If I understand it correctly, the weight of the car doesn't really matter. What matters is the PSI (pounds per square inch) of the tires. For example, if your tires are inflated to 42 PSI, the inner tire walls are experiencing 42 PSI of air pressure, which then exerts the same 42 PSI (plus the small weight of the tire) onto the floor. The weight of the car is spread over the surface area of the tires that are in contact with the floor. The heavier the car, the more contact surface area. So if your tile can withstand 42+ PSI which I don't see why not, then you're fine.
Greetings
I'm setting up my dream garage and thinking of tile. I'm a little nervous about it the weight of my Model X.
Any input on this, or what you did to prep the floor? Did you use a membrane?
I have ProSlat in my garage and love it. My setup doesn't look as slick as that picture though.I've attached a picture of my dream garage shortly after it was finished, but before I had a chance to clutter it! The floor tile is called SwissTrax, a.k.a. TraxTile. Everything you need to know about it is here:
https://www.swisstrax.com
With regard to your specific questions ... I applied it over raw concrete with no treatment or preparation. Since the tile doesn't adhere there is no need for surface preparation. Other than two screws at the front corners it is a 'floating' system ... the shape is an exact match to the garage so it can't go anywhere.
There is an infinite possibility of colors and patterns, but if I had to do it over I would go with more dark gray or black and less silver. Inevitably, hot tires discolor lighter colors so decide where your tires will park and put dark tiles there!
P.S. The same tile is used in fire stations under fire trucks and in airport hangers under aircraft ... not to worry about the weight of your Model X.
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Maybe I'm missing something here. If I put on wheels that are twice as wide as the standard Model X tires and pump them up to 42psi, are you saying that this will have no effect on the pressure seen under the tires compared to skinnier ones? Large truck manufacturers started to reduce the number of tires on rigs by using much wider single wheels to replace dual wheel setups. I assume the single, wider wheel exerts that same psi on the road as the dual setup to meet code. If what you say is true, then these wider wheels would use half the air pressure inside as the narrower versions. (I have no idea what psi trucks use btw).I'm not wrong. Let's do a quick math:
MX weight = ~6000lbs
Tire pressure = 42PSI. This pressure is the EXACT pressure being exerted onto the floor (plus a small amount to account for the weight of the tire itself).
6,000pounds / 42 pounds per square inch = 142 square inches of contact. That means the 6000lbs is spread over 142 square inches at a pressure of 42 pounds per square inch. In other words, the tires will deform at 42 PSI to make contact with the floor with a total surface area of contact of 142 square inches.
142 square inches / 4 tires = 35.5 square inches of contact per tire.
Each tire is about a foot wide. Each tire is deformed enough to contact the floor at 12 inches wide x 3 inches long. Look at how your tire is contacting the ground.
The total surface area of contact is a rectangle 12in x 3in at 42PSI, times 4 tires, = 6000lbs.
If your tile is not made out of mud and can handle 42PSI, you are absolutely fine.
We put heavy refrigerators+food on 4 small wheels onto tiles all the time without much thought. The contact pressure under the refrigerators are a lot more than 42PSI due to the small contact area involved.
Maybe I'm missing something here. If I put on wheels that are twice as wide as the standard Model X tires and pump them up to 42psi, are you saying that this will have no effect on the pressure seen under the tires compared to skinnier ones? Large truck manufacturers started to reduce the number of tires on rigs by using much wider single wheels to replace dual wheel setups. I assume the single, wider wheel exerts that same psi on the road as the dual setup to meet code. If what you say is true, then these wider wheels would use half the air pressure inside as the narrower versions. (I have no idea what psi trucks use btw).
You are a bit confused between the weight under each tire versus the weight under each tire over an area. Weight is different than weight/area.Maybe I'm missing something here. If I put on wheels that are twice as wide as the standard Model X tires and pump them up to 42psi, are you saying that this will have no effect on the pressure seen under the tires compared to skinnier ones? Large truck manufacturers started to reduce the number of tires on rigs by using much wider single wheels to replace dual wheel setups. I assume the single, wider wheel exerts that same psi on the road as the dual setup to meet code. If what you say is true, then these wider wheels would use half the air pressure inside as the narrower versions. (I have no idea what psi trucks use btw).
I'm not wrong. Let's do a quick math:
MX weight = ~6000lbs
Tire pressure = 42PSI. This pressure is the EXACT pressure being exerted onto the floor (plus a small amount to account for the weight of the tire itself).
6,000pounds / 42 pounds per square inch = 142 square inches of contact. That means the 6000lbs is spread over 142 square inches at a pressure of 42 pounds per square inch. In other words, the tires will deform at 42 PSI to make contact with the floor with a total surface area of contact of 142 square inches.
142 square inches / 4 tires = 35.5 square inches of contact per tire.
Each tire is about a foot wide. Each tire is deformed enough to contact the floor at 12 inches wide x 3 inches long. Look at how your tire is contacting the ground.
The total surface area of contact is a rectangle 12in x 3in at 42PSI, times 4 tires, = 6000lbs.
If your tile is not made out of mud and can handle 42PSI, you are absolutely fine.
We put heavy refrigerators+food on 4 small wheels onto tiles all the time without much thought. The contact pressure under the refrigerators are a lot more than 42PSI due to the small contact area involved.
I'm not wrong. Let's do a quick math:
MX weight = ~6000lbs
Tire pressure = 42PSI. This pressure is the EXACT pressure being exerted onto the floor (plus a small amount to account for the weight of the tire itself).
6,000pounds / 42 pounds per square inch = 142 square inches of contact. That means the 6000lbs is spread over 142 square inches at a pressure of 42 pounds per square inch. In other words, the tires will deform at 42 PSI to make contact with the floor with a total surface area of contact of 142 square inches.
142 square inches / 4 tires = 35.5 square inches of contact per tire.
Each tire is about a foot wide. Each tire is deformed enough to contact the floor at 12 inches wide x 3 inches long. Look at how your tire is contacting the ground.
The total surface area of contact is a rectangle 12in x 3in at 42PSI, times 4 tires, = 6000lbs.
If your tile is not made out of mud and can handle 42PSI, you are absolutely fine.
We put heavy refrigerators+food on 4 small wheels onto tiles all the time without much thought. The contact pressure under the refrigerators are a lot more than 42PSI.
According to your formula, if my tire pressure is 20PSI, I'll have 300 sq inches of contact. Or 75 sq inches per tire - that is quite large tires now Now things can get quite out of control if my pressure drops below 5 PSI
See image below. 6000lb steamroller.Curious as to if this theory holds out if the tires were solid rubber or, say steel as there would be no internal tire pressure? Also, if you double the tire pressure to 84 pounds per square inch, are you sure that the contact area is half?
I think the confusion lies in the post that seemed to say that the PSI in the tire would be the same as the pressure on the tiles. I understand the contact area and weight calculation to calculate exact pressure.
I think the confusion lies in the post that seemed to say that the PSI in the tire would be the same as the pressure on the tiles. I understand the contact area and weight calculation to calculate exact pressure.