Tire Pressure/Contact Area/Weight

answer to Dough

I also suspect that generalisation, mayby they determined the surface at zero pressure and forgot that this surface gets smaller when tire is stiffer at higher pressure.

But first assumed that we did not have to do the work of determining surface for this tire, because it was so neatly given at the bottom.
I expect the deflection give to be right and staying the same over the whole range for a sertain speed. At least i expect that being the goal of the tire makers.

To determine that surface you could use more then 2 sheets of paper.
2 for the sides , 2 for the front and back and again 4 wich you put a bit in the rounding on front and back.
Glue it to the flat underground ( asphalt or concrete) and drive away and measure the area within the sheets more accurate.

I have been trying to tie contact area to pressure and load. Looking at the chart, at "0" Km/h, surface area does not back-calculate as a constant 1128 cm2 and, in fact, varies considerably. So this effort does not hold up unless that contact area is just a generalization.

jadatis wrote:
The picture of the footprints from capri-racers article give that much more pressure on the ground at the given pressures, that you can explain it by the overall surface minus groves and the sidewals press a bit more to the ground.
Even at the 16 psi most of contact is about 20 psi but even at the edges at the sidewals some spots up to 90 psi.

I also have a pressure loadapacity list from Michelin wich has 3550 kg at 6 bar with 777cm2 contact surface given .
this thoug would mean when supposing bar to be the same as kg/cm2, wich it is almost.
3550/6= 591 cm2 so this would imply that my theory of a part of the load being bare by the construction of tire to be wrong.
But how accurate is this given?

But if you would put the tire on a blunt nailbed it would give the same bending so total surface included the grooves .

I also calculated in that list for 1400 kg 1.8 bar so can be because of rounding 1.7 to 1.8bar. given
Then the 777 cm2 surface comes closer. Asuming again 45kg bare by the construction gives 1355kg/777cm=1.75 kg/cm2 <> 1,75bar.

That is more to the idea I have about that higher pressure makes tire stiffer so at same deflection lesser surface(length) on the ground. For 3550kg at 6 bar it would give 3505 kg//5,95kg/cm2=590cm2 to be the surface on ground inclusiv grooves.

Here a picture I made for it to show how the surface lenght can be smaller for a stiffer tire, even if the deflection is the same.
The smaller surface at same deflection becomes because of the larger overgoing curve from unloaded tire radius to flat on the ground.



OK also the picture of the list I got it from , notice the 777 at the bottom.

DougE wrote:
Yea! A real world example. Thank you Gdetrailer! It sounds as if it would be reasonably accurate to determine the loading on each tire and could help with determining the load on the rear tires with and without the trailer attached or the actual weight of the trailer when it's not practical to get to a scale.

I didn't invent this "method", somewhere I had seen it in a science magazine for kids (we had several different kids type subscriptions for our DD).. Forgot about it then sort of remembered it sometime later and had to do some Internet searches to find the exact info so I could try my curiosity out on the theory..

Anyways, I would use it as a way of "guesstimating" since the formula does not take into account that the contact patch is NOT 100% rectangular and does not account for gaps in the tread that are not contacting..

So, I did another search and found a decent website explanation that might be most helpful for visualizing how it is done.

Tire contact patch step by step instructions plus diagrams..

HERE

Yea! A real world example. Thank you Gdetrailer! It sounds as if it would be reasonably accurate to determine the loading on each tire and could help with determining the load on the rear tires with and without the trailer attached or the actual weight of the trailer when it's not practical to get to a scale.

DougE wrote:
Not real life numbers. Just a mathematical example to test the hypothesis and start a discussion on a cold winter day. If true, could be a way of figuring weight on each tire, knowing the pressure and contact patch. I know of a high school science teacher that slipped sheets of paper under the front, rear and sides of a tire to enable measuring the tire contact patch to come up with the weight on the tire as stated. Force (weight) = pressure x area.

I have done this many years ago and it DOES "work" to a certain point.

It will get you in the ball park of how much weight is "resting" on the ground UNDER the tire.

The key to accuracy is getting accurate measurements of the actual contact area.. It also requires tires on concrete or asphalt so you can get the sheets of paper snugly against the tire without interference from dirt, rock or gravel.

Then you have to make sure the paper is SQUARE to the tire, any out of square measurements will affect the result..

I did this on a pickup truck and the overall weight I measured was within 100 lbs of the door sticker unladen weight. Not bad for a truck weighing 5500lbs!

The picture of the footprints from capri-racers article give that much more pressure on the ground at the given pressures, that you can explain it by the overall surface minus groves and the sidewals press a bit more to the ground.
Even at the 16 psi most of contact is about 20 psi but even at the edges at the sidewals some spots up to 90 psi.

I also have a pressure loadapacity list from Michelin wich has 3550 kg at 6 bar with 777cm2 contact surface given .
this thoug would mean when supposing bar to be the same as kg/cm2, wich it is almost.
3550/6= 591 cm2 so this would imply that my theory of a part of the load being bare by the construction of tire to be wrong.
But how accurate is this given?

But if you would put the tire on a blunt nailbed it would give the same bending so total surface included the grooves .

CapriRacer - At some point in the discussion the force acting downward at the axle must equal the force acting upwards at the tire contact patch. Your reference seems to detour around that basic effect. I.e., Newton's third law.

never mind, not worth commenting on.

msiminoff wrote:
......and if that is the pressure inside your tire then 50 is also the average PSI on the contact patch.

Cheers,
-Mark

Sorry, Mark, that is incorrect. The inflation pressure doesn't equal the average contact pressure - except by coincidence.

I talk about that here:

Barry's Tire Tech - Air or Tire?