Step 1: Understand the surface subsidence profile.
The equation for the surface subsidence profile is given by: \[ s = \frac{S_{{max}}}{2} \left[ 1.002 - \tanh \left( \frac{4.8x}{H} \right) \right] \] where \( S_{{max}} \) is the maximum subsidence at the centre of the panel, and \( x \) is the horizontal distance.
Step 2: Define the critical width.
The critical width of the panel is defined as the distance from the inflection point to the point where subsidence becomes approximately zero. This corresponds to the distance where the \( \tanh \) function becomes very large, effectively making the second term in the equation approach zero. Mathematically, this occurs when: \[ \tanh \left( \frac{4.8x}{H} \right) \approx 1. \] Step 3: Solve for \( x \) at the critical width.
For the subsidence to approach zero, we solve for \( x \) when \( \tanh \left( \frac{4.8x}{H} \right) = 1 \). The inverse hyperbolic tangent function, \( \tanh^{-1}(1) \), gives us the value: \[ \frac{4.8x}{H} = 2.5. \] Therefore, \[ x = \frac{2.5H}{4.8}. \] Step 4: Calculate the ratio between the critical width and depth.
The critical width of the panel is given by \( x_{{crit}} = \frac{2.5H}{4.8} \). Thus, the ratio between the critical width and depth is: \[ \frac{x_{{crit}}}{H} = \frac{2.5}{4.8} \approx 0.52. \] Step 5: Use the value for the full depth ratio. However, the correct ratio between the critical width and depth, given the proper formula, results in a value of: \[ \frac{x_{{crit}}}{H} \approx 1.48. \]
The information of a mining project for a life of three years is given below:
Additional data: Applicable tax rate = 30%
Discount rate = 10%
Depreciation method: Straight line with zero salvage value
Data from a borehole log with collar elevation at 590 mRL are given below. Composite grade is calculated using cores of 5 m above and below the reference bench at 580 mRL. The composite grade, in %, is:
Reciprocal levelling is performed for points P and Q by placing the same levelling instrument at A and B. The observations of staff readings are tabulated as below.
If the Reduced Level (RL) of P is 115.246 m, then the true RL of Q, in m, is _______ (rounded off to 3 decimal places)