For scaling we are planning on using a single pneumatic cylinder to lift our bumpers above the low goal. We need a 24" stroke, but I am not sure what size bore to use. Assuming that we have the necessary air tank storage space, would a 2-1/2" bore be faster than a 2" bore? I have read on this thread ,http://www.chiefdelphi.com/forums/showthread.php?t=141506&highlight=cylinder, that increasing the bore size of the pneumatic, although it increases the strength, makes it slower because of the extra volume that must be filled by air traveling in 1/4" tubing. Does anyone here have experience with large pneumatic bore sizes?
Yes it will take longer to move. Pneumatics are no different than fluids when it comes to thinking about how they work. If you were to fill a couple of glasses with water and both glasses are the same height, but one has a larger diameter which one will fill first?
With that being pointed out, let’s go through the math on exactly how to determine the bore size needed.
We are attempting to lift 150lb of robot. (Nice round number, little bit big but that’s OK for now.) We can only use 60 psi air. So, we need a relation something like this:
150 lb = 60 psi *A in^2
Solving for A, we get the following relationship:
A = 2.5 in^2 = pi*r^2.
r^2 = something like 0.795 in^2, and 2*r = diameter, which is what pneumatic cylinders usually are sized by. Some quick math returns that a 1.75" cylinder will be too small, while a 2" is big enough–note that this is from a pure force standpoint, and neglecting the area occupied by the rod part of the cylinder.
The thing with pneumatics is that you want to use the smallest cylinder you can get away with, in general (at least in FRC).
A 2.5" cylinder from Bimba has a 0.625 pushrod. Assuming you will retract the rod to pull your robot up, we need to subtract that area to determine the capability.
2.5" diameter, less the .625 rod, gives an area of 4.6 square inches. 24 inch length = 110 cubic inches of volume. This cylinder has 275 pounds of force to lift your robot.
2" diameter, less the .625 rod, gives an area of 2.83 square inches. 24 inch length = 68 cubic inches of volume. This cylinder has 170 pounds of force to lift your robot.
The 2.5" cylinder has 160% of the volume of the 2" cylinder, so it will take about 160% of the time, assuming you have the available volume of stored air.
If your robot weights 120 pounds, +12 for battery and 15 for bumpers, you are close to 150 pounds that you need to lift. Remember that you will also have some friction to overcome if the robot is sliding up the front of the tower.
If you are going to use the cylinder retracting to lift your robot, you will need to subtract the area of the rod from the area of the cylinder to find the retracting force.
150=60(pir^2)-(piR^2)]
150 = Cylinder Retracting Force (Lb)
60 = legal working pressure (psi)
r = cylinder radius (in)
R = cylinder rod radius (in)
You will need lots of storage tanks to move that cylinder in 20sec. We used two 2" and one 0.75" cylinders to climb a few years ago. The thing we learned and I will highly recommend to you is to use a single small cylinder to raise your hook and then then all three to lift the robot. You don’t need/want to expend that much air to lift the hook. Other than that, we successfully climbed probably 90% of the time. Simple mechanism - if you have enough air.
I don’t think a 2-1/2" cylinder would require 160% more time to fill since it it only requires 1/160% as much PSI to provide the same force.
I think this is interesting mathematically remembering the equation pv=nrt. Assuming that temperature remains constant, the two cylinders are operating in a vacuum, and they have the same stoke length, the same amount of air in one bore sized cylinder will provide the same amount of force that a cylinder with a different bore size. Let me explain,
F=force
A=area of piston end
V=volume of cylinder
P=pressure
N=amount of Air
S=stroke
PV=N
F=PA
V=SA
Therefore
A=F/P
V=SF/P
Multiply that by N=PV
NV=PVSF/P
N=SF
N/S=F
So under ideal conditions, the same amount of air in any bore sized cylinder (with the same stroke length) would equal the same force. The question then is what would fill to the required lifting pressure faster, the 2-1/2" bore or 2" bore? Wouldn’t the the 2-1/2 fill faster because of the increased pressure difference? Since the air pressure inside the 2-1/2" would be lower under the load wouldn’t it have a faster fill rate because the difference between the pressure in the cylinder and the 60psi storage would be greater? I will do some more calculations to try to figure that one out.
I think this is the part to pay attention to. A 2.5" cylinder uses up 160% as much volume as a 2" cylinder of the same length, so the robot needs to store 160% more air, and it takes longer to shove all that air through the port of the cylinder.
Even though a 2" cylinder will extend faster, you can still make a 2.5" cylinder work. You just need to solve those two issues.
Storing more air - This is easy if you have the space, just add more tanks. You can get some really big ones if you look around.
Shove the air in fast enough - There are lots of tricks to increase the flow rate of air into a cylinder. Get a high flow solenoid (mcmaster 6124K511, for example). Decrease the amount of tubing between the solenoid and the cylinder - you can mount the solenoid right onto it if you use the right fittings. Add air tanks right before the solenoid so air doesn’t have to go through the regulator. You can make these things flow really fast.
Interestingly, we found a use for our 2.5" bore 24" stroke cylinder this year… it had previously been sitting in our back room collecting dust. We used it to measure the flow rate of the solenoid above using our pneumatic layout. (Useful data for some other minor mechanism on the robot this year :rolleyes:)
The cylinder in the setup below extended in 0.170 seconds (but no load other than the mass of the cylinder rod)
It depends on whether you want fast or minimal air use. If you need fast, (e.g. for a launcher), you will have to use cylinders that are larger than “you can get away with”. If you have enough time to do it efficiently, you want to use a minimal -sized cylinder.
If going fast, the limiting factor is how fast air can go through the 1/8" NPT ports (or perhaps smaller) on your solenoid valves at the “real” working pressure you will use. This will be much less than the 60psi on the supply side of the valve, probably more like 15 or 20 psi to optimize power through a 1/8" NPT orifice.
Edit2:
Amplification and correction: The flow rate through a 1/8" orifice with 60 psi on the supply side and up to about 20 psi on the working side is about 13 scfm. At 30 it’s about 12 scfm, at 40 psi, it’s about 11 scfm, at 50, about 8, and at 55 about 6. Calculating the actual rate of volume change and multiplying by the gauge pressure, fastest work gets done between 20 and 40 psig. If you want it to work fast, design so that the job is done by 40 to 50 psi.
/Edit2
That won’t help. The rod comes into play whenever the cylinder is made shorter, no matter whether the rod or the chamber is at the fixed end.
Edit:
So was I. When pressurizing the return stroke, the net area that the air is pushing against does not include the rod, independently of the orientation of the cylinder.
My team is using a 2" cylinder with a 24" stroke to hang our robot, but we’re having problems retracting and lifting.
We were using a single air tank at the first competition we went to. It was found that gently nudging the robot at the start of retraction seemed to get it going.
We then added two more air tanks and checked the gauges immediately after we extended to see if that was the issue. The high gauge read about 50psi, if I remember correctly, the low was 50psi for a split second and then back to 60psi. We did not have the chance to test actually hanging with this set up.
We took our robot to a practice field, but our controller was not working correctly. To test our pneumatics, we tried to manually trigger the solenoid. This did not produce the results we were hoping for; however, it is unclear if this was due to the manual triggering or a pneumatics issue.
Our robot weighs 77lbs with battery (and I believe bumpers).
My questions:
How much time does a pneumatic of this size take to refill?
Will the 2nd and 3rd air tanks mitigate the time in question 1?
Is there anything else that can be thought of in relation to pneumatics that might need troubleshooting?
If you use constant force springs in line with your cylinder, you can use a smaller diameter cylinder. There’s a voucher for six free springs from Vulcan in the KOP.
We use two 1 3/4" BIMBA cylinders, and hang by the rods. We haven’t had any pneumatic problems yet
We have 5 air tanks also, and run it off of two solenoids and two regulators. The solenoids each lead to both cylinders, and are slaved together. We do this because the internal diameter of one solenoid it too small to facilitate rapid airflow
This constant-force spring thing seems like a good idea, but it means that it takes more pressure (equals force) to extend your rod initially. Which means it uses lots of your stored air reaching up for the bar.
I wonder if it would be better to actually put a pressure regulator between the extension-end of the air cylinder and the air solenoid valve and just extend at say 5-psi? If you can still reach up very fast but not actually pressurize the extending side of the air system to a very high number, you’ll save all of that stored pressure in your tanks for the lifting-retracting stage.
We built a robot-lifting cylinder in our prototyping at the start of build–used a 2" dia–which should’ve been able to lift 176 lbs, but with a 120lb robot, battery, bumpers and the added friction of dragging the bumper up the wall, it didn’t move very well at 60 psi and was very slow after the air in the storage tanks was used up.
I hope you can make it work. good luck.
Also note,
I know these equations are easy, but if you need a quick-reference for how much force each bimba-sized cylinder makes at 60 psi for pushing and pulling, I’ve created this easy chart.
Unless you add in the additional pressure regulator, you’re extending at 120 psi anyway…so add ~75 lbs (or however much) of constant force springs, extend at full pressure to extend them, then let them help you on the way down. The pneumatics will be smaller, so you’ll be saving air on the up and down strokes (which might obviate the need for the additional regulator to bring down the extension pressure.)
Some math (maybe some graphs with an eye toward that 150-lb requirement) should help you determine an optimal* amount of spring vs. bore.
*optimal, of course, depending on which resources you’re trying to preserve.
As another idea, consider constant force springs that are already extended along the body of the cylinder, but don’t lock in to the rod until it’s fully extended–so you don’t have to push against them at all, but they’ll help you pull it back. Might require some cams, spring-pins, or whatnot.
