What is the formula for the glug point? When you pour water out of a bottle, normally you have a smooth stream. However, if you pour it too fast it glugs, which is to say, comes out in quantized bursts. What is the formula for calculating the glug point angle, and what is the formula for calculating the size of the glug quanta and glug quanta frequency?
 A: Glugging is determined by pure geometry--- in your chosen tilt, as the water pours out, will the water completely cover the opening, or not? If the water doesn't cover the opening, air will stream in to equalize the pressure in the bottle to the pressure outside, and there will be no glug. When the water covers the opening, the water will still pour out, expanding the air trapped in the bottle. When the air pressure inside is small enough, the atmospheric pressure outside will push the water up more than gravity pulls it down, and it will not allow the water to leave. But the water's inertia streaming out will overshoot this point, so that the pressure will be less than this magic value, and air will force its way into the water.
This air will push a bubble into the water, recompressing the air trapped in the bottle until the pressure inside rises past the magic value. At this point, the water will fall back down through the opening again, allowing the air bubble to rise to join the air inside the bubble. This is an oscillatory process because of the inertia of the water--- both the outflow of water and the inflow of air overshoot the equilibrium.
The process is not "quantized", because it depends on the exact amount of air trapped in the bottle--- the more air there is, the further apart and bigger the glugs will be (you can see this effect in an emptying water-cooler bottle), and on the exact amount of extra water released before each glug, which depends on the inertia of the water and its exact exit velocity and opening geometry. In a good model, it should be possible to calculate the exact amount of glugging from the condition that the air is always in equilibrium, and the water is flowing out according to Bernoulli's law, with inertial effects leading to a certain lag in time of the response to the pressure.
An interesting case is when the water is forced to go out through a straw. In this case, the glugging doesn't happen, because the water's viscosity dominates and there is no lag. When you have an oscillatory system with friction, generically there is a critical value of friction which kills oscillations (see critical damping) and leaves only exponential decay. This allows a stable equilibrium, where the pressure of the air inside is just low enough to hold the water up. So the water doesn't leave the upside-down bottle. If you arrange for the bottle to be completely full, you can do this without a straw too, although this is often unstable, because if a certain size bubble enters, it will allow the oscillations to begin with its volume of trapped air expanding and contracting.
