### Water to the Edge

Not a long while ago, I was in a situation in which I had to sit in a chair in one place for eight hours a day, fourteen days straight. Obviously, such extended periods of stationary inactivity can make a man bored down to his core, so at times I played with my water bottle, tossing it back and forth. It was probably due to this that I started asking myself – if I open the bottle and tilt it enough, the water will spill. How does the angle at which the liquid starts pouring out change as a function of its volume, or height?
Let us deal with a relatively simple model: we look at a rectangular cuboid container from the side, so that it has height H and width d. We do not care at all about the depth of the cuboid – the calculations can be done entirely in two dimensions. When it is standing, it contains an amount of water that reaches a height of h.

The container is tilted until the water just reaches its upper edge; the angle θ relative to the floor is what we are looking for. We ask: what is θ(h)? It soon becomes clear that it is necessary to distinguish between two different cases: (I) in which there isn’t a large amount of water, and (II) in which there is. The following picture shows what happens when you tilt both cases by some angle θ: in (I), a triangle is formed by the water, in (II), a trapezoid.

In the former case, when the water starts pouring out, the angle θ is very small. Thus, the water forms a right triangle within the container, with one leg being the container’s left edge, the second leg being a section of the container’s bottom edge, and the hypotenuse making up the line separating the water and the air. It is not difficult to see, that the largest triangle possible within the rectangular container has volume equal to half of the container itself (when the hypotenuse is the diagonal of the container); thus, this situation will take place for all volumes smaller than half the container, or in other words, h < H / 2.
In the latter case, the water forms a right trapezoid when tilted, as can be seen in the picture. This happens for all h > H / 2.

We shall now figure out what the general angle is for each case.

I) Triangle:

The total surface of water cross section is v. Since it makes up a triangle, it can be written as:

$v = \frac{kH}{2}$

From basic trigonometry, we know that

$k = H \tan{\theta}$

Combining the two, we get:

$v = \frac{H^2 \tan{\theta}}{2}$

Remembering that v = d∙h :

$\tan{\theta} = \frac{2d\cdot h}{H^2}$

$\theta = \arctan{\left ( \frac{2d\cdot h}{H^2} \right )}$

Let us stop for a second, and verify that we don’t have it extremely off. At very low h, we get very small angles, which is good. At h = H / 2, the upper limit for the triangle case, we get tanθ = d / H, which is indeed what we supposed – the water makes up the largest triangle possible, with the hypotenuse as the diagonal.

II) Trapezoid

While we could start doing the same mathematical analysis as we did in the triangle, there is a neat trick here. Looking at the picture, we can imagine that the container is in fact smaller, by ignoring the whole bottom part which is filled with water. The remaining water forms the largest triangle possible within the remaining upper part of the container. Thus, our problem becomes as follows: we need to solve the triangle case, only with a container that has a smaller height. Let us look at the container as it is standing: in order to form a triangle, the right edge of the water level needs to go up by H-h. Consequently, the left edge goes down by H-h. This means that total water level in this “new” container is h’ = H-h. Remembering that the water needs to form the largest possible triangle, we know from the previous calculations that h’ = H’ / 2.

We are therefore dealing with a new container, with the following dimensions:

d’ = d; h’ = H-h; H’ = 2(H-h)

Of course, this is valid only for h > H / 2. All we need to do is replace these new figures in the formula we got from the triangle case, and we get:

$\tan{\theta} = \frac{2d(H-h)}{4(H-h)^2}$

$\theta = \arctan{\left (\frac{d}{2(H-h)} \right )}$

Unlike last time, the considerations we used in order to get to the results are not trivial. Let us again check that we are not entirely mistaken by checking the extreme cases. At h = H / 2, we expect exactly the same result as the triangle case, and this is indeed what we get. For h nearly equal to H, we get $\frac{d}{2(H-h)}\to\infty$, which means that arctan approaches 900, which is just what we would expect – when the container is nearly filled to the top, any small tilt will cause the water to spill.

We now have a theoretical result which looks rather ok. However, in order to really see that we have a good model, we need to verify it against empirical data.

Experiment – first try

A round glass cup of dimensions d = 7.1 [cm], H = 8.5 [cm] was chosen as the container. During the experiment I filled it with a certain amount of water and measured its height with a ruler. I tipped it with my hand, until the water came to the rim of the cup. Then, holding it with one hand, I used a ruler to measure the height of its rim, y, and the horizontal distance between the rim and the bottom of the cup, x. The quotient of these two values gives $\tan{\theta} = \frac{y}{x}$. This was repeated several times with different amounts of water, and was compared to the theoretical model, graphed in excel.

The experiment suffered from several technical shortcomings. The cup was made of thick glass, which distorted the ruler readings and did not affirm well to the model, which used only infinitely thin lines. I used my hand in order to balance the cup when tilted, which induced more measurement errors, and because of the cup’s thickness, I found it quite hard to precisely measure the horizontal distance between its base and its rim. Also, water tension meant that I was never completely sure when I should stop tilting the cup and start measuring. Finally, the cup upper cross section was circular, and not rectangular; nothing guarantees that it can be reduced to a two dimensional model the same way a rectangular cuboid can. All of these amounted to data which, while somewhat compliant with the model, left space for uncertainty:

[The green line is the predicted angle; the red dots are the empirical data. The other dots are the breaking up into triangle and trapezoid cases. It is not clear from this graph, that the red dots agree with the green data set, and not the blue, or something in between]
I therefore decided to perform a second try, which would correct the faults of the first one.

Experiment – second try

This time, I picked a milk carton as of dimensions d = 7.1 [cm], H = 17.3 [cm] as the container of choice. It was much thinner than the cup, which allowed me to take more precise measurements. In order to be certain as to the amount of water in the carton at each time, I put in an exact volume each time, and could thus acquire h with both a measurement with a ruler and by calculating from the volume of water I put in. Furthermore, in order to limit the uncertainty in the measurements, I no longer measured the horizontal distance x, but rather only the vertical distance y to the rim of the carton, relying on the fact that $\theta = \arcsin{\frac{y}{H}}$. The whole contraption was also tied to a string and a pulley, so I could fix the carton in place the moment the water reached its edge.

Although surface tension still occurred, this much more reliable experiment yielded excellent results:

[The experimental data is in much better agreement with the predicted angle. Notice that the graph looks different from the previous one, because H is different]

And there you have it – we now know the angle at which liquid reaches the edge of rectangular cuboids, and have also learned about the importance and significance of well thought out and accurate experiments. Isn’t that a good way to pass the time when you have nothing to do but sit around?