Question 2 of 6: Quarry Blast Redesign – Diameter Up-sizing and Vibration Compliance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, 09-Mmp-B2 Rock Fragmentation, May 2016, 3 hours, closed book
(one double-sided aid sheet permitted). Question 1 plus four (4) of Questions 2-6
constitute a complete paper; every question (1-6) is answered in
full as a complete study resource.
Reference texts: Persson, Holmberg & Lee, Rock Blasting and
Explosives Engineering; C.J. Konya & E.J. Walter, Rock Blasting and Overbreak
Control (FHWA); ISEE, Blasters' Handbook, 18th ed.; W. Hustrulid,
Blasting Principles for Open Pit Mining; SME Mining Engineering Handbook,
3rd ed., Ch. Drilling and Blasting; W.I. Duvall & C.F. Fogelson, USBM RI 5514
(cratering theory).
Given. Current: 3 rows × 12 holes, De₁=102 mm, H=12 m,
emulsion ρ=1.25 g/cm³, B₁=2.5 m, S₁=3.5 m, T₁=3.0 m, distance to
nearest structure R=300 m. Proposed De₂=165 mm.
Find. A pattern (burden, spacing, collar/stemming) at 165 mm that
preserves the current fragmentation and stays flyrock-safe at 300 m.
Approach. Fragmentation is governed by the geometric similarity of the
pattern to the hole diameter (Kuznetsov's V/Q and the powder factor are unchanged if every
linear pattern dimension scales exactly with diameter), so scale B, S and T by
De₂/De₁ and verify the resulting stiffness ratio and stemming-confinement ratio
are still adequate; separately check the larger diameter's worst-case flyrock throw
(Lundborg's empirical formula) against the 300 m stand-off.
Linear scaling to preserve fragmentation.
$$\frac{De_2}{De_1}=\frac{165}{102}=1.618$$
$$B_2=2.5(1.618)=4.04\ \text{m},\quad S_2=3.5(1.618)=5.66\ \text{m},\quad
T_2=3.0(1.618)=\boxed{4.85\ \text{m}}$$
Stiffness-ratio check (SR=H/B, want SR≥2 for good toe breakage).
$$SR_1=\frac{12}{2.5}=4.8\ (\text{very stiff}),\qquad SR_2=\frac{12}{4.04}=2.97$$
SR₂ is lower than SR₁ (fewer, larger-burden holes are inherently less "stiff"
relative to the fixed 12 m bench) but still comfortably above the SR=2 minimum – the
165 mm pattern remains adequately relieved and fragmentation is not expected to coarsen from
loss of stiffness.
Stemming/confinement ratio preserved.
$$\frac{T_1}{De_1}=\frac{3.0}{0.102}=29.4,\qquad \frac{T_2}{De_2}=\frac{4.85}{0.165}=29.4$$
– identical by construction of the linear scaling, so the collar is exactly as well
confined against gas venting/flyrock at 165 mm as it already is at 102 mm.
Worst-case flyrock throw vs. the 300 m stand-off (Lundborg).
$$L_{max}(\text{ft})=260\,D(\text{in})^{2/3}$$
$$L_{102}=260(4.02)^{2/3}=657\ \text{ft}=200\ \text{m},\qquad
L_{165}=260(6.50)^{2/3}=908\ \text{ft}=\boxed{276\ \text{m}}$$
The current 102 mm design throws at most ≈200 m – a wide margin below 300 m. The
165 mm design's worst-case throw rises to ≈276 m – still under the 300 m limit,
but with only ≈24 m of margin instead of ≈100 m.
Quantity
102 mm (current)
165 mm (proposed)
Burden
2.5 m
4.04 m
Spacing
3.5 m
5.66 m
Collar/stemming
3.0 m
4.85 m
Stiffness ratio SR=H/B
4.8
2.97 (still > 2, OK)
Stemming ratio T/De
29.4×De
29.4×De (unchanged)
Lundborg max throw
≈200 m
≈276 m (300 m limit)
Check: the 276 m worst-case throw is under the 300 m limit but the
margin has narrowed from ≈100 m to ≈24 m – recommend keeping stemming at or
above the scaled 4.85 m (never trim it to save drilling cost), using quality stemming material
(angular crushed rock, not drill cuttings), and, if any hole shows poor stemming confinement
or a joint/void intersecting the collar, adding 0.5–1.0 m of extra stemming on that hole
specifically as a field safety margin.
(b) Vibration-compliant loading at R=300 m, limit 12 mm/s
Given. $$PPV=745\left(\frac{R}{\sqrt{W}}\right)^{-1.3}$$ PPV limit
= 12 mm/s, R = 300 m, 165 mm pattern from part (a): B₂=4.04 m, T₂=4.85 m, H=12 m,
emulsion ρ=1.25 g/cm³.
Find. The maximum charge mass per delay Wmax that meets the
12 mm/s limit at 300 m, and how to load/time each 165 mm hole to comply.
Approach. Invert the given attenuation law for W at PPV=12 mm/s, R=300 m;
compute the full-column charge of one 165 mm hole; compare and, if it exceeds
Wmax, split it into decks fired on separate delays.
Solve for Wmax.
$$12=745\left(\frac{300}{\sqrt{W}}\right)^{-1.3}\;\Rightarrow\;
\left(\frac{300}{\sqrt{W}}\right)=\left(\frac{12}{745}\right)^{-1/1.3}=23.94$$
$$\sqrt{W}=\frac{300}{23.94}=12.53\;\Rightarrow\;\boxed{W_{max}=157\ \text{kg per delay}}$$
Full-column charge in one 165 mm hole. Area = 0.02138 m²; charge
length = H−T₂ = 12−4.85 = 7.15 m.
$$Q_{hole}=7.15\times0.02138\times1250=\boxed{191\ \text{kg}}$$
– a single hole fired as one instantaneous charge (191 kg) exceeds Wmax
(157 kg), so the current per-hole loading is NOT vibration-compliant at 300 m.
Deck the charge. $$n_{decks}=\left\lceil\frac{191}{157}\right\rceil=2,
\qquad \boxed{95.5\ \text{kg per deck}}$$
Two roughly-equal decks (95.5 kg each), separated by an inert stemming plug, each on its own
delay number, both stay comfortably under the 157 kg ceiling with margin for delay scatter.
Quantity
Value
Max charge mass per delay, Wmax
157 kg
Full-column charge, one 165 mm hole
191 kg (exceeds limit)
Recommended loading
2 decks per hole, ≈95.5 kg each, separate delays
Recommended inter-hole/inter-deck delay
≥17 ms (comfortably above the 8 ms
simultaneity threshold, Q1(h))
Timing. Sequence the round so that no two decks anywhere in the 36-hole
pattern (3 rows × 12 holes) ever share, or fall within ≈8 ms of, the same delay
number – e.g. row-by-row and deck-by-deck stepping at a nominal 25 ms interval, which
keeps the instantaneous per-delay charge at the single-deck value (95.5 kg, well under the
157 kg ceiling) for every event in the round, while also giving each row time to relieve
into the void left by the row ahead of it before the next row fires.
