(I couldn’t see if water was being thrown up inside the container, however, because it wasn’t transparent.)

kids get really dirty these days.

the inverted cup has vertical walls. While we take it downwards, a little amount of water comes in to balance the pressures outside and inside. So by the time we take the cup to bottom surface of tub, the cup has some amount of water in lower part and above that, the compressed air, which was initially uncompressed there at the time of cup’s first contact with the water.

Now, we take the cup above the open plughole. The plughole has air with atmospheric pressure. And above that, in the cup, are the compressed water and air, combined pressure of which is greater than atmospheric pressure. Therefore, water in the cup starts draining downwards, with compression of air and water in the cup releasing slightly. Thus pressure inside the cup decreases slightly.

Now, as the pressure of water inside the cup is lower than the pressure of water outside the cup; the outside water starts coming in. By symmetry, it should come in as circular ripples, starting from the edge of the cup, gathering towards it’s centre.

The speed of inflow will surely depend on how fine the seal is between edge of the cup and the plughole. Because the finer the seal, the lesser space for inflow, therefore, for the same pressure difference, faster should the molecules move.

Now, the water molecules are moving towards the centre, in the form of circular ripples approximately in the same plane as that of the circular edge of the cup. I don’t know how to explain their further motion with enough rigor. But from intuition, symmetry and the direction of flow, we can say that once such a ripple collides in itself nearly at the centre, the molecules will get mostly verticle / oblique directional velocities in expense of their previous horizontal velocities. The molecules can’t go in backward direction again in the form of circular ripples, because if they went, the inflow motion of water will stop, leading to instability in pressures inside and ouside.

The extent to which the water molecules get their verticle / oblique directional velocities depends on how vigorously the ripple collides in itself. i.e. it depends on the the ripple’s initial speed of inflow. Therefore higher the speed of inflow, better the fountain.
Thus finer the seal between the cup and the plughole, the better the fountain.

Now, again the outside pressure decreases because of lowering of outside water level. (But it being sum of pressures of outside water level and atmosphere, is still greater than atmospheric pressure.) Thus the inflow slows momentarily. Again the inside pressure decreases, with outflow greater than inflow. And again, the difference between outside and inside pressures is increased (effectively, maintained as previous). The inflow is again rejuvenated. These processes happen so fast and continuously that, effectively, we see that, there is constant speed for inflow, and constant fountain, even if the water level outside is decreasing. Thus the fountain should continue till the water in the tub nearly drains out.

And now, whether the height of outside water level makes any difference to the fountain or not.
Well, it should make a difference if the outside water level’s pressure is of the order of atmospheric pressure. Then the fountain will be very vigorous initially, and then decrease in it’s vigorousness. But for, smaller heights, such as the the height comparable to cup ( lesser than 15 cm = 0.15 m), water level’s pressure would be nearly,
pgh = 1000*9.8*0.15 = 1470 Pa << 101 kPa = 1 atm.
This is nearly 1.5 % of difference in the total pressure.
So this little difference should not affect the fountain's vigorousness.

Yes, that follows from the bernoulli equation. I guess that also in this case the speed of the waterflow rises under the edge of the cup. Maybe that fountain is caused by the shape of that metal thing around the hole. But water flowing just above the bottom of the bath that is (I assume) going to the hole is kept on the bottom surface by the stickness of the water, which makes it follow the surface below.

Secondly, I tried it with a different plastic cup that was narrower and more or less perfectly cylindrical on the inside. At first it didn’t work as well, but when I moved it about I found I could get it to work very well.

Finally, if I tried lifting the cup very slightly (which was difficult to do while keeping it steady) then the water just flowed more or less normally and there was no fountain.

My son’s baths are rather shallow, so all I know about the effect of depth is that the phenomenon continues until the water has almost gone. I don’t know whether a very deep bath (say with the cup completely submerged) would work better, worse, or about the same. I would guess the third.

When the cup is lowered into the water away from the plug hole the air inside is compressed to the water pressure at the bottom of the cup, so this pressure in the top of the cup is higher than the pressure outside the top of the cup. This creates the lifting force.

When the cup is placed over the plughole and held down the air pressure in the cup reduces to around atmospheric which is less than the water pressure so the cup is held down. With a perfect seal this would be a stable position.

But there could be another stable situation where the pressure in the cup is equal to the average pressure of the water outside the cup so it floats without moving up or down just above the plug. The water pressure at the bottom of the cup is higher outside than the pressure inside the cup so water is flowing in.

The tricky part is to understand why this would be stable as the experiment seems to indicate. If the cup moves up the speed of the water going in will increase and this must decrease the pressure inside to draw it back down.

It would help to know if this only works in certain depths of water. Does it stop working when the water gets deeper, or when it gets shallower?

If there is a plughole, water is drained out, and the air in the cup can expand and take up its original volume. Depending on the configuration of the plumbing, it may have to take up a larger volume (the cup plus part of the pipes), but it is generally safe to say that the pressure inside the cup will not be above atmospheric. Therefore the column of water above presses the cup against the bottom of the tub, without having to resort to any fancy dynamical phenomenon.

As for the fountain-like squirting into the cup, although something along the lines of Philip Gibbs’ theory would be really, really cool, I don’t think it can really happen: water flowing under the rim will have a lower pressure than still water at the same depth, and so would only further press the cup down, not keep it up. So the fountain must be a product of a poorly sealed rim. The dynamics of such an inward flowing disc shaped jet are probably fascinating…

The continuous outflow and inflow of water, as you mentioned, should be generated due to the pressure differences between 1) the atmospheric pressure of air inside plughole and the combined pressure of air and water in the cup, and 2) combined pressure inside the cup and outside water pressure. I guess the fountain is generated because of these dual processes.

I’m not sure this is so. Water is entering the cup from under the rim, and leaving it through the plughole. Air neither enters the cup nor leaves it, so there is no pressure drop in the air.

I think the solution is quite simple: the sides of the cup are (close to) vertical. Therefore, the water pressure on the sides of the cup acts only horizontally, and has no effect on raising the cup. If the cup were pressed down against an area of the bath floor with no plughole, and was a good enough fit that no water leaked in, it would stay on the floor of the bathtub, with the air inside at atmospheric pressure; the horizontally-acting pressure from the surrounding water would have no effect on the vertical force balance.

Now in practise, a cup pressed over a flat area of the bathtub will likely leak slightly, and the water which gets into the cup will raise the pressure of the air inside, eventually enough to raise the cup. When the cup is over the plughole however, the water that leaks around the edges is rapidly removed from the cup down the plughole, and the air pressure in the cup is therefore never raised sufficiently to overcome the weight of the cup.

My first guess as to what happens was this: the flame expands the air trapped beneath the glass, and then when it cools and shrinks it pulls the water in behind it. The problem with this is that there are no bubbles visible while the match burns, although the volume of the air at the end is quite a bit less than the volume when you first put the glass down.