This is a really interesting question. There are no doubt people who know the answer in great detail, but I can at least work out a first answer approximately.
The first thing to ask is what keeps the water from dripping down? The partial vacuum on the upper part of the tube can't be the full answer, because it doesn't explain why little droplets don't simply detach from the bottom side and fall without changing the size of the vacuum. The key physical ingredient of the explanation must be surface tension, the extra free energy that's needed per area of water-air interface. Formation of a droplet increases the surface area, so unless the gravitational force is strong enough, it will take extra energy for a droplet to form, and the water will stay stuck in the tube.
How does that limit the tube radius, R? I'll do a crude calculation, not keeping track of small numerical factors, i.e. using some dimensional analysis. The extra energy required to form a drop goes as its area times the surface tension s, very roughly sR2 for the biggest drop that would form from a tube of radius R. The energy lost by lowering the water in that drop by a distance comparable to its size is roughly ρgR4, where ρ is the mass density and g is the gravitational acceleration. So for the bottom surface of the water not to shed drops we need roughly sR2>ρgR4 or R2< s/ρg.
The surface tension of water near room temperature is about s= 70 ergs/cm2, ρ=1 gm/cm3, g=980 cm/s2, so we end up with roughly R2< 0.07 cm2. That would correspond to a tube with diameter of about 5mm. Given that I haven't tried to solve the problem exactly, that answer could easily be off by a factor of 3. Is it anywhere near the measured cutoff between tubes that hold the water and those that don't?
p.s. A first test showed that ~3mm diameter gave stability and ~ 10 mm diameter was unstable, so this calculation seems to be pretty close.
(published on 06/02/12)
This is a very interesting question! We might have seen someone doing this "magic" by turning a glass full of water upside down with a card-stock.
As you have mentioned, the basis of this "magic" is that the air pressure in the atmosphere is strong enough to hold the weight of the water inside the glass. However, you might have found out that this "magic" only works if your glass is completely full of water and you'll have to use a card-stock or something similar to cover the glass.
The reason for needing a full glass, even though the air pressure is strong outside the glass, is that there won't be any net force from the atmosphere to the water unless there's less air pressure inside the glass. That is to say if there's air in the glass, the effect due to the atmosphere outside the glass is just the same as the air inside the glass, hence there will be no excessive force due to the air pressure that can hold the water against its own weight.
Getting back to your original question, if we don't have a card-stock covering the glass the water can break up into drops which fall off without having to pull water into the top of the glass. (see the answer above) The surface tension of the water (the water molecule stick to each other) can hold it back in a little crack, but not over a big area like the whole surface of the glass. The card stock handles that.
FYI, this explanation of the pressure difference also explains why our body doesn't get crushed by the humongous atmosphere pressure. Because there is also air inside our lungs and other parts of the body, the air pressure inside is just enough to balance out that outside of our body. Hope this helps.
Lingyi (mods be mw)
(published on 02/26/13)