.2 .4 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 The above graph illustrates that a piston with a constant load of 59.75 pounds and a boundary face of 23.953 IN² requires a pressure of 69.12” WG (2.494 PSI), maintaining equilibrium at all points in its stroke. If another piston actuator receives an identical volume of fluid as the above example, with a boundary face twice as large (2 x 23.953 IN² = 47.906 IN²), its force is doubled; however, the stroke is one-half. Travelling one-half the distance of the above piston actuator, while producing double the force, generates exactly the same amount of work as the illustrated piston The fluid equation W = PV applies to conventional piston actuators regarding work, pressure and volume. The relationship is linear. The work input required for each .5” stroke for the illustrated piston is: W = PV = 2.494 PSI x (.5” x 23.953 IN²) = 2.494 PSI x 11.9765 IN³ = 29.87 in/lb EQUILIBRIUM PRESSURE REGARDING CONVENTIONAL PISTON ACTUATORS WITH BOUNDARY FACES OF 23.953 IN² AND A CONSTANT LOAD OF 59.75 POUNDS 69.12“WG (2.494 PSI) PRESSURE INCHES OF LIFT AT EQUILIBRIUM APPLIED TO PISTON’S 23.953 IN² BOUNDARY FACE AT EQUILIBRIIUM LOAD 59.75# LOAD 59.75# LOAD 59.75# PISTON FORCE < 59.75# PISTON FORCE = 59.75# PISTON FORCE > 59.75# FULLY RETRACTED EQUILIBRIUM FULLY DRIVEN FLUID PRESSURE FLUID PRESSURE FLUID PRESSURE < 69.12 “ WG (2.494 PSI) = 69.12”WG (2.494 PSI) > 69.12”WG (2.494 PSI) LOAD FORCES PISTON TO ANY POINT IN PISTON’S FLUID LIFTS LOAD TO PISTON’S MINIMUM STROKE STROKING RANGE PISTON’S MAXIMUM STROKE FLUID FLUID 13.268
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