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Is the X too heavy for a tile floor garage?

Discussion in 'Model X' started by Tube Guy, Nov 1, 2019.

  1. Tube Guy

    Tube Guy Member

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    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?
     
  2. teethdood

    teethdood Member

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    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.
     
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  3. pereiks

    pereiks Member

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    This sounds wrong, PSI is to measure the pressure within the tire. I don't understand how this would apply to tile.

    What he needs to know is the area each tire is applying weight to the garage floor. And then divide car weight by 4 and this will be the weight applied to this area. Divide more to get a weight per cm or inch and see what is the rating of the tile and will it take this kind of weight.
     
  4. teethdood

    teethdood Member

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    #4 teethdood, Nov 1, 2019
    Last edited: Nov 1, 2019
    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.
     
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  5. SaniDel

    SaniDel Member

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    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.

    IMG_0932.jpeg
     
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  6. VT_EE

    VT_EE Active Member

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    I have ProSlat in my garage and love it. My setup doesn't look as slick as that picture though.
     
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  7. VT_EE

    VT_EE Active Member

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    #7 VT_EE, Nov 1, 2019
    Last edited: Nov 1, 2019
    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).
     
  8. cypho

    cypho Member

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    The tire may be 2x as wide, but if the PSI is the same, then the area in contact with the ground will not change.

    So instead of a 12" x 3" strip of contact, you would have a 24"x1.5" strip of contact under each tire.
     
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  9. teethdood

    teethdood Member

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    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.

    In your example, if you put tires that are twice as wide (call it 24 inches) and pump it to 42 PSI, the pressure being exerted onto the floor is STILL 42PSI (understand what pounds per square inch means) and the contact area is STILL the same, but the dimensions are different. Instead of 12inx3 in rectangle, it is now 24inx1.5in rectangle.
     
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  10. teethdood

    teethdood Member

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    Cypho understands the math. Apparently a few people disagreed with my math. I tried my best to explain.
     
  11. pereiks

    pereiks Member

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    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
     
  12. SDRick

    SDRick Active Member

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    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?
     
  13. cypho

    cypho Member

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    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.
     
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  14. SaniDel

    SaniDel Member

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    We're making a mountain out of a molehill. The compressive strength of cured concrete can be as low as 2,000 PSI to as much as 10,000 PSI, but typically 2,500 to 3,000 PSI in residential construction:

    https://shop.iccsafe.org/media/wysiwyg/material/9090S15-Sample.pdf

    We're orders of magnitude away from worrying about crushing a concrete garage floor! The only issue is whether a tile covering the concrete is suitable. A ductile material such as the PVC in SwissTrax/TraxTile is not a problem, but I wouldn't recommend brittle sun-dried mud tiles. Porcelain or ceramic might work, but why bother ... imagine the labor cost for a two-car garage!
     
  15. teethdood

    teethdood Member

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    See image below. 6000lb steamroller.

    download.jpg
     

    Attached Files:

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  16. VT_EE

    VT_EE Active Member

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    This topic is something I’ve never really thought out before. Fascinating and thanks to those who explained it!
     
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  17. mswlogo

    mswlogo Well-Known Member

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    So you are saying that steamroller weighs as much as Model X? That weighs more like 10 tons !!

    BTW, I think your math is correct for an estimation.
     
  18. shinytop

    shinytop Member

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    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.
     
  19. mswlogo

    mswlogo Well-Known Member

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    It is the same ~PSI hitting the tiles that’s in the tire. Approximately. If you test that idea at the extremes other factors come into play.
     
  20. teethdood

    teethdood Member

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    #20 teethdood, Nov 1, 2019
    Last edited: Nov 1, 2019
    There's no confusion. The PSI in the tire is the same PSI that's hitting the tile. The tire serves to hold the air in, that's all. The tile is being subjected to the same 42PSI (plus a negligible weight of the tire). The tire is deformed at 42PSI over an area (36 square inches) to counter 1/4 the weight of the 6000lb vehicle.

    Let's say you stack another MX on top of your MX to make it 12K lbs and no air is added to the tire, so the tire pressure remains at 42PSI, the tire deforms under increased load to increase the contact area with the tile. The contact area is now:
    12,000lbs / 4 tires = 3000lbs/tire
    3000lbs / 42 PSI = 143 square inches
    143 square inches over a 12-inch wide tire means a contact rectangle of 12inx12in. The tire has now deformed or flattened to 12inx12in in contacting the tile underneath it. The contact area is increased under increased load, but the actual PSI is still the same, 42PSI.

    (Whether or not the tire can deform to make the 12"x12" contact is a different issue. If it can't, it will explode)
     
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