Sculpin, I was reviewing lots of posts tonight, and I like this one you did from last night. I cut & paste the last sentence of your post:

One gram of gold is not very big, and all you have to see is one tiny visable piece in the core every 1 m to give you the 1 g/t that Bow uses in his calculations.

Just really imagine how small this would be?? 1g/T (and 1m3 would be 2.5T approx)

So, in a 1M X 1M X 1M (1M3) you'd have 2.5g of gold

1g/T = 1g/1,000kg = 1g/1,000,000g = not very much!

Now think of only 1 of our main anomallies. Only the 1 anomally (1800M X 1700M X 1300M, and possibly deeper, but who cares)

Now the unknown, is what do you use for a scaling factor? Do you use the Ovoid Formula, do you scale it like G & S report, which was 68%?? I like the G & S best, due to the fact that it is the most conservative (most scaling) To scale it by 68%, you must multiply by 0.32

I know we've seen this all before, but lately, I've had many PM's about Math on the Anomally.

So, here is what I estimate we might have in only the 1 anomally only @ 1g/T, scaling it by 68%

1800M X 1700M X 1300M X 2.5T/M3 X 1g/T 1OZ/31.1g X 0.32 = 102 Million OZ

We all know that seems high, but, remember, we put in some pretty conservative numbers!

Math is Math, and that's only 1 anomally!

BOW2U