I actually agree with you and apologize
I got a bit confused but after reading a bit more into topic the math is correct for approximation.
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Is the X too heavy for a tile floor garage?
- Thread starter Tube Guy
- Start date
I actually agree with you and apologize
I got a bit confused but after reading a bit more into topic the math is correct for approximation.
I took my time to post the reason and the math/physics in this thread to explain why there should not be any problems with your porcelain tiles unless they are made out of mud. In simple terms, the weight of the vehicle does not matter because the weight is spread out over an area by the flattened tires. The heavier the vehicle weight, the more the contact area made by the tires. It spreads the load out. My other posts went over the physics behind it. Vehicles on rubber tires put a lot less pressure per surface area than a refrigerator+food on four small wheels. So the tile-cracking potential of a heavier weight of a Model X is not so different than a normal much lighter vehicle. (Think about what PSI - pounds per square inch - means).
Now if you put your car onto a jack stand with a small contact area like @mattack4000 in post #22, the weight is now concentrated on a small area and so it is much more likely to crack. So don't put your small jack stand directly on porcelain tile.
It astounds me when people mark posts "Disagree" when they don't post the reason why they disagree.
SaniDel
Member
The TraxTile in the appended photograph was installed in December 2014. Five years later it looks the same with one exception. Hot tires tend to discolor the lighter tiles where the vehicle is parked. Unfortunately, the pattern I chose has our vehicle parked on the light silver in the center ... when the garage is empty there are tire marks where it was parked.
We have our garage floor professionally cleaned annually. The process is similar to steam cleaning carpeting ... a truck mounted boiler injects steam and detergent under pressure followed by scrubbing with a floor polisher and finished by vacuuming. Anything other than the tire marks is dissolved and removed. I could easily replace the four discolored tiles, but the problem would recur ... perhaps before we sell!
If I had to do it over I would pattern the tiles differently so that darker tiles were under the tires, but otherwise I am pleased with the product. If "looking like plastic" is a problem there are alternative products that "look like rubber" ... would that be better?
In my experience there hasn’t been an issue with our X or our S in our tiled garage. No cracking of the tiles. We did build up the transition into the garage with a grout ramp so that stress wouldn’t be applied to the leading edge of tile at the garage door. That was the spot I was most worried about. One item to note is that the tiles we chose, like most these days, had a printed pattern to make them look like stone. The tires appear to be wearing that pattern down a little bit. Hard to see from the picture but my guess is that over the course of a number of years it will become noticeable.
SaniDel
Member
Is the slight deformation of the tile pattern where the vehicle was supported on a jack?
SwissTrax claims a compressive strength of 5,120 PSI which likely exceeds the strength of the underlying concrete!
https://www.swisstrax.com/garage/
I once had to beat something ... not someone ... with an elastomeric deadblow hammer on my TraxTile.
Welcome to Sears.com
Mission accomplished with no damage to the tile, but probably not a good idea with a steel hammer.
SwissTrax claims a compressive strength of 5,120 PSI which likely exceeds the strength of the underlying concrete!
https://www.swisstrax.com/garage/
I once had to beat something ... not someone ... with an elastomeric deadblow hammer on my TraxTile.
Welcome to Sears.com
Mission accomplished with no damage to the tile, but probably not a good idea with a steel hammer.
SaniDel
Member
This begs the question about what would have happened without the tile? If the compressive strength for the tile is 5,120 PSI which is more than typical concrete in residential applications ... would the jack have crushed the 'crete? We all know what happens when jacking on asphalt ... big dents!
I didn't mark disagree even though I don't agree with your math.
It is a discussion. No one is right all the time. Anyone took the time to post, explains and help out others with good intention should be appreciated.The PSI inside the tire is NOT the same as the PSI on the tile. However, the force exerted by the entire wheel to the tile and the force received by the tiles is obviously the same.
P=F/A ... let's look at A. The area of the tire internal surface area should be used to calculated the tire PSI, not the area of the contact point. The contact point of the tire to the tile should be used to calculated the tile PSI. So A is different.
For the force F, the force of the tire PSI is the force of air inside exerted on the internal tire wall. Not the weight of the car. It is the volume of the air. (U can skip to the 2nd last line of this paragraph if you agree.
) It is like squeezing a ballon will not affect it's PSI as long as the ballon can expand (PV=K). F is the force created by the difference between air pressure inside and the outside atmospheric pressure. F or the PSI of the tire will increase with higher attitude due to lower atmospheric pressure, but ironically, not related to the weight of the car. The weight of the car will increase the F indirectly only if it decreases the volume of the tire (not the shape). That explains why the tire is 45 PSI when it is on or off the car (P1 or P2). With heavy load, it is true that the contact point will increase and makes the wheel looks smaller, but the tire side wall as well as the entire shape of the wheel will also expand to maintain the internal air volume constant up to certain limit. (P1V1=P2V2, if P1=P2, V1=V2) With extreme heavy weight when the tire cannot expand anymore to compensate for the volume, the PSI will increase. In other words, the elasticity of the tire material (summer tire vs winter tire vs sports tire) in combination of the weight of the car will affect the tire PSI. Not the car weight alone. Meanwhile, the F on the tiles is simply total weight of the car divided by 4 not affected by attitude or temperature. So, F is different.If the F and A of the tire and tile PSI formula are different, their PSI cannot be the same. Going back to the original question: "Is the X too heavy for a tile floor garage?" ... I don't know
I didn't mark disagree even though I don't agree with your math.
The PSI inside the tire is NOT the same as the PSI on the tile. However, the force exerted by the entire wheel to the tile and the force received by the tiles is obviously the same.
P=F/A ... let's look at A. The area of the tire internal surface area should be used to calculated the tire PSI, not the area of the contact point. The contact point of the tire to the tile should be used to calculated the tile PSI. So A is different.
For the force F, the force of the tire PSI is the force of air inside exerted on the internal tire wall. Not the weight of the car. It is the volume of the air. (U can skip to the 2nd last line of this paragraph if you agree.
If the F and A of the tire and tile PSI formula are different, their PSI cannot be the same. Going back to the original question: "Is the X too heavy for a tile floor garage?" ... I don't know
For the record, in post #2, I said "if I understand correctly..." I was very open to discussion and learn. You obviously took the time to contribute and I appreciate that, whereas flyby "disagrees" by non-contributors don't enlighten anyone one bit.
Let's discuss, as I want to learn from you.
Say you pump a perfectly round balloon up to 42PSI. Everywhere on the inside, the internal wall of the balloon is experiencing an outward pressure of internal air of 42PSI. Everywhere on the external wall of the balloon, it must also experience an inward pressure from the external air of 42PSI to maintain a static system. When placed on the ground, at the point of contact, the ground itself must exert a force equal to 42PSI perpendicular to the point of contact. Do you agree with this statement?
The tire is basically a very rigid balloon. The internal pressure must be met with the same external pressure, otherwise the tire (or balloon) will keep expanding until the pressures equalize. So the PSI internally and externally are of the same vector.
As to the area of contact, how else do you propose we calculate the area? The tire will flatten a certain amount (you are right, depending on the characteristics of the tire) to distribute the load onto the ground while maintaining the same PSI. I did my math above in the other posts.
How would you calculate the PSI/area of contact experienced by the ground? 6000lb vehicle, 4 tires, 42PSI each tire.
No, that’s not true. First of all, even before the ballon touches the ground, the PSI outside and inside are not the same. The force is equal but the pressure is not. The external surface area is large than the internal surface area because of the thickness of the balloon. I know for ballon the thickness is neglect-able, but it’s important since for the tire which is much thicker, it will be a little bit more significant. P=F/A, if the F is the same but A is different so P is different. The thicker the balloon, the more different it will be because the outside area A is much more than the inside area A.
Secondly, when the balloon touches the ground, it is true that the ground must exert the same exact force perpendicular to the point of contact but it’s not 42PSI. It’s weight of the ballon in N or kg, not in PSI. The PSI of the ground is the weight of the ballon divided by the area of the contact point while the psi of the ballon is 42. Don’t forget there’s also air at the non contact point excerting force on the ballon to keep it in shape. If it’s in a vacuum, the balloon will expand.
To calculate the PSI of the ground, we just need to know of the weight of ballon and area of the point of contact which can be measured with applying paint on the ballon outside before it touches the ground then measure it. For the model X, you can spread the water on dry concrete around the tires then move the car away to measure those 4 spots. The weight and the contact area is all we need for the formula, and the PSI of the tire is irrelevant.
As far as rigid ballon that you mentioned, the extreme case would be a metal drum which is a very rigid ballon. Whether the metal ballon inside is vacuum or 1000PSI wouldn’t affect the PSI on the ground.
Without reading the whole thread carefully it seems to me that what you all are missing is that part of the weight on each tire is supported by the sidewall - not the internal air pressure. Draw the freebody diagram of the patch in contact with the ground. Acting up on it is 1/4 (approximately) the weight of the vehicle. Acting down on it is the a force equal to the internal pressure times the area and two additional forces through the sidewalls.
No, that’s not true. First of all, even before the ballon touches the ground, the PSI outside and inside are not the same. The force is equal but the pressure is not. The external surface area is large than the internal surface area because of the thickness of the balloon. I know for ballon the thickness is neglect-able, but it’s important since for the tire which is much thicker, it will be a little bit more significant. P=F/A, if the F is the same but A is different so P is different. The thicker the balloon, the more different it will be because the outside area A is much more than the inside area A.
Secondly, when the balloon touches the ground, it is true that the ground must exert the same exact force perpendicular to the point of contact but it’s not 42PSI. It’s weight of the ballon in N or kg, not in PSI. The PSI of the ground is the weight of the ballon divided by the area of the contact point while the psi of the ballon is 42. Don’t forget there’s also air at the non contact point excerting force on the ballon to keep it in shape. If it’s in a vacuum, the balloon will expand.
To calculate the PSI of the ground, we just need to know of the weight of ballon and area of the point of contact which can be measured with applying paint on the ballon outside before it touches the ground then measure it. For the model X, you can spread the water on dry concrete around the tires then move the car away to measure those 4 spots. The weight and the contact area is all we need for the formula, and the PSI of the tire is irrelevant.
As far as rigid ballon that you mentioned, the extreme case would be a metal drum which is a very rigid ballon. Whether the metal ballon inside is vacuum or 1000PSI wouldn’t affect the PSI on the ground.
Ground pressure - Wikipedia
"The ground pressure for a pneumatic tire is roughly equal to its inflation pressure."
The PSI of the tire is relevant. It is roughly equal to the ground pressure. If you read my other posts, I explicitly put the weight of the tire in between parentheses to account for the slight difference.
MichaelP90DL
Active Member
I doubt it would dent, but anyway I don't know. My house was purchased new, the tiles have been on the floor since day 1. I have the Racedeck. I use them on my small deck and my front entrance as a welcome mat. They are not perfect, but they do serve their purpose. I guess I am suppose to take them out often to clean underneath
And besides ignoring the part of the load taken up by the sidewalls you are all failing to recognize that your calculations should be done with psia, not psig. Where the tyre is sealed to the tile the atmosphere cannot exert its normal 1 bar of pressure.
The important thing is, of course, that the force exerted by each of the vehicle's tyres is about 1/4 of the weight of the car.
The important thing is, of course, that the force exerted by each of the vehicle's tyres is about 1/4 of the weight of the car.
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