Okay I have built a spreadsheet for the estimation of nickel. It's actually pretty easy to do given the standard drill results you see in any NR.
I am having a problem seeing where he gets his numbers for the assumption in step one?
The estimate methodology:
1) multiply core length (m) x 625 (assume each core represents an area of 25m x 25m). Gets the volume of the core sample material. Why 25 x 25? This is where I am having an issue. How can we know what are the standards he for the drillbit etc? Where and how does 25 meters x 25 meters come in here? HELP.....LOL
The rest I understand.............................
2) multiple volume above x 3.5 (tonnes per m3) .I assume this is to calculate tonnes from the volume calculated above?
3) multiply tonnes x 2204 to achieve lbs. of Ni containing rock. Gets you to LBS from Tonnes
4) multiply (lbs. Ni containing rock) x (avg. Ni grade) to obtain lbs. Ni per core sample. Gives LBS.Ni. per core sample
5) add all lbs Ni. from all core data, to obtain total lbs. Adding up the Nickel per core gets you total Ni for a given set of drill samples/cores in any given area
I did one old drill hole just for fun to see if the spreadsheet was correct. I can't see how this can be right? Seems like a lot of Ni. for a hole thats 72 feet long and has average 0.228% Ni??
Drill Hole |
From (M) |
To (M) |
Core Length (M) |
Core Length (FT) |
Average Ni (%) |
Volume |
Tons |
LBS |
LBS Ni / Core |
LN05-10 |
59 |
81 |
22 |
72.18 |
0.228 |
13750 |
48125 |
106,067,500 |
241,834 |