ôMuzzle velocityö 8

[Veemax]

I have a question about the “muzzle velocity” of projectiles fired from a gun barrel! I’ve always assumed that the projectile is still accelerating- providing the rate of burn and energy of the charge is sufficient to continue to overcome friction and inertia as it leaves the muzzle- that the projectile velocity is still increasing, even when it’s left the barrel. Am I correct in saying that only the rate of acceleration decreases, ultimately of course- the velocity decreases.

I’ve read several articles in various shooting magazines, that this is not the case! They appear to suggest that the velocity decreases as soon as it leaves the barrel and that maximum velocity is at the muzzle!

Are we to say that the charge and weight of projectile are perfectly balanced- so that the velocity is constant as it leaves the last section of barrel? If this is not the case, then the projectile could still be accelerating at many tens of g’s, as it leaves the barrel.

 

[CH5OH]

phase 1:the bullet is placed in the chamber, the chamber is sealed off:a vast quantity of solids (gunpowder) and air is present behind the bullet.
phase 2:the bullet is fired:the oxigen (only 21% and the gunpowder undergoes a exothermal reaction which propagate with mach 5 until either the gun powder (vapourised) or the oxigen is gone or both, when the mixure is perfect.while the oxigen and the gun powder undergoing this exothermal reaction the pressure and temperature of the combustion gasses go extremely high.
phase 3:the gasses not undergoing this exothermal reaction are pushed asside and hit the bullet's back, the barrel sides and the barrel back end.the barrel deformes and pushes the shooter back, the bullet deformes and starts to accelerate.(initial shock wave)
phase 4:while the bullet travels through the barrel,the heat of the combustion gasses is transferred to the nitrogen,the excess of oxigen or the excess of gun powder vapour and they start to expand.Although the volume of the barrel increases, while the bullet travels, the pressure in the barrel still increases as a result.IMPORTANT:this expansion does not result in a shock wave, its a homogeneous pressure increase.this pressure increase result in the bullet accelerating even further.
phase 5:the bullet reaches the muzzle and gasses start to leak out.the amount of pressure behind the bullet rapidly decreses so the accelaration of the bullet decreases
phase 6: the bullets back is now at the muzzle and the leaking gasses have changed direction from somewhat foreward to somewhat radially, resulting in a reaction force towards centerline of barrel (cancelling each other out).As the leakage increases, the acceleration of the bullet decreases.
phase 7: the bullet has left the muzzle and remaining gasses leak out
conclusion:
The acceration of the bullet is maximum, when the pressure in the barrel is maximum or when the pressure increase as a result of the gasses heating up is cancelled out by the volume increase of the barrel behind the bullet and the amount of leakage of the gasses asside of the bullet.Its for sure a point slightly before the bullet leaves the muzzle.When the bullet leaves the muzzle, thrusting forward and leaking gasses thrusting radially, it leaves somewhat of a vacuum behind the bullet, so acceleration drops to zero or even become negative .

 

[eadwine]

It appears that the problem to be confronted is, do we need an answer that is exact or close enough? Whether the projectile is still acclerating upon exit form the muzzle or is deacclerating. The issue becomes is this variable significant enough in the overall value of the muzzle velocity too include it? If we're looking for an "exact" value, yes. If not, no, and the "close enough" value is accurate enough for the needs at hand.

Remember, all physical and mathematical analysis is dependant on the needs at hand. Even the Calc. recognizes this and makes allowances for the dropping of insignificant values.

 

[Ltwine]

Curios mind does not know significance, but it is always the driving force behind wiping out the insignificant values due to our incapability in the past. All science are accumulation of insignificant matters.

No pun intend, just personal feeling, as I truely enjoy this discussion.

 

[FeX32]

Thanks for the star SNORGY.
Our race as a whole can benefit tremendously by coinciding our thoughts without "jerking" each other around.

Some strong points in the post. Particularly detailed is CH5OH.


Fe

 

[FeX32]

bang


Fe

 

[desertfox]

Hi Veemax

I found this article which actually suggests that the bullet could be accelerated just after leaving the barrel.
Its also interesting to note that muzzle velocity is usually measured several metres away from the actual muzzle so that the measurements taken are not influenced by muzzle blast.

http://www.quarry.nildram.co.uk/ballistics.htm

another article on measuring muzzle velocity.

http://www.ojp.usdoj.gov/nij/training/firearms-training/module07/fir_m07_t07.htm

regards

desertfox

 

[Veemax]

Sorry- I’ve been out of the discussion for quite some time! I have a habit, like the Wise Old Owl, of standing back and just listening.

Thank you to everyone who has helped out with my question, especially SNORGY, yes- we must have shared the same beer! Perhaps I should have drunk the beer after and not before writing!

In my original post, I was more concerned about whether acceleration could or could not drop to zero in zero time. I was in fact trying to determine if the velocity after leaving the muzzle was greater, no matter how small or insignificant. SNORGY quoted-

“My confusion arises from trying to get my head wrapped around the idea that at some point in time, acceleration is some non-zero quantity, and then in a *mathematical instant*, it is zero. Similarly, the net unbalanced applied force goes from "something to nothing" in a mathematical instant. The only way this can happen, unless I am missing something, is if the acceleration is non-zero and zero at the same time; similarly, the force is non-zero and zero at the same time. It has to be at the same time because there is no change in time.”

In my original post I wasn’t that concerned about any residual force after the muzzle.
FeX32 attached a link to a technical report on muzzle blasts, showing quite clearly that pressure beyond the muzzle rises quite sharply for a number of calibre lengths, then drops eventually to zero. I’ll have to assume then that there is still some pushing force.

Yes- I know we have clearly defined formulas, rules if you like. We can transpose numbers into boxes and arrive at answers. It’s good to question the answers, if only refresh our memories and sharpen our minds. Sometimes it’s good to park slightly over the lines, just outside of the boundaries or what we would normally regard as wrong- we might get a parking ticket though!!.

Veemax

The more he saw the less he spoke
The less he spoke the more he heard.

 

[IRstuff]

Phase 1 is not technically correct. There is no air behind the bullet. This is specific feature of high explosives and rocket fuels, which is that the oxidizers are mixed directly into or already incorporated in the fuel. In the case of a cartridge shell, the propellant contains explosive and oxidizer. so no air.

Most chronographs are sonic transducers, so the muzzle blast effects must be minimized.

Suggestion and reality are two different things. The literature and the photo posted Fex32 show that the gas expands laterally as much as it does forward. Based on that, and the decreasing acceleration due to the expansion of the gas down the barrel, the maximum increase in a 7.62 mm round muzzle velocity of 874 m/s using the data I posted earlier is only about 1.3 m/s. Lots of assumptions are made, so I won't claim any "truth" to that.

TTFN

FAQ731-376

 

[SNORGY]

handleman said it best.

"When you stop contacting and applying force, acceleration stops. Velocity continues, but acceleration stops."

This made sense to me when I put everything in space at zero gravity and zero air friction, and the only external force left to consider was the "propelling" force. The point of non-contact was coincident with the point of the decay to zero of the unbalanced "propelling" force and, hence, zero acceleration.

On Earth, the application of the "propelling" force simultaneously comes with the application of all of the the opposing forces. The "propelling" force decays to a point where it is in balance at some non-zero value with the equal opposing forces. The "unbalanced" force (which is the *net* propelling force, not the *gross* propelling force) is zero at this point, but all of the individual forces are themselves each still non-zero and still "in contact" with the bullet. At some further point during deceleration, the "propelling" force decays to zero and "breaks contact" with the bullet, and then from there, air and gravity do the rest.

In the end, "instantaneous rate of change in acceleration" is not important. If the unbalanced force can be applied instantaneously, then acceleration will change instantaneously; if not, then it won't.

Regards,

SNORGY.

 

[IRstuff]

"When you stop contacting and applying force, acceleration stops. Velocity continues, but acceleration stops."


Huh, and I've gone through my entire life thinking that Newton had said it best? and first...

"A body persists in a state of uniform motion or of rest unless acted upon by an external force"



"Instantaneously" is a very relative term. For us humans and even the bullet itself, 75 microseconds to 3568 G's, as listed in the simulation file I posted, is close to instantaneous.

TTFN

FAQ731-376

 

[SNORGY]

What I meant is that handleman said it best within the context and domain of this thread.



Regards,

SNORGY.

 

[btrueblood]

Irstuff, you beat me to it, a star for "In the case of a cartridge shell, the propellant contains explosive and oxidizer. so no air."

 

[IRstuff]

BT,

Thanks, wasn't even sure anyone noticed, or cared

TTFN

FAQ731-376

 

[btrueblood]

It is definitely a lesson in winnowing to read this thread.

 

[rb1957]

appreciated the Jethro Tull reference (Wise Old Owl) ... if it was meant as such ...

 

[chicopee]

Has anybody considered in these 97 postings the effect of gravitational force, hence a downward acceleration of 32ft/sec^2.

 

[GregLocock]

rb1957 did in the third post.



Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[IRstuff]

The effects of gravity at the distances traveled that we're talking about is negligible; everything occurs within the first 5 ms out of the barrel. The math is trivial: 1/2gt^2 = 122.6 um of drop

TTFN

FAQ731-376

 

[SomptingGuy]

Orders of magnitude...

My CFD professor was worried about the effect of gravity on the flow of diesel on a piston bowl until I pointed out the obvious. Always fun to get one over the boffins.

- Steve

 

[GregLocock]

One of Feynman's continual obsessions is exactly why gravity is such a weak force. Good thing it is otherwise I'd be even shorter and fatter.



Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight