Hello Jaromir,

I was intrigued by this algorithm as I'm not entirely convinced it will
work!

Certainly using the grid slope multiplied by the area of the grid is an
approximation unless the slope purely in the x or y axis. However, perhaps
the results are reasonable.

The thing I'm not sure about is using cos(). Cosine yields a value between
-1 and 1 which means the area of each cell is either the same, or reduced
(assuming the -ve is ignored for an area calculation). However, doesn't the
effective surface area of a grid cell get bigger with increasing slope?.
E.g. consider a cell with a slope of 45 degrees on the horizontal axis...
would that not yield an area of A*A*sqrt(2)? With sqrt(2) being about twice
cos(45), the calculation will be well out.

Let me know if I've missed something... it wouldn't be the first time.

Regards,
Warren Vick
Europa Technologies Ltd.
http://www.europa-tech.com

-----Original Message-----
From: Jaromir Svasta [mailto:[EMAIL PROTECTED] 
Sent: 21 January 2004 04:25
To: Lafleche, Marc; '[EMAIL PROTECTED]'
Subject: Re: MI-L Surface Area


An approximate value can be calculated with the following formula (my 
guess):

AREA = SUM(COS(SLOPE) * A * A),

where

A is the lenght of the side of a grid cell,
SLOPE is slope angle at each cell

You have to calculate it cell by cell. Can be easily done by, for 
instance, Grid Calculator module of Grid Analyser (www.terris.sk), which 
can calculate slopes as well. Calculate COS(SLOPE) * A * A first and than 
look at SUM in STATISTICS tab, which should show the value you are looking 
for.

Hope this helps,

Jaromir SVASTA



On Tue, 20 Jan 2004 13:27:50 -0700, Lafleche, Marc 
<[EMAIL PROTECTED]> wrote:

> Hello all!
>
> I want to find out how to calculate the surface area of a grid. I know 
> how to get the volume of a grid but I need the "3D" Surface Area.  If
> I double click on the polygon outline I get a value but it does not take 
> the
> elevation into account...  Is this possible at all?
>
> Thanks,
>
> Marc
>
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