4 ms·

All tyres are in a state of slip when you are accelerating or braking, it's unavoidable.

jmpman 7y ago
Isn’t the purpose of traction control to keep the tires in contact with the ground in a static coefficient of friction, and thus not slipping?
- TylerE 7y ago
  If there was zero slippage there would be zero acceleration.
  - Robotbeat 7y ago
    Not quite right? The tires have strain, but they do not have to slip on the surface.
  - adrianN 7y ago
    That's not very intuitive. Can you provide a longer explanation for this?
    - sokoloff 7y ago
      Here's a pretty decent, graspable explanation: https://physics.stackexchange.com/questions/142463/does-a-tire-need-to-slip-to-generate-force https://physics.stackexchange.com/questions/142463/does-a-ti... Tires that are not slipping are not creating any force against the road surface (so they don't have to slip when coasting, but will slip while accelerating longitudinally [braking or accelerating fore/aft] or laterally [cornering]).
      - jmpman 7y ago
        Would you say that a rubber belt on a pulley is slipping? I’m sure at a microscopic level there are some places where the belt isn’t making perfect contact and the rubber is slipping across the pulley, but at a macro level, the belt remains in contact with the pulley in the same place from when it initially contacts until it disconnects.
        jmpman 7y ago
        https://youtu.be/ZVW9uMeFXK8 https://youtu.be/ZVW9uMeFXK8 Yet another video about static coefficient of friction on tires that aren’t slipping. Electric cars are more easily able to keep their tires in this maximum static coefficient of friction range, maximizing acceleration.
        sokoloff 7y ago
        At a micro level, I would say that a rubber V-belt on a pulley which is transferring significant force is slipping as an unavoidable element of delivering that force. What traction control/ABS does is keep the the tire slip ratio in the right range to optimize acceleration. https://www.researchgate.net/figure/Typical-tire-longitudinal-friction-m-l-curves-for-several-road-surfaces_fig2_257774905 https://www.researchgate.net/figure/Typical-tire-longitudina... What we intuit about whether or not we're slipping our tires every time we brake for and then drive away from a traffic light "without slipping them" is not actually correct at a micro level. With respect to this thread, it is the micro level that is relevant to tire wear and resulting particulate emissions from that slippage.
  - jmpman 7y ago
    My understanding is that tires, even when accelerating, can operate in the static coefficient of friction range. And, you want them to operate in the range, because it provides the greatest acceleration. But, maybe my mental model is wrong. The following video is from someone who has considered it much more in depth. https://youtu.be/iyeLXkacocA https://youtu.be/iyeLXkacocA
    - sokoloff 7y ago
      That is a "driver's view" of tire slip (for best results, don't spin or slide the tires too much) rather than a chassis or tire engineer's view of tire slip (if you don't slip the tires, you can't generate any force).

Isn’t the purpose of traction control to keep the tires in contact with the ground in a static coefficient of friction, and thus not slipping?

TylerE 7y ago
If there was zero slippage there would be zero acceleration.
- Robotbeat 7y ago
  Not quite right? The tires have strain, but they do not have to slip on the surface.
- adrianN 7y ago
  That's not very intuitive. Can you provide a longer explanation for this?
  - sokoloff 7y ago
    Here's a pretty decent, graspable explanation: https://physics.stackexchange.com/questions/142463/does-a-tire-need-to-slip-to-generate-force https://physics.stackexchange.com/questions/142463/does-a-ti... Tires that are not slipping are not creating any force against the road surface (so they don't have to slip when coasting, but will slip while accelerating longitudinally [braking or accelerating fore/aft] or laterally [cornering]).
    - jmpman 7y ago
      Would you say that a rubber belt on a pulley is slipping? I’m sure at a microscopic level there are some places where the belt isn’t making perfect contact and the rubber is slipping across the pulley, but at a macro level, the belt remains in contact with the pulley in the same place from when it initially contacts until it disconnects.
      - jmpman 7y ago
        https://youtu.be/ZVW9uMeFXK8 https://youtu.be/ZVW9uMeFXK8 Yet another video about static coefficient of friction on tires that aren’t slipping. Electric cars are more easily able to keep their tires in this maximum static coefficient of friction range, maximizing acceleration.
        sokoloff 7y ago
        At a micro level, I would say that a rubber V-belt on a pulley which is transferring significant force is slipping as an unavoidable element of delivering that force. What traction control/ABS does is keep the the tire slip ratio in the right range to optimize acceleration. https://www.researchgate.net/figure/Typical-tire-longitudinal-friction-m-l-curves-for-several-road-surfaces_fig2_257774905 https://www.researchgate.net/figure/Typical-tire-longitudina... What we intuit about whether or not we're slipping our tires every time we brake for and then drive away from a traffic light "without slipping them" is not actually correct at a micro level. With respect to this thread, it is the micro level that is relevant to tire wear and resulting particulate emissions from that slippage.
- jmpman 7y ago
  My understanding is that tires, even when accelerating, can operate in the static coefficient of friction range. And, you want them to operate in the range, because it provides the greatest acceleration. But, maybe my mental model is wrong. The following video is from someone who has considered it much more in depth. https://youtu.be/iyeLXkacocA https://youtu.be/iyeLXkacocA
  - sokoloff 7y ago
    That is a "driver's view" of tire slip (for best results, don't spin or slide the tires too much) rather than a chassis or tire engineer's view of tire slip (if you don't slip the tires, you can't generate any force).