It is known that bullets can ricochet off a body of water. Is surface tension responsible for this or is this the same behavior we see when an asteroid ricochets off the atmosphere? I don't think surface tension has anything to do with it but I'm arguing with someone who disagrees. I think the major factor is the density of water relative to air and the density of the bullet.

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    $\begingroup$ Oh good ol' dam busters... $\endgroup$ – Pygmalion Apr 16 '12 at 17:43
  • $\begingroup$ One can also make flat pebbles skip on the water at the beach. I rememember measuring 14 skips for a shard from a roof tile. I think it has to do with all: angle of incidence, velocity and density of material ( gas in the case of the asteroids but they go very fast). $\endgroup$ – anna v Apr 16 '12 at 17:51
  • $\begingroup$ Hi John to Physics SE! My guess is that this will be hard to calculate (as all question involving bullets hitting something) but intuitively with the high density and sound velocity the surface tension is most likely not important. The experiment will be easy though, just take some detergent and try it. $\endgroup$ – Alexander Apr 16 '12 at 17:51
  • $\begingroup$ When dam busters experimented with bouncing their bombs, there were two conclusions: the relative speed (water-projectil) must be large enough and angle must be small enough. They actually spinned their bombs before throwing. However, I am not sure how is this possibly related to surface tension. $\endgroup$ – Pygmalion Apr 16 '12 at 17:55
  • $\begingroup$ @Pygmalion If there were no surface, which is what surface tension makes sure exists, there could be no ricochet? $\endgroup$ – anna v Apr 16 '12 at 17:58

The mechanism is explained, e.g., in W. Johnson, Int. J. Impact Engng, Vol.21, Nos 1-2, pp. 15-24 and 25-34. 1998.

The following main assumptions are used to derive the approximate Birkhoff formula for the critical ricochet angle for a spherical projectile:

(i) The pressure $p$ on a spherical surface element along its outward drawn normal is $\rho u^2/2$; u is the forward speed of the sphere resolved along the normal.

(ii) The pressure applies only to those parts of the sphere which are immersed below the undisturbed surface of the water. The effect of the splash on the sphere is considered not to contribute any pressure.

Thus, I believe, surface tension is negligible.


It's nothing to do with surface tension (art least for large objects).
It's simply the force needed to accelerate the water out of the way to allow the object to sink.

Imagine a bullet bouncing off another bullet, or metal armour. No problem accepting that, it's just Newton's laws and momentum. well water also has mass and needs a force to accelerate it in exactly the same way - the only difference in bouncing a bullet, or a stone, or a bomb, is the speed and angle and how much water you need to move and how fast.

I'm not sure at what speed/pressure the viscosity becomes a factor, has anyone tried skimming stones off super-fluid helium?

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    $\begingroup$ To reinforce your excellent point about displacement inertia, Myth Busters did an episode once comparing modern rifles to civil war muskets for shooting people swimming underwater. The unequivocal result: The civil war musket was deadly for swimmers, the modern rifle harmless. Why? Because the modern bullets moved so fast that the water by comparison moved more like a solid than a liquid, causing the bullet to self-destruct. The much slower civil war bullet gave the water in front of it enough time to move out of the way, allowing the bullet to go much farther. (Nice He-4 question, BTW!) $\endgroup$ – Terry Bollinger Apr 16 '12 at 23:30
  • $\begingroup$ When I studied fluid dynamics (which I've mostly forgotten) there was something called Reynolds Number, relating inertial to viscous forces. $\endgroup$ – Mike Dunlavey Apr 17 '12 at 12:41
  • $\begingroup$ I think a problem with this answer is this concept of water moving "out of the way" and "how fast." If you throw a baseball at a very thick piece of glass and it bounces it's not accurate to say the glass molecules couldn't get out of the way fast enough. It seems more of an issue of the elasticity of the collision. $\endgroup$ – John Apr 17 '12 at 14:35
  • $\begingroup$ @John - I think an elastic collision with a window is different to a recoil from a fluid. At some very high speed, or with a non-newtonian fluid the recoil could be elastic and behave very like glass - but I think at skimming stones speed it's more useful to think of in momentum terms, liek a newtons-cradle toy $\endgroup$ – Martin Beckett Apr 17 '12 at 15:03
  • $\begingroup$ @MartinBeckett - I agree. My point was that this concept of particles not being able to get "out of the way" fast enough seems incorrect. Given enough energy a particle will move out of the way at nearly the speed of light. It doesn't seem like a very scientific explanation. $\endgroup$ – John Apr 17 '12 at 16:00

As a particle physicist I tend to see this as a semi elastic scatter, where the velocity and the angle of incidence and the medium's cohesion must enter the solution.

If it is a solid, which has high cohesion, there is high probability of ricochet/semi-elastic-scatter .

An asteroid skimming the top of the atmosphere needs a very high velocity and small grazing angle.

Liquids are in between, depending on the variables stated.

I expect that at the microscopic level, the electrons of the projectile at a given angle and speed see the projection of the electrons of the surface as an impenetrable continuum,comparable to the the one presented normally by solids.

  • $\begingroup$ Would a single electron refract as it enters a medium that impedes its velocity? Maybe a group of electrons behave like a pulse of individual electrons. Some would scatter diffusely and some would refract. But because they are bound, instead of scattering you have water molecules scattering and the electrons in the bullet refracting. Does that make sense? $\endgroup$ – John Apr 17 '12 at 14:03
  • $\begingroup$ @John More or less. They scatter collectively as part of the solid projectile. The water molecules must for a delta(time) seem like a solid.And it is reflection, not refraction. $\endgroup$ – anna v Apr 17 '12 at 16:13
  • $\begingroup$ annav, I'm wondering if a single electron behaving like a wave would refract as it travels from air to water. And maybe the bullet could be seen as a group of electrons (pulse) behaving as a wave being reflected when the angle of incidence equals the angle of refraction. $\endgroup$ – John Apr 17 '12 at 16:31
  • $\begingroup$ refraction is when the beam enters the water. Reflection when it is scattered out. A single electron would, quantum mechanically, have some probability of entering the water, refracting, and some of reflecting. Again it would depend on the angle of incidence, the velocity of the electron and the density of the medium it impinges on. The electrons on the surface of the projectile will see the collective field from the surface of the liquid, and the projectile will either ricochet, or penetrate. Are you confused by "total internal reflection"? en.wikipedia.org/wiki/Total_reflection $\endgroup$ – anna v Apr 17 '12 at 17:09

It is probably easiest to understand if you think of the bullet moving in two separate directions, horizonal and vertical. The bullet moves slowly up or down into the water, while at that depth it moves horizontal a great distance at speed, it will encounter a significant amount of water mass which will be ejected as a reaction, the total momentum of this mass results in the trajectory being reflected. Hence the water is imparting momentum required to deflect the slower vertical component.


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