Number of Pneumatic Air Tanks

[Dr.Gusta]

My team does not have much experience with pneumatics and this year we are planning to have pneumatics play a major part in our robots function. We are planning on using four 5/16" bore and 14" stroke single action (spring return) that will open and close up to 25 time but let's say 50 times to be safe. Also there will be another smaller cyllender of unknown bore but about 5" stroke dual action and would probably only open and close 5 times. We would also have the standard Viair compressor. We were thinking 6 of these tanks (http://www.andymark.com/product-p/am-2649.htm) pressurized at 120psi. Do you think that will be sufficient?

[Mike Marandola]

I like to use Paul Copioli's Useful Calculations spreadsheet for this. The only thing is that it doesn't factor in the compressor flow rate.

[MrForbes]

My general feeling this year is that it's probably going to work OK to use pneumatic stuff to "grab", but not to "lift". If you try to do too much work with pneumatics, you run out of air pressure, and the compressor can't keep up.

[Nate Laverdure]

By the combined gas law the quantity (P*V)/T remains constant throughout a closed system in steady state. If we can also assume that the temperature of the system remains approximately constant over the duration of interest, then the quantity P*V remains approximately constant too.

P*V ≈ constant

This is a pretty powerful tool. We can use it to equate the energies of two different volumes of gas at two different pressures.

P1*V1 ≈ constant ≈ P2*V2
P1*V1 ≈ P2*V2

The quantity P*V will be in units of energy. This number will in fact be a estimate of how much stored energy is contained in the pressurized system. (But it's not quite a measure of how much energy is available in the system to do useful work!)

Here's an example. First let's look at how much energy is contained by the storage tanks.

P1 = 120 psig storage pressure
V1 = two 16 in^3 tanks = 32 in^3
P1*V1 = (120 psig)(32 in^3) = 3840 in*lbf

Now, how many times can you use that energy to actuate an example cylinder?

P2 = 60 psig point-of-use pressure
V2 = (n actuations)(5/16 in bore x 18 in stroke) = n(.0767 in^2)(18 in) = n(1.38 in^3)
P2*V2 = n(60 psig)(1.38 in^3) = n(82.8 in*lbf)

P1*V1 = P2*V2
3840 in*lbf = n(82.8 in*lbf)
Solve for n...
n = 46 actuations

[John Retkowski]

Depending on what you want to be doing with your pnuematic cylinder you may want to rethink/recalculate what you want to use. I've found this calculator very quick and useful. http://www.pneumaticsonline.com/calc2.asp

If you plan on picking up more than one tote, you'll probably need a much larger bore. This is assuming you want to use pnuematics to pickup totes of course.

Also you may want to take a look here http://www.chiefdelphi.com/forums/sh...anks+exploding This was last year and I don't know if they've changed the tanks at all.

[Jon Stratis]

Originally Posted by Nate Laverdure

By the combined gas law the quantity (P*V)/T remains constant throughout a closed system in steady state. If we can also assume that the temperature of the system remains approximately constant over the duration of interest, then the quantity P*V remains approximately constant too.

P*V ≈ constant

This is a pretty powerful tool. We can use it to equate the energies of two different volumes of gas at two different pressures.

P1*V1 ≈ constant ≈ P2*V2
P1*V1 ≈ P2*V2

The quantity P*V will be in units of energy. This number will in fact be a estimate of how much stored energy is contained in the pressurized system. (But it's not quite a measure of how much energy is available in the system to do useful work!)

Here's an example. First let's look at how much energy is contained by the storage tanks.

P1 = 120 psig storage pressure
V1 = two 16 in^3 tanks = 32 in^3
P1*V1 = (120 psig)(32 in^3) = 3840 in*lbf

Now, how many times can you use that energy to actuate an example cylinder?

P2 = 60 psig point-of-use pressure
V2 = (n actuations)(5/16" bore x 14" stroke) = n(0.342 in^3)
P2*V2 = n(60 psig)(0.342 in^3) = n(20.5 in*lbf)

P1*V1 = P2*V2
3840 in*lbf = n(20.5 in*lbf)
Solve for n...
n = 187 actuations

Nate, can you go over how you calculate V2 in a little more detail? Less than 1 cubic inch for one of those cylinders seems really small to me, and doesn't match my simple math (but I'm not a much e, so I could very well be missing something). 187 actuations also seems to be really high, based on practical experience with FRC pneumatic systems.

Does it take into account what happens when storage pressure drops below 60 PSI, it does it assume you can go down to 0 psi storage while maintaining 60 psi working? When your storage drops below 60 PSI, the equations have to change a bit, as you can no longer have the desired working pressure.

Perhaps a better approach would be to do the math from the other direction - what volume of your stored air (at 120 psi) would be needed for a single actuation (at 60 psi)? Since we know PV is roughly a constant, we know we have roughly half our total storage volume available to us before we drop below 60 psi storage. If it takes 2% of the total storage volume for one actuation, then we would get 25 actuations before dropping below 60 psi, for example.

[MrForbes]

5/16" bore cylinders are really small! they will only produce a few pounds of force. But they won't use much air.

[Jon Stratis]

Originally Posted by MrForbes

5/16" bore cylinders are really small! they will only produce a few pounds of force. But they won't use much air.

The volume of the cylinder would be ((5/16)/2)^2 * pi * 18, right? Pi * r^2* h. That comes out to 1.38 for me, about 4 times larger than in Nates equation... What did I miss?

I'm just trying to make sure I understand the math right so I can guide. my kids through it later when we need to do it... I don't want some mistake or something forgotten in the 10 years since I last really used this stuff to screw up my team!

[Michael Hill]

Originally Posted by Jon Stratis

The volume of the cylinder would be ((5/16)/2)^2 * pi * 18, right? Pi * r^2* h. That comes out to 1.38 for me, about 4 times larger than in Nates equation... What did I miss?

I'm just trying to make sure I understand the math right so I can guide. my kids through it later when we need to do it... I don't want some mistake or something forgotten in the 10 years since I last really used this stuff to screw up my team!

Several things....Nate forgot to multiply by pi, and you used an 18" stroke length where he used 14".

[Jon Stratis]

Originally Posted by Michael Hill

Several things....Nate forgot to multiply by pi, and you used an 18" stroke length where he used 14".

Oops, my team had been looking at 18" stroke, I just have transposed it to this discussion by mistake. Thanks for the help!

[Michael Hill]

Originally Posted by Jon Stratis

Oops, my team had been looking at 18" stroke, I just have transposed it to this discussion by mistake. Thanks for the help!

Sure, also, don't forget that the pushing on a cylinder is greater than the pulling force because of the rod (you lose usable surface area). It's generally not that much, but it can bite you if you're borderline.

[JesseK]

Originally Posted by Mike Marandola

I like to use Paul Copioli's Useful Calculations spreadsheet for this. The only thing is that it doesn't factor in the compressor, but that can be easily figured out by using the flow rate (0.88 cfm).

Raul from Wildstang did a pneumatics presentation at Championships '09. In it he showed data that 111 collected on pneumatic compressor performance. The data showed that the compressor rating (0.88cfm for the small one) is correct when the tanks are empty. As the tank fill gets close to capacity, performance was reduced by non-trivial amount for pneumatically-dependent robots.

This was especially important if a robot used a lot of air with a lower storage volume, as the compressor would start to re-fill the tanks when they were already close to full, thus not re-filling the tanks at the 0.88cfm rate. Back in the day, we weren't allowed unlimited air tanks - so maybe these days that performance hit isn't as big a deal.

[Nate Laverdure]

Originally Posted by Jon Stratis

...about 4 times larger than in Nates equation... What did I miss?

Fixed.

Was hoping you'd look more at the thermodynamic concept at the beginning of my post, not the geometry at the end.

[MrForbes]

Originally Posted by JesseK

As the tank fill gets close to capacity, performance was reduced by non-trivial amount for pneumatically-dependent robots.

This is the killer. Many compressor spec sheets list CFM at a few different pressures, and you'll see with small compressors like the ones we can use, the difference in flow between 0 PSI and 60 PSI is substantial. As is the difference between 60 PSI and 120 PSI.

[Jon Stratis]

Originally Posted by MrForbes

This is the killer. Many compressor spec sheets list CFM at a few different pressures, and you'll see with small compressors like the ones we can use, the difference in flow between 0 PSI and 60 PSI is substantial. As is the difference between 60 PSI and 120 PSI.

Yeah, it's a huge difference. For the viair 90c for example (which I think is the one from FIRST Choice):
PSI CFM
0 0.88
10 0.71
20 0.67
30 0.64
40 0.60
50 0.57
60 0.53
70 0.48
80 0.45
90 0.43
100 0.39
110 0.36
120 0.34

Numbers pulled right off their website.