Libya: Ballistics bullet derail

[egbert]

MODENOTE: Moved from this topic: http://rationalia.com/forum/viewtopic.p ... 43#p779189

Every time I see some rag tag Libyan "freedom fighters" being interviewed for TV news, they're inevitably whooping it up by firing bursts of automatic weapons fire into the air. What goes up inevitably comes down - how many have been killed by their own bullets?
And they complain of a lack of weaponry - shouldn't they be saving those bullets for when they really need them? Just wondering...

[JimC]

Every time I see some rag tag Libyan "freedom fighters" being interviewed for TV news, they're inevitably whooping it up by firing bursts of automatic weapons fire into the air. What goes up inevitably comes down - how many have been killed by their own bullets?
And they complain of a lack of weaponry - shouldn't they be saving those bullets for when they really need them? Just wondering...

It is a catalytic reaction to being observed and filmed by western media...

They know their roles, and they play them well...

[epepke]

I don't want to do anything, and I don't want to pay for it. Let the French do something and get into a quagmire, and I hope we don't try to help them and get nailed for it, as happened with Vietnam.

Oh, and....

Every time I see some rag tag Libyan "freedom fighters" being interviewed for TV news, they're inevitably whooping it up by firing bursts of automatic weapons fire into the air. What goes up inevitably comes down - how many have been killed by their own bullets?

Mythbusters did a story on this. Basically, it doesn't happen, because by then the round is tumbling too much to be going very fast.

And they complain of a lack of weaponry - shouldn't they be saving those bullets for when they really need them? Just wondering...

That's a good point. I've noticed that a lot of extremely poor places seem to have enough guns and explosives.

[Coito ergo sum]

Every time I see some rag tag Libyan "freedom fighters" being interviewed for TV news, they're inevitably whooping it up by firing bursts of automatic weapons fire into the air. What goes up inevitably comes down - how many have been killed by their own bullets?

Well, firing a bullet straight up is like dropping the bullet from the highest point it goes. It won't be coming down fast enough to kill anyone. It has to be fired at an angle sufficiently off the y axis (verticle) to have enough velocity to kill someone.

And they complain of a lack of weaponry - shouldn't they be saving those bullets for when they really need them? Just wondering...

They're the civilians we're protecting. They hardly know what they're doing.

[Coito ergo sum]

Mythbusters did a story on this. Basically, it doesn't happen, because by then the round is tumbling too much to be going very fast.

It's not really the tumbling, it's that if you fire the bullet straight up, then the force carries the bullet up until gravity stops it. For a brief millisecond, it's in equilibrium at the top, and then its descent is governed by the calculation F=MA - Force = Mass x Accelleration. As I recall from physics the accelleration due to gravity is 9.8 meters per second squared, and googling the mass of an AK-47 bullet I get 8 grams for a round number on the heavy side. So, that's F=8 grams x 9.8 m/s2. So, the force of the bullet fired straight up on its descent is 78.4 gm/s2. Now, one Newton of Force is, as I recall, 1 kg m/s2, so to estimate, the force of the little AK-47 bullet on its free-fall is about 1/12th (on the high side) of one Newton.

To put this into real terms - one Newton is less than 1/4 pound of force. That's not much.

Of course, the last time I took physics was 25 years ago, so I could have this wrong. Anyone good with physics? Am I close?

[Hermit]

Can't help you with the mathematics, but there is also such a thing as terminal speed as far as freefall is concerned. The wind resistance would ensure that a freefalling bullet travels not nearly as fast coming back down than when it leaves the muzzle on its way up. Thus the force when it impacts is lower. I imagine that being hit by one of those would be akin to being hit by an 8.5 gram pebble dropped from 20 floors up: It'll sting, but it won't kill.

[Sisifo]

Can't help you with the mathematics, but there is also such a thing as terminal speed as far as freefall is concerned. The wind resistance would ensure that a freefalling bullet travels not nearly as fast coming back down than when it leaves the muzzle on its way up. Thus the force when it impacts is lower. I imagine that being hit by one of those would be akin to being hit by an 8.5 gram pebble dropped from 20 floors up: It'll sting, but it won't kill.

In some countries I've lived it was customary to shoot to the air in celebrations. Actually, tracing ammunition was very much demanded for those celebrations. In one of those, a falling bullet went completely through a child's arm. This is not hearsay, or urban legend. I personally witnessed it. Wether it was angle or it's composition, I don't know, but falling bullets kill.
I have also taken to the hospital a man with a 20cm wooden stick half buried in his shoulder, from a firework.
Human bodies are fragile enough to make them more vulnerable than invulnerable to high speed solids. Even a hand thrown stone can hole a cranium in unlucky circumstances (for the cranium owner).

[Feck]

Mythbusters did a story on this. Basically, it doesn't happen, because by then the round is tumbling too much to be going very fast.

It's not really the tumbling, it's that if you fire the bullet straight up, then the force carries the bullet up until gravity stops it. For a brief millisecond, it's in equilibrium at the top, and then its descent is governed by the calculation F=MA - Force = Mass x Accelleration. As I recall from physics the accelleration due to gravity is 9.8 meters per second squared, and googling the mass of an AK-47 bullet I get 8 grams for a round number on the heavy side. So, that's F=8 grams x 9.8 m/s2. So, the force of the bullet fired straight up on its descent is 78.4 gm/s2. Now, one Newton of Force is, as I recall, 1 kg m/s2, so to estimate, the force of the little AK-47 bullet on its free-fall is about 1/12th (on the high side) of one Newton.

To put this into real terms - one Newton is less than 1/4 pound of force. That's not much.

Of course, the last time I took physics was 25 years ago, so I could have this wrong. Anyone good with physics? Am I close?

NO .

Neglecting friction that bullet accelerates at 32ft/s/s air resistance is not going to slow it down much as it's a dense metal in a smooth lump .

Wiki is your friend http://en.wikipedia.org/wiki/Celebratory_gunfire

[Coito ergo sum]

Mythbusters did a story on this. Basically, it doesn't happen, because by then the round is tumbling too much to be going very fast.

It's not really the tumbling, it's that if you fire the bullet straight up, then the force carries the bullet up until gravity stops it. For a brief millisecond, it's in equilibrium at the top, and then its descent is governed by the calculation F=MA - Force = Mass x Accelleration. As I recall from physics the accelleration due to gravity is 9.8 meters per second squared, and googling the mass of an AK-47 bullet I get 8 grams for a round number on the heavy side. So, that's F=8 grams x 9.8 m/s2. So, the force of the bullet fired straight up on its descent is 78.4 gm/s2. Now, one Newton of Force is, as I recall, 1 kg m/s2, so to estimate, the force of the little AK-47 bullet on its free-fall is about 1/12th (on the high side) of one Newton.

To put this into real terms - one Newton is less than 1/4 pound of force. That's not much.

Of course, the last time I took physics was 25 years ago, so I could have this wrong. Anyone good with physics? Am I close?

NO .

Neglecting friction that bullet accelerates at 32ft/s/s air resistance is not going to slow it down much as it's a dense metal in a smooth lump .

Wiki is your friend http://en.wikipedia.org/wiki/Celebratory_gunfire

How does that negate what I said? 9.8 meters per second squared is the same thing as 32 ft per second squared. My calculation was without reference to friction, but friction will only slow it down, if only a little bit.

One Newton of force is 1 kilogram-meter per second squared. Yes? And, the 123 grain bullet masses about 8 grams. So, it's not coming down with a lot of Force, is it?

If I wasn't close, then what's the force calculation you came up with?

[Coito ergo sum]

Can't help you with the mathematics, but there is also such a thing as terminal speed as far as freefall is concerned. The wind resistance would ensure that a freefalling bullet travels not nearly as fast coming back down than when it leaves the muzzle on its way up. Thus the force when it impacts is lower. I imagine that being hit by one of those would be akin to being hit by an 8.5 gram pebble dropped from 20 floors up: It'll sting, but it won't kill.

Well, if fired straight up, the muzzle velocity is only important as a determiner of how high it goes. The Force at any point on the downward fall is calculated F=MA - Force = Mass x Acceleration.

Acceleration is 9.8 meters per second squared, or as Feck pointed out 32 feet per second squared.
Mass of the bullet is 8 grams.
So, the force is easy to calculate, I would think.

We can ignore friction because whatever value of F we calculate will be slightly higher than the real F because friction hinders the bullet's fall.

But, the Force of a bullet hitting a person on the ground after being fired up in the air has to be the same as the force of a bullet dropped from the highest point the bullet goes. That's just basic Newtonian physics.

[Feck]

you are calculating the force on the bullet that causes it's acceleration ,this it not the force it delivers on impact .

depending on the calibre bullets tend to hit with about 10% the force they went up with ... I think a 150gr bullet reaches about 300fps that's not fast compared to rifle velocities BUT That's likely to be a head injury and at an energy of about 30 ft/lbs , The legal limit for an air rifle in the UK is 12 ft/lbs at those energies I've put holes through deer skulls . If a rifle bullet his the top of your head at 300fps I'm almost sure you would be dead or at least fucked up .

read the wiki article and the references

[JOZeldenrust]

Mythbusters did a story on this. Basically, it doesn't happen, because by then the round is tumbling too much to be going very fast.

It's not really the tumbling, it's that if you fire the bullet straight up, then the force carries the bullet up until gravity stops it. For a brief millisecond, it's in equilibrium at the top, and then its descent is governed by the calculation F=MA - Force = Mass x Accelleration. As I recall from physics the accelleration due to gravity is 9.8 meters per second squared, and googling the mass of an AK-47 bullet I get 8 grams for a round number on the heavy side. So, that's F=8 grams x 9.8 m/s2. So, the force of the bullet fired straight up on its descent is 78.4 gm/s2. Now, one Newton of Force is, as I recall, 1 kg m/s2, so to estimate, the force of the little AK-47 bullet on its free-fall is about 1/12th (on the high side) of one Newton.

To put this into real terms - one Newton is less than 1/4 pound of force. That's not much.

Of course, the last time I took physics was 25 years ago, so I could have this wrong. Anyone good with physics? Am I close?

NO .

Neglecting friction that bullet accelerates at 32ft/s/s air resistance is not going to slow it down much as it's a dense metal in a smooth lump .

Wiki is your friend http://en.wikipedia.org/wiki/Celebratory_gunfire

Coito, your math/physics is rubbish. The force isn't all that interesting: it's more about kinetic energy. Force and contact surface determine penetrating power. Kinetic energy determines the damage done. Also, you're calculating the force exerted on the bullit by gravity, not the force exerted by the bullit on a body when it decelerates upon hitting the target. For that, you'd need the actual velocity of the bullit and how long it takes for the bullit to stop. IOW, you fail Newtonian physics forever.

Let's do this properly, it's not that hard. An object moving against the direction of gravitational acceleration will follow a parabolic trajectory, and ignoring friction, the speed of return is the same as the speed of departure. That's because all of the initial kinetic energy (1/2 mass x velocity2) is converted into a height potential (mass x gravitational constant x height). Whn the projectile hits the ground again, the height potential is once again converted into kinetic energy. An AK-47 has a muzzle speed of 710 m/s, with a projectile weighing 7.8 g, so the kinetic energy of an AK-47 round is 1/2 x 0.0078 kg x 504100 m2s-2 = 1966 J (1 Joule is 1 kgm2s-2, so dimensions check out). Transferring about 2000 J to a human body in a fraction of a second is enough to be lethal (an AK-47 wouldn't be very effective if it wasn't). An AK-47 round sometimes passes through the body, sometimes it stays inside the body. That means a round generally travels about 0.2 m inside a body. The terminal speed insode the body is 0 m/s, so the average speed is 355 m/s, so it takes about 0.0006 seconds for the round to stop. That means the transfer of energy happens at a power of about 3.5 MW. That's far from gentle.

However, with these kinds of velocities, air resistence isn't negligible. That's why an AK-47 has an effective range of about 400 m. Shooting straight up, the bullit will reach a terminal velocity. As Mythbusters showed, this velocity isn't high enough to kill a person reliably. But it's next to impossible to actually shoot straight up. A bullit fired up at an angle will follow a ballistic trajectory, keeping much of its kinetic energy. It will endanger people in a radius quite a bit over 400 m around the shooter, probably more like 1000 m. People standing really close to the shooter are quite safe, though. The round is almost guaranteed to travel more than a few dozen metres horizontally. So celebratory shooting in a compact crowd surrounded by dessert is relatively safe. In an urban environment, however, there are quite a few people in danger.

[Hermit]

Can't help you with the mathematics, but there is also such a thing as terminal speed as far as freefall is concerned. The wind resistance would ensure that a freefalling bullet travels not nearly as fast coming back down than when it leaves the muzzle on its way up. Thus the force when it impacts is lower. I imagine that being hit by one of those would be akin to being hit by an 8.5 gram pebble dropped from 20 floors up: It'll sting, but it won't kill.

Well, if fired straight up, the muzzle velocity is only important as a determiner of how high it goes. The Force at any point on the downward fall is calculated F=MA - Force = Mass x Acceleration.

Acceleration is 9.8 meters per second squared, or as Feck pointed out 32 feet per second squared.
Mass of the bullet is 8 grams.
So, the force is easy to calculate, I would think.

We can ignore friction because whatever value of F we calculate will be slightly higher than the real F because friction hinders the bullet's fall.

But, the Force of a bullet hitting a person on the ground after being fired up in the air has to be the same as the force of a bullet dropped from the highest point the bullet goes. That's just basic Newtonian physics.

I think you may have lost the point of this tangent, which was Egbert's inquiry concerning how many people may have been killed by bullets that have been short more or less straight up in the air. Sisifo provided anecdotal evidence that they are quite dangerous, and Feck provided a link indicating that about 30% of people who get hit by one of those projectiles finish up dead. Case closed.

Back to Libya: If the resistance fighters finish up winning, I predict factional fighting to the extent that the majority of Libyans will regret the removal of their dictator.

[JOZeldenrust]

Mythbusters did a story on this. Basically, it doesn't happen, because by then the round is tumbling too much to be going very fast.

It's not really the tumbling, it's that if you fire the bullet straight up, then the force carries the bullet up until gravity stops it. For a brief millisecond, it's in equilibrium at the top, and then its descent is governed by the calculation F=MA - Force = Mass x Accelleration. As I recall from physics the accelleration due to gravity is 9.8 meters per second squared, and googling the mass of an AK-47 bullet I get 8 grams for a round number on the heavy side. So, that's F=8 grams x 9.8 m/s2. So, the force of the bullet fired straight up on its descent is 78.4 gm/s2. Now, one Newton of Force is, as I recall, 1 kg m/s2, so to estimate, the force of the little AK-47 bullet on its free-fall is about 1/12th (on the high side) of one Newton.

To put this into real terms - one Newton is less than 1/4 pound of force. That's not much.

Of course, the last time I took physics was 25 years ago, so I could have this wrong. Anyone good with physics? Am I close?

NO .

Neglecting friction that bullet accelerates at 32ft/s/s air resistance is not going to slow it down much as it's a dense metal in a smooth lump .

Wiki is your friend http://en.wikipedia.org/wiki/Celebratory_gunfire

Coito, your math/physics is rubbish. The force isn't all that interesting: it's more about kinetic energy. Force and contact surface determine penetrating power. Kinetic energy determines the damage done. Also, you're calculating the force exerted on the bullit by gravity, not the force exerted by the bullit on a body when it decelerates upon hitting the target. For that, you'd need the actual velocity of the bullit and how long it takes for the bullit to stop. IOW, you fail Newtonian physics forever.

Let's do this properly, it's not that hard. An object moving against the direction of gravitational acceleration will follow a parabolic trajectory, and ignoring friction, the speed of return is the same as the speed of departure. That's because all of the initial kinetic energy (1/2 mass x velocity2) is converted into a height potential (mass x gravitational constant x height). Whn the projectile hits the ground again, the height potential is once again converted into kinetic energy. An AK-47 has a muzzle speed of 710 m/s, with a projectile weighing 7.8 g, so the kinetic energy of an AK-47 round is 1/2 x 0.0078 kg x 504100 m2s-2 = 1966 J (1 Joule is 1 kgm2s-2, so dimensions check out). Transferring about 2000 J to a human body in a fraction of a second is enough to be lethal (an AK-47 wouldn't be very effective if it wasn't). An AK-47 round sometimes passes through the body, sometimes it stays inside the body. That means a round generally travels about 0.2 m inside a body. The terminal speed insode the body is 0 m/s, so the average speed is 355 m/s, so it takes about 0.0006 seconds for the round to stop. That means the transfer of energy happens at a power of about 3.5 MW. That's far from gentle.

However, with these kinds of velocities, air resistence isn't negligible. That's why an AK-47 has an effective range of about 400 m. Shooting straight up, the bullit will reach a terminal velocity. As Mythbusters showed, this velocity isn't high enough to kill a person reliably. But it's next to impossible to actually shoot straight up. A bullit fired up at an angle will follow a ballistic trajectory, keeping much of its kinetic energy. It will endanger people in a radius quite a bit over 400 m around the shooter, probably more like 1000 m. People standing really close to the shooter are quite safe, though. The round is almost guaranteed to travel more than a few dozen metres horizontally. So celebratory shooting in a compact crowd surrounded by dessert is relatively safe. In an urban environment, however, there are quite a few people in danger.

By the way the force exerted by the bullit on the body is about 0.0078 kg x (710 m/s / 0.0006 s) = 0.0078 x 1183333 ms-2 = 9230 N, or a little over 2000 lbs.

[Coito ergo sum]

you are calculating the force on the bullet that causes it's acceleration ,this it not the force it delivers on impact .

depending on the calibre bullets tend to hit with about 10% the force they went up with ... I think a 150gr bullet reaches about 300fps that's not fast compared to rifle velocities BUT That's likely to be a head injury and at an energy of about 30 ft/lbs , The legal limit for an air rifle in the UK is 12 ft/lbs at those energies I've put holes through deer skulls . If a rifle bullet his the top of your head at 300fps I'm almost sure you would be dead or at least fucked up .

read the wiki article and the references

So, what formula do you use to calculate the force of the falling bullet?