on the rod side formula

F₁ = π / 4 (d₂² - d₁²) P₁

where

d₂ = piston diameter (inches)

d₁ = rod diameter (inches)

F₁ = rod pull force (lb)

P₁ = pressure in the cylinder (rod side) (lf_f/in²)

The force produced opposite the rod formula

F₂ = π / 4 d₂² P₂

where

P₂ = pressure in the cylinder (opposite rod) (lf_f/in²)

F₂ = rod push force (lb)

psi (lb/in2) = 144 psf (lbf/ft2) = 6,894.8 Pa (N/m2) = 6.895x10-3 N/mm2 = 6.895x10-2 bar

1 lbf (Pound force) = 4.44822 N = 0.4536 kp

1 in (inch) = 25.4 mm