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Unread 27-01-2005, 22:57
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Re: pneumatics lifting forces

You can look it up in a table, but here is the calculation so you understand.

F=PA, or Force = Pressure x Area.

If you are working with a 2" dia. cylinder, the area = pi(r)^2
or 3.14*(1)^2 =3.14 square inches.
(remember to use the radius, not the diameter)

If you are working at 60 psi, the force is 60 psi x 3.14 sq in = 188.4 pounds of force in the extending direction.

To determine the force in the retracting position, you do a similar equation, except you have to subtract out the area of the rod, because the air pressure cannot act on it.

Same 2" dia. cyclinder = 3.14 square inches. Assume the rod = 1/2" diameter.
Area of the rod = pi(.25)^2 = .196 sq in.
Effective area of the retracting end = 3.14 - 0.196 = 2.945 sq. in.

If you are working at 60 psi, the force is 60 psi x 2.945 sq in = 176.64 pounds of force in the retacting position.

You have more force in the extending direction than the retracting direction, because of the smaller area.

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