what is the limit on pneumatics this year?
2 air tanks.
Same as last year.
Sorry? could you elaborate more
and by cylinders do u mean tanks?
The Reservoir is the tank
The Cylinder is the piston
Hope that answers your question
two pneumatic reservoirs (same as last year) and 10 pneumatic pistons.
hope this answers your question.
Just curious, where is the limitation on cylinders? I must have missed it.
Pretty sure you ( @1815Delta ) are making the “10 cylinder” rule up or misunderstanding it???
Only limits you to two reservoirs, does not specify anything about cylinders. If I’m wrong, please state the exact rule.
I think the confusion may be coming from the following rule. But it does not say only 10 cylinders. It says 10 motors or servos AND a legal pneumatic system (as per R18)
There is no limit on pistons. This rule was exactly the same last year, and I know teams that have used more than 10 pistons.
The only limit on pistons is a practical one: each solenoid requires a digital port (unless they’re y-cabled), of which you have 12 minus however many you use for sensors, and each piston fire uses a fair amount of air.
Just wondering, is there any way to “calculate” the number of actuations you would get? IE, I need 5 lbs of force or whatever, how much air is expended?
It would be useful to be able to discard impractical applications of pneumatics before building the entire prototype, but as of yet the only method I know of is counting the number of actuations by hand
Here’s some design stuff on pneumatics:
First, in industry, we call the actuator (what people on this forum call a “piston”) a “pneumatic cylinder”, either single-acting, if you push it out with air and it retracts with a spring, or double-acting if you use air to push it both directions. The reservoir, although it is made from a modified pneumatic cylinder, is not a cylinder, but rather just an air tank.
The vex pneumatic cylinders have about a 10mm bore (.375"). So, find the cross sectional area (square inches), then multiply the systems psi (pounds per square inch) by the cylinder area (inches cancel) and you will have the force exerted by the pneumatic cylinder. The pressure is that of the regulator, not the pressure in the reservoir. And once the pressure in the reservoir drops to the working pressure, the force will of course start dropping, but it will be constant up to that point.
For those who are nit-picky and using double-acting cylinders, the cross sectional area to extend the cylinder is a littler larger than the cross sectional of the retract side because of the diameter of the cylinder rod itself. You can calculate that area and subtract it from the total area if you wish, but it really isn’t that critical for this application.
If you do the calculations correctly, your force number will be in the range of 8-12 pounds.
As for how many actuations of the cylinder you’ll get, the air consumption can be calculated (P1V1=P2V2), but I wouldn’t bother, as there are some extra variables, especially the length of tubing between the solenoid valve and the cylinder. In testing, you’ll get 30-40 cylinder actuations, depending on the working pressure. Remember that a double-acting cylinder take 2 actuations to cycle it in and out.
Wow, thanks for the explanation @kmmohn ! I’ll keep that estimate in consideration
So, if I had two tanks, could I get 60-80 actuations?
Don’t think you will ever get 60 to 80 actuations. Of course technically or theoritically it is possible. But you will need to regulate the air to such a small amount that it will render it almost useless.
More precisely, does doubling the number of tanks double the number of actuations?
Assuming everything else remain constant - then it should be almost doubled.