I had originally intended to use an off-the-shelf check valve to use in my pneumatic soft-close mechanism, but the more I played around with the overall idea, the less suitable it appeared to be. It would have to hang off the side of the cylinder. The piston needed to be "special" to accommodate the need to put a hole in the cylinder for the air to pass through. And so on.
So I was thinking about integrating my own check valve into the piston itself. The original design used the aluminum stem for the air flow (it is a tube), with a ball bearing to act as the seal. The bearing would be pushed against the end of the tube with a small spring. The tube, bearing and spring would be installed in the piston.
But my piston is only .75" long, and it was hard to find the right itty-bitty spring so I decided to replace the spring with yet another magnet that would pull the bearing onto the end of the tube. The magnet would be a ring magnet that the aluminum tube slides through, and would be glued to the top (or outer) side of the piston. I liked this approach because it's based on a simple physical mechanism (magnetism) and should work OK for a very long time. The only thing that might mess it up is dust and dirt between the ball and end of the tube; and if that becomes a problem I could glue a small air filter to the bottom of the piston to junk out. Maybe I'll be pro-active and just do that up front :).
But the question arose: would there be enough force acting on the bearing ball to pull it into the end of the tube? Or would the force be too great so the check valve wouldn't permit the damper to easily open (and therefore cause the drawer to not easily open)?
This looked like another simulation to perform using FEMM, Finite Element Modelling, so I could answer these questions. Long story short: it looks good. The attractive force is small, on the order of 3.9 grams; but since the check valve will be on the horizontal plane it won't take much force to pull the ball toward the end of the tube. And 3.9 grams over an area of .049 square inches (the cross-sectional area of the tube) indicates that the check valve should open with a pressure differential of just .17 pounds per square inch.
Here's a screen shot of the simulation:
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