A little physics problem

A few months ago I accidentally spotted an amusing phenomenon during my son’s bathtime. He has some plastic cups that he likes playing with in the bath, and as the water was running out, I took one of them, turned it upside-down, and pushed it down so that it was full of air and just above the plughole, surrounding it. If you want the hard version of the problem it is to work out what happened next. The easier version of the problem is to read past the fold, where I will say what happened, and then to explain it. I am hopeless at this kind of problem, so I don’t know the answer myself, and I also don’t rule out that the answer is too easy to be interesting. If that is the case, then apologies in advance.

What happened is (i) that the cup stayed down rather than being pushed up as it would have been if it had been above some other part of the base of the bath, and (ii) that there wasn’t a perfect seal between the rim of the cup and the surround of the plughole, so some water got in, and (iii) that water was thrown up in a kind of fountain inside the inverted cup (which I could see because the cup was slightly transparent), that lasted until almost all the water had run out.

Part of a satisfactory explanation would include an answer to the following basic question: was what I saw (and have seen several times since) a fluke that depended on the precise shape and degree of airtightness of the cup and its join to the bath, or would pretty well any plastic cup have done the job? If what one observes depends on the cup, then what is the range of possible behaviours?

Just in case it helps, and so that I can continue my series of photographs of drinking vessels, here are pictures of the plughole with and without the cup over it, but without the bathwater.

32 Responses to “A little physics problem”

  1. Neil Budko Says:

    That’s an interesting phenomenon, indeed. OK, we understand that the air is sucked out of the cup by the water flowing down the pipes, which works pretty much like a bicycle pump. The little fountain inside the cap is a hint of that – the water is trying to fill the whole available de-pressurized volume. If I am right, then you should start feeling the downwards force on the cup a certain distance away from the sink. Would be interesting to predict (and measure) that distance based on the diameter of the sink and the cup shape. Sounds like a problem for IgNobel πŸ™‚

  2. Sarang Says:

    Parts (i) and (ii) are intuitive — water is draining through the plughole so (assuming air can’t easily leave the inverted cup) the cup has to move downwards to keep the air in the cup from expanding. The fact that there wasn’t a perfect seal is probably generic, but I imagine you could make cups and tub geometries for which this is not true. Part (iii) is interesting and is presumably closely related to the fact that when you push an inverted cup into the bathtub the water gets in all at once. (Here the pushing is self-sustained.) I would speculate that this is a surface tension effect: air can only leak out in bubbles, and when an air bubble enters water it sort of disrupts the surface of the water where it enters and this leads to splashes / fountain effects. I’m not sure though, seems like a problem for Sid Nagel and the other James Franck people at Chicago.

  3. John Sidles Says:

    Experienced urinators will recall with pleasure Chapters 3 and 6 of Patrick O’Brian’s novel Treason’s Harbor, which recount (among numerous other topics) the adventures of the Astronomer Royal, Edmond Halley, and his Diving Bell, as described in Dr. Halley’s celebrated article of 1716, The Art of Living Under Water; non-urinators can look forward to increasing their store of knowledge relating to the now-sadly neglected technology of Diving Bells! πŸ™‚

  4. Laurent Jacques Says:

    Interesting … Other people seem to like similar experiments in order to prove the existence of … “white holes”. See here:

    “White Holes And Kitchen Sinks”


    Funny to see the kind of high level physics we can do with just an opened tap in your kitchen sink πŸ˜‰

  5. Philip Gibbs Says:

    This is just my guess…

    Although the air is at first sucked out of the cup so that it is held down, the plumbing is designed so that the water drains out and air at atmospheric pressure can then come back in. The cup then rises slightly above the metal rim of the plug hole allowing water to flow at a rate that keeps it in equilibrium at that position. The water is flowing past the rim of the cup in all directions so it jets towards the middle of the cup where it collides and is goes up to form the fountain.

    I think the details to show that such a stable position is possible could be complicated and unintuitive, but the description suggests that it happens.

    According to this theory you would get this effect even with a perfectly smooth cup rim.

    • Theo Says:

      the plumbing is designed so that the water drains out and air at atmospheric pressure can then come back in.

      Plumbing is designed so that _nothing_, and in particular _not air_, can come up the pipe. If air can go up the pipe, then with it can come the awful smells that are in the waste-water pipes below (or, worse, in the septic tank if you are off the grid). This is accomplished by putting a bend (the “P-trap”) in the pipe below the tub which holds a little bit of water in it.

      If memory serves, there’s a little bit of extra downward pressure caused by a sort of Bernouli effect from the flowing water in the lower-down septic pipes. It’s not strong enough to suck out the water in the P-trap unless there is also some water weight above it. That’s why drains don’t feel like their sucking the air into them, but do suck water in fairly quickly.

    • Philip Gibbs Says:

      Theo, you are right that air does not come back through the u-bend, but what I mean is that there is a vent beyond the u bend that allows bad air out and which keeps the presure in the plus at around atmospheric. Otherwise there would be a stronger suction from the plug. Here is a picture of one system http://www.inspectapedia.com/plumbing/1620ss.jpg

      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?

    • Juan Says:

      Hello Mr.Gibbs,Im doing research on virtual particles for my 12th grade advanced physics class and was really hoping if you can answer some questions i have?

  6. obryant Says:

    I learned a possibly-related bar trick while I was living in Hungary that I never became comfortable with. After some pleasant banter about the difficulty of putting a shallow plate of water into a beer glass without spilling any, you wow the gathered crowd as follows. You float a beer-bottle cap in the water (in the shape of a boat), and then rest a lit match on the bottle cap. Then, put the glass upside down over the match/bottle cap. Of course, the match burns for a couple of seconds and then goes out. After it goes out, the water is quickly sucked into the inverted glass, where it stays.

    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.

    • svat Says:

      Isn’t the simpler explanation (which I’m not sure is correct) that the burning match consumes oxygen in the air (combines with it to turn into ash) and thus decreases the volume of air in the glass?

    • gowers Says:

      I remember being wowed by an uncle of mine when I was a child, who did something similar with a banana. If my memory serves me correctly, he peeled just the top, and placed the peeled end in the top of a bottle that had a match burning inside. Suddenly and very dramatically, the banana was in the bottle and the peel remained outside. If anyone manages to repeat this I’d be delighted to hear of it.

    • Harshwardhan Says:

      @ svat
      While burning of a matchstick, the oxygen get’s consumed, but burning also releases other gases such as CO2, CO, etc. Therefore, I doubt whether the consumption of oxygen will have major effect. (this was an experiment taught at school level to show nearly 20% proportion of oxygen in atmosphere, but now i think it showed that fact incorrectly.)
      obyrant’s analogy of this experiment seems much more accurate.

    • obryant Says:

      I don’t have a beer handy to re-test, but the mystery is deepened by the fact (if I recall correctly) that the water doesn’t flow into the cup until the match goes out.

    • Tim Silverman Says:

      I think it’s relevant that carbon dioxide is much more soluble in water than oxygen is.

  7. Chris Johnson Says:

    -[Parts (i) and (ii) are intuitive β€” water is draining through the plughole so (assuming air can’t easily leave the inverted cup) the cup has to move downwards to keep the air in the cup from expanding.]-

    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.

    • Harshwardhan Says:

      Nice.. Also to note that even if a cup pressed over a flat area of the bathtub will leak slightly, it won’t let the water come in continuously. Because while we move a cup downwards, the water comes inside the cup in certain amount, so as to balance the pressures of compressed air inside the cup, water in the cup and water level outside the cup. So after a limit is reached for pressure of compressed air inside the glass, the water inflow will stop.

      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.

  8. Jaime Says:

    When you push the cup underwater and start pushing it to the bottom, the pressure of the air within starts rising, to meet that of the water near the bottom of the cup. Air is compressible, and therefore the level of the water will go slightly over the rim of the cup. If you press such a cup against the bottom of the tub, even with a perfect seal, there will be enough water inside the cup to keep the pressure of the compressed air equal to atmospheric plus the column of water above. Basically, there is no difference here from there not being a wall below the cup, and so according to Archimedes the cup floats up.

    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…

  9. Dylan Wilson Says:

    My favorite thing about this question is that everyone has an intuitive reason why something should be happening, and everyone has an intuitive reason why something someone else said couldn’t be happening. πŸ™‚

  10. gowers Says:

    After giving my son another bath this evening, I have obtained some more data. It seems that the best fountains are obtained if the seal between cup and plughole is very good without being perfect. In that respect it is perhaps a bit like putting your finger over the end of a garden hose: the water goes furthest if you cover the end of the hose almost completely but not completely.

    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.

  11. Gc Says:

    “It seems that the best fountains are obtained if the seal between cup and plughole is very good without being perfect. In that respect it is perhaps a bit like putting your finger over the end of a garden hose: the water goes furthest if you cover the end of the hose almost completely but not completely. ”

    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.

  12. Erik Says:


  13. Harshwardhan Says:

    Here’s a proposed mechanism for generation of the fountain, i hope it matches with your observations:

    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.

  14. LifeIsChill Says:

    Is there a polynomial time algorithm known for NP-complete problems where the word sizes one has to operate are exponentially large ? Say you do quadratically many multiplications to solve an NP-complete problem but on words of size n^n bits. If such an algorithm is found, would that be p=np? or at least would that be of any interest to the community?

  15. gowers Says:

    Yet more data. I looked more carefully, and I now think that the size of the fountain decreased as the bathwater ran out. I also noticed that the diameter of the cup was very slightly larger than the diameter of the metal rim round the plughole, which is set into the enamel of the bath but sticking slightly out of it, so the cup fits very neatly round the rim and it may be that the water is moving in a slight upward direction when it comes in at the bottom. This would explain why it worked less well with the narrower cup.

    • Philip Gibbs Says:

      That would give a simple explanation. The water leaks through just because the seal is not perfect. Does the bath have an overflow hole above the normal water level? These are usually plumbed to the outlet which means the pressure in the cup over the plughole will always be at atmospheric. As the water level goes down the pressure difference decreases and the flow slows down.

  16. ddd Says:

    why don’t you give your son a quick series of baths and try out all the variations people suggest?

    kids get really dirty these days.

  17. Rea Says:

    Is the blog dead or it was taken off online somewhere else?…

  18. Olof Sisask Says:

    I can confirm the effect, with slight variations in setup. First, I used a rectangular-based container instead of a glass; this was about 7cm by 10cm, and about 6cm high. Second, this was after my daughter’s bath, so it appears that the effect is insensitive to the sex of the child.

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

  19. observer Says:

    Are you using a ramp fluid?

  20. science and math Says:

    Interesting experiment.
    I will do this at my home.
    Thanks for sharing.

  21. Delilah Deitsch Says:

    You may store for poker cards from the local store. It is no question why so many players turn to poker tournaments to progress their seat and their successful purse. Gambling addicts make poor choices and endure the implications.

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