Libya: Ballistics bullet derail

[Gawd]

Use terminal velocity, Coito. You can then calculate the kinetic energy of the bullet falling down. Divide that by the cross sectional area of the bullet to get force. Your lack of understanding of basic high school physics shocks me.

This is so wrong. You don't use terminal velocity, because the bullits generally aren't in free fall, but rather in a ballistic trajectory. Also, kinetic energy divided by the area of the cross section doesn't give you a measure of the force: dimensions don't check out: kinetic energy is kgm2s-2. Area is m2. That would mean the division you want to use would yield kgs-2, but force is kgms-2.

Here's how you do calculate the force of impact:

You take the velocity of the bullit at the point of impact, v0, let's say half the muzzle velocity of the bullit, so for an AK-47 that would be 355 ms-1.
You take the end velocity of the bullit, vend, which is often 0 ms-1, as quite often the bullit stays inside the body. That gives a change in velocity dv of 355 ms-1.
You take the distance traveled by the bullit during impact, ds, about 0.2 m, as that's about as far as a bullit in the centre body mass can travel before exiting on the other side of the body.
Assuming that the acceleration of the bullit inside the body is linear, the average speed vav of the bullit is 1/2 v0, so 177.5 ms-1.
With that average speed, you can calculate the duration of the impact, dt, as ds/vav = 0.2 m / 177.5 -1 = 0.0011 s.
So the bullit accelerates from 355 ms-1 to 0 ms-1 in 0.0011 s. That's an acceleration a of dv/dt = 355 ms-1 / 0.0011 s = 322727 ms-2.
Now you use "force equals mass times acceleration" or f = ma. The mass of a bullit from an AK-47 is 7.8 grams, so 0.0078 kg, so the force is 0.0078 kg x 322727 ms-2 = 2517 N, or 566 lbs.

You could divide this figure by the surface area of the impact, giving the pressure applied by the bullit. Let's say the area of impact is about 1 cm2, or 0.0001 m2. That would give a pressure of about 25170000 Nm-2, or roughly 250 times atmospheric pressure. That's enough to penetrate a body.

Seriously, this is all basic high school physics. Didn't you guys pay attention in class?

As I said above, multiply it by the physical length of the acceleration phase towards the ground and you will get units of force. It's not rocket science. And your attempt at treating the problem as an elastic collision without a suitable model of the physical bullet or the physical body is absurd. There was a reason that high school teachers didn't make the mistake you are making because they know that you need computational calculus for such scenarios.

[Gawd]

Coito, you couldn't calculate your way out of a straight line. I see you've been talking to colubridae. And yet you are still wrong. Simply follow my steps and multiply by the length of the acceleration phase towards the ground. The quantity you get is the force of the bullet in flight subject to a non-elastic collision. I do hope to your high school teachers that you know what I just said.

This is a blatant lie. Coito has not had any contact with me.

I explained to you why your physics was wrong and even offered you the opportunity to be a man and admit your mistake.
Predictably you failed to do the honourable thing and then post a conspicuous lie. You bring shame and dishonour on yourself.


How much of all your other posting is shameful lies?

allah will not look favourably upon you.

You can add that to your notes, Doctor Freud.

[JOZeldenrust]

Use terminal velocity, Coito. You can then calculate the kinetic energy of the bullet falling down. Divide that by the cross sectional area of the bullet to get force. Your lack of understanding of basic high school physics shocks me.

This is so wrong. You don't use terminal velocity, because the bullits generally aren't in free fall, but rather in a ballistic trajectory. Also, kinetic energy divided by the area of the cross section doesn't give you a measure of the force: dimensions don't check out: kinetic energy is kgm2s-2. Area is m2. That would mean the division you want to use would yield kgs-2, but force is kgms-2.

Here's how you do calculate the force of impact:

You take the velocity of the bullit at the point of impact, v0, let's say half the muzzle velocity of the bullit, so for an AK-47 that would be 355 ms-1.
You take the end velocity of the bullit, vend, which is often 0 ms-1, as quite often the bullit stays inside the body. That gives a change in velocity dv of 355 ms-1.
You take the distance traveled by the bullit during impact, ds, about 0.2 m, as that's about as far as a bullit in the centre body mass can travel before exiting on the other side of the body.
Assuming that the acceleration of the bullit inside the body is linear, the average speed vav of the bullit is 1/2 v0, so 177.5 ms-1.
With that average speed, you can calculate the duration of the impact, dt, as ds/vav = 0.2 m / 177.5 -1 = 0.0011 s.
So the bullit accelerates from 355 ms-1 to 0 ms-1 in 0.0011 s. That's an acceleration a of dv/dt = 355 ms-1 / 0.0011 s = 322727 ms-2.
Now you use "force equals mass times acceleration" or f = ma. The mass of a bullit from an AK-47 is 7.8 grams, so 0.0078 kg, so the force is 0.0078 kg x 322727 ms-2 = 2517 N, or 566 lbs.

You could divide this figure by the surface area of the impact, giving the pressure applied by the bullit. Let's say the area of impact is about 1 cm2, or 0.0001 m2. That would give a pressure of about 25170000 Nm-2, or roughly 250 times atmospheric pressure. That's enough to penetrate a body.

Seriously, this is all basic high school physics. Didn't you guys pay attention in class?

As I said above, multiply it by the physical length of the acceleration phase towards the ground and you will get units of force. It's not rocket science. And your attempt at treating the problem as an elastic collision without a suitable model of the physical bullet or the physical body is absurd. There was a reason that high school teachers didn't make the mistake you are making because they know that you need computational calculus for such scenarios.

Hahaha, no.

My approach has a few imperfections, sure. It doesn't factor in the heterogenous composition of a human body, i.e. if the bullit impacts hard tissue the force exerted will be much greater, and if it only hits soft tissue the force will be quite a bit smaller. It also doesn't factor in the deformation of the bullit, including the possibility that the bullit fractures at some point, which is a very real possibility. However, as an approximation, my approach is completely correct. A bullit impact is an elastic collision. If it wasn't, firearms would make very ineffective weapons.

Your criticism of my method is nonsensical. Exactly what mistake did I make here?

[sandinista]

Now That it funny!

You honestly think Gawd knows what he's talking about here?

No, just that you don't.

[JOZeldenrust]

Because F=MA is not Newtonian physics and the 9.8 m/s2 is not the acceleration due to gravity. I already told you you were right.

It is Newtonian physics, and 9.8 ms-2 is the acceleration due to gravity, but it's not the relevant acceleration in a bullit impact. The relevant acceleration is the slowing down of the bullit once it hits a body, which is much greater then 9.8 ms-2.

You're actually suggesting that acceleration (A) increases when a falling body hits another body beneath it? Or, do you want to rephrase that? I imagine you must not have meant what you typed.

The bullit accelerates in a direction opposite to the direction it's moving in. That's what deceleration is. For the purpose of calculating the force of impact, the direction of acceleration is irrelevant.

And, of course it's the "relevant acceleration" -- that's how fast a body's speed increase while it's falling down, bullets included. So, if you drop a bullet from 10,000 feet, it will start at 0 meters per second, and then it will increase in speed at about 9.8 meters per second per second as it falls (until it reaches terminal velocity). The force it has when it hits the ground is equal to the mass of the bullet times the acceleration at that that time.

No, that's the force the earths gravitational pull is exerting on the bullit, or more precisely the net force of the earths gravitational pull and the air resistence. At terminal velocity, that net force is 0 N.

The bullit doesn't "have force", it has velocity. That velocity changes because the body and the bullit exert force on one another. That force os the force of impact, causing the bulliit to decelerate. In my previous post I've explained how to calculate the force of impact.

You ridicule a perfectly correct statement, Coito. Have some intellectual integrity, and admit you screwed up basic high school physics.

Use terminal velocity, Coito. You can then calculate the kinetic energy of the bullet falling down. Divide that by the cross sectional area of the bullet to get force. Your lack of understanding of basic high school physics shocks me.

This is so wrong. You don't use terminal velocity, because the bullits generally aren't in free fall, but rather in a ballistic trajectory. Also, kinetic energy divided by the area of the cross section doesn't give you a measure of the force: dimensions don't check out: kinetic energy is kgm2s-2. Area is m2. That would mean the division you want to use would yield kgs-2, but force is kgms-2.

Here's how you do calculate the force of impact:

You take the velocity of the bullit at the point of impact, v0, let's say half the muzzle velocity of the bullit, so for an AK-47 that would be 355 ms-1.
You take the end velocity of the bullit, vend, which is often 0 ms-1, as quite often the bullit stays inside the body. That gives a change in velocity dv of 355 ms-1.
You take the distance traveled by the bullit during impact, ds, about 0.2 m, as that's about as far as a bullit in the centre body mass can travel before exiting on the other side of the body.
Assuming that the acceleration of the bullit inside the body is linear, the average speed vav of the bullit is 1/2 v0, so 177.5 ms-1.
With that average speed, you can calculate the duration of the impact, dt, as ds/vav = 0.2 m / 177.5 -1 = 0.0011 s.
So the bullit accelerates from 355 ms-1 to 0 ms-1 in 0.0011 s. That's an acceleration a of dv/dt = 355 ms-1 / 0.0011 s = 322727 ms-2.
Now you use "force equals mass times acceleration" or f = ma. The mass of a bullit from an AK-47 is 7.8 grams, so 0.0078 kg, so the force is 0.0078 kg x 322727 ms-2 = 2517 N, or 566 lbs.

You could divide this figure by the surface area of the impact, giving the pressure applied by the bullit. Let's say the area of impact is about 1 cm2, or 0.0001 m2. That would give a pressure of about 25170000 Nm-2, or roughly 250 times atmospheric pressure. That's enough to penetrate a body.

Seriously, this is all basic high school physics. Didn't you guys pay attention in class?

If you don't get it yet, I'm quite willing to explain in mre detail or even smaller steps. However, you claim to have taken college level physics classes, and your misunderstanding of the concept of accelleration just doesn't jive.

You ridiculed me for being right. I think you owe me, and yourself if you wish to claim any intellectual integrity, an explanation.

[Gallstones]

I heard, that if you drop a penny off the top of the Empire State building and it hits someone on the ground, it will go right through them and they will be dead.

[JOZeldenrust]

I heard, that if you drop a penny off the top of the Empire State building and it hits someone on the ground, it will go right through them and they will be dead.

Which is bullshit. If the ESB was in a vacuum it would likely be true, but in reality a penny would hardly have enough velocity to break the skin.

[JOZeldenrust]

Quote from here: http://rationalia.com/forum/viewtopic.p ... 14#p810714

By the way, Coito, when are you going to admit you were talking out of your arse on the subject of ballistics?

http://rationalia.com/forum/viewtopic.p ... 75#p810675

The thing is, I wasn't "talking out of my arse." http://rationalia.com/forum/viewtopic.p ... 20#p802628 Moreover, in the very post where I brought up the issue I invited folks with a greater knowledge of physics than myself to clarify the issue. I wrote this:

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?

http://rationalia.com/forum/viewtopic.p ... 55#p802392 You're correct, as far as I can tell, as to the physics.

But, I am also correct insofar as a bullet fired straight up will not hit a person with sufficient force to kill them (that much was proved/confirmed in the Mythbuster's episode). I was very clear in my explanation that once you veer off the vertical and enter a non-vertical ballistics trajectory. This is exactly what I typed:

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.

http://rationalia.com/forum/viewtopic.p ... 55#p802383

I will also point out that I never argued the point with you, or accused you of being wrong about the physics.

I was immediately attacked by several people, not content with discussing the issue, they had to hurl rhetorical barbs and personal attacks. I was happy to discuss the physics, but most folks weren't doing that (at least not without complementing their posts with enough digs and jabs to get their jollies).

Also, in the post you linked to above, you accused me of "ridiculing" you. I did not at any point ridicule you (or anyone else).

I have no qualms about the suggestion that my math was wrong. I said as much in my very first post on the topic. What I'm not going to play ball with, though is the personal nonsense. That's why I lost interest in the thread.

The rhetoric devices you used in this passage:

You're actually suggesting that acceleration (A) increases when a falling body hits another body beneath it? Or, do you want to rephrase that? I imagine you must not have meant what you typed.

can only reasonably be interpreted as ridicule. The rhetorical question of the form "Are you actually suggesting..." hinges on an implied ad hominem through a false dichotomy: either I don't actually mean what I wrote, or I do, which makes me an idiot for meaning it.

The propositional content of the message is "You're an idiot, because either you think P or you think -P but you wrote P, which makes you incredibly inarticulate".

The dichotomy is false, however. There is the possibility that what I wrote is what I meant to write, and completely correct, as was the case here.

I think this qualifies as ridicule. If you hadn't wanted to imply the ad hominem, you could've rephrased:

Are you sure that acceleration (A) increases when a falling body hits another body beneath it? I don't see how it could. Maybe you could rephrase that to avoid misunderstanding? Am I missing something here, or did you maybe mean something else then you wrote?

Also, you missed an essential bit from the Mythbusters item: we're discussing celebratory gunfire, and those shots are never fired straight up. Firing straight up requires a specially constructed rig to hold the weapon, as well as near windless conditions. The experiments by the Mythbusters aren't relevant for these real life examples.