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, <

[hidden email]> 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>

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