AREA = SUM((A/COS(SLOPE)) * A) ----> IT REPORTS NEGATIVE VALUES
IF I DO A/(Cos(Grid2)*A) the sum of values is negative!!
grid2 contains slope not %
----- Original Message -----
From: "Jaromir Svasta" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; "'Lafleche, Marc'" <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Wednesday, January 21, 2004 11:03 AM
Subject: Re: MI-L Surface Area
> On Wed, 21 Jan 2004 09:47:47 -0000, Warren Vick, Europa Technologies Ltd.
> <[EMAIL PROTECTED]> wrote:
>
> > 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.
>
> Of course, I made a mistake. Using Cosine is right, but the equation
> should look like this:
>
> AREA = SUM((A/COS(SLOPE)) * A)
>
> Accept my apology, please.
>
> Jaromir
>
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