RE:A math calculation to blow your mind.Hello Jefe,
Just wanted to "paraphase" what you discussed below, starting with a formula that most people, including Eric Sprott, would use, along with some discussion of key parameters. Eric's wrap-up is in link below, starting @ ~10:00min for WM discussion (Eric' napkin cals: 400mx500x200x2.7 = 100 M tonnes rock x 1gpt/31 = ~ 3 M ozAu, (6.6 MozAu for 2gpt, etc..).
https://www.youtube.com/watch?v=LXndtIFXczA
Resource Estimate (RE) Formula: L(2,500m strike) x D (400m) x W (200m) x sg (2.7) x grade (1 gpt) /31 g/oz The math results for 1gpt uniformly distributed across: 540M tonnes of rock x 1gpt/31 =
17 M ozAu (which is almost the same as 3M ozAu x 6 =
18M ozAu to account for the strike length and small difference in depth. In other words, the 17M ozAu is consistent with Eric's napkin calculation).
1. Strike Length: 2.5km = 2500m, this is optimistic since the volume of the ore deposit is assumed to be fully mineralized at the same uniform grade, asay at 1 gpt, as an example. (Eric used 400m strike length, i.e. a factor of 6 lower than 2.5km strike length)
2. Depth: 400m is a reasonble number since they have drill results to that depth (Note: Eric used 500m depth)
3. Width: Same as #2 above (got drill results to back up the 200m value)
4. s.g. of rock (2.7): basic info on gold-bearing rock at reasonable grade (bonanza grade would change everything)
5. Grade: 1gpt is just an example used as a base, which can be scaled to some other value (suggest scale UP, since the last grade for the bulk samples was around 20gpt). Grade is also expected to increase with depth.
In summary, on the above list,
- #4 is the basic rock fact, and the variation is not significant;
- #2 and #3 are to be proven by the drill bits;
- #1 and 5 seem to be key left over parameters. Strike length could increase another 3km to the northern boundary (Hi Ho Silver would be happy) . But, let's stick with 2.5km strike length and assume some kind of waste rock/ore ratio, say 10:1. This would bring down the RE by a factor of 10 (to 1.7M ozAu). However on the other hand, grade could be a much more important factor. Grade of 10gpt would bring the RE back up by a factor of 10, to 17MozAu, (or a factor of 20, to 34MozAu, if grade is proven to be 20gpt). In other words, #1 and #5 could cancel each other out, but there could be surprises as well.
Let's just assume, arbitrarily, that the RE = 5MozAu. At a production of 200,000 oz/yr (LOM: 25 yrs) and a net profit of ~C$400/oz would bring in C$80M profit per year, which is not shabby (= $20M/yr profit for Eric with his 25% of the company, or $16M with20%). Note: It's the same situation, we just looked at it from a different perspective. It's mind boggling, even with 5MozAu, but it would seem a reasonable situation.
Cheers,
GH11
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jefedeoro wrote: So I watched Sprott's weekly update and was thinking about his comments on the size of the potential ore body.
we know the strike length is 2.5 kilometers.
we know its 200 meters wide and currently 400 meters deep.
So the billion dollar question is how much of the 2.5km corridor will be gold bearing and what grade?
if we know those items we can do an easy calulation of total gold. Of course the problem is we dont know. However we can do a 'scenario' .
so lets say only half of the 2.5 km is gold bearing, (1.25 km) and it continues to stay 500 meters deep, and 200 m wide. And lets say the average gold is low, only 5 grams per ton.
with that very low esitmate lets calculate.
1250 m length X 200 m wide X 500 meters deep = 125 million cubic meters of ore
although rock density does vary, lets use a gold bearing rock called "Gabbro". Given this is what the main deposit is called "Main Gabbro zones" in Wallbridge's photo from the last press release. see link.
https://www.wallbridgemining.com/i/maps/20190423-Fig-1-Fenelon-Plan-final-2.pdf
therefore a simple internet search reveals that ingnaous rock, specifically Gabbro, has a density of 2.7 to 3.5 tones per cubic meter. see link.
https://www.google.ca/search?ei=DJPDXPulEoXIsQXW65eADw&q=density+of+igneous+rocks&oq=density+of+ig&gs_l=psy-ab.1.0.0j0i22i30l9.39869.45957..49011...0.0..0.287.2605.0j17j1......0....1..gws-wiz.......0i71j0i70i251.uqZEPXMAJac
so lets split the differnce and call it 3 tonnes per cubic meter.
so now that we have all the needed information for our modest calculation.
we have 125 million cubic meters. With 3 tones for each cubic meter, that makes
375 million tones of rock. Now we times that by our (low) estimate of 5 g/t
and we get 1,875,000,000 (one Billion 875 million grams of gold)
given there are 30 grams per oz, that translates to 62,499,999 ozs of gold.
times that by gold price of $ 1300.00 per oz, and we get $ 81,250,000,000
I had to triple check this number as I did not believe it.
Its 81 Billion 250 million dollars.
so to conclude, if even half of the strike zone is only 5 grams per tone, we are sitting on over
80 billion in revenue.
Minus all the costs of running a mine, and you are left with.....
(I really dont know, but lets be super generous and just say half.)
that means we are left with 40 Billion dollars ----- divided by current market cap of
450,000,000 shares, and that gives you 80 dollars for a share price.
Now this is all TOTALLY hypothetical, but I also dont think it is a huge stretch that half the strike zone could average 5 grams per ton.
so everytime they drill across that strike zone, and come up with gold hits all the way across, we get another step closer, and we only need HALF of it to produce gold.
SO, after I made these calculations, my mind was litterally blown. I am almost conviced the math is wrong somewhere, so if anyone can correct me, or point out any errors, please do.
I'll be sitting here drooling. :)
Jefe de Oro