Quote:
Originally posted by TJ:
The above psi and diameter rule of thumb does not always apply.

For example, to take it to the logical extreme to have proof of concept...

I have a 27" bicycle 1" wide racing tire tire, and a 27" truck tire...i put 30 psi in each, and load on 2,000 lb.

The bike tire contact patch is as flat and wide as its going to get, with a full inch of tread width by about a full inch of tread length, or, about 1 square inch of tread contact.

The truck tire will have perhaps a 8-10" wide patch, and it wil be MORE than 1/8" long...disproving the concept.

If you LOOK at the tread patterns of typical truck tires...and hold them side by side.....the taller tire has a longer tread patch than the shorter tire, and the fatter tire has a fatter tread patch.

Now, when we compare different aspect ratios...the tread change does do the length to width orientation flip shown.
If the bike tire is as wide as it will get then there are two possibilities
  • the PSI in the tire is perfectly balanced with the weight it is supporting and has the maximum width possible without the rim hitting the ground
    Or
  • The rim of the bike tire is supporting some of the weight because there is not enough air (aka high enough psi) to support the load.

Hint: the simpler solution is the answer. And because bikes with 1" wide tires only have 1" of sidewall...

If the bike tire has a contact patch of 1 in^2 then the air in the tire is only supporting ~30lbs of the weight, the rim of the wheel is supporting the rest. in order to support 2000 lbs with 1 in^2 of surface area, the psi needs to be ~2000psi = 2000lbs/1 in^2 so really the example with the bike tire doesn't apply at all. of course it all depends on how many tires you have, I'm assuming you mean the 2000 lbs to be placed on one tire.

If you had a load of 2000lbs in a tire with 30psi you would need 66 2/3 in^2 , if you have a contact area that is 8-10" wide you'd have a contact length of 8 1/3"- 6 2/6". Now if the 27" truck tire had 2000psi in it then the contact patch would be 1/8"-1/10" long

If the air in the tire is not supporting all the load placed on it (remember psi is a function of weight and surface area) then something else other than the air in the tire is supporting the load. All force vectors must cancel out for an object to be stationary.

BTW if we really wanted to send this thread to ALR or never let it die you only need to mention 1 person and one 1 club...