Suppose a hole is drilled perpendicularly into the side of the beaker which is full to the brim with a fluid (say water). This will result in water spurting out, travelling in a parabolic trajectory and hitting the solid, flat surface (say, a table) the beaker is resting upon.

For a cylindrical beaker, the distance the water lands from the base of the beaker depends on the height at which the hole is drilled. It turns out that drilling the hole at half the total height of the water column maximises the distance the water spouts.

Suppose now that, instead of a beaker with vertical sides; its vertical cross section is curved (like a wine glass). The distance from the base that the water lands is now harder to calculate, because it leaves the beaker travelling at an angle.

Does a beaker exist which the property that, regardless of where the hole is drilled, the water spout will hit the table at the same distance away from the base of the beaker? If so, what algebraic equation (if any) defines the vertical cross section of the beaker?

Note: for the purposes of this problem, ignore fluid dynamics of the water and air resistance. I'm envisaging a beaker with a circular horizontal cross section. The problem pertains only to the point the water spout initially hits.

  • $\begingroup$ The question is mathematically incomplete without giving the expression for the outgoing flow velocity at the hole. According to Torricelli's law, it is $\sqrt{2gh}$, where $h$ is the depth of the hole from the top of the water. I assume the direction is perpendicular to the beaker surface. $\endgroup$ – Rahul Jun 4 '16 at 6:42

Given your assumptions (no air resistance, etc) and that flow velocity follows Torricelli's law and is perpendicular to the beaker surface (thanks Rahul), such beakers can exist. Below is an example, where the side of the beaker is shown in red and examples of water spouts are shown in dashed blue.

Beaker with water spouts

Strictly speaking, the beaker only fulfills the requirements in 95% of its height as the top 5% were just kept vertical. The reason for this is that at the very top the flow velocity becomes 0 and this would result in a much less interesting problem.

The shape of the beaker's side was arrived at by setting up the equations for Torricelli's law and projectile motion and then manually "massaging" angles until the correct results occured. I don't know if a function for this curve can be derived analytically, but Excel provides the following "best fit" polynomial (note that it has the height $Y$as input): $$X=-0.0000270647*Y^6 + 0.0006428505*Y^5 - 0.0057240747*Y^4 + 0.0232746382*Y^3 + 0.0122432012*Y^2 - 0.1534098297*Y - 2.8075911701$$

One last comment: Probably the most important basis for determining the shape of the beaker (aside from the equations mentioned above) is deciding at what distance from the beaker you want all the spouts to hit. The limiting factor here is clearly the weak (low velocity) spouts at the top of the beaker. I decided to make the spot to hit equal to the distance a horizontal spout at $5$% below the top could muster (see top most spout in diagram). The range of the spouts at the top can be extended by angling them above horizontal, but not by very much.


You mentioned in your comment that you would have liked to see the beaker's shape when even the very top fulfills the requirements, so here it is. Best fit polynomial equation is available upon request.

Entire beaker fulfills

  • $\begingroup$ Thanks Jens. Yes, I mean flow velocity following TorricilIe's law. I had envisaged a beaker which curved around at the top to be directly overhead the point it hits, such that even though the water drops at an infinitesimally small velocity, it can just drop straight. I was hoping for an analytical derivation, but I realise that would be very tricky. Thanks for the equation. Cheers $\endgroup$ – Thomas Delaney Jun 5 '16 at 2:14

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