pneumatics lifting forces

[akshar]

my team was just wondering how much you could possibly lift with the pneumatics provided in the kits, does anyone know and would it be effective in using it for the arm?

[BoyWithCape195]

Well, for how much you could possibly lift, you can look that up in the 2005 Pneumatics manual. It has charts that show how much force in and out. And make sure you don't forget, you can order two free cylinders

link http://www2.usfirst.org/2005comp/Man...ticsManual.pdf

[UsmanB]

page 12 of pneumatics manual http://www2.usfirst.org/2005comp/Man...ticsManual.pdf

[Chris Fultz]

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.

[greencactus3]

Originally Posted by Chris Fultz

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

my english teacher would be ripping at you now i think. or maybe me if youre correct and im not. wouldnt it be
You have more force in the extending direction than the retracting direction, because of the larger area? as the area is reffering to the extending direction rather than the retracting? ugh. i hate english. i really shouldnt be trying to act smart like this cuz im prolly gonna look like an idiot again and yea. well back to the topic.
dont forget pulleys/levers/cams/etc will increase or decrease max lifting weight.
double the range and half the max weight liftable (is that a word? )
or double the weight liftable and half the range. so whether its usable on your arm or no, how much stroke do you have? how much range do you need? how big is your cylinder inside radius? how many cylinders will you use? how much pressure will you be running? how much air will you be storing and how fast can you pump it up again? and most importantly, do you like using pneumatics? personally i loveum. never had any problems withum while motors bug me

[ngreen]

Yep. F=PA. Double the cyclinders, double the area, double the force. Most people use the cyclinder as levers so know how much you want to lift and how far you want to move it. Then you can find where you need to place cyclinder(s), how many cyclinder, length of stoke, size of bore, the pressure you need to run at, etc. With pneumatics you only need to use what you need. Just figure how much you need to lift with respect to where you are lifting if from and you'll be fine.

[Andy Baker]

When sizing a pneumatic application, another thing to think about would be side loads and efficiency losses.

For instance, working from the numbers Chris posted above, there are many situations where you will not get your entire 176 lbs (retracting) for a 2" cylinder at 60 psi. If the rod of that cylinder is experiencing any side loads, then more friction is created inside of the cylinder. A certain amount of force is needed to overcome this friction. This amount may be up to 1/4-1/3 of the force of the cylinder, if the side load is significant. Therefore, cylinder output forces in these situations could be only 70% - 80% of your calculated force.

Andy B.

[Raul]

Added note to "Save as"

Originally Posted by Andy Baker

When sizing a pneumatic application, another thing to think about would be side loads and efficiency losses.

For instance, working from the numbers Chris posted above, there are many situations where you will not get your entire 176 lbs (retracting) for a 2" cylinder at 60 psi. If the rod of that cylinder is experiencing any side loads, then more friction is created inside of the cylinder. A certain amount of force is needed to overcome this friction. This amount may be up to 1/4-1/3 of the force of the cylinder, if the side load is significant. Therefore, cylinder output forces in these situations could be only 70% - 80% of your calculated force.

Andy B.

I will go further and say that there is no legal way to move anything at the rated max forces for these cylinders. The cylinders can HOLD the max rated force, but they cannot move with that much force. Consider these the stall forces.

Do not forgot the time dimension part of the power equation. Cylinders also have a force versus time curve for each given pressure . My team has done some experiments and have proven that these curves are not linear. I have attached a sample of our experimental results. It is difficult to characterize pneumatics because the entire system affects the available airflow volumes (think of a current limiting analogy from electronics), so these curves are very dependent on your set up. I'm too busy right now to go into any more details but I may do a white paper on the power curves of our pneumatic systems during the off-season.

Don't forget to do a "Save As" on the XL file.

[rswsmay]

And to add further....

There may also be a derating of force based upon the angle that the force is applied by the piston. For a good overall explanation, go to http://pneumaticsfirst.org and down load the powerpoint presentation dealing with pneumatics.

http://pneumaticsfirst.org/media/PneuAndFIRSTFinal.ppt

Originally Posted by Raul

Added note to "Save as" I will go further and say that there is no legal way to move anything at the rated max forces for these cylinders. The cylinders can HOLD the max rated force, but they cannot move with that much force. Consider these the stall forces.

Do not forgot the time dimension part of the power equation. Cylinders also have a force versus time curve for each given pressure . My team has done some experiments and have proven that these curves are not linear. I have attached a sample of our experimental results. It is difficult to characterize pneumatics because the entire system affects the available airflow volumes (think of a current limiting analogy from electronics), so these curves are very dependent on your set up. I'm too busy right now to go into any more details but I may do a white paper on the power curves of our pneumatic systems during the off-season.

Don't forget to do a "Save As" on the XL file.