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Essays: Pavement Temperatures under a Skidding Tire

This issue has been bugging me since it came up on one of my mailing lists a while back. The
common wisdom in A/R circles says that skid marks are a result of oils coming to the surface as the skidding tire heats the pavement while passing over it, and perhaps bits of tire grinding being trapped in a molten roadway surface layer which cools immediately after the tire passes. I have been skeptical of both explanations  if I swing a torch over pavement (open propane flame being clearly hotter than a skidding tire) I don't get a black stripe. And what about concrete? why do we get similar marks on concrete.....So.
Having not learned from that curious cat, I decided to calculate the energy dumped into the ground by a skidding car, and see how much it would heat the pavement.
Let's start with a 2500 lb car travelling 60 mph, assume a drag factor of 0.8. For severity and simplicity, let's assume that the rear wheels are in the air, and the fronts do all the
work  this means all the heat input on one side is through one contact patch in one event, so there's no cooling between tires, and no second event to mess with.
If the brakes are locked, we will dissipate all the kinetic energy [(mass*v^2)/2] in the skid. this gives us 300,600 ftlbs. Each skid gets half of the energy, or 150,300 ftlbs.
The total expected skid length works out to be:
Distance = v^2 / (2 * g * mu) = 150.3 feet (which makes sense, yay!)
So, if we assume even distribution of the energy along the
skid length, we get:
(150,300 / 150) = 1002 ftlb energy per foot of skid
If the skid is 5 inches wide, that's
1002 ftlb / (5 in * 12 in ) = 16.7 ftlb
of energy dumped into each square inch of pavement. This is our energy in, Q.
According to the Asphalt Institute (asphaltinstitute.com) the heat capacity (Cp) of pavement is about 0.22 cal/gramdegC which is equivalent to 0.22 BTU/lbm  F which equals
171.2 ftlb / lbm  F
Typical wearcourse pavement density is 14.20 in^3/lb , or 0.071 lb/in^3
(see http://www.highwaysmaintenance.com/Bitdata.html for
this info)
If the energy dumped into that onesquareinch area of pavement heats a layer say 0.01 inches deep (which is on the order of pavement roughness acording to several websites regarding texture and mean profile depth, see http://www.greenwood.dk/html/profilo/macro.htm, http://www.vti.se/nordic/100mapp/100dk1.html, and http://www.tfhrc.gov/hnr20/rosan/rosandoc.htm), it's heating 0.01 square inches, or about 0.00071 lb of asphalt, recalling that
we can estimate the change in temperature of our thin layer to be
delta T = Q / (Cp * m) =
=16.7 ftlb / (171.2 ftlb/lbmF * 0.00071 lbm)
= 137 degrees F.
Now, I'm no worldfamous heattransfer guru, so I may be way off here,
but it looks right to me. I would love to review ANY documentation of
the "skids are more superheated pavement material than rubber"
theory. I continue to be skeptical of the old theory. There does
not seem to be enough energy to significantly heat the pavement. Even if the thermal layer is halved, we only have a 260 degree elevation in temperature...That's not too different from pouring boiling water on the roadway, which we all know does NOT bring
"bituminous materials" to the surface.
It has been suggested to me that the fact that a brake disk is hot after a long (nonlocked) stop is proof of this theory. I disagree. A disk or drum is absorbing the entire 190or so BTUs (presuming the rear brakes suck) in the several seconds that it takes to
stop, with a mass of around 8 pounds, and a relatively small surface
area (yielding at best modest convection cooling during the event and only modest heat conduction through the disk thickness). This is NOT even close to being the same as 150 foot long piece of pavement, with the car passing over each oneinch length
in a tiny fraction of a second (about 0.001 seconds at about 60mph).
Mechanical Forensics Engineering Services, LLC.
This page created on 21NOV2001 and last modified 21NOV2001
