Calculating rifle movement as bullet reaches muzzle

[wwbrown]

I’ll throw another piece of spaghetti at this wall and see if it sticks.

When we ignite the combustible in the cartridge, why does the rifle as a whole, not attempt to move down or up, left or right? Inarguably the gas is expanding in all those directions, right?

The forces the gases put on the chamber perpendicular to the bore (transverse forces) all cancel out because they are all of equal size and are radially outward in direction. In other words say there is a force of X in the upward direction there is a force of X in the downward direction and they cancel each other out and the same happens in every radial direction.

It only attempts to move the gun in the direction opposite of the nozzle (muzzle). This is not a minor detail.

So what if it didn’t have a vent? Is it still going to move rearward? If we cap off a jet engine, is it going to provide any thrust whatsoever, even a tiny amount? Or is it just going to increase internal pressure that constantly attempts to equalize itself in all directions?

If the expanding gas is unvented and it exceeds the ability of the vessel to contain that pressure it will leak out in a catastrophic fashion (Boom!) and if you add up the momentum of all of the pieces as vectors the sum will be zero.

So that isolates gas expansion and pressure as being directionally net neutral. There are only two components at play, gas and the bullet.

The other part of this is a moving solid. So, does the weight of part of the “system” traversing the system end to end cause dislocation?

If under the power of that part of the system I’d say yes. I cannot walk across a platform on ball bearings without pushing against it. To traverse it, I will invariably move the platform the opposite direction, somewhat.

If the platform had electromagnets affixed that propelled a levitating mass, then yes, again. But a rifle doesn’t propel the bullet, and neither does the bullet propel itself. Rather, this would be like inserting a zerk fitting between the bullet and the bolt and slowly introducing oil pressure. Do they both move? I don’t know, but as the force isn’t being applied by either the gun or the bullet, I don’t think it necessarily* invokes the equal and opposite action, principle.

Tons and tons of water careening down a water slide under the influence of gravity do not “push” the physical structure in the opposite direction of the water’s motion. A 000 buck ball in a barrel that is tilted down and starts rolling toward the end does not push the barrel backward whatsoever, so there is nothing inherent about rearward motion in a bullet moving forward.

A bowling ball rolled down a lane is not under its own power, it’s exerting force down, and slightly forward in the direction it’s going as the lane slows it down. Certainly no equal and opposite force, which is often applied generally here.

Combine the two concepts, - a drag racing car attempts to curl the asphalt under and behind the rear tires, to the rear. If I extend a winch’s hook to the car and pull it with a cable, the car exerts no lateral pressure on the asphalt. ** if we pull a bullet out of the barrel from the front, the gun isn’t going to try to move backward, I don’t believe, under the equal and opposite principle.

Obviously the bullet is not self powered. The gas, as explained, seeks out lower pressure with the net effect of neutrality in moving the object containing it in any particular direction. The bullet is simply acting as an enlargement of the expansion chamber.

At all relevant times to the question, pressure is contained by the gun, so where am i missing something here? Where does the net rearward force come from when there is no “vent” until the back of the bullet passes the muzzle?

The bullet has momentum when it travelling down the barrel which will create the gun to have momentum in the opposite direction, the bullet causes recoil before it leaves the barrel. I am ignoring the momentum of the gas.

 

[jelenko]

The change in distance is then half the barrel length minus half the height of the initial powder column.

You've got to stop thinking that you can eliminate time because it's the same "dimension".
If we were able to do that, ALL the terms would drop out.

Momentum is mass times velocity; momentum is conserved.
Mass times distance is not conserved.

 

[wwbrown]

You've got to stop thinking that you can eliminate time because it's the same "dimension".
If we were able to do that, ALL the terms would drop out.

Momentum is mass times velocity; momentum is conserved.
Mass times distance is not conserved.

Well stated!

 

[davidjoe]

The forces the gases put on the chamber perpendicular to the bore (transverse forces) all cancel out because they are all of equal size and are radially outward in direction. In other words say there is a force of X in the upward direction there is a force of X in the downward direction and they cancel each other out and the same happens in every radial direction.

If the expanding gas is unvented and it exceeds the ability of the vessel to contain that pressure it will leak out in a catastrophic fashion (Boom!) and if you add up the momentum of all of the pieces as vectors the sum will be zero.

The bullet has momentum when it travelling down the barrel which will create the gun to have momentum in the opposite direction, the bullet causes recoil before it leaves the barrel. I am ignoring the momentum of the gas.

Keep reading down the posts. I gave example of where a moving part of the system with momentum imparts no equal and opposite reaction to the remainder of the system, offering an explanation of why in those examples. The bowling ball, as one is actually trying to push the lane with it, forward, not backward, as the lane tries to slow it down. When the bowler released it, he went backwards, a small amount.

Gas: All expansive forces in other directions do cancel out, which is my point, the vent is the only difference.

Let’s look at a recoiless rifle fired from the shoulder. It has two vents, and just sits there when fired. The round has momentum, and is an example of momentum not necessarily transferring to the gun.

So, if it’s not momentum, and if my last post of the evening shows gas does the same work on both directions for a net neutral push on the barrel, then another explanation is needed. Possibly the explanation lies in an honest look at all the videos where the gun doesn’t move, and try to explain that.

 

[Varminterror]

When we ignite the combustible in the cartridge, why does the rifle as a whole, not attempt to move down or up, left or right? Inarguably the gas is expanding in all those directions, right?

Functionally and literally, NO. The gas is not expanding in all directions, because it is bound by the vessel - the case and chamber. Quite simply, the force (pressure exerted over a unit area) pushing upward on the case wall and rifle chamber by the pressurizing (not at this point expanding) gas is canceled over the scope of the entire rifle by the equal force pushing downward on the case wall and rifle chamber. In exactly the same way a pressurized air bubble simple sits idle - there is no net force accelerating the entire rifle in any particular direction without the vent, in other words, without something MOVING due to the force.

Certainly, we can calculate the stretch of the rifle barrel and action due to the increased internal pressure during the firing cycle, but we’re not talking about anything near the magnitude of the movement caused by the dynamic portion of the system which is the motive direction of that force moving the mass of the bullet.

 

[TaperPin]

If we estimate the rifle movement based only on the final velocity of the bullet, factoring its distance traveled as the barrel length less the nominal seated base position, then we have grossly underestimated the total dwell time.

I agree, if final velocity is used, it wouldn’t give us good results.

What I’m proposing is using an algebra trick so we don’t need to know velocity at all.

If momentum is mass x velocity and the gun is going one direction while the bullet/powder goes the opposite direction then:

mass(gun) x velocity(gun) = mass(bullet) x velocity(bullet) + mass(powder) x velocity(powder)

Before we put any numbers in this equation, just an algebra trick, anything that is equal on both sides of the equation that can be factored out goes away. Velocity is distance/time and while the distance moved of the gun and bullet/powder are different, the amount of time is the same, so velocity is simplified to only distance.

 

[Varminterror]

Let’s look at a recoiless rifle fired from the shoulder. It has two vents, and just sits there when fired. The round has momentum, and is an example of momentum not necessarily transferring to the gun.

Momentum is absolutely conserved in a recoilless rifle, simply the momentum is “transferred” (not apt, but we’ll use it here) to the static air mass around the rear vent AND to the shoulder of the rear breach vent. Plainly, we have more than two bodies and more than 2 directional forces acting in a recoilless rifle, but without question, momentum is conserved.

 

[Varminterror]

What I’m proposing is using an algebra trick so we don’t need to know velocity at all.

The fact this “algebra trick” does not work is the very foundation for why the differential calculus required to solve this system was invented.

What you have is largely a cat chasing its tail, and have caught it. To get an accurate barrel time, such as using Quickload, you must have used some integration of a transient system - your “time constant” you’re applying is using differential equations to solve for t.

In a manner of speaking, to have t, you already solved for dx/dt (the variable velocity), so what you’re doing is saying that you know 4-2=2 because you solved 2+2=4. It’s trivial, because the “trick” to simplify the math relies upon the solution of the higher math…

What you’re ending up with is an apparent ratio of relative displacements based on the ratio of the masses, however, because you used a constant for dt, you’re assigning linearity to dx/dt, which isn’t apt.

 

[davidjoe]

Functionally and literally, NO. The gas is not expanding in all directions, because it is bound by the vessel - the case and chamber. Quite simply, the force (pressure exerted over a unit area) pushing upward on the case wall and rifle chamber by the pressurizing (not at this point expanding) gas is canceled over the scope of the entire rifle by the equal force pushing downward on the case wall and rifle chamber. In exactly the same way a pressurized air bubble simple sits idle - there is no net force accelerating the entire rifle in any particular direction without the vent, in other words, without something MOVING due to the force.

Certainly, we can calculate the stretch of the rifle barrel and action due to the increased internal pressure during the firing cycle, but we’re not talking about anything near the magnitude of the movement caused by the dynamic portion of the system which is the motive direction of that force moving the mass of the bullet.

Exactly, no net force on the barrel absent a vent. I know the answer to that guys, it’s posed to set up and require that a harder question be answered about whether a pressure vessel moves or not, that is the real issue.

So, you guys follow the thinking on why this is so important to the question, right? The bullet cannot anticipate a vent, and start the system recoiling, before it gets there. If it indeed does make the rifle recoil before then, reconcile that conclusion with the hypothetical I posed of two opposing bullets of different weight, above, being pushed at different rates by increasing pressure between them, with no net force on the barrel left or right, because the same amount of work is being done on both sides of the pressure bubble.

This example might answer my whole thread from years ago, and may be the key to the whole thing.

This OP is a question independently mirrored from several years ago I posed, so it’s fresh on my mind, and so, then and now I broke it down on purpose into discreet logical steps to walk toward the answer of whether the barrel should move or not, prior to exit.

This is an approach taken to determine where agreement crosses over into disagreement, in analyzing a progression of events. Skipping straight to, “this particular gun” appears to move in a video, doesn’t advance the ball. There are videos of heavy guns that clearly do not, where movement after exit, is discernible, so it’s not as if they cannot move.

If it does recoil first, then there is an explanation. If there is an explanation, one of us should be able to find it, or deduce it.

 

[jelenko]

What I’m proposing is using an algebra trick so we don’t need to know velocity at all.

You didn't do well in physics, did you?

 

[dryhumor]

Expanding gas in a pressure vessel (barrel) overcomes the metal to metal contact, friction, and weight of the bullet. Air resistance ahead of the bullet is static, with little or no effect while in the barrel.

As the contact, friction, and mass of the bullet is overcome, the pressure still building, the two sidewalls of the bullet and barrel slip past each other. At peak pressure the strongest influence is available. 50k PSI can influence the mass of the bullet and the mass of the rifle.

As the pressure moves past peak, it’s direct influence lessons but does not dissipate. As the pressure curve flattens, however, momentum has influenced both the bullet beginning flight and the rifle rearward motion.

The bullet’s momentum ends at impact when momentum’s energy is dissipated. Recoil’s momentum ends when it is dissipated into the shoulder.

It appears, that the expanding gases as they leave a barrel still perform a linear influence on moving the rifles mass rearward. Adding a break, disrupts the linear flow and dissipates it radially, this removes the linear influence of the gases leaving the barrel and mitigate the mass of the rifle moving rearward.

That’s how my brain sees it anyway.

 

[BoydAllen]

We should all know that if we have a rifle positioned in bags such that a sling swivel stud touches one of the bags during recoil that the group will suffer. Would this happen if the rifle did not move before the bullet exited?

 

[xr650rRider]

Anyone that owns quickload, should open the help file and starting around page 90, there is a discussion of interior ballistics. You can also under "Options", "Output window settings" then click on "Progress of combustion" that shows a complete picture of the projectile distance, time and pressure while in the barrel.

 

[TaperPin]

You didn't do well in physics, did you?

I did take off too much time to go hunting. Lol

 

[TaperPin]

You've got to stop thinking that you can eliminate time because it's the same "dimension".
If we were able to do that, ALL the terms would drop out.

Momentum is mass times velocity; momentum is conserved.
Mass times distance is not conserved.

Stop me where the math is incorrect.

mass(a) x velocity(a) = mass(b) x velocity(b)

Do we at least agree this is the basic conservation of momentum equation?


mass(a) x velocity(a) = mass(b) x distance(b)/time(b)

Do you agree velocity is a distance/time?


time(b) x mass(a) x velocity(a) = mass(b) x distance(b)

Multiplying both sides by time(b) is algebraically correct?


time(b) x mass(a) x distance(a)/time(a) = mass(b) x distance(b)

Still algebraically ok if velocity(a) is expressed as a distance/time?


If we are only looking at a point in time, do you agree time(a) is the same as time(b) ?

When time(b) is divided by time(a) do we agree, no matter what value, the result is 1, because they are equal numbers, and thus are no longer a measure of time?

That leaves: mass(a) x distance(a) = mass(b) x distance(b)


There is obviously a specific time for any given barrel/bullet/powder combination to rush down the barrel, but the time doesn’t matter for the algebra manipulation of the conservation of momentum equation to get the distance.

 

[TaperPin]

You didn't do well in physics, did you?

I really have no desire to defend something if it’s incorrect - I’ll gladly use whatever someone else comes up with if it produces a good answer.

 

[CharlieNC]

Stop me where the math is incorrect.

mass(a) x velocity(a) = mass(b) x velocity(b)

Do we at least agree this is the basic conservation of momentum equation?


mass(a) x velocity(a) = mass(b) x distance(b)/time(b)

Do you agree velocity is a distance/time?


time(b) x mass(a) x velocity(a) = mass(b) x distance(b)

Multiplying both sides by time(b) is algebraically correct?


time(b) x mass(a) x distance(a)/time(a) = mass(b) x distance(b)

Still algebraically ok if velocity(a) is expressed as a distance/time?


If we are only looking at a point in time, do you agree time(a) is the same as time(b) ?

When time(b) is divided by time(a) do we agree, no matter what value, the result is 1, because they are equal numbers, and thus are no longer a measure of time?

That leaves: mass(a) x distance(a) = mass(b) x distance(b)


There is obviously a specific time for any given barrel/bullet/powder combination to rush down the barrel, but the time doesn’t matter for the algebra manipulation of the conservation of momentum equation to get the distance.

No, it's instantaneous velocity. Velocity is always changing so as previously mentioned differential equations are necessary to describe this. Example, the velocity of the bullet leaving the muzzle is not the barrel length divided by the barrel time as the behavior is like a drag racer with acceleration.

 

[Straightshooter1]

Stop me where the math is incorrect.

mass(a) x velocity(a) = mass(b) x velocity(b)

Do we at least agree this is the basic conservation of momentum equation?


mass(a) x velocity(a) = mass(b) x distance(b)/time(b)

Do you agree velocity is a distance/time?


time(b) x mass(a) x velocity(a) = mass(b) x distance(b)

Multiplying both sides by time(b) is algebraically correct?


time(b) x mass(a) x distance(a)/time(a) = mass(b) x distance(b)

Still algebraically ok if velocity(a) is expressed as a distance/time?


If we are only looking at a point in time, do you agree time(a) is the same as time(b) ?

When time(b) is divided by time(a) do we agree, no matter what value, the result is 1, because they are equal numbers, and thus are no longer a measure of time?

That leaves: mass(a) x distance(a) = mass(b) x distance(b)


There is obviously a specific time for any given barrel/bullet/powder combination to rush down the barrel, but the time doesn’t matter for the algebra manipulation of the conservation of momentum equation to get the distance.

Doesn't there still have to be a resistance, friction, gravity component for anything being held or being supported? Or did someone already cover that and I missed it?

 

[wwbrown]

Stop me where the math is incorrect.

mass(a) x velocity(a) = mass(b) x velocity(b)

Do we at least agree this is the basic conservation of momentum equation?


mass(a) x velocity(a) = mass(b) x distance(b)/time(b)

I assume in you equation time(b) is the elapsed time between the bullet just started moving to when it leaves the barrel, any other time is arbitrary since the bullet itself has no timer onboard that we can read directly. Your equation would only be correct if the speed of the bullet was constant during the elapsed time. When the speed is not constant calculus rears its ugly head and you need to integrate the equation of motion.

If you really want to go down the rabbit hole including time you would need to do an analysis similar to the following:


Long story short: Working in the time domain adds a complexity to the problem that is not pleasant that is why we stick to conservation of momentum when we lay out the problem.

Don't worry about the calculus aspect but believe me when I say your equation with t(b) is only true if the speed of the bullet is constant and I think we can all agree it is not. I can't make it any clearer than that.

Trust me people use to pay me a fair amount of money to do physics that was much harder than that (one area of work was internal ballistics although sometimes we all find we bit off more than we can chew). In addition, I taught much harder (and correct) physics for 11 years at the college level before venturing into the defense industry. These topics have been well known and demonstrated for about 350 years and I don't think you are going to change that body of knowledge.

TaperPin, step away from the keyboard as your insistence in using t(b) in your equation don't not foster the idea of your wise/keen knowledge of this subject.

 

[TaperPin]

I assume in you equation time(b) is the elapsed time between the bullet just started moving to when it leaves the barrel, any other time is arbitrary since the bullet itself has no timer onboard that we can read directly. Your equation would only be correct if the speed of the bullet was constant during the elapsed time. When the speed is not constant calculus rears its ugly head and you need to integrate the equation of motion.

If you really want to go down the rabbit hole including time you would need to do an analysis similar to the following:

View attachment 1501625
Long story short: Working in the time domain adds a complexity to the problem that is not pleasant that is why we stick to conservation of momentum when we lay out the problem.

Don't worry about the calculus aspect but believe me when I say your equation with t(b) is only true if the speed of the bullet is constant and I think we can all agree it is not. I can't make it any clearer than that.

Trust me people use to pay me a fair amount of money to do physics that was much harder than that (one area of work was internal ballistics although sometimes we all find we bit off more than we can chew). In addition, I taught much harder (and correct) physics for 11 years at the college level before venturing into the defense industry. These topics have been well known and demonstrated for about 350 years and I don't think you are going to change that body of knowledge.

TaperPin, step away from the keyboard as your insistence in using t(b) in your equation don't not foster the idea of your wise/keen knowledge of this subject.

Your credentials and experience are quite impressive. The calculus definitely looks like calculus. I’m sure thousands of your students are better off having taken your classes.

So riddle me this - how is this not a simple linear center of mass problem? You know, like a person walking in a boat - the boat moves only enough to keep the center of mass at the same place as the initial state? Until the person steps off the boat, or the bullet leaves the barrel these seem like simple internal systems.

What my use of that equation has done is simply quantified the movement of the mass of the bullet and mass of powder, and a movement of the mass of the rifle to retain the original center of mass, which answers the original question. It may not be the most elegant way to go about it, but it seems to work. No?

If I’m so far off base, living in never never land, why then does my method match Quickload to within .0028” of the screenshot in message #48?