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Is the X too heavy for a tile floor garage?
- Thread starter Tube Guy
- Start date
This is your first mistake. The internal pressure when the gauge reads 42 PSI is 57 PSIA. Where the tire is in contact with the air there is a force of 57 pounds on each square inch of the inside with a force if 15 lbs on the outside for a net of 42 pounds termed 42 PSIG with the G standing for "gauge" because that's what a gauge would measure. There is no air under the patch so the only force countering the air pressure from the outside of the patch is 1/4 the weight of the car divided by the patch area.
6,000pounds / 57 pounds per square inch = 105.25 square inches of contact. That means the 6000lbs is spread over 105.25 square inches at a pressure of 57 pounds per square inch. In other words, the tires will deform at 57 PSIA (42 PSIG) to make contact with the floor with a total surface area of contact of 105.25 square inches.
But the area of contact will be less that that because the upward force on the car is the sum of the patch force plus the sidewall load. Following the original line of reasoning here if we remove the Schraeder valve the tyre pressure goes to 0 psig and the patch area would have to go to infinity to keep the car from sinking into the earth. But we know that doesn't happen. As the air escapes the tyre the patch size does increase somewhat but the load no longer carried by the air pressure is assumed by the sidewalls (and, eventually, the rim). The sidewalls are, thus, clearly carrying some of the load. This means that the patch size is even smaller than the 105 square inches. So it appears that the patch size may be about 25 in^2 per tyre. My tyres are 7" wide. If that full width is part of the patch then the patch length would have to be about 3.6" long. Is that how long it actually is? Well, yes, it's about that long but to be honest about it I don't have much confidence in the measurement.
To summarize, you are wrong with respect to two factors:
1)Failure to appreciate the difference between PSIG and PSIA
2)Failure to allow for the part of the load taken by the tire sidewalls.
SDRick
Active Member
I also am unconvinced of a direct one to one relationship with tire pressure. As stated by ajdelange, sidewalls are a factor. If you have a stiff run flat tire and let the out, the internal tire pressure becomes meaningless as also would be the case with a solid rubber car tire.
In any case, if installed properly, tile should be fine.
In any case, if installed properly, tile should be fine.
While plane tires aren't exactpy the same, Boeing recommends same formula to calculate tire contact zone https://www.google.com/url?sa=t&sou...FjAAegQIARAB&usg=AOvVaw368Aofsn9R5QJkhmlUFcKo
I was at first surprised that Boeing would not be smart enough to know about the PSIG/PSIA distinction but then looked at the pressures involved. At 215 psi neglecting the difference would imply an error of only 7% and these calculations are pretty rough anyway. I was fascinated by the elliptical patch info though. I could, in trying to measure patch dimensions, see that the patch was not diagonal but using the numbers in the Boeing bulletin I calculated that the major axis would be 7.1" and for my car it is more like 5".
1) Refuting your contention that I failed to account for PSIG/PSIA:
Disregard the weight of the tire for now.
The tire gauge itself is at the same altitude as the tire and the ground. The tire gauge's PSIG is 0 and PSIA is 14.7PSIA (sea level). The tire uninflated is 0 PSIG and 14.7PSIA. The ground where the tire sits on is 0 PSIG and 14.7PSIA. We are talking about pressure diffferences and so you need to either pick PSIG throughout or PSIA throughout. The uninflated tire's pressure relative to the external atmospheric pressure is 0. When we say the tire is inflated to 42PSI, we mean the pressure inside the tire is 42PSI higher than the external atmospheric pressure outside the tire.
If you're going to use PSIG: Internal tire pressure 42 PSIG, external atmospheric pressure 0 PSIG, difference of 42PSI.
If you're going to use PSIA: Internal tire pressure 42+14.7PSIA, external atmospheric pressure is 14.7PSIA, difference of 42PSI.
2) Refuting failure to account for the load taken by the sidewalls:
The load (aka weight of the vehicle) is NOT taken at all by the sidewalls. The load is directed straight down 100% to the contact area. In other words, the contact area is supporting 100% the weight of the vehicle. If you're saying the sidewall is somehow taking on a load, the 6000lb vehicle would no longer weigh 6000lbs.
If you don't believe what I say is true, please read the entry below:
Ground pressure - Wikipedia
"The ground pressure for a pneumatic tire is roughly equal to its inflation pressure."
Roughly because there's the weight of the tire as well.
Forget all the math that was used to prove the point. If the tire is inflated to 42 PSI, the ground pressure is also ~42PSI. It does not matter what the weight of the vehicle is, because the tire will flatten out (until it structurally can't) to support it at the same PSI.
Ground pressure - Wikipedia
"The ground pressure for a pneumatic tire is roughly equal to its inflation pressure."
Roughly because there's the weight of the tire as well.
Forget all the math that was used to prove the point. If the tire is inflated to 42 PSI, the ground pressure is also ~42PSI. It does not matter what the weight of the vehicle is, because the tire will flatten out (until it structurally can't) to support it at the same PSI.
SDRick
Active Member
Then you are going to have to take into account the characteristics of the material, since it is the material and NOT the air that is supporting the weight of the vehicle
Vinc
Member
Tangentially related. I have EpoxyShield and it does not like EVs whatsoever. Maybe it was not applied correctly, but 2 years of use with an Honda Civic and Mazda Tribute left no mark. First time I parked a 500e and i3 with wet tires and let them sit overnight, the paint underneath the contact area stuck to the tires and came right off. Same when I replaced the Fiat with a Model S.
Maybe it is not the weight but the tires themselves but, there it is...
Refuting your refutation: The tire surface under the patch is subject to no pressure from the air as it in not in contact with the air.
So now let's refute this refutation with the following gedanken experiment: Take a hollow cylinder and cap it at both ends. Drill into the side of the cylinder, tap it and install a valve and pressure gauge. Using the valve pressurize the cylinder to 50 psig. Set it on its end on a rigid surface. Now put a heavy weight on the top cap and read the pressure gauge. What happens? Well depending on the elasticity of the cylinder material it may go up a fair amount or very little. Were the cylinder steel it would not go up noticeably as the entire load would be taken up by the steel. It would compress but a wee bit so that the length and thus volume of the cylinder would not change much at all and the air pressure, nRT/V, wouldn't change appreciably. But now suppose the cylinder be made of rubber. It is much more elastic and would, under the top load, compress considerably more than the steel one. The volume would become smaller and the pressure go up noticeably. Thus, we see, or should be able to see, that the forces holding the weight up are shared between the air column and the wall material depending on their relative moduli of elasticity. For some reason you think that the wall carries no load - that it is all carried by the air column.
To make things even clearer let's now evacuate the cylinder through the valve assembly. There is now no air to carry any of the load associated with the added weight. Does the cylinder collapse? Certainly not so the load must be being borne by the wall. The steel cylinder will shorten only a wee bit under vacuum and is presumably not going to collapse radially either under -15 psig (30" vacuum) but the rubber one will so we assume that we have put steel rings inside separated by some distance. Even so under -15 psig it will both shorten and rubber will be sucked in between the rings but it won't collapase. Note that the rubber is in compression - pushing back against the air pressure on the top cap and the sides. When we put the weight on the top cap now we increase the vertical load and the rubber compresses more thus producing the hookian force necessary to support the weight.
In both cases the load is supported entirely by the sidewall material as there is no air.
Well yes, the contact area supports 100% of the load but only part of the load is transmitted to the contact area by the air column. The rest (or in the case of an evacuated experimental cylinder or tyre all of it) is transmitted through the mechanical structure of the container which, in the case of the tyre, is the sidewall.
That's exactly what I am saying. For example lets say 5400 lbs are supported by the air column and 600 by the sidewalls (I have no idea if those are reasonable proportions). It's apparent you aren't familiar with how these calculations are done but the usual starting point is a "free body" diagram. This isolates a part of the system from the entirety and then the forces acting on this piece are evaluated. At equilibrium the vector sum of these forces is 0. The free body here would be a piece of the the tyre with the dimensions of the patch. The entirety of the vertical forces on this free body are:
1)Down: Air pressure (PSIA times area of the patch)
2)Up: Air pressure (0 as the bottom surface of the patch isn't in contact with the air)
3)Up: Substrate reaction (up approximately 1/4 the weight of the car per tyre).
4)Down: Shear forces (from the interface between the patch rubber and the rubber in the rest of the tire.)
With respect to the last item we have carried on here as if all these shears are transmitted through the sidewalls but the rubber in front of and behind the patch deforms too and thus also imposes hookian forces on the patch.[/QUOTE]
I think maybe you are beginning to get the idea. You have a couple of structures sharing the load. One is the air column and the other is the rubber of the tyre. How the load is shared depends on the elasticities of the materials and, of course, the geometries.
TL;DR
I give up. Please google "ground pressure" and see how the ground pressure is related to the tire inflation pressure, how the contact area is determined and how the load is transferred to the area. It is not just me but people and companies much smarter than me that's using the fact that ground pressure ~ tire inflation pressure in their calculations. I'd like to see your refutation to that fact.
I give up. Please google "ground pressure" and see how the ground pressure is related to the tire inflation pressure, how the contact area is determined and how the load is transferred to the area. It is not just me but people and companies much smarter than me that's using the fact that ground pressure ~ tire inflation pressure in their calculations. I'd like to see your refutation to that fact.
No, I'm not beginning to get any idea. We are talking about pneumatic tires with air in it and how air is supporting the weight in its entirety, 100%. The run-flat tire example is entirely different, as it is not supported by air but by the tire material 100%. Entirely different animals.
That would probably be best. A couple or posters here clearly understand the physics better than you do. Wikipedia is a great resource for people qualified to interpret it. Yes, ground pressure is APPROXIMATELY the same as tire pressure but as has been shown to you here the approximation can be pretty rough.
I was pretty sure you wouldn't be able to but we are always hopeful. That's why we take the time to explain these things (plus it's good mental exercise)
So what happens if I let the air out of the tire?
It is not just Wikipedia but a simple search would result in multiple references to that fact.
I'm not so sure people have shown me why the approximation can be rough. Case in point, the PSIG/PSIA confusion on your part and its resulting calculation to show the rough approximation is wrong. We are talking about pressure differences which is the same whether you use PSIG or PSIA. The approximation is actually very close and not rough at all. The thing that is missing is the weight of the tire, which is negligible and I have repeatedly stated as such.
@cypho in post #13 already covered it:
"There is a limit to how much the tires can flatten out. If you let the pressure drop too low, the wheels will touch the ground and the PSI the floor sees will go through the roof ( since the wheels will not flatten at all).
All of a sudden you will have all 6000lbs pressing down over 0.5 sq inch of wheel for 12000psi onto the floor. That is when your tiles will crack."
What is your point?
I have shown you, and confirmed by measurements, rough as they are, that there can be considerable error.
There is no confusion with respect to PSIA/PSIG. It is very clear that PSIA must be used because there is no air pressure acting on the bottom of the patch. It is not in contact with the air. The air pressure difference across the patch is, when tyre pressure is 42 psig, 57.7 pounds per square inch. I have explained this more than once and you are either not reading what I wrote or don't understand the basic physics well enough to comprehend.
What I'd really like to understand is how you came up with this idea that in a pneumatic tire the air column carries all the load "entirety, 100%". So I ask questions like what happens if I let the air out and you ignore the question. Do you have any technical background at all? Just curious and it's really none of my business.
There is no confusion with respect to PSIA/PSIG. It is very clear that PSIA must be used because there is no air pressure acting on the bottom of the patch. It is not in contact with the air. The air pressure difference across the patch is, when tyre pressure is 42 psig, 57.7 pounds per square inch. I have explained this more than once and you are either not reading what I wrote or don't understand the basic physics well enough to comprehend.
What I'd really like to understand is how you came up with this idea that in a pneumatic tire the air column carries all the load "entirety, 100%". So I ask questions like what happens if I let the air out and you ignore the question. Do you have any technical background at all? Just curious and it's really none of my business.
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