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.
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.