Tire contact patch physics question
 

Ok, I'm in a bit of a pissing match on a motorcycle forum with a guy (who is actually pretty smart and a good guy) about this subject. He swears that the amount of tread on the ground has nothing to do with the tire size, or in the case of motorcycles the profile of the tire. He claims that the only factors that effect how much tread is on the ground is the weight on the tire, and the tire pressure - regardless of how big or small the tires are.

I've used the Corvette as an example. I've asked him why GM would put a 325 tire on the Z06 vs the 285 on the base car. According to him, if the weight and tire pressures are the same then the contact patch area is the same for both tires.

I've contended that the physics theory that he subscribes to is taken out of context when making that argument. What do ya'll think?

Sorry, the other guy is correct, as far as thread contact area is concerned.

If he is right, we should still have bicycle tires on our cars.:D

I think you mean "tread"?

:D

That's exactly the point I've made.

So you are saying that the 325's on the Z06 really are just for show, and that all these wide-ass tires we are seeing on ever faster cars are a waste of money, and that manufacturers are willing to add more unsprung weight just for looks?

Here is an interesting read I admit to only skimming over it, but it would appear to support whichever theory you want it to, but leaning toward a conclusion that wider tires do put more rubber on the road.

http://performancesimulations.com/fa...on-tires-1.htm

325 Is a bigger tire surface to the ground causing better traction and I will tell you, the best advice is to let the guy believe what he believed, Gm did that because the car had more power and they were trying to get the power to the ground so the car can be more enjoyable, the car is also lighter than the base C6, so if weight was the factor, maybe the 285 would have done the job.
Motorcycle is a different factor unless its a straight line bike because a bike mostly uses the side curve of the tires because of the leaning process of riding.

Tire width vs friction vs heat vs stability
 

A wider, lower profile tire gives more stability, as it is lower and wider (duh on me). Try to roll/flip a manhole cover sidways versus a ladder. A lower profile tire also does not generate heat like a taller sidewall tire. Each time the tire rolls around, and touches the ground, it flexes the sidewall, doing this generates heat, the faster you go, the more heat is generated, as the sidewall flexes.

In statics, the normal force, times the friction coefficient, equals the total force to the road a tire can generate. There is nothing in the equation concerning area (contact patch). A wider tire distibutes the load differently than a narrower tire, in physics, pressure equals force(weight) divided by the area. The wider a tire the less pressure per square inch it forces onto the road. Although none of this really helps with tire dynamics, its too generic. I think contact patch size is mostly irrelevant to grip.

Grip has other factors, like total vector sum of all forces in the plane of tire contact with the road. If you roll into a corner, no power, no braking, the vector sum is only the load holding the car into the corner, F=ma. When you brake, or power on, then you add a small vector in the direction of the acceleration. If the new voctor sum is greater than the total grip the tire can withstand, you will slip, with now a lower kinetic friction coefficient. The new sum is the hypotenuese of the triangle, which is the greater of the 3 sides of the vector summation. That is why you never gas, or brake in the corner, but before.

Ole Tad Juechter (Corvette Chief Engineer) says that tires, wheels and shocks account for only 10% of a cars handling abiity anyway. Tire and steering technology is too in depth to ask simply"Why wider tires", see Hans B Pacejka, the magic formula; it is a very complex look into the physics of tires, slip angle, steering angle and tire load.

But.......big fat tires are the shiznizel!!!!!!!!!!!

This is not my area of expertise, it is aircraft structural integrity, but a physics weenie told me long ago,"Tire width has nothing to do with grip, it has to do with heat buildup", I have been curious ever since, so I like to chime into these threads about treads.

Doug Jones
Aerospace Engineer
BS/ME Physics/Mechanical Engineering

You both are right. If the two tires were mounted on the same size rim then I could see the two contact patches being the same. If the wider tire is mounted on a wider rim it will not curl up on the sides causing the contact patch to stay the same.

Another point is why do drag cars run skinnies up front? Less contact patch = less friction= less drag.

That being said wider is better to an extent. A 345 on a 12" rim on my Z06 works out but that same setup on say a prius would suck at life. Too much width can cause too much drag or friction. So what I am trying to say is a stretched out 315 on a 12" rim thats super sticky would be more effective than a 345 thats mildy sticky.

The physics of of a smaller and thinner rubber will not hold to that of a stronger more reliable size. As the saying the cup is only half full not half empty, your counter part will seem to endlessly indulge in a war against facts of this matter. The wider tire will eliminate more negatives due to more positive (345 vs 285) wider in the case does prove to be the better route. Physically provides the better "compound" for or equal to the power to weight ratio. All in all your friend seems to wanna win a never ending course of rebellion to facts, but lacks the knowledge to sustain the truth above all.

The other guy is correct. For a given tire pressure, the contact patch AREA is the same. What changes when you go to a wider tire is the shape of the contact patch. A narrower tire will have a more square contact patch (fore/aft and side to side). Now if you go to a wider tire the fore/aft dimension is reduced but the side to side dimension increases. As you can envision, for lateral handling a contact patch that has a wider dimension will have better lateral adhesion.

regarding the OP: to a good approximation, the other guy is correct. f = sum(p*da) (assumes constant pressure)... assuming most of your force is caused by the air pressure in the tire, it will conform to have an area such that it produces enough force to hold the car up. so if you have the same pressure in the wider and narrower tire, then each will conform such that enough area is facing the ground to hold the car up... which is to say the contact area for each will be the same. in the real world, other things like sidewall stiffness probably also make a non-trivial difference. however... if the wider tires have a stiffer sidewall, this suggests that the narrower tires would actually put more rubber to the road.

it's an interesting concept; i haven't really considered it before.

this equation is only an approximation... for the most part it's a decent approximation, however in addition to what you mentioned, i believe there are also going to be significant contributions due to molecular forces which are also affected by the rubber's deflection under various forces and also due to macroscopic roughness, both of which will be affected by the tire's width, tread shape, and the way its pressure is distributed during various exercises. i think another benefit of wider tires is that they can put down a 'better' patch; among other things, the skinnier tires will probably down a more uneven pressure due to the tire resisting being flattened. also, i think the wider tires are better able to control their temperatures during spinning/sliding situations.

i think the OP has waded into a very, very complex subject :D

It's not so complex.

I used to believe this too (which is why I had the same questions as the OP long ago)... the problem is that the coefficient of friction is not a constant, it actually depends on pressure. The old high school/college physics approach of assuming that surface area and contact pressure do not affect static friction forces is false. The *lower* the contact pressure between two surfaces, the *higher* the coefficient of friction, thus the more "grip" you will get. Also, with racing tires, there is an adhesive, glue-like component to the friction force between the tire and the road, and adhesive strength is essentially directly proportional to contact surface area. Think of the tire's interaction with the track surface as being something like duct tape; even with zero normal force the tire would have to be 'pulled' away from the asphalt.

The pressure in the tire, and the weight (force) on the tire, are the only things that affect the size of the tire patch in contact with the road. If maximum grip is your goal, you want the tire pressure as low as possible, which has the side effect of causing the tire patch to be as large as possible. The wider the tire (and the lighter the car), the lower the pressure you can run in the tire (without deforming the sidewall and contact patch into something that looks like an oval footprint pointing in the direction the car is traveling). Lower pressure = good for static friction. Larger contact patch = also good for static friction, assuming there is an adhesive component going on.

To sum up: a larger tire allows you to run a lower pressure without deforming your tire in a bad way. Running at lower pressures increases the effective coefficient of friction, increasing grip.

Other upsides to larger tires (perhaps even more important than the grip advantage) include the ability to wear longer and handle heat better (more thermal mass).

The big disadvantage of larger tires is the additional weight, which alone can be a killer for lap times. The big disadvantage of using the lower pressure available to you with larger tires, especially super sticky tread compound tires, is increased rolling friction.

Interesting responses (let's leave it at that for now) and I appreciate them. But let's throw another factor/question in the mix. What about the thread pattern of the tire? Why then are pure racing tires slicks if they don't put any more rubber on the road?

I also go back to my earlier point. We all know that larger/heavier tires & wheels are not good in that they add unsprung weight and the weight of the rotating mass hurts performance and economy (this is a known fact). So again, why would manufacturers be putting such wide tires on vehicles?

I get that on some vehicles it truly might be a marketing decision, But on many high performance cars I seriously doubt they are going wider if there is not some real benefit.

Me thinks you are right. I may regret this one :leaving:

KISS;

Same rubber compound.
Solid rubber to eliminate a pressure question.
Same O.A. diameter.
Correct rim width for tire.

Tell me how a 26" bicycle tire has the same size contact patch that a 26" 285/35/19 has.

a 26" bicycle tire at 30 PSIG would deform until the car was supported by the rim, removing pressure from the model. for a bicycle tire to support the car by air pressure, the pressure would need to be very high... and therefore the contact patch would be small.

Dear God,

Please forgive me for the sins I have committed since we last spoke. Please continue good health to mom, dad and the rest of my family. Please continue looking over me as you have in the past.

Oh, and one more thing, please forgive me for starting the thread on the Corvette forum about tire contact physics. You apparently were working with someone else when I started that thread. I forgive you.

OK, so you guys are saying a Corvette with 275 width rear tire at 30psi has the same contact patch as the same Corvette with a 325 width rear tire at 30psi? No friggin' way! :hide:

Way to go.:rofl:

Yes friggin way. Reread what I wrote earlier. This isn't rocket science folks, it's not that bad.

Having a tread pattern decreases 'grip' for two reasons.
1) the tread blocks will deform and fold over under lateral force, this can't happen with a full racing slick.

2) putting grooves in the tire effectively reduces its width, which brings me back to what I said earlier. Narrower tires have the same contact patch area at the same time pressure, but you have to run higher pressures to prevent them deforming the sidewall and distorting the contact patch into a forward/rearward stretched ellipse.

In addition, a tread pattern, by virtue of having lower surface area, will wear out faster.

The only disadvantage to slicks is they cannot channel standing water, which will cause your car to hydroplane. :cheers:

Your buddy is correct, but only in the simplest form. Tire dynamics have a lot more to do with tire pressure, size.

This is why people lower their tire pressure when they drag race and why people who run slicks run a very low tire pressure. Tire PRESSURE and vehicle weight, vehicle downforce is what causes contact patch area (and sidewall stiffness). Tire width will give the tire it's contact patch geometry (wider tires will have a rectangle with the long side going from side to side, narrow tire will have a rectangle with long side going from front to back). The geometry effects the tire dynamics though, especially heat. If the rectangle has the long side front to back, then the sliver of tire that first contact the ground will touch the ground for more duration, creating heat. Think of a particular part of the tire touching the ground, it has to make its way from the front of the rectangle, all the way to the back.

There is a ton about tire dynamics that are beyond my knowledge, here is an article from an engineering forum with lots of good information.

http://www.eng-tips.com/viewthread.cfm?qid=102250

Oh, I forgot. You wrote it on an internet forum. That makes it true. My bad....

Does the stiffness of the sidewall have an effect that could counteract this effect?

BTW, I fully get that lowering or raising pressure effects the patch, but that does not necessarily tell me that the tire width is not a factor.

Thanks for your thoughts. Have not had a chance to read the link yet. Pesky work is getting in my way today.

super back of the envelope...

our 'vettes will put ~750 pounds of normal force on the tire. ignoring runflats (that's an interesting concept :think: ), the tire will probably push back with a force that is a nonlinear function in sidewall deflection; i'm going to speculate that at the deflections we see, the tire's sidewalls are pushing back with somewhere between 10 and 50 pounds of force. so probably not totally neglegible, but at the same time not a huge effect.

when i get home, i'll grab an extra old TA gforce i have for my hatch, let its air out and press down until it looks like it has a car on it... then report back. considering other factors, this will also be inaccurate, but better than nothing... (i.e., i'll find out if it'll hold my weight before collapsing)

Hi Jim missed ya when we came through Texas; hope I had the correct number. Anyway stirring up some thoughts on this one are you. It's a dynamic solution were both of you can be correct at some point or points. Fundamentally just start by thinking two wheels solid as in steel of the same diameter but one is one inch thick and the other is 10 inches thick. That example will lend itself to the one having 10X the contact. Now one can extrapolate that with all sorts of variables as in pressure, weight, tire compound, sidewall strength and anything else you can dream up and answers will be all over the map.

Jim - yes, you had the correct number. I was out of town when you called, then I screwed up and deleted your message and number. My bad!

I respect your opinion - thanks for any thoughts on this. It's one of those topics that the more answers I get, the more questions I have. I have to be honest and tell you that based upon everything I've heard, I'm going to believe what I want to believe :crazy:.

I'm not convinced that the physics of the issue are as black and white as some would like us to believe. That's not to say I'm not open minded to the responses - I am and I appreciate the responses.:cheers:

Simple physics makes it true. P=F/A. The internet's just here to teach you. But if free instruction on the internet doesn't work for you, you can come by my class and pay for it instead.

You don't have to take my word, or anyone else's word for it, you can look up what's been said here and verify for yourself if you like. At the end of the day, it's not that complicated. A larger tire can be run at lower pressure, and lower pressure increases the coefficient of friction, which increases grip. That's pretty much it. There are other secondary effects, including the adhesive nature of the tire at high temperatures, the sidewall stiffness, etc. but they are less important. Larger tire --> lower pressure = more grip.

I get that if you lower the pressure the contact area increases. That has never been in doubt. The point I'm questioning is that if you take a 285 tire, and a 325 tire - both identical except for the width, they will have the same contact area at the SAME pressure. That is what I've been told.

Unless pressures start getting really low, in which case the sidewall actually starts supporting a large part of the car's weight, this will be true. It has to be true, Pressure = Force/Area. If the force (weight) is the same, and the pressure is the same, the contact area will have to be the same. There's no way around it. The narrower tire will have to achieve this area by deforming more of the sidewall and providing a footprint that points in the same direction as the car, while the wider tire will achieve this with less deformation and have a footprint that points perpendicular to the direction of the car.

Yup. But the narrower tire will have to deform more to achieve this (i.e. become more flat where it is touching the ground - becoming less round)

Here is a simple sketch to kind of show it (very simple sketch). You will also notice that an instant line across patch on the lefts tread surface will only touch the ground for half the time as the one on the right. The 100 lbs on the top is constant (cars weight and assuming car is not moving - when car accelerates, moment transfers more weight to back, less weight to front which would in effect change the 100 lbs to be more for that period of time). Again, this is extremely simplified. There is a lot more to it than this, especially when the car is moving.

http://i436.photobucket.com/albums/q...suresketch.jpg

I didn't say that. I said your friend is correct that a wider tire has the same contact area as a norrower tire.

Some of you made two mistakes. One, you assume a wider tire gives you a larger contact patch. Two, you assume a larger contact patch is the reason why a tire has more mechanical grip. Both of those assumptions are wrong.

It is correct that wider tires (assuming all other parameters are the same) do provide better performance, but it is not via the size of the contact patch. The following explanation was given in a seminar by a friend of mine who is a tire engineer. He started his career with Goodyear and is currently with one of the import brands.

During the seminar (attended by mainly amateur racers), the question was asked if a wider tire is always faster. My friend's answer was "yes". His expalantion was that a wider tire provides an elongated contact patch shape compare to a less elongated shape from a narrower tire. Rubber, being a non-rigid material does not transfer lateral stress in a linear uniform fashion. The free edge (parameter of the contact patch) will experience higher stress than the inner area of the contact patch. As a result, mechanical grip generated by a unit area (say one square inch) of rubber at the free edge is lower than a square inch of rubber further behind the highly stressed free edge. Also, the parameter around the contact patch does not see the same stress, different sections of the parameter (free edge) see different level of stress depending what the vehicle is doing (cornering, braking etc). What an elongated contact patch does is that it will provide a larger percentage of "less stressed" rubber compare to a more square shaped contact patch, hence providing a higher level of mechanical grip, especially in lateral loading (mid corner)or a combination of lateral/acceleration (corner exit and corner entry) situation. Basically, what he said was that an elongated contact patch shape is a more optimized shape for high performance applications.

The reason why slick tires are faster (even if tire compound and carcass construction is the same) is it minimizes the free edges, hence provide a larger percentage of less stressed rubber, similar concept as above.

Some of you guys can believe whatever you want to believe. My recomendation is that next time you go to a MAJOR race event and you see one of those large transporters from a tire manufacturer, go up there and ask for the tire engineer. If you are lucky, he might just answer some of your questions.

Thanks for the first reply that actually addresses the issue in a sense that explains why a wider tire provides better traction.......I think ;)

Thanks hoefi great explanation.

When you talk about the less stressed areas of the contact patch, for the wider tire. Would less stress mean lower tire tread temperatures for the wider tire with track use than the narrower tire?

:thumbs:

Interesting on how a question on how big a patch area vs. how wide a tire has morphed into all sorts of traction issues.

Agreed, that if this were simple physics it would be true. But this isn't simple physics. A spherical balloon is simple physics, a tire ain't so simple. This is a discussion about a very complex engineered product that is designed to transmit and absorb forces far beyond your simple formula.

So, to understand complicated stuff my simple practical mind likes to simplify things even more. There was mention earlier of a steel cylinder on a flat plate. By definition, the contact patch in that scenario is a line without width. The "275" cylinder would create a line about 10" long. The "325" cylinder would create a line about 12" long. Hmmm, wider tire, oops, I mean cylinder = longer line by 20%.

Let's now assume that the "275" and "325" cylinders are designed for similar purposes and manufactured by similar processes of similar materials. The load on each is also similar. And the smart engineers designed some flex into them so that they would spread out a bit and provide friction against a rough surface. I think it's safe to assume that they would flex in a similar fashion in response to the similar load. That said, the line would gain some width which equates to area, or contact patch. Assuming both lines expanded by a similar amount, I'm gonna go with the "325" line equating to a larger contact patch by virtue of it starting out as a longer line. So much for my back woods Texas math.

If this weren't the case, why (all else being equal), does a Z06 or GS with a 325 tire have better skid pad performance that a C6 with a 275 tire? Why would F1 reduce the width of the tire to reduce grip and make the series more competitive? Why does my pals GT1 car have front tires wider than my Z's rear tires, and rear tires that are wider still? Simple answer, more contact patch equals more grip and wider tires create more grip so there must be more rubber in contact with the road. Same reasons slicks provide more grip in the dry - more rubber in contact with the road.

Now, some smarty pants is going to chime in with the formula that says that sliding is independent of area because the coefficient of friction times normal force and all that. Yup, I took physics and I don't argue that at all - for uniform surfaces. Tires though, depend on mechanically locking the tire to the road by flexing in and out of irregularities in the road surface. Like gear teeth or a chain and sprocket. Here again, more gear teeth meshed together spreads out the load and increases the ability to transmit force. So, the larger the contact patch, the more teeth are meshed together, the more force can be transmitted.

That's why I like my 325's on my Z06.

That's my story and I'm sticking to it :D

By the way, I'm in the electrical business - have been for 37 years. I'm not an engineer, but just from osmosis I've learned a little over the years. What I have learned is that you don't have to be able to see something move to know what the results are. Just saying.... :woohoo:

hey Jim, maybe this will help.

http://www.boeing.com/commercial/air...ontactarea.pdf

Just because it's simple, and the problem your trying to solve seems complicated, doesn't make a fundamental formula untrue. P = F/A is true in all situations, uniform or not, moving or not, whatever. That's one of the nice things about a good formula, it keeps us from making incorrect assumptions, especially in situations where our "gut instincts" lead us to incorrect conclusions. This is one of those situations.

The only reason wider tires give you more grip is because you can run them at lower pressure. A wide tire and a narrow tire, made of the same compound and run at the same temperature, will essentially provide the same contact patch size and the same grip (assuming no run flats so we can ignore what should be relatively insignificant effects such as sidewall compression and things like that).

Need an example? Look at my formula car. It runs very wide tires for its size, but at widths of 8" up front and 10" out back they are still smaller than the Z06 tire sizes. Nonetheless, because my car is so light I can run them at only 15 psi (hot), and that's why, even though the formula car has much smaller tires than the Z, it gets way more mechanical grip. WAY MORE.

Mechanical grip from tires is controlled by the following three things, in order of importance:

1. Compound
2. Tire pressure (lower pressure, more grip)
3. Tread pattern (lower profile/more widely spaced tread blocks, more grip)

Larger tires do not, in and of themselves, increase contact patch size or grip. What they do allow is for you to run lower pressures, and it's the lower tire pressures that increase grip.

If you run wider tires and do not lower the tire pressure, you will not see any grip advantage, but you will see a weight penalty to performance. Running larger tires at high pressures only provides advantages for tire wear and heat management.

My Z06 weighes right at 3200#, and has 325/30-19 tires at 26.7 " diameter,and with 30psi.

My 1964 coupe weighes right at 3200 # and has 205/70-15 tires at 26.3 " diameter, and with 30 psi.

I did a rough measurement on both. The contact area on the Z06 was 4.06" length X 10.38" width which equals ~42 sq in of contact area.

The 64 contact area was 5.06" length X 5.06" width which equals ~26 sq in of contact area.

Those meausements aren't exact but both tires/cars were measured in the same fashion so the relationship to each other should be close.

Now, who believes that if I installed the 205/70-15 wheels/tires (making believe the 15 " wheels would fit over the Z06 calpiers) that I would have the same traction as the 325/30-19 wheels/tires? And how did my contact patch area gain 73% in size since I have both the same weight and the same pressure in both size tires?

Remember, we have the same weight on each size tire and they are both at the same pressure and almost the same diameter and the tire pressure is what I actually drive both cars with.

Just had to spoil all the longwinded nitpicking theory eggheads didn't you. The thread started about contact patch size, not traction.

Common sense and practical proof have no business on this thread.

:rofl:

my texas math disagrees with yours. excepting the force held by the tire's resistance to deformity, in a static situation the load is carried by the tire pressure. and for that F=p*A is a pretty good approximation. i don't know about you, but excepting runflats, i can pretty easily tell a full tire from a low tire from a flat tire.

the above says nothing about why a 325 is better, it only makes a statement regarding the area of the contact patch.

In your first example, assuming you have about 800 lbs of load on the tire and a contact area of 42 in^2, the average pressure on the surface of that tire would have to be 800/42 = 19 psi. This is too low to make sense.

On your other car, you have 800 lbs of load distributed over an area of 26 in^2, so the average pressure on the contact surface of that tire will be 800/26 = 31 psi.

The second example is pretty close to what would be expected, but your first is not. If I had to take a guess, I would reexamine your measurement technique, it's pretty easy to incorrectly measure the "length" of a wide tire


Ah, anti-intellectualism at its finest. Just because you don't understand something doesn't mean you have to hate it. Why don't you just let the 'eggheads' do all the thinking, and when we're done we'll let you know. In the mean time, check out the flat earth society, they base all of their beliefs on "common sense" just like you, you'd love it.

Runflat

Runflats, by virtue of a stiff sidewall, will *decrease* contact area, not increase it. That's why you can drive on a runflat when you have no tire pressure.

sure so how did you determine the comparative tire presure and mock him.

What are you talking about? I calculated the average pressure at the contact patch of the tire directly from the weights and areas he provided. Pressure is defined as force over area. Then I said it works out with the tire *air* pressure he provided for one of his cars, but not the other (the contact patch pressure and tire air pressure should be equal in both cases). In the case where it doesn't work out (the Z06), he's either overestimated the area, or underestimated the inflation pressure in the tire, because the contact patch is too big. I suppose he could have grossly underestimated the weight of the car, which would also cause this to happen, but unless he's filled it with lead weights that's unlikely. You then responded with "runflats", which doesn't make sense, since runflats would cause the contact area of the Z06 tire to be *smaller* than expected, not larger than expected (which is what we are seeing here).

You calculated 19psi you could have that patcch with 0psi

Enough already, my brain hurts.:D
One thing I know for sure is that the wider the tire, the cooler the look.:thumbs:

There's no question that if his contact patch measurements are correct, and his weight estimate is correct, that the average pressure at the interface between the tire and the floor is 19 psi, it's defined that way. With a runflat, yeah, you might be able to get that patch size at "0" *gauge* pressure in the tire. That's because runflats, especially at lower gauge pressures, decrease the contact patch size. He stated that his gauge pressure was 30 psi, and I said he either underestimated the gauge pressure or overestimated the contact patch size. 0 gauge pressure would be an underestimate of 30 (albeit a crazy underestimate), and it could very well have caused the error we're seeing. But all things being equal, a runflat tire will have a smaller contact patch than a 'normal' tire, not a larger contact patch, which is what we are seeing here... therefore the error is not being caused by the tire being a runflat.

Your equation is not valid with a stiff sidewall, remove the 19pse term and replace it with zero and try and verifly the load weight with the so stated patch size. The patch size remains almost the same regardless of tire presure. ( I just did it) I feel like I'm talking to myself here.

I doubt your 64 has the same fore/aft weight distribution as the new Z. Just because the vehicles weight the same, that doesn't mean the corner weights are the same. I don't even believe the left rear is the same as the right rear.

What you should do is to put the two front tires of the Z on the rear and then try to measure them for comparison. At least the tire construction is similar in this case. Don't be lazy and only put one front tire to the rear because it will put a wedge into the chassis and jack weight into the diagonal corners, hence distorting the weight on the tire you want to measure.

Have fun changing the four tires. If you want to argue, at least you have to sweat a bit and do some work.

Cheer.

Okay, I'm going to say this AGAIN - but understand I'm still not sure which side to believe. If the answer is that you can run lower pressures, then why isn't the recommended pressure on the 325's on Grand Sports and Z06's lower than the 285's? Why do they put bigger tires on cars that have more power even though they don't recommend lower pressures?

C6 Z06 has 49.3% front/50.7% on the rear, weight distribution. That means each rear tire has a 811 # load.

1964 Coupe with auto transmission has 48% front/52% rear, weight distribution. That means each rear tire has a 832# load.

I measured the footprint on the rear tires of both cars and the measurements are very close as they were measured in the same fashion. Both are side by side in my garage. Where are the Z06 and the 64 coupe that you are measuring located? Or, are you saying my measurements are wrong, even though I have made actual measurements, or are you saying your measurements are correct even though you have not made any measurements?

Well that's a good question. I don't know what guides the recommended tire pressures from the factory, but I think they are more concerned with safety and meeting fuel mileage standards than they are with performance. Whatever it is, there is no doubt that by using a lighter car and/or wider tires, you can run much lower tire pressures... that's the whole point. I know it's an extreme example, but the tire pressures (and resulting improved grip levels) of a formula car easily show this. Or you can look at my 245 width van tires, they are narrower than the vettes 325's AND they have to support a crazy heavy van, and that's why they run near 80 (!) psi. For tires with the same profile, the optimal pressure for a tire is determined by the weight it has to bear and the width of the tread. Lower weight, lower pressure. The lower the pressure, the higher the coefficient of friction, and therefore the greater the grip.

But don't take my word for it, here's an article written for road bike guys that spells out pretty much the same thing, the graph at the bottom of the first page is worth looking at:

http://www.bikequarterly.com/images/TireDrop.pdf

The graph shows how, for a given weight on the tire, you need less and less pressure as the tire gets wider and wider. For example, trace up from the 40kg line on the x-axis; as you go up, you see that for the 37mm wide tire you only need an inflation of about 40 psi, but with a 32mm wide tire you need to inflate to 50 psi, and when your tire is only 23mm wide you need to inflate to 90 psi.

You might feel like you're talking to yourself because we're not talking about the same thing, I don't know exactly what you're getting at here. The equation P = F/A is always valid, it's the DEFINITION of pressure. The average pressure on the outer surface of the tire must, BY DEFINITION, be equal to the weight it is supporting divided by the contact patch area. Now the pressure between the tires outer surface is not necessarily equal to the gauge pressure inside the tire, especially in an extreme case such as a runflat at 0 gauge pressure. But pressure, regardless of where you get it from, always always always equals force/area. Since the sidewall can support some (or all) of the outer pressure on the tread of the tire, for a given gauge pressure inside the tire, you may very well see a *smaller* contact patch than you would expect, but not a *larger* contact patch like we were told. A runflat will NEVER cause the contact patch to increase in size; the whole point of a runflat is to decrease the contact patch (when the tire is flat) to a manageable size.

You made the accusation that he misread his pressure gauge, not sure how you drew that conclusion. This example can’t be explained with simple linear equation, yes the equation as it stands is correct but not for the example described. Most all the weight is distributed via the side wall and not the psi term you used. Also to further the explanation in describing grip with tire pressure is also a dynamic and not a simple three member equation. For example you have an inverse curvature which changes the cross section weight distribution and depending on tire composition it could have less grip. Again this has morphed to God only knows what, all Jim wanted is what happens to the patch size.

Score;

JoesC5;
2 posts with actual measurements showing a larger contact patch with a wider tire.

HerculesRockefeller;
Multiple long doubletalk posts of theory and equations.

JoesC5 = WIN

HerculesRockefeller = Fail

Reading all this I agree that the contact patch will be the same at a given pressure. Afterall, 30 PSI for a 3000 lb car means that 100 square inches (if equally distributed) would be supporting the entire weight. I have to believe that tire construction plays a big role in that distribution as in the case of runflats. The few inches of surface of the tirewall contacting the ground are supporting more than the 30 lbs each in this example, aren't they? This would skew the calculation of contact area. The same would be the case of steel cylinders of varying width where the line touching the ground would still have area and that area would have a rated PSI. Solid rubber tires were also mentioned but did not take PSI into consideration.

But, I only slept in a Holiday Inn, so really I'm just asking.

--Dan

Yes, the runflat tire could screw the results, by decreasing the size of the contact patch. Imagine the extreme version of the runflat, a tire with solid steel sidewalls. In this case, most of the weight of the car will be supported by the sidewalls, not the air pressure in the tire, and so the contact patch will be much smaller than you would expect for a non-runflat tire (at the same gauge pressure). But I'm not familiar enough with the latest runflat designs to know just how much the sidewall supports the tire when the tire is properly inflated... it could be very small fraction. Either way, the largest possible contact patch area you can get with a tire is given by P = F/A, where P is the gauge pressure in the tire and A is the contact patch area. Any stiffness in the sidewall will decrease (but never increase) the contact patch area below the value predicted by the equation. Therefore, there is NO WAY that a round tire, inflated at 30 psi gauge pressure, supporting a load of 800 lbs, could ever have a contact patch greater than 27 in^2. If someone states that a tire, inflated at 30 psi, has a contact area of 42 in^2, somehow, somewhere, a mistake must have been made.

Agreed. I'm also thinking that PSI, as related to contact area, must be measured in virtually a balloon constrained in various widths. There would need to be no discernible support other than the air. When sidewall stiffness is introduced, that area will support a higher PSI which then reduces the PSI needed to be carried by the remaining contact area. The same would seem true for tread design as certainly the voids or grooves in the tread are not supporting weight and must be subtracted from the contact area measurement. Wouldn't a super thin-wall race slick yield the most accurate reading? In a normal street tire the PSI is read as it relates to the pressure exerted on the interior wall (a slick-like surface) of the tire contact area which may have nothing to do with the actual exterior amount of rubber on the road providing useful grip surface.

I guess after all this I'm inclined to believe that the equation that may be valid under idea circumstances may not apply exactly when it comes to tires that have a wide variety of sidewall constructions, tread patterns etc.

I'm not very smart when it comes to math, but I'm pretty good at thinking through common sense kind of stuff, and this just doesn't pass the common sense test. I like to think of things in extremes to try to understand these things. I just can't comprehend that if you put say a 345 width slick on a car, and a drag skinny with a very thick and stiff sidewall construction on the same car (same tire location) and put the same pressure in them that they have the same contact patch.

I'm guessing that the physics theory behind this is based upon two flexible spheres that have identical constructions and sidewall heights. I'll never swear that I know I'm right nor will I tell anyone else they are an idiot for not agreeing with me. I'll just continue to believe it knowing that there is some chance I'm wrong.

As I stand in my garage I see my bicycle is next to a spare rear Z06 tire and I'm trying to understand why the patch isn't what I think it is then I go and look at my wall and try and understand my life’s progressions and I say what a lousy way to start my day..:smash::smash::smash:

If you want to see your motorcycle friend go really bonkers, tell him a car tire on the rear wheel of a cruiser bike performs better than the OEM motorcycle tire.

See how he reacts, then refer him to the "Darkside" forum dedicated to running car tires on motorcycles.

Cw

There is much more then simple math to determine the contact patch of a tire. The tires construction being the one with the most variables.

Take a 6.70-15 tire(original size on my 64 coupe and the originalmounted spare is in my basement) is of a bis belted costruction. The diameter is ~27 inches and the tread width is 4". The aspect ratio is ~.88. Inflating that tire to 30 psi, on my car, will have a very small footprint. Smaller then the footprint of the 205/70-15 now on the car. Why is that? The stiff sidewall and an almost round tire. If anyone is old enough to remember when Michelin first introduced the radial tire to the USA, you will remember that they always looked like they were flat, even with 30 psi. The super flexible sidewall and the plies running at 90 degrees meant the sidewalls flexed way over what a bias belted tire would. That original radial tire had a large footprint when directly compared to the replaced bias belted tire. Same size tire, same weight car, same tire pressure, different size footprint.

Then look at a wrinkle wall drag slick. The side wall is constructed to allow the tire to distort to create a huge footprint when under rotational load. You could take that same tire, make the sidewalls a super stiff 20 ply and, with the same weight and pressure, have near zero sidewall deflection, thus a smaller footprint.

I could make a tire from steel, internally braced,so that the tire is perfectly round. . I could mount it on my car and the contact area would be nil. A narrow line, a coupe of thousands wide X the width of the steel tire. I could inflate it to 15 psi or 30 psi, and the contact patch(footprint) would remain the same. What would change the contact area would be changing the width of the tire. A 10" wide tire would have twice the contact area of a 5" tire, because the line of a couple of thousands would not be halved when the width is doubled.

My point is that there are to many variables in the various tire's designs that affect contact area other then straight psi. As poined out, the 325/30R19 runflat on my Z06 will have an extremely small change in contact area at 0 psi or 15 psi or 30 psi, all because of the tire's design.

:iagree: No argument here. All I'm saying is that the *maximum* patch size area "A" tire can have is given by A = F/P(gauge), where F is the load on the tire and P(gauge) is the air gauge pressure in the tire. Increasing the sidewall stiffness will decrease the contact patch size. But, with respect to the OP question, increasing the width of the tire does not, in and of itself, increase the patch size. But since wider tires allow for lower operating pressures, you can get a larger contact patch with a wider tire by running the tire a lower pressure.

:thumbs::thumbs:

That's a different forum that we have that debate. That's the Goldwing Forum - I know that forum and I've gotten into that debate once. I saw no point in spending much time debating with the darksiders. The Concours forum is where my debate is.

I mentioned one extreme of getting a maximum contact patch....the wrinkle wall drag slick. Here is an extreme example on the other end of the the spectrum. When I built my house, I had the 18" X 60" strips of sod installed. After the crew had them all laid, the guy rolled a farm tractor off his trailer. He had installed rolled steel hoops on the rear tires so he could mash the sod level. The hoops were 1/2" steel plate X 18" wide, rolled, and butt welded at the ends. He had then around the rubber tire and the tires were inflated to hold the steel hoops in place. Of course this type of tire construction would have a constant size contact patch no matter what the air pressure was.

Just another example of how a tire's design affects the contact patch.

As I mentioned earlier a simple 3 member linear algebraic equation can not explain this at all. Example transpose P=F/A where it becomes F=PA since A is now constant with a run flat you can now change the weight of your car by changing air pressure. :willy::willy::willy::willy:

Good points Joe. Your posts throughout this thread have been excellent.:cheers:

I think you are confused... you don't seem to understand that there can be a difference between the gauge pressure in the tire, and the contact pressure between the outer surface of the tire and the ground. If you are going to start talking about the effect of runflats and stiff sidewalls, you are going to need to be clear which "P" you are talking about. In your example, the area remains constant AND the pressure remains constant, because, whether you realize it or not, you're talking about contact patch pressure, NOT gauge pressure.

The pressure at the contact patch interface between the tire and the ground will always be equal to or greater than (but never less than) the gauge pressure in the tire. OR, if you prefer, the contact patch size between the tire and the ground will always be equal to or less than (but never greater than) the area predicted by A = F/P, where P is the GAUGE pressure of the tire.

Increasing the width of a tire will not increase the size of the contact patch. There are ONLY 3 realistic ways to increase contact patch size (and in some extreme cases only 1 of the following will have an appreciable effect):

1. Decrease Gauge Pressure in the tire.
2. Decrease Sidewall stiffness.
3. Increase weight on the tire.

Changing the size of the contact area automatically changes the PSI exerted on the ground by whatever is touching as that is simple math. Air pressure isn't an accurate measurement due to tire construction and that it is an internal measurement of force and contact area. As with the steel "tire" example increasing the width changes the external PSI of the contact area but the internal PSI can be zero. The A=F/P works for contact area pressure. A tire's contact patch has many different pressures.

Are you in sales? :skep::skep::skep:

No, but am I wrong? I'm just trying to understand.

I hate to be a jerk here but if you think two different tires of two different sizes on the same weight car will make the tire patch the same you're high.

Lets stop being nerds here and think about this in REAL simple terms.

You have a tire that has a 8" designed contact patch and you pressurize that tire it should be darn near flat and you should get around 7.** inches of good contact. If you take a 12" tire and pressurize that (same car same PSI) the tire will flatten out as it was designed and you will get 11.** inches of contact.

I will say that your tire pressure, weight on tire, and RPM of tire at the time (effecting weight and pressure) will effect the contact patch. However, trying to say if you take a 8" tire and a 12" tire slap them on a Z and inflate them both to 30PSI they will have the same contact patch is INSANE. If you don't believe me take your front tire off slap it on the rear. Inflate both the stock rear and the stock front (on rear) to 30PSI and drive through a puddle. Then take a picture and send it to the tard. Ignorance is bliss. I suppose Top Fuel runs those massive tires for looks and added rotational weight?

:iagree:

Some people open the door very gingerly and cautiously step inside hoping they have come to the right house for Thanksgiving dinner. On the other hand, some folks throw the door open and with a loud, confident voice say "I'm here, where the hell is dinner" not really caring whether they are even at the right house or not.

Welcome to the debate! :thumbs:

Confused, no sir you have no clue as to this discussion, I made reference to your consistent reference to air pressure and the equation you used as being invalid. You acuse people of being ignorant and I find that offensive.

Quote: A = F/P(gauge), where F is the load on the tire and P(gauge) is the air gauge pressure in the tire

i'm on my third 'vette, but i'm pretty new to actually spending much time on this forum. what i see of it...

C6Z owners, educated or not, appear to be unable to be wrong about an engineering/physics problem. or anything else for that matter.

i'm getting a :lol: @ both the guys who think their 'street smarts' > engineers and/or physicists when it comes to a physics/engineering question and the egghead set that condescend to the former while oversimplifying the problem.

i'm still of the opinion that for a 'standard' tire, the internal air pressure carries the bulk of the load, so within reason, the OP's nemesis's claim has merit. not to say that it's exact. and as i said several posts back, for runflats, all bets are off.

regarding measurments... IMO, any attempting eyeballing is going to be too inaccurate to get a decent comparison.

probably a good way would be to drive the car on glass, take a pic from the bottom, then integrate the area on a computer. but that's not gonna happen... other methods... maybe put down some water-soluble finger paint or chalk, park on it, wash all finger paint/chalk away, then jack up the car?

my measurement of my C6Z's contact patch was radically different from julyc5's results. more like 2.5"-3.5" in the short direction. i could make my measurement change by sliding the paper i was using more or less far under the car. which is to say, the patch is probably not such a nice rectangle. but i'm running the stock GY runflats, so the comparison is invalid anyway.

again, i think that for non runflat tires and in the static regime, probably >80% of the weight is carried by the air pressure so long as you stay away from extreme deformity and/or rim... which suggests that this effect isn't as negligible as many would think. to cater to the 'street smarts' set, you can tell when a tire is going flat.....

but i don't claim to know everything, and the true results are where the rubber meets the road. (thank you thank you, i'll be here all week. or maybe not...) so take and document some accurate and precise measurements if you want to put this one to bed. i.e. something where the sigma isn't bigger than what you're trying to measure to begin with. personally, i'm not curious enough to do it, but that could change when a convenient opportunity arises.

Your thoughts are good ones - but let's consider what you are saying. You say that with runflats "all bets are off". Carrying that further, there is really nothing magical about runflats, they just have a stiffer sidewall to keep them from collapsing when deflated. But that is exactly my point about using the physics equation to support the notion that pressure and weight are the only factors in determining the contact patch.

All tires are constructed somewhat differently than each other. The sidewall stiffness will vary even among non-runflats. Your point basically supports my contention when it comes to tires the contact patch is not such a simple equation as pressure and weight.

As I've said more than once - I'm not saying I can support my stance and prove I'm right so you won't see me being overly arrogant in claiming I know more than the guys who disagree with me. I'm just stating what I consider to be my common sense opinion on the matter.

jschindler -- i agree that sidewall matters; i've included it from the beginning. so we agree there :) IMO, it's a question of how much. i single out runflats because my guess is their sidewalls are significantly much more stiff than that of a regular tire. either way, there is a contribution. my guess is it's something less than 20% of the static force, but that's a guess. and i keeps staying with statics to try to simplify the engineering/physics problem some...

one thing nobody's really mentioned... what about the centripetal force that has to be put on the rubber at speed... :D :leaving:

I think the obsession with sidewall stiffness is distracting folks from the original question, which asks if the tire contact patch will increase in size when increasing the width of the tire. And the answer is no, increasing tire width, in and of itself, does nothing to the contact patch area. The only way you can increase the contact path area is to lower the gauge pressure in the tire, soften the sidewall, or put more weight on the tire.

As I've said before, there are to many variables for a simple math equation to work.

http://www.nhtsa.gov/staticfiles/saf...HS-810-561.pdf

Read chapter 7. If the contact area was so simple to determine, chapter 7 would be one sentence long.

sweet.... tensors :cool:

notice how the flat tire on p 267 has a larger contact area than the full tire :p

but seriously, i'm DL & saving that pdf. lots of good stuff there.

Wow. I'm almost afraid to toss in my two cents here, but here goes, simply put...

For the same tire construction, at the same pressure, a wide tire, and a narrow tire, will have the same contact patch area. The shape of the contact patch is what changes, the narrow tire having a long/narrow patch vs. the wide tire having a short/wide patch.

The short/wide patch distorts more favorably during cornering loads, providing more grip.

My suggestions to the doubters:

http://ecx.images-amazon.com/images/...SH20_OU01_.jpg

http://ecx.images-amazon.com/images/...500_AA300_.jpg

or if you have the time:

OptimumG 3 Day Seminar

Dave, I think you nailed it with key being the distortion of the patch area, and your references are THE classics in the field. My opinion is the top racing teams basically try to "manage" patch distortion (for forward bite, or speed through the center, or to compromise both).

Looks good. All I would add is that I think patch distortion in a wider tire is less important than the advantage of being able to run lower pressures in a wider tire, since lower pressures are known to increase the coefficient of friction. Then again, maybe it isn't... it's just a theory, but it makes sense to me. It is known that by running *really* wide tires, you start running really low pressures, and that increases friction. But how much that friction coefficient increases, I don't know. :thumbs:

And when the tire pressure gets too low, the sidewall flex can become to much, and it's just not contact patch distortion, but tire distortion, and grip goes away...usually this is called a puncture ;)

Tires, being made of material that is not infinitely flexible and being of a somewhat fixed geometry, will not follow the pressure versus contact patch relationship exactly. For example, let's say you have 1,000 lbs on a wheel at 33.3 psig tire pressure and it has a contact patch of 30 square inches. 1000/33.3=30. Now let's say we let the pressure down to 1 psig. The equation would suggest that the contact patch is now 1000/1=1000 square inches. We know that won't happen.

Once we agree that the above equation is for a theoretical ideal case, not real world, we can see that a wider tire does have a bigger contact patch. So, let's take the pressure and weight vs. contact patch relationship out of the discussion completely in order to make a point. As an example, let's take two front wheels off of your car. Stand one of the wheels on the garage floor. It will a contact patch of X. Now to approximate a tire that is twice as wide, stand the second tire right next to the first one. The contact patch size just doubled to 2X. I'm not saying doubling the tire width necessarily doubles the contact patch in the real world, but it does go up for the wider tire.

Like Dave I hesitate to get into this thread b/c there is so much misinformation - your mistake is that you completely took the tire structure out of the anaylsis and you appear to be saying that only the pressure within the tire is carrying the full 1000 lb load, which it is not. Let all the air out of any tire and the tire carcass still carries some portion of the load (meaning the rim is not actually physically on the ground, although it might be close with some tires with very soft sidewalls). The fact is that the tire and the air is a system and both carry the load, so you can't simply use the equation in an extreme case and call it valid. As a matter of fact I recently read a highly theoretical tire design technical paper where they used numerical analysis to compare theoretical results with real world patch measurment data (for aircraft tires actually) and they found that at some point when you lower pressure enough the center tread portion actually buckles and is no longer in contact with the ground - only the outermost portions near the shoulder were carrying load - again this would validate that the patch area changes very little and that it in reality distorts (changes shape) rather than changes size.

When I said, "Tires, being made of material that is not infinitely flexible and being of a somewhat fixed geometry..." I was commenting about tire structure.

But you can't take the weight out of the equation, since the corner weight is going to determine how much the tire compresses. Any additional weight from the wider tire/wheel, is going to be a negligible amount to the total weight.

Increased cornering grip comes from the change in contact patch shape, not contact patch size, for the same internal air pressure and tire construction.

:lurk:

very interesting arguments. someone needs to do a real world test. take a car put a wider tire on one side and a narrow tire on the other side. jack it up, wet them down and lower it. the water will leave the contact patch. jack the car back up and measure the wet spot. the tires should be the same dia and air pressuer equal. it might make sense to use suggested pressures.

What some fail to realize is this >>>>A = F/P<<<< is a real world test. Without the "Real" world, these figures would mean nothing. The hard part for most people to wrap their heads around is that you have to take all differences out of the equation, such as tire material differences & tire construction etc, etc. Remember this is about contact patch and nothing else.





The equation can be moved around, and yes you will change the weight of the vehicle... On that one tire. It is called weight bias. You can move 50 to 100 pounds around on a car by simply changing tire pressure. I was really hesitant to even say anything, at all, but this is VERY basic physics, and I'm not sure why some are having a hard time understanding it.

Correct on weight but can't use that equation without limits, therefore no longer 3 member linear expression.