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I am currently reading “Fundamentals Of Vehicle Dynamics” by Thomas D. Gillespie.

enter image description here

The author mentions how the asymmetric pressure distribution across the length of the contact patch of a tire while rolling results in Rolling resistance. This, then inevitably creates a moment about the axis of the wheel which resists its movement.

The author states, however, that this asymmetric pressure distribution only occurs when a tire is rolling. There isn’t any asymmetry when it’s in a dead-stop position.

What I want to know now is the reason for this asymmetric pressure distribution. What causes it? How does Rolling generate asymmetry?

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I came across a site that, among others, tries to explain exactly the phenomena in your question. It goes quite into depth about the different components of the tyre and how they interact with each other and the effects it has on the contact patch and rolling resistance. I also especially enjoyed the part speaking of the standing wave.

Just as a post script:

I really like @Mark's answer and I've drawn a sketch showing how I understand it: enter image description here

Basically, the moment when an element of the tyre touches the ground it has to stop to a standstill (relative to the ground). In order to have this happen, a force in the opposite direction needs to be applied to decelerate it. fx isn't shown in the diagram but that's the required forward force to decelerate the element, also causing rolling resistance.

The opposite obviously applies to the trailing edge, needing an upward force to accelerate it again.

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    $\begingroup$ Thanks for the source my friend 👌...it really has a lot of invaluable information. $\endgroup$
    – RedHelmet
    Commented Jan 9, 2018 at 13:31
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At the leading edge of a tire, I would presume that as the tire tip has a velocity with a component heading in the negative z direction, the resulting change to a "deformed patch" will result in it having an acceleration to stop this velocity, which would require a force. This force would take the form of additional stress on this leading edge.

We would have a reverse scenario on the trailing edge, where the tire tip will gain velocity in the positive z direction, and the resulting force will reduce the stress in the tire on the trailing edge.

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    $\begingroup$ That seems like a very logical explanation. Can you please cite the source from where you read this? $\endgroup$
    – RedHelmet
    Commented Jan 9, 2018 at 2:48
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    $\begingroup$ I wouldn't have a source, that's just how I'd set up the free body diagram. $\endgroup$
    – Mark
    Commented Jan 10, 2018 at 16:47
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When the tire rolls forward, there is gradual loading of the weight of the car on the tire that causes it to deflect and set in to a small flattening motion.

This creates bending on the bottom face threads and the walls buckling. This constant folding of the tire creates heat and consumes energy and resist the rolling forward.

That is one of the reasons that having fully pressured tires saves gas.

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