1. Surface Mining Ore Body Geometry ©2011 Dr. B. C. Paul Note- Concepts given here are considered to be common knowledge to those involved in early stage ore estimation. Geometric concepts used are found in most books on geometry. Acknowledgement is given to SME for model P.E. exams that guided the choice of some materials chosen for review.

2. Need to Define a Size of an Orebody In more advanced stages of mining orebodies are often modeled as millions of tiny blocks that are analyzed by computer In more advanced stages of mining orebodies are often modeled as millions of tiny blocks that are analyzed by computer In early stages approximate magnitude of orebody size is often determined by comparison to basic geometric objects. In early stages approximate magnitude of orebody size is often determined by comparison to basic geometric objects.

3. Example – The Coal Seam Coalstrip Coalmining Company has obtained the rights to 6 sections of coal bearing land measuring 10,560 ft X 15,840 ft. The coal seam is approximately horizontal and is about 7 feet thick on average. It is bituminous coal with a typical density of 80 lbs/cubic foot. If the company wants a 20 year reserve and will recover 90% of the coal, what mining rate would they use? Coalstrip Coalmining Company has obtained the rights to 6 sections of coal bearing land measuring 10,560 ft X 15,840 ft. The coal seam is approximately horizontal and is about 7 feet thick on average. It is bituminous coal with a typical density of 80 lbs/cubic foot. If the company wants a 20 year reserve and will recover 90% of the coal, what mining rate would they use?

4. Step #1 – Get the Coal volume This is a big box (but not a very tall one) This is a big box (but not a very tall one) Volume for a 3D box is Length*Width*Height Volume for a 3D box is Length*Width*Height Apply 15,840 * 10,560 * 7 = 1,170,892,800 cubic feet Apply 15,840 * 10,560 * 7 = 1,170,892,800 cubic feet

5. Step #2 Convert the Volume to a Weight We know that a cubic foot of coal weighs 80 lbs – therefore the weight is We know that a cubic foot of coal weighs 80 lbs – therefore the weight is –1,170,892,800 * 80 = 93,671,424,000 lbs We also know there are 2000 lbs in a short ton (the kind used in US coal mining) – therefore the weight in tons is We also know there are 2000 lbs in a short ton (the kind used in US coal mining) – therefore the weight in tons is –93,671,424,000 / 2000 = 46,835,712 tons

6. Step #3 Adjust for Recovery Only 90% of the coal in place will actually be recovered by mining Only 90% of the coal in place will actually be recovered by mining –46,835,712 tons * 0.9 = 42,152,141 tons

7. Step #4 Get the Average Mining Rate The reserve will be mined over 20 years time The reserve will be mined over 20 years time 42,152,141 / 20 = 2,107,607 tons/year 42,152,141 / 20 = 2,107,607 tons/year As a practical matter one would say mining rate is about 2.1 million tons per year As a practical matter one would say mining rate is about 2.1 million tons per year

8. Why is this information Important? Early in the definition of a deposit you will need to know how you might mine it Early in the definition of a deposit you will need to know how you might mine it –Knowing how big something is will help you to see what is practical A 2.1 million ton per year surface mine makes sense A 2.1 million ton per year surface mine makes sense A 50,000 ton per year surface coal mine does not make a lot of sense A 50,000 ton per year surface coal mine does not make a lot of sense –Total amount of recoverable coal may give you an idea of how much “profit” you could make. The amount you can invest in the deposit must be less than this amount.

9. Lets take this coal deposit one step further Suppose we assume that the coal will pick up about 7% material that is not really coal (out of seam dilution) Suppose we assume that the coal will pick up about 7% material that is not really coal (out of seam dilution) Suppose we figure the preparation plant will have a yield of 70% Suppose we figure the preparation plant will have a yield of 70% How many tons of clean coal will we market and how big had the coal preparation plant better be. How many tons of clean coal will we market and how big had the coal preparation plant better be.

10. Feed to the Preparation Plant The 42,152,141 tons of coal will be diluted by 7% more out of seam material The 42,152,141 tons of coal will be diluted by 7% more out of seam material –42,152,141 * 1.07 = 45,102,791 tons –Over 20 years average feed per year is 45,102,791 / 20 = 2,255,140 45,102,791 / 20 = 2,255,140 About 2.25 million tons per year About 2.25 million tons per year Incidentally our coal loading equipment at the mine better be able to move that tonnage too. Incidentally our coal loading equipment at the mine better be able to move that tonnage too.

11. Size Up the Prep - Plant About 2,250,000 tons per year About 2,250,000 tons per year A typical coal prep plant will run 250 days per year, 2 shifts per day, with about 7 hours of real run time on a shift A typical coal prep plant will run 250 days per year, 2 shifts per day, with about 7 hours of real run time on a shift –250 * 14 = 3500 operating hour per year Size the Plant Size the Plant –2,250,000 tons/year / 3500 hours/year = 643 tons per hour –About 650 tons/hour

12. More Observations With the scale of operations approximated we can make good guesses on our capital investment and operating costs With the scale of operations approximated we can make good guesses on our capital investment and operating costs –We are set up to do early economic evaluation to determine if we should invest in more exploration and planning.

13. Lets Try a Dipping Vein Next – This might be a phosphate deposit We have popped three holes in the deposit on a line every 30 meters We have popped three holes in the deposit on a line every 30 meters –A top 30 m bottom 56 –B top 68 m bottom 98 –C top 106 m bottom 140 What is the thickness of the vein of layer? What is the thickness of the vein of layer? (Note we cannot really get the dip of the deposit because we would need to know the holes ran perpendicular to the strike of the deposit – which we do not know). (Note we cannot really get the dip of the deposit because we would need to know the holes ran perpendicular to the strike of the deposit – which we do not know).

14. Depth to deposit midpoint A is at 43 meters A is at 43 meters B is at 83 meters B is at 83 meters –It went down 40 meters for 30 meters over C is at 123 meters C is at 123 meters –It went down 40 meters for 30 meters over We are going down 40 meters for 30 over We are going down 40 meters for 30 over 40 30 Arctan(40/30) = 53.1 degrees of Apparent dip

15. Look at Our Drill Hole Intercept Distance Hole A 26 meters Hole A 26 meters Hole B 30 meters Hole B 30 meters Hole C 34 meters Hole C 34 meters Average Intercept is 30 meters Average Intercept is 30 meters But we Hit it like this But we Hit it like this Our intercept is Clearly not the seam thickness

16. Correcting to Get Seam Thickness 53 30*cos(53) = thickness = 18.05 meters About 18 meters

17. A Dipping Seam Tonnage The strike of a phosphate bed runs for 1000 meters The strike of a phosphate bed runs for 1000 meters It dips at 53 degrees for a distance of 500 meters It dips at 53 degrees for a distance of 500 meters It is 18 meters thick It is 18 meters thick The density is 2.45 tonnes/cubic meter The density is 2.45 tonnes/cubic meter

18. Step #1 Get the Volume This time the box is dipping This time the box is dipping 1000 * 500 * 18 = 9,000,000 cubic Meters 1000 * 500 * 18 = 9,000,000 cubic Meters Step #2 – convert to weight Step #2 – convert to weight –9,000,000 cubic meters * 2.45 tonnes/cubic meter –22,050,000 metric tonnes