Re: MI-L Surface Area

[Jaromir Svasta]

Take care of parenthesis in the formula, it should read:

(A/Cos(Grid2))*A

(assuming Grid2 is the slope grid)
Then look at the sum of all values of the newly created surface area grid.
Jaromir

On Wed, 21 Jan 2004 19:23:07 +0100, Samuel E.R <[EMAIL PROTECTED]> 
wrote:

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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