Libya: Ballistics bullet derail

[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.

[Coito ergo sum]

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.

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.

[Gawd]

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.

Coito, what did you major in? Because it sure as hell ain't science.

[Coito ergo sum]

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.

Coito, what did you major in? Because it sure as hell ain't science.

Funny that someone like you who knows so little can convince himself that he actually knows something.

I'm still waiting for you to post the correct formula to calculate the force of the bullet shot straight up in the air. Naturally, you're all hat and no cattle, as usual.

[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.

[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?

[colubridae]

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.

[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.

[sandinista]

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.

Coito, what did you major in? Because it sure as hell ain't science.

Funny that someone like you who knows so little can convince himself that he actually knows something.

Now That it funny!

[Coito ergo sum]

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.

I'll let you present the formulas, if you can.

Since I actually took college level physics, I'm aware of inelastic (not "non"-elastic) vs. elastic collisions. The force of the falling bullet (which was fired straight up) is measured by multiplying its mass times its acceleration due to gravity. Are you suggesting that is wrong? Really?

[Coito ergo sum]

Now That it funny!

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

[Gallstones]

Soooooo.........we aren't talking about Libya anymore?

[JOZeldenrust]

Soooooo.........we aren't talking about Libya anymore?

Nope, we're talking about Newtonian mechanics. How about a split to Science and Technology?

[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.

I'll let you present the formulas, if you can.

Since I actually took college level physics, I'm aware of inelastic (not "non"-elastic) vs. elastic collisions. The force of the falling bullet (which was fired straight up) is measured by multiplying its mass times its acceleration due to gravity. Are you suggesting that is wrong? Really?

Yeah, you "took college level physics". Doesn't mean you understood any of it.....

[Coito ergo sum]

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.

I'll let you present the formulas, if you can.

Since I actually took college level physics, I'm aware of inelastic (not "non"-elastic) vs. elastic collisions. The force of the falling bullet (which was fired straight up) is measured by multiplying its mass times its acceleration due to gravity. Are you suggesting that is wrong? Really?

Yeah, you "took college level physics". Doesn't mean you understood any of it.....

Yawn - still waiting. The Joos must've stolen your physics.