RE:RE:Monster target at Albert Lake
Amber12 wrote: Mega can you help me with the calculation of the size of the target.
Length 450m
Width 150m
Thickness 100m Assume.
Thats 450 X 150 X 100 equals 6.75m cubic meters.
Do we then apply a specific gravity ratio to arrive at tonnage.
What ratio............ say 2.8. Thats 6.75m X 2.8 equals 18.9m
Is that calculation correct. Allbeit numbers assumed.
I am just a beginner trying to work it out.
See link below for calculations
https://undervaluedequity.com/mineral-deposit-value-how-to-calculate-the-potential-value-of-a-mining-project/
1. Calculate the Tonnage of the Mineral Deposit
| Formula | Example |
| Strike Length x Depth x Width x Specific Gravity = “X” (in tonnes) | 500 x 200 x 100 x 2.5 = 25,000,000 tonnes |
So you are using a specific gravity of 2.8 which is in the ballpark i would think. And you are assuming a thickness of 100 meters
volume of rock is 450 x 150 x 100 = 6750000
6750000 x 2.8 = 18900000 tonnes
So using your example with a specific gravity of 2.8 that deposit would be roughly 19 million tonnes.
Keep in mind the specific gravity of each individual mine would vary depending on the type of host rock and of course the percentage of metals in the host rock.