Forgot to put a parking tag on them, sorry. These are mine. I'm going to saw the tops off and make recycling containers for aluminum cans.

We go through so many cans at HF that it would be nice to capture that as revenue and feed it back into the organization - perhaps as a way to offset the many discounted memberships that we give to students and unemployed folks.

I've got one can crusher and I know there's another in the office. I'll be building plywood tops with crusher on-board (literally!) and a slot for crushed cans. I want to discourage putting uncrushed cans in these, as they'd fill up almost overnight.

Someone should build a 2-stage can compressor; I mean crusher. Those wimpy peasant crushers only get them down to about 3/4" tall. I feel that we can do much better; I mean denser.

edit:
By my quick math, a 4" pancake cylinder on 90psi should create about 1100 pounds of force. That should squish 'em flatter than a cancake.

If we somehow built a dual-rail air supply to connect -90psi to the other side of the cylinder, would that double the force? (This thing will probably need finger guards and/or a breathalyzer on the ignition...)

what would it take to get the same flatness by decompressing the cans?

I have this image in my head of something that would rapidly create a vacuum inside the can, and that sucks.

Riley

I sorta figured there'd be hacking and wild inventions to crush cans. That's part of the fun. I'll set these up with "traditional" can crushers and hackers are more than welcome to develop their own competing can crushing technology as upgrades.

Great idea!

Maybe some kinda diabolical pneumatic can crusher. Something that just looks like its going to take your hand off. (its a plus if it doesnt), but I'd use the heck out of that!

Here are some basic calculations for fluid power cylinders.

F=AxP Force equals area (in square inches) multiplied by Pressure. (psi)
Area for cylinder (according to fluid power standards) is D(squared) times .7854
V=AxL volume equals area (in square inches) times length

1 gal to 233 cubic inches

soo Here is a basic cylinder example.

You have a 2" outside dia cylinder with a 1 1/2 rod. Your Cap end area (the area without the rod) is 4 is 3.14. Your rod area is 1.77. Your rod end area is 3.14-1.77 = 1.37. Soo your extend force is 3.14 * 90 psi is 282.6 lbs. Your retract force is 123.3.

I would recommend a single action cylinder. It is spring loaded on one end so you don't need a directional control valve, just a solenoid valve. 90psi +90psi is still 90 psi. Its volume is added together. i.e. if two compressors are running at the same psi and you put them together in a manafold you are adding the two volumes together. If one is higher pressure then the other it will back flow and equalize both tanks. I don't know how a negitive pressure on a double action cylinder will act. But you are probably pushing the cylinder past the specs of the side walls.

For those having to reread the area calculation a couple of times (like me): D =2r, therefore using substitution, A=πr^2 == A=D^2(π/4)

Yeah...it confused me too. That's what is in the parker training manual though...I'll try to get my hands on a fluid power lightning guide. (the end all be all of fluid power...so I'm told)

(No disrespect whatsoever is intended towards using traditional crushers, in case that was unclear)

Isn't there a pancake cylinder sitting on a hackable parts shelf somewhere? If it's still around, we should bust it out on a Wednesday night and jury-rig up a finger cruncher; I mean can crusher prototype.

A hypothetical -90psi vacuum supply would indeed double the force of the cylinder. The 90psi in the top of the cylinder is relative to 0psi (atmospheric pressure) in the bottom half. 90 on top an -90 on the bottom is the same as having 180psi on top and the usual nothing on the bottom. As to how and why someone would actually create a large scale -90psi vacuum supply to pipe around the shop, I'll leave an exercise for the reader...