ôMuzzle velocityö
ôMuzzle velocityö
(OP)
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.
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.





RE: ôMuzzle velocityö
Yes.
How can it accelerate without any pressure behind it? (to put it simply)
Fe
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Once the bullet as left the barrel there's no force to push it, prior to it leaving the barrel expanding gas accelerates the bullet, but once its left everything is returned to atmospheric pressure.
Here is an online calculater you might find interesting:-
http://www.zknives.com/bali/bvtengy.shtml
desertfox
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
- Steve
RE: ôMuzzle velocityö
but seriously, free body of bullet in barrel ... expanding gas behind it, friction acting on it, produce acceleration. as the bullet emerges from the barrel friction is removed so there'll be an increase in acceleration towards the end of the barrel. the instant the bullet emerges, the expanding gas is still (momentarily maybe) pushing the bullet (as the pressure falls to atmospheric) so there'd be another small increase in acceleration. i accept this is like "how angels fit on the tip of a needle" but ...
RE: ôMuzzle velocityö
Rocket propelled projectiles are a whole different ballgame.
RE: ôMuzzle velocityö
I suppose that it could either increase or decrease during the time the bullet is traveling through the barrel, dependent upon barrel length and bore and the amount and type of powder used.
RE: ôMuzzle velocityö
David
RE: ôMuzzle velocityö
Barrel pressure being constant is irrelevant to maximum velocity, only maximum acceleration. Maximum acceleration is not the question here.
The only question is whether the bullet continues to accelerate at some rate after leaving the barrel.
The bullet may continue to accelerate until the back end completely clears the barrel. It may accelerate slightly faster as the contact area with the barrel is reduced. However, as soon as the bullet ceases to seal the barrel, the gas propelling the bullet disperses into the atmosphere.
One might think at first that the exiting gases might continue to "blow" on the bullet for a very short distance (1-2 inches) after it leaves the barrel. I seriously doubt this is possible. Certainly not measurable.
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
If you draw analogies, consider throwing a curling rock or pushing a child on a bicycle or pushing a bobsled. If you let go quickly at maximum push speed, it is likely the object will still accelerate further away from you, but probably what is happening is that you are giving that last little nudge or "oomph" and releasing more energy stored in your arms. If you let go very gradually, maintain your speed, and reach out, then chances are you will still be able to touch the object because it has not accelerated away from you.
Pictured that simplistically, it becomes easier to believe that instantaneous cessation of applied force produces instantaneous cessation of acceleration. The only question that then remains is whether or not it is physically possible to instantaneously commence or cease an applied force. If the bodies involved are perfectly rigid, then yes; if however there is any elasticity whatsoever, then no. The conclusion is that, since everything has at least *some* elasticity, there can be no "cusp" at all - only very very short duration, steep curves.
Coffee time.
Regards,
SNORGY.
RE: ôMuzzle velocityö
Excepting when a bullet in in freefall, under the influence of gravity, it usually ceases to accelerate after it leaves the barrel of the gun.
The force that accelerates the bullet is the pressure built up in the chamber, and within the barrel, caused by the combustion of the propellent charge (gunpowder, to use the vernacular - although it is not normally what is traditionally known as gunpowder, which is now regarded as black powder).
Exactly where the peak acceleration happens depends on the way the propellent burns, which depends on factors such as the grain size of the propellent (smaller grains burn faster, but you can also change the shape of the grains, so the they start by burning faster, and then get slower, or visa versa). Generally, the longer the barrel, the more distance you have in which to build up speed, so you will use slower burning propellent.
In general, you want to be as efficient as you can in the conversion of propellent energy to building up speed in the bullet; so you want to make sure that when the bullet leaves the barrel, there is as little excess pressure behind the bullet as possible (as that excess pressure may make for a loader bang, but is otherwise wasted energy). Since the acceleration of the bullet will be proportional to the pressure built up behind the bullet, it therefore follows that you wish to minimise the amount of acceleration that happens just as the bullet is about to leave the barrel (ofcourse, with short barrelled guns, you may not have too much choice in the matter, but if that is the case, then you are wasting a lot of energy).
RE: ôMuzzle velocityö
{bullet = -rocket, bullet inside barrel}
{bullet <> rocket, bullet outside barrel}
Sorry...don't know how to do a "not equal to" sign...
Regards,
SNORGY.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Why do we make such simple things so annoying for the OP.
Fe
RE: ôMuzzle velocityö
After the bullet leaves the barrel there will be some "blowing". The muzzle gas has velocity and mass - thus it has inertia. Some of the muzzle gas will be impacting on the back of the bullet, and impart some of its inertia to the bullet.
So the bullet may continue to accelerate for some small distance after it leaves the barrel. I would expect this acceleration to be very much less than the acceleration while it is in the barrel.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
The gases continue to exert force on the bullet even out of the barrel until their pressure equals atmospheric pressure (this for a very short distance). Internal ballistic, intermediate ballistic and external ballistic govern the bullet motion in the barrel in the transient zone and out of muzzle.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
=/=
--Scott
http://wertel.eng.pro
RE: ôMuzzle velocityö
rmw
RE: ôMuzzle velocityö
Fe
RE: ôMuzzle velocityö
Actually, this would be the minimum acceleration.
If the bullet is stopped in less distance than it spent accelerating through the barrel, it would experience its maxumimum absolute acceleration. This all assumes that the acceleration through the gun barrel and the target are approximately constant.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
bullet inside barrel:
case 1:pressure increases,volume increases (rapid combustion)>kinetic energy (velocity) bullet increases
case 2:pressure remain constant, volume increases (slow combustion)>kinetic energy (velocity) bullet increases
bullet leaving the barrel (seal expanding gas breaks):
pressure drop,gas expanding all directions.As this will create a chock wave travelling at the speed of sound
the bullet speed when leaving the barrel is beyond that speed.So the chock wave can never give the bullet an extra push as it is to slow.
RE: ôMuzzle velocityö
Projectile motion has been studied for over a century and volumes have been written. For anyone interested - htt
RE: ôMuzzle velocityö
V shaped wave in front caused by bullet pushing the air asside
curly shaped wave behind the bullet caused by drag and rotation of bullet
circular shaped wave behind bullet caused by sonic boom
RE: ôMuzzle velocityö
The trajectory analysis showed it losing about 4 m/s in velocity with 5m of the muzzle.
TTFN
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RE: ôMuzzle velocityö
speed bullet:869m/s (refering to your attachment)
speed sound:330m/s
so no push,not even hypothetically
RE: ôMuzzle velocityö
That V shaped wave is not so much the bullet pusing the air. The V shaped wave is actually an attached shock wave.
The wave is attached for any pointed objects, and will unattach for blunt nose objects. Attached waves have alot less drag, so most supersonic rounds will have a pointed nose, rather than a round nose as seen in subsonic small calibre handgun rounds. The waves on the back side of the bullet are mach waves not shock waves (although I supose a mach wave is a type of shock wave).
The photo you have is a schleiren photo, showing the density varriations in the test fluid.
RE: ôMuzzle velocityö
Star.
Regards,
SNORGY.
RE: ôMuzzle velocityö
Since the pressure is rapidly decreasing from its peak value, there's not a whole lot of oomph left, but there's going to be some.
TTFN
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RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Fe
RE: ôMuzzle velocityö
The reply from Ltwine, I thought was really good- "Pop open a bottle of Champaign and observe the change in velocity". I suppose this is the same as when a Ping Pong ball can be balanced at the top of a vertical air stream flowing from a tube. If there were no force beyond the open end of the tube, then the ball would rest at the open end, not in mid air. Anyone who has fired blank ammunition will know that it can produce pressure waves on quite distant objects. Anyone who is willing to place their hand a few inches from the muzzle, whilst firing a blank round- I would consider very foolish! To say that a projectile is not being pushed beyond the muzzle opening- I would consider wrong!
The reply from SNORGY was very good indeed. IRstuff has stated that- "Output from an interior ballistics simulation puts peak linear acceleration at 161409 g for 7.62 mm round at 30.5 mm travel in a 609.6 mm barrel. Acceleration at the muzzle is only 17022 g. SNORGY has mentioned "instantaneous energy conversion". If also "instantaneous acceleration" is indeed possible and likewise instantaneous deceleration, then in an instant the projectile would reduce in rate of acceleration from 17022 g to zero g. In theory it seems this is possible, in practice, it may be very different. Let's look at ""instantaneous acceleration". If it were possible to accelerate a projectile from zero g to 17022 g when time is zero, we would need an infinite amount of energy! If this were put into practice then we could have barrel of zero length producing phenomenal velocities.
If instantaneous deceleration is not possible after the projectile leaves the muzzle, then it must take time- no matter how short. Lets take the figures of 17022 g to zero g, then between these two figures, the projectile is still accelerating, although the rate of acceleration is reducing, if we have time and acceleration, then we have an increase in velocity- no matter how small.
RE: ôMuzzle velocityö
Do you mean instantaneous deceleration, or instantaneous transition to deceleration? The acceleration will drop to zero or a negative value just as fast as the force does, unless the mass of the bullet changes.
RE: ôMuzzle velocityö
I think you are mixing up velocity, acceleration and jerk here (or force, momentum, energy, etc).
"accelerate a projectile from zero g to 17022 g" doesn't make sense. You can apply 17022g instantly if you can apply enough force instantly. Energy isn't required until the projectile starts moving.
- Steve
RE: ôMuzzle velocityö
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: ôMuzzle velocityö
a ping pong ball floating on air or a hand in front of riffle fired with a blank.....
mmmm... a bit difficult to compaire those objects with a bullet leaving the barrel (speed of the former=0m/s,the latter=supersonic)
a detonation can only excist in a combustible mixture:
to little gunpowder:the gun powder is burned even before the bullet leaves the chamber
to much gun powder:the excess of gun powder is vapourised and burns when entering the air (lack of oxigen when bullet seals the barrel)
however most of the kinetic energy is derived from the thermal expansion of the gas, not from the detonation of the gun powder(hence longer barrel,more bullet speed).The accelaration is a derrived unit (compairing speed over period of time)it would be impossible for the bullet speed to drop to zero instantly (newtons law), the acceleration however does, the moment you seize the energy supply (when the bullet leaves the barrel)
RE: ôMuzzle velocityö
Fe
RE: ôMuzzle velocityö
Perhaps I'm on the wrong track here!! Maybe someone has used a high speed camera to see exactly what happens in practice?
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: ôMuzzle velocityö
Did we share the same thought or the same beer tonight?
Maybe the energy of *motion* can instantaneously change provided that the conversion to other forms of energy occur over some duration of time. These might include:
* plastic deformation
* heat
* light
* sound
Maybe that's where one can reconcile the paradox of "infinite energy required" for instantaneous acceleration / deceleration. My college introductory physics books seem to read this way: they explain that the energy that is apparently "lost" when one tries to apply conservation of energy to problems involving plastic collisions is not "lost" at all - it goes into other things that aren't as readily quantified.
Regards,
SNORGY.
RE: ôMuzzle velocityö
However to try and wrap it up, as the shell leaves the barrel it is being pushed out by a pressurised gas that is capable of travelling faster than the shell (beacsue it is pushing on it). When it leaves the barrel the shell will be surrounded by an expanding jet of gas that will initially be faster than the shell, and so will accelerate it further. However very soon the gas will slow down, and so slow the shell.
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: ôMuzzle velocityö
I should have mentioned earlier that "Veemax" derives from Vmax (Maximum Velocity), I'd liked to have used that, but it had already gone- quite quickly!!
RE: ôMuzzle velocityö
http:/
Fe
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Just looking for a starting point here... both of these posts reveal fundamental lack of understanding basic concepts.
Have either of you guys ever taken a Calculus or Physics course?
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
My knowledge is apparently decelerating very rapidly.
Perhaps someone ought to shoot me, since it probably wouldn't hurt me, given that I evidently don't know how to stop a bullet.
Regards,
SNORGY.
RE: ôMuzzle velocityö
SNORGY, going back to your earlier post, it made a bit more sense. Due to elasticity in everything, it is almost impossible to instantaneously stop applying force to an object. However, when the force does stop, the acceleration stops. Acceleration does not continue due to "inertia" or something. In fact, acceleration would not even occur if you instantaneously removed friction. All that would happen is that deceleration would cease to occur. Velocity would then remain constant until some other force was applied. In your example of pushing/throwing things, objects will most certainly not continue to accelerate away from you after release. When you stop contacting and applying force, acceleration stops. Velocity continues, but acceleration stops.
Force causes acceleration.
The continued application of force over distance transfers energy to an object during its acceleration.
Infinite force for zero time over zero distance is zero energy. Infinite acceleration for zero time is zero energy. There is no paradox here anywhere. Zero force over infinite time and infinite distance is zero energy.
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
"Infinite force for zero time over zero distance" would, I think, mathematically, be *undefined*, not zero.
i.e., F(infinite) x [{0 time}/{0 distance}] = undefined
That said...
F = ma
dF/dt = d(ma)/dt
dF/dt = m(da/dt)
[1/m]dF/dt = da/dt m > 0, m = constant
So, if you can *instantaneously* remove a net applied unbalanced force, you can have an *instantaneous* rate of change in acceleration. 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.
But...I suppose if you can draw a cusp, it's possible to have one.
Consideration of some measure of elasticity and, in a convoluted way, the dynamics of energy conversion and transfer, are perhaps being improperly brought into the discussion in an attempt to understand things a bit better.
My brain (the part that hasn't yet been killed by beer) hurts.
Regards,
SNORGY.
RE: ôMuzzle velocityö
Excuse my ignorance. How about motion after impact. I need to check on impulse & momentum.
RE: ôMuzzle velocityö
Now raise two balls and drop them together. Why do two rise from the other side? Why not one, twice as far? When you can comfortably explain that, there is hope.
- Steve
RE: ôMuzzle velocityö
You don't need to explain the "uncomfortable" phenomenon but my question - motion/acceleration after force is stopped.
RE: ôMuzzle velocityö
"In fact, acceleration would not even occur if you instantaneously removed friction."
??? Really don't get it. Can anyone explain?
RE: ôMuzzle velocityö
Interesting how this turned into a physics lesson.
I suggest not poking each other.
Fe
RE: ôMuzzle velocityö
As far as a force being instantaneously removed, as I said I agree with you that it's impossible in the real world due to elasticity. Whether or not you need to account for these infinitesimals or they can be ignored depends on your analysis needs.
I'm guessing you misunderstood my context. This thread is branching away from the bullet and into elementary physics.
What I intended to say was that if a body is in free motion and is losing velocity due to friction, if you remove that friction the body will not accelerate, it will simply cease to lose velocity.
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Star.
We're all here to help each other in instances where we don't fully understand something - for whatever reason - regardless of how complex or how simple.
We all went to college or university, we all graduated in engineering and got a degree, and we all have had at least some success in staying employed and getting at least some things right. It stands to reason, therefore, none of us are stupid - or whatever adjective or innuendo one wants to read in place thereof.
"Poking" is counterproductive and doesn't help anything or anybody. I know from experience in *every* aspect of life that the best way to start an argument or to have someone disagree with you is to say or do something.
I am assuming that all of the "pokers" out there all went through scoring 100% on every exam, never needing technical advice from a mentor, never having been faced with a drawing revision due to an oversight during detailed design, and so forth. If that is true, they have every right to continue to poke away with impunity.
For myself, I simplify things:
(1) Try to help ehere I can;
(2) Try to learn where I need.
Well...OK...I still might bash the odd MBA, but that's a separate issue.
Regards,
SNORGY.
RE: ôMuzzle velocityö
- Steve
RE: ôMuzzle velocityö
This is indeed a discussion on basic physics pertaining the opening question. What a refreshment for many of us whose college days were hidden in remote memory.
RE: ôMuzzle velocityö
Damn. Never thought about that.
RE: ôMuzzle velocityö
If you want to argue with science, why not start a seperate thread for that.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
m1*v1=m2*v2
and
(1/2)*m1*v1^2=(1/2)*m2*v2^2
The same amount of mass (same number of balls) must move and they must have the same kenetic energy (minus sound and other nonconservitive losses).
RE: ôMuzzle velocityö
Coming back to the OP, in short, it is my belief (right or wrong) that maximum bullet velocity occurs at some point in space away from the exit from the muzzle of the gun, not *right at* the exit from the muzzle of the gun.
But...I think I have contributed to the decay of this thread, and offer my apologies for that.
Regards,
SNORGY.
RE: ôMuzzle velocityö
Regards,
SNORGY.
RE: ôMuzzle velocityö
I would guess for a well designed gun with the proper grain in the shell, the max velocity will be just before exiting the barrel. If upon exiting the bullet was not faster than the shock expansion out of the muzzle, it would not be as accurate, even with its rotational inertia stabalizing it.
That is my take on this.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Almost all modern firearms have significant excess pressure left in the barrel before the projectile exits. Carrying the excess weight around to achieve the condition you describe would be pointless (why purposely make a gun with a built-in braking mechanism for the bullet), and even if built...
" If upon exiting the bullet was not faster than the shock expansion out of the muzzle, it would not be as accurate, even with its rotational inertia stabalizing it."
The friction loss for the bullet being squeezed down the rifled barrel means you must have enough pressure in the barrel to drive and accelerate the projectile. The speed of sound in the driving gases must be significantly higher than the speed of the projectile, or you won't be able to accelerate the bullet at all (the pressure wave(s) from burning propellant wouldn't catch up to the bullet). In a typical firearm, the speed of sound of the gas mixture in the chamber are around 4,000 to 5,000 fps. Typical firearm muzzle velocities are about 60% of those sound speeds, and that ratio of speeds holds for almost all guns.
The net result is that even if the bullet's acceleration had dropped to near zero just prior to exiting the gun, the blast wave from escaping gas will still be there, and possibly add a slight net thrust and acceleration as the chamber empties of gases moving at speeds higher than that of the bullet itself.
In most high speed video and film of guns firing, you cannot see the bullet due to blow-by gases escaping the muzzle before the bullet exits. The cloud of smoke obscures the muzzle for several frames, and the bullet then appears out of the expanding smoke cloud.
Maybe a high-speed xray would discern the slight nudge given the bullet by escaping gas. It would need a frame rate somewhere in excess of 100000 per second to do so, which might be pushing current technology.
RE: ôMuzzle velocityö
Fire the gun - or throw an object - in space in the absence of fluid friction and in the absence of gravity (for all intents and purposes). So, if an astronaut throws a baseball, there comes a point in time or, easier to conceptualize, a position in space where the baseball is no longer incontact with his hand, which is the only source of unbalanced applied force. Beyond that point in time / space, acceleration must immediately be zero; otherwise, the ball would gain speed indefinitely. Relativistic mechanics aside, that won't happen - can't happen, so the acceleration *must* instantaneously become zero.
One could describe the firing of a gun in space in similar fashion. There will come a point in time coincident with a position in space where the "thing" providing the propelling force pushing on the bullet is no longer in contact with the bullet, at which point acceleration must immediately be zero.
Do the same thing on Earth firing the gun from left to right. The propelling force on the left driving the bullet to the right is counteracted by the retarding forces on the right pushing to the left back on the bullet, with the imbalance producing acceleration. There comes an instant in time coincident with a position in space where the propelling force on the left is equal to (in balance with) the retarding forces on the right. It is at that point in time / position in space where acceleration has been reduced to zero and velocity reaches a maximum. Beyond that point, velocity decreases and acceleration becomes deceleration.
So, the problem was never one involving instantaneous rate of change of acceleration. It was one of trying to pinpoint the position in space where the summation of forces being applied to the projectile reach equilibrium.
Feeling dumb now.
Regards,
SNORGY.
RE: ôMuzzle velocityö
TTFN
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RE: ôMuzzle velocityö
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 .
RE: ôMuzzle velocityö
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.
RE: ôMuzzle velocityö
No pun intend, just personal feeling, as I truely enjoy this discussion.
RE: ôMuzzle velocityö
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
RE: ôMuzzle velocityö
Fe
RE: ôMuzzle velocityö
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
regards
desertfox
RE: ôMuzzle velocityö
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.
RE: ôMuzzle velocityö
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
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RE: ôMuzzle velocityö
"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.
RE: ôMuzzle velocityö
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
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RE: ôMuzzle velocityö
Regards,
SNORGY.
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Thanks, wasn't even sure anyone noticed, or cared
TTFN
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RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: ôMuzzle velocityö
TTFN
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RE: ôMuzzle velocityö
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
RE: ôMuzzle velocityö
Cheers
Greg Locock
I rarely exceed 1.79 x 10^12 furlongs per fortnight
RE: ôMuzzle velocityö
This video shows a muzzle blast exiting ahead of, and out-pacing a (admittedly subsonic round) 45-cal pistol bullet.
ht
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
I suggest that you run this through your calculator again:
1/2*386.0886*0.0052
It comes out to 0.0048 inches on mine, or 0.1226 mm. You seem to calculating velocity, and not distance:
386 in/s2 * .005 in = 1.93 in/s
TTFN
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RE: ôMuzzle velocityö
Please check your understanding of the basic equations of motion, then post your English units answer.
Bonus to restore engineer cred:
Calculate the start time of the 5ms period in which the bullet actually does drop 1.98 inches.
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
RE: ôMuzzle velocityö
Without, it's about 101 ms
TTFN
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RE: ôMuzzle velocityö
I didn't ask how long it takes for the bullet to drop 1.98 inches from the start. Worded slightly differently, how long does it take for the bullet to accelerate to the point where, during the next 5ms (and taking into account acceleration during that 5ms) it drops 1.98 inches.
Yeah, and without lift.
-handleman, CSWP (The new, easy test)
RE: ôMuzzle velocityö
Now, if F=ma and F/m=a, then if "m" is held constant or even if it changes and "F" goes to zero it must follow that "a" goes to zero. Unless, of course, we're using the "New Math" and the Time/Space Continuum isn't a factor here.
RE: ôMuzzle velocityö
TTFN
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RE: ôMuzzle velocityö
Most chronographs operate from photodetectors noting the passage of the projectile. They still need to be placed well away from the muzzle of the gun barrel, though, or muzzle gases will affect accurate readings.
The propellant ("gunpowder") is a variable. Fro a given caliber, projectile weight and velocity, one can chose powders with differing burn characteristics, reaching peak pressures sooner or later in the firing sequence. This used to be particularly important with certain models of semi-automatic firearms that used gases tapped from the barrel to operate their mechanisms. The characteristic is also a variable in achieving maximum accuracy.
Aside from those two points, and a hundred-odd posts, the short answer is that the bullet starts slowing when the force behind it pushing it faster is equal to the force in front of it slowing it down. Giving the gases a little credit for mass and velocity, it would appear to my electrical mind that this point would be a very short distance in front of the muzzle, on the order of a couple of bullet diameters.
old field guy