That is correct.

The ground is pressing upwards on the tire with 800lbs of force. The tire is pressing down on the ground with 800lbs of force. And we know the tire is pressing down with a pressure of 30 lbs per square inch...


WHY is my science flawed? You say that, but you aren't telling me why.

The ONLY evidence you have offered is your extreme case of 1psi, which I have explained. With a properly inflated tire, the pressure is FAIRLY even across the contact patch at 30psi. (there may be some slightly higher pressure on the sidewalls, but for the most part, it is spread pretty evenly). Knowing there are approximately 800lbs pressing down and there are 30lbs per square inch, then there are approximately 26.7 square inches of contact patch. This is required by Newton's third law.

Since the width of the tire is fixed, then the front-to-back patch is what must change depending on tire pressure and weight.

If the ONLY THING YOU CHANGE is width (the 6inch wide tire) then the front-to-back distance of the contact patch MUST lengthen to keep approximately the same area touching the ground.

Where is the flaw in that logic?

"The shape of a tire's contact patch or "footprint" greatly influences its performance and is dependent on its profile or "aspect ratio". Low profile tires (most performance tires) have a short and wide contact patch that is effective in converting the driver's input into very responsive handling, cornering stability and traction...especially on dry roads.

High profile tires (light truck and most passenger tires) have a long and narrow contact patch which helps to provide predictable handling, a smooth ride and especially good traction in snow."

They say The shape of a tire's contact patch or "footprint" greatly influences its performance and a short and wide contact patch that is effective in converting the driver's input into very responsive handling, cornering stability and traction.

They NEVER mentioned the contact patch's SIZE. They discuss the SHAPE!

My argument specifically addressed aspect ratio as I kept it identical to wash it out as a variable. It is a four inch constant in my model.

Here is the original model from an earlier post in this thread:

"Let's assume two different tires: One is 24 inches tall, 6 inches wide. The other is 24 inches tall, 12 inches wide.

The aspect ratios and rim sizes are different in a way that leaves them both with the same amount of rubber between the tire and rim. Let's say 4 inches from bead to tread.

Now, with my extremely unusual tires, we have identical sidewall properties. They are the same overall height, they have the same amount of rubber between the rim and the tread. The only difference is that one is twice as wide as the other and they have different size rims."

Assuming they are inflated to the same pressure and have the same amount of weight on them then I believe the contact patch will have just about the same area. But the 295 tire will have a slightly shorter (front-to-back) contact patch, since it is wider and has the same overall area touching the ground.

The area should not change. You have 30 psi and you have 800 lbs. That is 26.7 square inches of area. The area becomes a constant, and the width is a variable. Given the two widths on the different tires, the wider one MUST have a shorter front-to-back contact patch.




[Modified by Tom Steele, 9:29 PM 8/5/2002]

 

All I'm going to say is that 30psi inside the tire doesnt mean that there is 30 lbs per square inch where the rubber meets the road.

Thats what I've been saying all along.

If you can prove otherwise I'll be happy to hear it. Everything you've said is based on this one assumption.

You got me curious enough so I went out to the garage and measured. Its not exact, but I put a piece of paper on the floor and pushed it into the tire. I then put another behind. Made sure they were parallel (eye ball) and measured the gap. Got on my hands and knees and did the same to measure contact width. Lets just say that I know my car doesnt weigh 1317 lbs at the left rear wheel. I have weighed it a number of times.

Eric

 

http://www.howstuffworks.com/tire4.htm

"The next time you get in your car, take a close look at the tires. You will notice that they are not really round. There is a flat spot on the bottom where the tire meets the road. This flat spot is called the contact patch...

If you were looking up at a car through a glass road, you could measure the size of the contact patch. You could also make a pretty good estimate of the weight of your car, if you measured the area of the contact patches of each tire, added them together and then multiplied the sum by the tire pressure.

Since there is a certain amount of pressure per square inch in the tire, say 30 psi, then you need quite a few square inches of contact patch to carry the weight of the car. If you add more weight or decrease the pressure, then you need even more square inches of contact patch, so the flat spot gets bigger."

I'm still looking for more resources and will post them as I find them. I'm going to try your experiment too, but I would say that if you came up with 1,317 that wouldn't surprise me too much, that is going to be a tough measurement to get exactly and a few square inches off is a lot of lbs...

 

I have to jump in here. Eric is right on. The air presure inside the tire has very little to do with the pounds per square inch load on the tire. A couple of quick examples and you will see how the two are almost completely unrelated. Tire on the car inflated to 30 psi with 800lbs load spread over a regtangular contact patch. remove the tire from the vehicle and stand it up, load is now reduced to the weight of just the wheel and tire, say 40 lbs. or 5% of the on the car situation. Contact patch is reduce by 95%. No way. Go more extreme, imagine a tire with the same diamiter as stock, but 20 feet wide and inflated to 30 psi. If the contact patch is only 2 inch long it would still be 240 inches wide and be 480 square inch. If the 30 psi would work it would be holding up 14,400 lbs. If the 800/30=contach patch to work the imaginary 20' wide tire would have a contact patch of less than an 1/8 of an inch long. Third idea; park the car level with 3 tires touching the ground normaly and the fouth tire resting on a steel beam only 1" wide. The tire will deform somewhat making the contact patch longer but not enough to equal the contact area of the other three tires. If they are say 2"x10" the one resting on the 1" wide beam will not be 20" long, therefore the contact patch will be smaller and the lbs. per square inch will go up. A large truck will have virtualy the same air pressure measure at the valve stem both empty and fully loaded.


[Modified by Terry Humiston, 11:54 PM 8/5/2002]


[Modified by Terry Humiston, 11:57 PM 8/5/2002]

 

I will try to keep this within your static paradigm.

The force from acceleration must deform the tire, that force is distributed on the contact patch, the toe of the patch is in compression.

Torque is force x distance or Lb-Ft. The distance is the Radius of the tire

If the contact patch on both the narrow and wide tire is the same then in your model both tires produce the same torque.

If you agree and reality supports the proposition that the wider tire provides more torque then the effective radius most be different under the applied loads.

Why??

The contact patch of the narrow tire is creating the torque further ahead of vertical, this effectively reduces the radius of the applied force. In reality I think the force vector has more vertical component and thus the torque is reduced.

In other words the narorw tire thinks it is going up a steeper hill than the wide tire, among other consideration such as buckling discussed previously.

 

As I rememer it the mag in question was Hot Rod and they didn't say it was impossible only that the car would have to average 1G of acceleration for the whole 1/4 mile to do 150.

 

Just a note, I haven't bailed here. I'm just doing some serious reading and researching to try and find more answers before posting. It is suprisingly hard to find good information on tires!

 

Real quick though...

Why not, it should be able to sit on an extremely short (front-to-back) and wide contact patch.

I don't have a problem with that. 1/8 inch is fine with me. 30 pounds per square inch is the key! :)

More interesting and a scenario that I believe takes you past the limit of a simple assumption of a homogenous 30psi across the tire. I freely admit that using 30psi is an approximation. There is no doubt that when called on to do so, the sidewalls start working harder and providing the extra psi of support.

I would say the sidewalls start taking up slack here. Any extremely low psi scenario begins to involve the sidewalls.

This is kind of like Newton vs Einstein. Newton was right at "real world speeds." Or right enough for most of us in most situations. There is no reason to bring relativistic physics equations into a drag race.

That is more-or-less where the properly filled tire situation exists. It isn't perfect to assume 30 psi exactly, but it is a good approximation. I am searching for the more exact formula, but haven't found it yet.

However, even when I do, I suspect it is going to wash out and even thought the contact patch may turn out to be (bigger, actually) than predicted by the simple 800 lbs/ 30psi model, I think that it is going to still leave us with the same scenario:

Similar (almost exactly the same) sized contact patches for a skinny tire and a fat tire, with different shapes.

Which will leave us with the same questions:

Why is wider better for handling, and why is wider better for drag racing?

So will your Corvette. Try jacking the car up and taking the tire pressure, then set the car down and take the pressure again. It shouldn't change.

 

My wife agrees! :D

 

Solution: 1) When tires are cool they do not perform optimally BUT when they are too hot they start to melt and break down.
2) For drag racing fat tires do two things: they increase the total potential contact surface of the tire aka the total area that comes in contact with the ground with any one revolution of the tire, thus decreasing the amount the tire heats up so the rubber wont loose its optimal cof. also they are deflated making more surface contact which DOES NOT increase the total friction force but it DOES increase the resistance for any portion of the tire to heat up or break off due to mechanical forces.
SO You dont use skinny tires because they heat up too fast and break apart. Thats why the people who enter the burn out contest and catch there tires on fire usually have skinny tires.

--this was the explination offered by my phisics professor all the way back in high school. :flag


[Modified by 1990bevile, 5:04 PM 8/8/2002]

 

I have three cars in my garage to use for some quick measurements. The actual contact patch is not quite a rectangle but almost. Use two large pieces of square thin cardboard. Slide one along the floor until the edge jams between the front edge of the tire and the floor. Do the same thing with the back edge. Keep the outside edges parallel to the side of the tire using a straight edge along both pieces of card board? Measure between the two edges and this will be the width of the contact patch. Now measure the actual tread print width. This is easy to see by the tire tread wear patterns. Using this method you will be able to measure the resting contact patch fairly closely.
Here are my results:
Neon with 205/50-15 weight 2581, 30 psi in all tires. 5.6”L * 6.5”W or 36.4 sq. inches
Honda Van 215/65-16 weight 4328, 32 psi in all tires. 6.4”L*6.0”W or 38.4 sq. inches
Z06 rear 295/35-18 weight 3044, 32 psi in all tires, 4.7”L*10.5”W or 49.3 sq. inches

If the Honda had even weight distribution (I measured a front tire) it presses the ground at a bit over 28 psi. (4328/4=1082/38.4=28.2)
The Neon would be only 19.6 lbs per square inch (2581/4=712.7/36.4=19.6)
The Vette would be even less at 15.4 lbs. per square inch (3044/4=761/49.3=15.4)

 

The answer here is subtle.

First off, ASSUME the sidewalls offer no resistance to 'bending' towards the middle. In other words, setting the tire on the narrow beam will deform the tire, pulling the sidewalls towards the middle. If they offer resistance, 'pulling' back, that will help support the tire, decreasing the remaining support needed from air pressure alone.

Now, the contact patch is the only thing supporting the tire on the beam. If the beam is very narrow (say 1/2"), preventing an appropriately sized patch based on 30psi (the tire isn't 40" long), then the tire is going to compress, raising the psi until it's high enough to support that tire with the amount of contact patch available to it.

If the contact patch is too small, like 1" square, the air pressure will either increase to 800psi, the tire will contact the rim so the rim is supporting the car not the air, or the tire will rupture.

Obviously, sidewalls DO offer resistance to flexing, but the argument is still valid. Contact path area IS directly proportional to the air pressure and the weight on the tire. Wider tires have shorter patches.

 

Simple: it has nothing to do with pressure. More surface area == more friction when pushing the tire SIDEWAYS.

 

Tom, one of the things I think that is wrong in your assumption is that when the tire width increases from 6 to 12 inches, the length of the contact patch decreases from 4 to 2 inches. The length of the contact patch does not decrease proportionally to the increase in the width of the contact patch.

If the width increases 100%, then the length of the contact patch may only decrease 22%, resulting in a larger total contact patch for the wider tire.

Assuming the pictures of the contact patches you posted fairly represent actual contact patches, and they look like they do, take some measurements, do some calculations and you'll see where I'm coming from with this idea.

 

:sleep: :sleep: :sleep: :sleep:

 

Alan,

Just to let you know where I'm getting that. I am assuming 30psi in the tire, and 800lbs weight above it. So, that means a constant of 26.7 square inches.

So, assuming that since weight and tire pressure remain constant, then the contact patch HAS to remain constant. That means that the front-to-back contact patch distance would have to change proportionally to the side-to-side contact patch distance. Assuming that the basic assumption is a good approximation.

So far, that has been difficult to prove or disprove and is where I am right now in the research. Trying to find out if the tire can be modeled as a simple "balloon-like" air-container, or if the sidewalls (from which the car hangs - right everyone? :) ) make a difference in the size of the contact patch.

I have also seen something that suggests that there may be 30psi at the center (front-to-back) of the contact patch, but then it lowers as you move out from that center-line, approaching zero at the edges. That would result in a larger than expected contact patch as the AVERAGE PSI would drop...

I haven't confirmed that yet.

It looks like some expensive SAE books may be the next step.

 

The reason I think the wider 12" tire works better for acceleration is that the sweap area of the tire is going to be twice that of a 6" tire while it rotates.

 

Tom, your problem with finding the solution remains in your original assumptions. 2 tires of the same rim size and tire height but have different widths will not have the same psi rating. The wider tire will have more volume and require more air, which will give it the same contact patch lenght but ofcourse will be wider, there is your larger contact patch. The only way I could even guess what would shorten the contact patch area would be to put the wider tire on a larger rim to stay at 30 psi. But ofcourse that would defeat the purpose of going to a wider rim.

 

Actually, my model does not have the same rims. I have made the model a little unusual, so as to keep the tire-sidewall height the same on both tires to try and take sidewall "scrunching" out of the equation.

I may be reading this wrong, but I don't think volume makes any difference in the equations, other than the amount of volume it requires to reach a certain psi. But since my model had both tires at 30 psi, then volume doesn't effect any of the equations.

800 lbs over each tire (approx)
30 lbs per square inch in the tires

800/30 = 26.7 square inches of contact patch, regardless of the width of the tire...

While I realize the skinnier tire might have a suggested air pressure that is different than 30psi, I could have made the pressure 36psi and we would have been WELL WITHIN the reasonable running pressures for almost ANY tire made for an automobile and street use.

Another example is to compare a Z06 tire to a C5 coupe/conv tire. Same AIR PRESSURE, similar weights, but the Z06 tire is wider... The wider tire on the Z06 doesn't seem to allow for any great pressure changes.

I still tend to lean toward a more vector-oriented answer.

With a short/fat contact patch, it may be easier to change the direction of the contact patch's vector, than if it is longer front-to-back. And it may be that the long side-to-side vector resists sliding better.

That would help answer the handling question, but not the drag racing question.

With drag racing, if it merely a matter of having more overall rubber (with each rotation) sweeping the pavement, then TALLER (diameter) tires would accomplish the same thing...

 

From an engineering and physics standpoint, all the replies on this subject has ignored the very important fact that tire traction is not only a component of friction, but also a component of the tensile strenght of the the tire's rubber in shear. When you reach the limits of friction traction, the tire starts to slide or spin and the rubber catches sharp points on the road and tears away from the tire surface. While doing this, it stretches and provides an additional traction component.
If friction were the only factor, a tire would have the same breakaway force on any road, rough or smooth. As we have all experinced, this is not true. The road surface is rough with many sharp edges that form a mechanical coupling with the rubber. When the tire spins, the rubber must be torn away from the tire and left on the roadway. That's why burnouts cost money.
The reason that a wide low profile tire provides better accelleration is that the physical shape of a two inch by twelve inch surface provides a better surface for thousands of stretched rubber bands to pull the tire forward than a four by six inch square which would be a much more stable surface and does not allow rubber deformation.
Tires that offer the best traction have a rubber blend that is soft to allow a good bite and with a high tensile strength to avoid tearing the stretched rubber loose from the tire.
If friction were the only criteria for tire traction, the shape of the tire would matter little.

Dennis