Physics of shooting a rifle

[mehavey]

...the bullet does not feel the movement of the volume of air it is in.

Huh?
Unless the air and the bullet are equilibrium as far as relative motion, both perceive the motion of the other.

In the case of forward motion, equilibrium never happens
In the case of lateral deflection, equilibrium would only happen if/when the bullet finally assumes the same lateral speed as the lateral wind speed vector.
(which for practical purposes, is again never)


(Am I missing something here?)

 

[davidsog]

(Am I missing something here?)

Yes. A background in Aerodynamics, LOL.

It is dismaying to find that mere wind - so flimsy, so intangible a
thing - can blow a heavy powerful machine about at will; but it is
even more dismaying to discover, by and by, that the effects of wind
are almost exactly what your common sense would not expect them
to be.

You almost let yourself see what is going on once the bullet is in the air here:

Unless the air and the bullet are equilibrium as far as relative motion, both perceive the motion of the other.

The bullet as a creature of the air is all about equilibrium to the Relative Wind. Since its thrust force is finite it cannot achieve equilibrium and will eventually hit the ground because of this but that does not keep its behavior from seeking that equilibrium. Once could say the only force the bullet feels is drag on the axis of the Relative Wind. That is present with or without the wind we perceive from the ground reference.

Just like an airplane does not feel "turbulence" as it is a torque on the CG and not a positional jump...the bullet is the same.

All aerodynamic forces and torques on the bullet as it flies through the air are caused by the velocity of the bullet relative to the air, that is, Vbullet relative to the air.

 

[davidsog]

Because there is no net force acting on an object in equilibrium, then from Newton's first law of motion, an object at rest will stay at rest, and an object in motion will stay in motion.

https://www.grc.nasa.gov/www/k-12/airplane/equilib.html

A very basic concept when dealing with forces is the idea of equilibrium or balance. In general, an object can be acted on by several forces at the same time. A force is a vector quantity which means that it has both a magnitude and a direction associated with it. Two forces with the same magnitude but different directions are not equal forces. The vector sum of all of the forces acting on a body is a single force called the net force . If the net force is equal to zero, the object is said to be in equilibrium. Because there is no net force acting on an object in equilibrium, then from Newton's first law of motion, the object continues to move at a constant speed.

https://www.grc.nasa.gov/WWW/k-12/airplane/equilib3.html

With a bullet, the only axis not in equilibrium is the drag axis. All other axis achieve equilibrium. Aerodynamically, it is exactly like a glider shot into the sky with a rubber band. Granted, a glider with very poor wings and powerful rubber band but the physics is the exact same.

 

[Unclenick]

We need to be careful about describing the bullet as moving together with the volume of air. If it did that, wind deflection would be calculated as the time of flight times the wind speed. Bullets don't deflect nearly that far. Instead, the fact there is constant drag on the bullet throughout its trip to the target shows it has a significant speed differential with respect to the air. It does, however, deflect in the direction of the air. In that sense, its slowing speed is a rearward deflection, of which wind deflection is a vector component.

 

[tangolima]

We need to be careful about describing the bullet as moving together with the volume of air. If it did that, wind deflection would be calculated as the time of flight times the wind speed. Bullets don't deflect nearly that far. Instead, the fact there is constant drag on the bullet throughout its trip to the target shows it has a significant speed differential with respect to the air. It does, however, deflect in the direction of the air. In that sense, its slowing speed is a rearward deflection, of which wind deflection is a vector component.

I concur.

But Mr Sog believes we have entered a brand new era that all the classical results in ballistics have become obselete.

-TL

Sent from my SM-N960U using Tapatalk

 

[davidsog]

We need to be careful about describing the bullet as moving together with the volume of air.

Mmm
Engineers at Sierra beg to differ...
They seem very comfortable describing the flight of a bullet in aerodynamic terms.

The following vector equation is the beginning point for every analysis of wind deflections:
Vbullet relative to the ground = Vbullet relative to the air + Vair relative to the ground

If it did that, wind deflection would be calculated as the time of flight times the wind speed. Bullets don't deflect nearly that far.

No, A bullet is spin stabilized and gyroscopic procession is a factor. It is a part of the forces the bullets seeks equilibrium at in the air.

The bullet turns because of its gyroscopic stabilization. When the bullet exits the muzzle, there is an initial misalignment between the bullet axis and Vbullet relative to the air. This misalignment disappears quickly due to the gyroscopic stabilization, and the bullet takes the “average” angular orientation.

That average angle formed by the forces of the aerodynamic forces and the gyroscopic forces is why you cannot directly just add the wind component. That does not mean the bullet does not travel at that average angle like any other object in the air and that WITH the volume of air it resides.

Once the bullet is orientated into the Relative Wind, it acts like any other aerial body and does not feel the wind. It moves with that Volume of Air.

But Mr Sog believes we have entered a brand new era that all the classical results in ballistics have become obselete.

Little Catty aren't we as I never said that. I said the traditional math was developed when we did not have the understanding of aerodynamics that we do today.
G series calculations were literally developed in 1881. The last improvement was the Sierra Bullet Company contribution to change the shape of the reference bullet to a more modern boat tail spritzer.

That does not mean that math does not deliver a good approximation of where the bullet will be in its flight path.

Those calculations tell you nothing about the actual forces acting on the bullet in flight or the behavior of that bullet.

That is a fact and it has nothing to with me.

 

[tangolima]

I apologize to have your feelings hurt.

Aerodynamics sounds cool, but it is not that modern, other than the computers that do the numerical analysis. The ballisticians use the same computers too. Some of the famous ballisticians were involved in recent rocket development. They must also have "aerodynamic background".

Some equations in classical ballistics are indeed empirical. The famous Greenhill equation on twist rate is one of those. But the wind deflection equation is not. It is solution to set of differential equations including forces in all 3 dimensions. Certainly it didn't include all possible factors, but its accuracy has been proven by real measurements.

W=Vw*(TOF-D/Vo)

and

W=Vw*TOF

cannot be possibly both correct. One has been proven and one has been ... talked about.

-TL




Sent from my SM-N960U using Tapatalk

 

[davidsog]

I apologize to have your feelings hurt.

You have not hurt my feelings at all. I am not the one who acted Catty, LOL.

W=Vw*TOF

This is your made up formula not mine. What you are telling me is that if I am flying in my airplane and I input some rudder to apply a sideforce on the airplane such that W = Vx*TOF no longer works out....

The physics of aerodynamics no longer applies. That is silly.

One has been proven

Sierra engineers proved it nicely and they have measured it. It is not my fault you don't understand it and have chosen to attack me.

You come from a long of folks who have the same reaction to things they are not grasping.

https://origins.osu.edu/milestones/...ited-copernicanism?language_content_entity=en

Unfortunately, I am not in any their leagues nor am I the founder of new theory. I just can read and understand what the engineers at Sierra are saying because my degree is in the field of Aerodynamics.

Did you read the Sierra Engineers explanation I posted?

https://thefiringline.com/forums/showpost.php?p=6962038&postcount=218

Please download it and read it.

 

[davidsog]

I will try one more time to explain what is going on with a bullet in the frame of reference of the air.

Frankly, making this personal is just low-IQ and I harbor no ill will towards anyone over this but instead have found the discussion an interesting problem.

Some basic concepts that there seems to be confusion about is the rotational movement about the CG any object in flight experiences.

Hypothetically, we can visualize what is going on by pretending we have the strength to go out to an airport and lift a 4000lbs airplane over our heads. We will lift it up directly underneath its Center of Gravity.

If we pick that airplane up and rotate it about that CG, raising the nose while lowering the tail and wagging the wings from side to side, it will require us to input FORCE to make those rotational movements.

We have now proven that airplane has the ability to absorb FORCE thru rotational movement about the CG.

It is foundational principle of aerodynamics that as long as an object rotates about its CG, it moves as a creature of the air WITH the air mass it is traveling in. That object does NOT feel the wind anymore than we feel the movement of the Earth as it hurdles thru space. That is a basic concept and starting point for our understanding of what is going on with our bullet.

If we push hard upon our airplane we are holding up and lift the entire craft up a few more inches that too requires FORCE. We have exceeded the Force the object can absorb and that represents a positional change in location of the CG.

To an creature of the air, that represents a LARGE magnitude change in the properties of air that it is traveling in such as an actual shear. When that positional change occurs, the object feels the effect just as you would feel the effect of the Earth suddenly stopping all motion.

Now, our bullet seems very easy to rotate it about its CG as it has very little mass and nothing but gravity acting upon it as we take it out of box. However, that bullet has massive amounts of force imparted upon it by the chemical reaction of the propellent.

F = ma

Our bullet has a lot of force to absorb using rotational moments as opposed to having to make a positional jump.

That is what the engineers at Sierra relate. There is NO positional jump of the CG during the bullets flight. It rotates about the CG picking an angle to equalize the force acting upon it from the 3 axis it can stabilized and that is its Relative Wind. Once it has established that, it moves as creature of the air and does not feel movement of the air mass it is traveling in but rather moves with it. In order for it "feel" the movement of that airmass, we would have to impart a LARGE magnitude change for the CG to change position.

 

[rickyrick]

Read the last few pages and the first 3 pages, but I didn’t find the question that I have. It may be somewhere in between or I scanned over it.

Anyway, I’ve always wondered when recoil actually starts.
I’ve watched many high-speed films and it seems as though the rifle doesn’t start to move until after the bullet leaves the barrel, I don’t know if this perception is correct or not.
I know it’s a change of subject from what the latest posts are discussing. I feel that it’s a simple answer from someone who knows.

 

[tangolima]

Read the last few pages and the first 3 pages, but I didn’t find the question that I have. It may be somewhere in between or I scanned over it.

Anyway, I’ve always wondered when recoil actually starts.
I’ve watched many high-speed films and it seems as though the rifle doesn’t start to move until after the bullet leaves the barrel, I don’t know if this perception is correct or not.
I know it’s a change of subject from what the latest posts are discussing. I feel that it’s a simple answer from someone who knows.

The rifle starts to accelerate backwards as soon as the bullet accelerates forward in the bore. That is recoil.

However, if we regard recoil as the force felt on shooter's shoulder, that becomes a it-depends story.

-TL

Sent from my SM-N960U using Tapatalk

 

[davidsog]

Interesting Question.

This US Army Study on Recoil says:

The impulse caused by the departure of the projectile is reasonably straightforward—it is the momentum of the projectile at shot exit.

Which would seem to say that recoil is not realized until the bullet exits the barrel. Now, that maybe just a starting point for the math as I would think the forces are imparted upon the weapon when the pressure begins to increase and the bullet begins to move.

In fact, the report goes on to discuss pendulum testing and reinforces the concept recoil impulse begins at the muzzle:

The impulse due to firing at a constant temperature is invariant given a cartridge and a defined weapon. The surest way to ascertain the value of the impulse is via ballistic pendulum tests. These tests will reveal the total impulse but not the rate at which impulse is generated (actually, the force as a function of time). In the case of a muzzle loader or hand-operated, bolt-action weapon, the impulse is delivered over a short time, on the order of a few factors times the time it takes the bullet to transit the bore of the weapon; this is, perhaps, subtly modified by the stock of the weapon and the material used to construct the stock (e.g., polymers are notoriously rate-sensitive materials). Everything changes when semiautomatic (one shot per pull of the trigger), automatic (continuous fire at a cyclic rate until the trigger is released or the weapon expends its ammunition), or recoil mechanisms are introduced. In the case of a gas-operated weapon, the bolt is unlocked when the bullet is close to reaching the muzzle, and the bolt is thrust rearward by the propellant gases against the action of a spring. The impulse generated before the bolt is unlocked passes through the bolt to the receiver and stock of the weapon. Some, if not all of the impulse generated after the bolt starts its rearward motion goes into the bolt and gets transmitted to the receiver and stock at a more leisurely pace. As a result, the impulse is spread over a longer time, and the average force associated with the impulse is reduced. Further, in some weapons, a buffer (spring/damper element) is put in series with the bolt spring to smooth out the impulse delivery. In a blow-back automatic weapon (the bolt is not locked, but reacts inertially against a spring caused by the pressure at the base of the cartridge case), virtually all the projectile momentum and much of the propellant gas momentum is transmitted to the motion of the bolt, dramatically altering the rate at which impulse is transmitted to the receiver and stock of the weapon.

 

[davidsog]

However, if we regard recoil as the force felt on shooter's shoulder, that becomes a it-depends story.

Different frame of reference so that makes sense. The US Army Study is about force felt at the shoulder and the impulse as a factor of time.

 

[JohnKSa]

Now, that maybe just a starting point for the math as I would think the forces are imparted upon the weapon when the pressure begins to increase and the bullet begins to move.

Correct. There is high-speed footage available that demonstrates conclusively that recoil begins the same instant that the bullet begins to move--certainly before it exits the muzzle.

That's also consistent with the basic laws of motion.

The total recoil force is based on the ejecta momentum--the total of the momentum of the projectile and any gases or unburned propellant that exits the muzzle.

For practical purposes, it's generally not necessary to worry about what recoil is doing before the projectile exits, but in some cases it can be very important.

 

[davidsog]

Correct. There is high-speed footage available that demonstrates conclusively that recoil begins the same instant that the bullet begins to move--certainly before it exits the muzzle.

That's also consistent with the basic laws of motion.

The total recoil force is based on the ejecta momentum--the total of the momentum of the projectile and any gases or unburned propellant that exits the muzzle.

For practical purposes, it's generally not necessary to worry about what recoil is doing before the projectile exits, but in some cases it can be very important.

Exactly. External vs Internal Ballistics with the math picking External Frame of Reference to calculate the forces.

Exactly like the concepts of Aircraft Performance in Aerodynamics and Ballistic Calculations of Trajectory. It does not tell you exactly what is going on but because Energy cannot be created or Destroyed still delivers a good approximation of where the object will be...

 

[rickyrick]

For practical purposes, it's generally not necessary to worry about what recoil is doing before the projectile exits, but in some cases it can be very important.

I seem to recall now that changing bullet weight in a handgun can cause a POI shift, up or down, because the gun has recoiled more before the bullet has left the muzzle. For example, heavier bullet, the barrel is angled a little more up because of the tendency of a handgun to rotate about its center of gravity. Not sure how correct my thinking on this is.
That’s assuming the shooter is very consistent in aiming I’m guessing.
I did take physics classes in college, but that was long ago

 

[JohnKSa]

It depends on the gun. In a revolver, or a handgun where the barrel is fixed to the frame, the recoil pushes the gun back and the shooter's hand, which is generally lower than the barrel, tries to stop the push. That results in the gun torquing upwards and the muzzle will rise while the bullet is still in the barrel as a result of the recoil.

This can be seen by taking a long barreled centerfire revolver and setting a yardstick across the sights and a dowell into the barrel. It's easy to see that the dowell angles downward compared to the sights. A little thought will reveal that if the barrel really is pointing lower than the sights, when the bullet exits the muzzle, the bullet would never be able to do anything other than to hit lower on the target than the sights showed.

What's going on is that the sights are compensated properly for the amount of barrel rise due to recoil before the bullet exits.

On the other hand, if the slide/barrel can initially recoil freely, as in many centerfire autopistols, then they can move nearly straight back while the bullet is in the bore. Eventually the barrel hits something and transfers momentum to the frame of the gun and then muzzle rise will occur, but, by design, that usually happens after the bullet is out of the bore. As long as the barrel/slide can recoil straight back (resisted only by a relatively weak spring) there's very little muzzle rise. Again, this can be seen with the yardstick and dowell. There's a marked difference between what will be seen with the autopistol and the revolver when the same test is run on both.

Here's the result of a test I did with a 9mm autopistol (non-fixed barrel) and a Ruger GP100.

The bore line clearly angles downward compared to the sight line for the revolver but the bore line and sight line are nearly parallel in the autopistol.

 

[mehavey]

However, countless SlowMo films of auto-pistols show bullets leave the muzzle before the slide ever moves/unlocks.
https://www.youtube.com/watch?v=HElQk2wEY5w

so there must be something else at work here.

 

[JohnKSa]

That's partially correct.

The bullet does typically leave the muzzle before the barrel unlocks--by design. However, if you find sufficiently clear high-speed footage, it is apparent that recoil (barrel/slide movement) begins at the instant that the bullet starts moving. The amount of motion is small and it happens during a very short interval, so it's not apparent in some videos.

Here's one excellent example where it is clear.

https://www.youtube.com/watch?v=kq-xcqs5NIk&t=133s

It shows the muzzle of a 1911 pistol starting at 2:13 in the video. The slide and barrel are definitely moving before the bullet exits. That's very comforting since that's what physics tells us must happen.

Actually in your video link, if you look at the closeup of the muzzle in the top frame at around 1:35 it is possible to see the slide/barrel move before the bullet exits. You may have to slow the viewing speed down to 0.25x real time to see it clearly, but it's there.

Doing the same thing at about 2:35 in your link will also show that the slide is moving before the bullet leaves the barrel.

It's not as clear as it is in the video link I provided, but it is visible. The places where it's visible in your video are when the frame rate is at 25,000fps. The slower frame rates are too slow to show the effect.

 

[rickyrick]

Yes, the logic indicates that the recoil starts when the bullet starts moving but most footage can lead to misconceptions.
I found the clip with the AR bolt interesting too, it clearly shows two impulses: a small one before the the bolt is fully to the rear and another larger one when everything “bottoms out.”

Thank you for taking the time to reply.