[ChuckMcM]

Hmm air can carry a few grams of water per kG of air [1] and 1kG of water is one liter, so if you can pull 1g of water out of each kG of air you need 1,000 kG of air. Stole this from wikianswers: "Each mole has a volume of 22.4 liters and a mass of 28.97g/mol at STP, therefore a cubic meter of air is 1.293 kg at 0o Celsius on the coast. An average mass of 1.2kg per m3 at room temperature and standard pressure is often used as a rule of thumb." so that would suggest about a thousand cubic meters of air flowing past this bottle's extractor area to get one liter of water out of the air.

No idea how long that would take though.

[1] http://en.wikipedia.org/wiki/File:Relative_Humidity.png

[Jabbles]

Good work. I imagine 1g/1m^3 is roughly what you'd expect - especially in a desert.

If the extractor area is the size of the bottle's cap (2cm x 2cm = 4e-4m^2) then to get a litre of water (1e3 m^3 air) requires 2.5e6m of air to flow past it - If you want it full in a day that's 30m/s... hmmm... doesn't seem likely.

However if the size of the extractor area is as large as the bottle (say 10cm x 10cm = 1e-2m^2) then a 1 m/s wind may do it.

Based on this I'm not confident in the "marathon runner" claim, but I'm sure techniques for extracting water from wind have been used before, but improvements could be made.

Scaling up the technology obviously increases the viability, yielding something like this: http://www.geek.com/articles/geek-cetera/prototype-wind-turb...

[andrewcooke]

i think you're confusing collecting area with airflow cross-section. they're completely different things, unless somehow the air is blowing through the side of the (solid) bottle.

[Jabbles]

Not really. This volume of air has to make contact with the bottle, whatever the collecting area's shape. Imagine loosely fitting a tube around the neck of the collecting area. All the air has to pass through that tube.

[andrewcooke]

no, i'm saying that only a small fraction of the incident air will touch the surface. most will flow around - probably laminar flow containing a stagnant area of higher pressure directly in front of the obstacle.

you are calculating the swept volume. but only a small fraction of the swept volume "makes contact" (ie is roughly within the mean free path of a water molecule) with the collecting surface (which appears to be necessary here for water to be extracted).

with netting your argument is closer to being correct. but this is an impervious bottle, not a net.

if you implemented the tube you suggest (and i understand that was simply an illustration to explain swept volume) then pressure would increase in the tube slowing flow and/or flow would be predominantly around the edges, with, again, a stagnant higher pressure region in the centre.

[primitur]

Well, think of the most efficient collector of water (and sunshine) you can. Water bottles have to look like that, or at least .. have a spigot you can attach a bottle to ..