is it true, or just myth?

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I won't disagree with anything you wrote, but maybe I didn't ask my question properly.


Would exploding gunpowder in Situation A be the same as the energy released in Situation B.

Situation A is a normal shooting condition, and Situation B is Tark's test.

I'm not asking what the energy being released does or does not do - just the total amount of energy being released.

(They may not be the same, because the way the powder burns in each is very different.)
 
That "energy" has to come from the exploding gunpowder. Would exploding gunpowder accelerating the bullet down the barrel be the same as the energy released in Tark's test?

My point is that there's more to these things than just "energy".

And to answer your question: no. The energy released from the exploding gunpowder accelerating a bullet down the barrel is NOT the same as the energy released in tark's test.

Don't get me wrong...the TOTAL energy of the SYSTEM is conserved. But energy comes in many forms and is converted from one to another. A moving bullet is taking energy away from the chemical reaction in the form kinetic energy. Some of that kinetic energy is, in turn, converted to heat energy by friction between the bullet and the barrel. Some of the energy from the chemical reaction of the burning powder is converted directly to heat, some of which is transferred to the metal of the brass and firing chamber. Some is converted to kinetic energy in the form of gasses which leak out around the brass and firing chamber walls.

There are two basic types of energy: potential and kinetic. Potential energy is stored energy. Kinetic energy is inherent in moving objects.

Energy comes in many forms and it can accomplish different things depending on the type of energy you have...and how it's converted into some useful form for a given task. Stored energy in chemicals, like gun powder for example, only becomes useful when it's released to create heat and pressure. That heat and pressure in turn only becomes useful from the standpoint of a firearm when it's converted to kinetic energy. Kinetic energy, however, only lies in objects that are moving...no motion means no kinetic energy.

Once a bullet starts moving, it has kinetic energy...which means the bullet is taking SOME energy AWAY from the energies that were released when the chemical reaction took place.

Once a mass is in motion is also has a quality called "momentum". Momentum is not energy. Momentum has a vector, kinetic energy does not. Momentum is always conserved...kinetic energy is not.

In a collision, momentum is always conserved , whether the collision is elastic or inelastic. It's a straight forward application of mass times velocity.

Kinetic energy, however, is only conserved in elastic collisions. In inelastic collisions, it gets converted into other forms...such as heat, sound, deformation of objects, etc.

When the bullet is held in place to prevent any relative motion between it and the rest of the gun, then there is no transfer of kinetic energy down the barrel. The bullet also develops no momentum which can be transferred down the barrel. With no relative linear motion imparted to the bullet, there is also no "recoil" (conservation of momentum).

This means that the energy released from the chemical reaction of the burning powder is felt solely on the immobile portions of the firing chamber region. If the pressure generated by this release of stored energy is not sufficient to overcome the fracture toughness of these components, then those components will not "break".

So the pressure has no choice but to be either contained, leaked out, or converted to other forms of energy, like heat, which can be absorbed by the surrounding material.


Does this make sense?


If not, think of it this way:

A gas cylinder can be thought of as a firing chamber of a gun in which the bullet is not allowed to move, if you simplify it.

If you force more gas into the cylinder, what happens? The pressure inside the cylinder goes up. It also gets hotter. So long as the pressure does not exceed the fracture toughness of the cylinder, the cylinder will not break. The energy used to force the gases in is converted to potential energy (static pressure) and heat. The excess heat is eventually transferred to the cylinder walls, where it radiates out until the cylinder comes to thermal equilibrium with the rest of the environment.

If the cylinder develops a leak, some of the gasses slowly leak out...converting potential energy to kinetic energy in the process. The lowering pressure also transfers heat away from the cylinder, and it cools down some.


Now, back to the firing chamber with the immobilized bullet:

Potential energy is converted to heat and pressure by the chemical reaction. If that pressure does not exceed the fracture toughness of the material, then nothing breaks. The heat gets transferred to the surrounding structural materials and the gases leak out, lowering pressure and removing more heat.
 
My mind is baffled by all of this but I am fascinated.

Thanks for the test Tark!

Now somebody needs to try this in a 9mm or 38 super 1911 and see if the results are similar so that we can prove if its just the low operating pressure of the 45acp that allows this phenomenon or the 1911 design itself.
 
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Is the TOTAL amount of energy released by the ignition of the gunpowder in normally firing the gun equal to the TOTAL amount of energy released in Tark's test firing?
A given amount of powder only has so much energy. Given that ALL the powder was consumed, then ALL it's energy was released. So the answer is, "yes."
 
.....Given that ALL the powder was consumed, then ALL it's energy was released.....

Yes, but that's also a question - how do we know that ALL the powder was equally consumed? In Tark's test, the pressure would have been much higher, which may or may not have influenced the energy from the powder.

I don't think we should make that assumption unless/until we know for sure. Presumably Tark still has the burnt powder, that can be examined...
 
From the web page I just linked to:

From a thermodynamic point of view, a firearm is a special type of piston engine, or in general heat engine where the bullet has a function of a piston. The energy conversion efficiency of a firearm strongly depends on its construction, especially on its caliber and barrel length. However, for illustration, here is the energy balance of a typical small firearm for .300 Hawk ammunition:[1]

  • Barrel friction 2%
  • Projectile motion 32%
  • Hot gases 34%
  • Barrel heat 30%
  • Unburned propellant 1%.
which is comparable with a typical piston engine.​


In Tark's test, this would change as follows:

  • Barrel friction zero %
  • Projectile motion zero %
  • Hot gases ? %
  • Barrel heat ? %
  • Unburned propellant ? %.


Total for the first scenario adds up to 99%
Total for the second scenario needs to be 99%
That means Hot gasses, Barrel heat, and unburned propellent have to increase significantly.

One way for that to happen is for the temperature to go up much higher, rather than hot gasses, and maybe more unburnt propellant.

If that were the case, it could explain Tark's results.
 
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how do we know that ALL the powder was equally consumed?
I don't understand what you mean by "equally consumed." But there was nothing left in the cartridge case, which means all the powder was consumed -- it burned, changed to gas, leaving no unburned grains in the case.
 
What I meant by "equally consumed" was that the powder was equally burnt, perhaps leaving residue - I don't know.

Tark's exact words were: "There was no powder left in the case."

Maybe the powder burned.
Maybe the powder was vaporized, and as the gas cooled off is now coating various places.

I don't know what I "mean", just that it is an unknown. When the temperature and pressure get so high, things might happen that we haven't yet considered.



Back to reality. Some people thought the gun would "explode". If all that energy being released went only into extremely high temperature and pressure, the gun might very well look just as Tark explained. Apparently the pressure wasn't enough to flatten the primer, if I understand correctly, so as others suggested, the energy might have gone mostly into heat.
 
You might want to try YouTube... There are a MASSIVELY amount of gun torture tests available....
I've seen some insane things done to attempt to kill a Hi Point and it just won't die!
 
I don't know what I "mean", just that it is an unknown. When the temperature and pressure get so high, things might happen that we haven't yet considered.
Remember we are dealing with a cartridge that generates o about a third of the pressure of a .308 Winchester. So even with a plugged barrel, we aren't going where no man has gone before. If the powder was gone, it burned and generated gas.
 
But....the mention of 'burned equally' kind of stuck in my brain. With double-base powders, perhaps there is some secondary form of chemical combination that happens under certain conditions, but might not have been present in the test.

Perhaps there is some threshold where the powder reacts differently? We all know that burning powder in open air has a much different burning rate than when contained, and some powders react in different ways when they start getting to their maximum safe loads. Some will continue to increase velocity at a fairly steady rate, while others become unstable and you can get either velocity spikes or even decreases. When you are adding more powder but the velocity is going back down...you need to stop!:)

I've not experienced these phenomenon in person, but some of the loading manuals described this happening sometimes...so what if when more powder makes the pressure go back down the same thing happend in the test gun? The thing that keeps ringing in my head is that the primer looked normal!! If pressure had spiked...but just not enough to fracture the chamber...wouldn't you think the primer would be flattened or otherwise show signs that it had experienced something out of the ordinary? Perhaps...it didn't.
 
Yes, but that's also a question - how do we know that ALL the powder was equally consumed? In Tark's test, the pressure would have been much higher, which may or may not have influenced the energy from the powder.

I don't think we should make that assumption unless/until we know for sure. Presumably Tark still has the burnt powder, that can be examined...

See post #98.

ALL the powder was consumed. The powder burns with its own oxidizers...what makes you think there's a case nfor partial combustion here?
 
It seems pretty clear that things didn't happen as a number of us EXPECTED them to happen.

I'll accept WHAT occurred. I'd like to know WHY it happened the way it happened.

The question now is WHY that is. There ought to be somone, somewhere that can explain the phenomenon (or -na) described, so that those of us, who were so skeptical, can better understand what happened.
 
If anyone here knows a gunpowder scientist/engineer at a propellant company, they should ask them what happens in this type of situation. A Technical Director or someone of that type title would most likely know.

Anyone here have a non-Customer Service contact that you've spoken with for tech support at any of the powder companies?
 
It seems pretty clear that things didn't happen as a number of us EXPECTED them to happen.

I'll accept WHAT occurred. I'd like to know WHY it happened the way it happened.

The question now is WHY that is. There ought to be somone, somewhere that can explain the phenomenon (or -na) described, so that those of us, who were so skeptical, can better understand what happened.
Use Occam's Razor -- do not unnecessarily create entities. The simplest explanation that takes into account all the observations is the best.

1. All the powder was consumed.
2. The gun was not damaged.
3. The cartridge is question is a low pressure design.

Therefore: The pressure did not exceed the yield limits of the steel.
 
Vern's right. Like I say at work, "let's not over-engineer the solution to this".

Powder burning is a self-contained, chemical reaction. It generates heat and pressure in the process. Confining the pressure with no release just means a momentary higher pressure, which in turn causes heat to increase...which would only help, and not hinder, the chemical reaction already taking place.

The gun, especially the barrel, was not damaged, by any apparent visual indications. This simply means that the pressure generated by the confined burning powder simply did not exceed the fracture toughness of the barrel. No more, no less.
 
Vern Humphrey said:
Use Occam's Razor -- do not unnecessarily create entities. The simplest explanation that takes into account all the observations is the best.

1. All the powder was consumed.
2. The gun was not damaged.
3. The cartridge is question is a low pressure design.

Therefore: The pressure did not exceed the yield limits of the steel.

Occam's Razor is a theoretical model -- a practical rule. It is not an answer. It says simply don't overcomplicate, but it doesn't EXPLAIN anything. And it isn't ALWAYS the best or correct answer. Practically speaking, it's a good rule to use. That the pressure did not exceed the yield limits of the steel is probably correct, but that's an educated guess.

I've been involved in a related discussion with another shooter who has also done this experiment with the same (surprising) results. He has chosen not to get involved in this discussion, but is reading comments here. Like you, this other reader feels that the pressures generated by the powder simply don't exceed the mechanical capabilities of the components. And that is probably the best explanation. Put simply, most guns are more stout than some of us have come to believe.

This other person also, as you did, noted that the .45 round isn't particularly high pressure; but he also cited another shooter who did the same experiment with a 9mm round, which IS higher pressure, and that higher pressure round behaved in the same manner.

I'm neither a ballistician nor a chemist -- and I know at least some of my limitations -- so I'll just watch and read. As I told the offline participant, I expected the primer area of the case -- which is a potential opening not entirely blocked -- to become a "vent" when the chemical reactions got fully underway. That did not happen.
 
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Occam's Razor is a theoretical model -- a practical rule. It is not an answer. It says simply don't overcomplicate, but it doesn't EXPLAIN anything. That the pressure did not exceed the yield limits of the steel is probably correct, but it's an educated guess.
Occam's Razor is not MEANT to explain anything. It's a tool to help YOU explain something, based on the evidence.
 
If pressures or temps get high enough, it will definitely impact the reaction. At high enough energy states, the reactants will break down into different compounds before they can burn as gunpowder, and the result may be an ultimately lower energy resultant. Pure conjecture, I am no chemist, but I do know bottling up some reactions does impact their efficiency.

TCB
 
Can you document those claims? Particularly in regard to smokeless powder?

Remember, rifles work at about 3 times the pressure of the cartridge in question. If pressure can somehow prevent combustion, how it is it possible for rifles to work?
 
If pressures or temps get high enough, it will definitely impact the reaction. At high enough energy states, the reactants will break down into different compounds before they can burn as gunpowder, and the result may be an ultimately lower energy resultant. Pure conjecture, I am no chemist, but I do know bottling up some reactions does impact their efficiency.

TCB

I'm having a hard time believing the energy release from the chemical reaction is going to generate plasma, much less generate plasma quickly enough to chemically break down the burning powder which is generating the plasma before the powder finishes its combustion.

:scrutiny:
 
I accept that the gun in question did not self-destruct, as some of us expected. (Even those doing the test weren't sure about the likely outcomes... or they wouldn't have needed to do the test.)

It seems there's two possible reasons for that, and either one of them seem plausible at this point, to me. I'm not a ballistician nor a chemist, so what seems plausible to me may not seem so to a more knowledgeable participant.

1) a properly locked-up semi-auto is much stronger than most of use think or expected. (Given that the chamber is thick, and the locking lugs typically hold things together pretty well, that's a quite rational explanation.)

2) there's a chance that confining the area of the chemical reaction -- which doesn't happen when a gun is actually fired (as the bullet leaves the case pretty easily, and the area available for the chemical reaction rapidly increases) -- changes the nature of the reaction. (If the combustion chamber -- probably the wrong term -- changes its size, it's not unreasonable to assume that the pressures generated are different than if the chamber size remains fixed. That doesn't mean the powder doesn't burn completely, but it could mean that it burns differently. The fact that powder is self-oxidizing, mentioned in the discussion, may or may NOT be a factor; we'll have to let a chemist address that point.

I'd like to know WHY things didn't happen as expected. Just saying, as some have said that it didn't self-destruct, therefore, it didn't self-destruct, DOESN'T EXPLAIN WHY it didn't.

It would be nice to know WHY. That incorporates understanding rather than blind acceptance. Using Vern's Occam's Razor approach, saying that #1 above is the answer seems good, but that's an inference, not a proof, and not an explanation.

I suspect THAT is the answer -- i.e., most semi-auto actions, when properly locked, are much stouter than we realize -- but I would like to know that this is the case, rather than just suspect or assume.
 
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