Chuck Holes!
Member
Joined: Jan 2008
Posts: 4,371
From: NC USA
his math leaves a lot out. The spring force works like a spring (duh) so the force exerted is directly proportional to the extension and has a high spring constant. The shock has a slowing "drag" effect proportional to either the vertical speed of the wheel or speed squared.
It's a differential equation to solve all that. Nasty enough that I'd need to pull out my dynamics book to solve it. Though to be fair, it's a pretty common equation.
All that said, I again think its about weight transfer and much more complex. If the wheel is at full extension and the rest of the car is unmoved, then the wheel can more easily come back up rather then deform on the edge. If the car has time to shift forward, then it's like hitting a bump. If the car has time to start shifting forward but not reach equilibrium, then the car has some downward energy pushing down on the wheel extra hard. Ie the wheel first has to push the car back up from falling some before it can start to go over the hump itself. Taking all that into account makes it hard to give a best answer.
It's a differential equation to solve all that. Nasty enough that I'd need to pull out my dynamics book to solve it. Though to be fair, it's a pretty common equation.
All that said, I again think its about weight transfer and much more complex. If the wheel is at full extension and the rest of the car is unmoved, then the wheel can more easily come back up rather then deform on the edge. If the car has time to shift forward, then it's like hitting a bump. If the car has time to start shifting forward but not reach equilibrium, then the car has some downward energy pushing down on the wheel extra hard. Ie the wheel first has to push the car back up from falling some before it can start to go over the hump itself. Taking all that into account makes it hard to give a best answer.
Quite right although I would say the force is proportional to the spring compression rather than extension. And shocks don't offer much slowing spring rates on rebound so the tire has a better chance of following the road surface. I did do a differential to come up with the average force of 300 lb then used the simple math to demonstrate that unless youre traveling at 200+ mph the tire will take a beating from the spring but then the horizontal speed generates enough force of its own to virtually guarantee tire blowout. And at speeds over 60 mph tire blowouts become a threat to life and limb. The other guys are right: avoid potholes first, slow down if traffic permits. I believe tire tests by manufacturers have shown that any time the tire falls below the rim height in a hole the tire hass only a 50% chance of surviving at 60 mph.
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