Originally Posted by bjoerges
Car tire: 4 * (4" x 6") squares of tire touching ground at 32psi: 4 * 24sq.in. = 384 sq.in of contact. 32psi * 384sq in = 12288 lbs of support.
Bike tire: 2 * (1.25" x 2") squares of tire touching ground at 60psi: 2 * 2.5sq.in. = 5sq.in of contact. 60psi * 5sq.in = 300 lbs of support.
Actually it's the other way around with the buoyancy equations. The tyre will deform and compress based upon the total load vs. pressure:
3200lb car / 4 tyres = 800lbs on each tyre (assuming 50/50 weight-dist F/R & L/R)
800lbs / 32lbs/sq.in = 25sq.in. contact-patch per tyre
4000lb car / 4 tyres = 1000lbs on each tyre
1000lbs / 32lbs/sq.in = 31.25 sq.in. contact-patch per tyre
Obviously a heavier car at the same 32psi pressure causes the tyre to compress more and have more of it touching the ground. The actual shape of the contact-patch will depend upon the width of the tyre. Narrow tyres have more square-shaped contact-patches while wider tyres have rectangular patches.
175lb rider + 25lb bike = 200lbs total load
200lbs / 2 tyres = 100lbs load per tyre
100lbs / 60lbs/sq.in. = 1.67 sq.in. contact-patch per tyre @ 60psi
100lbs / 120lbs/sq.in. = 0.833 sq.in contact-patch per tyre @ 120psi
So by pumping up the tyre to a higher-pressure, it compresses and sinks less for any given load. This prevents pinched flats where the tyre sidewall compresses completely and the rim-edge cuts through the tyre & tube (results in snake-bites pattern of holes). There's diminishing returns with higher-pressure though and you end up with lots more road-shock and discomfort on larger tyres.
Someone posted a great equation last week on finding optimum tyre-pressure ranges based upon weight and tyre-width.