4 ms·

It could, but it could also be a function of the car's weight (batteries are heavy) and the tires it comes equipped with from the factory. Poor tires will yiel

by magduf 8y ago

It could, but it could also be a function of the car's weight (batteries are heavy) and the tires it comes equipped with from the factory. Poor tires will yield poor braking performance due to lack of grip.

dragontamer 8y ago
But it can't be poor tires, because the Model 3's brakes were proper in the first test (~130 feet stopping distance). The same car with the same tires then had different braking behavior in every test after the first test. That's just odd.
- londons_explore 8y ago
  Most vehicles detect a sharp press to the brakes, and add extra assistive braking force. Thats from the observation that most humans do not apply enough braking force in an accident (potentially because they have never applied that much force before, and don't know how much is 'enough'). It's possible that the tesla software either isn't applying the extra assistive force, either due to incorrect tuning or a bug.
- cc439 8y ago
  I can almost guarantee the problem is caused by the brake pads glazing over. The working surface of a brake pad will become work hardened, which drastically lowers their friction coefficient, once pushed past their designed temperature limit. The Model 3 (3,800lbs for the long range model) is a fat pig of a car compared to other, conventionally powered cars in its class (3,300lbs for a BMW 3 Series) but this is countered by Tesla fitting them with relatively enormous brakes (14" rotors up front, 13.5" in the rear, larger rotors have more heat absorbing capacity for the pads). If I had to guess the root cause for this problem, I'd say it is likely down to Tesla's focus on minimizing aerodynamic drag causing a lack of airflow within the wheelwell and across the brakes out the wheel itself. Either that or they chose an absolutely terrible brake pad compound but that just doesn't seem likely in this modern era. Either way, the brakes are getting so hot after a single stop that CR's cooling procedure (a reasonable one I might add) is inadequate to keep the brake pads from drastically overheating. This also explains why braking performance was still compromosed after letting the car cool overnight. The work hardened surface will perform poorly until it is literally ground off from further heavy use of the brakes.
  - dragontamer 8y ago
    > I can almost guarantee the problem is caused by the brake pads glazing over. That's actually a very reasonable explanation. I can't say I'm a car expert, but your logic at least sounds correct.
  - justin66 8y ago
    Can't brake pad material also adhere to the surface of the rotor as well, requiring that they be turned to remove the material? I think that's what people mean when they talk about the rotor being "glazed," a related problem to what you're talking about. This is all a potentially expensive repair.
    - magduf 8y ago
      Usually, when pad material adheres to the rotor, people call that "warped rotors" because it feels like the rotor is warped when braking. It's not usually uniform, I think. As for expense, brakes are simple and the parts generally cheap, at least for mass-market cars. You can get a full set of pads and rotors for less than $100 from Autozone.
      - justin66 8y ago
        "Warped" rotors, yes, I've heard that term. Not something that comes up in conversation as much as it used to. They are a lot cheaper today but the cost of rotors varies tremendously depending on the manufacturer and their volume. I gather they turn rotors a lot less than they used to because they're so darn inexpensive to replace, but I'm pretty sure it's still economical for some of the higher-end stuff.
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.

by magduf 8y ago

dragontamer 8y ago

But it can't be poor tires, because the Model 3's brakes were proper in the first test (~130 feet stopping distance). The same car with the same tires then had different braking behavior in every test after the first test. That's just odd.

londons_explore 8y ago
Most vehicles detect a sharp press to the brakes, and add extra assistive braking force. Thats from the observation that most humans do not apply enough braking force in an accident (potentially because they have never applied that much force before, and don't know how much is 'enough'). It's possible that the tesla software either isn't applying the extra assistive force, either due to incorrect tuning or a bug.
cc439 8y ago
I can almost guarantee the problem is caused by the brake pads glazing over. The working surface of a brake pad will become work hardened, which drastically lowers their friction coefficient, once pushed past their designed temperature limit. The Model 3 (3,800lbs for the long range model) is a fat pig of a car compared to other, conventionally powered cars in its class (3,300lbs for a BMW 3 Series) but this is countered by Tesla fitting them with relatively enormous brakes (14" rotors up front, 13.5" in the rear, larger rotors have more heat absorbing capacity for the pads). If I had to guess the root cause for this problem, I'd say it is likely down to Tesla's focus on minimizing aerodynamic drag causing a lack of airflow within the wheelwell and across the brakes out the wheel itself. Either that or they chose an absolutely terrible brake pad compound but that just doesn't seem likely in this modern era. Either way, the brakes are getting so hot after a single stop that CR's cooling procedure (a reasonable one I might add) is inadequate to keep the brake pads from drastically overheating. This also explains why braking performance was still compromosed after letting the car cool overnight. The work hardened surface will perform poorly until it is literally ground off from further heavy use of the brakes.
- dragontamer 8y ago
  > I can almost guarantee the problem is caused by the brake pads glazing over. That's actually a very reasonable explanation. I can't say I'm a car expert, but your logic at least sounds correct.
- justin66 8y ago
  Can't brake pad material also adhere to the surface of the rotor as well, requiring that they be turned to remove the material? I think that's what people mean when they talk about the rotor being "glazed," a related problem to what you're talking about. This is all a potentially expensive repair.
  - magduf 8y ago
    Usually, when pad material adheres to the rotor, people call that "warped rotors" because it feels like the rotor is warped when braking. It's not usually uniform, I think. As for expense, brakes are simple and the parts generally cheap, at least for mass-market cars. You can get a full set of pads and rotors for less than $100 from Autozone.
    - justin66 8y ago
      "Warped" rotors, yes, I've heard that term. Not something that comes up in conversation as much as it used to. They are a lot cheaper today but the cost of rotors varies tremendously depending on the manufacturer and their volume. I gather they turn rotors a lot less than they used to because they're so darn inexpensive to replace, but I'm pretty sure it's still economical for some of the higher-end stuff.

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.