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.
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.
GRADE/TONNAGE CURVES
So, we now know how to calculate the function F for one, two and three
dimensions, and hence can state the difference between the ‘point’ variance and
the ‘regularised’ variance of regular shaped areas and volumes. This will give
us a numerical quantity for the reduction in the variance, but unless we make
some assumptions about the distribution of the samples, we cannot actually
quantify the change in the ‘tonnage above cutoff’ and so on. There are two ways
to approach the problem:
i.
Assume that the histogram of the samples represents the
whole deposit accurately.
ii.
Assume that the histogram represents a set of samples
from the whole deposit, and as such contains some random variation from the
‘population’ distribution.
The first approach declares that the samples are ‘typical’
of the whole deposit, and leads to graphical anamorphisms and transfer
functions. The second approach declares the belief that if we could measure the
grade at every point in the deposit we would end up with a smooth curve of a
fairly simple form. This is a much simpler approach, and generally seems to be
sufficient for most deposits.
To start
with a simple example, let us consider an iron ore deposit which is known to
follow a Normal distribution with a mean of 48%Fe and a standard deviation of 5%Fe. This distribution
has been established on samples small enough to be called ‘points’. We also
know that the deposit follows a point semi-variogram model which is spherical
with a range of influence of 400ft. Now, suppose that the mine plan is to be
constructed on blocks which are 100ft by 100ft by 50ft. What will the
distribution of these blocks look like. The first thing we can say is that it
will probably be Normally distributed. It will certainly have the same mean
(48%Fe) as the ‘points’. The only change
will be in the standard deviation of the distribution. We need to evaluate the
function F(100,100,50) for a spherical model with a=400 and C=25. To use Table 3.5 we must
‘standardise’ the situation so that the range of influence becomes 1. That is,
F(100,100,50) for a=400 is the same as F(0.25,0.25,0.125) for a=1. Table 3.5 gives a value of 0.209 for these arguments, but this is for a model
whose sill is 1. For our model the required value is 0.209´25=5.225(%Fe) ². This is the difference between the point
variance and the block variance. Therefore the variance of the block values
will be 25-5.225=19.775(%)² leading to a block standard deviation, sv, of 4.45%Fe. This is slightly over 10%
less than the point standard deviation, as would be expected with such a ‘small’
block. Thus we have two distributions to be considered, both
i.point distribution: =48%Fe;;s=5%Fe
ii.block distribution: =48%Fe;;sv =4.45%Fe
These two distributions are shown in Fig. 3.12, and the reduction
in spread for the block distribution can be clearly seen.
Fig. 3.12. Comparison of the
distributions of points and blocks within a hypothetical iron ore deposit.
To see how the difference in the
distributions will affect the grade--tonnage calculations, let us take as an
example a cutoff grade of 44%Fe.
The
proportion (P) of the distribution which is above
cutoff is given by:
P = Pr{g
> c}
where g denotes the grade in general, and c the cutoff grade. Tables are widely available for the
Standard Normal distribution, which tabulate the proportion of the Standard
Normal distribution which lies below a given value z. For any other Normal distribution the z value is determined by taking the value of interest,
subtracting the mean of the distribution, and dividing by the standard
deviation. In our example, we are considering the cutoff grade, c, so that
The
Thus, if we consider the distribution of the ‘point’
values, we obtain the following:
Consulting a Standard Normal table gives F(z)=0.212 so that P=0.788. That is, about 79% of the deposit will lie above
a cutoff of 44%Fe. The second question is the average grade of this ore. For
the Normal distribution, the grade above cutoff is given by:
where c denotes the grade above cutoff, and f(z) is the height of the standard normal curve at
the value z, i.e.
For our example, then, f(z)=0.290 so that:
To summarise, 78.8% of the ore lies above a cutoff of
44%Fe, and this ore has an average grade of 49.8%Fe.
Let us repeat this exercise, but now taking into account
the ‘selective mining unit’ of 100ft x 100ft x 50ft. We have:
The Standard Normal table gives f(zv)=0.184 so that P is now 0.816, and the average grade above cutoff is
Although the differences in this example are not substantial,
it can clearly be seen that if the selection is applied to the average grade of
100ft by 100ft by 50ft blocks, then the tonnage mined (the proportion of the
deposit) is larger and
the grade of the ore is lower than would be expected from the original samples. If the cutoff
grade chosen were above the mean grade of the deposit, the position would be
reversed. This is not merely an academic observation, but is borne out by
experience on many mines in existence.
Now let us turn to a much more common situation, in which
the grade distribution is log-normal. Take as an example a lead/zinc deposit, where the ‘combined
metal’ percentage is the economic variable. The samples are known to be
log-normally distributed, and the mean and standard deviation of the ‘point’
samples are 12% and 8% combined metal respectively. The semi-variogram is
spherical with a range of 15m. The ‘selective mining unit’ is a block 10 by 10
by 5m. Using Table 3.5, we can find that F(10,10,5) when a=15 and C=64 is given by 0.516´64=33.034. This produces a block standard deviation of
5.56% combined metal. The two distributions are compared in Fig. 3.13.
Fig. 3.13. Comparison of the distributions of
combined metal percentages within a hypothetical lead/zinc deposit.
To calculate the proportion above cutoff and the average
grade of the ore for a log-normal distribution requires an extra step in the
proceedings. By definition, if a variable has a log-normal distribution, then
the logarithm of that variable has a Normal distribution. It is necessary to
calculate the mean and standard deviation of this Normal distribution before we
can perform any calculations. If we write y for the logarithm of the grade, then the mean and
standard deviation of the y values are given by:
Once these parameters have been calculated, then the
following may be evaluated:
where P is again the proportion of the ore
above cutoff. The average grade is found by the following process:
where Q=1-F(z-sy).
In the lead/zinc example above, 4% combined metal is a
feasible cutoff grade. Considering the ‘point’ distribution then:
The original sample data informs us that 93.4% of the
deposit lies above 4%(Pb+Zn) and that this ore has an average value of 12.62%
combined metal. Now, considering the distribution of 10 by 10 by 5m blocks, the
following results are found:
Once again, selection made on a mining block unit, this
time of size 10 by 10 by 5m, produces a larger tonnage and a lower grade than
the small samples would suggest. Table 3.6 shows the resulting values when a set of possible cutoff values is chosen.
Figure 3.14 shows the resulting grade/tonnage curves in the normal manner
employed in mining reports. The one minor difference here is that ‘proportion’
above cutoff is given rather than tonnage, to keep the example general. The
‘bias’ introduced by considering ‘point’ samples rather than the true selective
mining unit can clearly be seen on this graph.
|
|
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|
Fig.
3.14. Comparison of grade/tonnage curves in the lead/zinc deposit. |
Table 3.6
Comparison of grade/tonnage calculations for point and block values for
combined metal (Lead/Zinc) deposit |
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Here is a very brief example with which to
finish off. A low grade uranium deposit has a log-normal distribution with mean
0.30% U3O8 and
a standard deviation of 1.05% U3O8. The spherical
semi-variogram has a range of 40m, and the selective mining unit has a size of
25 by 25 by 10m. Using Table 3.5 gives an F function of 0.477´1.1025=0.526, producing a standard deviation for the
blocks of 0.76% U3O8. Table 3.7 shows the results of
applying cutoffs of 0.05, 0.10, 0.15 and 0.20% U3O8 to
(i) the ‘point’ samples and (ii) the block distribution.
Table 3.7. Comparison of grade/tonnage
calculations for point and block values for uranium deposit
|
|
POINTS |
BLOCKS |
||
|
|
proportion |
average grade |
proportion |
average grade |
|
Cutoff |
above cutoff |
|
above cutoff |
|
|
0.05 |
0.622 |
0.47 |
0.712 |
0.41 |
|
0.10 |
0.452 |
0.62 |
0.527 |
0.53 |
|
0.15 |
0.355 |
0.75 |
0.414 |
0.64 |
|
0.20 |
0.291 |
0.88 |
0.337 |
0.75 |
Figures 3.15 and 3.16 illustrate these
results. Notice how the skewed nature of the original distribution, and the
relatively large size of the block combine to produce an ever widening gap
between the point curve and the block curve.
|
|
|
|
Fig.
3.15. Comparison of point and block grade/tonnage curves in a uranium deposit
--- cutoff grade versus proportion above cutoff. |
Fig. 3.16.
Comparison of point and block grade/tonnage curve in a uranium deposit ---
cutoff grade versus average grade. |
CONCLUSION
In this chapter we have considered the problems introduced by the volume, size,
shape etc. of both samples and selective mining units. We have seen how the
‘point’ semi-variogram model may be derived, even though the experimental
semi-variogram may have been constructed on samples with a definite support. We
have also seen how, for certain regular shapes, the distribution of values
changes according to the support of the selective mining unit. The construction
of ‘theoretical’ grade/tonnage curves can be achieved providing the following
information is available:
i.
the distribution of the ‘point’ values;
ii.
the semi-variogram of the ‘point’ values;
iii.
the size and shape of the selective mining unit.
In this way more ‘realistic’ first estimates of the
grade/tonnage calculations may be produced at a very early stage in any study.