## September 14, 2018

### Elevation IX: Rivers IV: Rolling Down the River

Someone recently pointed out that my river code will probably not be satisfactory until I add deposition. I tend to agree. My code so far is merely erosion, and the models I'm modifying don't account for deposition either.

All in all, deposition is pretty important for smaller scale modelling, but it's worth considering if it could have an effect. But it turns out it's pretty complicated, and you need to know some things like the type and size of sediment each river is carrying. So I need a way to simplify the model.

However, the first thing I need to know is how fast the river is flowing. If it slows down too much, it may deposit sediment, which could end up correcting its own course.

Part of the problem I'm dealing with is again one of scale. To affect an average of 1 ft of change in a 20 mile hex, you'd need to remove (or deposit) about half a billion tons of soil. So in the real world, these things are studied on much smaller scales. Throughout this process, there are a couple of places I have to disengage from reality, if only because I'm limited in the amount of information I can provide and my sampling density.

To determine velocity, I generally need to know the discharge amount of the river, its depth and width, and the slope and roughness of the channel. None of these are easy.

There are many models relating drainage area (the only variable I really have besides slope) to the discharge amount, but they all yield numbers that are much higher than I want to use. Remember that a single hex covers 346.6 sq.mi. That's a lot of water to collect. The following equation is often offered for width in terms of drainage area (the coefficients vary across studies but are usually in this range):
$w\textrm{ [ft]} = 13.23 \left(A\textrm{ [mi^2]}\right)^{0.446}$
But this yields a channel width of 180ft for a single hex. This is probably real world realistic, but it's not what I need to use. I just don't have enough detail at this scale to say for sure if all 400 hexes are draining out at a single point on the edge of a hex. But if I just use the number of hexes (1 in this case) instead of the area, I get $w = 13$ ft. This is reasonable, even if it's not strictly accurate.

The depth of the river is another matter. In the real world, this is easy to measure, and so it's usually offered as an input rather than a result. However, I can use the same process to arrive at the following:
$d\textrm{ [ft]} = 0.9951 \left(A\right)^{0.3012}$
This yields a depth of a little less than a foot. That's still a fairly substantial stream - it would present a significant obstacle to trade, to say nothing of travel on foot. My small hiking experience has crossed many streams of just a few inches depth, and anything more than a foot is a noticeable, but not insurmountable, obstruction. The width to depth ratio is about 14, which is reasonable, but it has a large variation depending on very local conditions.

I considered using the Rosgen Classification, but I don't think I have enough data to make it worth it.

I need a few more geometrical parameters, the cross-sectional area, and the wetted perimeter. For now, I'll just use an ellipse to model it, although this isn't really ideal. It's not ideal because I have to use the following equations to find the perimeter:
$\varepsilon = {\sqrt{w^2 - (2 d)^2} \over w}$
$P_W = w \pi \left(1 - \sum_{i=1}^\infty {(2 i)!^2\over(2^i i!)^4} {\varepsilon^{2 i}\over 2 i - 1}\right)$
It was at this point I wondered if it's all worth it. The area is simply(er) given by:
$A_R = {\pi w d \over 2}$
These can be used to determine the hydraulic radius:
$R_H = {A_R \over P_W}$
And then the Manning Equation is used to finally determine the flow velocity:
$v = {1 \over n}R_H^{2 \over 3}S^{1\over2}$
Where $S = {\Delta y \over \Delta x}$, and $n$ is the Manning roughness factor, which varies depending on topography. I may use a random function to determine it in future, but for now I've set it at $n = 0.05$.

For testing, I picked a stream path at random. This once again demonstrates the scale problem. A drop of 400 ft over 20 miles is nothing. Generally speaking, walking along a slope is can be perceived at around 2.5%, which corresponds to a height change of 2640 ft. That's a lot, but of course the local terrain will be much different. All the slopes here are much less than 2%.

That yields the following profiles for slope, accumulated drainage area, and velocity in mph.

Sedimentation will be related to the river velocity; if it drops below a certain threshold, it may end changing the whole course of the river.