CHAPTER 3: The Volume -- Variance Relationship
VOLUME--VARIANCE
CALCULATIONS
This process of the semi-variogram changing with different ‘support’ is usually
known in the literature as ‘regularisation’ --- on the basis that the samples
get more regular as the sample size increases. We have seen that we can handle
experimental semi-variograms for core samples, and still derive the supposed
point model. However, this leads us on to another problem of the
‘volume--variance’ relationship and the influence of sample size on the sort of
distribution encountered. Suppose at the pre-feasibility stage of investigating
a deposit the management requests a grade/tonnage calculation.
That is, given an economic cutoff grade
(or list thereof) can we evaluate (i) the tonnage of ore in the deposit which
is above cutoff and (ii) the average grade of that ore. Suppose we take an
example to illustrate the problem which arises. A hydrothermal tin vein has
been sampled by means of nine development drives approximately 100ft apart in
the plane of the lode. Chip samples are taken every 10ft along these drives.
The sampling setup is shown in Fig. 3.7.
Fig. 3.7. Typical sampling situation
in Cornish tin example.
These chip samples may be considered as ‘points’
since they have a very small volume. Figure 3.8 shows a histogram of the 2730
chip samples taken from the development drives in this lode. Suppose we now
specify a ‘cutoff grade’ of 25lb/ton for this lode. The histogram shows that
about 44% of the chip samples lie below 25lb/ton. We could (possibly) make the
statement that we therefore believe that 44% of the ore in the lode lies below
25lb/ton.
Fig. 3.8. Histogram of chip samples
taken from the drives in the cassiterite vein.
Now, the usual method of estimating the
value in the stopes is to delineate a block (say 125ft long) between the drives
and allocate to that block the average of all the peripheral development
samples. It is this estimate which determines whether a stope block enters ‘reserves’
or not. Figure 3.9 shows the corresponding histogram of the estimates of 125ft
by 100ft stoping blocks, i.e. the averages of the drive samples over two
lengths of 125ft each.
Fig. 3.9. Histogram of estimates of
stope values in the cassiterite vein.
We have seen from the previous exercise
that we expect averages over lengths to be somewhat less variable than ‘point’
samples. This is adequately borne out by the behaviour of these estimates.
Whereas the point values range up to 300lb/ton or more, the drive averages
seldom exceed about 150lb/ton. Whilst 44% of the point samples lie below
25lb/ton something less than 8.5% of the ‘block estimates’ do so. Should we now
say that 8.5% of the ore lies below 25lb/ton? What we really need to do is to
redefine the phrase ‘of the ore’. In the first case what we meant was that if
the deposit were divided into chip samples, we could reject 44% of these as
being below cutoff. In the second it was 8.5% of the drive averages below
cutoff. That is, if the deposit were divided into pairs of 125-ft strips 100ft
apart, 8.5% of these would be below cutoff. Or alternatively, by my
estimate 8.5% of
the stope blocks would be below cutoff. In other words, we cannot define how
much ore we have after selection unless we define a unit of selection in
terms of size and shape. The real question is ‘how many 125 by 100ft stope
panels are below cutoff?’ To answer this question we must determine what sort
of distribution these panels would follow. The full answer will depend on (i) the
distribution of the original samples and (ii) the semi-variogram of the
deposit.
Let us make a general statement of the
problem and see how it leads to a solution. The original sample data has a
‘support’ of, say, l; it has a semi-variogram gl(h) with a sill Cl; it has a distribution of grades which
can to some extent be characterised by the histogram and which has a mean 
l and variance Cl. The panels or blocks being estimated
will have a support of, say, v; a semi-variogram gv (h) with sill Cv; a distribution with mean v and variance Cv. The first thing we can say is that l and v should be the same, since both describe
the average grade of ore over the whole deposit. Thus we can replace them both
by , the average of ‘point’ samples. The second thing we can say is
that if we have a model for the point semi-variogram we can state the
relationship between the point sill C and the ‘core’ sill Cl, and between C and C v
for any defined volume v. Suppose we take the simple
example of a core of length l which can be represented as a
straight line (since the diameter is very much smaller than the length).
Fig. 3.10. Derivation of the
variance of grades within a ‘line’ segment.
This is illustrated in Fig. 3.10. Consider
two points on this line, M and M’. We could calculate from the model
semi-variogram the ‘difference’ between the grades at these two points. Now
suppose we took all possible pairs (M,M’)
which exist within the line --- including the case when M=M’. In this
way we could get a measure of the ‘variability’ of the grades within the line.
If we take the average of the semi-variogram values g (M-M’) over all possible pairs,
then we obtain the variance of the grades within the length l.
This is
the variance which is removed from the system if we only consider the average
grade over the length l, i.e. the difference between the point
sill and the regularised sill, C-C l.
Mathematically:
where F(l) defines the variance of grades
within the length l. Although this looks fearsome, it reduces
to:
for the
linear model
for the
exponential model, and for the spherical model:
These, of course, correspond exactly with the difference
between the point and regularised semi-variograms. Now suppose we want to
consider a two-dimensional panel such as that shown in Fig. 3.11.
Fig. 3.11. Derivation
of the variance of grades within a panel.
The F function now becomes F(d,b) to show that it has two dimensions. This would
be a quadruple integral, since the points M and M’ can now move throughout the
whole panel. The formulae get complicated, but not impossible, and for example
of the type of values encountered, Table 3.4 has been produced. This table
shows the F(d,b)
function for a spherical model with range equal to 1 and sill of 1. This
is a ‘standardised’ spherical model --- in the same sense as a ‘Standard’
Normal distribution. This table can be used to produce the corresponding value
of the F function for any spherical model, as follows:
i.divide the lengths of the sides of the panel by the range of influence a;
ii.read off the corresponding entry in the table;
iii.multiply this value by C.
Examples of
such calculations are given later in this chapter. Similar tables may be
produced for the linear and exponential models.
In three dimensions the problem of calculating the F(l,b,d) function analytically appears to be
insurmountable. It is necessary to resort to a numerical approximation using a
computer. The easiest way to do this is to go back to the definition of the F function: we take pairs of points (M,M’) within the
block; consider all such pairs; calculate the semi-variogram
value between M and M’; sum all these values and average them --- this gives
the F value. Now, suppose we do not take all of the pairs but only a few
‘representative’ ones. That is, instead of considering the block as an infinite
number of points we consider it to be a ‘grid’ containing a finite number of
points, say on a 5 by 5 by 5 grid. Some authors suggest taking ‘randomly’
distributed points, but there seems little sense in that. Using such a method,
Table 3.5 was produced for the ‘standardised’ spherical model. In order to
produce only one table, it has been necessary to insist that two sides of the
block have the same length. This table is used in the same way as the
two-dimensional one.