CHAPTER 3: The Volume -- Variance Relationship
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) ˛. 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(%)˛ 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
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POINTS |
BLOCKS |
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proportion |
average grade |
proportion |
average grade |
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Cutoff |
above cutoff |
|
above cutoff |
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0.05 |
0.622 |
0.47 |
0.712 |
0.41 |
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0.10 |
0.452 |
0.62 |
0.527 |
0.53 |
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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.
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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. |
