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
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 4mm. 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 05/31/2012)