[GRASS-user] Re: r.param.scale curvatures: are really OK
Marcello Gorini
gorini at gmail.com
Tue Nov 2 15:53:54 EDT 2010
Hello,
I cannot really answer your question, but I have been working with a
combination of your fuzzy modules for GRASS and r.param.scale in order to
accomplish a fuzzified version of Wood's morphometric feature extraction
method.
In doing so, I noticed that the results of "r.param.scale param=feature"
differ very much from the results of this same module in Wood's original
software Landserf, by using the same slope and curvature thresholds. Do you
think this can be an associated issue to what you mentioned?
All I can say is that the formula you described (the one used in GRASS) is
exactly the same that is in Wood's Ph.D. thesis, which in turn is based in
Evans (1979) work, as you probably already know.
By the way, great fuzzy modules!
Best regards,
Marcello.
Jarek Jasiewicz wrote:
>
> Hi
> while I modified r.param.scale for my purposes I noticed that some
> formulas for curvatures differs from that I know:
>
>
> for example:
>
> plan curvature: is
>
> (2.0 * (b * d * d + a * e * e - c * d * e) /
> pow(e * e + d * d, 1.5));
>
> probably shall be:
>
> ( (b * d * d + a * e * e - 2.0 *(c * d * e)) /
> pow(e * e + d * d, 1.5));
>
>
> profile curvature: is
>
> return (-2.0 * (a * d * d + b * e * e + c * e * d) /
> ((e * e + d * d) * pow(1.0 + d * d + e * e, 1.5)));
>
> probably shall be:
>
> return (- (a * d * d + b * e * e + 2.0* (c * e * d)) /
> ((e * e + d * d) * pow(1.0 + d * d + e * e, 1.5)));
>
> Of course threre could be reasons why little different formulas are used
> but I do not see any.
>
> This formulas is in grass 6.4, 6.5 and 7.0 and I assume in older version
> too.
>
> regards
> Jarek
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>
>
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