Consider a case where a 20-metric-ton capacity, articulated rear-dump truck hauls ore to the surface. Ore is placed in the truck by a 6.1-cubic-meter capacity, remote LHD near the entrance of the stope. The truck hauls the ore 550 meters along a nearly level drift, then 1,450 meters up a 10% gradient to the surface. Once on the surface, the truck travels another 200 meters to the mill, where the ore is dumped into a crusher feed bin.
First, the speeds of the machine must be estimated over the various segments of the haul route. Using information from technical manuals supplied by equipment manufacturers (in this case the Wagner Mining Equipment manual) approximate speeds over the following gradients are--
• Up a 10% gradient, loaded @ 6.4 kilometer/hour
• Over a level gradient, loaded @ 16.1 kilometer/hour
• Over a level gradient, empty @ 20.3 kilometer/hour
• Down a 10% gradient, empty @ 15.8 kilometer/hour
These values account for a rolling resistance equivalent to a 3% gradient. The estimated time for this haul profile is then--
HAUL
550 meters (16.1 kilometers/hour &##8734; (1,000 meters/ kilometer) &##8734; 60 minutes/hour = 2.05 minutes
1,450 meters (6.4 kilometers/hour &##8734; 1,000 meters/ kilometer) &##8734; 60 minutes/hour = 13.59 minutes
200 meters (16.1 kilometers/hour &##8734; 1,000 meters/ kilometer) &##8734; 60 minutes/hour = 1.04 minutes
550 meters (20.3 kilometers/hour &##8734; 1,000 meters/ kilometer) &##8734; 60 minutes/hour = 1.63 minutes
1,450 meters (15.8 kilometers/hour &##8734; 1,000 meters/ kilometer) &##8734; 60 minutes/hour = 5.51 minutes
200 meters (20.3 kilometers/hour &##8734; 1,000 meters/ kilometer) &##8734; 60 minutes/hour = 0.59 minutes
TOTAL TRAVEL TIME = 24.41 minutes
Evaluators may wish to further tune the above estimate by accounting for items such as altitude duration, acceleration, and deceleration. However, the effort spent should be
proportionate to the purpose of the estimate and the reliability of the available information. Specifically, if acceleration and deceleration increase the overall cycle time by half a minute (or 2%), but the mill has not been firmly sited and the overall haul distance may yet change by as much as 10%, then the effort spent fine-tuning the estimate would be futile because it would not increase the reliability of the results.
In addition to traveling, the truck expends time as it's loaded, as it dumps its load, and as it maneuvers into position for each of these tasks. In this example, the cycle time for the LHD must also be estimated to figure the time needed to load the hauler. Assume that the volume capacity of the LHD is 2.7 cubic meters and the weight capacity is 5.44 metric tons. On any given load, the bucket is typically 85% full, and the material in its blasted condition weighs 2.85 metric tons per cubic meter. If the round trip from the dump point to the stope and back takes 2.5 minutes and the LHD takes 0.4 minutes to dump the load, then the following series of calculations
provides the time necessary to load the truck.
First, the volume capacity of the LHD is examined.
2.7 cubic meters/load &##8734; 2.85 metric tons/cubic meter &##8734; 0.85 = 6.54 metric tons.
Since the weight capacity of the LHD is 5.44 metric tons, the load is limited by weight. Consequently, the time required to load the truck is--
20 metric tons 5.44 metric tons/load = 3.68 loads.
3.68 loads &##8734; 2.9 minutes/load = 10.67 minutes to load truck.
Assuming that the truck dump mechanism takes 0.4 minutes to cycle and 2.25 minutes are spent turning and maneuvering during each cycle, the total cycle time estimated for the truck is--
LOAD = 10.67 minutes, TRAVEL = 24.41 minutes, DUMP = 0.40 minutes, MANEUVER = 2.25 minutes TOTAL CYCLE TIME = 37.73 minutes
If the mine operates 2 shifts per day, 10 hours per shift, then one truck can deliver--
2 shifts &##8734; 10 hours/shift &##8734; 60 minutes/hour 37.73 minutes/cycle &##8734; 20 metric tons/cycle = 636 metric tons/day.
If the mine production rate is to be 4,000 metric tons per day, then the required number of trucks would be--
4,000 metric tons/day 636 metric tons/truck = 6.29 or 7 trucks,
And if the production rate per truck is--
20 metric tons/cycle 37.73 minutes/cycle &##8734; 60 minutes/ hour = 31.80 metric tons/hour.
Then total daily truck use is--
4,000 metric tons/day 31.80 metric tons/hour = 125.79 hours.
Now, for the number of operators. Typically, operators work noticeably less than the total number of hours for which they are paid. When time spent at lunch and breaks (in addition to time lost in traveling to and from the work face) is factored into the estimate, an average of 83% of the operator's time is actually spent working. Consequently, the total amount of labor time required to meet production is--
125.79 hours 0.83 = 151.55 hours.
required:
151.55 hours 10 hours/shift = 16 workers
Looking back to the number of trucks initially selected, it can be seen that (after accounting for worker efficiency) more trucks will be required.
16 workers 2 shifts/day = 8 trucks
So, in examining the truck as it operates over the designed haul profile, several cost parameters unfold. These include the number of trucks, the number of operators, daily truck use requirements (hours per day), and the number of hours that the drivers must work. Note that the two latter values differ. Each value is used to determine a different cost, so each must be estimated separately.
Properly determining the hourly workforce goes a long way toward ensuring an acceptable level of reliability. Wages often account for over half the total operating cost, so if the
workforce estimate is solid, the cost estimate is probably more than halfway home. Conversely, the cost of operating
underground equipment typically represents only a minor portion of the total underground cost. However, since the size and configuration of the workforce is closely related to the equipment requirements (and since equipment purchase costs can be significant), the importance of determining those
requirements should not be minimized. The information directly impacts the reliability of estimated costs.
Other cyclic operations (drilling, mucking, loading, hauling, hoisting, etc.) are modeled in a manner similar to the truck example above, with the same cost parameters developed for each. As is evident, the actual cycle time calculations are quite simple. The most difficult task is usually finding the rate
(speed) at which the machine operates (drill penetration rates, mucker transport speeds, hoist velocities, etc.), but even this information is often readily available. The most common sources include literature from the manufacturer, references such as this book, information databases such as Mining Cost Service (Western Mine Engineering, Inc., 2000), or statistical compilations such as the Canadian Mining Journal's Mining Sourcebook (Southam 2000). Exclusive of these, speed is very often easy to estimate through observation. With the machine
speed and a bit of imagination, the evaluator can provide a more than reasonable estimate of the cost parameters associated with almost any cyclic operation.
Most noncyclic underground operations are based on the continuous movement of materials (ore, waste, air, water, workers, etc.), and the cost parameters can be estimated accordingly. Conveyors, generators, pumps, and ventilation fans all fall under this category. In the following example, the parameters associated with draining the mine and pumping the water to the surface are determined to illustrate the process as it applies to continuous-flow operations.