Question 6 of 6: Vertical Crater Retreat (VCR) Blast Design

Question 6: Vertical Crater Retreat (VCR) Blast Design (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Charge selection, optimum burden/lift and pattern

Given. Borehole De=160 mm; Figure 1 peak at scaled depth of charge N≈0.70 m/kg^1/3 (scaled volume ≈0.535 m³/kg).

Find. An explosive, a spherical(-ish) charge that fits the 160 mm hole, and the resulting optimum burden (=vertical lift height per blast), spacing, and loading/ timing sketches.

Approach. Livingston crater theory says the crater/breakage volume per unit explosive is maximised at the scaled depth of burial N=d/W^1/3 where the given curve peaks – that peak N is exactly the design ratio between burden (depth of charge below the current free face) and the cube root of the charge weight: B=N·W^1/3. VCR uses a compact, near-spherical charge (length/diameter ≤6) so the cratering theory (derived for point/spherical charges) applies; choose the largest such charge the 160 mm hole can carry, then the peak-N relation converts that charge weight directly into the achievable lift height.

1. Explosive selection. VCR holes are drilled vertically upward or downward and commonly encounter groundwater; a bulk emulsion (assumed ρ=1.2 g/cm³, VOD≈5000 m/s) is selected over ANFO because it is water-resistant (reliable even in wet holes, unlike ANFO which desensitises when wet) and pumps/gasses readily to the required density in a large-diameter vertical hole – both properties that matter more for VCR than for a dry, horizontal production bench.
2. Largest near-spherical charge in a 160 mm hole (L/D≤6). $$L_c=6(0.160)=0.96\ \text{m},\qquad Area=\frac{\pi}{4}(0.160)^2=0.0201\ \text{m}^2$$ $$W=L_c\times Area\times\rho=0.96\times0.0201\times1200=\boxed{23.2\ \text{kg}}$$
3. Optimum burden = lift height, from Figure 1's peak (N=0.70). $$B=N\,W^{1/3}=0.70(23.2)^{1/3}=0.70(2.85)=\boxed{2.00\ \text{m}}$$ – a realistic VCR lift height (field practice is typically 2–3 m per retreat), which is good corroborating evidence the charge/curve combination is being read correctly.
4. Spacing and charge placement. The cratering mechanism is radially symmetric around each charge, so a square grid with $$S=B=\boxed{2.00\ \text{m}}$$ is used. The charge is centred at depth B=2.00 m below the current (horizontal) free face – from 1.52 m to 2.48 m down the hole – with stemming filling the 1.52 m above it up to the collar.