The Mining Play

Auryn/Medinah - 2022 - 2nd Half General Discussion

Thats why DOC brings down his estimate from 60gpt to 30gpt.

And you saying the grades thus far do not support the opinion… is true but they also dont support your opinion. We haven’t assayed the actual DL vein. Stay balanced my friend.

I am just hoping before December shareholder meeting that we will have good news that supports hitting DL vein and high grades
Maybe the 3rd quarter update will give us a lot of this info and if so I might travel to Santiago if great news comes out


How many tonnes per day did they produce? 2 or 3?

Just for perspective:
This is exactly 1 one ton of river rock. About the same weight as our ore I would speculate. This is washed and sorted- sized rock, verses ore blasted from hard rock. Sized rock =less air space then larger chunks of ore.
So size may vary a bit. But not that much one would notice.

Now if we had a couple hundred loads of that at the mill.
They could be writing us a fat check :writing_hand: .
:index_pointing_at_the_viewer: so lets get at it.

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And this is what a yard of material looks like.

A cubic yard of rock, or anything else, is 3 ft x3 ft x 3ft.or 27 cubic feet. In Chile they use the metric system, cubic metres and metric tons or tonnes. A cubic metre is 35.3145 cubic feet. The photo shows a pile of rock containing multiple cubic metres.

Right . That picture shows how much material a try axle dump truck holds. Roughly 16 yards of material depending on material. And ours is rated at 20 ton.

Yes, but you stated that that was a picture of a cubic yard of material-I assume in error…

For what it’s worth; [(5X17X17= 1445 cu ft) / 27 cu ft per cu yd] = 53.5 cu yd :confused:

Assuming the 5 ft dimension in the photo is accurate the 17 ft is incorrect. If that is actually actually a 16 cu yd pile, the other linear dimensions are 9.3 ft.
20 tons is a lot of rock to haul! :smile:


Easy You are assuming that the pile is rectangular, which it isn’t. It is more conical or pyramidal in shape-different formulae to calculate the volume. I know of no mine using 20 t haul trucks. Minera Escondida and Chuquicamata and many more use 400 tonners.

Yes Carlos ,My Mistake. Did Not Want To Lead Anyone Astray.
That’s Ol Less’s Job.


CF, you are quite correct. From the information given, the 17X17 ft dimensions are of little use in approximating the volume. A more useful dimension if assuming a conical shape would be the radius, which in the instance of a 5 ft height must be equal to 9.1 ft, not much different from the incorrect square thinking cubical approximation that also gives the 16 cu ft answer. Approximations can use different models without too much complexity. Can you tell me how one would use the 17 X 17 ft dimension given in any model determining the volume? (A pyramid model approximation would not give the desired 16 cu yd answer.)

I’ll take a wild-ass guess and say that there may be a problem with overloading.

Why is the 16 cu yd figure of importance. You are assuming that it is 16 yards? The volume must be calculated, not wished for. The pile shown is an irregular, neither a cone or a pyramid. True dimensions are required to calculate the exact volume. If very irregular one would have to measure the cross sectional area at several points and calculate the volumes of these using one of several different formulae. For rough results, use the Average End Area, for more exact ones, use the Prismoidal Formula.

CS stated the pile is 16 cu yd, the pile in no way appears to be prism shape. Use whichever model is the simplest given the facts that are given. (Prism modeling yields 17.8 cu ft using the 17 ft dimensions and 5 ft height given.) I merely presented the simplest mathematical model most non engineers or math majors would be quite familiar with. Complex shapes are best modeled with greater dimensional information and measurements. Why make a mountain out of a mole hole? :rofl:

I have no doubt our mountain is much more than a mole hill and will require some very complex mathematical modeling with more detailed measurements than I care to think about! :laughing:


Not sure what is the prism model that you are referring to. I said the PRISMOIDAL Formula.

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Sorry Carlos,
Now you are being ridiculous! With the information that is given, CS states as a fact (not an assumption) the volume is 16 cu yds. The diagram has only 3 dimensions given to present a model that arrives at a volume of 16 cubic yds. I’ve already stated that a prism model using the measurement information presented will not give you volume of 16 cu yds and is a poor and inappropriate model to use.

The prismoidal model requires many more measurements than the 3 that are given:


Apologies if my two representations do not display properly as I’m away on vacation and only have my cell phone to post a reply. Let’s give it a rest and state the only assumption that I’m making is that CS showed an image representing 16 cu yds. material which is less than the load capacity of AURYN’s truck to haul ore.
Have a nice day everyone and enjoy the weekend!

(Being a former math teacher) … Closest approximation would probably be conical :

Diameter = 17 ft.
Radius = 17 ft / 2 = 8.5 ft = 2.833333333 yds

h = 5 ft. 1.666666667 yds

cone volume formula : volume = (1/3) * π * r² * h

V = 3.1415926535 / 3 * 2.833333333 * 2.833333333 * 1.666666667 = 14.01111538 cubic yards.

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Once again, I said that the Prismoidal Formula (NOT THE PRISM FORMULA) would work. Google it. Obviously there is not enough information in the photo to calculate this with any real accuracy using any formula. With the prismoidal formula the volume is divided into several sections, the volumes of these sections are calculated and then summed to give the total volume.