CHAPTER
3: The Volume -- Variance Relationship
In the
previous chapters we have discussed semi-variograms, calculated experimental
semi-variograms and fitted models to these as if the samples had no characteristics
other than ‘position’. We have ignored the size and shape of the sample, the
way in which it may have been taken and/or measured, and so on. We have
effectively assumed that the sample values were located at ‘points’ within the
deposit. In this chapter we will see what effect those other characteristics --
collectively called ‘support’ -- have on the sample value itself, and hence on
the semi-variogram.
Let us consider the lead/zinc example which was discussed
in Chapter 2. Although the cores were actually 1.52m long, we ignored this fact
and calculated the experimental semi-variogram as before. Suppose, however,
that the cores had been sectioned into 3.04m lengths instead of 1.52m -- what
effect would this have on the sample values and on the semi-variogram? Table
3.1 shows the ‘borehole log’ for 1.52 and for 3.04m samples. The experimental
semi-variogram was also calculated for the 3.04m cores, with the results shown
in Table 3.2.
Fig. 3.1. Experimental
semi-variograms constructed on various lengths of core --- lead/zinc example.
Figure 3.1 shows the ‘new’ experimental
semi-variogram alongside the 1.52m one for comparison. For good measure, both
tables and the figure also show the resulting values for cores of 4.56m. It can
be seen immediately that the 3.04m semi-variogram is always lower than the
1.52m, and the 4.56m one is considerably lower than both. Let us return to the
basic assumptions of Geostatistics and try to explain this behaviour. We must
recall two facts from Chapter 1. The first is the basic definition of the
semi-variogram: it is the average square of the difference in grade between two
samples a given distance apart. If those samples were ‘points’ then the grade
is assumed to be measured ‘at a point’; if they are cores then the grade
measured is the average grade over the core length. Thus we are not comparing
two individual grades, e.g. g1 and g2, we are comparing two average grades 
1 and 2. We cannot reasonably expect the average
grade over 1.52m of core to have the same behaviour as the grade of a
‘teaspoonful’ of ore. Similarly, if we take the grade and average it over 3.04m
we would expect different behaviour again. The question is how to characterise
this difference in behaviour.
The second fact to recall from Chapter 1
is that the sill of the semi-variogram --- if one exists --- is equal to the
ordinary sample variance. If we are dealing with ‘point’ samples, then we can
estimate the sill of the semi-variogram, and compare this value with the sill.
That is, C=s² (ideally).
Now, if
the samples are cores of a certain length l (e.g. 1.52m) and we measure the
average grade over that length, then we have smoothed out some of the ‘point’
variation. We have replaced a large number of individual ‘points’ with one
average value. The variance of the averages will therefore be less than the
variance of the ‘points’, so that
In a similar way C3.04 will be less than C1.52 and so on. If we have a model for the
semi-variogram for the point samples we could produce the model for any other
size of sample, by employing the mathematical relationship between the point
model (g ) and the model for samples of length l (gl). Since we are only using a
limited number of simple models for the point semi-variogram, it is not too
difficult to state this relationship.
If we have a linear model for the point samples, g(h)=ph, where p is the slope of the semi-variogram
line, then the semi-variogram for samples of length l is given by:
This is illustrated in Fig. 3.2, with the point model for
comparison.
Fig. 3.2. Regularisation of a linear
semi-variogram by core lengths.
In practice, we generally have an experimental
semi-variogram for samples of length l, that is gl*, and we need to find the model for
the point samples (g ) for use in
the later chapters. Since the slope, p, of the core model is the same as
that of the point model, simply measuring the slope of our experimental gl* will give a value for p, and hence the point model, g. One complication arises if the point
model is actually a linear model plus a nugget effect. Taking core samples
lowers the line, but a nugget effect will raise it again. From the above
formula, if no nugget effect is present, extending the line of the core model back until it intersects the
semi-variogram axis should produce an intercept of -pl/3. Once an estimate of p has been made this can easily be checked, and if
necessary a nugget effect C0 added to the model.
Now suppose our deposit followed an exponential model, with
sill C for ‘point’ samples, i.e.
For cores of length l, the theoretical model becomes:
with a rather more complex form for distances less than the length of the core (h<l). Since we are unlikely to have
values of an experimental semi-variogram for distances less than the sample
length, the form of it seems rather academic. Figure 3.3 shows a point
exponential model and the corresponding ‘regularised’ curve for a sample of
length l. It can easily be seen that C l is lower than C. In fact:
so that a sample which was, say, one-fifth of the range of influence,
would produce a sill:
That is, the new sill will only be 94% as high as that of the point
model. It will also be noticed from Fig. 3.3 that extending the ‘linear’ part
of the core model (close to the origin) until it intersects the sill produces
an estimate of the range of influence for the cores a l, which is longer than that of the points, a.
Fig. 3.3. Regularisation of an exponential
semi-variogram by core lengths.
In fact, a l =a+l. This seems quite sensible if you remember that cores will
have to be just that bit further apart before they become independent.
The above arguments and formulae apply to the situation
where you know the ‘point’ model and you wish to find the ‘regularised’ model.
In practice the situation is generally reversed. We usually have an
experimental semi-variogram which has been calculated on cores of a given
length, and we need to find the point model for use in the estimation
techniques. Suppose, then, that we have a graph of the experimental
semi-variogram g l*, and we have decided that our
deposit follows an exponential model. The first step is to guess the two
parameters Cl
and al
. Since the model is exponential, the sill Cl
will be greater than most of the experimental points on the graph.
Having guessed Cl, produce a line up through the
first two or three points on the graph until it cuts the sill. This will give a
first guess at al. We know that a=al-l, so we have a first estimate of a. Using this in the above formula for Cl, we can reverse the equation and produce
a value for C, the point sill. We now have guesses at
the values of a and C which govern the point model. The
next question is whether these are ‘good’ guesses. We have already stated that
if we know the point model, we can produce the corresponding model for cores of
any given length, i.e. gl(h). If our guesses are good ones then this
theoretical model for gl(h) should match the experimental
semi-variogram, gl*(h). Substituting values for h, l, a and C produces a smooth curve like the lower one in Fig.
3.3, and this can be compared to the data. If necessary, a and C can be altered until the ‘model’
values become a good fit to the ‘data’ values. In effect, this is the same
procedure as was used in Chapter 2, but with an additional consideration of the
sample length.
Let us now turn to the most common model --- the spherical
model. This will be influenced in the same sort of way as the exponential. The
sill for the cores will be lower than that for the ‘points’, and:
The formula for the semi-variogram of the cores is
extremely complex because of the ‘discontinuity’ in the model but an example is
shown in Fig. 3.4. A subroutine to evaluate the formula has been published. If
the calculations are to be done by hand (or hand calculator) then it is easier
to use tables such as Table 3.3.
Fig. 3.4. Regularisation of a
spherical semi-variogram by core lengths.
This
table shows the form of the ‘regularised’ semi-variogram for a core of length l if the original point semi-variogram had a range of
influence a, and a sill of 1. The use of this table
is best illustrated by an example. We can now return to the example shown in
Fig. 3.1 of the zinc values measured over core lengths of 1.52m. In Chapter 2
we guessed that the sill lay at about 10.5(%)². This is our first approximation of Cl. Producing the line through the first two
points on the experimental semi-variogram gives 2al ¸3=9.6m (approximately). That is, al =14.4m, and hence a=12.9m. Using the formula:
The first estimates, then, for the parameters of the point
model are a=12.9m and C=11.2(%)². We must find the row in the Table 3.3
which corresponds to our value of a/l, i.e. 8.5. The entries along this
line correspond to multiples of the sample length l. That is, h/l=1 means h=1.52m, h/l=2 means h=3.04m and so on. We see that at h/l=1 the table gives a value of 0.116. This would
be for a semi-variogram with a sill of 1. Since we have a sill of 11.2(%)², the value we require is 0.116´11.2=1.30(%)². This is now a ‘model’ value for the
semi-variogram of cores of length 1.52m and can be plotted on the graph next to
the ‘observed’ value of 1.33(%)².
A second point on the model would be at h=3.04m, i.e. h/l=2. The table gives a value of
0.288 for C=1, so that our model value is 0.288´11.2=3.23(%)². This can be compared with the
experimental value of 3.09(%)². This process is repeated until we have a model value
to compare with each observed value. The resulting model curve has been plotted
in Fig. 3.5. This seems to be rather a good fit to the experimental
semi-variogram, if we accept the sill at 10.5(%)².
Fig. 3.5. Fitted regularised model
to the lead/zinc example --- 1.52m cores.
Adjustments could be made if the sill was
thought to be too low, by raising C and a. Suppose we accept this ‘point’ model with a=12.9m and C=11.2(%)². We can run a secondary check by
comparing the models for core lengths 3.04m and 4.56m. For the former, a/l=4.25 so that we must interpolate in the
table between a/l=4.00 and a/l=4.50. Linear interpolation is generally
sufficient for this sort of exercise. Figure 3.6 shows the experimental and
model curves for each sample length, and the point model for comparison.
Fig. 3.6. Fitted models to the
lead/zinc example --- 3.04m and 4.56m cores and the ‘point’ model.
The model
seems to be a good fit to the 3.04m semi-variogram, especially to the first
four points. However, after the first point on the 4.56m semi-variogram the
model here is consistently considerably higher than the experimental
semi-variogram until h is about 41m. This could perhaps be
neglected in view of the fact that each of these experimental values is
calculated on 15 or fewer pairs. All in all, the spherical model as estimated
seems to be a pretty good fit.