smeggy wrote:
If a 1800kg vehicle is at 60mph, braking at 0.5G, instantaneous power transfer will be 240kW.
4 4 square inch contact patches will result with 600W per square cm, or about 2.5J over 3.8ms per square cm. Assuming in that time the heat conducted down only 0.1mm (10nM^3 volume per square cm) and thermal mass is 4kJ per litre (1mM^3) per degree (so 40mJ per degree), my really dodgy maths would suggest a rise of surface temp of 62 degrees. I guess 90C will melt tarmac.
I made a few assumptions in there, can someone check?
If the car is braking at 0.5G, a lot more of its weight will be on the front tyres than the rear. Exactly how much depends on its initial weight distribution and its centre of gravity height. At 0.9G, on most modern front wheel drive road cars, I think you could (for the purposes of THIS exercise anyway) neglect the weight on the rear tyres and assume it's just about all on the fronts!
That said, it would be easy to establish whether or not the tar melted by experiment. Just go out, jump on the brakes and lock up, then get out and see if the tar has melted! I doubt it will have re-solidified in the few seconds it takes you to get out of the car and look!
We have a stretch of road not far from my house which lies in a sun trap. On very hot summer days it starts to melt. When braking (even only moderately hard) on it I have noticed wheel lockup - very like aquaplaning, but it's pretty obvious where that has occured looking at the road surface because there are big grooves in it where my tyres have been! OK, it's not quite the same situation - the sun with have heated the tar through to a much greater depth than tyres ever could, but the point is that the surface is very visibly damaged if it melts and should therefore be very easy to see.