Tire width and bead holding ability: 15x8” vs 15x10” Proposed test and discussion

Vtx531

Vtx531

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If the bead is being held by air pressure...
And with a 15x8 and a 15x10, having the same amount of air pressure and same surface area pressing the tire bead against the bead seats, I fail to see how the distance between the two plays any factor.

If I had the means, I would set up a test rig.

1. Pair of wheels, 15x8 and 15x10. Same brand, same style, new with perfectly clean bead seats and flanges to eliminate that playing any factor.

2. Tires - size 33x12.50 mounted on the wheels with the same brand/type/amount of mounting lubricant

3. Some sort of hydraulic shop press with a force gauge. A blunt curved piece of metal fashioned as a bead breaker attached to the press head.

4. Mount the tires on the wheels, multiple times, test different air pressures and what force it takes to break the bead. Someone could come up with a nice set of data plots.

My hypothesis: There would be no statistical difference between the force required to break the bead on the 8” rim and the 10” rim for a given air pressure.
 
CodaMan said:
Physics tells me that a narrower wheel would allow for more side to side tire roll before the bead is forced off.
Click to expand...
Physics and common sense tells me that it takes more air pressure to hold the beads of a 12.5" wide tire seated against wheel edges 10" apart than when they are 8" apart.
 
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CodaMan said:
Physics tells me that a narrower wheel would allow for more side to side tire roll before the bead is forced off.
Click to expand...
I am imagining the air pressure similar to a bead lock. One uses bolts for clamping force and one uses air pressure but the physics should be the same. Does side to side tire roll have anything to do with it?
 
Vtx531 said:
I am imagining the air pressure similar to a bead lock. One uses bolts for clamping force and one uses air pressure but the physics should be the same. Does side to side tire roll have anything to do with it?
Click to expand...
As you turn, especially with lower tire pressure the tire rolls left to right. At some point the tire will be rolled off the bead, but the narrower wheel should allow for more roll before being forced off.
 
Vtx531 said:
My hypothesis: There would be no statistical difference between the force required to break the bead on the 8” rim and the 10” rim for a given air pressure.
Click to expand...

The 8" vs. 10" wheel width debate has been raging on the Internet since its earliest days. Neither side has been able to prove its point with anything but anecdotal evidence and conjecture. The discussion has devolved into who speaks loudest and most often, which is all that one sees when googling the subject.

I'm afraid this thread will turn out the same way.
 
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Mr. Bills said:
The 8" vs. 10" wheel width debate has been raging on the Internet since its earliest days. Neither side has been able to prove its point with anything but anecdotal evidence and conjecture. The discussion has devolved into who speaks loudest and most often, which is all that one sees when googling the subject.

I'm afraid this thread will turn out the same way.
Click to expand...
So who wants to conduct the experiment and put this debate to rest?? :D
 
The wider wheel will add x number of cubic whatevers which will add to the number of air molecules inside the tire.

Based on my statement we could change this thread to either debate whether or not I'm full of $h!+ or how less molecules would do a better job. Heck, we could even argue about both for a while but I'm not going to take a side, not even my side.
 
Vtx531 said:
If the bead is being held by air pressure...
And with a 15x8 and a 15x10, having the same amount of air pressure and same surface area pressing the tire bead against the bead seats, I fail to see how the distance between the two plays any factor.

If I had the means, I would set up a test rig.

1. Pair of wheels, 15x8 and 15x10. Same brand, same style, new with perfectly clean bead seats and flanges to eliminate that playing any factor.

2. Tires - size 33x12.50 mounted on the wheels with the same brand/type/amount of mounting lubricant

3. Some sort of hydraulic shop press with a force gauge. A blunt curved piece of metal fashioned as a bead breaker attached to the press head.

4. Mount the tires on the wheels, multiple times, test different air pressures and what force it takes to break the bead. Someone could come up with a nice set of data plots.

My hypothesis: There would be no statistical difference between the force required to break the bead on the 8” rim and the 10” rim for a given air pressure.
Click to expand...

add to your measurements the distance the press moved before breaking the bead and you probably have something. I'd probably use a 10.5" wide tire instead of a 12.5, too. And get a long ass lever for your press so you can stand safely away from it. There's a lot of energy stored up in that compressed air tank that you're purposely rupturing.

The only difference in the physics from a pressure standpoint is how much net force is pushing the tire bead against the wheel flange, and that's determined by the air pressure against the inside of the sidewall and the force of the tire structure wanting the bead to sit in it's relaxed position (how far apart are the beads when the tire is dismounted?).

When dismounted, if the beads are narrower than the rim, then the tire structure is naturally pulling away from the bead and your net force on the bead will be the tire structures force subtracted from the air pressure force. If the beads are wider than the rim when relaxed, then the net force on the bead will be the tire structures force added to the air pressure force.

The only thing that remains is to determine whether the force from the tire structure pulling the bead to it's natural position is significant compared to the force from the air pressure. Given that a sidewall on a 33" tire mounted to a 15" wheel has about 650 square inches, you can do the math for how much force is being applied by an 8psi inflation pressure and compare that to how easy it is to pull the beads to be 10" apart. I suspect the tire structures role is not significant with a 12.5 tire between 8 and 10" rims. It might be measurable trying to stretch a 30x9.5" tire onto a 12" rim running 5 psi, but no remotely serious offroader is doing anything like that.

The remaining factor is the aforementioned tire roll, which is why I suggested adding the press displacement to your test. I don't have a good way to predict that analytically, but I suspect the difference will be nothing if pressed at the bead, but might give a couple more inches at the shoulder.
 
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freedom_in_4low said:
add to your measurements the distance the press moved before breaking the bead and you probably have something. I'd probably use a 10.5" wide tire instead of a 12.5, too. And get a long ass lever for your press so you can stand safely away from it. There's a lot of energy stored up in that compressed air tank that you're purposely rupturing.

The only difference in the physics from a pressure standpoint is how much net force is pushing the tire bead against the wheel flange, and that's determined by the air pressure against the inside of the sidewall and the force of the tire structure wanting the bead to sit in it's relaxed position (how far apart are the beads when the tire is dismounted?).

When dismounted, if the beads are narrower than the rim, then the tire structure is naturally pulling away from the bead and your net force on the bead will be the tire structures force subtracted from the air pressure force. If the beads are wider than the rim when relaxed, then the net force on the bead will be the tire structures force added to the air pressure force.

The only thing that remains is to determine whether the force from the tire structure pulling the bead to it's natural position is significant compared to the force from the air pressure. Given that a sidewall on a 33" tire mounted to a 15" wheel has about 650 square inches, you can do the math for how much force is being applied by an 8psi inflation pressure and compare that to how easy it is to pull the beads to be 10" apart. I suspect the tire structures role is not significant with a 12.5 tire between 8 and 10" rims. It might be measurable trying to stretch a 30x9.5" tire onto a 12" rim running 5 psi, but no remotely serious offroader is doing anything like that.

The remaining factor is the aforementioned tire roll, which is why I suggested adding the press displacement to your test. I don't have a good way to predict that analytically, but I suspect the difference will be nothing if pressed at the bead, but might give a couple more inches at the shoulder.
Click to expand...
100%!!!!!

I think the inward force exerted of the natural relaxed state of the tire is negligible on an 8” or 10” rim. <1psi to compensate would be my guess
 
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Vtx531 said:
So who wants to conduct the experiment and put this debate to rest?? :D
Click to expand...

I've already run 8" and 10" wheels with 12.50" wide tires. Many years experience with both. I have found no practical difference in terms of lost beads at typical trail pressures that can be ascribed with any reasonable certainty to the difference in wheel width.*

There are advantages and disadvantages to each wheel width. Pick your poison.

Or split the difference with your next set of tires and go for 17x9 wheels. After all, 17's are the new 15's. ;)





_________________________

* Running at 3 psi or less in deep snow doesn't count - no bead is safe on any wheel without beadlocks at that pressure.
 
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after having switched from x10 wheel to a x8 wheel with the same 12.50 tires being swapped over, I found the ride to be improved noticeably.

I prefer the bulging look of the tire on the narrower wheel as well.

can't speak on debeading as I haven't had that happen yet but that is the input I'm able to provide
 
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