Originally Posted By: Hokiefyd
Originally Posted By: Vikas
If the tire is inflated to 16 psi, how is the 90-100 psi [yellow/red] being shown on the edge of footprint [bottom right picture]? Anybody knows the scientific explanation for it?
That's the
contact pressure of the tire on the scale that was used for the test. That's not inflation pressure. That image makes a lot of sense: as you lower the pressure, the outer area of the tread will receive the most wear and the center of the tread will tend to want to lift up off the ground. As you increase the pressure, the tire will become more balloon-like, with increasing
contact pressure in the center compared with at the edge.
I understand that. But the contact pressure times the contact area will equate to the total load supported by that tire. Actually, if you integrate the pressure over area, you will get the total supported weight. I suppose I am making lot of simplifying assumptions here but at least for a range of pressures that should be true. If the inflation pressure is way too high, it would seem that the contact patch would be thinner as the weight supported is always going to be contact pressure times area and if you increase the inflation pressure, contact area has to decrease. If the inflation pressure is too low, it would make sense that shoulder of the tire i.e. rubber would be supporting the weight of the car rather than the air in the tire and thus it would make sense that the shoulder would be exerting lot more pressure on the pavement. What was surprising is that even at the recommended inflation pressure, the shoulder edges were exerting lot more pressure in that picture!
Now, in ideal condition contact pressure and inflation pressure will be the same because that part of the tire is not accelerating up or down and thus on that patch of the rubber contact pressure exerted by the pavement has to be same as the inflation pressure exerted by the tire on the pavement. Newton's second law and all that
I believe the real answer to my apparent dilemma is that tire shoulder have NO air supporting it and thus none of my explanation applies there!