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

[Tomislav Hengl]

Haa! Thanks for spotting this Edzer. But I am only talking about the 
predictions in this example. The predictions from any sp prediction 
method do not go through points (plus you do not have the coordinates of 
the points). So the question is - is it a derivative work or not?

On 27-11-2014 15:44, Edzer Pebesma wrote:
> 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 In case the nugget in m would have been replaced by a measurement error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,
>>
>
>
>
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