re: Ann: Nabanga Gold Project - Extended High... FGSTR, From the Nabanga report,
L(length)=2300m
Average down-hole intersection is 3.90m, but the drill is at 60 deg and the ore body is not vertical. From the cross section in the report, the ratio of width W ( at a right angle across the ore body) to down hole intersection is about 8 to 15, i.e. 0.533333, so W=3.90x0.533333=1.95m.
Reported depth 200m. – so taking the ore gradient as about -75 deg, the distance down along the ore body is 200/sin(75 deg) = 200/.965926 . So D=207m
Density taken as 2.65 t/cubic m
Grade – average grade reported as 6.27 g/t
The answer comes out in g so need to multiply by 0.035273 to convert to Oz
The 20% reduction for the gaps in the ore body along strike is a guess from the diagram in the report. A more accurate estimate could come out larger.
The calculation is very rough but what else can you do with the data provided? Note: the number of decimal points in the answer is not an indication of the level of accuracy. Use one or at most two significant figures. i.e., my estimate is about 0.5 million Oz.
Open to suggestions/discussion. What is your estimate?
re: Ann: Nabanga Gold Project - Extended High... FGSTR, From the...
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