Resource CalculationI posted this on another site in response to a question and thought I would ask for comments/critique/editorial on my calculations.
Thanks
In researching other mining results/resource calculations and even the Chet/Arequipa/Barrick deal for the Pierina Gold Deposit in Peru, I have an extremely hard time calculating LESS THAN 10MM oz of gold.
I would appreciate someone else critiquing my result as I present them here.
First some premises must be made, so I looked at the Arequipa/Barrick deal back in '96 (as well as other recent smaller deposits/deals). This development was a 270 km land deal. The reports when Barrick purchased it stated "it had no proven or probable reserves" but resources in '96 were set a 3.5 M oz. with a potential of 10M oz. A subsequent 1999 Deposit Summary (from a metalurgist for Barrick) calculated "220 tonne" of contained gold based upon a grade of only 1.92 g/t which yields 6.5M to 7M oz. A prophetic buy by Barrick.
Not knowing the exact geological composition of the earth (Arequipa or Africa), I was extremely conservative in calculating the density of the earthen material - thus I used only 2 tonne per cu. meter - for typical comparisons - gravel is 1.92t/cu.m; granite 2.72; iron ore 4.
Back calculating Arequipa's 270 km area, resulted in an approx. dig depth of 208 meters (reasonable) which gets to the metalurgist report that there was 112.5 M tonne of resource material - yielding, at 1.92 g/t, the 220 t of Au (which is 7M oz).
Utilizing this similar process, Sabodola is a 230km project with grade results way above the 4g/t that I am using here in my calculations. I also figure a depth of 200m.
This results in 46M cu.meters - 92M tonnes - of resource material and thus 12.9M oz at a grade of 4 g/t. At 2 g/t it gives 6.5M oz.
Granted this makes some assumptions, and not all (100%) of the site is resourceable, BUT, the average grade across all of the drills is way, way,way over 4 g/t.
Appreciate any comments, corrections, editorials.
hoosier1