Bullets usually fragment upon impact with surfaces, water is no different. Though if you're asking the street at which you would certainly be far from a bullet fired while underwater and also resistant to fragmentation, the is a separate question.

You are watching: How far does a bullet travel in water

say girlfriend fire a total underwater, how far away would i need to be for this reason i can plainly capture it without damaging myself or the bullet. Cause im assuming that regardless of the water, the bullet isnt gonna fragment mid-air. You re welcome correct me if anything is wrong.

As others have actually mentioned, the Mythbusters walk a good episode on this.

From Wikipedia: "All supersonic bullets experiment (up to .50-caliber) degenerated in less than 3 ft (90 cm) of water, however slower bullets, prefer pistol rounds, require up to 8 ft (2.4 m) that water to sluggish to non-lethal speeds. Shotgun slugs call for even more depth; the specific depth could not be determined due to the fact that one of your tests broke the rig. However, as many water-bound shots room fired indigenous an angle, less actual depth is needed to create the necessary separation."

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· 7y · edited 7y

How would certainly you deal with this mathematically? fine the system is an extremely complex, gift able come perfectly model the system would it is in impossible. Especially since a many kinetic power of the bullet is lost into generating the unstable eddies.

But, over there is a an excellent equation for determining viscous traction in laminar flow. That is known as stokes law (http://en.wikipedia.org/wiki/Stokes%27_law). This will be our starting equation for determining the drag pressure exerted by the fluid.

-Fd = -ma = 6PimuRv OR -md2x/dt2 = 6PimuRdx/dt

This deserve to be solved by approximating the nature of water and also the bullet. Take it note, if we usage the info that the cartridge traveled about 2m, we could even calculate what the early velocity indigenous the barrel must have actually been.

Of food this is assuming the there is no power lost to turbulent flow, so the calculate will constantly be off. Friend could include a 'turbulent coefficient' to comprise for the increased energy dissipation.

By addressing the Diff EQ using laplace transforms, I acquired x(t) = x'(0)/C*(1-exp(-C*t))

Where x'(0) is the early velocity that the bullet, x(t) is the place of the bullet, and C = 6Pimu*R/m

At t=inf, y = y'(0)/C, and knowing the distance traveled through the bullet was around 2 m and at 5 Celcius C= 0.0144 for water. This predicts that the cartridge traveled in ~ 0.0288 m/s initially.

Interesting. Obviously, the stokes equation walk not use to solution that generate so lot turbulence. It appears that there demands to it is in a fudge factor of around 10,000 come account for every one of the energy lost come turbulence. If C = 144 then the y'(0) would be 288 m/s. The number is an ext reasonable. With this new value that C and y'(0), it would take around 25ms to reach 2m. This seems to correspond well v the video.

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It is also interesting to keep in mind that the C coefficient has actually a very solid dependence top top temperature. In ~ T=5C the C element is 1.5x bigger than the C variable at 20C. The difference in between water in ~ 0C and also 100C is nearly a variable of 20x. That dependence will certainly linearly impact the distance traveled through the bullet.

Finally, to answer your question, due to the fact that the place equation is one exponential v respect come t, capturing a bullet will certainly be very challenging because over there is together a little window in between the bullet going really quick to the cartridge stopping. But, if we do experiments to identify the fudge factor much more accurately, you can find the perfect an ar to capture a bullet. This would certainly be a an extremely fun video game indeed. At the very least you will be underwater therefore no one have the right to hear girlfriend yell when the bullet accidentally goes with your hand!