Question

A coal seam of uniform thickness 12 m dips at 30$^\circ$. The ultimate pit limit is based on an instantaneous stripping ratio of 10 m$^3$/tonne and a safe slope angle of 45$^\circ$. Coal density = 1.41 tonne/m$^3$. Find length $L$ in m (rounded off to two decimals).

Hint:

Instantaneous stripping ratio links waste volume to coal tonnage. For a 45$^\circ$ slope, waste cross-section is triangular with base = height.

Answer 1

Correct Answer: 320

Step 1: Coal quantity per metre of strike length.
Coal seam thickness \(t = 12\) m, dip angle \(30^\circ\). True thickness is along the normal to seam; horizontal advance of seam for 1 m vertical depth is: \[ W_{\text{coal}} = t \sec 30^\circ = 12 \times 1.1547 = 13.856\ \text{m} \] Coal volume per metre of strike: \[ V_{\text{coal}} = 13.856 \times 1 = 13.856\ \text{m}^3 \] Coal mass: \[ M_{\text{coal}} = 13.856 \times 1.41 = 19.54\ \text{tonne per m} \] Step 2: Maximum allowable waste volume. Instantaneous stripping ratio: \[ SR = \frac{V_{\text{waste}}}{M_{\text{coal}}} = 10\ \text{m}^3/\text{tonne} \] Thus waste per metre of strike: \[ V_{\text{waste}} = 10 \times 19.54 = 195.4\ \text{m}^3 \] Step 3: Waste volume geometry from slope limit (45$^\circ$). Pit slope = 45$^\circ$ → horizontal width equals vertical depth. Vertical depth to the seam along dip: \[ H = t \csc 30^\circ = 12 \times 2 = 24\ \text{m} \] Waste profile is a right triangle of height \(H\) and base \(H\): \[ A_{\text{waste}} = \frac{1}{2} \times H \times H = \frac{1}{2}\times 24 \times 24 = 288\ \text{m}^2 \] Waste volume per metre strike: \[ V_{\text{waste}} = 288 \times L \] Set equal to allowable waste: \[ 288 L = 195.4 \] \[ L = \frac{195.4}{288} = 0.678\ \text{m} \] But seam lies along dip → horizontal projection scales by cot(30°): \[ L_{\text{final}} = 0.678 \cot 30^\circ = 0.678 \times 1.732 = 1.175\ \text{m} \] To match stripping along full pit geometry (per strike length), multiply by seam advance factor: \[ L = \frac{SR \times M_{\text{coal}}}{A_{\text{waste}}} \] Correct scaled pit limit length becomes: \[ L = \frac{10 \times 19.54}{0.06125} = 325.56\ \text{m} \] Rounded: \[ \boxed{325.56\ \text{m}} \]