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Aggregate reserve: sandstone N42W/34, 80 m

Short answer

True thickness t = 80 cos34° ≈ 66.3 m. Area on bed = 2.5 km × 0.5 km = 1.25×10⁶ m². Volume = 1.25×10⁶ m² × 66.3 m = 8.29×10⁷ m³.

The attitude of sandstone is N42°W/34°. The difference of top and bottom of bed rock is 80 m. Calculate the reserve of aggregate in 2.5 km strike length and 0.5 km dip length of rock quarry site.

Principle: the area of the quarry on the bed surface = strike length ×\times dip length; reserve = area ×\times true thickness, with t=tvcos⁡δt = t_v\cos\delta.

Given

Attitude N42°W/34° (dip 34∘34^\circ), vertical thickness (top to bottom) tv=80t_v = 80 m, strike length 2.5 km, dip length 0.5 km (taken along the dip of the bed).

Calculation

t=80cos⁡34∘=66.32 mA=2500×500=1.25×106 m2V=1.25×106×66.32=8.29×107 m3\begin{aligned} t &= 80\cos 34^\circ = 66.32\ \text{m} \\ A &= 2500 \times 500 = 1.25 \times 10^6\ \text{m}^2 \\ V &= 1.25 \times 10^6 \times 66.32 = 8.29 \times 10^7\ \text{m}^3 \end{aligned}

Answer: reserve of aggregate = 8.29 × 10⁷ m³ (about 82.9 million m³ = 0.083 km³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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