Vids wrote:Here is an example to help things along (using round numbers for simplicity's sake):
The only way the internal tire psi affects the psi exerted on the ground is by changing the contact area (less psi = bigger contact area).
But the support of a properly inflated tire is precisely the dominant mechanism for supporting the weight of the bike and rider (with a much lesser, but nonzero, influence of the tire's rigidity. Consider a tire closer to a balloon (almost no rigidity) and a steel tank (very much), to use examples earlier in this thread; also consider how a 20lb bike sinks to the ground on a flat tire, to get an idea of the degree of support from the sidewalls alone).
A cushion of air in a tire -- if sufficient pressure exists to keep the rim off the ground -- is responsible for nearly all of the support of the weight above it.
A tire is a container for holding sufficient air to support the weight above it. (Bridgestone).
The volume of the tire remains nearly constant (sinks lower and stretches fatter); the mass of the air in the tire remains constant; the pressure remains nearly constant.
The issue with your example is that the numbers are both simplifed and made up
But if those were real weight and contact patch sizes, the pressure in the tire to yield those results would be close to 11psi.
See mtnfiend's actual experiment earlier in the thread. It was a great match of empirical observation to theory -- and he began the experiment being dubious of the theory!
In very low psi examples, when the pressure isn't enough to hold up the entire weight, gravitational forces are sufficient to overcome all of the air pressure in the tires, and any remaining unsupported weight is supported by the structure of the bike. Maybe this is the disconnect between people saying "internal psi != external psi" -- because it's an example when not all of the support is given by the air pressure.
Perhaps it helps to think of an inflated tire as a crude scale: the more weight pushing down on it, the larger the contact patch is, up until the internal pressure of the tire is overcome. In that case, the analogy would be a fat person standing on a spring-loaded scale, and bottoming out the scale! All we could conclude is that the weight was at least heavy enough to overcome the support of the air (or spring).