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This is intended for people interested in the subject of military guns and their ammunition, with emphasis on automatic weapons, particularly in larger calibres (12.7+mm).

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Energy vs. Momentum   General Military Discussion

Started 5-Aug by JesseH1234; 3286 views.
RovingPedant

From: RovingPedant

7-Aug

 

JesseH1234 said...

I just want to know WTF "1,000 ft-lbs of energy" really means, because it sure doesn't seem to be the ability to move 1,000 pounds a foot. 

1,000 lbf applied over the distance of 1ft.

which is equivalent to 1lbf applied over the distance of 1,000ft

When you are talking about moving 1,000lb 1ft you are missing a great deal about the conditions. In space, with no other forces acting on a 1000lb (mass) object, then any transfer of kinetic energy, 1ft-lbf or 1000ft-lbf, will result in a 1000lb mass travelling an infinite distance, provide infinite time is available.

On the surface of the earth, you’ve got friction, air resistance and gravity acting against you.

For transferring velocity from one item to another, the properties that are important are momentum and the coefficient of restitution.

 

EmericD

From: EmericD

7-Aug

Since the 19th century, it is known that there is a relationship between the kinetic energy of shell and the volume of the cavity made in the soil (Elie's formula).

The same relationship was found between kinetic energy of a bullet and the volume of the cavity in ballistic gelatin.

At a time, the US Army used an interesting criteria which was M.V^(3/2) [so, midway between momentum and kinetic energy]. I think that the idea was that penetration (for non-deforming bullets) is a function of M.V^(1/2) / S [M being the weight, V the velocity and S the frontal area] and material displacement is proportional to S.V (big frontal area and high speed means more instantaneous displaced materials during impact).

So, if you "integrate" the instantaneous volume displaced by the bullet along it's path, you end up with something proportional to M.V^(3/2).

With the exception of some gun-writer's theories that tried to make the .45 ACP looks better compared with the 9x19 mm, I don't know how you could correlates momentum with "power".

QuintusO

From: QuintusO

7-Aug

Maybe you should work on comprehending what you wrote, first:

unless one is designing a round specifically for those first two instances, the projectile's momentum is a much more useful measure of the relative "power" of the round.

Given the context of the conversation, we can assume you did not mean the physical "power" of the round as power is a different quantity than momentum and I try not to assume that my dancing partners are idiots, without proof. So you then meant some kind of other figurative "power", and you certainly didn't mean "market power" or "religious" power, so we can assume you meant something like "stopping power", because there simply isn't any other relevant option (perhaps "penetrating power", but you never mention it in reference to penetration specifically).

It isn't my fault that your initial point is so half-baked and nonsensical that it's barely readable, and I would make damn sure that I was communicating my ideas properly (which you're not) before I went slinging around "muh readin comprenshun" at people.

  • Edited 07 August 2020 11:58  by  QuintusO
QuintusO

From: QuintusO

7-Aug

And none of that changes the fact that you're extremely scientifically illiterate, and you need to read the Wikipedia articles on all these physical phenomena and then do a bunch of math problems as practice, at the very least, before you try talking about this stuff.

  • Edited 07 August 2020 8:42  by  QuintusO
RovingPedant

From: RovingPedant

7-Aug

EmericD said...

 

With the exception of some gun-writer's theories that tried to make the .45 ACP looks better compared with the 9x19 mm, I don't know how you could correlates momentum with "power".

As anyone with an understanding of physics knows, Power is energy over time.

In this instance I think that the OP is looking for a measure of the ability to knock over metal plates, which is why he favours momentum as the term to maximise.

Following this methodology, however, one would have to conclude that the most effective projectile would be a dense rubber that can rebound from the target. This would result in maximum target velocity, but would not cause much in the way of damage. Probably why rubber bullets are favoured for less lethal applications.

 

 

 

stancrist

From: stancrist

7-Aug

JesseH1234 said:

...unless one is designing a round specifically for those first two instances, the projectile's momentum is a much more useful measure of the relative "power" of the round.

I'm really curious if someone can point out where my thinking is technically flawed...

IMO, your biggest flaw is trying to oversimplify the matter by comparing military cartridges on the basis of relative power, regardless of how "power" is defined.

There are so many factors that are much more important than either momentum or energy, in determining bullet performance against soft and/or hard targets.

Mustrakrakis

From: Mustrakrakis

7-Aug

JesseH1234 said:

I went down this road in my head asking myself why 5.7 hasn't blown all other autopistol calibers out of the water; I ascribed it to just the archaic nature of the gun world, but when I saw someone hitting plates with a 5.7 and then a 9mm a big ol question mark appeared. 

If the question is, "Why hasn't 5.7 blown all other autopistol calibers out of the water," we can certainly answer that.  However, it will consist of several answers (as the concerns of citizens and armies don't always overlap), and while they don't necessarily have to be painfully long and dry, they likely will wind up that way.  Additionally, several of us will argue bitterly over the significance of largely unimportant minutiae while agreeing on the overall conclusions.  It's not good or bad; it's just how we do things here.  However, since I'm willing to be the change that I want to see, I'll take a swing at this with two short answers.

Short answer: (Civilian) - Too expensive/Meme round/Fuck FN/Grip's too big/USPSA doesn't allow it/I tried reloading once-fired brass and my gun jammed/I tried reloading once-fired brass and my gun exploded/iTs JuSt A .22WMR u NuB. 

Short answer: (Military) - Given that the service pistol's impact on an army's overall combat effectiveness lands somewhere between the color of their dress uniforms and the brand of boots that they're wearing, the correct pistol to issue is whatever chambers the cartridge that's already in your inventory and costs the least while meeting generally arbitrary standards...in other words, probably a Glock.  (If the selection process is rigged, compromised, or otherwise biased, it will be something other than a Glock but it will work just as well as one, assuming that it's drop-safe from the factory and won't go off while holstered.  Not every manufacturer gets that last part right unfortunately.)

In reply toRe: msg 27
nincomp

From: nincomp

7-Aug

Since part of this thread is about the observed performance of different cartridges on knocking down steel plates, how much does the construction of the bullet affect the amount of energy transferred to a hard surface such as a steel plate?  For example, a slow-traveling lead projectile may merely deform whereas one traveling at a much higher speed will disintegrate into many fragments that travel at high speed laterally after impact.  Also, something like a solid brass or steel-cored projectile will tend to bounce. 

JesseH1234

From: JesseH1234

7-Aug

Emeric: "So, if you "integrate" the instantaneous volume displaced by the bullet along it's path, you end up with something proportional to M.V^(3/2)."

See, I knew there would be a good answer out there, and it would be complex.  Thanks!  I didn't actually expect momentum to be the only thing that matters (hence "working theory").

I think I get it now, the "kinetic energy" is the mathematical sum total of all the different energetic effects we see, such as friction, heat, vibration, sound, shockwaves, projectile deformation, material displacement, etc., which explains why it doesn't have a vector; it's definitely not "1,000lbs ke moves 1,000lbs 1 foot). Momentum is the directional motion through air and material.  You'll always have both but the ratio will determine the effects you see.

QuintusO

From: QuintusO

8-Aug

JesseH1234 said:

I think I get it now, the "kinetic energy" is the mathematical sum total of all the different energetic effects we see, such as friction, heat, vibration, sound, shockwaves, projectile deformation, material displacement, etc.,

No. You could just look up what these things mean, but it's obvious that you're not interested in doing that.

JesseH1234 said:

Momentum is the directional motion through air and material.  You'll always have both but the ratio will determine the effects you see.

No. This isn't that complex. Energy is the capacity to do work. To penetrate and kill a target you need to do work, which means you need to have energy. That's why energy and work use the same unit (Joule, or ft-lb). Momentum is the product of mass and velocity, and has a direct relationship with force. If you actually DO THE MATH on kinetic systems, you find these things actually shake out! A Newtonian collision results in a change in both momentum and energy, and work done. This is not hard to understand, you only need to put in the effort.

JesseH1234 said:

which explains why it doesn't have a vector; it's definitely not "1,000lbs ke moves 1,000lbs 1 foot).

Multiple people have explained to you that it's not this, you're just not listening to them.

 

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