## AI-GEOSTATS: Ripley's K

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• I am using Ripley s K to analize distribution patterns of trees in a forest plot. Here some basic information. The plot is 1000 x 1000 m. There are 7765 trees
Message 1 of 2 , Nov 27, 2002
I am using Ripley's K to analize distribution patterns of trees in a forest
plot. Here some basic information. The plot is 1000 x 1000 m. There are
7765 trees that were measured throughout the plot. This is the first time I
am unsing Ripley's K and I run into several questions / problems.

I am using Crime Stat 2.0 and SPP (by Peter Haase) to calculate the
statistcs but get very different results from the two packages. Both
packages are supposed to output SQR(K(t) / Pi) - t for the data set and the
confidence intervalls. The attached image showes the output from both
packages. Both were run with 19 iterations and Rectangular Bordercorrection
was used in Crime Stat. The third graph showes the Coefficient of
Dispersion CD=SD^2/M for diffferent quadrant sizes.

Does anyone have an idea what the problem could be? Why do the confidence
intervals show a curved shape and are not around 0? Is there any other free
and easy to use software to calculate Ripley's K (I know that there is R
but I don't have the time right now to get familiar with it).

Mathias

[Non-text portions of this message have been removed]
• I am using Ripley s K to analize distribution patterns of trees in a forest plot. Here some basic information. The plot is 1000 x 1000 m. There are 7765 trees
Message 2 of 2 , Nov 27, 2002
I am using Ripley's K to analize distribution patterns of trees in a forest
plot. Here some basic information. The plot is 1000 x 1000 m. There are
7765 trees that were measured throughout the plot. This is the first time I
am unsing Ripley's K and I run into several questions / problems.

I am using Crime Stat 2.0 and SPP (by Peter Haase) to calculate the
statistcs but get very different results from the two packages. Both
packages are supposed to output SQR(K(t) / Pi) - t for the data set and the
confidence intervalls. The following link shows the output from both packages.

http://www.botanypages.org/Mathias/ripleysk.jpg

Both were run with 19 iterations and Rectangular Bordercorrection was used
in Crime Stat. The third graph shows the Coefficient of Dispersion

Does anyone have an idea what the problem could be? Why do the confidence
intervals show a curved shape and are not around 0? Is there any other free
and easy to use software to calculate Ripley's K (I know that there is R
but I don't have the time right now to get familiar with it).