Some basic Pneumatic Design Questions

[ttldomination]

Originally Posted by DonRotolo

Cubic inches * # of actuations = air needed. Multiply by (output pressure / storage pressure) (usually 60 and 120) and that's how much storage you need "theoretically". (Tanks have a cubic inch capacity specified) In reality, pressure drops as you draw against it, so it becomes a differential equation... So just round up and add a tank, empirically it will be close. Then test.

So, suppose a piston uses 1 cubic inch of air, and is fired thirty time. As per your math, it would be:

1 cu in * 30 actuations * (60 output psi / 120 psi) ~> 15 cu inches of "theoretical" storage?

And as far as the design question is concerned, my question, which I now understand was slightly misleading, was more in regards to CADing the piston.

Are there nice models with the appropriate motion constraints built in, or do people just model in a pole and restrict its range of motion?

Thanks,
- Sunny G.

[DampRobot]

Originally Posted by ttldomination

Are there nice models with the appropriate motion constraints built in, or do people just model in a pole and restrict its range of motion?

Thanks,
- Sunny G.

There are definitely some. Check the Bimba website. Usually, when you import them as a .step, you have to add in the motion constraints yourself.

[Michael Hill]

Actually, Bimba has a really cool CAD generator that can hook into your CAD software and watch it be made in front of your eyes. I can't recall if motion constraints are placed or not.

[IndySam]

<up on soapbox>

Please they are not called pistons they are called cylinders.

<off of soapbox>

[DampRobot]

But would "wow so cylinder" look as good in our shooter?

[KrazyCarl92]

Keep in mind the difference between gauge and absolute pressure, and your calculations will work out pretty well (or they have for us in the past). 60 psi gauge pressure is 74.7 psi actual pressure, and 120 psi gauge pressure is 134.7 psi actual pressure. This means each cu. in. of storage air corresponds to roughly 1.8 cu. in. of air in max working pressure. Another source of error in these calculations is assuming that the air is behaving as an ideal gas, but this is close enough for determining the number of storage tanks needed on an FRC robot, especially considering it makes sense to include a safety factor anyway.

[tr6scott]

A couple of finer details to consider.

If you are working with the constraints of only using tanks, remember you have to feed the cylinder on both the extend and retract if it is a repetitive motion. In these situations, it may be favorable to look at cylinder that has a spring to return for one of the motions, so you don't have to supply air from the tank to return the cylinder.

Also there is what is called rod end of the cylinder. This is the side the working arm attaches to the piston in the cylinder. The working area on the rod end, is the area of the cylinder bore, less the area of the rod bore. So the rod end has less force at the same psi. Also the volume of air is less on the rod end of the cylinder.

[RyanCahoon]

Originally Posted by ttldomination

So, suppose a piston uses 1 cubic inch of air, and is fired thirty time. As per your math, it would be:

1 cu in * 30 actuations * (60 output psi / 120 psi) ~> 15 cu inches of "theoretical" storage?

This is how we calculate air needs, but we usually multiply by 2, which would mean (using the simplified linear model) that you end the match with full working pressure, so that your actuators don't lose force. Basically this also becomes a safety factor - you can continue using the system for more actuations if necessary, but the force will reduce with continued usage.

Originally Posted by ttldomination

And as far as the design question is concerned, my question, which I now understand was slightly misleading, was more in regards to CADing the piston.

Are there nice models with the appropriate motion constraints built in, or do people just model in a pole and restrict its range of motion?

We pretty much exclusively use Bimba cylinders - between the free product voucher in the KoP and the fact that McMaster carries them. Bimba has configurable CAD models of all their cylinders on their site.

If you use the Direct Insert method (which controls your Solidworks/Inventor/ProE remotely using a Java app and effectively generates the part on the fly) I think it may add the constraints. (It's been a while since I've used it, though, so I may be wrong.)

The Direct Insert process takes quite a while on my computer, so I usually opt for the STEP file download. STEP files don't have constrains, but if I need motion simulation, then I group the parts into either two subassemblies: cylinder and piston, then apply a single constraint to enforce axial motion.

I use Inventor - if you use Solidworks or ProE, it looks like there are application-specific formats which may include constrains.

[DELurker]

We pretty much exclusively use Bimba cylinders - between the free product voucher in the KoP and the fact that McMaster carries them. Bimba has configurable CAD models of all their cylinders on their site.

Actually, you are not guaranteed to get a Bimba from McMaster-Carr. We have gotten several non-Bimba (but compatible) cylinders from them in the past two years.

[DonRotolo]

Originally Posted by ttldomination

So, suppose a piston uses 1 cubic inch of air, and is fired thirty time. As per your math, it would be:

1 cu in * 30 actuations * (60 output psi / 120 psi) ~> 15 cu inches of "theoretical" storage?

Yes. That 15 in^3 would drop to zero if it remained at 120 PSI until the last actuation. In reality, it would drop about 4 PSI (1/30 * 120) each actuation, so after 15 actuations your storage pressure would be 60 PSI. You can do the math to see where you'll end up. Ryan's "double it" is reasonable.

Be sure to take into account all leaks, which can be non-trival. Also, storage is often not at 120, but a bit less.

An AndyMark AM-2478 ($15) is 30 in^3, weighs 0.56 pound, two of those puppies will keep you moving 1 in^3 well above 30 actuations. On the other hand, a 12" cylinder with a 2" bore would work maybe twice at full pressure.

[ttldomination]

Originally Posted by DonRotolo

Yes. That 15 in^3 would drop to zero if it remained at 120 PSI until the last actuation. In reality, it would drop about 4 PSI (1/30 * 120) each actuation, so after 15 actuations your storage pressure would be 60 PSI. You can do the math to see where you'll end up. Ryan's "double it" is reasonable.

In the drop calculations, I'm assuming that the 4 PSI is dependent on the cylinder size? Or is that more of a rough constant?

- Sunny G.