How much water would be needed to stop a .22 bullet?

[savonnn]

I'm trying to figure out how deep a column of water I would need to stop a .22 bullet. I'm interested in recovering bullets from a series of guns and trying to see those 'characteristic striations' each gun leaves as part of its 'fingerprint.'

For what it's worth, Mythbusters tried to address this but didn't use .22 caliber, only larger bullets. Oddly enough, if we extrapolate down in caliber, it would seem to me that the .22 will penetrate farther than the higher caliber bullets.

I don't know anything about fluid dynamics, which might help me understand how a small projectile traveling through water would dissipate its energy.

It would appear in this site that a 10-foot tank is used to obtain control bullets from test fires.
http://www.firearmsid.com/A_BulletID.htm


Thanks!

 

[lightarrow]

I'm trying to figure out how deep a column of water I would need to stop a .22 bullet. I'm interested in recovering bullets from a series of guns and trying to see those 'characteristic striations' each gun leaves as part of its 'fingerprint.'

For what it's worth, Mythbusters tried to address this but didn't use .22 caliber, only larger bullets. Oddly enough, if we extrapolate down in caliber, it would seem to me that the .22 will penetrate farther than the higher caliber bullets.

How did you extrapolate it? The larger bullets were exactly the same shape, same density, same speed but with all dimensions scaled of the same factor?

 

[savonnn]

No, I didn't do any formal extrapolation, but it seemed in the Mythbusters that the high velocity, high caliber guns tended to shatter upon contact with the water, while the lower-velocity 9mm didn't. I'm assuming the small & relatively slow .22 cal will not shatter, and might actually penetrate farther than the 9mm. Similar bullet head shapes, smaller diameter for the .22.

 

[Danger]

Welcome to PF, Savonnn.
You need to be a bit more specific with your wording around here. When you say '.22', my first assumption is a .22 LR. There are, however, a lot of other .22 calibre cartridges, including some ridiculously powerful wildcats.
If you are indeed referring to a .22 LR, you don't need anything like 10 feet of water. An oil drum would suffice.
As with anything involving firearms, please make sure to take all appropriate precautions. Wear hearing protectors and shooting glasses, and make sure that the zone is clear of innocent bystanders. Also, ensure that the tank enclosure is bullet-proof. A bullet can skip off of water if fired at the wrong angle.

 

[DaveC426913]

... make sure that the zone is clear of innocent bystanders...

... the guilty ones can fend for themselves ...

 

[Danger]

Damn straight. As the sign on my lawn says: 'Trespassers will be executed.'

 

[savonnn]

Yes, .22 LR. Great- I'll try to round one of those up..or some comparable column of water.

Thanks

 

[DaveC426913]

Damn straight. As the sign on my lawn says: 'Trespassers will be executed.'

And the sign said
people caught trespassin'
will be shot on sight

So jumped up on the fence
and I yelled at the house
"Hey what gives y--"
*whack* *thump*

 

[ok123jump]

Savonnn,

I think that this question is a difficult one to answer. For simple calculations, ignoring most of the thermodynamics effects of the bullet traveling through the water, we can assume that the water is the medium of consideration, mathematically model the geometry of the slug and use the equation for aerodynamic drag to calculate the distance traveled before the bullet slows below a certain threshold.

http://hypertextbook.com/physics/matter/drag/

The site that I reference is a good tutorial on calculating Aerodynamic Drag; however, it does not talk about how to model the geometry of the item which you want to calculate the drag for. In this light, I will suggest some assumptions that you can make, as well as point you in a good direction (which I hope is the right direction):

- The slug (if it is a normal bullet) is a composed of a cylinder, capped by a hemisphere. If the total length of the bullet is L and the radius of the hemisphere R, then the length of the cylinder is L' = L-R.

We have a fair understanding of the drag coefficient C_d for both of these shapes. For the cylinder, I suggest you approximate it as if it were a flat plane of length \ell = 2\pi \cdot R then the drag coefficient will be C_{d-plane} \cdot \gamma -- where \gamma is a number between 0 and 1 which indicates the amount of contact between your cylindrical plane and the passing fluid (1 = 100% contact; .5 = 50% contact; 0 = no contact). Water does not make perfect contact with the sides of the bullet, the contact is chaotic - thus, by tinkering with \gamma you can simulate the average amount of contact with the sides of the bullet.

Note that \gamma changes with velocity - if the bullet is traveling slow, \gamma is large - if the bullet is traveling a medium speed, the shockwave effects of the water in front of the bullet will push the water away from the sides and \gamma will be small - if the bullet is traveling very fast (faster than the shockwave), the shockwave will not have time to form and \gamma will be large. If we imagine that there is some perfect velocity v_p at which the fluid does not touch the side of the bullet, then the shape of the graph of gamma with respect to velocity looks like a Bell Curve, if the Bell was upside down - so that at only one spot C_d = 0 for the sides of the bullet.

That is all the insight I have for now. Can anyone else add to or modify these assumptions? I think that I have stated the easiest case, but I'm not sure.

 

[Danger]

And the sign said
people caught trespassin'
will be shot on sight

So jumped up on the fence
and I yelled at the house
"Hey what gives y--"
*whack* *thump*

Man, I've always hated that song. Yours is a much better version.

 

[Chaos' lil bro Order]

Mythbusters did two episodes on this. The 2nd episode showed them firing all sorts of guns into a swimming pool. The one conclusion was that the 9mm handgun penetrated the water the furthest (if I recall 3-4ft. The larger caliberered guns' bullets tended to shred when they went from aim medium to water medium. Heck, the .50 cal Barret fired round exploded on water impact. I'd suspect its because the faster and longer bullets have less time to react to the stresses they face when crossing the medium boundary. Its kind of like for a fraction of a second, the .50 cal's leading edge slows when entering the water, but the back of the bullet is still traveling super fast.

 

[lightarrow]

Mythbusters did two episodes on this. The 2nd episode showed them firing all sorts of guns into a swimming pool. The one conclusion was that the 9mm handgun penetrated the water the furthest (if I recall 3-4ft. The larger caliberered guns' bullets tended to shred when they went from aim medium to water medium. Heck, the .50 cal Barret fired round exploded on water impact. I'd suspect its because the faster and longer bullets have less time to react to the stresses they face when crossing the medium boundary. Its kind of like for a fraction of a second, the .50 cal's leading edge slows when entering the water, but the back of the bullet is still traveling super fast.

And, maybe (but I don't really know), that it's also water which has less time to be displaced and so the faster bullet impact on it as if it were solid (water).

 

[Danger]

Well, as any hydroplane driver will tell you, water is like concrete when you hit it fast.

 

[Mech_Engineer]

You should keep in mind that not all .22LR bullets are slower than the 9mm Luger. A 115gr http://en.wikipedia.org/wiki/9_mm_Luger_Parabellum" goes about 1130 ft/s, a "high velocity" 38gr copper plated lead can go about 1260 ft/s, and the 36gr high velocity variety can go 1380 ft/s.

 

[Danger]

True, but... and this might be a cultural thing... up here, the high-velocity rim-fire rounds tend to be hollow points. That makes a big difference upon entering a medium.