Shovels, Excavators & Loaders

Loading equipment digs broken rock and fills trucks. Learn electric rope shovels, hydraulic excavators, front-end loaders and LHDs, bucket fill factor, swell and cycle time, productivity in bcm/h, passes to fill a truck, and matching loaders with trucks.

The loading tool is the heart of the mining fleet: if the shovel waits for trucks or trucks queue at the shovel, money is lost. The largest shovels carry 100+ t per pass, filling a 360 t truck in 3–4 passes.

An analogy

A shovel and its trucks are like a waiter and a kitchen: the waiter (truck) should arrive just as the plate (bucket) is ready — not queue, not leave the kitchen idle.

Machine types

Electric rope shovels (very large, low cost per tonne, less mobile); hydraulic excavators (face shovel or backhoe, selective digging); wheel loaders (mobile, blending, stockpiles); LHDs (load-haul-dump) underground.

Productivity factors

Bucket size and fill (fragmentation, digging conditions), swing angle, operator skill, truck exchange time and job efficiency (≈ 50 min per hour). Poorly blasted muck lowers fill factor and raises cycle time.

Pass matching

Ideally 3–5 passes per truck. Too few passes risk overloading; too many waste shovel time. Payload monitoring systems measure each bucket.

Selectivity and automation

Hydraulic excavators dig selectively along ore/waste contacts to reduce dilution. Shovel automation and payload/fragmentation cameras provide feedback to blasting and dispatch.

Formulas

• Productivity: Q = 3600 V_b F E / (t_c S_f) (bcm/h)
• Bucket payload: m_b = V_b F × ρ_bank / S_f (t)
• Passes per truck: n = truck payload / m_b (round sensibly)

Worked example: Shovel productivity

Given: 40 m³ bucket, fill 0.9, efficiency 0.83; Cycle 30 s, swell 1.35

Find: bcm/h.

• Q = 3600 × 40 × 0.9 × 0.83/(30 × 1.35) = 2656 bcm/h

Answer: ≈ 2660 bcm/h (≈ 7200 t/h at 2.7 t/bcm)

Worked example: Passes per truck

Given: Bucket 40 m³ × 0.9 fill, ρ_bank 2.7, swell 1.35; Truck 220 t

Find: Passes.

• m_b = 36 × 2.7/1.35 = 72 t
• 220/72 = 3.06

Answer: 3 passes (≈ 216 t)

Common mistakes

• Wrong: Using heaped capacity without fill factor. — Right: Apply fill factor and swell.
• Wrong: Overloading trucks with an extra pass. — Right: Respect the 10/10/20 payload policy.

Applications

• Fleet selection: Matching shovel and truck classes.
• Selective mining: Excavators on narrow ore zones.
• Underground mucking: LHDs in stopes and development.

Key takeaways

• Q = 3600 V_b F E/(t_c S_f).
• Bucket payload and passes per truck.
• Fragmentation drives loader performance.
• Match loaders and trucks.

Going further

• Payload monitoring: Bucket weighing and truck load optimisation.
• Shovel automation: Assisted digging and swing automation.
• Selective mining units: Matching SMU to excavator size.