Title: How to Calculate the Force Output of a Retracting Cylinder
Course: Fluid Power Pneumatics (Sandbox) > Section: Module 6: The Principle of Pneumatic Pressure and Force - 2 Hours

Retracting Instead of Extending

We now know how to calculate the force of the piston if the cylinder is extending. But what if the cylinder is trying to retract a load? If that is the case, the air pressure is exerted on all inside surfaces of the rod end of the cylinder.

Look at the following diagram:

Parker Hannifin Corporation. (1980). Industrial Pneumatic Technology. Parker Hannifin Corp. DOI: 1557690154

As with the cap end, the only area where the pressure acts to move the load is on the piston. In this case, however, the pressure does not act on the full area of the piston because the rod is attached to it. This means that to get the true area of the piston so that we can calculate the force, we must subtract the area of the rod from the area of the piston.

Since the rod side of the piston will have less area (Piston Area - Rod Area), it will take more pressure (psi) to retract the same load as opposed to extending it.

Formulas:

Let's calculate a pneumatic system. I want to know what force is exerted with the following cylinder retract system:

F = P x A [Force (lbs) = Pressure (psi) x Area (square inches )]

D^2 x 0.7854 = square inches

Piston Area - Rod Area

Example:

- Piston Diameter = 2"
- Cylinder Piston Diameter = 2"
- Rod Diameter = .5"
- psi = 100

Piston: 2^2 = 4

4 x 0.7854 = 3.14 square inches

Rod: .5^2 = .25

.25 x 0.7854 = 0.196 square inches

Now we take the square inches of the piston (3.14) and the square inches of the rod (.196) and we subtract: 3.14 - 0.196 = 2.94 square inches.

So, we now know that the total surface area that 100 psi is pushing against is 2.94 square inches. We can put that into our formula:

F = 100 psi x 2.94 square inches

F= 294 lbs.

Extra Resources

Watch the following video for additional help.

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