2 
contact_patch_area = weight / tire_pressure

Is this true?

  • Wikipedia says "mostly".
  • This guy analyzed some automobile tire manufacturer, and although the info was self-conflicting, he concludes that it very much depends on tire construction, but the equation is for sure false. However, he bases some of the conclusions on the assumption that rubber acts as a constant-rate spring, which it doesn't.
  • This answer assumes it is true.
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    In which units would the area be represented?
    – ebrohman
    Commented Mar 17, 2016 at 18:26
  • How are you defining weight?
    – Batman
    Commented Mar 18, 2016 at 0:52
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    It's approximately correct. For a perfectly flexible tire it will hold as exactly as you can measure, but for a tire that has any inherent stiffness that will affect things (generally by reducing the contact area). Ie, it's going to be pretty darn close for a thin, high-pressure road tire, not nearly so close for a low-pressure, heavily lugged off-road tire. And throw the equation out entirely for a 4" "fat tire" bike.
    – Daniel R Hicks
    Commented Mar 18, 2016 at 3:05
  • With car tyres the whole idea can't work, because they are designed to be cylindrical rather than toroidal, and a round cross-section is normally a symptom of over-inflation. Bike tyres are much closer to being a long skinny balloon tied into a circle. A tubular is about as close to that as you can get while still being able to ride the bike.
    – Móż
    Commented Mar 20, 2016 at 20:53

1 Answer 1

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Assuming a 200 pound rider+bike+clothes+load and a tyre pressure of 100 PSI, and a nice even 50/50 weight distribution, then each wheel is supporting 100 pounds and should have a contact patch size of 1 square inch.

You could empirically measure this using a scale and a dry piece of paper, and a slightly dampened tyre. Moisten the tyre and use its "footprint" on the paper. Do the same with the rear wheel. And then draw around the boundary with a pen. Perhaps a dusting of talc powder on the tyre would work better than water.

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    Apologies for using quaint, antique and obsolete imperial units, but pounds per square inch fits the simple maths better than bar.
    – Criggie ♦
    Commented Mar 18, 2016 at 0:14
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    You are not excused for using barbaric units of measure. However, the idea is solid. I have been looking for a reason to take the unicycle out of storage and this one is perfect. Will report back with results.
    – Vorac
    Commented Mar 18, 2016 at 9:15
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    @Vorac - It's science! Don't forget to include some replication so that the effects of measurement error can be accomodated. If you are statistically comparing different tires your stats model will also need to accommodate the subsampling. I can help if needed.
    – Rider_X
    Commented Mar 18, 2016 at 19:09
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    I actually want to try this at home, with two equally sized tires, one very supple, the other with a very stiff casing. I am willing to bet the contact patch is not identical.
    – Rider_X
    Commented Mar 18, 2016 at 19:10
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    @Criggie I don't think the surface area will necessarily be the same. Imagine a spectrum of casing stiffness from steel to very supple. A perfectly stiff casing will have virtually no defection (regardless of pressure) and infinitely small contact patch.
    – Rider_X
    Commented Mar 19, 2016 at 12:51

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