Note, the "common" approximation formula for friction has NOTHING todo with pounds per sq. inch ... it simply is the normal force and the coefficient of friction ...
Here is a quote from the wikipedia article you pointed me to:
Traction between two surfaces usually depends on several factors including
Material properties of each surface.
Macroscopic and microscopic shape or "roughness".
Force of contact.
Area of contact.
Contaminants at the material boundary including lubricants and adhesivesNotice the third factor there where it says force of contact, that is PSI
To spell it out ...
Quote
... If the normal force is increased, per given area of contact patch, the coefficient of friction decreases ...
SO ... 500 pounds on one square inch will have a LOWER coefficient of friction. call this U1
if I have 1 of these I have a normal force of 500 lbs
Quote
... and as the normal force decreases, the coefficient of friction increases.
SO ... 5 pounds in one square inch will have a HIGHER coefficient of friction. call this U2
if I have 100 of these I have a normal force of 500 lbs
SO ...
U2 * 500 lbs > U1 * 500 lbs
WHAT?Look it is very simple, 4,000lb vehicle 1,000lbs on each tire. If the contact area is 10 square inches then the force is 100lbs per square inch. If the contact are is 20 square inches then the force is 50lbs per square inch.
Now back to ice. when a vehicle is driving on ice many of the components of traction are compromised, One aspect of traction which is not affected is (force of contact) see above. The greater the force between the tire and the ice the more traction you are going to have. Period, end of story.
Look what is your point with all this, If you are saying that a tire with a smaller contact patch on ice does not provide more traction than a larger patch when all other factors, vehicle weight, tread, rubber compound, etc are equal then you are just plain wrong.