Non-aligned Data

This category can be reduced to one of the former two.

(a)

By defining approximately rectilinear pathways passing through the available data locations. Each of these approximate alignments is then treated separately with a possible grouping into distance classes, cf. Figure 4.13. The disadvantage of this method is that all the available data are not used, and it is difficult to program.

Figure 4.13: Computation of the Variogram: Possible Paths.

(b)

By grouping the data into angle classes followed by distance classes, cf. Figure 4.14. To construct the variogram in the direction ${/boma /alpha}$, each data value $z({/boma x}_0)$ is associated with every other value located within the arc defined by $[{/boma /alpha} /pm /delta({/boma /alpha})]$. Within this angle class, the data can be grouped into distance classes $[r /pm /varepsilon(r)]$. The effect of smoothing, of course, is again larger.

Figure 4.14: Computation of Variograms: Grouping into Angle Classes.

In the two-dimensional case, the first step will be: Compute the variogram in 4 directions, namely $/alpha /pm /pi /8.$
units <1.0cm,1.0cm> x from -1.0 to 1.0, y from 0.0 to 1.0

<0.15cm> [0.35,0.7] from 0.0 0.0 to 1.0 0.0 <0.15cm> [0.35,0.7] from 0.0 0.0 to 0.75 0.75 <0.15cm> [0.35,0.7] from 0.0 0.0 to 0.0 1.0 <0.15cm> [0.35,0.7] from 0.0 0.0 to -0.75 0.75

In case different variograms in different directions are indicated, one should decrease the number of classes. (Anisotropy: This can often be solved by some corresponding transformation.) Alternatively, one could try to obtain better variograms by grouping: Suppose we have computed $K$ variograms $/hat/gamma_k / (k=1,/ldots,K)$ with corresponding $n_k$ pairs of data values. Then a mean variogram with all pairs is simply found by