Thank you for your post Bow, I am continuing to learn more each day. For those that aren’t familiar with the term scaling factor, (as I am, and had to look it up) it refers to “r”, a “classification factor” pertinent to an arbitrary mathematical value assigned to the probability for recovery of metal ores. (see original post by Bow2U)

I apologize for posting values and not being familiar with mining terms and methods, and only able to use simple math. Nevertheless it is important to try and ascertain the relative size of the red center of the anomaly as it represents the high conductivity values and reflects the very high grade surface veins (1-3+ounces per ton) Again the drills will confirm the levels.

Using a weight per ton of 2.5 rather than 2 and then multiply x .32 scaling factor

For the strawberry coloured volume, using 2.5 x 137.8m3= 344.5 million x .32 and this calculation is 110 million tonnes

The pink volume is, using 2.5 x 116 = 290 million tonnesx .32 = 92.8 million tonnes

Once the cross sections are published then a more accurate value of the volumes can quickly be determined. In the meantime I think using the model gives an insight to the volume of the of the higher grade ore depicted in the website picture.

I think it is obvious that one has to assign a substantially higher value for the strawberry portion, because there is such high grade surface veins with known values. One might assign a one ounce per tonne value for the 110 million tonnes, for example.