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ôMuzzle velocityö
8

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.
 

RE: ôMuzzle velocityö

2
"velocity decreases as soon as it leaves the barrel and that maximum velocity is at the muzzle"
Yes.
How can it accelerate without any pressure behind it? (to put it simply)

peace
Fe

RE: ôMuzzle velocityö

you could fire downwards, like off a mountain ...

RE: ôMuzzle velocityö

Hi Veemax

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

Max. Acc is at the instant the bullet leaving the muzzle with pressure and sudden drop of friction.

RE: ôMuzzle velocityö

Bullet ≠ Rocket

-handleman, CSWP (The new, easy test)

RE: ôMuzzle velocityö

This kind of question appears often in E-T forums.  Max Acc is only at the intant the bullet leaves the muzzle if max force is also at that point.  It's likely that max velocity is at the instant the bullet leaves the muzzle (hopefully), but velocity won't increase after it's left and the accelerating force is gone.

 

- Steve

RE: ôMuzzle velocityö

ok, i tried to have a little fun (if you fire angled downwards, there'll be a component of gravity accelerating the bullet).

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

I have a theory that there is a slingshot effect at the moment the projectile is freed from the friction of the barrel (inertia+sudden reduction of friction).  Chronometers are generally set up around 3 feet from the muzzle, so I believe any readings you get are going to be maximum anyway.  Don't know if there is a method to getting an actual measurement of the projectile speed inside of the barrel.  Even with high speed photography, you would not know the exact timing, in nanoseconds, that the projectile exits the casing and begins its travel down the barrel.  There are milliseconds of delay from when the striker falls against the primer to when the primer actually detonates to when the charge detonates and builds sufficient pressure to put the projectile in motion.

Rocket propelled projectiles are a whole different ballgame.

RE: ôMuzzle velocityö

Everyone seems to be assuming that barrel pressure is constant.  I doubt that it is.

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

I don't think that it matters if the pressure in the barrel is constant or not.  As long as there is enough greater pressure behind the bullet than in front of it to overcome friction the bullet will accelerate.  At the end of the barrel, if there is still a "positive" dP then you get the "slingshot" or "spring release" effect for a final burst of acceleration, but that is a very short duration.  Within a very short distance from the end of the barrel dP drops to zero and the bullet starts slowing due to friction with the air (this all assumes that the bullet was fired horizontally, any inclination will either cause positive or negative acceleration that may be greater than the negative acceleration caused by friction).

David

RE: ôMuzzle velocityö

OK, inertia plus sudden reduction of friction cannot increase the bullet velocity.  That "theory" has no grounding in any physical law.

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

Pop open a bottle of champaign and observe the change in velocity :)

RE: ôMuzzle velocityö

of the cork.

RE: ôMuzzle velocityö

What it comes down to is whether or not one accepts that the change in acceleration with time follows a "cusp" or a "curve" (regardless of how steep or short) from maximum acceleration to zero acceleration, and whether or not one accepts the concept of "instantaneous" energy conversion / transfer versus "extremely fast" energy conversion / transfer.  If one believes the laws of classical Newtonian mechanics and the resulting algebraic expressions, then one is compelled to believe the former in each case.  Therefore, continued acceleration beyond muzzle exit is only possible if there is some reserve of unconverted stored energy at that instant in time; the slingshot effect and the exiting, rapidly dissipating gas effect described above probably most accurately capture this phenomenon, since recovery and expenditure of elastic strain energy stored in the projectile would be unlikely for - for example - a plastically deformed lead bullet.

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

Saw this on another forum:

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

If what TenPenny states is true, which it probably is, then one must modify the inequality postulated by handleman:

{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

The pressure drop at the muzzle should be small compared to drop on friction. The bullet will acc more in a very short distance then decrease. Only test can tell.  

RE: ôMuzzle velocityö

F=m*a dude, unless there is still gas exiting the barrel and hitting the projectile - and I suppose there is a possibility of that happening for a very short amount of time - there is no force in the equation, hence acceleration must also be zero.

RE: ôMuzzle velocityö

like firing downwards will give you any significant added accel...... how about we go into space, seal the barrel, take into account the effect of combustion with no O2 in space (while the GP has O2, as an oxidizer is set within the bullet casing)? ...
Why do we make such simple things so annoying for the OP.  

peace
Fe

RE: ôMuzzle velocityö

In the very short time while the bullet is exiting the end of the barrel friction against the barrel wall is being replace with parasitic drag from the air.  This will result in a net reduction in retarding force, so assuming a constant barrel pressure, there will be an increase in acceleration as the bullet exits the barrel.

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

Agree with the above.

RE: ôMuzzle velocityö

When the bullet reaches the muzzle, the propellant gases have still quite high pressure (100 bar approx.). Once the bullet exits the barrel, the gases are free to move past the bullet and expand in all directions.
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

Gee, I thought that the peak acceleration of a bullet is when/if it hits an impenetrable target wink

RE: ôMuzzle velocityö

I would immagine the acceleration out side of the barrel is neglegable and occurs over such a short distance that it really makes no difference.  Part of the loud noise from a gun is due to the rapid expansion of the air out of the barrel (the expansion is almost instant and in all directions, specially those with the lease resistance ie the directions where there is no bullet to obscure the flow).  For all practical cases, measuring the "muzzle" velocity even several feet infront of a gun will give you accurate velocity values.  Sure this doest account for any reduction in speed due to the air in the foot or so after the barrel, but a speed measurement at this point is accurate enough for any calculation you would be doing.  The uncertainty in your flight path calculations (due to bullet spin, ballance, air density, air currents etc) is going to be greater than any uncertainty in your velocity.

RE: ôMuzzle velocityö

You might say that any beneficial effect of the energy of the expanding gasses behind the bullet outside of the barrel lasts about as long as the noise of the blast, maybe less.

rmw

RE: ôMuzzle velocityö

!=

peace
Fe

RE: ôMuzzle velocityö

Quote (CRG):

Gee, I thought that the peak acceleration of a bullet is when/if it hits an impenetrable target   

   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.   

               JHG

RE: ôMuzzle velocityö

Is the OP still with us? :)

RE: ôMuzzle velocityö

huh, basic thermodynamics
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

Wow, 30 responses in one day.  Here's mine: Yes, a projectile is accelerated a negligible amount beyond the muzzle.    Muzzle exit pressures are somewhere on the order of ~10 ksi.  When the propellant gasses are no longer restricted, they rapidly expand and outrun the projectile.  For a distance of several calibers, the projectile is actually 'flying' backwards relative to the surrounding gasses.  This is a region of aerodynamic instability and can contribute to increased yaw, which damps out quickly in well designed projectiles.  Fin stabilized rounds (i.e. tank and smooth bore ammo) are more sensitive than spin-stabilized.  

Projectile motion has been studied for over a century and volumes have been written.  For anyone interested  -   http://books.google.com/books?id=F3q59-hcGDoC&pg=PA263&lpg=PA263&amp;dq=gun+strength+curve&source=bl&amp;ots=tqb-kPcy2N&amp;sig=7aSAGcA-YHQkLdySbVJrf6vTmY8&hl=en&ei=PqdYS-CwI42W8Aa428DDAw&sa=X&oi=book_result&ct=result&;resnum=2&amp;ved=0CAoQ6AEwAQ#v=onepage&amp;q=gun%20strength%20curve&amp;f=false  Some other good general references are Hayes – Elements of Ordnance, Rheinmetall – Handbook on Weaponry.
 

RE: ôMuzzle velocityö

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.  Hypothetically, one could imagine some miniscule bit of acceleration within a couple of calibers of the muzzle, but that would be about it.


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

irstuff,
speed bullet:869m/s (refering to your attachment)
speed sound:330m/s
so no push,not even hypothetically

RE: ôMuzzle velocityö

CH50H,
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

I am glad I read Bestwrench's post before I uploaded my simplistic viewpoint about the "blast gas" being hot and therefore likely having its corresponding speed of sound greater than the muzzle velocity of the bullet, which at least in part would permit the phenomena that Bestwrench very clearly explains and raises some doubt in my mind about the bullet being beyond the influence of the trailing shock immediately upon exiting the barrel.

Star.

Regards,

SNORGY.

RE: ôMuzzle velocityö

Bear in mind that the mere fact that the muzzle pressure is well above atmospheric pressure, prior to muzzle exit, basically means that the pressure front is indeed moving at supersonic speed.  That's the basic nature of explosive detonations, and that's why there's still 11000 g of acceleration at the muzzle.

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

FAQ731-376: Eng-Tips.com Forum Policies

RE: ôMuzzle velocityö

Of course there is gas still exiting the barrel at the moment of the projectile exiting the barrel. I would agree with figures somewhere around the 10ksi figure for modern shouldered centerfire cases fired from an average length barrel.  If this were not so, special forces would not be using suppressors on rifles.  The speed of detonation is something like 5000 m/s ? in modern small arms propellant, a little bit higher than the speed of sound, yes?  With a suppressed weapon, the intended target(s) or other bystanders cannot precisely determine point of origin of the shot.  The sonic boom of the bullet by itself will not tattle.  Therefore, it makes incredible sense that there is still acceleration occuring at the exit of the muzzle, even if decceleration begins within an inch or so.  If acceleration is still occuring, and there is the instantaneous removal of friction, voila', slingshot effect.  I challenge any of you to take a new projectile and force it through a rifled barrel.  There is more than a little bit of friction.  This is why modern centerfire rifles are developing 50,000 + CUP's in the chamber.  Anyway, I've rambled enough, someone else's turn.

RE: ôMuzzle velocityö


cannon      wiggle

peace
Fe

RE: ôMuzzle velocityö

(OP)
Thank you everyone, who has contributed in trying to getting to the bottom of my answer. I'm always very grateful to forums such as this, not only to answers to my own personal questions, but also to the creation of searchable information accessible to other Web users, myself included. It's a very valuable resource, although sometimes you have to "weed out" the correct information for yourself.

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

Quote (Veemax):

...

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.  

   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.
 

               JHG

RE: ôMuzzle velocityö

Veemax,

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

Instantaneous changes in acceleration or deceleration are possible, if you ignore elastic waves in the body. That's what happens when you stop applying a force.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

RE: ôMuzzle velocityö

still remain difficult to convince:
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

Many good replies here. Although the OP seems to be in left field. Often is the case.

peace
Fe

RE: ôMuzzle velocityö

(OP)
Let's say that we are moving the projectile from rest to an acceleration of 1 metre per second per second at the end of a distance of 1 metre, over the next metre let's say that with increased energy we're able to produce 10 metres per second per second, next- let's say 100 metres per second per second. If we were to try to achieve the same acceleration, with the same projectile, in a shorter time, we'd obviously need more energy, as time approaches zero, then we would need infinite energy to produce the same result. I'll have to assume that the reverse is also true- that if acceleration went from its maximum to zero in an instant, then infinite energy would need to be dissipated. If this cannot be achieved in an instant, there must be a time when the projectile is still accelerating, with a corresponding increase in velocity.

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

(OP)
Are inertia forces fictitious?

RE: ôMuzzle velocityö

Depends on your frame of reference.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

RE: ôMuzzle velocityö

Veemax:

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

In reality if you apply a force to a solid body it deforms locally and series of stress waves pass backwards and forwards and sideways through the body. Eventually this damps out and the body ends up accelerating as a whole. If the speed of vibration in the body length L is a, then any discussion of events that last less than  a time of the order of L/a can no longer afford to ignore this.

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

(OP)
Thank you very much once again SNORGY. I'm afraid I don't know whether I'm correct on the understanding of this problem, I have my point of view and would like to think that it is correct. Socrates was well know for his inquiries and debates between individuals with opposing viewpoints based on asking and answering questions to stimulate rational thinking and to illuminate ideas. I'd like to think that I've learned a lot from this debate!

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

The added boost (acceleration) from propellant gases to the bullet, after it exits the muzzle, should be lost approx 10 feet from the muzzle and so it is of very little practical interest, anyway it exists. Doppler radar is probably the best way to get the bullet's peak velocity.

RE: ôMuzzle velocityö

Quote (Veemax):


over the next metre let's say that with increased energy we're able to produce 10 metres per second per second

Quote (SNORGY):


Maybe that's where one can reconcile the paradox of "infinite energy required" for instantaneous acceleration / deceleration

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

Yes...but it was long, long ago.

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

Haha!  Fortunately for the universe, its laws continue to operate regardless of our ignorance of them!  smile

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

Actually...

"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

"However, when the force does stop, the acceleration stops."

Excuse my ignorance. How about motion after impact.  I need to check on impulse & momentum.

RE: ôMuzzle velocityö

If you want a refresher on impulse, momentum, kinetic energy, etc, get hold of a Newton's Cradle and play with it.  Everyone knows that raising and dropping one ball at one end will result in one rising at the other.  It seems amusing, but not too challenging.

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

Somp:

You don't need to explain the "uncomfortable" phenomenon but my question - motion/acceleration after force is stopped.

RE: ôMuzzle velocityö

Another amusing question:

"In fact, acceleration would not even occur if you instantaneously removed friction."

??? Really don't get it. Can anyone explain?

RE: ôMuzzle velocityö

Some of you are trying to simplify a problem that should not be so.
Interesting how this turned into a physics lesson.
I suggest not poking each other. poke
cheers

peace
Fe

RE: ôMuzzle velocityö

When I said "force over distance," I meant that the force is applied over a distance.  Not force divided by distance.  Work is f(dot)d.  Force divided by distance is undefined, whether or not the distance is zero.  winky smile

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.  

Quote (Ltwine):


Another amusing question:

"In fact, acceleration would not even occur if you instantaneously removed friction."

??? Really don't get it. Can anyone explain?

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

After the impact your mass is subjected to a new force which is the difference between the force which produced the acceleration before the impact and the reaction force produced by the impacted mass on the impacting mass.

RE: ôMuzzle velocityö

FeX32:

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

I suppose not everyone "gets" dynamics.  It's too easy to assume that what seems natural to you will also seem natural to someone else.
 

- Steve

RE: ôMuzzle velocityö

It is difficult to cover a deep involved problem like this in a short but concise manner. Misunderstanding is inevitable, therefore, I would say chasing for clarification rather than poking should be considered normal.

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

Quote (Somptinguy):

Why do two rise from the other side?  Why not one, twice as far?

Damn.  Never thought about that.
 

RE: ôMuzzle velocityö

It seems like you guys need to start a new thread on basic physics/engineering.  The debate over instantaneous acceleration (acceleration being a mathmatical term describing the change in velocity) is a seperate issue from the bullet problem.  Science says acceleration can change instantly (with an instant aplication of force, which is possible unless you are looking on a molecular level with most materials), so for the point of the bullet analysis that is the truth.

If you want to argue with science, why not start a seperate thread for that.

RE: ôMuzzle velocityö

I guess it can be explained by conservation of mass & energy.

RE: ôMuzzle velocityö

The newtons cradle?  That is momentum and energy.

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

A similar thread occurred a while ago - on a totally unrelated topic - wherein energy, inertia and elasticity (in various forms) all got tied into the discussion.  Of the phenomena discussed was the "flywheel".  At some point in the thread, I think, a landing was arrived at in which "instantaneous change in acceleration" might not necessarily occur in an elastic system, but that some elasticity would have to exist in order for there to *not* be instantaneous change in acceleration upon release of an applied force.

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

I also apologize for obviously completely blowing it in my attempt to attach the thread.

Regards,

SNORGY.

RE: ôMuzzle velocityö

The point of max velocity can not be general said to be at any specific point in general.  There are too many variables.  If you dont have enough grain in your shell, the bullet will be slowing before it even exits the barrel, not to mention the possiblity of it not even making it out.  Conversely if you pack the bullet casing with a high explosive and have a short barrel, the shock wave could easily accelerate the bullet for a distance after the muzzle.

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

if you charge with high explosive, creating a shock wave, even if it travels at mach 5 , when it hits the bullet, the bullet need to accelerate first. so the initial shockwave comes to a halt, remaining energy deforms the bullet, deforms the barrel and creates a shock wave traveling back into the barrel.what doesn't happen is a mach 5 shock hitting the bullet and therefore slows down to the speed of the bullet,and then reaccelarate to mach 5 after bullet leaves the barrel

RE: ôMuzzle velocityö

CH5OH, good call.  Not to mention the horrible shock reflections in the barrel, even without a bullet.

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."

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

I think I have come to a realization of my errors in thought.

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

There is a significant difference between a hand and an expanding gas.  The hand is finite object with defined boundaries, while the gas expanding from the shell has no boundaries of its own, although it can be contained by the barrel of the gun.  

 

TTFN

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RE: ôMuzzle velocityö

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 .

RE: ôMuzzle velocityö

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.    

RE: ôMuzzle velocityö

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.

RE: ôMuzzle velocityö

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.  

peace
Fe

RE: ôMuzzle velocityö

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

RE: ôMuzzle velocityö

(OP)
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.
 

RE: ôMuzzle velocityö

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

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RE: ôMuzzle velocityö

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.

RE: ôMuzzle velocityö

"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

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RE: ôMuzzle velocityö

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

 

Regards,

SNORGY.

RE: ôMuzzle velocityö

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."

RE: ôMuzzle velocityö

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

RE: ôMuzzle velocityö

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

RE: ôMuzzle velocityö

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

RE: ôMuzzle velocityö

rb1957 did in the third post.

 

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

RE: ôMuzzle velocityö

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

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RE: ôMuzzle velocityö

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

RE: ôMuzzle velocityö

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

RE: ôMuzzle velocityö

5 ms will equate to a 1.98 inch drop.  It's still 33 ft/sec regardless of forward momentum.

RE: ôMuzzle velocityö

It's actually 32.174 ft/sec2, but I'll give you a pass on both the value and the units.  

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

Norsk:
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

Yep, a little too quick punching the keys.  Feeling kinda stupid right now :>(

RE: ôMuzzle velocityö

IRstuff:

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.  smile

-handleman, CSWP (The new, easy test)

RE: ôMuzzle velocityö

When it comes to hitting a target, it's all about windage. elevation, charge and holding the weapon steady on target. Unless of course you're using a projectile with a built in thrust & guidance system.  

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

Former highpower rifle competitor and handloader of ammunition here.  Couple of points:

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

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