Interpreting drill results;

To summarize the math involved:
1) Start out with the length of the intersection of the mineralization.
2) Determine/guess the percentage of the intersection which is the true width.
3) Cube the true width of the intersection to calculate the estimated mineralized volume.
4) Determine/guess the rock density (normally 10 to 15 cu ft per ton).
5) Multiply the mineralized volume by the rock density to compute estimated tonnage.
6) Finally multiply the tonnage by the grade to determine the estimated resource.

Some useful numbers for performing these calculations :
1 meter = 3.283 feet
1 foot = 0.305 meters
1 troy ounce = 31.1 grams
1 gram = 0.032 troy ounces
1 cu yd = 27 cu ft
1 cu m = 35.4 cu ft
1 mton (tonne) = 1.1 short tons

I do not profess to have any expertise in this field at all. The leaders will take some time to assess the results. It would appear that the drills are angled to discover the grades and it appears to me that the rock has more than viable grades to the depth drilled. This means that the material removed to get down to the anomaly will be profitable. Since the grades tend to get better with depth, if the massive anomaly comes in with an average of 4 gms per ton, then the 105 million tons in the anomaly would be at least 4 times larger than the 1 gm per ton conservative estimate, would it not?