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Subject:How to Calculate Pneumatic Cylinder Force

How to Calculate Pneumatic Cylinder Force

Postdate:2009-06-12 06:56:40   Hits:15756

How to Calculate Pneumatic Cylinder Force

 

Step 1

Measure the size of the air cylinder to calculate the total area that the air will be pushing against to create the force. All pneumatic cylinders are classified by the diameter of the cylinder and the length of the stroke. The most important measurement we need in calculating force is the diameter of the cylinder.

Step 2

Find the total area in the pneumatic cylinder if the diameter is 4 inches. To find the area of a round circle for any object, the formula is A= (pi) x (Rsquared). Where A is equal to area, pi is equal to 3.1416 and Rsquared is the radius of the circle squared.

Step 3

Calculate the area of the 4-inch diameter circle by first multiplying the radius times itself. This would equal 2 inches as the radius, times itself or 2 X 2 is equal to 4. Next multiply pi (3.1416) times the 4 and the answer is 12.5664 square inches. So the total area of the air cylinder is equal to 12.5664 inches^2, read as inches squared.

Step 4

Find the total force from the air cylinder if the air pressure is equal to 100 PSI. Since PSI is equal to pounds per square inch, all we need to do is multiply the total area in square inches of the pneumatic cylinder times the air pressure in PSI. The total force delivered from the 4-inch diameter air cylinder with 100 PSI of air pressure would equal 12.5664 inches^2 times 100 PSI, which is equal to 1,256.4 pounds of force.

Step 5

Understand that the 1,256.4 pounds of force would be on the push stroke only and not on the retraction end of the cylinder. The diameter of the push rod must be subtracted from the overall area, as the connection of the push rod interferes with the force measurement.

Step 5

Understand that the 1,256.4 pounds of force would be on the push stroke only and not on the retraction end of the cylinder. The diameter of the push rod must be subtracted from the overall area, as the connection of the push rod interferes with the force measurement.

 

 


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