# Thread: Calculating pounds pressure on air cylinder

1. Member
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## Calculating pounds pressure on air cylinder

The Speedy Air vise that I use says that 150psi air pressure will result in 1500 pounds of holding pressure.

The project I am doing works just fine at that, but I would like to come at the die from four sides instead of one.

Would I need half the 1500 pounds if I had an air cylinder on either side?

How is force calculated for air cylinders?

Thanks,
Chris

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I can't answer the four sides part.
If I remember correctly the the force applied by a piston is given by this formula. .7854 times the diameter of the piston squared times the gas pressure (air, steam, or whatever), applied.
For a 2" piston at 150 psi that would be .7854 times 2 times 2 times 150 = 741.24 pounds of push.
The unit clamping pressure (psi or whatever units you are working with), will depend on the area of contact between the jaws and the workpiece, small area = high unit force, larger area = lesser force.
If I am all wet on this, will someone please enlighten us both.

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Thanks, Jim. I'll check the diameter of the piston on the Speedy Air Vice against the formula. I appreciate your help.
Chris

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## pressure

Take the area of the cyclinder times the air pressure.

Example: (3 inch cyclinder with 100 psi applied) would be the radius squared time pie.
1.5 x 1.5= 2.25 x3.14= 7.065
7.065 x 100= 706.5 .lbs of force
The would be on the side oppositive the piston.
The pistion side would be the area of the piston substracted from the area of the cyclinder then that amount times the psi applied.

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Originally Posted by rebel54
Take the area of the cyclinder times the air pressure.

Example: (3 inch cyclinder with 100 psi applied) would be the radius squared time pie.
1.5 x 1.5= 2.25 x3.14= 7.065
7.065 x 100= 706.5 .lbs of force
The would be on the side oppositive the piston.
The pistion side would be the area of the piston substracted from the area of the cyclinder then that amount times the psi applied.
That means the rod end of the cylinder. There's always less force in that direction than the piston side without the rod. So, on a double acting cylinder (pressure port on both ends) there's more push available than pull.

In your application make sure to distinguish between pounds force and "pounds per square inch". The fluid has psi which is converted to pounds of force by the size of the piston. If the piston rod is pressing against a workpiece, you'll again have a pounds force to pound-per-square-inch factor but it won't be material unless you're denting the workpiece or some such.

Having cylinders on both sides of the part doesn't increase the force, that's still limited by the air pressure/cylinder size relationship. If you had a smaller cylinder on one side it would simply be pushed back by the larger one. To increase the pounds force on one side, you'll need to either increase the air pressure or increase the size of the piston. Having two pistons on one side counts, since your effective piston area is the sum of those two.

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You still need 1500 lbs on both sides, if you come from both sides.

With a static jaw, the vise provides, in itself, the "opposing" 1500 lbs by virtue of it not moving. Every force, equal and opposite, etc. If it didn't, the vice jaw would take off at a rapid clip toward the back of the shop.

If another air cylinder is providing the counter-force, you must have 1500 in that cylinder to oppose 1500 lbs coming the other way. The part ONLY sees 1500 lbs, just like it does in the vise.

If you draw the free body diagrams for both the vise, and the dual cylinder situation, you will see that with a 1500 lb applied force, they come out the same - that is, the other cylinder MUST have 1500 lbs in it to get it to not move, and still apply 1500 lbs to the part.

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Originally Posted by toastydeath
You still need 1500 lbs on both sides, if you come from both sides.

With a static jaw, the vise provides, in itself, the "opposing" 1500 lbs by virtue of it not moving. Every force, equal and opposite, etc. If it didn't, the vice jaw would take off at a rapid clip toward the back of the shop.

If another air cylinder is providing the counter-force, you must have 1500 in that cylinder to oppose 1500 lbs coming the other way. The part ONLY sees 1500 lbs, just like it does in the vise.

If you draw the free body diagrams for both the vise, and the dual cylinder situation, you will see that with a 1500 lb applied force, they come out the same - that is, the other cylinder MUST have 1500 lbs in it to get it to not move, and still apply 1500 lbs to the part.
So, if I could fixture the part in the center I would then have 1500 pounds coming at both sides, no?
Chris

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Originally Posted by Hollowbuilt
So, if I could fixture the part in the center I would then have 1500 pounds coming at both sides, no?
Chris
Could you say something more about the fixture? A picture or sketch would help. I don't know quite how you anticipate clamping, but when you say you're coming from four sides with air cylinders, there's no certainty of positioning.

Oversimplified perhaps, but if you have opposing clamping forces applying 1500 lbs, you can be assured that the part is securely gripped (almost typed groped) but could be just as securely held anywhere within the strokes of the cylinders. If you have something like a casting that might have variation in its outside dimensions but you want it pretty well centered, you'll need some mechanical means of centering.

If you're locating on something like a pin or boss, a clamp force from two adjacent sides is all you need, provided the pin can withstand the clamp pressure.

If you're dealing with something more like a hold-down force, that's different.

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"So, if I could fixture the part in the center I would then have 1500 pounds coming at both sides, no?
Chris "

If you're pushing it against a fixed jaw that isn't moving, you already have 1500 lbs coming at both sides. You gain no more force by replacing that fixed jaw with another 1500lb air vise.

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## Denting?

If the piston rod is pressing against a workpiece, you'll again have a pounds force to pound-per-square-inch factor but it won't be material unless you're denting the workpiece or some such.

What if I wanted to dent the piece? If the Speedy vise dents it at about 1500 psi, would two pistons at 750 each dent the piece to the same degree?

Thanks,
Chris

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