Pneumatic Actuation Getting it Straight

FOR IMMEDIATE RELEASE
(Free-Press-Release.com) January 22, 2010 --

Pneumatic actuation plays a major role in today's world of computerized automation. It's reliable, economical, and surprisingly easy to use. Understanding pneumatics is a matter of physics. When air inside a container builds up, pressure magnitude is the same at all points within the fluid. The air also pushes out on its vessel uniformly. Pneumatic actuators leverage this fact of fluid dynamics to provide clean, quiet motion, with less waste heat and electromagnetic interference than their electric counterparts. They also excel in applications involving fast repetitive moves, heavy loads, and very smooth motion profiles.

Here's how it works: Compressed air enters an opening in a cylinder and pushes against the interior, including the one wall that can move: the piston. If the difference in force across the piston is larger than the total attached load plus frictional forces, the piston floor drops out. The resulting net force (proportional to the force to mass ratio) accelerates the load, converting pneumatic to linear mechanical power. With air power as the driving force, pneumatic actuators are safe for hazardous environments where electric sparks must be avoided.

Calculating force
Single-rod double-acting pneumatic actuators — those with air ports on both sides of the internal piston — are the most common in industry. We'll now explore the physics behind pneumatic motion, using this type as our example.

All pneumatic force depends on two things: air pressure and piston area. Let's say air pressure is set at 50 psi. If the diameter of the piston is 7 mm, then it has an effective surface area of 38 mm2. Force, then, is the product of pressure and area:

Force = Pressure × Area

Converting from psi to N/mm2, pushing force equals 13.1 N. However, notice that the piston rod reduces the effective area on its side, meaning that pull force is not as great as push force. Specifically, for pull force:

Effective area = Piston area - Rod area

If we assume the rod is 3 mm in diameter, then the rod area is 7.069 mm2. The effective area becomes 30.931 mm2 for a pull force of 10.66 N.

In contrast, double-rod double-acting actuators have equal push and pull forces and can be used for both actions. Rodless variations too have equal push and pull force, but higher than that of double-rod types for the same bore. Since its output linkage isn't contained, the internal piston drives an external slide as it moves. Similarly, cable air actuators have an external slide tied with a cable that's wrapped around a pulley at each piston end.

No matter what cylinder type is used, correct sizing is essential. Bigger cylinders — though they increase system natural frequency and allow faster accelerations — require larger, more expensive compressor pumps to power the system. As with any motion technology, oversizing is something to avoid.

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