For the purpose of this question, let us define “flat” as meaning “having a surface which is a plane”.

Clearly, the Earth being round, water is not flat. If you take a sheet of water of length $2l$, the middle of it will bulge above a straight line joining the two ends, the height of the bulge being $$\frac12\frac{l^2}R$$ where $R$ is the radius of the Earth.

Here are a few specific examples:

  • The centre of an Olympic swimming pool ($l=25$m) is $\frac1{20}$mm above the line joining the two ends.
  • The centre of a flooded football pitch ($l=50$m) is $\frac1{5}$mm above the straight line from one goal line to the other.
  • In a $1$km long conduit half full of water, the water at the centre will be $2$cm above the path of a laser beam from the water surface at one end to the water surface at the other.
  • In real money, this translates to $2”$ in a $1$-mile conduit.

The question is how and to what extent these bulges are observable in real life - and, indeed, to what extent they need to be taken account of in civil engineering. If you built a $10$-mile laser-straight tunnel, it could have a puddle sixteen feet deep in the middle and still be dry at either end…

The most interesting challenge is probably the swimming pool: while $\frac1{20}$mm is clearly measurable, confounding factors such as draughts and differential thermal expansion might make accurate determination impossible.

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    $\begingroup$ You defined what you mean by flat in the question, so you already know how flat these things are compared to that definition. Your other questions relate to engineering. $\endgroup$ Jan 29 at 10:59
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    $\begingroup$ See the Bedford Level experiment. $\endgroup$ Jan 29 at 12:18
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    $\begingroup$ I remember hearing that the Stanford Linear Accelerator (3.2 km long) had to account for the curvature of the Earth in its construction. $\endgroup$ Jan 29 at 14:19
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    $\begingroup$ Float glass might be a better example of a liquid with a flat surface. Water is very fluid and easily gets ripples. Molten glass is more viscous. $\endgroup$
    – mmesser314
    Jan 29 at 18:21
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    $\begingroup$ The "eight inches per mile squared" (which is just a less formal version of your $\frac{\ell^2}{2R}$) is of course only an approximation valid up to a certain point, after which "sphere" becomes a much better description of the Earth surface than "parabola" $\endgroup$ Jan 29 at 18:56

2 Answers 2


One answer to the question "to what extent are these bulges observable in real life" is the well-known observation that ships at sea, when seen from an observer far enough away (on a clear day), seem to be half-submerged in the water. Also if you are approaching land then (again on a clear day) you will see the mountain-tops first before you can see the shore.

Typically engineers and architects have made much use of pendulum-bobs, from ancient times to modern. So they defined the horizontal by means of local gravity, not by looking along a notionally horizontal line. Consequently large structures such as long walls and canals follow gravity rather than laser beams. That is to say, whatever slight hump there may be in the water surface, relative to a laser beam, there will be that same hump in the bottom and sides of a canal, or in any given layer of stones in a wall.

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    $\begingroup$ I read somewhere that the surface of the Earth is smoother than a billiard ball, if they were scaled to the same size. Thus, all water is flat wrt. the center of gravity of Earth, within a margin of error ! $\endgroup$
    – Criggie
    Jan 29 at 18:45
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    $\begingroup$ Be careful with interpreting the Earth's curvature in terms of ground-level optical observations, laser beams, etc. -- this is notoriously difficult, because atmospheric gradients of temperature (hence density, hence refractive index) produce bending of light rays that is typically of the same order of magnitude as the Earth's curvature. $\endgroup$
    – nanoman
    Jan 30 at 5:49
  • $\begingroup$ @Criggie that's level, not flat. $\endgroup$
    – Rodney
    Jan 30 at 5:51
  • $\begingroup$ @Criggie the comparison with billiard ball is more about height of mountains and depth of ocean trenches. If the billiard ball were somehow enlarged to Earth size, keeping all its scratches to scale, then they would appear as mighty canyons, and any slight bumps would be huge mountains. $\endgroup$ Jan 30 at 10:51

Since this is quite an engineering question, I'll try to give a civil engineering answer.

The answer of how important is Earth curvature in engineering is that it is not very important. For works up to a few kilometres it is negligible because in civil engineering a fraction of a millimetre (or even a few millimetres) in a few kilometres doesn't matter much. Earth moving equipment and forms for concrete structures have very lower precision. That may be different for very special facilities like the linear accelerator mentioned in comments.

For larger structures, like long canals, Earth curvature is already included in the way geodesic measurements are made. For practical purposes, an horizontal is defined as the shape of the surface of water. If distances are large, that is not an straight line as defined by optical means as an optical level or a theodolite, but the curvature of Earth is dealt with as an error of the measure. That means just that we see our theodolites as having a systematic error (which we can correct for) to get a horizontal (e.g. spherical) surface. Furthermore, atmospheric refraction causes a similar and opposite error that is dealt with in the same way.

Nowadays large scale geodesy is not done with optical means but GPS, thus making it even easier to deal with Earth's curvature. You just need to read three UTM or geographical coordinates for any given point and the shape of Earth has already been included in the computation. From here you can act as if those coordinates where orthogonal coordinates on a flat Earth and the difference won't matter - assuming that your works are confined in a few kilometres height and don't cover a whole continent.


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