> A perfect cylinder has a line of contact with the road: the width of the wheel. A perfect sphere has a point of contact.
And both have an area of zero. And the area is what matters here.
> but the perfect examples here are indicative of cylinders having more surface area with higher air pressure
This is not true. It's physically impossible. (What you have done is the geometrical equivalent of dividing by zero to prove 1 = 2.)
Weight of car / Contact patch / number of wheels = PSI + strength of sidewall.
This equation is exact. The geometry of the tire makes no difference. You can not create pressure out of nothing. The pressure on the ground must exactly equal the weight of the car.
And the pressure on the ground must exactly equal the pressure in the tire adjusted for the size of the area of ground contact.
Do you see how for any given PSI (including the strength of the sidewall) the contact patch is an exact figure? The forces must all be equal, it's a basic law of physics.
There are certainly engineering issues, I'm not arguing about that. But the size of contact patch is not one of them. Put the same PSI in a cylindrical or spherical tire (neglecting the PSI contribution of the rubber) and the contact patch will have an identical size.
Typically wheels are made of metal wires and are stiff - so they don't like large contact areas because it means lots of flex. But it doesn't have to be that way. You can make a material that doesn't care about flex - if would be harder to make obviously, but it's not an impossible obstacle.
And both have an area of zero. And the area is what matters here.
> but the perfect examples here are indicative of cylinders having more surface area with higher air pressure
This is not true. It's physically impossible. (What you have done is the geometrical equivalent of dividing by zero to prove 1 = 2.)
This equation is exact. The geometry of the tire makes no difference. You can not create pressure out of nothing. The pressure on the ground must exactly equal the weight of the car.And the pressure on the ground must exactly equal the pressure in the tire adjusted for the size of the area of ground contact.
Do you see how for any given PSI (including the strength of the sidewall) the contact patch is an exact figure? The forces must all be equal, it's a basic law of physics.
There are certainly engineering issues, I'm not arguing about that. But the size of contact patch is not one of them. Put the same PSI in a cylindrical or spherical tire (neglecting the PSI contribution of the rubber) and the contact patch will have an identical size.
Typically wheels are made of metal wires and are stiff - so they don't like large contact areas because it means lots of flex. But it doesn't have to be that way. You can make a material that doesn't care about flex - if would be harder to make obviously, but it's not an impossible obstacle.