Simple block model problem
Simple Flat Vein Block Model problem
A uniformly 2-meter thick, horizontal gold vein is drilled and cuttings assayed. The geologist on the project seems to have been enthusiastic about closing off the limits of the deposit, but oddly cost-sensitive when drilling the mineralized zone, so the sample spacing is non-uniform. The geostatistician on the project is very conservative, which frustrates the promotion department, but you have the block model shown here.
Question 1: What is the average grade of the vein?
Question 2: If the density of the vein is 2,000 kg/m3 (it’s really vuggy silica!), how many grams of gold are in the deposit?
Calculating average grade
Because the blocks are different sizes, we cannot just add the grade numbers together and divide. We need the weighted average.
1. Multiply each grade value by its corresponding area to calculate the grade×area products
40 pmm ×100 m2 = 4,000 ppm•m2
102 pmm ×100 m2 = 10,200 ppm•m2
60 pmm ×100 m2 = 6,000 ppm•m2
25 pmm × 400 m2 = 10,000 ppm•m2
75 pmm × 400 m2 = 30,000 ppm•m2
22 pmm × 900 m2 = 19,800 ppm•m2
2. Add up the grade×area products to get the total grade×area product
4,000 + 10,200 + 6,000 + 10,000 + 30,000 + 19,800 = 80,000 ppm•m2
3. Divide the total grade×area product by the total area to get the weighted average grade
80,000 ppm•m2 ÷ (400 × 500 m) = 40 ppm Au
Calculating total gold
You (should) know that 1 ppm = 1 g per metric ton
1. Calculate total volume of ore by multiplying area × thickness
400 m × 500 m × 2 m = 4,000 m3 ore
2. Calculate total mass of ore by multiplying volume of ore by ore density, then convert to metric tons (because our grade is given in that unit)
4,000 m3 × 2,000 kg/m3 = 8,000,000 kg ore ÷ 1,000 kg/ton = 8,000 tons ore
3. Multiply total mass by grade (making sure your units are consistent!) to get total mass of gold
8,000 tons ore × 40 g Au/kg ore = 320,000 g Au
(at $75/gram, that’s a 24 million dollar vein with near-zero mining costs! It’s nice to fantasize sometimes. 😊)