4 ms·

For those wondering about the source of the "4th power of weight per axle", I found this after some short, but intense, Googling: http://co-asphalt.com/wp-cont

For those wondering about the source of the "4th power of weight per axle", I found this after some short, but intense, Googling: http://co-asphalt.com/wp-content/uploads/2015/03/TrashTruckDamage.pdf http://co-asphalt.com/wp-content/uploads/2015/03/TrashTruckD... Reference 1 lead me to this: (Page I-11) https://habib00ugm.files.wordpress.com/2010/05/aashto1993.pdf https://habib00ugm.files.wordpress.com/2010/05/aashto1993.pd... From that: "The load equivalency factor increases approximately as a function of the ratio of any given axle load to the standard 18 kip single axle load raised to the fourth power." That is also cited here, on page 14: https://wisconsindot.gov/Documents/dmv/agri-eq-veh/ag-study/rpt-phase2.pdf https://wisconsindot.gov/Documents/dmv/agri-eq-veh/ag-study/...

andys627 8y ago
I am bad at math... could you help me understand. A 80,000 truck and 5,000 car drive down the road... how much more damage does the 80,000 truck do? Thank you.
- SilasX 8y ago
  If the 80k truck has 5 axles (as is typical of an 18 wheeler) and a 5k car has two, then the truck has 16k on each axle and the car has 2.5k on each axle. Because the damage scales with the 4th power, then the damage ratio is the 4th power of the axle weight ratio: (16k/2.5k) ^ 4 = 1677x as much damage. Edit: Sorry, that oversimplifies it a bit. For the same vehicle, the damage scales that way, but for a fair comparison, you'd have to account for the fact that the weight is distributed over more and bigger tires on each axle. Still, that should show you how the math works.