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Is Driving on the beach harmful to the undercarriage?
- Thread starter RidgeRunner
- Start date
Tire contact area = Vehicle weight/tire pressure. Tire width and tire height is irrelevant. (Except that the taller the tire the lower tire pressure you can run.) A vehicle that weighs 4400 pounds and is running 20 psi tire pressure has 55 square inches per tire of contact area. It doesn't matter if it's a Jeep, a truck or a Model Y. (4400/4 tires)/20 psi tire pressure = 55 square inches contact area per tire.
The truck's advantage over the Y is increased ground clearance. The truck can sink more before getting stuck--if driven well.
That would depend on the truck. In fact, if I had a lift and some paper and a few trucks to compare with I suspect the MY would come out with a larger area than most trucks.
That formula would work for balloons. For tires, maybe not....
Tire Contact Patch | Lowering Pressure Gain Traction
How to increase tire traction by controlling the contact patch of the tire based on size, rubber compound, width, height, and weight of vehicle for better performance
www.enginebasics.com
The most accurate way to calculate it would be with a lift, ink up the bottom of the tires, apply brakes and lower tire onto paper until all vehicle weight is on the tires, then lift it back up and measure the resulting stains on the paper.
Hmmm. OK, yes I left out sidewall loading because under most circumstances the difference between sidewalls isn't going to be that much on passenger tires. The whole discussion just serves to muddy the water for most, but let's go ahead and delve into it. First and foremost, the data listed is from Avon tires, a motorcycle tire maker. I think it's obvious right off the bat that sidewall strength is going to play more of a role in a vehicle weighing much less. So the study quoted really doesn't apply. But let's go ahead and modify the equation, getting closer to the real world application. Assuming four identical tires for simplicity: Total tire contact patch = (vehicle weight - load supported by the sidewalls)/ tire pressure. If you understand physics even a little bit you understand that this is simply true for a static condition. So a truck with E load rated tires is going to have a smaller contact patch than the same truck with C rated tires. Run flat tires will not have a linear relationship between pressure and contact patch, because the sidewall will take more of the load the lower the pressure. For our purposes, the sidewall of our cars--barring run flat tires, is not going to be carrying an appreciable load at low tire pressures. Once the sidewall is bent over, it loses a lot of it's weight carrying capacity. Since we're talking about a low pressure scenario, the original equation will be very close to correct even if we ignore sidewall load.
TLDR: For our purposes my original equation is close enough as we aren't talking about motorcycles here.
gnuarm
Model X 100 with 72 amp chargers
I was going to argue that the area depends on the tire width, but I think Nakk is right. To a reasonable approximation the tire contact area is defined by the vehicle weight and tire pressure. The sidewall force is minimal compared to the tire pressure. That's why a tire clearly shows bulging side walls with only a few pounds of pressure loss. Reduce the tire pressure by 20% and the contact area has to increase by 20%. Depending on how much the sidewall force contributes this is reduced.
I've worked mounting tires before and I can bend a sidewall flat just by leaning part of my weight on the tire. This is not much weight compared to the car weight. If you want a more accurate estimate come up with a number for sidewall force and subtract that from the vehicle weight in the equation. As Nakk said, the direct estimate is good enough.
I've worked mounting tires before and I can bend a sidewall flat just by leaning part of my weight on the tire. This is not much weight compared to the car weight. If you want a more accurate estimate come up with a number for sidewall force and subtract that from the vehicle weight in the equation. As Nakk said, the direct estimate is good enough.
Avon makes far more than motorcycle tires (though they may under the Cooper name now most of the time since they were acquired by them at some point), and if you looked at the charts and tire sizes in the article those aren't motorcycle tires. I also don't think you can simply discount sidewall loading as insignificant for our cars. Different tires and rubber formulations will perform differently. I'm not sure why you assume that the sidewall loading, or we should more accurately state the sidewall load bearing ability, will not vary much in passenger cars. You could be right. I'm not a tire expert and don't claim to be. I just don't know what your basing that statement on. Regardless, the simple version of the formula could be inaccurate at times depending on the tire, and car, involved and I think that's an important fact to know before you go risking your daily driver car off road using it as a basis for decision making about tire pressure.
Of course it's not entirely accurate. The point is it doesn't need to be. We know that cutting our tire pressure in half is going to roughly double our contact patch size. It doesn't matter that it really is 183.7% of the original contact patch area. Who cares? That number, if we could calculate it exactly, is meaningless. It's enough to know approximations, and that wide tires really have no advantage over skinny tires. In fact, knowing that surface area is roughly the same between the two we can surmise that skinnier tires might just be better in the sand, as they don't have to push through as wide a trough. And guess what? Experience in Africa backs that up. Take a look at what tires those guys off roading in African deserts run, who do this more than anyone else. Yeah sure, you could go to the effort of measuring the surface area. You won't get any useful information from that though.
BTW, I know that the sidewall load is irrelevant for several reasons. One: I see the sidewall bend when I air down the tires. Once bent, the sidewall loses a considerable amount of load bearing capability. Two. I can't vary sidewall strength, other than bending it by lowering air pressure, which I'm doing anyway. Three: I know that we'll actually get a slightly better improvement in surface area airing down than trucks do, because truck tire sidewalls are generally far stronger and stiffer than ours. The truck's advantage is ground clearance, not surface area. (Except where they take advantage of taller sidewalls and air down dramatically, like to 5 psi.)
You're fixating on nit picky details that are utterly irrelevant. What you're coming up with is not "an important fact". If it was, you could do something with it. You can't. What you can do is lower your air pressure to the lowest point possible that doesn't risk tire/wheel damage under current conditions, and increase air pressure when those conditions warrant it. Smaller wheels are better because you can run lower air pressure with taller sidewalls.
Same on assateague Island in Virginia and Maryland.
Wider tires, all else being equal, would be better on the sand for the same reason as the larger contact patch at lower air pressures is better.
Xenoilphobe
Well-Known Member
The Point at Assateague Island with a couple of friends.
Attachments
I made it to the point on the Virginia side once, and had to hurry back to keep from getting cut off by the incoming tide.
gnuarm
Model X 100 with 72 amp chargers
So you are telling me that when the tires start to sink into the sand, they don't make any more contact with the sand than they do when sitting on top?
I think your rule only applies to asphalt and other solid materials.
Jim R
Member
Then why is it that the drivers who drive in deep sand more than anyone else use skinny tires? Wider tires do NOT regulate contact patch, sidewall strength and tire pressure do. Period. Physics, you know? Contact patch = (weight - sidewall load)/tire pressure. Notice that there is no place in that equation for tire width. In fact, Skinny tires are better because you have less frontal area to "plow" through. Paddle tires are an entirely different story of course. With paddle tires wider is better because you get more paddle in contact with the sand. If you read the whole thread you'll see this has already been discussed.
Physics, yes? unless the car is accelerating up or down, the weight of the car EXACTLY equals the force applied to the four tires. If the car weighs 4,000 pounds--assuming a 50-50 weight distribution for simplicity--each tire is supporting exactly 1,000 pounds. That makes sense, right? So let's assume the sidewall is supporting 100 pounds, and the tire pressure is 15 psi. The contact patch is EXACTLY 60 square inches--regardless of tire width. (1000-100= 900. 900/15=60.) Now, decreasing the tire pressure to 5 psi will roughly triple the contact area--assuming the tires are tall enough that you don't start riding on the rims. I say "roughly" rather than "exactly" because we don't know what the change in sidewall loading will be. It may increase as we increase the flex on the sidewall, or it may decrease as the increased flex increases leverage and actually reduces the weight carried by the sidewall. Regardless, we can't control this except to install tires with less sidewall strength. Run flat tires or truck tires with high load ratings--i.e. "E" rated tires--will have stronger sidewalls and consequently have less benefit to reduced air pressure.
Taller tires will have the benefit of reduced approach angle as you sink into the sand. That and increased ground clearance are why our Ys can never be as capable as a 4WD truck in the sand.
gnuarm
Model X 100 with 72 amp chargers
You are confusing pressure inside the tire to the pressure on the outside of the tire. A tire is not a soap bubble.
Tires aren't soap bubbles? Really?! Damn! LOL. I guess that's why I include load carried by the sidewall in my equation.
You are confusing a lack of understanding of basic physics with an understanding of basic physics. I'd like to see your equation for what happens to the shape of the bottom of the tire--you know, the part carrying the load--when the pressure on the outside of the tire is different than the pressure inside the tire, LOL. You do understand that the part of the tire not carrying a load isn't part of the equation, and that if there is unequal pressure the tire will move to equalize that pressure? No? Maybe not make such a silly post if you aren't really sure what you're talking about? Better yet, take physics again.
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gnuarm
Model X 100 with 72 amp chargers
The missing part of physics that you are ignoring is the fact that inflation is not the only part of the tire that carries weight. You can roll tires along the ground with zero inflation. So clearly, there is something going on other than the pressure inside the tire. It's not zero.
