Crude Estimate with MANY AssumptionsLike the heading of my post suggests, this is a very crude estimate that I've made with very large assumptions, there are only 2 holes and we don't even know the grade of the second hole so the second hole doesn't even tell us all that much. A third hole either stepped back 30 meters at the same angle as the first hole, or moved along strike some distance will give way more info as we'd have a third dimension.
I briefly plotted the holes in cad software to give me a good picture of what things are looking like.
The picture of mineralization as we see it now is a 2 dimensional triangle with the narrowest corner closest to surface and the mineralization widening at the base of the triangle. A hole drilled at 90 degrees to the ground through the Apex of this triangle down to the same level as mineralization ends on the first hole would encounter approx. 100 meters of mineralization. Extrapolating the width at the bottom of the triangle is approx 50 meters (fairly close to being a right triangle). This gives us a triangular prism with height 100 meters and width 50 meters and length 400 meters.
Total Volume = 1 million m3
1 million m3 * a Specific Gravity of 4 (which is almost guaranteed to be higher than actual) would give 4 million tonnes.
The neighbouring deposits are small with Liberty's at 0.275 million tonnes and McWaters at close to half a million tonnes ore. I'd be surprised if my estimate was ever matched or surpassed just from the history of the area as they are generally small deposits.