RE:RE:Jay Taylor on NFG
To produce 12 tons of gold per year, you would need 240 tons of ore per year if the ore contains 5 grams of gold per ton. This is because 12 x 20 = 240.
This is really just a math's exercise, with a little twist, in that you need to account for recovery. No matter what process you use to recover the gold from the ore, you will never achieve 100% recovery, therefore you must allow for it in your calculation.
Also, for simplicity I am going to assume that you meant 12 tonnes (i.e. metric) of gold.
12 t = 12,000,000 g
Assuming that your recovery is 90% and your feed grade is 5 g/t, then you will recover 4.5 grams from every tonne mined.
So your calculation is:
12,000,000 / 4.5 = 2,666,667 tonnes of ore must be mined.
You haven’t said what type of deposit you are mining, or whether it is open pit or underground, but you will most likely need to allow for mining waste in order to access the ore, either striping overburden or driving tunnels in waste rock to access the ore. So you will need to move considerably more tonnes than 2.7 Mt. Of course 2.7 Mtpa underground mine is a significantly big mine. Even 2.7 Mtpa surface mine is a big mine!
The usual way to approach the selection of a sustainable mining rate is to determine the total reserve and then apply something known as Taylor’s Rule.
The format for Gold & Silver Open Pits is t = a.Reserve^b
Where
t = tonnes per day
a = 0.416
b = 0.5874
So, for example if you have a reserve of 4 million tonnes the calculation would be:
t = 0.416 x 4,000,000 ^0.5874 = 3,141 tpd
Assuming 350 operating days per annum your annual ore mining/processing rate should be around 1.1 Mtpa. ( 3,141 x 350 = 1,099,350).
Since this company refuses to put a reserve number on any portion of the deposit, the calculations are useless, meaningless, and misleading.