Shafts, Declines & Development
Before ore can be mined, it must be reached. Learn access options — vertical shafts, inclined shafts, declines (ramps) and adits — hoisting systems and capacity, decline design, the development cycle (drill, charge, blast, muck, support), and development rates.
Underground mines need openings for people, equipment, ore, air, water and power. The choice between a shaft and a decline depends mainly on depth and production rate: declines let trucks and loaders drive in, shafts hoist ore efficiently from great depth (the deepest, Mponeng, reaches ≈ 4 km).
An analogy
A shaft is the mine's lift, a decline is its spiral car-park ramp, and the development drives are the corridors to every room.
Types of access
Adit: horizontal tunnel into a hillside. Decline/ramp: inclined road (1 in 7 to 1 in 9) for trucks — flexible, cheaper to start. Vertical shaft: circular, concrete-lined, with headframe and winder — most economic for deep, high-tonnage mines. Inclined shafts follow dipping orebodies.
Hoisting
Friction (Koepe) hoists use ropes over a wheel with counterweights — efficient for deep mines. Drum winders coil the rope — suitable for very deep single-stage shafts. Skips carry ore, cages carry people and materials. Capacity is set by payload, speed and depth.
Development cycle
Drill a face (jumbo, 3–5 m rounds) → charge with explosives → blast → re-entry after ventilation → scale and muck (LHD) → install ground support (bolts, mesh, shotcrete) → survey. Typical single-heading advance is 3–6 m/day; multiple headings run in parallel.
Advanced access
Shaft boring (raise boring, blind boring) removes drill-and-blast from shaft sinking. Tunnel boring machines and mechanical cutters speed development. Conveyor declines replace trucks for high tonnages. Hoisting from > 3 km requires sub-vertical shafts.
Formulas
- Hoist cycle:
t_c ≈ D/v + v/a + t_dwell - Hoisting capacity:
Q = m_s × 3600 / t_c (t/h) - Decline length:
L = depth × gradient ratio (1 in 7 → L = 7 × depth)
Worked example: Daily hoisting
Given: Skip payload 20 t, cycle 90 s; 20 hoisting hours/day
Find: Daily capacity.
- Cycles/day = 20 × 3600/90 = 800
- Capacity = 800 × 20 = 16 000 t/day
Answer: 16 000 t/day
Worked example: Decline length
Given: Target depth 400 m; Gradient 1 in 7
Find: Decline length.
- L = 400 × 7 = 2800 m
Answer: 2.8 km (at 5 m/day ≈ 1.5 years single heading)
Common mistakes
- Wrong: Using only travel time D/v for cycle time. — Right: Include acceleration and loading/dumping dwell.
- Wrong: Underestimating development rates. — Right: Plan realistic metres per day including support time.
Applications
- Deep gold mines: Multi-stage shafts in South Africa.
- Mid-size metal mines: Truck declines to 800–1000 m.
- Block caves: Conveyor declines and large production shafts.
Key takeaways
- Access: adit, decline, shaft.
- Q = payload × 3600/t_c.
- Development cycle: drill–blast–muck–support.
- Deep, high-tonnage → shafts.
Going further
- Shaft boring systems: Mechanised sinking without blasting.
- Rope safety: Rope condition monitoring and statutory factors of safety.
- Rapid development: Automation of drilling, charging and bolting.