Any of you running 235/85/16 tires?

[TheRoadie]

I think you're comparing apples to asteroids here.

Still looking for the difference.

The "PSI" of the tires has nothing to do with how much weight the vehicle is carrying. It's just a measurement of the air pressure inside the tire.

PSI of the tires is the only thing that pushes against the ground. If you double the weight on the vehicle, assuming the volume of the tire is relatively constant, the tires will automatically double the contact patch area. Otherwise what stops the vehicle from collapsing the tires onto the rim as if they're flat (no PSI)?

I agree that if you increase the weight of the vehicle that the tire pressure should be independent and relatively constant as long as the tires are deforming. But I'm looking for a better explanation about why narrow tires "perform" better, and many articles including Scott's, say it's because they push against the terrain with more pressure because of a smaller contact patch. I don't see how, and I'm an engineer (albeit electrical, not mechanical).

 

[98roamer]

Would you prefer a lady wearing high heels shoes to step on your big toe with the heel or with the front part of her shoes? Her weight never changed.

 

[7wt]

Would you prefer a lady wearing high heels shoes to step on your big toe with the heel or with the front part of her shoes? Her weight never changed.

As long as she is wearing fishnets, I don't really care.

 

[TheRoadie]

That illustrates the flaw in the analogy exactly. Tires deform. Shoes don't. Tires deform just enough to balance their internal pressure with the external pressure.

Surely you see that's why airing down helps. Lowering the internal pressure increases the contact patch.

If a high heel were allowed to deform it would spread out too. But it isn't, so the PSI under the heel is in exact inverse ratio of its size, compared to the size of the front of the shoe. But tires deform, so the physics are different.

 

[bigreen505]

I'm looking for confirmation that my understanding is correct that for snow a narrow tire is preferred. I'm not talking about the Arctic Trucks but for normal winter conditions.

I run 235/85 studded snows on my Trooper and they are fantastic.

 

[Schattenjager]

As long as she is wearing fishnets, I don't really care.

U da man!
:victory:

 

[valerieO]

When we went sand driving my husband explained it to me that the contact patch only lengthens. So the contact patch or driveable part of the tread of a narrow tire doesn't expanded sideways. Even thought the sidewalls bulge you don't drive on them. Due to the build of a narrow tire the patch can only expand so much.
98Roamer knows about high heels on the toe

He'll be so proud of me.

That illustrates the flaw in the analogy exactly. Tires deform. Shoes don't. Tires deform just enough to balance their internal pressure with the external pressure.

Surely you see that's why airing down helps. Lowering the internal pressure increases the contact patch.

If a high heel were allowed to deform it would spread out too. But it isn't, so the PSI under the heel is in exact inverse ratio of its size, compared to the size of the front of the shoe. But tires deform, so the physics are different.

 

[Martinjmpr]

Still looking for the difference.PSI of the tires is the only thing that pushes against the ground. If you double the weight on the vehicle, assuming the volume of the tire is relatively constant, the tires will automatically double the contact patch area. Otherwise what stops the vehicle from collapsing the tires onto the rim as if they're flat (no PSI)?

I agree that if you increase the weight of the vehicle that the tire pressure should be independent and relatively constant as long as the tires are deforming. But I'm looking for a better explanation about why narrow tires "perform" better, and many articles including Scott's, say it's because they push against the terrain with more pressure because of a smaller contact patch. I don't see how, and I'm an engineer (albeit electrical, not mechanical).

It sounds like you're mixing up ground pressure and PSI. The ground pressure should be the area of the contact patch / the weight that the wheel is carrying. In our hypothetical a 5000lb vehicle (1250lb per wheel) with a 200 square inch contact patch should have a ground pressure of 6.25lb/square inch. You can increase the side of the contact patch by airing down or by putting on wider tires. If you double the size of the contact patch obviously you'll halve the ground pressure. That's why tracked vehicles that weigh 25 tons can go places where wheeled vehicles that wiegh 5 tons can't - the contact patch is huge and so ground pressure is low (the "contact patch" on a tracked vehicle is the entire portion of the track that is on the ground. An M1 Abrams tank weights ~60 tons but can go places where my Tacoma can't because the contact patch is probably 15 - 20 square feet.)

But the ground pressure doesn't have much to do with the PSI inside the tire. Imagine our hypothetical 5000lb vehicle with 7" tires aired down to 15psi. Now imagine it with 15" wide tires at 35psi. The contact patch is going to be bigger with the 15" wide tires because the tires themselves are bigger. The bigger tires = less ground pressure.

Surely you're not saying that the 7" wide tire is going to have the same contact patch size at 15psi as the 15" wide tire? If that were the case, why would anybody buy fat tires? The whole benefit of fat tires is that it gives you a bigger contact patch which means less ground pressure which means you can go into softer terrain without sinking.

 

[TheRoadie]

It sounds like you're mixing up ground pressure and PSI.

Yes, exactly. I'm trying to understand why they aren't identical. The ground pressure in PSI is separated only by the tire from the internal pressure in PSI. One mental model I think of says that they need to be equal - 15 PSI on the ground is generated by 15 PSI in the tire. Because I'm modeling the tire tread as flexible, and the ground surface as inflexible. So in that model, the contact patch expands to make the internal and external pressure equal, like the tire is a balloon pressed against a wall.

The ground pressure should be the area of the contact patch / the weight that the wheel is carrying. In our hypothetical a 5000lb vehicle (1250lb per wheel) with a 200 square inch contact patch should have a ground pressure of 6.25lb/square inch. You can increase the side of the contact patch by airing down or by putting on wider tires.

My problem is thinking about HOW a 200 square inch contact patch could come into existence with 15 PSI internal pressure. Why would it expand so far if the vehicle could be supported by a smaller patch pressing on the ground with 15PSI?

If you double the size of the contact patch obviously you'll halve the ground pressure. That's why tracked vehicles that weigh 25 tons can go places where wheeled vehicles that wiegh 5 tons can't - the contact patch is huge and so ground pressure is low (the "contact patch" on a tracked vehicle is the entire portion of the track that is on the ground. An M1 Abrams tank weights ~60 tons but can go places where my Tacoma can't because the contact patch is probably 15 - 20 square feet.)

Understood. The tank's contact patch is fixed because its tread is not flexible. You can't "air down" a tank. That creates the fundamental difference with tires that I'm struggling with. If you inflate a tire to 300 PSI, and make it essentially inflexible, then on hard ground, the small contact patch will have very high PSI, like a trail on steel track. If the terrain is flexible, IT will deform to increase the contact patch, but just to the point that its area can generate enough PSI to balance the vehicle. Ahhhhh, I'm getting a better mental model. The terrain deforms, causing essentially a FLOTATION force that presses back against the tires with a lower PSI than what's inside the tires. It doesn't have to be at equilibrium because the terrain deforms.

But the ground pressure doesn't have much to do with the PSI inside the tire. Imagine our hypothetical 5000lb vehicle with 7" tires aired down to 15psi. Now imagine it with 15" wide tires at 35psi. The contact patch is going to be bigger with the 15" wide tires because the tires themselves are bigger. The bigger tires = less ground pressure.

That's what I was trying to work out. I didn't see how wider tires automatically made the contact patch bigger. I thought it could be like the displacement of a boat in water. Think about that situation - a boat's displacement is directly related only to its weight. Not shape. Long and skinny canoes will displace the same amount of water as a washtub of the same weight. That's the situation with a deformable terrain (lake) and a non-deformable boat.

Surely you're not saying that the 7" wide tire is going to have the same contact patch size at 15psi as the 15" wide tire?

That's what I was struggling with, yes. It's intuitive that the wide tire's contact patch is larger, but it was also intuitive to some folks that bricks might fall faster in a vacuum. I was trying to get the physics behind the intuition.

If that were the case, why would anybody buy fat tires? The whole benefit of fat tires is that it gives you a bigger contact patch which means less ground pressure which means you can go into softer terrain without sinking.

And it looks like that's the issue - it's a flotation-based force that resists the vehicle "sinking." Fat tires for performance road cars make for a wider contact patch that improves directional change when the tire is steered in a new direction. Fat tires for offroading changes the way the sidewall interacts with the tread.

http://auto.howstuffworks.com/question506.htm brings up a couple of good points. At the edges of the contact patch, the PSI drops way off, so the PSI isn't a constant unless you're on liquid terrain. And that the PSI inside the tire isn't the only thing resisting tire flexibility - the sidewall and tread stiffness also interact and add force. More things to add to the force model.

 

[njtaco]

Not trying to complicate things for ya...but check out post 46 in this thread, where Joaquin Suave says "Just an FYI: I run Casa with 385/65R22.5's on the Pismo Dunes with 0 PSI and have not even come close to rolling a bead."

Just to put in percpective (maybe wrong word?) regarding inside air pressure vs. pressure on the ground, holding the truck "up". Seems like the stiffness of the tire alone is enough to hold up that heavy truck. Perhaps this contributes on smaller/light trucks as well?

 

[Martinjmpr]

Roadie: The size and shape of the tire will affect the size of the contact patch.

All the air pressure inside the tire does is affect the size and shape of the tire and that's going to be different depending on how big the tire is, what it's made of, how it's put together, etc. If the tire had zero air pressure but the same contact patch size the ground pressure would be the same. Or if it had 1000 psi but the same contact patch size it would be the same ground pressure.

To put it another way, you could have two tires that have the exact same air pressure inside but completeley different contact patches because one tire is physically larger than the other tire. A tire that's twice as wide is going to have twice the contact patch, all other things being equal.

OTOH, it would also be possible to have tires that have different PSI inside but the same contact patch (and thus the same ground pressure) on the outside.

 

[DiploStrat]

New Numbers, Old Size

The is the modern listing for the old 7.50x16 tire. The Brits built them as cross ply tires with about 8 plys - they were great on rocks and, at least on a Landrover, death on your kidneys.

The French got smart and built them as radials, the most famous being the old "Saharien" XS. The trick with the radial was that the contact patch it was not wide, but very, very long as the tire aired down. This is still the hot ticket for the sand monsters of Africa. I remember reading a post by an Australian who sang the praises of this tire in a radial. Because of the high load rating, he had to air down, but once that was done his performance was equal or better to that of the wide, American style "High Flotation" tires.

We run 7.50's or 235/85's on all of our vehicles with good results. At the end of the day, diameter is more important than width.

As always, YMMV.

 

[TheRoadie]

To put it another way, you could have two tires that have the exact same air pressure inside but completeley different contact patches because one tire is physically larger than the other tire.

This is helping, but I'll amplify where my curiosity is coming from.

The force exerted by a tire on the terrain comes from two sources. Internal pressure and sidewall stiffness. That's why a zero PSI tire might still hold up a vehicle and not be sitting on the rims. If the sidewalls flex enough, then the rims pressing through the tread are a third source of force. Let's ignore that scenario.

My first modeling error was neglecting the sidewall force. I also assumed that the force per unit area of the contact patch would be uniform, but that's clearly not true. The internal pressure force is applied uniformly because it's a fluid (gas) inside the tire, but the sidewall force is higher near the sidewalls, naturally. But the big difference is on the terrain side. Sand/rocks clearly resist compression with a force that's dependent on deflection, so the patch under the lowest part of the tire has more ground pressure than at the edges.

A wider tire has more of its contact patch in line with the lowest part of the tire (perpendicular to the vehicle centerline), and may approach a constant pressure over the patch. A longer, skinnier patch from a narrow tire will have more of a falloff in pressure from highest to lowest.

But because the up-force and down-force have to be in equilibrium (if they weren't, then the vehicle is still sinking into the terrain or rising up), the force integrated over the contact patch has to be equal. The force is some number of pounds (vehicle weight divided by 4, assuming equal distribution). The terrain and the tires ONLY interact at an interface defined by the contact patch. That's a certain number of square inches.

I think it's a given from the laws of physics that the down force and the up force have to be equal and opposite, or else the system's not at a rest state.

So the force from the terrain upwards is going to be the average ground pressure times the contact patch area. (pounds per square inch times square inches gives the final result unit in pounds - this is OK.) The force downwards from the tire is the sidewall force plus the internal PSI times the contact patch area. These totals have to be equivalent.

But there can be a significant ratio of the peak ground pressure to the least, and narrow tires will maximize that ratio, which can give you some benefit. In non-uniform terrain, the peak pressure might "stick" rocks of a certain size into the substrate (sand/dirt) like pressing blueberries into pancakes and then the tire can push against that non-moving rock for more traction. A wider tire with lower peak/minimum ground pressure might not press that rock into the substrate, then the tire will slip on the rolling./moving rock as it moves forward. This effect might even work down to 1/8" size gravel, and I'm thinking it might even be the main benefit of narrower tires.

Even if the rock isn't pressed into its substrate, the narrow tire's peak/minimum ground pressure ratio will force the tire to "flow" over the rock more when the pressure is lower (on approach and departure), and therefore gain more traction.

This reminds me a bit of how pen plotters (not pin feed) can move paper with such precision over long distances because the feed rollers have microprotrusions that engage the paper like miniature pin feed teeth. As if anybody uses pen plotters nowadays.

 

[Bogo]

You also have belt flexibility and elasticity, and sidewall stiffness and elasticity added into the mix. The thickness and composition of the tread also matters. PSI is the only variable easily changed so most people try to talk about it as the only thing that matters. You can change the others but it takes changing to a different tire or modifying the tire.

And yes, with increased diameter you get a shallower slope to drive up when traversing soft ground.

I'm not positive on this, but with a narrower tire it should compact the ground more so it may bear more weight and or provide better traction through keying into the ground. At least that is what my intuition says. This rapidly fails when the ground easily flows like mud or is highly rounded and smoothed ground materials like some sands, beach stones and river pea gravel.

 

[Scott Brady]

Contact area the same?

Roadie and others.

The traditional formula for CP (in2): CP= Fv (lb vertical load) / P (lb/in2)

This formula of course assumes that the Torodial Membrane on a Cylindrical Rim has no influence on CP, which is entirely impossible. The tire/rim combination is what contains P and aids in supporting Fv. The tire must also have enough carcass integrity to resist punctures, high temperatures, etc.

Contact patch (CP) varies between narrow and wide tires, not as a result of the variable of PSI, which does remain the same, but as a result of how the carcass bears the load.

During my testing with Michelin, we conducted a series of imprint tests to measure how the carcass deforms (i.e. length and width) and total contact area. I still have the imprint boards, as the results were quite interesting.

The testing centered around a 235/85 R16 tire and a 265/75 R16 tire, both of the same load rating and the same carcass composition. Not surprisingly, the contact area at 35psi for the 265 tire was about 10% more than the 235 tire. Mathematically, the 265 is 11% wider than the 235.

So, while the load force (v) is the same in both tires, that is not a correlation to contact area being the same, as the wheel and carcass provide a structure. That structure is under tension from the weight of the vehicle and the PSI inside the assembly. Both the membrane in the the wide and narrow tire deform at the same rate (given all other variables are the same, other than width), but since the wider tire is already "wider" by 10%, that percentage will continue as PSI is decreased.

The gas pressure inside the tire is the most meaningful variable, but the influence of the carcass structure, load rating, construction, and environment temperatures all influence final CP.

I have seen a lightweight Jeep with HD Michelin 52" tires running no valve cores. Does that mean there is no CP because there is no internal gas PSI? Of course not. It is just that the carcass is sufficient to bear the Fv load.