Haul Trucks & Fleet Sizing

Haul trucks move most material in open pits. Learn truck classes, rolling and grade resistance, rimpull and speed, cycle time components, match factor, fleet size, productivity, fuel/energy and trolley assist, and truck dispatch.

Ultra-class trucks carry 220–400 t, cost over $5 million each and burn ≈ 100–200 L of diesel per hour. Haulage is often the largest single mining cost, so accurate fleet sizing and efficient roads matter enormously.

An analogy

A haul truck climbing a ramp is like cycling uphill with a heavy backpack: the grade and soft ground both resist you, and your legs (engine power) set the top speed.

Resistance

Grade resistance ≈ 1 % per 1 % of slope; rolling resistance ≈ 2 % on good roads, 4–5 % on poor roads, more on soft ground. Every 1 % of rolling resistance on a ramp acts like an extra 1 % of grade.

Rimpull and speed

Manufacturer rimpull–speed curves give maximum speed for a given total resistance; retarding curves limit downhill speed. Travel times per segment are summed to get cycle time; simulation (TALPAC, FPC) automates this.

Fleet sizing

Trucks per loader = cycle time ÷ loading time (including spotting). Match factor ≈ 1 balances fleet; < 1 means the shovel waits; > 1 means trucks queue. Dispatch systems (Modular, Wenco) assign trucks dynamically.

Energy and electrification

Trolley assist powers diesel-electric trucks from overhead lines on ramps (faster, less fuel). Battery-electric and hydrogen trucks are being trialled to decarbonise haulage — the largest source of mine Scope 1 emissions.

Formulas

• Resisting force: F = GVW × g × TR
• Power at speed: P = F × v
• Truck cycle: t_c = load + spot + haul + dump + return
• Trucks per loader: N = t_c / (load + spot)
• Match factor: MF = N_trucks × load time / (N_loaders × truck cycle)

Worked example: Force and power on a ramp

Given: GVW 380 t (160 empty + 220 payload); Grade 10 % + RR 2 %, speed 12 km/h

Find: Resisting force and power.

• F = 380 × 9.81 × 0.12 = 447 kN
• P = 447 × 3.33 = 1490 kW

Answer: 447 kN, ≈ 1.5 MW

Worked example: Fleet for one shovel

Given: Load 1.5 + spot 0.5 + haul 10 + dump 1.5 + return 6 min; Payload 220 t

Find: Trucks and production.

• Cycle = 19.5 min; loading slot = 2 min → N = 9.75 → 10 trucks
• Per truck = 220 × 60/19.5 = 677 t/h
• Fleet ≈ 6770 t/h (shovel-limited)

Answer: 10 trucks, MF ≈ 1.03

Common mistakes

• Wrong: Ignoring rolling resistance. — Right: Poor roads add grade-equivalent resistance.
• Wrong: Sizing fleet to average cycle only. — Right: Cycles grow as pits deepen — plan year by year.

Applications

• Fleet planning: Truck numbers for each year of the mine plan.
• Dispatch: Real-time assignment to reduce queuing.
• Decarbonisation: Trolley assist and battery trucks.

Key takeaways

• F = GVW g (GR + RR).
• Cycle = load + haul + dump + return.
• N = cycle/load time; MF ≈ 1.
• Plan fleets as pits deepen.

Going further

• Haulage simulation: TALPAC/FPC with rimpull curves.
• Dispatch optimisation: Linear programming for truck assignment.
• Battery trucks: Charging strategies and regenerative braking.