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This is partly a physics question, but has anyone come across any calculations for the ideal following distance at highway speeds to prevent rock strikes?

I assume rocks are lobbed upwards with negligible forward velocity, such that they follow a parabolic trajectory with respect to the car windshield. There may be two safe spots, one near the forward vehicle where the rock may clear your car (especially if it's a low sports car), and one far away where the rock has time to fall back to the ground before striking the following vehicle.

That's probably an oversimplification though, thoughts are appreciated!

Edit: adding some constraints - let's limit the discussion to pebbles which can chip windshields.

Edit 2: Perhaps we can treat this as an optimization problem and assume that the upper and lower limits of following distance are both safe. I suspect there is a sweet spot in the middle where you can safely pass under most rocks that are launched.

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    It seems like there are so many variables that this question becomes hard to give a single answer for. Are you talking about tiny pebbles and pieces of grit that will slowly sandblast the paint and windshield? Or actual rocks that will cause damage? Do we have any indication that rocks even follow a consistent trajectory? Couldn't they go higher or lower depending on how they were tossed or hit by the vehicle in front? And at any rate, aren't there other, more important factors that should influence following distance - i.e. safety?
    – dwizum
    Jan 9 '20 at 19:44
  • That 2 second rule was probably invented for a reason...
    – Solar Mike
    Jan 9 '20 at 20:07
  • Surely the safest distance to follow to prevent this completely is 1mm because any rocks would hit the underside of your vehicle. Obviously this isn't practical but it's a pure physics answer. Jan 10 '20 at 12:41
  • @SteveMatthews Treating this as an optimization problem, I think there are at least three minima. One, as you point out, is only optimal if you don't consider following safety. I suspect there is a second maximum, dependent on speed and vehicle size, which is centered under the maximum height of the parabola(?) traced by the rock. And then of course there is a third which is anywhere sufficiently far away that the rock comes back to the road before intersecting with the following vehicle. Jan 10 '20 at 22:31
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The obvious answer is "far away". Rocks get thrown up with variable height and speed. If a rock doesn't go high enough to clear your car in the first place, the only way to avoid it is to be far enough back. A position that avoids rocks launched high will get pelted by rocks launched low. A position that's safe from rocks launched low may get hit by ones launched higher. The only safe place that avoids both is far back. Trying to find the "sweet spot" behind a car that will allow flying debris to sail harmlessly over your head is likely an exercise in futility, due to the highly variable nature of how the rocks get launched.

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In a system of two cars moving at the same constant speed one behind the other we can consider a rock thrown by the first car (wheel) to be a body thrown at an angle to the horizon. The initial velocity of a body in such system would be the road speed of the car. Given the distance which the body travels horizontally:

which maxes at a 45 degrees angle:

where V is the speed in [m/s] and g is the acceleration due to gravity (9.8 [m/s^2]) that would be the minimal safe distance [m].

For example, 65 mph yields the distance of 85 meters, and 30 mph - 18 meters.

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    Good analysis under these assumptions, but I'm not sure how well this reflects the initial trajectory of rocks kicked up by a car. On its face, it makes sense that the rock would shoot out of the tire at the same speed it's spinning. But if you think about it, launching the rock at 65mph at an angle gives it a horizontal velocity of less than 65mph (relative to the car). Since the car itself is moving at 65mph, that means the rock is moving forward relative to the ground - how could a tire pushing backwards accelerate the rock forwards? Jan 10 '20 at 14:18
  • I like this answer as well, though I believe using 45° is not viable for most vehicles. This is because there's not 45° of angle between the rear tire and the body of the car. Each vehicle is going to be different as to a rocks escape trajectory, and then if they have a mudflap, there even less angle there. Jan 10 '20 at 14:28
  • I like this approach but building on @Paulster2's answer, I'm also not sure if 45° is a valid assumption. My intuition is that relative to vehicle road speed (say, 65mph) the lateral velocity of a rock is negligible. In which case the trajectory should still be parabolic as in your answer. I think the primary variables here are the speed of the follower, the mass of the rock (which determines launch velocity), the height of the follower, and the initial trajectory, which may deviate from vertical according to some distribution. Jan 10 '20 at 22:38
  • @NuclearWang as for the less horizontal speed - yes, you're right, that's why there's a sine of an angle in the first formula. What about the second part it's not quite right, coz we're considering two cars in a frame of reference which moves along the horizontal axis with the cars. So the cars are at rest in such a frame and rocks are shot with car's ground-relative speed in all directions. If we consider a frame of reference which is not moving relative to the ground, the rocks are never thrown backwards, only forward.
    – max
    Jan 12 '20 at 16:07
  • @Pᴀᴜʟsᴛᴇʀ2 sure, 45° is the worst case scenario.
    – max
    Jan 12 '20 at 16:14

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