Checking for Traction
Problems
©Dr. B. C. Paul 2000 revised 2008
Note – The methods outlined here are common practice in industry and can be found in a wide range of books. The slides specifically include tables from the Caterpillar Performance
Handbook and the book Surface Mining from SME

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Gradability and Retarder Charts give peak speed that engine can transfer power to wheels for
Does not guarantee the tire can transfer to ground without spinning
Max transfer of force
• F = coefric * Weight * cos(pheta)

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Coef of friction depends on road material primarily
• Generally get from table
Cos(pheta) term is usually almost 1
• for 10% grade cos(pheta) = 0.995
• Common practice to take as unity
Weight is weight on Drive
Wheels (in lbs) - important distinction for two wheel drives

Crushed Rock or
Gravel Would Be
Common in a Quarry

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Cat on the Hill
• Empty weight is 87000 lbs
• Distribution empty is 53% to rear wheels or 46,000 lbs
• Loaded weight is 204,000 lbs
• Distribution Full is 67% to rear wheels or
137,000 lbs
• Note that this truck was weight limited and reached gross weight limit - volume limited truck would not.

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It’s a Quarry so road is gravely sometimes rains so worse case may be wet gravel
• From Table 0.35 is coef of friction
Could have considered icy conditions after freezing rain
• But would you run that way?
• Quarries often close in winter because of slabbing from face

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137,000 lbs * 0.35 * 0.995 =
48,000 lbs loaded
46,000 lbs * 0.35 * 0.995 =
16,000 lbs empty
Like Wow Man - Awsome
Numbers. Now What do I do with them?

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Answer comes in lbs force that can be transferred between the tire and the ground
This has the form of Rimpull
Compare the number to the
Rimpull required to propel the truck (or keep it from sliding down hill out of control)

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Rimpull to hold back empty truck
• 43.5 tons * 115 lbs/ton = 5,000 lbs
Wait! Hold Everything - That’s downhill so 115 lbs/ton must be negative
• The tire doesn’t know who’s pulling who - only whether the friction is enough to balance the force

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For Empty Truck Going Downhill
• Rimpull Required is 5,000 lbs
• Traction Force Available is 16,000 lbs
• 5,000 < 16,000 so I’m OK
For Loaded Truck
• Rimpull Required is 29,000 lbs
• 29,000 < 48,000 lbs

Gee This is So Much Fun I
Can’t Wait to Do it for Every
Segment on My Haul Road
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Actually - You Don’t Have to
We are comparing two numbers
• Rimpull Needed = lbs/ton * Total
Tons Weight
• Traction Available = coefric * Drive
Wheel Weight
In Simple Terms
• RN = K
1
Dweight
* Tweight TA = K
2
*

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Both Equations are a Linear
Function of Weight
• On a ton for ton basis the rimpull needed is the same loaded or unloaded
• Traction depends on portion of that weight on drive wheels
• The Traction Value is minimum when the truck is empty

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If the Empty Truck won’t spin the loaded Truck won’t
Look for points of maximum total grade - gives maximum rimpull
Look for points of low coef of friction - minimizes Traction
Check Those Points

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Our haul road is uniform so we have no minimal coef of friction
Our haul road is uniform so we have no higher rolling resistance
Thus maximum grade will be the worst case coming out where the two resistances are additive

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Loaded Truck Coming Out and
Empty Truck Going in - We’ve already checked
There is a non routine point what if we have to take a truck out empty?
• In checking truck and road compatibility usually check things that are realistic

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43.5 tons * 285 lbs/ton = 12,400 lbs
Compare
• Rimpull Needed 12,400 lbs
• Traction Available 16,000 lbs
I’m Still Ok

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You have entered haulage profile and frictional forces
You have picked trucks and all tables including weight distributions are loaded
FPC checks all segments for spinning
• If a problem occurs it gives an error message
FPC only checks regular haul segments
• What if you need to bring an empty truck out of the pit when a loader break-down occurs?