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MINERAL RESERVES
ESTIMATION BY USING
TRIANGULAR METHOD
BY
MD.NUMAN HOSSAIN
DEPT. OF PME ,SUST.
MINERAL RESERVES :
• A Mineral Reserve is the economically mineable part of a Measured and/or
Indicated Mineral Resource.
• It is described as total workable or probable working which depending on
thickness of deposit , it’s depth , quality , geological factors etc.
• In general mining practice , there are two types of mineral reserves .
• These are: (1) probable reserve & (2)proven reserves
:Proven Reserves Probable reserves
Proven reserves constitute mining assets with
greater confidence than any other mining assets
In other word , reserves in fully explored deposits
Probable reserves constitute mining assets with
lesser confidence than the proven reserves
In other word , reserves in estimated and fully
explored deposits
Mineral recovery :
 In mining , the process of valuable mineral extraction in known as mineral recovery
 It is the measure of mining or extraction efficiency
 For mineral recovery it is very essential to estimate ore reserves,know the mineable reserves ,
pit limits , cutoff grade etc .
 In this slide we will discuss only about different methods for estimating ore reserves.
Methods of ore reserves estimation
Triangular method
Polygonal methods
Method of sections (plans); longitudinal, transverse
Regular grid, random stratified grid
Inverse distance weighting (1/d, 1/d2, 1/d2.7, 1/d3, etc.)
Contouring methods
Here we only discuss about triangular method to estimate ore reserves.
Triangular method
By using triangular method, the area of the ore body is calculated
After that , this calculated area is multiplied with average thickness of the ore to
calculate the volume of the ore body
The average thickness can be calculated by
 Using known thickness of the ore body
 Using grade and the length of ore body
 Using weight age average thickness
After calculating volume , multiplying this value with ore density to estimate the
ore reserves.
• Here ,the area of the rectangle abcd = ( X3 – X2) (Y3 –Y1)
• Area of triangle A1 = ½ ( X2 - X1)(Y2-Y1)
• Area of triangle A2= ½ (X3 –X2) (Y3 –Y1)
• Area of triangle A3= ½ (X3 –X1)(Y3 –Y2)
• So the area of ore A = area of rectangle abcd – area of triangle (A1+A2+A3)
Calculate A for the following
Easting ( x )m northing (y) m
1100 1200
1500 1200
1100 800
Triangular method
• Using above equation
• The area of rectangle =(1100-1500)(800-1200)=160000 m2
• The area of triangle A1+A2+A3 = 1/2 (( 1500-1100)(1200-1200)+(1500-1100)(1200-
800)+(1100-1100)(800-1200)) =80000 m2
• The area of ore body ,A=160000-80000=80000 m2
Calculation of average thickness of the ore
body:
1. If thickness is givenC1= 3 m, C2=5 m,C3= 4 m , then ave. thickness =(C1+C2+C3)/3
=(3+5+4)/3 =4 m
2. If the length of the ore body (m) and the gradient(%) is given then the ave. thickness
is calculated by t(ave) = ∑tigi / ∑gi , for example we calculate ave thickness for
following case :
ymbol
The ave thickness , t(ave) =(0.6*0.59+1.4*0.48+1.4*0.6+1.4*0.56+1.3*0.32)/(0.59+0.48+0.6+0.56+0.32)
=0.503 m
3.calculation of reserves using weight age average thickness . For this case the following equ is used:
weight age ave. thickness, tw =( (t1Ѳ1+t2Ѳ2+t3Ѳ3)/60 )÷3 = (t1Ѳ1+t2Ѳ2+t3Ѳ3)/ 180 .an example
problem related with this is given bellow : where thickness t1=40 m, t2 =60 m ,t3 =50 m.
Length of the ore body (m) grade(%)
0.6 0.59
1.4 0.48
1.4 0.6
1.4 0.56
1.3 0.32
Now, ore volume calculation:
• Ore volume = area of the ore body * average thickness
• For , first case, t(ave)=4 m, so, ore volume = 80000*4 =320000 m3
• For , second case , using length of the ore body and grade, ave thickness = 0.503
So, for this ore volume = 80000*.503 =40240 m3
• For third case, weight age ave thickness , tw =50.77 m then ,
ore volume = 80000*50.77 =4061600 m3
N.B.here, I use one example problem for several ave thickness ,so don’t be confused with
evaluated ore volume .This indicates just calculation procedure . Here, the ave thickness
data had not been obtained from same ore reserve but I use here only to show the
evaluation procedure ,and so, ore volume can’t be obtained same for all time.
Ore reserve estimation:
• Ore reserve(in tons) =ore volume(m3) * density of ore (tons/m3)
• For example ,if the density of the ore = 1.35 tons/m3.
• Then ,for 1st case, ore reserves =320000*1.35 =432000 tons.
• For 2nd case, ore reserves = 40240*1.35 =54324 tons.
• For 3rd case, ore reserves =4061600*1.35 =5483160 tons.
Exercise Problem :
• For 1st fig the following parameters are given for a ore deposit :
• If the thickness for first layer C1=4 m, C2=8m ,C3=3m and for 2nd layer
C1=5m,C2=7m,C3=6m .And the density of the ore is 2.5 tons/m3 .Then estimate the
reserves for this ore deposit.
Easting (x) m Northing (y) m
300 700
600 300
100 0
Exercise Problem :
• Estimate the reserves by using triangular method for the following mineral
deposit(having density =1.024 tons/m3:
Length of ore body (m) grade (%)
.9 .62
1.2 .56
1.2 .61
1.8 .49
2.1 .68
2.8 .65
2.8 .61
2.5 .57
2.2 .52
1.9 .48
x (m) y(m)
3500 600
1365 450
350 46
N.B. use fig :1
Exercise Problem : Ѳ2
Ѳ1
Ѳ3
t3(800,1000)
t1(200,500)
t2(600,1500)
Calculate the reserve of the ore shown in above fig having density 2.65 tons/m3 by using the weight
average thickness , when t1=80 m, t2 = 125 m, t3 = 98 m.
Exercise Problem
• Calculate the reserve of the ore shown in above fig having density 2.65 tons/m3 by
using the weight average thickness where others information has give
References :
• Open Pit Mine Planning and Design,TwoVolume Set & CD-ROM Pack
-William A. Hustrulid, Mark Kuchta, Randall K. Martin
• Applied Mineral Inventory Estimation - ByAlastair J. Sinclair & Garston H.
Blackwell
• Internets & others
THAT’S ALL
THANKS……..

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The mineral reserves & reserves estimation using triangular methods

  • 1. MINERAL RESERVES ESTIMATION BY USING TRIANGULAR METHOD BY MD.NUMAN HOSSAIN DEPT. OF PME ,SUST.
  • 2. MINERAL RESERVES : • A Mineral Reserve is the economically mineable part of a Measured and/or Indicated Mineral Resource. • It is described as total workable or probable working which depending on thickness of deposit , it’s depth , quality , geological factors etc. • In general mining practice , there are two types of mineral reserves . • These are: (1) probable reserve & (2)proven reserves
  • 3. :Proven Reserves Probable reserves Proven reserves constitute mining assets with greater confidence than any other mining assets In other word , reserves in fully explored deposits Probable reserves constitute mining assets with lesser confidence than the proven reserves In other word , reserves in estimated and fully explored deposits Mineral recovery :  In mining , the process of valuable mineral extraction in known as mineral recovery  It is the measure of mining or extraction efficiency  For mineral recovery it is very essential to estimate ore reserves,know the mineable reserves , pit limits , cutoff grade etc .  In this slide we will discuss only about different methods for estimating ore reserves.
  • 4. Methods of ore reserves estimation Triangular method Polygonal methods Method of sections (plans); longitudinal, transverse Regular grid, random stratified grid Inverse distance weighting (1/d, 1/d2, 1/d2.7, 1/d3, etc.) Contouring methods Here we only discuss about triangular method to estimate ore reserves.
  • 5. Triangular method By using triangular method, the area of the ore body is calculated After that , this calculated area is multiplied with average thickness of the ore to calculate the volume of the ore body The average thickness can be calculated by  Using known thickness of the ore body  Using grade and the length of ore body  Using weight age average thickness After calculating volume , multiplying this value with ore density to estimate the ore reserves.
  • 6.
  • 7. • Here ,the area of the rectangle abcd = ( X3 – X2) (Y3 –Y1) • Area of triangle A1 = ½ ( X2 - X1)(Y2-Y1) • Area of triangle A2= ½ (X3 –X2) (Y3 –Y1) • Area of triangle A3= ½ (X3 –X1)(Y3 –Y2) • So the area of ore A = area of rectangle abcd – area of triangle (A1+A2+A3)
  • 8. Calculate A for the following Easting ( x )m northing (y) m 1100 1200 1500 1200 1100 800
  • 9. Triangular method • Using above equation • The area of rectangle =(1100-1500)(800-1200)=160000 m2 • The area of triangle A1+A2+A3 = 1/2 (( 1500-1100)(1200-1200)+(1500-1100)(1200- 800)+(1100-1100)(800-1200)) =80000 m2 • The area of ore body ,A=160000-80000=80000 m2
  • 10. Calculation of average thickness of the ore body: 1. If thickness is givenC1= 3 m, C2=5 m,C3= 4 m , then ave. thickness =(C1+C2+C3)/3 =(3+5+4)/3 =4 m 2. If the length of the ore body (m) and the gradient(%) is given then the ave. thickness is calculated by t(ave) = ∑tigi / ∑gi , for example we calculate ave thickness for following case : ymbol
  • 11. The ave thickness , t(ave) =(0.6*0.59+1.4*0.48+1.4*0.6+1.4*0.56+1.3*0.32)/(0.59+0.48+0.6+0.56+0.32) =0.503 m 3.calculation of reserves using weight age average thickness . For this case the following equ is used: weight age ave. thickness, tw =( (t1Ѳ1+t2Ѳ2+t3Ѳ3)/60 )÷3 = (t1Ѳ1+t2Ѳ2+t3Ѳ3)/ 180 .an example problem related with this is given bellow : where thickness t1=40 m, t2 =60 m ,t3 =50 m. Length of the ore body (m) grade(%) 0.6 0.59 1.4 0.48 1.4 0.6 1.4 0.56 1.3 0.32
  • 12.
  • 13. Now, ore volume calculation: • Ore volume = area of the ore body * average thickness • For , first case, t(ave)=4 m, so, ore volume = 80000*4 =320000 m3 • For , second case , using length of the ore body and grade, ave thickness = 0.503 So, for this ore volume = 80000*.503 =40240 m3 • For third case, weight age ave thickness , tw =50.77 m then , ore volume = 80000*50.77 =4061600 m3 N.B.here, I use one example problem for several ave thickness ,so don’t be confused with evaluated ore volume .This indicates just calculation procedure . Here, the ave thickness data had not been obtained from same ore reserve but I use here only to show the evaluation procedure ,and so, ore volume can’t be obtained same for all time.
  • 14. Ore reserve estimation: • Ore reserve(in tons) =ore volume(m3) * density of ore (tons/m3) • For example ,if the density of the ore = 1.35 tons/m3. • Then ,for 1st case, ore reserves =320000*1.35 =432000 tons. • For 2nd case, ore reserves = 40240*1.35 =54324 tons. • For 3rd case, ore reserves =4061600*1.35 =5483160 tons.
  • 15. Exercise Problem : • For 1st fig the following parameters are given for a ore deposit : • If the thickness for first layer C1=4 m, C2=8m ,C3=3m and for 2nd layer C1=5m,C2=7m,C3=6m .And the density of the ore is 2.5 tons/m3 .Then estimate the reserves for this ore deposit. Easting (x) m Northing (y) m 300 700 600 300 100 0
  • 16. Exercise Problem : • Estimate the reserves by using triangular method for the following mineral deposit(having density =1.024 tons/m3: Length of ore body (m) grade (%) .9 .62 1.2 .56 1.2 .61 1.8 .49 2.1 .68 2.8 .65 2.8 .61 2.5 .57 2.2 .52 1.9 .48 x (m) y(m) 3500 600 1365 450 350 46 N.B. use fig :1
  • 17. Exercise Problem : Ѳ2 Ѳ1 Ѳ3 t3(800,1000) t1(200,500) t2(600,1500) Calculate the reserve of the ore shown in above fig having density 2.65 tons/m3 by using the weight average thickness , when t1=80 m, t2 = 125 m, t3 = 98 m.
  • 18. Exercise Problem • Calculate the reserve of the ore shown in above fig having density 2.65 tons/m3 by using the weight average thickness where others information has give
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  • 21. References : • Open Pit Mine Planning and Design,TwoVolume Set & CD-ROM Pack -William A. Hustrulid, Mark Kuchta, Randall K. Martin • Applied Mineral Inventory Estimation - ByAlastair J. Sinclair & Garston H. Blackwell • Internets & others