Question

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:

 

Hint:

To compute composite grades, use a weighted average based on core length. Carefully match the elevation range to include appropriate samples above and below the target level.

Answer 1

Step 1: Determine elevation of each core.
Starting from 590 mRL and moving downward:
Core 1: 590–587 mRL (3 m) → Avg. elev = 588.5 m
Core 2: 587–584 mRL (3 m) → Avg. elev = 585.5 m
Core 3: 584–582 mRL (2 m) → Avg. elev = 583.0 m
Core 4: 582–579 mRL (3 m) → Avg. elev = 580.5 m
Core 5: 579–577 mRL (2 m) → Avg. elev = 578.0 m
Core 6: 577–574 mRL (3 m) → Avg. elev = 575.5 m
Core 7: 574–571 mRL (3 m) → Avg. elev = 572.5 m
We take 5 m above and below the reference bench at 580 mRL:
- 5 m above = from 580 to 585 mRL → Cores 2 and 3 (partial), Core 4
- 5 m below = from 580 to 575 mRL → Core 5 and 6 (fully), Core 4 (shared)
Step 2: Identify relevant cores and their valid lengths:
{Core 2: Full 3 m (grade 42)}
{Core 3: Full 2 m (grade 41)}
{Core 4: Full 3 m (grade 43)}
{Core 5: Full 2 m (grade 42)}
{Core 6: Full 3 m (grade 41)}
Step 3: Compute composite grade (weighted average):
\[ {Total length} = 3 + 2 + 3 + 2 + 3 = 13\, {m} \]
\[ {Weighted grade} = (3 \cdot 42) + (2 \cdot 41) + (3 \cdot 43) + (2 \cdot 42) + (3 \cdot 41) = 126 + 82 + 129 + 84 + 123 = 544 \]
\[ {Composite grade} = \frac{544}{13} = \boxed{41.85 \approx 42.0\%} \]