The resource model was divided into a 3D block model framework. The block model has

720,000 blocks that were 5m in the X direction, 5m in the Y direction and 5m in the Z direction.

There were 100 columns (X), 120 rows (Y) and 60 levels (Z). The block model was rotated 20

degrees clockwise. Separate block models were created for rock type, density, percent, Ni, Cu,

Pt, Pd, Au, Ag and NSR.

So Eagle 1 a had more thickness and width then depth according to this. I am not suggesting that we will end up with more thickness and width then depth on lens b and c but if we can maintain the same thickness and width, that would be wonderful!

allow me this calculation again please:

volume = xyz

991237 =100Y x 120Y x 60Y

Y=1.11

Then you would have for Eagle 1 that thicknes is 111m, width is 133 meters and depth is 67meters. Now there is the factor of a 20 degree rotation which I cannot do an interpretation on. However if the thickness and width of lens a is consistent in lens b (and there is absolutly, absolutly nothing to remotly suggest this but I still expect someone to remind of this fact) then volume of lens b would be

178*111*133=2 627 814. This deposit has lower grades and as such would probably have a density closer to 3 tons per meter cubed which would give us...7 883 442

For lens C we would have:

149*111*133=2 199 687. The grades here are better so I would expect density to creep up to 3.1 give or take. That would give us...6 819 030 tons

So all toll, this very unscientific calculation would give us 17.6M tons for lenses a b and c.

More fun math which I plan on getting abused with but I like to know so I figure others may like to know as well!

Glorieux