How much force do you REALLY get?

[David Fort]

We understand the math (and the tables) for calculating force from a pneumatic cylinder, but we seem to not be getting it.

We have a 2" cylinder at 60 PSI and expect to get 170 pounds of Retract Force not counting whatever friction and stuff.

In a fairly unscientific test, we are getting less than 100 pounds of Force.

(The test is more or less connecting a rope to the Yoke, and having it pull down on a person standing on a bathroom scale).

There are a couple pulleys in the setup, just to guide the rope to the right place (no mechanical advantage/disadvantage). Supposedly high quality, but small, pulleys, which put a Z bend in the rope.

Does anyone know how much less than "theoretical" is the force typically achievable with a pneumatic cylinder?

Any suggestions for things we might be doing wrong?

Thanks!

[Christopher149]

Are you accounting for the lower surface area when retracting caused by the cylinder rod? Sure you're at 60 psi? Is it 2" inside diameter, or outside?

That's what seems obvious to me, since I'm not an expert in all the details.

[Deke]

I'm no pneumatic expert, someone else could have input, but two things come to mind. Are you sure you have 60 psig? Another is if the force is not directly inline with the scale, you will not get the maximum amount of force. You would need to account for the angle, even if it is small.

Thinking farther along, is there a target force you are looking for, or another reason for your testing?

[Owen Makin]

Try subtracting the surface area that is being taken up by the connection point of the shaft to the piston (for lack of a more accurate term). The 2" diameter is for pushing force, but when pulling the psi is acting on a surface area that looks like a washer.

[Derek Bessette]

Were you able to stop the cylinder mid stroke during your test? If not that could be affecting your results. Flow restrictions can prevent pressure from building up during the stroke. Full pressure can only be assure when the cylinder's piston stops moving.

[DonRotolo]

Theory says Pounds per Square Inch, multiplied by Square Inches, equals Pounds.

A cylinder with a 2" Outside diameter might have a 1.75" piston in the cylinder. If the rod is 1/2" diameter, you get 2.4 square inches in the extend direction and 2.2 SqIn for retract.

At 60 PSI this is 144 pounds of force in extend, and 132 in retract. I would expect about 140 and 130 in reality.

Theory and Reality should be quite close in Pneumatics, so I'd be questioning the scale and your measurement techniques.

[David Fort]

Thanks everyone for the suggestions and questions. Here is some additional info.

First, as I said, the test was unscientific, and that is a profound understatement. We are pulling a catapult back with a rope. We can pull on that rope while standing on a scale, and see that we pull the rope with about 50 pounds of force.

The direction of pull is not perfectly normal to the scale, maybe 10 degrees off.

There is one pulley in that system to turn the rope up away from the floor so it can be pulled up, pushing our feet down into the scale.

In our catapult system, we have a 2" bore cylinder, and thus think it should have an area of pi.

We are pulling, rather than pushing. The rod has diameter 0.625.
The effective area is the annular area, the cylinder diameter minus the rod cross section area.

Thus, the effective area for retracting is 0.307 square inches.

(all the dimensions in this discussion come from the FIRST Pneumatics Manual. We haven't actually measured them).

The conclusion that we are "getting less than 100 pounds" of force is based on the highly rigorous observation that we can pull the catapult back by hand with 50 pounds of rope tension, as measured by the bathroom scale, and that when we use the 2" cylinder, we can't pull it back. We did not measure the force exerted by the cylinder. Not even with the bathroom scale. (we couldn't figure out how to do that. An idea comes to mind now, but at the time, not).

(the additional 2x is from a 1:2 pulley system which divides the cylinders force in half to get twice the rope travel. Assuming massless frictionless pulleys and all that).

here is a chart of our force/area calculations (which match those listed in the FIRST pneumatics manual):
CylinderDiameter RodDiameter CylinderArea RodArea AnnularArea PushForceLbs RetractForceLbs
0.750 0.250 0.442 0.049 0.393 26.507 23.562
1.500 0.437 1.767 0.150 1.617 106.029 97.030
2.000 0.625 3.142 0.307 2.835 188.496 170.088

There are 3 (fixed position) pulleys in the rope between the cylinder and the catapult to route the rope over the river and through the woods as required to get the whole thing to fit.

Final disclosure: It wasn't a bathroom scale. It was pinched from the wrestling team.

Thanks for your thoughts.

[MrBasse]

Can you post a picture of the set up as it is currently run? The forces in pneumatics are kind of strange as there are a lot of efficiency issues involved. The angle of attack on the cylinder is a major one. You should get more than enough force to pull 50 lbs. We have used a 2" cylinder to lift people before to demonstrate the forces involved.

Thus, the effective area for retracting is 0.307 square inches.

Check your math on that one, don't trust someone else with it every time.


You'll lose some in friction, efficiency, and actual air pressure available. But in no way should you have an issue creating 50 lbs of force.

Check your plumbing, check your air availability, and check the angle you are applying force at. If all of those check out, try a different cylinder as that one isn't doing what it should.

[Tem1514 Mentor]

A very simple jig to get some accurate results is to make a 'dead lift' system using those muscle man weights as the load.

Set the cylinder either vertical or horizontal with a pulley.

Just a word of caution. Releasing the load back down will always be very fast unless you are using flow controls. Make sure you are using very strong cording as the tension/shock loads can be very high.

[David Fort]

Can you post a picture of the set up as it is currently run?


sorry. I didn't think to get a photo while at school. Interesting how frustration hinders thinking.


Thus, the effective area for retracting is 0.307 square inches.

Check your math on that one, don't trust someone else with it every time.


apologies - I mis-typed. 0.307 is the Rod diameter. the effective area is the piston area minus that,
3.14 sq in - 0.307 sq in = 2.8 sq in.

2.8 sq in * 60 p/sq in = 170 pounds

(I will update my previous post to correct this)

I wonder if this is as stupid as having a bad pressure gauge. It says 60 psi, but this is from the old parts bin and we have had similar disappointments in "Actual Force" on our pneumatics before. Maybe we have a bad gauge lurking. I think that would be a good thing to check, just in case.

[mentorDon]

Try getting another air pressure gage and place it with a 'T' connector close to the cylinder to verify your air pressure is close to 60 psi.

[David Fort]

Update:
We checked with another pressure gauge - they agree we are working with 60 PSI.

We have straightened out some pulleys and doubled the piston area. Not happy, but seems to be working.

Thanks for the assistance everyone.