[R-sig-Geo] Question about derivative work - what is the license for map derived using e.g. spatial "predict" function?

[Edzer Pebesma]

Tom, in your example below, x contains the kriging variance; points with
zero kriging variance must be observation locations, with the predicted
value equal to the observation.

In case the nugget in m would have been replaced by a measurement error
component (Err in m and m1 below), you would not have this effect, and
also have no discontinuity in the interpolated surface at observation
locations:

> m = vgm(1, "Sph", 900, Err = 1)
> v = variogram(log(zinc)~1, meuse)
> m1 = fit.variogram(v, m)
> m1
  model      psill    range
1   Err 0.05066243   0.0000
2   Sph 0.59060780 897.0209
> m2 = fit.variogram(v, vgm(1, "Sph", 900, 1))
> m2
  model      psill    range
1   Nug 0.05066243   0.0000
2   Sph 0.59060780 897.0209
> as(krige(log(zinc)~1, meuse, meuse[1,], m2), "data.frame")
[using ordinary kriging]
       x      y var1.pred     var1.var
1 181072 333611  6.929517 1.110223e-16
> as(krige(log(zinc)~1, meuse, meuse[1,], m1), "data.frame")
[using ordinary kriging]
       x      y var1.pred   var1.var
1 181072 333611  6.884401 0.03648868
# note identical predictions but different variances
# for other locations:
> as(krige(log(zinc)~1, meuse, meuse.grid[1,], m2), "data.frame")
[using ordinary kriging]
       x      y var1.pred  var1.var
1 181180 333740  6.499624 0.3198084
> as(krige(log(zinc)~1, meuse, meuse.grid[1,], m1), "data.frame")
[using ordinary kriging]
       x      y var1.pred var1.var
1 181180 333740  6.499624 0.269146


On 11/27/2014 02:35 PM, Tomislav Hengl wrote:
> 
> Dear list,
> 
> I have a question about licensing the data that is produced by spatial
> prediction from point data. My ideas is that a map produced by using
> e.g. geostatistics from point data is a new data product and as such
> does not falls under the regulations of the original license used for
> the point data (so if the license for the point data is restrictive, the
> license for the output maps does not have to respect this). Consider for
> example:
> 
> R> library(gstat)
> R> library(sp)
> R> demo(meuse, echo=FALSE)
> R> m <- vgm(.59, "Sph", 874, .04)
> R> x <- krige(log(zinc)~1, meuse, meuse.grid, model = m)
> 
> The produced map "x" can be considered a new data product. There is
> absolutely no way that one could reproduce the original input points
> from this map, hence it should be considered "a non-derivative work".
> Only if we would derive a map using interpolation technique that allows
> re-constructions of points (e.g. Thiessen polygons) the license would
> need to be respected.
> 
> Or am I mistaken? (I know this is a type of a question for lawyers in
> fact, but any experience / opinion you have is welcome)
> 
> http://www.publicdomainsherpa.com/derivative-work.html
> 
> "To qualify as a derivative work, the derivative must use a substantial
> amount of the prior work’s expression. How much? Enough so that the
> average person would conclude that it had been based on or adapted from
> the prior work"
> 
> thank you,
> 

-- 
Edzer Pebesma,  Co-Editor-in-Chief Computers & Geosciences
Institute for Geoinformatics (ifgi), University of Münster
Heisenbergstraße 2, 48149 Münster, Germany. Phone: +49 251
83 33081 http://ifgi.uni-muenster.de GPG key ID 0xAC227795

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