CLASSICAL ORE RESERVES
CLASSICAL ORE RESERVES
ESTIMATION METHODS
ESTIMATION METHODS
GE
GEOS
OSTA
TATI
TIST
STIC
ICS
S CO
COUR
URSE
SE
Dr. Arifudin Idrus Dr. Arifudin Idrus
Department of Geological Engineering
Department of Geological Engineering
Gadjah Mada University
Gadjah Mada University
E-mail: [email protected]
Definition of terms
Definition of terms
Global reservesGlobal reserves
refers to the mean grade of the reserves to refers to the mean grade of the reserves to be mined over the life time of the mine.
be mined over the life time of the mine.
Local reservesLocal reserves
refers to the mean grade of reserves to be refers to the mean grade of reserves to be mined over short time
mined over short time increaments eg. yearincreaments eg. year by year.
by year.
Local ore reserves are used to
Local ore reserves are used to produce theproduce the mining schedule.
Definition of terms
Definition of terms
Global reservesGlobal reserves
refers to the mean grade of the reserves to refers to the mean grade of the reserves to be mined over the life time of the mine.
be mined over the life time of the mine.
Local reservesLocal reserves
refers to the mean grade of reserves to be refers to the mean grade of reserves to be mined over short time
mined over short time increaments eg. yearincreaments eg. year by year.
by year.
Local ore reserves are used to
Local ore reserves are used to produce theproduce the mining schedule.
Definition of terms
Definition of terms
ResourcesResources
usually based on
usually based on geological interpretationgeological interpretation only. Mining parameters have not (or only only. Mining parameters have not (or only partly) applied.
partly) applied.
ReservesReserves
incorporate all aspect of the impact of incorporate all aspect of the impact of mining on the
mining on the geological interpretageological interpretation,tion, especially ore loss and dilution.
Requirements of an ore reserve
Requirements of an ore reserve
method
method
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They should be robust. The method should yield the correct answer over a wide renge of data and should not be such that small changes in the data can yield
On the philosophical side
.
They should be understandable. If the method cannot understand what is happening then this can create uneasiness with the method and result in uncertainly.
They should de defendable when confronted by peers. Or reserve calculation must be defendable with conviction.
They should be consistent with the data density. If only limited data is available the method should reflect this.
They must reflect themining method realistically.
“It is only in comics and films that superman mines
On the philosophical side
ore o es
Most orebodies are mined by mortals.
On the practical side the computations should be:
1. Rapid 2. Reliable
3. Easily checked
The reliability of reserve computations depends
on:- The accuracy and completeness of our knowledge of the orebody under study.
The data density and reliability of the base data.
The assumptions for interpreting the variables under study.
The relevance of the mathematical methodology used.
Selecting an Ore Reserve Method depends
on:- The geology of the mineral deposit.
The densit of data.
The purpose of computations.
At the mine design stage
More complete calculations are required. (Global Estimates)
More complete calculations are required. (Local Estimates)
Types of Ore Reserve Method
Classical method.
CLASSICAL ORE RESERVE METHODS
In carrying out classical ore reserve calculations we work from the highest data density to the least data density.
Consider an orebody drilled on section. The
sectional data is densest. The interpretation should be carried out on section first.
All methods are based on computing solids with their bases in the plane.
For sectional data
(and vertical longitudinal sections)
Interpret data in the plane of the section.
Measure the area in the plane.
Use a horizontal distance for extending to a volume.
For plan data
1. Interpret data in he plane of the plan. 2. Measure are in plan.
3. Use a vertical distance for extendig to a volume.
Often the vertical distance is the bench height.
Often ore reserves are calculated by generating longitudinal sections in the plane of the dip of the mineral body.
For inclined longitudinal sections
interpret data in the plane of the sections.
use true thickness for he inclined plane.
Generally in reserve calculations the true strike and true
If data in not presented with true dip and strike it needs to be corrected to allow for this. This is achieved by carrying out geometrical corrections.
All ore reserve methods involve.
1. Weighting data (ie. Weighting assay data). 2. Extending data to obtain volume.
VOLUME CALCULATONS
1. Trapezoidal rules
The trapezoidal rule – assumes the area consists of a sequence of trapezoids a1 a2 a3 a5 a4 h Area = a7 a6 a8 a10 a9
•
•
•
+
+
2 2 3 2 2 1Where the area of a simple trapezoid is
(
)
2 h b a S=
+
2. Simpsons rule
Assumes the boundaries of each strip are best represented by parabolas passing through
consecutive points. Area =
Volume calculations
Volume = Area x thickness.
(
a odd even an)
h + 2
∑
+ 4∑
+ 31
Tonnage Calculations
Tonnes = area x thickness x tonnage factors.
This is an improtant factor in the calculation
of tonnage.
WEIGHTING OF DATA
This is done by various methods
including:-
Simple arithmetic methods.
Weighting by width or thickness, length,
area, specific gravity.
1. Simple Arithmetic Methods mean =
n
g
g
g
g
1+
2+
3+
••• nSome weighting examples to calculate the mean grade.
Assumption all blocks are equal.
2. Thickness weighted
All blocks are equal in area and have the same SG. mean = n n n t t t t g t g t g t g t ... ... 3 2 1 3 3 2 2 1 1 + + + + + +
3. Area weighted
All blocks have constant thickness and weight factor but different area.
mean = n n g A g A g A g A ... 3 3 2 2 1 1 + + + 4. Volume weighted
The assumption here is that all blocks have SG. mean = n A A A A1 + 2 + 3 +... n n n V V V V g V g V g V g V
+
+
+
+
+
+
3 2 1 3 3 2 2 1 1 ...5. Tonnage weighted
the assumption is that the tonnage and grade of blocks are different.
mean = n n n
T
T
T
T
g
g
g
g
+
+
+
+
+
+
3 2 1 3 3 2 2 1 1 ...CASE 1A
An averaging method (simplest case)
Consider a plan view of 14 drillholes
7 8 6 4 5 10 11 12 13 14 9
Drillhole Thickness Grade 1 t1 g1 2 t2 g2 3 t3 g3 4 t4 g4 . . .
Measure the area A Then tonnes = t A SG grade = G . . . . . . 14 t14 g14 14
∑
= t i t 14∑
= g i GIn this calculation thickness is not considered important
This method is accurate in uniform deposits, where there is a very small
CASE 1B
Consider the previous example of the 14 drillholes. In the folowing calculation it isDrillhole Thickness Grade Product
1 t1 g1 t1g1
2 t2 g2 t2g2
3 t3 g3 t3g3
thickness varies from point to point. An average method (thickness weighting). 4 g4 4g . . . . . . . . . . . . 14 t14 g14 t14g14 Σti Σtigi
Average grade
∑
∑
=
i i it
g
t
G
Tonnes are calculated by using local thickness. If the blocks are all the same area, and only the thickness changes, then
Tonnes = A [t
1+ t
2+ … t
14] SG
A PLAN METHOD USING POLYGONS
Drillhole
Area of influence polygon boundary
Ore zone limit
Line segment between
drillholes (construction aid)
If the polygons have different tonnages. Then use a tonnage weighted method.
CROSS SECTIONAL METHODS
The orebody is devided into geological section along the lines of drilling. Two methods are used.
.
A step change.
In plan
b. Step change
The calculation of volumes may use the following formulae. (1) End Area Formula
L A A V + = 2 2 1 L A A V = 1 + 2
For several sections
(2) Wedge Formula – where one end tapers to a line.
2 ... 2 2 3 2 1 L A A A A V n + + + = L A V 2 =
(3) Cone formula – where where one and tapers to a cone L A V 3
=
(4) The frustum formula
Note
The frustum formula is inaccurate in wedge
like orebodies 1 A2
L
(
1 2 1 2)
3
A
A
A
A
L
(5) The prismoidal formula
(
)
6 4 2 1L
A
A
A
V
=
+
m+
This is better for ore bodies which pinch and swell. Am = mean area between section i.e., auxiliary.
COUNTOURING METHODS
(Isoline methods)
Contours are curved lines which join all
points of equal value.
Data is used to construct contours by
interpolation
between
point
of
known
values. Various techniques of interpolating
data may be used. (Specific techniques of
interpolation are discussed later).
Contouring method
As an example of the
interpolation technique consider the method of finding the
volume of the following:-The volume is calculated by measurin each area within the
Section
contour interval and using
volume calculation procedures previously discussed.
The average ore grade can be computed by constructing
contour maps and by weighting each area by its contour grade.
To achieve this use
2 1
0
0 2 2 ...
2 A A A A
g g g n + + + + =
Contouring method
where go is the minimum grade of the ore
* g is the constant grade interval between contours * Ao is the area of the body with grade go plus g and
higher
* A1 area of orebody with grade go plus g and higher
* A2 area of the orebody with grade go plus 2g and higher etc
The method reqiures data which has 1. A sufficient number of dat.
2. Appropriate data density
3. Appropriate distribution of data.
Contouring method
When data is unvenly distributed there can be problems. (These problems will be discussed later).
The map produced shows the areas of rich and poor ore.
Contouring method
As an example consider the data ( ) ( ) ( ) 0 32 31 22 21 1 0 0 0 2 2 2 A A A A A A A g a g G + + + + + + = A2 A3 A3 g1 A0 g0 g2 g3 g3 A2 1 1 2 2The method of contouring should be used only in deposits of orderly changing thickness and grade.
It is not useful in very complex, discontinous
Contouring method
orebodies.
It is partucularly useful in orebodies where
thickness and grade decrease from the centre to the periphery
THE METHOD OF POLYGONS
In this method all factors determined for a certain point of a minerals body extend half way to the adjoining and surrounding points forming an area of influence.
2. Stagged drillholes (face centered)
A note on the use of the FRUSTUM FORMULA
A1 = area M1 = metal = grade x area A2 = area M2 = metal = grade x area The volume is
(
A1 A2 A1A2)
L
V = + + For the prismoidal method
The metal
(
1 2 1 2)
3 1 M M M M M = + + v m G =(
1 4 2 3)
6
A
A
A
V
=
+
+
(
1 2 3)
6 1 M M M M=
+
+
V M G = ThenEXAMPLE OF POLYGONAL ORE RESERVES
The area of influence method of calculating ore reserves is as following:
1. Difine for each drillhole, a boundary enclosing the area closest to that drillhole. This is done by constructing lines which are
-to the line segment between the two drillhole location points. 2. Each area so defined is treated as a polygon of constant grade
and thickness, ie. The grade and thickness of the single drillhole inside the polygon.
3. The reserves are determined simply by adding the tonnes and metal derived for each polygon.
Drillhole
Area of influence Polygon boundary This method is ussualy applied on a plan basis as shown below:
Ore zone limit
Line segment between
In some instances, the same method may be apllied on a cross Sectional basis. Each polygon is assigned the average grade Of samples inside the polygon. The polygon thickness is the Cross-section width (ie. Mid-way to adjoining section).
EXAMPLE OF POLYGONAL ORE RESERVE
Area of influence Polygon boundary Ore zona limit
INVERSE DISTANCED METHODS
Consider the problem of estimating the grade of a block from the surrounding data.
Eg.
Ore method of solving this problem is to use a method based on the distance of the samples from the block.
The most common distance weighted methods
INVERSE DISTANCED METHODS
are:-1.Inverse distance.
2.Inverse distance squared. 3.Inverse distance cubed.
The following examples show the application of these techniques.
The general formulae
Inverse Distance n n g d g d g d G 1 ... 1 1 1 ... 1 1 2 2 1 1 + + + + =Inverse Distance cubed
3 2 3 2 1 3 1 1 1 1 1 ... 1 1 n n g d g d g d G + + =
Inverse Distance Squared
n 2 1 2 2 2 2 1 2 2 2 2 1 2 1 1 ... 1 1 1 ... 1 1 n n n d d d g d g d g d G + + + + = 3 3 2 3 1 ... n d d d + +
λi
=
d −i r n −( )
x
i
Z
Z
n i V=
λ
*Estimation of block grade:
Where:
INVERSE DISTANCED METHODS
∑
i=1 n λi=
1 i= 1 d i n=1Example-Inverse distanced square (IDS) 0,48% 0,64% 0,69% d4=78m d3=66m 0,43% 0,75% 0,53% d5=92m d6=64m d2=52m d1=32m
