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If aerographite is lighter than air, but porous, would it be possible to create a "filtered balloon", and blow aerographite dust into the ballon, allowing air to be displaced by traveling through the filter, which would block the aero graphite dust. If a fan kept the resulting "dusty air" moving around, more air would escape through the filter, and the remaining "atmosphere" would consist of a higher proportion aerographite, and a lower proportion air, thus acquiring lifting force.

Am I thinking about this correctly, or is there a flaw in the logic?

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The flaw is that the given density for aerographite (about 180g/m3) is based on the air in it being sucked out of it and replaced by vacuum. It's a foam and though theory says it's something like 6x lighter than air, practice has it typically full of air and thus heavier than air since the graphite component is heavier than air. The "lighter than air" part is based on its volume as a foam and not some property of graphite itself.

Grinding it into dust would essentially turn it into something resembling graphite powder and not foam. The density would go up dramatically as volume diminishes and it would be even less capable of creating a lifting force than standard aerographite is.

If you want to create a balloon of it you'd need to suck all of the air out, cover the outside of the structure with a very light skin of, say, graphene, and if atmospheric pressure didn't squish the whole thing, you should get some lifting.

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  • $\begingroup$ Thank you Jake. I understand that the problem with a theoretical "vacuum balloon" is that it gets crushed by atmospheric pressure. So, one follow n question: If we took your suggestion above, is there a way to calculate how big a "graphene/aerographite ball" could be made before it would be crushed if the air were sucked out of it? i know how to calculate the resultant lift if it's form holds, but don't know how to calculate the "squishing point" based on it's strength when under compression. $\endgroup$ – Dave Feb 8 '17 at 3:51
  • $\begingroup$ Compression isn't a liner equation. TBH, it's a lot of head scratching to calculate and depends on knowing a lot more about the material than I do. $\endgroup$ – Jake Watrous Feb 8 '17 at 4:23

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