Csdrhrt

I have two elevation models that I subtract from each other, to find out how the topography changed over time. For the elevation models I have given the RMSE error of the horizontal and vertical uncertainty/error.

I want to calculate the error of the resulting elevation model, that represents the differences between the two models.

I know that I have to calculate a error propagation, but I am very unsure how to do it. The method does not need to be very exact or complex, I need a simple method to state that the resulting elevation model has an error of round about xy.

The first model has a resolution of 0.05 m with an error of 0.03 m horizontally and 0.05 m vertically. The second model has a resolution of 1.0 m and an error of 0.15 m horizontally and 0.10 m vertically.

So I would say, I have four variables:

1. Variable 1: 0.05 m +/- 0.03 m


2. Variable 2: 0.05 m +/- 0.05 m


3. Variable 3: 1.00 m +/- 0.15 m


4. Variable 4: 1.00 m +/- 0.10 m

Is this correct?

How can I further calculate the error?

EDIT:
I am not sure about the output variable but I think I would have two output variables.

1. Variable 1: output resulting from Variable 1 and Variable 3. Maybe something like XX m +/- XX m or a percentage value.




2. Variable 2: output resulting from Variable 2 and Variable 4

Here is what I would do.

First, convert everything into the vertical error. If your terrain has slope about $s$ near the point $(x,y)$, then the vertical RMS error corresponding to the horizontal error $h$ is about $sh/sqrt 2$. The slope factor should be clear: if you move one unit horizontally, then you move at most $s$ units vertically. The $sqrt 2$ comes from the fact that you, most likely, moved not straight in the direction of the steepest descent, but somewhat sideways as well. So, if your horizontal error is 0.03m and the slope around $(x,y)$ is, say 1:2 (which is already quite a lot unless you are in the mountains), then the corresponding vertical error is just about 0.01m. (I approximated $sqrt 2$ by $3/2$ to do the calculations in the head) Then, assuming that your original vertical and horizontal errors were independent, you can find the total vertical RMS error as the square root of the sum of squares: $sqrt{0.05^2+0.01^2}approx 0.051m$.

Similarly (for the same 1:2 slope estimate and the second mesh), you get approximately about $sqrt{0.10^2+0.05^2}approx 0.11m$.

As you can see, if your terrain is reasonably flat, the horizontal errors can essentially be neglected, but if the slopes are more like 2:1, they may provide the leading contributions.

The interpolation between the nodes (which you'll, probably, have to do since your meshes are of different sizes) doesn't make the errors any worse and the RMS error of the difference is just the square root of the sum of the squares of the individual RMS errors, as usual. So in the example case you get the final answer of about 0.125m.

This is done under the standard assumptions that your measurements are the exact values contaminated with mean zero noise and the noise measurements at different nodes are independent.

I hope it makes sense.