4 ms·

I'm not sure how true this is. As the car's weight increases so too does the grip, as it's a function of down-force multiplied by the roughly constant coefficie

by gnode 8y ago

I'm not sure how true this is. As the car's weight increases so too does the grip, as it's a function of down-force multiplied by the roughly constant coefficient of friction of the tires. Conceivably, the car's weight may mean that aerodynamic down-force produces a relatively smaller impact on grip, and air resistance a relatively smaller deceleration. But I'm not sure that's enough to account for the difference.

dragontamer 8y ago
Excellent racing tires have coefficient of friction > 1. That is, a 2-ton car can have more than 2-tons of force applied from the tires. But most consumer tires have coefficient of friction of less than 1. ~.9 for higher end stuff, but it could be as low as ~.5 or so. Which mean 2-tons of car will lead to LESS than 2-tons of acceleration / deceleration / turning force. In effect, "more weight is more advantage" is only true if you have a coefficient of friction greater than 1.
- londons_explore 8y ago
  The last line of your post isn't clear. Since F=MA, A=F/M. Assuming the braking force (F) is μ (coefficient of friction) * Mass * Gravity, thats F=μMG. Combining those equations you get: A = μMG / M And the M's cancel, giving: A = μG Therefore the braking distance due to road friction alone (and ignoring aerodynamic downforce effects) is independent of the mass of the vehicle.
  - magduf 8y ago
    Isn't this neglecting the forward velocity and momentum? According to theory, the width of a tire has zero effect on tire grip. Race cars and high-performance cars prove this is incorrect: there's a reason they have wide tires (which also cause poorer fuel economy through extra friction). Tires and their interaction with the road are a lot more complex than some simple physics equations.
  - dragontamer 8y ago
    Hmm, that makes sense. Its like the whole "area" doesn't matter as well. I can't help but think that maybe the Newtonian friction model is overly simple sometimes. Surely weight and surface area would have an effect? But perhaps not really. Or at least, not in a way that's very measurable.
    - magduf 8y ago
      I think it's definitely overly simple. If it were correct, high-performance cars wouldn't have wide tires. According to Newtonian friction, the contact patch is the same size regardless of the size of the tire. Obviously that's silly; a Lamborghini is not going to get sufficient grip out of economy-car tires, nor would an 18-wheeler. That model completely neglects the complexities of the construction of the tires; they're not like balloons, they're actually semi-rigid.

by gnode 8y ago

dragontamer 8y ago

Excellent racing tires have coefficient of friction > 1. That is, a 2-ton car can have more than 2-tons of force applied from the tires. But most consumer tires have coefficient of friction of less than 1. ~.9 for higher end stuff, but it could be as low as ~.5 or so. Which mean 2-tons of car will lead to LESS than 2-tons of acceleration / deceleration / turning force. In effect, "more weight is more advantage" is only true if you have a coefficient of friction greater than 1.

londons_explore 8y ago
The last line of your post isn't clear. Since F=MA, A=F/M. Assuming the braking force (F) is μ (coefficient of friction) * Mass * Gravity, thats F=μMG. Combining those equations you get: A = μMG / M And the M's cancel, giving: A = μG Therefore the braking distance due to road friction alone (and ignoring aerodynamic downforce effects) is independent of the mass of the vehicle.
- magduf 8y ago
  Isn't this neglecting the forward velocity and momentum? According to theory, the width of a tire has zero effect on tire grip. Race cars and high-performance cars prove this is incorrect: there's a reason they have wide tires (which also cause poorer fuel economy through extra friction). Tires and their interaction with the road are a lot more complex than some simple physics equations.
- dragontamer 8y ago
  Hmm, that makes sense. Its like the whole "area" doesn't matter as well. I can't help but think that maybe the Newtonian friction model is overly simple sometimes. Surely weight and surface area would have an effect? But perhaps not really. Or at least, not in a way that's very measurable.
  - magduf 8y ago
    I think it's definitely overly simple. If it were correct, high-performance cars wouldn't have wide tires. According to Newtonian friction, the contact patch is the same size regardless of the size of the tire. Obviously that's silly; a Lamborghini is not going to get sufficient grip out of economy-car tires, nor would an 18-wheeler. That model completely neglects the complexities of the construction of the tires; they're not like balloons, they're actually semi-rigid.