How to calculate potential traction loss on a haul road

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SUMMARY

This discussion focuses on calculating potential traction loss for trucks operating on a haul road in a mining context. The user is analyzing a scenario involving a 10% gradient and a coefficient of friction of 0.25, which raises concerns about safety due to potential traction loss. Key calculations include total resistance factors, stopping distances, and the necessary number of trucks to maintain operational efficiency. The user seeks a definitive formula to quantify traction loss, particularly under adverse conditions such as downhill travel with full loads.

PREREQUISITES
  • Understanding of mining engineering principles, particularly haul road operations.
  • Familiarity with truck performance metrics, including speed and load capacities.
  • Knowledge of traction concepts, including coefficients of friction and rolling resistance.
  • Ability to apply physics equations related to motion and stopping distances.
NEXT STEPS
  • Research "Kaufman and Ault stopping distance formula" for detailed applications in mining scenarios.
  • Explore "traction loss calculations" specific to heavy vehicles on inclines.
  • Investigate "braking force distribution" equations to assess uneven braking impacts.
  • Study "grade resistance factors" and their implications on vehicle performance in mining operations.
USEFUL FOR

Mining engineers, safety analysts, and logistics managers involved in haul road operations and truck performance optimization will benefit from this discussion.

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Homework Statement


This is a mining engineering question, but the section I need help with is more general: I am working through a long question regarding a given mine. In previous sections of this question, I have selected equipment based on required machinery outputs etc.
I have now come to the last question which is "ensure trucks are capable of carrying out this mining safely". Which is a bit of an open-ended one!
In previous questions I have figured out that I need between 2 and 3 trucks to keep the excavator swinging without stopping. The question didn't ask for it, but to give a decent answer I need more data, so I selected an appropriate truck from the market and used its data (I listed that in assumptions). I feel that I have partially answered the question, but there is data given that I have not used much (ie coefficient of traction) that makes me think the lecturer is looking for calculations showing traction loss. I have been hunting for days and searching through forums and engineering text-books but can't seem to find a clean formula for this purpose. I feel from experience that a coefficient of traction of 0.25 (similar to snow!) would lead to safey problems from traction loss but I can't seem to prove it. I thought stopping distances would do it (see my work below) but unless I messed up the math (quite possible) the stopping distance was not great but still acceptable.
The data given in the question:

28. A pit with a straight ramp of 800m, with a gradient of 10%, supplies a ROM pad 1500m away via a level haul road all of loose gravel. Mining of a dipping ore body, with a specific weight of 2.5 t/m3, is required to be carried out at a rate of 10,000 tonnes/8 hour shift to feed the mill. Since truck speed is restricted to 50kmph, assume the average speed of a truck during a round trip is 45kmph, including reversing to load. Waste material of specific weight = 1.85 t/m3 is also to be mined, and deposited on a waste dump 200m from the pit. The waste dump already has a height of 10m, and an area of 100m x 60m from pre-stripped overburden. Excluding the overburden, the strip ratio is 1:4.5.

Other factors that are working assumptions are:

· Availability of excavator 83%,

· Swell factor of 30%

· Bucket fill factor of 88%

· Excavator cycle time 45 seconds

· Rolling Resistance Factor (RRF) = 15 kg/t

· Coefficient of Friction, m = 0.25

· Grade Resistance Factor = 20 kg/t

Homework Equations


As below for equations I have finished. I have not been able to find an equation to calculate if I will lose traction while driving into/out of the pit

The Attempt at a Solution


What I have so far for this question:
Total resistance = grade + rolling resistance% = 10% + (15kg/1000kgx100) = 11.5% uphill (Or 8.5% downhill)
Truck Performance (I can attach this chart if requested, but I don't think you will need it):
Using the Hitachi retarder performance data for a EH4500-2 dump truck (using total resistance numbers), a downhill 10% grade for an empty truck would allow a speed of up to 55kh/h with safe retarder use. Using the rim pull chart and the total resistance, a full truck could obtain a maximum speed of up to 13km/h going up the ramps, and 55km/h on the flat (limited to 50km/h as given).

From this data the following would be within manufacturer recommendations:
Minimum cycle times:
Waste:
13km/h roads = total uphill ramp distance = 800m pit ramp + 100m dump ramp = 900m
50km/h roads = 2130m-900m = 1130m
Minimum cycle time for trucks = (0.9km/13km/hx60)+(1.13kmx50km/hx60)+3mins loading = (4.15+1.36)+3 = 5.51mins haul time + 3 mins loading = 8.51mins
Rom/Ore:
13km/h roads = total uphill ramp distance = 800m pit ramp
50km/h roads = 4600m-800m = 3800m
Safe cycle time for trucks = (0.8km/13km/hx60)+(3.8kmx50km/hx60)+2.25mins loading = (3.69+4.56)+2.25 = 8.25mins haul time+2.25mins loading = 10.5minsTrucks required to do waste circuit safely = 5.51/3 = 1.84
Trucks required to do rom/ore circuit safely = 8.25/2.25 =

Rolling resistance = 15kg/t
Grade resistance factor = 20kg/tCoefficient of Friction, m = 0.25
Truck gross vehicle mass: downhill empty 200t, uphill full = 470t
Truck downhill speed: 50km/h = 13.89m/s

Stopping distance for trucks, using Kaufman and Aultd 1977 SAE formula
SD= 0.5gt² sinø + volt + [ (gt sinø + Vo)/(2g{Umin-sinø})]²

g= 9.81m/s²
t = 6 seconds (combined reaction time of truck and operator)
sinø = sin10% (slope)
Vo= 13.89m/s (vehicle speed of 50km/h)
Umin = 0.25

SD=0.5x9.81²sin10%+13.89x6+[( 9.81x6sin10%+13.89)/(2x9.81x{0.25-sin10%)}]²
SD= 0.5x96.2361x0.1736+13.89+[1.703+13.89)/(19.62x0.0764)]²
SD= 22.2432+[15.593/1.4990]²
SD= 130.5m
 
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Usually traction problems are most likely to show up in increased stopping distances when going down hill.

Try considering a worst case scenario: downhill at the steepest angle, full load, unequal distribution of braking forces, sliding friction, etc.
 
Dr Courtney, thankyou for the reply. In my answer I have calculated the worst-case scenario, which is a 10% downward grade with the low traction (SD=130m). I have no idea how I would calculate uneven brake force though with the data I have, are you able to point me in the direction of an equation that I can use to calculate that? The formula I used did not take vehicle weight into consideration, although any vehicle traveling down an incline would be empty.
Is there an equation that can be use to figure out a what angle of slope a vehicle would lose traction, that takes the coefficient of friction into account?

To calculate the stopping distance I used a Kaufman and Ault formula based on the SAE stopping distance limitations, its a bit hard to decipher from my answer so I have put it below:

SD= ½ gt²sinø + volt + [ gt sinø + Vo ]²
2g(Umin-sinø)
SD=Stopping Distance
g = Gravatational acceleration (9.81m/s²)
t = elapsed time between perception of need to stop and actual frictional brake contact (given by authors as t= 4.5s for the truck + 1.5s for the operator)
ø = angle of descent in º
Umin = coefficient of friction
Vo= vehicle speed in m/s
 
I found an error in my work in the math. Stopping distance actually = 372.9m :wideeyed: With that horrible number I think I can show why its not safe! If anyone has a formula for traction loss, however, it would be very appreciated.
 

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