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I am trying to make sense of the mining process of minerals or metals.

  1. I found two measures related to waste: the stripping ratio (which is usually larger than 3:1 for copper, i.e. 3 tonnes of waste for one tonne of copper ore) and the ore grade (which is <5% for copper, i.e. out of a tonne of copper, you can extract only 50kg copper). Is my interpretation correct?
  2. How do they stack up in the mining process? Using a 1% ore grade and 3:1 stripping ratio, does that mean that to get 1 tonne of copper, we need to move 400 tonnes of earth, as we need 100 copper ore that comes along with 300 tonnes of waste?
  3. Does the same logic apply to rare earth mining as well? I found the following numbers for the Bayan Obo mining district in China (see [1]):
    • Neodymium content: 18.5%
    • REO(rare earth oxide) grade: 4.1%
    • Recovery rate: 50%
    • Tonne mined/tonne REO: 49

So according to my understanding, to get one tonne of neodymium, we need 49/(18.5%*4.1%*50%) = 12920 tonnes. Is that correct? That would be a lot of mining for a tonne of neodymium.

Ultimately, I am interested in the amount of earth that has to be moved/mined to get one tonne of copper/other metals (which of course depend of course on the mine and the ore grade).

Thanks a lot :-). I am trying to make sense of the numbers to get a better understanding of the footprint of certain technologies (e.g. given that there are 2 tonnes of neodymium in a modern wind generator, we need to remove 26,000 tonnes of earth to get the neodymium for this (if my numbers are correct).

Best regards, ProGeologist.

[1] Rare Earth mining: https://ui.adsabs.harvard.edu/abs/2013JOM....65j1327T/abstract - tables on page 1329 and 1331

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  • $\begingroup$ This could vary from mine to mine (and perhaps locally, within one mine, too). Imagine the ore be covered by just few meters of sand vs. an other spot where the same layer of veins is covered by a dozen of meters already because today's arrangement of the layers do not need to be parallel to the surface. Which still may be independent to the content (both in terms of amount, as well as in concentration) of the ore within the very layer. $\endgroup$ – Buttonwood Apr 3 at 17:24
  • $\begingroup$ Thanks for the comment. I am more interested in the "finding the removed earth given ore and stripping ratio" than the actual ore and stripping ratio. So assuming I have a fixed ore and stripping ratio, would my calculation be correct? Or am I missing something? $\endgroup$ – ProGeologist Apr 3 at 23:40
  • $\begingroup$ Remember that in some cases you're extracting more than one metal from a given mass of ore. This is especially important for the rare earths, because you're getting the entire series. $\endgroup$ – Gimelist Apr 5 at 4:08
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Your questions regarding stripping ratio only apply to open pit mines. They produce much more waste rock than underground mines. Typically they mine more waste rock than ore.

Underground mines are more surgical and depending on circumstances some of the waste rock they produce is used as backfill in mined out stopes.

the stripping ratio (which is usually larger than 3:1 for copper, i.e. 3 tonnes of waste for one tonne of copper ore) and the ore grade (which is <5% for copper, i.e. out of a tonne of copper, you can extract only 50kg copper). Is my interpretation correct?

You are partially correct. In this scenario the amount of copper that would be delivered to the processing plant (also known as a mill or concentrator) in mined ore is 50 kg, i.e the amount of metal in the mill feed is 50 kg.

To get usable copper metal from the ore, after mining, it needs to treated in a processing plant to concentrate the copper bearing minerals and to remove the gangue minerals. The copper concentrate is then sent to a smelter to produce copper metal. The waste from the concentrator is sent to the tailings dam/pond/storage-facility.

If you account for the recoveries of these two processes, typically 80% for concentrating and 90% for smelting, the amount of copper metal produced from 50 kg of copper in mill feed would be 36 kg (50 x 0.8 x 0.9).

How do they stack up in the mining process? Using a 1% ore grade and 3:1 stripping ratio, does that mean that to get 1 tonne of copper, we need to move 400 tonnes of earth, as we need 100 tonnes copper ore that comes along with 300 tonnes of waste?

Yes. Such a scenario would give you 1 tonne of copper in mill feed.

Unfortunately, the paper you refer to was most likely written by primary metallurgists, or possibly chemical engineers, and not mining engineers. The emphasis of the authors is on the amount of metal recovered from mining and the amount of energy required to produce usable metal from various mining projects, not the overall quantities of material mined.

Regarding your question about the Bayan Obo mining district in China your calculations are not correct. Also, from the paper you refer to you cannot obtain the total amount of earth mined, ore and waste, because there is no reference to stripping ratio.

As you state, from Table I of your reference, the proportion of neodymium in REM (REM being rare earth metal)is 18.5%. Table II states Tonnes mined/Tonnes REM is 59. This usually refers to the tonnes of ore mined and supplied to the mill, not the total tonnes of earth mined (ore and waste). Additionally, as stated in Table II and reiterated in Table IX, the recovery from the primary processing plant is 50%.

Now, 1 tonne of REM supplied to the mill contains 0.185 t (185 kg) of neodymium. With 59 tonnes of ore mined per tonne of REM, the amount of ore mined and supplied to the mill to get 1 tonne of neodymium is 59/0.185 = 318.9 tonnes.

The mill only recovers 50% of the metal supplied to it. Therefore, for the mill to recover 1 tonne of neodymium 637.8 (318.9/0.5) tonnes of ore must be supplied to the mill.

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  • $\begingroup$ Perfect, that makes sense. Thank you very much. Just one quick question: Where do the 0.0925 come from? Is this the amount of neodymium we get out of 638 tonnes or Rare Earth mining (excluding the stripping rate of course)? $\endgroup$ – ProGeologist Apr 4 at 19:15
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You need to be specific. In the US , I think copper content of ore is typically 0.03 %. Overburden is irrelevant as the open pit mines do not open new areas. In some areas of the old Belgian Congo ( Zaire ?) copper content is 4 % on the surface of rich ore bodies. They are not generally mined because of political instability. I searched Bagdad ( AZ ) copper mine and found various numbers for copper content. The 0.03 % is probably the poorest ore ; better deposits may be up to 0.5 %.

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  • $\begingroup$ Thanks for the comment. I am more interested in the "finding the removed earth given ore grade and stripping ratio" than the actual ore and stripping ratios. So assuming I have a fixed ore grade and stripping ratio, would my calculation be correct? Or am I missing something? $\endgroup$ – ProGeologist Apr 3 at 23:41
  • $\begingroup$ The Bagdad mine apparently just digs down with no new stripping . In Zaire , there is no stripping as good ore is at the surface. $\endgroup$ – blacksmith37 Apr 4 at 1:26
  • $\begingroup$ Belgian Congo? Zaire? Do you mean, DRC? $\endgroup$ – Gimelist Apr 5 at 4:08
  • $\begingroup$ I hate when the world keeps changing. I became familiar with it in about 1972 when it was Zaire ; Now it is DRC. $\endgroup$ – blacksmith37 Apr 6 at 18:37

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