"I noticed that no one answered my ? statement? question about had anyone figured out how come we had big numbers in the OVM but,sub 2 gram dirt -- which meant to me that - noone had figured it our and/or it didn't suit their bashing agenda."
The only reason to explain this is that the gold in the resource is laterally dispersed.
Take a simple example of a 1mx1mx1m=1m3 block of ore that contains 31.1 grams =1 troy oz of gold sitting at surface.
For simplicity assume the block of ore has a density of 1gram per cubic metre =1g/cc (it would actually be more like 3 g/cc if it was all fresh dolerite) then its grade would also be 31.1 g/t =1oz/t.
This block of ore would contribute 1oz to the overall OVM at its depth (ie in the depth range 0m to 1m).
Now if the same block of ore had one tenth of the grade ie 3.11g/t you would need 10 of these blocks at the same depth to contribute the same 1oz of gold to the OVM.
The OVM's here are high because they have captured large areas into their pit shell models.
At 2g/t and a density of 3g/cc you have about 6 grams per 1m3 or 0.19ozs per 1m3 so for a OVM of 3000oz/m you would need 3000/0.19 = 15789 of these blocks if all blocks have the same grade. That's about an area of ~16,000m2 or 160m x 100m or 1600m x 10m.
Given their cut-off grades are 0.5g/t and the density will probably be a bit lower than 3g/cc, due to weathering in the shallow parts of the deposit and and other rock types, the actual area will be much larger than what I've just calculated but this gives you a rough idea of how it works. The high OVMs are no mystery and aren't calculated on high grade pockets, they are a result of integrating the vertical gold endowment over a large grade range and over large areas. Esh
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