5 ms·

> What is true is that the 50mph collision with a brick wall is equivalent to a 100 mph collision with a car that is standing still, but that wasn't the claim.

by dools 2y ago

> What is true is that the 50mph collision with a brick wall is equivalent to a 100 mph collision with a car that is standing still, but that wasn't the claim. Ummmm ... what? I mean, sure if you assume that the dynamics of a brick wall and a car are different, then they won't be identical but if you just take a parked car and a brick wall to both be "immovable objects at rest" which for all intents and purposes is true for the driver, then I don't see how this statement could possibly be true.

chmod775 2y ago
As a driver you mostly care about the amount of deceleration you will personally experience. If you have 1m of hood to work with, you're decelerating 100km/h to 0km/h over 1m in the wall case. In the hood-to-hood case, depending on reference frame, you're either decelerating 50km/h to 0km/h over 1m or 100km/h to 0km/h over 2m (both hoods). Either way it should be trivially obvious that running into a wall is going to be a much more violent experience. Likely the wall crumpling up the front of your car won't be enough to bring you to a stop before you personally impact something.
- dools 2y ago
  Not true, because when you exert a force on the vehicle coming the other direction, an equal and opposite force is exerted on your car. Then you also have the force exerted by their car attempting to accelerate you backwards. For each car, there are 2 forces: the force that you exert due to your trying to accelerate them in your direction of travel and the opposing force. What you're saying is the same as saying that if you fall from the roof of an elevator at rest, the effect is the same as if you fall from the roof of an elevator that travels upwards at the same time as you fall. That's obviously not true, but the physics you need to prove it are F = ma and Newton's 3rd law.
  - chmod775 2y ago
    This would make sense as a refutation if this was perfectly elastic rigid body physics, which wasn't precisely what we were doing here.
    - dools 2y ago
      So sum the forces: For Car A: - Force of Car B accelerating Car A opposite to direction of travel - Force of Car B opposing force of Car A trying to accelerate Car B And vice versa for Car B. EDIT: Like if you hit a wall, the wall feels the force of your car. The force that crushes the car is the force opposing that, which is the wall acting on the car. EDIT2 (because I can't reply): the force acting on the driver will be the sum of the force of Car B acting on Car A opposing the force of Car A acting on Car B plus the force of Car B acting on Car A. There are 2 forces in each direction of travel that are exerted by the cars. The force felt by the driver is the sum of these 2 forces, exerted by the dashboard on the driver's face. If the car hits a brick wall, the only force opposing the direction of travel of the car is the force of the wall acting on the car, which opposes the force of the car acting on the wall. As such the force felt by the driver of the dashboard acting on their face is halved. EDIT3 (still because I can’t reply): the dynamics of the materials is not the issue. Yes bricks behave differently than cars, but that’s not the basis of the author’s argument.
      - chmod775 2y ago
        In a perfectly elastic collision between rigid bodies, any impact forces acting on the cars would have to also be transferred to the driver in both instances, but cars are neither perfectly elastic, nor do they behave like rigid bodies in a collision. > EDIT2 (because I can't reply): the force acting on the driver will be the sum of the force of Car B acting on Car A opposing the force of Car A acting on Car B plus the force of Car B acting on Car A. There are 2 forces in each direction of travel that are exerted by the cars. The force felt by the driver is the sum of these 2 forces, exerted by the dashboard on the driver's face. If the car hits a brick wall, the only force opposing the direction of travel of the car is the force of the wall acting on the car, which opposes the force of the car acting on the wall. As such the force felt by the driver of the dashboard acting on their face is halved. If this was how it worked, crumple zones would do nothing. Next you're going to model the airbag as a rigid body too? Ouch.
      - antod 2y ago
        > The force felt by the driver is the sum of these 2 forces Correct me if I'm wrong, but I think you've added non existent extra forces. In the car/car situation, there are still only two forces - each the reaction of the other depending on which cars point of view you're taking. There aren't two separate forces to sum. Each car experiences a decelerating force - the reaction of which is the decelerating force for the other car and the reaction of that is the original deceleration force of the original car.
kazinator 2y ago
We would never treat a parked car as an immovable object. It only makes sense to use the "brick wall" symbolism for am idealized, immovable barrier that suffers no damage. A parked car will move when struck. The combination of the two cars continues to have kinetic energy, which is lost by friction.
- dools 2y ago
  That’s not the argument he is making. He is making the argument that if you imagine a sheet of paper in between the 2 cars into which they crash, then since they both stop since they are perfectly symmetrical then the fact that there is another car on the other side of the sheet is immaterial. It’s complete garbage.
  - kazinator 2y ago
    Pardon me, I seem to have made exactly the same argument myself, completely with crude pictures. https://news.ycombinator.com/item?id=40629971 https://news.ycombinator.com/item?id=40629971 It is entirely unassailable; all you can argue with is the realism of my assumptions. (Real head-on collisions won't be precise mirror images.) When two identical objects are moving toward each other with a relative speed v, the collision is like hitting an ideal barrier at speed v/2.
    - dools 2y ago
      CROSS POST: I posted this in reply to your linked explanation. Okay yep I get it now. The 4mv2 did it for me mathematically but I still couldn't get it intuitively until I imagined a car going 50mph rear ending a car going 40mph. If they wind up going the same speed as a result of the collision then the deceleration felt by the car in the back would only be 10mph if the car in front didn't change speed because it was like, an ocean liner. If the cars are the same size and the collision is perfect and all that then the car in front would gain 5mph and the car behind would lose 5mph. Since in a head on, they are both changing speed then the deceleration must be half the combined relative deceleration. The way I was imagining it was as if the car collides with another car going 50mph in a head on collision and the other car doesn't slow down due to the collision. So it appears that it is I who am confidently wrong about physics (although I still hold that Milton Friedman is confidently wrong about economics!).

by dools 2y ago

chmod775 2y ago

As a driver you mostly care about the amount of deceleration you will personally experience. If you have 1m of hood to work with, you're decelerating 100km/h to 0km/h over 1m in the wall case. In the hood-to-hood case, depending on reference frame, you're either decelerating 50km/h to 0km/h over 1m or 100km/h to 0km/h over 2m (both hoods). Either way it should be trivially obvious that running into a wall is going to be a much more violent experience. Likely the wall crumpling up the front of your car won't be enough to bring you to a stop before you personally impact something.

dools 2y ago
Not true, because when you exert a force on the vehicle coming the other direction, an equal and opposite force is exerted on your car. Then you also have the force exerted by their car attempting to accelerate you backwards. For each car, there are 2 forces: the force that you exert due to your trying to accelerate them in your direction of travel and the opposing force. What you're saying is the same as saying that if you fall from the roof of an elevator at rest, the effect is the same as if you fall from the roof of an elevator that travels upwards at the same time as you fall. That's obviously not true, but the physics you need to prove it are F = ma and Newton's 3rd law.
- chmod775 2y ago
  This would make sense as a refutation if this was perfectly elastic rigid body physics, which wasn't precisely what we were doing here.
  - dools 2y ago
    So sum the forces: For Car A: - Force of Car B accelerating Car A opposite to direction of travel - Force of Car B opposing force of Car A trying to accelerate Car B And vice versa for Car B. EDIT: Like if you hit a wall, the wall feels the force of your car. The force that crushes the car is the force opposing that, which is the wall acting on the car. EDIT2 (because I can't reply): the force acting on the driver will be the sum of the force of Car B acting on Car A opposing the force of Car A acting on Car B plus the force of Car B acting on Car A. There are 2 forces in each direction of travel that are exerted by the cars. The force felt by the driver is the sum of these 2 forces, exerted by the dashboard on the driver's face. If the car hits a brick wall, the only force opposing the direction of travel of the car is the force of the wall acting on the car, which opposes the force of the car acting on the wall. As such the force felt by the driver of the dashboard acting on their face is halved. EDIT3 (still because I can’t reply): the dynamics of the materials is not the issue. Yes bricks behave differently than cars, but that’s not the basis of the author’s argument.
    - chmod775 2y ago
      In a perfectly elastic collision between rigid bodies, any impact forces acting on the cars would have to also be transferred to the driver in both instances, but cars are neither perfectly elastic, nor do they behave like rigid bodies in a collision. > EDIT2 (because I can't reply): the force acting on the driver will be the sum of the force of Car B acting on Car A opposing the force of Car A acting on Car B plus the force of Car B acting on Car A. There are 2 forces in each direction of travel that are exerted by the cars. The force felt by the driver is the sum of these 2 forces, exerted by the dashboard on the driver's face. If the car hits a brick wall, the only force opposing the direction of travel of the car is the force of the wall acting on the car, which opposes the force of the car acting on the wall. As such the force felt by the driver of the dashboard acting on their face is halved. If this was how it worked, crumple zones would do nothing. Next you're going to model the airbag as a rigid body too? Ouch.
    - antod 2y ago
      > The force felt by the driver is the sum of these 2 forces Correct me if I'm wrong, but I think you've added non existent extra forces. In the car/car situation, there are still only two forces - each the reaction of the other depending on which cars point of view you're taking. There aren't two separate forces to sum. Each car experiences a decelerating force - the reaction of which is the decelerating force for the other car and the reaction of that is the original deceleration force of the original car.

kazinator 2y ago

We would never treat a parked car as an immovable object. It only makes sense to use the "brick wall" symbolism for am idealized, immovable barrier that suffers no damage. A parked car will move when struck. The combination of the two cars continues to have kinetic energy, which is lost by friction.

dools 2y ago
That’s not the argument he is making. He is making the argument that if you imagine a sheet of paper in between the 2 cars into which they crash, then since they both stop since they are perfectly symmetrical then the fact that there is another car on the other side of the sheet is immaterial. It’s complete garbage.
- kazinator 2y ago
  Pardon me, I seem to have made exactly the same argument myself, completely with crude pictures. https://news.ycombinator.com/item?id=40629971 https://news.ycombinator.com/item?id=40629971 It is entirely unassailable; all you can argue with is the realism of my assumptions. (Real head-on collisions won't be precise mirror images.) When two identical objects are moving toward each other with a relative speed v, the collision is like hitting an ideal barrier at speed v/2.
  - dools 2y ago
    CROSS POST: I posted this in reply to your linked explanation. Okay yep I get it now. The 4mv2 did it for me mathematically but I still couldn't get it intuitively until I imagined a car going 50mph rear ending a car going 40mph. If they wind up going the same speed as a result of the collision then the deceleration felt by the car in the back would only be 10mph if the car in front didn't change speed because it was like, an ocean liner. If the cars are the same size and the collision is perfect and all that then the car in front would gain 5mph and the car behind would lose 5mph. Since in a head on, they are both changing speed then the deceleration must be half the combined relative deceleration. The way I was imagining it was as if the car collides with another car going 50mph in a head on collision and the other car doesn't slow down due to the collision. So it appears that it is I who am confidently wrong about physics (although I still hold that Milton Friedman is confidently wrong about economics!).