I recently had some body work done on my car, so the body shop had my car for 2 weeks. I also recently put new tires on it (about a month before that).
When I finally picked up my car and drove around locally for a few miles at neighborhood speeds. Later in the evening, I drove on the highway and shortly got a flat tire.
While I think it would be too much of a stretch to blame the body shop, I wonder if there was a nail or something from the body shop that was already wedged in the tire, but perhaps it the pressure that the tire experienced on the local roads wasn’t enough to make the tire go flat. And once I got on the highway, it was high enough to have the air leak out.
First, are you ssure it was a nail or something? I had new tires installed once, and it turned out the dealer had a batch of defective valve stems. At some point all four tires went flat (two within 72 hours) and none of them showed any sign of a nail or road damage.
You could have punctured the tire on the highway. The right type of hole only takes a minute to go flat. The small additional pressure from highway driving might dislodge a nail and speed up the process, but so could regular operation on the side streets. Hitting bumps in the road temporarily increases tire pressure much more than the small amount in highway driving.
There certainly are alternative, perhaps more likely, explanations but there is a factual question being asked.
Standard passenger vehicle hits a routine bump in the road at say 55 mph, what is the transient increase in tire pressure assuming tire is filled to recommended pressure ahead of time?
I do not have the answer myself but it seems tractable.
ETA driving flat highway is just the heat increase, maybe 4 to 5 psi.
Not much. It takes a really big bump to change the enclosed volume of the tire by a significant amount. The thing that causes most of the increase in contact force when hitting a bump is the change in the area of the contact patch. I’m not in a position to do the math right now, but I’d wager that moving the wheel’s axis of rotation a quarter inch closer to the pavement might cause the contact patch area to increase by 30% or so. I’ll post some calculation results later this week.
Curious if that’s load-dependent. My truck shows 7-8 psi difference after a few hours of highway towing. I haven’t checked it unloaded, as I rarely travel in it that way. Now I’m curious and will pay attention next time I head to my Dad’s house (often go empty 1 way on that trip).
If you have a non contact thermometer, try taking the tire temp before and after the trip. Both empty and loaded. I never tried that, but after a hi-way trip the tires feel significantly warmer to the touch.
I wonder if it really does? The assumption here seems to be that the air in the tire compresses in direct proportion to the deformation of the tire when it hits the bump. But the tire is somewhat malleable. Perhaps the non-bumped portion of the tire expands somewhat during the bump, thereby reducing the change in psi.
That’s exactly what I’m finding trying to determine what increase in pressure occurs. It can be enough to cause the bead to separate from the wheel momentarily allowing a lot of air to escape, but at the same time the tire is being deformed by the bump so the air can start escaping from that alone. I haven’t found anything that settles the matter. I suspect that bumps can create pressure waves within the tire of very high pressure but very brief and rapidly dissipating.
I did some math and some simple 2-D CAD work on this, and now I’m pretty sure bumps don’t meaningfully increase tire pressure.
I used my own car as a model. The tire is a 225/55-R17, and with a curb weight of ~3700 lbs, the per-tire load is 925 pounds.
With the tire filled to 35 psi, it needs a contact patch area of 26.4 in^2 to bear that 925-lb load. Based on the geometry of that tire - most importantly, the tread width and tread diameter - the tread has to deform just 0.084 inches away from its circular profile to create a contact patch that’s 2.98 inches from front to rear and 8.86 inches wide (that’s the 225mm in the tire spec).
Viewed from the side, the projected area of the whole tire’s sidewall is 334.6 in^2. This is the area of a circle the same diameter as the tread OD (26.74"), minus the area of a circle the same diameter as the tire bead (17"), and minus the area of the circular segment whose chord is the front-to-rear length of the contact patch (2.98").
Now suppose instead of 925 pounds, I want the tire to bear twice as much load, 1850 pounds. Assuming no change in tire pressure, I need twice as much contact patch area to make this happen. The width of the contact patch can’t change, so the contact patch needs to become longer from front to rear. To make this happen, the tread has to deform 0.337" from its circular profile. This changes the projected area of the sidewall from 334.6 in^2 to 333.4 in^2. If you assume the tire volume is proportional to sidewall projected area, then the tire volume has decreased by just 0.36%, with a commensurately tiny increase in tire pressure. However, that assumption is suspect, because when the pavement deforms the tread away from its circular profile, the sidewalls bulge outward visibly: the width of the enclosed tire volume increases a bit, offsetting the decrease in height and therefore mitigating the volume decrease (and pressure increase) to something less than 0.36%.
Suppose instead of twice the usual static load, I want the tire to bear four times the norm. Instead of 0.084 inches, the tread is displaced 1.41 inches from its usual circular profile. The sidewall projected area is reduced from 334.6 in^2 to 323.5 in^2. Assuming no sidewall bulge at all, and also assuming adiabatic compression, the tire pressure goes from 35 psi to 36.2 psi. This is less than the temperature increase that occurs during normal driving due to viscoelastic heat production. It’s also not enough of a difference to dramatically affect leakage rates, unless that tire distortion during bumps also increases the size of the leak path (which can happen).
It’s dependent on how fast and how much the sidewalls and tread flex. These are a function of inflation pressure, load, and road speed. Taken to its logical extreme, a tire that carries zero load will not heat up at all. For a non-zero load, if the sidewalls are flexing too fast/too much, they can generate enough heat to compromise structural integrity of the tire carcass. You get around this by reducing the load, reducing the speed, or increasing the inflation pressure (or moving to a cooler climate). The problem in the late '90s with the Ford Explorer and its Firestone Wilderness AT tire failures was attributed in part to Ford specifying too low an inflation pressure in the Explorer’s owner manual:
A tire bead can be unseated from its installed position on the rim if you drive into a curb or other object at a “bad” angle (here “bad” means somewhere between straight on, which will just smash your rim, and almost-parallel so that the curb merely redirects your car’s path of travel instead of peeling the tire out of its seat on the rim). Unseating the bead results in instant 100% deflation, and you’ll need to call AAA for help. But I’ve never heard of an ordinary vertical bump that was small enough to not bend the rim, but somehow large enough to cause a transient leak path to form.
Relative to the tire, the bump is moving at a small fraction of the speed of sound. Any “pressure wave” coming off of it will manifest as a bulk dislocation of air that quickly equalizes pressure throughout the rest of the tire volume.
There might be a pressure wave from the impacts … The tyre moving up suddenly compresses the air next to it… much more significant than any change in contact patch size… but hard to calculate or measure… maybe the noise volume can be used as a measure ? But the noise can be outside the tyre as well as in …
But certainly a screw or nail could be thrown out of the tyre by highway speeds and the higher pressure waves that hit it. Its plausible but not the common experience.
I mentioned pressure waves but didn’t intend that to mean they would likely push a nail out of a tire. A pressure wave could greatly increase pressure at one location inside the tire very briefly but maybe not even close to the tire. @Machine_Elf description of that effect sounds pretty good to me.
I think leaking around the bead does occur with poor seals. A shock to the tire doesn’t have to result in total deflation, the bead will close up again, usually not well and leaking more over time. But again, that happens when the tire is deformed from impact with an object, not from any momentary increase in pressure if there is any, and your calculations make that sound unlikely.