Calculating rifle movement as bullet reaches muzzle

[wwbrown]

So, if that doesn’t apply to this situation, how would you propose to estimate the amount a rifle recoils at the time the bullet exits the muzzle?

Turning off reality isn’t an option in this case. Influences that are small can can be ignored as a practical matter, no?

Your use of time does not apply to the problem in the way you think it does. The conservation of momentum says that the momentum of the gun, bullet and powder (I will treat the powder in burnt and unburnt the same as it does not effect the analysis)

Before you pull the trigger we assume that the the momentum of the system is zero. So if the conservation of momentum is applied when you pull the trigger the momentum of the system is zero, objects with velocities in opposite directions will have momenta of opposite sign since they are vectors. At the instant the bullet and gases leave the barrel and before the rifle impacts the shoulder (free-recoil is easiest to picture) the momentum of the system. I say before it impacts your shoulder to keep the external forces at zero. The speed of the bullet is easy to measure I am not sure what the speed of the burnt powder is but I am pretty certain some or most of it is moving at a pretty good clip when it goes from the high pressure environment inside the barrel to the low pressure atmospheric pressure outside the barrel. I don't know what the speed is but I am pretty sure that not all the gas is moving at the same speed which complicates the situation. There are assumptions made about the gas speed in every recoil calculator.

The speed and mass of the powder needs to be included as the mass of the powder is anywhere from 20%-40% of the bullet's mass. If you don't think the burnt powder impacts the recoil tell me "How do muzzle brakes work?"

The amount of time between when the bullet left the case to when it leaves the barrel has no impact so using time directly in the equation as you tried is not physically sound. I hope my explanation above is clear I don't have any more time to devote to this right now (plus I imagine I have bored 90% of the readers if they even got this far) so if you still think your use of time is valid we will have to agree to disagree and move forward. Trust me when I say your use of time is incorrect, I have spent more time working in this area than many and have the physics background such that most would accept what I say in this area on the subject discussed.

 

[okie]

I'm pretty sure i must have missed the day at school where they went over this...

 

[jimmymac]

I'm pretty sure i must have missed the day at school where they went over this...

But at least you can read this thread when you get insomnia.

 

[davidjoe]

Your use of time does not apply to the problem in the way you think it does. The conservation of momentum says that the momentum of the gun, bullet and powder (I will treat the powder in burnt and unburnt the same as it does not effect the analysis)

Before you pull the trigger we assume that the the momentum of the system is zero. So if the conservation of momentum is applied when you pull the trigger the momentum of the system is zero, objects with velocities in opposite directions will have momenta of opposite sign since they are vectors. At the instant the bullet and gases leave the barrel and before the rifle impacts the shoulder (free-recoil is easiest to picture) the momentum of the system. I say before it impacts your shoulder to keep the external forces at zero. The speed of the bullet is easy to measure I am not sure what the speed of the burnt powder is but I am pretty certain some or most of it is moving at a pretty good clip when it goes from the high pressure environment inside the barrel to the low pressure atmospheric pressure outside the barrel. I don't know what the speed is but I am pretty sure that not all the gas is moving at the same speed which complicates the situation. There are assumptions made about the gas speed in every recoil calculator.

The speed and mass of the powder needs to be included as the mass of the powder is anywhere from 20%-40% of the bullet's mass. If you don't think the burnt powder impacts the recoil tell me "How do muzzle brakes work?"

The amount of time between when the bullet left the case to when it leaves the barrel has no impact so using time directly in the equation as you tried is not physically sound. I hope my explanation above is clear I don't have any more time to devote to this right now (plus I imagine I have bored 90% of the readers if they even got this far) so if you still think your use of time is valid we will have to agree to disagree and move forward. Trust me when I say your use of time is incorrect, I have spent more time working in this area than many and have the physics background such that most would accept what I say in this area on the subject discussed.

This is a bedeviling puzzle. It consumed a great deal of my thoughts regarding the old thread a couple of years ago, that I referenced, and several others too.

I like your thought process but it causes me to wonder if you would tend the conclude that until gas or the bullet is expelled, it is contained in a sealed system, and as such, its momentum is part of the system?

Imagine that instead of expanding gas, there is a compressed, free floating spring between the bullet and the bolt face, such that when the trigger is pulled, the spring pushes against both surfaces. As a spring, its energy gets equally divided and expended, between the bullet and the bolt face.

Further, imagine that the muzzle is capped and the bullet simply stops when it hits the cap. Naturally so does the spring.

In this scenario, does the gun recoil at all? Is that movement of a small component of the system, from one end to the other, “recoil” in the sense we understand if?

Years back I postulated that a grandfather clock on ball bearings will move slightly back and forth to counter to the pendulum. It will move the same amount regardless of how fast or slow the pendulum moves. In other words, if the pendulum was on oil dampened shocks, the movement would get to the same slight place, just not as quickly.

You are actually the first person to engage the thought I had, against great resistance, that time/speed is not part of this consideration.

Further, I suggested that if all of this is sound and true, or even some parts of it, then a magnum is NOT necessarily less accurate than a standard, by reason of “recoil” before the bullet leaves, being greater.

 

[jelenko]

So, if that doesn’t apply to this situation, how would you propose to estimate the amount a rifle recoils at the time the bullet exits the muzzle?

Turning off reality isn’t an option in this case. Influences that are small can can be ignored as a practical matter, no?

First, conservation of momentum applies. The issue is what is the 'system' under consideration. It's also important to recognize that, in this discussion, we're not trying to land a module safely on the moon. So, we can ignore or make simplifying assumptions - provided we don't expect them to affect the first order results.

All he's saying is that the actual conservation of momentum calc needs to consider all the forces.
As I remember, we started the discussion assuming the rifle was on a rest on a bench.

If the rifle is in your shoulder, and you wanted an 'accurate' answer, you'd need to model the shoulder - maybe in terms of a shock absorber or just a spring.

 

[jelenko]

Your use of time does not apply to the problem in the way you think it does. The conservation of momentum says that the momentum of the gun, bullet and powder (I will treat the powder in burnt and unburnt the same as it does not effect the analysis)

Before you pull the trigger we assume that the the momentum of the system is zero. So if the conservation of momentum is applied when you pull the trigger the momentum of the system is zero, objects with velocities in opposite directions will have momenta of opposite sign since they are vectors. At the instant the bullet and gases leave the barrel and before the rifle impacts the shoulder (free-recoil is easiest to picture) the momentum of the system. I say before it impacts your shoulder to keep the external forces at zero. The speed of the bullet is easy to measure I am not sure what the speed of the burnt powder is but I am pretty certain some or most of it is moving at a pretty good clip when it goes from the high pressure environment inside the barrel to the low pressure atmospheric pressure outside the barrel. I don't know what the speed is but I am pretty sure that not all the gas is moving at the same speed which complicates the situation. There are assumptions made about the gas speed in every recoil calculator.

The speed and mass of the powder needs to be included as the mass of the powder is anywhere from 20%-40% of the bullet's mass. If you don't think the burnt powder impacts the recoil tell me "How do muzzle brakes work?"

The amount of time between when the bullet left the case to when it leaves the barrel has no impact so using time directly in the equation as you tried is not physically sound. I hope my explanation above is clear I don't have any more time to devote to this right now (plus I imagine I have bored 90% of the readers if they even got this far) so if you still think your use of time is valid we will have to agree to disagree and move forward. Trust me when I say your use of time is incorrect, I have spent more time working in this area than many and have the physics background such that most would accept what I say in this area on the subject discussed.

While time doesn't enter the calculation for conservation of momentum, it would be needed to definitively answer the question: How far does the rifle move from a shot?

While I gave a possible approach to estimating that started with a conservation of momentum calc, I think the definitive answer requires an equation of motion that includes all the forces acting on the rifle. Once the equation of motion is built, you'd solve for x when dx/dt =0, yes?

ETA. I think I misused the word 'definitive'. Since we would need to model both the friction forces and the effect of the shoulder, a better word might be 'really good'.

 

[davidjoe]

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?

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?

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?

If you can isolate both variables, expanding gas and the bullet moving forward, as not inherently moving the rifle rearward, until there is a vent, then there needs to be an explanation why, together, they would move the rifle rearward prior to a vent, when they don’t do it, alone.

I’m not saying there isn’t an explanation, but no intrepid person has offered one, that I have ever read, and I don’t know it myself.

If I had to guess, I’d surmise that to an extremely small extent, a bullet going down a barrel is no longer part of a single bullet-gun sealed system, but two objects that have properties of both being connected and separate, until they are completely separated, that are both resisting something expanding between them, thus influencing each other.

 

[Infrequent Shooter]

There are indeed a ton of variables involved here; bullet weight, powder charge, rifle weight, stock shape, friction and more. All of this ignores shooter influence. Today, most of my shooting is off-hand, so the rifle is always moving (in my case, it's moving a lot! ). Back when I was shooting more BR, I always felt the biggest influence was what I did between the time that the trigger tripped and the rifle fired, rather than during the bullet's travel up the barrel. The truth was, everything mattered. I can remember one year when I started out, testing a new rifle, and was having a heck of a time with the occasional high shot. I finally figured out that I was relaxing at the moment the trigger tripped and letting the butt drop (I was squeezing the rear bag for elevation). My shooting style was such that I had to follow through consistently; sometimes, I didn't! Even if I did hold until ignition, I could still get a shot which was an eighth inch high. Bench technique has changed a little over the last forty years or so but I think consistency after the trigger breaks is still just as important; even more so when shooting rimfire or BPCR. WH

...the truth was, everything mattered... excellent

 

[Infrequent Shooter]

best site in town.

 

[Bottom Fodder]

Well I thought I knew how to shoot but after reading all this I have determined that I don’t and will need more college than I have before I can.

 

[Dgd6mm]

I have found pleasure in reading this thread from all participants.

 

[J_idaho]

Since google is not as much in the business of providing answers as much as selling products and keywords, it’s surprising difficult to find something as simple as how far a rifle moves before the bullet leaves the barrel. Where I’m going with this is just adding a good, even if rough, value to the movement in discussions - rather than explaining to a kid the gun moves an amount that nobody knows, it would be awesome to be able to say in free recoil it might move .0xx” and with a good two handed hold and shoulder pressure that changes it .00x”.

Engineers among us will find this quite simple, so humor me

From memory of physics 101 years ago (feels like 101 years ago), I’m tending to think it would be a simple conservation of momentum? Roughly, if powder expands equally, would the distance of the center of mass of the powder be about half the barrel length, ignoring volume of cartridge case above and beyond the bore diameter?

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

My gut says that’s the easy part and describes the rifle in free recoil at the moment the bullet exits. Has anyone seen, or wish to take an educated guess, as to what kind of range a two-handed hunter hold combined with pressure from the shoulder adds to the mass of the rifle, reducing movement?

Somewhere, someone has high speed camera footage of free recoil and different holds as the bullet comes out, but I’ll bet it’s been dumbed down into an average fudge factor in an equation in a footnote of a ballistics book.

Thanks for your thoughts! This is one of those things that’s been in the back of my mind for decades and I wait for something to come up, but I was either busy and didn’t notice the conversation, or it was too vague to be of much help.

 

[Straightshooter1]

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?

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?

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.

However, 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.

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 a 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?

If you can isolate both variables, expanding gas and the bullet moving forward, as not inherently moving the rifle rearward, until there is a vent, then there needs to be an explanation why, together, they would move the rifle rearward prior to a vent, when they don’t do it, alone.

I’m not saying there isn’t an explanation, but no intrepid person has offered one, that I have ever read, and I don’t know it myself.

If I had to guess, I’d surmise that to an extremely small extent, a bullet going down a barrel is no longer part of a single bullet-gun sealed system, but two objects that have properties of both being connected and separate, until they are completely separated, that are both resisting something expanding between them, thus influencing each other.

This was pretty much my thoughts regarding the expanding gases, not moving in any one direction, having equal pressure in all directions and its mass not really contributing anything inside the bore until is escapes at the muzzle. It's when the gases finally escape the muzzle that produces additional force to that which the projectile's movement produces and a break simply redirects some of that force as the projectile passes by inside the break. The projectile's force as it's being accelerated by the pressure creates an opposite force to that direction through the bore, then you have the gases exiting at the muzzle adding additional rearward force the same way a solid fueled rocket booster engine works.

The math is beyond me now, but think I still understand the principle physics at play. . .???

 

[davidjoe]

This was pretty much my thoughts regarding the expanding gases, not moving in any one direction, having equal pressure in all directions and its mass not really contributing anything inside the bore until is escapes at the muzzle. It's when the gases finally escape the muzzle that produces additional force to that which the projectile's movement produces and a break simply redirects some of that force as the projectile passes by inside the break. The projectile's force as it's being accelerated by the pressure creates an opposite force to that direction through the bore, then you have the gases exiting at the muzzle adding additional rearward force the same way a solid fueled rocket booster engine works.

The math is beyond me now, but think I still understand the principle physics at play. . .???

A twist in the question might help answer the questions. Again, I’m not hiding the ball with the right answer, this has bugged me off and on for years.

Let’s modify the bullet / bolt hypothetical. Let’s take a 50 inch barrel and put two bullets an inch apart in the very middle of the barrel, facing opposite directions.

Install the zerk fitting. But instead of exactly equal bullets, have the longest bearing surface bullet, and the shortest, facing opposite directions.

Slowly start applying oil or air pressure. My theory is that the shorter bearing bullet will creep forward and out the barrel before the longer bullet starts moving much or at all even though it is not fixed in place.

If the longer bullet holds its ground such that it doesn’t move, or only does a little, then we really don’t need an entire gun to move from absorbing an equal and opposite force of the bullet moving down the barrel. The barrel acts more like a mere pressure vessel with two ends that have different resistance, but let the same “work” be done.

If the longer bullet moved a tiny amount, but the same work was being done on each side of the zerk fitting, then the barrel would not need to move at all, right? It would just sit there perfectly still while one bullet moved out toward the end, much faster than the other one.

And you see where I’d go with this thinking, if a barrel would sit perfectly still, then why wouldn’t a whole gun? Work is undoubtedly being done on the gun end, just look what we do to brass.

 

[mauser284]

I am sure a lot of us grew up with slow motion once it was a thing in the 1970's and 1980's and especially today.

Most hunters after seing how an arrow moves once released think they have a handle on things but that sort of movement is toddler level. I do not say that to be insulting either.

I assume most guys on this sight have seen how ripples move out fromt he point of impact when one tosses a rock into a pound. Take that sort of ripple and expnasion of the wave now make that pond out of metall and rotate it 90 degrees to be perpendicular tot he origanal pond. Now extend that meteal pond so it is the length of a rifle barrel. When the cartridge is ignited along the path from the breach to the muzzle you get ripples that transverse the length of the barrel and the metal actualy expands in a progressive waveform from tip to butt but to tip .....LOL People think that the barrel is just wipping but that is not the case it is also expanding and contracting in a like manner. Imagine if you will visualy that you are pushing a Brazil Nut through something rigid but also flexible maybe a chunk of garden hose or a sausage casing sort of fixed at each end. As the nut travels the length of the material you see a bulg or wave front occur that rises and falls along the length.

I have not done themath is decades and I might be rembering this wrong but since the speed of sound for instance changes with density of the materil sound travels through a brass trumpet bell something like 7X faster than it travels through at at standard condition which in aviation a standard day is 59°F, Sea Level @29.92. ANyone that has delt with steel or metals in general undertands how they behave strangely at high speed and pressure impacts almost behaving like a liquid instead of solid when all that kinetic energy is dumped into them think of how cars behave saw when two of them hit at say 75MPH to 120MPH or if huge mass is involved like a tractor trailer going 55mph hitting a on coming car at 75MPH. It might split as if cut with a pair of giant shears or it might fold together and fuse as if it was liquid and suddenly became a sold or it might crumble in a very expected way. Even radial stress might be prsent but not possible to measure int he moment!

So more than you think is going on. That is even before you start thinking on a molecualr or quantum levels.

Even with super high speed X-Ray photography ther is more than meets the eye. Maybe when super high speed scalar wave scanning technology becomes available we might be able to see what we do not know!

 

[TaperPin]

Your use of time does not apply to the problem in the way you think it does. The conservation of momentum says that the momentum of the gun, bullet and powder (I will treat the powder in burnt and unburnt the same as it does not effect the analysis)

Before you pull the trigger we assume that the the momentum of the system is zero. So if the conservation of momentum is applied when you pull the trigger the momentum of the system is zero, objects with velocities in opposite directions will have momenta of opposite sign since they are vectors. At the instant the bullet and gases leave the barrel and before the rifle impacts the shoulder (free-recoil is easiest to picture) the momentum of the system. I say before it impacts your shoulder to keep the external forces at zero. The speed of the bullet is easy to measure I am not sure what the speed of the burnt powder is but I am pretty certain some or most of it is moving at a pretty good clip when it goes from the high pressure environment inside the barrel to the low pressure atmospheric pressure outside the barrel. I don't know what the speed is but I am pretty sure that not all the gas is moving at the same speed which complicates the situation. There are assumptions made about the gas speed in every recoil calculator.

The speed and mass of the powder needs to be included as the mass of the powder is anywhere from 20%-40% of the bullet's mass. If you don't think the burnt powder impacts the recoil tell me "How do muzzle brakes work?"

The amount of time between when the bullet left the case to when it leaves the barrel has no impact so using time directly in the equation as you tried is not physically sound. I hope my explanation above is clear I don't have any more time to devote to this right now (plus I imagine I have bored 90% of the readers if they even got this far) so if you still think your use of time is valid we will have to agree to disagree and move forward. Trust me when I say your use of time is incorrect, I have spent more time working in this area than many and have the physics background such that most would accept what I say in this area on the subject discussed.

I haven’t included time in any calculations. You insisted that velocity should be used and that consists of a distance/time. Perhaps I didn’t do a good job of defining the problem and that has caused our confusion.

The question we’re trying to answer is how much the rifle will move at the point the bullet is about to leave the barrel, not total movement or recoil forces before and after the bullet leaves.

I don’t think considering the velocity of escaping gasses is correct, because at the moment the bullet is about to exit the barrel no gases have escaped yet.

If the gases are assumed to be homogenous to simplify calculations, the center of mass of the unburned powder is the initial state. At the point where the base of the bullet is even with the crown of the barrel, the powder has essentially filled the bore evenly from end to end, so the center of mass is then the middle point of the barrel. The change in distance is then half the barrel length minus half the height of the initial powder column.

 

[mauser284]

This is also why I laugh at most peoples simpleton idea that a short action is stiffer than a long action as it relates to bolt action rifles. The action is either stiff at the point of the cartridge chamber or it is not. This is especially true of rifles with locking luggs up front. If the action is stiff at the point of lock up nothing behind that point matters at all. Now if the locking lugs are at the rear of the bolt everything up to the point of lock up matters and then length would come into play. No one uses locking lugs today at the rear of the bolt because it makes it harder but not impossible to extract maximum accuracy. It can still be done but why make life diffacult if you have a choice?

Scientific reality versus 99% of peoples perception of reality or the world completly different things. The problem is that stuipd people and ignorant people think they are smart when infact most of them are not.

To many people are stuck in Euclidian 2D models and old school analog ways of thinking but the world is a digitial world. Just like people get stuck in Einsteins world view or Newtonian world view. The got us to this point but now they are holding everyone back. You need to instead look at Maxwell and Tesla and expand on it and take Euclidian math to at the very least a 3D model.

Just some things to reframe thoughts.

 

[Straightshooter1]

Just to lighten things up a bit :

 

[Straightshooter1]

The question we’re trying to answer is how much the rifle will move at the point the bullet is about to leave the barrel, not total movement or recoil forces before and after the bullet leaves.

I don’t think considering the velocity of escaping gasses is correct, because at the moment the bullet is about to exit the barrel no gases have escaped yet.

I agree that the escaping gasses is not a consideration as I was essentially just addressing this along with the idea of powder mass moving within the bore before leaving the muzzle:

The speed and mass of the powder needs to be included as the mass of the powder is anywhere from 20%-40% of the bullet's mass. If you don't think the burnt powder impacts the recoil tell me "How do muzzle brakes work?"

 

[Varminterror]

The value for time on both sides of the momentum equation are equal and can be taken out, leaving mass times distance.

With all due respect intended: This really doesn’t work, because what we have as the conservation of momentum, isn’t including a constant velocity. 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.

It has been mentioned - simulating a Quickload estimate for in-bore velocity, we can get a velocity curve for the rifle, and it’s relative instantaneous acceleration curve, and resultingly, we can integrate its position in time, with a calculated velocity at the end of the bullet dwell time. Free recoil energy and free recoil velocity obviously also work to calculate final rifle velocity, but again, that will grossly underestimate the recoil distance, as it assumes too short of dwell time (it assumes the final velocity is the mean, constant velocity).