It's not true width of course but consider the blue line I've...

It's not true width of course but consider the blue line I've drawn that follows the plunge of the high grade ore to the south.

You have pierce points along this line in the plane of [email protected]/t, [email protected]/t and now 11m@54g/t and the line is about 85m long from the furthest north pierce point, to furthest south pierce point.

Assume the gold is contained in a elliptical pod along this plunge line with a major axis of 85m, minor axis 8m (estimating a little for true width) and vertical axis of 16m (ie the ore extends vertically for twice its true width at the "tallest" point of the ellipsoid) you get approx. 5,700 m3 of ore, say grading at an average of 28g/t (if my calculation is correct). Using the density of a mafic rock like basalt of 2.9t/m3 you get ~16,500 tonnes at 28g/t or 462,000 grams of contained gold which is about 14,900ozs, worth ~$28 million in the ground at today's gold price and that's just this small section of the deposit.

I can make even more conservative assumptions but the above gives an idea of what this intersections could be worth.

I'd be starting to get excited and I'd start accumulating if those grades extend much further down plunge.

That deepest unassayed hole is slightly shallower than the plunge line I've drawn but if it hits similar g x m as either of the other 3 intersections on that line it opens the plunge of the mineralisation up by another 140m. If you then assume the gold is also continuous you could scale up the value above by at least 2 times. You could then be talking an in ground value of gold based on my estimate of $60 million based on only those 4 holes. DYOR IMO. Esh