Usign a 2-1/2 bore, 24" stroke pneumatic cylinder.

[BotDesigner]

Hi all,

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/sh...light=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?

[dradel]

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?

[EricH]

Just for reference, 2.5" > 2".

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).

[Chris Fultz]

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.

[Everett33]

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[(pi*r^2)-(pi*R^2)]

150 = Cylinder Retracting Force (Lb)
60 = legal working pressure (psi)
r = cylinder radius (in)
R = cylinder rod radius (in)

[dradel]

I was going to suggest mounting the cylinder up side down, or backward from how they are typically mounted. That way the rod won't come into play

[EricH]

Originally Posted by dradel

I was going to suggest mounting the cylinder up side down, or backward from how they are typically mounted. That way the rod won't come into play

That's a possibility, sure... but it will require slightly different mounting methods and somewhat longer air lines to pull off.

[dradel]

This is true, but more than doable.

[BryceKarlins]

Originally Posted by dradel

I was going to suggest mounting the cylinder up side down, or backward from how they are typically mounted. That way the rod won't come into play

Originally Posted by EricH

That's a possibility, sure... but it will require slightly different mounting methods and somewhat longer air lines to pull off.

Mount the cylinders upside down, and have them push into the floor, then attach a static hook, and retract.

[MrBasse]

Originally Posted by BryceKarlins

Mount the cylinders upside down, and have them push into the floor, then attach a static hook, and retract.

That would cause some bumper zone issues...

[gpetilli]

Originally Posted by BotDesigner

Hi all,

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/sh...light=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?

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.

[BotDesigner]

Originally Posted by Chris Fultz

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.

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.

[s_forbes]

Originally Posted by Chris Fultz

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.

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.

1. 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.
2. 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 )

The cylinder in the setup below extended in 0.170 seconds (but no load other than the mass of the cylinder rod)

[GeeTwo]

Originally Posted by EricH

The thing with pneumatics is that you want to use the smallest cylinder you can get away with, in general (at least in FRC).

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

Originally Posted by dradel

I was going to suggest mounting the cylinder up side down, or backward from how they are typically mounted. That way the rod won't come into play

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:

Originally Posted by dradel

I was referring to the rod affecting the power not the speed.

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.

[dradel]

I was referring to the rod affecting the power not the speed.