symptoms of wrong tire size

Jul 18, 2003 48 Replies

Just to throw a little more into the mix for everyone, here's an engineer's view on contact patch area and tire pressure. Note that he says the air gives rigidity to the tire, and that rigidity supports the weight (which follows my bulk modulus argument in the sci.physics thread). The air doesn't hold the weight according to this view.

Quote-

---------------------- "Bill Hussey wrote:

Not so. Heavy equipment uses HUGE bearings (the rollers that support our 8 million pound dragline are about 2' in diameter) because they spread the load over a much larger area than would rollers of the same length but a smaller diameter (imagine needle bearing rollers 1/8" in diameter but 3' long). The smaller rollers will lead to spalling (pitting) of the races and rollers because of the much higher contact stresses generated. Higher stress is due to smaller contact area.

A bicycle tire at 36 psi holding up 300 lbs does NOT flatten to that extent. (I should know, I can wheelie a bicycle rather well, and with a good-sized backpack of the sort that engineering students tend to carry, I can top 300 lbs.) The contact patch is nowhere near as large as the contact patch of my VFR. In order for the 1/2" wide bicycle tire contact patch to have the same area as the VFR contact patch (2.25" long, assume 2.25" diameter, therefore figure about 4 square inches), it would have to have a contact patch 4.5" long. Clearly it doesn't.

Remember, it's not simple hydraulic theory. The air gives rigidity to the tires, and the tires support the load. The air doesn't support the load itself.

Granted, different inflation pressures will give you different contact patch sizes. However, this is due to the fact that less pressure in the tire will give you less rigidity. The size of the contact patch does not have any sort of linear relationship with the inflation pressure. Physics, folks, physics. ;-)

By hydraulic theory, the area of the contact patch on my Land Cruiser (1100 lbs on each fron tire, tires aired down to 8 psi for off-road conditions) would have to be 138 square inches. Since I run tires that are 9.5" wide, the contact patch would have to be 14.5" long! Considering that this is very close to half the diameter of the tire, I think I would have noticed it by now. :-)

Dave Aley | mailto: snipped-for-privacy@cia-g.com DoD#454 Maintenance Mechanical Engineer | ASME #6363402 TLCA #8419 Lee Ranch Coal Company |

formatting link
, New Mexico, USA ---------------------" Granted, a comparison with bicycle tires probably is not as valid, but his point with his Land Cruiser seems to back up this view.

Todd Wasson Racing Software

formatting link

Hi Todd,

Yes I did note your post but given that the data is taken as a function of deflection and not load it isn't possible to draw any conclusions from it that would apply to the current debate. The wider tyres may well have required a larger load to obtain a given deflection at a given pressure and this would explain the longer contact patch. However that's pure speculation and without the load data we'll never know. Otherwise I don't see anything there that explicitly contradicts my post.

Dave Baker - Puma Race Engines

formatting link
"How's life Norm?" "Not for the squeamish, Coach" (Cheers, 1982)

Oh, ok. I understand. Thanks.

I'm curious, does this load over pressure equal the contact patch area then? I.e., instead of air pressure could you use bulk modulus (combination of air pressure and additional structural stiffness) and load to get the patch area within reason? But then it seems this would result in the same relation where you'd double the patch length by doubling the load.. Err... Maybe I need to read some more :-)

Todd Wasson Racing Software

formatting link

Yes, you're right of course.

I wish the guy had posted the load numbers to see how that works...

Todd Wasson Racing Software

formatting link

It doesn't. It has 32 PSI pushing OUT, from the rim, in all directions.

You have four tires, each at 32 PSI. Does that mean the four tires can only support 128 pounds?

Read the excellent text at titles "How does 95 POUNDS support 5,400?"

Okay, here it is algebraically and exactly.

If four P205-75R14s have X square inches of contact patch, say four P215-75R14s would have X+N. If the car weighs L pounds, and it's sitting on X square inches, there's L/X pounds on every square inch of contact patch. Widen the tires, to L pounds on X+N square inches, and you now have L/(X+N) pounds per square inch of contact patch.

For all X greater than zero, and all N greater than zero, L/X is greater than L/(X+N). Therefore a wider tire will have fewer pounds per square inch of pressure on the contact patch than a narrower tire on the same car.

So when you roll over a three inch rock, and your contact patch is suddenly only two or three square inches, your tires autoinflate to several hundred PSI for that instant? Or maybe they explode every time? Or your car gets suddenly lighter?

Pressure inside the tire has nothing to do with the amount of pressure the tire exerts on other objects (e.g. the road). Pressure inside the tire only influences the shape of the tire as it presses on other things.

It doesn't. It has 32 PSI pushing OUT. You can prove the 32 PSI isn't pushing against the ground by lifting the tire off the ground and observing that it still has 32 PSI in it.

The pressure inside the tire only influences the shape of the tire as other things put external forces on it.

From experience: I put wider tires on my 1976 Torino. It lost grip. Thinking the model of tire made a difference, I put still wider tires of the same model on the car. It lost more grip. I finally put the manufacturer-recommended size of the same model tire on the car, and it got its grip back.

With the widest tires on the car, on snowy pavement I could not touch the accelerator at a standing stop without spinning the tires. The traction was so bad that I had to let off the brake and let the car get a rolling start before I could touch the accelerator. The tires would still spin if I did more than a slow, gradual acceleration from there.

Reducing contact pressure can never increase grip. Increasing pressure increases friction.

Food for thought...

To get the maximum contact patch area, racers use "slicks", tires with no tread at all and wide as the vehicle can stand. They have good grip off the line... but imagine how they'd be on cornering, wet pavement, snow/ice, and braking.

What a strange comment. A rubber molecule generates its grip equally in all directions. Slicks are used on race cars because they corner, brake AND grip under acceleration better. They don't just do the latter well and be crap at the first two. That's absurd.

Even if the argument were to be confined just to very specialised wrinkle wall drag slicks (which I suspect is all you have experience of) you can hardly improve grip off the line without the tyre working just as well under braking.

Dave Baker - Puma Race Engines

formatting link
"How's life Norm?" "Not for the squeamish, Coach" (Cheers, 1982)

What rubbish. You clearly don't have a clue. Decreasing the load on a tyre increases the coefficient of friction and therefore grip. Whether that's done by reducing weight or increasing tyre width and therefore contact patch area for a given weight the effect is the same. It's a cornerstone of race car chassis tuning and anyone who thinks otherwise shouldn't be sounding off in a technical thread. Exceptional cases like snow/mud/water have been covered in detail earlier on.

Dave Baker - Puma Race Engines

formatting link
"How's life Norm?" "Not for the squeamish, Coach" (Cheers, 1982)

"As an approximation, the number of square inches of footprint are is about equal to what you get when you divide the load by the inflation pressure. (It's not perfect, but it's close enough for our purposes.)"

Do you know what PSI means?

Maybe you should too.

Quote from your source:

"As an approximation, the number of square inches of footprint are is about equal to what you get when you divide the load by the inflation pressure. (It's not perfect, but it's close enough for our purposes here.)"

It's amazing you actually thought I didn't know this already... ;-)

Todd Wasson Racing Software

formatting link

You're talking about snow? That's entirely different from dry pavement.

On snow or a wet surface, this is right. For dry pavement it is not. If you need some papers on experiments that verify this, let me know.

Todd Wasson Racing Software

formatting link

Whoops, I spoke to soon in agreeing with you. The bit about cornering and braking is wrong. A tire will, for the most part, be capable of producing a given force in any direction. I've yet to see force data on a tire that shows it can produce more acceleration force than braking force. I won't go into why this occurs. The greatest deviation between forward/rearward and lateral force capability is about 20% at high loads. Most tires are much closer to the same in all directions.

As for cornering, I'll agree with Dave and say that's so far from the truth you should be evicted ;-)

Wrinkle wall drag tires are probably an exception in the cornering case because under lateral strain the contact patch might buckle and do other bizarre things, but I doubt anyone's measured this. Anyway, have a chat with someone that's driven road racing slicks through corners. Big grip...

You're right about the lousy grip on snow/ice with wider tires or slicks though.

BTW., when you switched to wider tires and lost grip, how much wider was the tread, exactly?

Todd Wasson Racing Software

formatting link

Let's try an experiment. You'll need an 8"x10" sheet of sandpaper, a plastic cup and a brick.

Put the sandpaper rough side down on a flat surface. Apply lateral force only and push the sandpaper. Note it doesn't have much grip.

Put the plastic cup on top of the sandpaper. Apply lateral force only and push the sandpaper. Note it has more grip with the increased pressure from the plastic cup.

Remove the plastic cup and put the brick on top of the sandpaper. Apply lateral force only and push the sandpaper. Note it has a whole lot of grip with the increased pressure from the brick.

What did we just do? We gradually increased the pressure between the sheet of sandpaper and the flat surface, and as we did, we noticed that the sandpaper gripped the flat surface better every time we increased the pressure. With very little pressure we had very little grip; with slight pressure we had good grip; and with lots of pressure we had a whole lot of grip.

Document this, please.

Let's try another experiment. For this one you'll need a car and a strong jack.

First, we'll use the car with lots of weight on the drive wheels (the car's own weight). Start the car, put it in gear and punch it. Note the car goes with comparatively little tire slippage.

Now, return the car to the starting point. Engage the jack and lift the car until the tires on the drive wheels just touch the ground. Start the car, put it in gear and punch it. Note the vastly increased grip caused by the decreased load on the tire, which increased its coefficient of friction and therefore its grip.

You're obviously thinking only of the case of grip for the drive wheels on sudden application of torque. The cases of cornering and braking are vastly different.

I went up one size each tire change.

Then all that was happening was the suspension, tyre pressure or some other factor was preventing the bigger tyres working properly because set up right there's no way a wider tyre produces less dry grip than a narrow one.

Dave Baker - Puma Race Engines

formatting link
"How's life Norm?" "Not for the squeamish, Coach" (Cheers, 1982)

As I see from comment in this thread yet clearly from first principles the relationship can't be linear for a semi rigid structure.

I'm only working with foam tires right

Fascinating (raises eyebrow a la Spock). So relative to the model car's weight these are still relatively stiff structures it appears. So this would imply that if mu falls with load that total grip per unit load also falls significantly? Did you do friction measurements on this or just contact patch area so far? Of course a real world tyre would only go through relatively small changes in load. Perhaps an order of magnitude at most for say a laden to unladen lorry or a rally car hitting the ground after a jump. An F1 car will experience downforce of about twice the static load and no doubt there will be data for changes of that sort of order at Michelin and Bridgestone. ( the current F1 suppliers)

(3

Even a balloon isn't a perfect example of an infinitely flexible structure of course as the internal pressure creates stiffness. Only when just the gas itself moves in response to load do we find agreement with the Universal Gas Laws.

Without knowing what empirical measurements they took to formulate their equation for fit it's hard to comment.

Also,

Even with a balloon tyre I doubt it would be a linear relationship. It was the closest analogy I could find to a flexible tyre structure in my earlier post though.

Then perhaps it would progress towards the volumetric distortion

I have a hunch, and that's all it is, that the relationship between load and contact patch area for a real tyre is not only non linear but also geometric. i.e. for small changes in load the relationship is different from that for large changes when the tyre carcass is forced to distort very significantly. As such maybe neither bulk modulus nor the Gas Laws would help model the relationship and perhaps only a completely empirical approach would suffice. Interesting though and I'm sorry I have no more answers for you. If you ever do measurements on a pneumatic tyre I'd love to know the outcome.

It may be that to develop a model car tyre that behaves somewhat closely to how a real one does that you need to do some load/area measurements on a real tyre and you might then find that the bulk modulus of the foam one is still far too stiff for the model car mass or that even any solid tyre structure behaves totally differently to an air filled one.

Dave Baker - Puma Race Engines

formatting link
"How's life Norm?" "Not for the squeamish, Coach" (Cheers, 1982)

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required