bottom area of a lake calculation
Sophie, I think a reasonable approximation might be obtained: (1) if you were to calculate the slope at each grid point using r.slope.aspect (2) use r.mapcalc to calculate A/cos(slope), where A is the pixel projected area (e.g., 100mx100m = 10000 m^2) -- lat-long coordinates will not work. Consequently, if the slope were approximately 0, cos(slope) approaches 1 and A/cos(slope) = A; as slope approaches 90 deg, cos(slope) becomes very small, so A becomes very large. This works where the slope direction is orthogonal to a pixel edge, so a correction to account for other slope directions may be needed. But, with smaller grids, the error may be negligible for you... Tom
On Wed, Aug 8, 2018 at 4:55 AM, <sophie.leblanc at neuf.fr> wrote:
<span style="font-family:arial,helvetica,sans-serif; font-size:12px">?</span>Dear all,<br> <br> I would like to calculate the bottom area of a lake. I have already used marmap to calculate projected area, i.e. surface area.<br> But i have not found anything to calculate a non projected area. I have the coordinates of points (longitude, latitude and depth) I can convert them into xyz. But then I don't know how to calculate the surface I have found polygonal area calculation but I have a 3D object, a polyheron and I think I can not use polygonal area.<br> Do you have any idea? do you know an R package that could help?<br> <br> thanks,<br> Sophie Leblanc<br> <br> <!-- +Signature --><font size="2"><em><a href="http://sophieleblanc6. wix.com/photographie">http://sophieleblanc6.wix.com/photographie</a></em></font> <!-- -Signature --> [[alternative HTML version deleted]]
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