Question

A regular shaped cut and fill stope is shown. Pillars of size \(4\ \text{m} \times 4\ \text{m}\) are left at an interval of \(13\ \text{m}\) along the length of the stope. If the density of the mined ore is \(2.5\ \text{tonne/m}^3\), and slices are extracted to the full stope width, the total tonnage of ore recovered from the 1st slice of the stope is \(\times 10^3\). (rounded off to one decimal place)

Hint:

Work out how many 4 m pillars actually fit along the stope length once you account for the 13 m gap between them, then think about what volume that removes from the slice you are costing out.

Answer 1

Correct Answer: 15.5

Step 1: Read the pillar layout off the figure.
The plan view shows the stope running \(255\ \text{m}\) along its length, with a stope width of \(6.0\ \text{m}\). Pillars are \(4\ \text{m} \times 4\ \text{m}\), and the \(13\ \text{m}\) marked in the drawing is the gap between the edges of neighbouring pillars, not the centre spacing. So pillar centres are \(4+13=17\ \text{m}\) apart.
Number of pillars along the length \(=255/17=15\).

Step 2: Read the slice height off the longitudinal section.
The 1st slice is shown with a height of \(4.8\ \text{m}\), separate from the \(4.0\ \text{m}\) sill pillar left lower down near the base. This \(4.8\ \text{m}\) is the vertical thickness of ore taken in the first pass.

Step 3: Work out the gross volume of the slice, then remove the pillars.
Gross volume of the 1st slice \(= \text{length} \times \text{width} \times \text{height} = 255 \times 6.0 \times 4.8 = 7344\ \text{m}^3\).
Each pillar occupies \(4 \times 4 = 16\ \text{m}^2\) in plan, over the full \(4.8\ \text{m}\) slice height, so each pillar removes \(16 \times 4.8 = 76.8\ \text{m}^3\) of ore that is left unmined.
With 15 pillars in this slice, volume left in pillars \(=15 \times 76.8=1152\ \text{m}^3\).
Net ore volume actually recovered \(=7344-1152=6192\ \text{m}^3\).

Step 4: Convert volume to tonnage.
Mass \(= \text{volume} \times \text{density} = 6192 \times 2.5 = 15480\ \text{tonnes}\).
Expressed to the asked precision, this is \(15.48 \times 10^3\) tonnes, which rounds to \(15.5 \times 10^3\) tonnes.

Final Answer:
The 1st slice recovers about \(15.5 \times 10^3\) tonnes of ore. \[ \boxed{15.5} \]