I would like to know the following:

  1. What is the angle at which water gets splashed when I ride my vehicle through a water on the road?

  2. How does angle of water varies with speed?

  3. What is the relation between the distance the water goes with the speed of the vehicle?


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    $\begingroup$ You can get a rough estimate relatively easily, but do you want a precise answer? That would be a major project. $\endgroup$ – Ron Maimon Sep 11 '12 at 4:02
  • $\begingroup$ I am fine in getting the approach in calculating the angle..if that is possible. Rough estimate based on logic is also fine. $\endgroup$ – meetpd Sep 11 '12 at 4:18
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    $\begingroup$ I doubt there is any non-numerical, non-experimental calculation of the angle that could be written in words and equations. $\endgroup$ – Luboš Motl Sep 11 '12 at 9:00
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    $\begingroup$ @RonMaimon how would you do the estimation? $\endgroup$ – Yrogirg Sep 11 '12 at 9:31
  • $\begingroup$ You should specify situation more, there is also water going backwards from the car tire (it can be also stones) - in this case that is a standart homework problem. People are trying to answer here more difficult situation when the water is splashed PERPENDICULAR to the direction in which car moves. $\endgroup$ – Asphir Dom Sep 11 '12 at 10:20

This seems a simple enough question, but a moderately intense Google failed to find any simple answers. Since you're happy with a "rough estimate based on logic" I'd point out that car tyres are flat along a line normal to the direction of motion, so the displacement of water is going to be comparable to pressing a rectangular slab into a film of water.


The initial motion of the water is horizontal, but as the displaced water hits the stationary water outside the tyre contact patch the water will rise up in the same sort of way a wave breaks (a wave breaks because the water at the top is moving faster than the water at the bottom).

The velocity of the displaced water is simply related to the volume displaced per unit time, and therefore to the car velocity. The difficult bit is working out the mechanics of the "wave breaking" and how far the spray will go. A quick look at the Wikipedia article on wave breaking suggests this is far from a fully understood problem.

At a guess, I would say that above some critical speed most of the displaced water is simply displaced upwards with a small percentage being lost to viscous losses in the puddle. Ignoring air resistance the distance travelled by the spray will be roughly proportional to the car velocity, however I'd guess air resistance is important for water spray so the range will be proportional to velocity at low velocities but won't increase as fast for high velocities. I would also guess that the puddle depth affects the angle of travel of the spray, though I hesitate to guess the form of this relationship.

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    $\begingroup$ Your simple model of how water is thrown outwards seems very plausible, but I think that, rather than wave breaking, you want to look at the hydraulic jump <en.wikipedia.org/wiki/Hydraulic_jump>. That would allow you to get some numbers, which may (or may not) have some realtion to the OP's question. I'll try to sketch a solution during lunch, if no one else does before... $\endgroup$ – Jaime Sep 11 '12 at 18:13
  • $\begingroup$ Agreed, a hydraulic jump is a better model than a wave, though I note that none of the examples in the Wikipedia article are high velocity enough to generate the spray you get fom a car tyre. If you can post a calculation of the spray distance in a hydraulic jump I'd be very intrested to se it. $\endgroup$ – John Rennie Sep 11 '12 at 18:23
  • $\begingroup$ I did not get too far with the hydraulic jump thing... My only contribution after about a quarter of an hour of juggling with the equations is that exit speed from the side of the tire, and therefore probably spray behavior, grows with the width of the tire, as there is more fluid to be evacuated through the same side area. $\endgroup$ – Jaime Sep 11 '12 at 19:47

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