Only an hydraulic engineer can tell us whether such a narrow stream of water follows mostly the Earth's shape or the shape of the canal's bed, since any place that is relatively high will make the water even more shallow and will make the flow of water even slower, creating a high point on the canal's surface.

This is really an interesting problem trig, and one that I doubt has been seriously considered in the discussion of this experiment. I did some work to try and reason out what might happen:

The first step, I think, is to consider the nature of a flow such as this. Assuming that we are looking at the canal during a period of steady-state operation, the principle of conservation of mass requires that the mass flow rate of water through any given cross-section of the canal will be identical with that at any other cross-section.

To help visualize this, consider drawing an invisible box around a section of the canal. Unless the volume of water in that box is changing (it is not, because we are working during steady-state), then the amount of time it takes for 1kg of water to flow into the box must be identical to the time it takes for 1kg of water to flow out.

To properly explain what comes next, I must introduce a few concepts related to fluid flow analysis.

The first is mass flow rate. In SI units, this is given in kg/s -- it is the quantity discussed above: the total mass of water that flows through a cross-section of the channel in a given time interval. Again, this quantity remains constant.

The next is the concept of an incompressible fluid. For 99% of all fluid flow applications involving liquids, fluids can be treated as incompressible. This means that density remains constant across all parts of the flow. There's a lot of unnecessary math that goes into the proof of this; the short version is, it takes an extreme amount of pressure to compress most liquids to a degree that will have any impact at all on measurements taken. For this application, we're talking a change in density that is many orders of magnitude below what relevant tools can measure (thus, negligible).

Finally, the concept of volumetric flow rate. This has several mathematical definitions, which will be discussed later. For now, it can be understood as the volume of water that flows through a given cross-section over a specified interval. This is separate from the mass flow rate, although similar and related. It is generally given in units of cubic meters per second, or m^3/s.

MATH TIME!

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It take a bit of math to relate these three concepts in a way that is useful to the present inquiry. First, some relevant equations:

MF = m/t (mass flow rate = mass / time)

d = m/vol (density = mass / volume)

VF = vol/t (volume flow rate = volume / time)

vol = A*L (volume = cross-sectional area * length)

v = L/t (flow velocity = length / time)

Solving the density equation for volume and substituting this result into the volumetric flow rate gives the following:

vol = m/d

VF = (m/d)/t = (m/t)/d

Since we know that MF = m/t, we can again substitute and demonstrate that:

VF = MF/d

This is a very important result. Because both mass flow rate and density are constant, we now know that volumetric flow rate is constant. Again, we can use this to get some more interesting results. Combining the equations for volume and volumetric flow rate, we can relate the volumetric flow rate to the channel's geometry:

vol = A*L

VF = vol/t = A*L/t

Further augmenting this, we can mix in the equation for flow velocity:

v = L/t --> L = v*t

VF = A*L/t = (A/t)*L = (A/t)*(v*t) = A*v

It is important to note here that the time term drops out. This makes sense mathematically, but it also should make sense intuitively; because we are working at a steady state, none of the properties of the flow are changing with time.

If we examine the equation we just derived (VF = A*v), there are a couple of other important conclusions that can be drawn. First, we can break down the 'A' term. For simplicity, I will treat the channel as rectangular. The precise equations derived might change a bit with different channel geometry (i.e. trapezoidal), but the result will be the same because we are dealing with a channel that is the same at one end as at the other.

A = w*h (area = width * height, where height is the height of the water from the bottom of the channel)

Going back to the previous equation to substitute, we now have our final result:

VF = w*h*v

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This equation is extremely important in applications of open-channel flow. Because we know that the shape of the channel is constant and the volumetric flow rate is constant, the equation demonstrates that there are two possible variables that can change and that these variables are inversely proportional. As flow velocity along the channel increases or decreases, the height of the flow must do the opposite. If the flow speeds up the height of the water will decrease, and if it slows the water will get deeper.

Now, things are finally starting to look like somewhat of a useful result when considering the question that was asked in the first place. All we need to do is decide whether the flow is faster in the middle or end than at the beginning, and we can make sense of what is happening.

To decide this, we can look at what forces are acting on the fluid. Most significant to the inquiry at hand is gravity. Because we are trying to discover the (potential) effect of a round Earth on this system, it is useful to draw a picture to visualize how gravity changes along a flat channel on a round Earth:

http://imgur.com/GBPwc (please excuse my poor MSPaint skills)

Now we must make a big assumption for this analysis. This *must* be checked before actually performing the experiment, or none of this is meaningful. We are assuming that the entire length of the channel (not the water) is flat. If you take a laser and shine it parallel to the channel at one point, it will remain parallel at every point.

we can see that along any distance the direction of gravity's pull will gradually change. Because most (or all) of the channel will not experience a gravitational pull that is exactly perpendicular to the channel floor, the gravitational acceleration of the water will have at least a small horizontal component.

Depending on what section you pick, the component will vary from with the flow to against it. It is easy to see, however, that as you move downstream in any case the amount of horizontal pull that is "helping" the flow will decrease. If it is initially positive it will become smaller or eventually become negative; if it is initially zero or negative, it will continue to become more negative.

The important part of this is that we have demonstrated that for any flat (as discussed above) channel, two possible cases exist. The flow will either have a positive acceleration at a downstream location, or it will have a negative acceleration. The relation here to the acceleration of the flow at the observer's location is unimportant, only whether the fluid downstream is accelerating or decelerating.

In the first case, because positive acceleration always means that the fluid is speeding up, the fluid flow downstream will have a greater velocity. As we discovered earlier, greater velocity corresponds to lower flow depth. Thus, in this case the flow downstream will actually be shallower than the flow at the observer's location. Because the fluid's acceleration is changing, its change in velocity will be nonlinear. Specifically, a graph of velocity vs. distance would have a positive slope, but a negative curvature. This means that the change in velocity (and depth) between two points near the observer will be greater than the change between two points downstream. Thus, although the fluid will be growing shallower it will be curving towards level with the channel (curving upwards, towards the observer but still getting shallower).

In the second case we see something similar happening. The flow's depth will be increasing because it is slowing down, but it will be getting deeper at an increasing rate because gravitational acceleration will have a growing horizontal component against the flow. Thus, we will again observe a flow that curves upwards toward the observer (and getting deeper).

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So here we have very interesting results. Although in neither case does the flow actually follow the Earth's curvature, observation of either one would demonstrate that such curvature exists. Without a variation in the gravitational pull (such as on a flat Earth) the forces acting on the flow would be uniform throughout, and the entire expanse would be perfectly even. Depth might change if the channel is slanted, but the rate of change in depth would be linear and no curvature would be observed.

It is interesting to note that the height of the water relative to the center of the Earth -- and correspondingly, the strength of gravity -- at any point along the canal doesn't really affect the results in this problem. In a stationary fluid situation this would be significant because the weight of the heavier fluid in the center of the canal would have to balance the combined pressure from the ends and we would likely observe a curvature that matches a round Earth (assuming the Earth is round). In this case, however, the more important factor is the direction of gravity's pull relative to the plane of the canal.

Finally, it should also be noted that I neglected friction throughout this analysis. Considering that three sides of the flow are in contact with a solid surface, friction will likely be a significant factor in the actual shape of the flow. Consider also, however, the sort of effect it would have. Friction will slow the flow and cause it to get deeper. In fluid applications, friction between the fluid and a containing surface is directly proportional to the velocity of the fluid. Thus, as the fluid slows the frictional losses will decrease and the rate at which the fluid slows will correspondingly decrease.

This will yield an observation that is quite opposite to our two previous results. The observed effect of this frictional loss will be a flow that is increasing in depth but at a decreasing rate, appearing to curve downward from the observer (though never below horizontal). If you wish to research this, a good place to start would be to google the Darcy-Weisbach equation.

This is a very complex problem, and the specific interaction between the frictional forces and the gravitational forces would likely have to be measured carefully to determine the net effect.

It is an unfortunate truth, but the nature of fluid dynamics is such that most interesting problems are too complex to solve analytically; instead they must be done experimentally, or computationally using a highly specialized simulation that would probably have to be specifically coded to work for your application. Oh, and you'd need one heck of a computer because to get the degree of accuracy needed here would require simultaneous computation of thousands, if not millions of tiny fluid elements.

Visualization:

http://imgur.com/OBaLinote: Greatly exaggerated. The actual effects would be extremely small over the distances discussed in this thread.