variograma trabajo
DESCRIPTION
VARIOGRAMATRANSCRIPT
NOMBRE: VICENTE DUEÑAS ANTONY ANGEL
CODIGO: 121.0802.452
PROBLEMA:
En la exploración minera de un yacimiento de oro, se dispone de 3 taladros en dirección vertical: DDH01, DDH02 Y DDH03 cuya información se muestra en el siguiente cuadro.
COORDENADAS DDH01 DDH02 DDH03E 373200 373250 373300N 8168100 8168100 8168200
COTA 5100 5100 5100
DDH01FROM TO Au (gr/t)
0.0 1.5 5.01.5 3.5 4.53.5 6.0 4.76.0 8.0 4.48.0 9.5 4.59.5 11.0 4.2
11.0 14.0 4.314.0 16.0 4.116.0 18.0 4.218.0 19.5 3.919.5 21.5 3.821.5 23.0 3.723.0 25.0 3.8
Con esta información realizar los siguientes cálculos:
a). Compositar los taladros a 5 m.
b). Construir y graficar:
Los variogramas de DDH01, DDH02 Y DDH03 Variograma Promedio
b) Construir y graficar:
1). Variograma del taladro DDH01 (Au)
γ (h=1 )= ( 4.47−4.71 )2+(4.24−4.47)2+(4.05−4.24)2+(3.77−4.05)2
2×4=0.028
γ (h=2 )=(4.24−4.71)2+(4.05−4.47)2+(3.77−4.24)2
2×3=0.103
γ (h=3 )=(4.05−4.71)2+(3.77−4.47)2
2×2=0.231
γ (h=4 )=(3.77−4.71)2
2×1=0.442
2). Variograma del taladro DDH02 (Au (gr/t))
γ (h=1 )= (3.01−3.34 )2+(2.34−3.01)2+(2.20−2.34)2+(1.72−2.20)2+(1.24−1.72)2+(0.72−1.24)2
2×6=0.109
γ (h=2 )=(2.34−3.34)2+(2.20−3.01)2+(1.72−2.34)2+(1.24−2.20)2+(0.72−1.72)2
2×5=0.396
γ (h=3 )=(2.20−3.34)2+(1.72−3.01)2+(1.24−2.34)2+(0.72−2.20)2
2×4=0.987
γ (h=4 )=(1.72−3.34)2+(1.24−3.01)2+(0.72−2.34)2
2×3=1.397
γ (h=5 )=(1.24−3.34)2+(0.72−3.01)2
2×2=2.414
γ (h=6 )=(0.72−3.34)2
2×1=3.432
Variograma del taladro DDH02 (Ag (Oz/T))
γ (h=1 )= (0.98−0.74 )2+(1.53−0.98)2+(1.82−1.53)2+(2.53−1.82)2+(3.16−2.53)2+(3.74−3.16)2
2×6=0.140
γ (h=2 )=(1.53−0.74)2+(1.82−0.98)2+(2.53−1.53)2+(3.16−1.82)2+(3.74−2.53)2
2×5=0.559
γ (h=3 )= (1.82−0.74 )2+ (2.53−0.98 )2+(3.16−1.53 )2+(3.74−1.82 )2
2×4=1.240
γ (h=4 )= (2.53−0.74 )2+ (3.16−0.98 )2+(3.74−1.53 )2
2×3=2.140
γ (h=5 )= (3.16−0.74 )2+ (3.74−0.98 )2
2×2=3.369
γ (h=6 )= (3.74−0.74 )2
2×1=4.501
3). Variograma del taladro DDH03 (Au (gr/t))
γ (h=1 )=(2.60−3.45)2+(2.25−2.60)2+(1.59−2.25)2+(1.24−1.59)2+(0.94−1.24)2
2×5=0.147
γ (h=2 )=(2.25−3.45)2+(1.59−2.60)2+(1.24−2.25)2+(0.94−1.59)2
2×4=0.488
γ (h=3 )=(1.59−3.45)2+(1.24−2.60)2+(0.94−2.25)2
2×3=1.170
γ (h=4 )=(1.24−3.45)2+(0.94−2.60)2
2×2=1.910
γ (h=5 )=(0.94−3.45)2
2×1=3.15