Physics of shooting a rifle

[stagpanther]

He still can see only the ground,
not the air. His sense of balance, his sense of motion, his nervous
system, tend to react simply and naively to his observed motion over
the ground, whether he wills it so or not. But, since his motion over
the ground includes both motion with the air and motion through the
air, his resulting control actions are bound to be wrong.

I can give a perfect real world example of this.

When flying a glider close to the surface on a mountain ridge the glider pilot often turns back into the ridge in order to work the the uplifted air next to the ridge. on the leg flying towards the ridge, the glider's airspeed will remain unchanged (unless the glider pilot deliberately changes it) but because the glider is flying with a tailwind, the pilot's reference to the ground rushing by can trick him into believing he is flying very fast, and because the ground reference is moving by fast it seems like the control input to roll the glider is having no effect. In reality, the glider is rolling and turning just as it would into the wind, but because the terrain is looming fast and a collision is imminent if the turn isn't completed, the pilot can erroneously increase pitch believing that it will slow the glider down relative to the ground and better coordinate the turn. That often just makes matters worse because this can put the glider or the glider's wing into an incipient stall. I've known pilots who have died because of this.

[davidsog]

I can give a perfect real world example of this.

Great example.

If you look at the math of the ballistic coefficient currently used to explain bullet behavior it is easy to see the enormous shortcomings of the system.

BC is the mathematical equivalent of hand tools and levers. That is not to say you cannot do great things with hand tools and levers, just look at the engineering works of Rome.

BC is accurate but not precise. As our wind tunnel technology advances into the realm of compressible aerodynamics and our Computational Fluid Dynamics models become more accurate I think BC will go the way of the sundial in predicting bullet behavior as CFD based computations will be more accurate and precise.

[davidsog]

You can see the concepts we discussed in this solver:

You can just run this in a solver. I did it for the Lapua match bullet (an email from Lapua confirms all their match ammo uses the same bullet except their Polar Biathlon ammunition, which has its own design with a tapered shoulder that I want to try). Basically, a 10 mph headwind or tailwind, as compared to zero wind, makes no practical difference at 100 yards, about 0.25 moa at 200 yards, about 3/4 moa at 300 yards, and about 1.5 moa at 400 yards. These are all elevation adjustments up for a headwind and down for a tailwind. Note that I used the same 1070 fps MV I used in the previous post. Most Lapua match is rated at 1073 fps MV, except their Polar Biathlon, and the new Long Range and Super Long Range that you can't get yet are rated at 1106 fps MV. The faster MV Long Range cartridges increase the drop differences by about 10.5% because drag is higher at 1106 MV.

https://thefiringline.com/forums/sho...&postcount=105

I am going to borrow Unclenicks excellent illustration:



The path of the bullet thru the air is the same regardless of wind. In this case, we are using a "zero value" headwind and tailwind which nicely illustrates the fact the bullet does not feel the wind in flight.

At 400 yards our bullet drop is 354.9 inches. In a zero wind situation if we move the target close along the ground our bullet drop will be less. Subsequently if we move the target farther away along the ground in a zero wind situation, our bullet drop will increase.

Since the bullet does not feel the wind but is inside a moving volume of air the movement of that air will effect the impact of an object that is not moving with that air and on the ground. We have a "Compound Frame of Reference" problem.

Our headwind is a volume of air moving AWAY from the target. The bullet moves with that air experiencing the same aerodynamic forces it felt in the zero wind situation. Just like the passenger on a train, the bullet moves with the volume of air that is moving away from a point relative to the ground that is the target. We see the result is a larger value of bullet drop at 361.1 inches. We would get the same effect in the zero wind situation of we moved the target on the ground farther away.

In the tail wind situation our volume of air is moving TOWARDS the target. The bullet moves with that air experiencing the same aerodynamic forces it felt in the zero wind situation. Just like the passenger on a train, the bullet moves with the volume of air the bullet is traveling in and is moving towards a point relative to the ground that is the target. We see the result is a smaller value of bullet drop at 349.2 inches. We would get the same effect in the zero wind situation of we moved the target on the ground closer.

[JohnKSa]

The path of the bullet thru the air is the same regardless of wind.

Isn't that sort of an odd usage of the word 'path' given that the trajectory is clearly different?

[tangolima]

Sorry I haven't been following the "aviation model" discussions till now. I wouldn't think it is applicable to flight of bullet though, i.e. the bullet does feel the wind.

Pilot flies his plane by maintaining certain air speed, regardless the wind. With that, the plane indeed doesn't feel the wind. The ground speed is vector sum of this heading and the wind direction, which is varying with the wind. This method makes sense as it makes the pilot's job much easier. The plane is assumed to have unlimited source of energy to maintain its flight path.

A bullet cannot maintain air speed. Actually its air speed is the apparent wind, which is vector difference of its velocity and the wind's velocity. So it always feels the wind.

I was a private pilot myself in my previous life. When I got into the more affordable hobby of firing guns, I also use this method, thinking the bullet is just a flying object moving with the air mass. I lost quite some number of hairs when I read books on external ballistics. It took a bit of efforts to knock the old habits.

Going back to wind deflection.

W=Vw * Tlag =Vw * (TOF - D/Vo)

W: wind deflection
Vw: cross wind speed
D: target distance
Vo: bullet's MV

If the bullet moves with the air mass, the wind deflection would have been

W=Vw*TOF

which is incorrect.

-TL


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[stagpanther]

I'm just playing devil's advocate, and not arguing the physics of it (way over my head), just a logic thing based on my "boots on the ground" experience.

Except when the value of crosswind is equal to 0--in nature it is almost never a constant velocity, nor at a constant angle, that it is being "felt" or "pushed" at. How does this model account for that?

[mehavey]

Which is why I always wondered how BC played a part in drift calculations
(since BC is frontal, not side angle)

[davidsog]

I wouldn't think it is applicable to flight of bullet though, i.e. the bullet does feel the wind.

I am editing this for clarity because it seems to be yesterday did not happen.

What I said yesterday is still true today:

What happens is the bullet rotates about its CG until those torque moments equalize the force and the bullet achieves equilibrium just like an airplane.

Yes, it is decelerating from the moment it is fired as a finite amount of thrust force has been applied by the powder charge but does not negate the principle of it achieving equilibrium in the side force yaw axis.

That is an almost instantaneous process as the force is applied. While it is interesting to examine what is going on in that tiny moment in time, it is a Tree and not the Forest. The POSITION of the CG remains the same despite it rotating about its axis.

Those moments about the CG rotate the bullet into what is termed the Relative Wind.

In the Relative Wind, the bullet does not feel the Ground Frame Reference wind and acts exactly like any other aerial vehicle.

The very Definition of Relative Wind is "The Speed and Direction of Air impinging on a body passing thru it. It is equal and opposite in direction to the flight path velocity"

A bullet does not feel the same wind you and I do on the ground, it only feels the Relative Wind and it moves thru a volume of air as a part of that volume of air until it encounters the target or the ground.

A bullet cannot maintain air speed.

No, it has finite amount of thrust imparted upon from the energy of propellent. Like any aerial vehicle is moves towards equilibrium of the forces acting upon. The result of that is a constant deceleration the moment it leaves the muzzle until it encounters the target or the earth. That movement is the result of that quest for force equilibrium and why the bullet does not just drop to the ground the moment it leaves the muzzle.

If the bullet moves with the air mass, the wind deflection would have been

W=Vw*TOF

It most certainly does. Your formulation just does a good job at hiding it from you and your formula is not about the forces acting on the bullet but rather a crude attempt to predict the location of the bullet on the target. It tells you nothing about what is going on during the flight. Does it account for lift? No. Bullets have lift due to the magnus effect and side force due to drag/magnus effects....

In an M24 SWS, One click to the right at a target 500 meters or more tells me where the bullet will be due to spindrift but it tells me nothing about what caused it to be there. That click is your math in action.

Your Tlag is a ratio of a theoretical flight thru a vacuum and the actual flight. So there is some very simplistic aerodynamics that gets covered up in late 17th Century math. Your formula lumps everything into a crude ratio to approximate a prediction not on what the forces are acting on the bullet are doing but where the bullet will be.....

It is not the math for telling you what is going on but rather the math to tell you where it will be....

The first is gravitational force; the other is aerodynamics. Several kinds of aerodynamic forces act on a bullet: drag, lift, side forces, Magnus force, spin damping force, pitch damping force, and Magnus cross force.

https://www.sierrabullets.com/exteri...bullet-flight/

Isn't that sort of an odd usage of the word 'path' given that the trajectory is clearly different?

Frame of Reference....

From the Frame of Reference of the ground, the trajectory is different.

From the Frame of Reference of the Air the trajectory is the same and since the volume of air is moving.... the bullet strikes the target at a different point along that same trajectory.

[davidsog]

since BC is frontal, not side angle

BC is just a modifier. It is the ratio of a theoretical bullet ability to cut thru the air vs a couple of measured data points of the actual bullets ability to cut thru the air.

It is a tool to cobble crude math and pound it into a shape that delivers results that approximate reality. It tells you nothing about the forces acting upon that bullet.

It fills a similar role as Equivalent Flat Plate does but the two are not directly comparable. EFP is much more accurate when used correctly.

[mehavey]

I still don't see (in any way) how a frontal BC has any effect on side impact fluid flow.

Perhaps someone can shed light on that ?

[davidsog]

I still don't see (in any way) how a frontal BC has any effect on side impact fluid flow.

Perhaps someone can shed light on that ?

I feel bad. It was my intention to clarify not confuse, LOL. Therefore I will make another attempt to help!

It is a reference area to approximate and represent the shape. The math corrects itself when describing motion as energy cannot be created or destroyed.

I know that sounds confusing. Aircraft Performance math uses the area of the wing to describe Lift, drag, and aircraft motion. External Ballistics uses a method first developed in the late 1600's of using a ratio of theoretical bullet behavior to measured bullet. Yes, it is a super confusing set of formulas that skirts the edges of actual Aerodynamics because it is rooted in a time when Aerodynamics was poorly understood.
It works because it is based on conservation of energy too. It is not exact but it gives an approximation.

[davidsog]

This might help.



Frame of Reference is a very important concept to understand if we are to understand what our bullet is doing.

As we sit at our computers, We are all passengers on a train moving at approximately 1.3 million miles per hour. If you add all the motion of the earth, the Solar system, and the Milky Way Galaxy we are in constant motion at over 1 million mph. We don't' perceive that motion because everything in our evolution has adapted us to be creatures of the earth. Once the bullet leaves the barrel, it becomes a creature of the air. The only thing it feels is the Relative Wind and that is NOT wind you and I experience. It does not feel the wind you and I experience.

[tangolima]

Originally Posted by davidsog

It most certainly does. Your formulation just does a good job at hiding it from you and your formula is not about the forces acting on the bullet but rather a crude attempt to predict the location of the bullet on the target. It tells you nothing about what is going on during the flight. Does it account for lift? No. Bullets have lift due to the magnus effect and side force due to drag/magnus effects....

W=Vw*(TOF -D/Vo) is wrong, and it should be replaced by W=Vw*TOF?

It is not my formulation. It is result of solving system of 3D differential equations, and has been proven by experiments of all sorts. We are entitled to our opinions, but I'm not quite ready to call generations of ballisticians obselete just yet, some of them are still with us today.

-TL

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[davidsog]

Isn't that sort of an odd usage of the word 'path' given that the trajectory is clearly different?

No, it was meant to be a precise use of the language. Trajectory is a path thru space. A path is series of vertices which is what our ground reference frame is experiencing or to the bullet in the air a vertex. In other words, the shape of the graph remains the same and the effect of wind is to shift that graph forward with a tailwind and backwards with a headwind along the LOS axis.

[davidsog]

It is result of solving system of 3D differential equations, and has been proven by experiments of all sorts. We are entitled to our opinions, but I'm not quite ready to call generations of ballisticians obselete just yet, some of them are still with us today.

It is a poor system for explaining the exact forces acting on the bullet and ballistic behavior in flight. It is a good system for approximating the location of a bullet along its flight path.

It is two different things and two different questions.

For almost half the history of Aviation we did not have to answer the same question. It was not until the 1940's that the speed and power of aircraft reached a point stability and control performance was necessary or even a science beyond the basics. Up until then, pilots flew aircraft routinely with stability and control characteristics that would be deadly in todays aircraft.

We haven't had the tools to dissect the external ballistic problem much further nor any real need too. That is changing as our tools become much more refined and our understanding more complete.

[mehavey]

Frame of Reference ....

So you're noting that there's a vector component to wind drift that lines up with a "frontal" aspect BC.

(But) that BC aspect will go as the cosine of wind impact angle -- effectively decreasing to zero/no role as wind reaches 90°, and defection maximizes.

[davidsog]

So you're noting that there's a vector component to wind drift that lines up with a "frontal" aspect BC.

The vector component is in the frame of reference of the air. I never denied that. It moves with the volume of air.

I see what you are asking. In regards to sideforce calculation the reference area remains the same "frontal aspect" just like it remains wing area most commonly in Aircraft Performance.

That sideforce is what rotates the bullet about its CG to align with the Relative Wind the moment it leaves the barrel. We don't need to know the exact area of the side profile of the bullet as the math will compensate. Energy cannot be created or destroyed and if we know the given force input we will get a good approximation of the output. It won't be the exact same forces on the actual aerial vehicle but the result of those forces will be the same. Understand? It's a confusing concept and one Aerodynamics Professors spend some time on when introduced in class, lol.

[davidsog]

That sideforce is what rotates the bullet about its CG to align with the Relative Wind the moment it leaves the barrel.

Sierra has a great explanation of what is going on to include measured results. Just like any aerial vehicle a bullet oscillates as it seeks equilibrium. Those oscillations are minor to the outcome.

When a bullet exits the muzzle of a gun, it immediately begins some angular pitching and yawing motions which have several possible causes including the crosswind. These angular motions are small, cyclical, and transient. They typically start out with amplitudes of a degree or so, and damp out, or at least damp to some very small residual values, after the bullet travels a relatively short distance. From our experience measuring ballistic coefficients, these motions damp within 100 yards or less of bullet travel downrange. Throughout the trajectory, including the initial transient period, the bullet has an “average” angular orientation which aligns the longitudinal axis almost exactly with Vbullet relative to the air. In this orientation the principal aerodynamic force on the bullet is drag, which acts in a direction opposite to Vbullet relative to the air. The side forces on the bullet are essentially nulled in this “average” angular orientation.

The bullet does not feel the wind. It rotates about the CG upon muzzle exist during its transition to exterior ballistics and becomes a creature of the air moving with but not feeling the movement of that volume of air.

[stagpanther]

I think what David is trying to say is that once an object becomes "detached" from the surface of the earth it becomes behaviorally influenced by the atmosphere--which really is dynamic and is much more like water in an ocean than a vacuum or uniform element. As shooters--or anyone on the surface--our frame of reference is "anchored" to observations and interpreted relative to our being on surface. But once we or any object is launched into the air the reference becomes "anchored" to the movement of the air masses it's entrained in.

[davidsog]

That's it Stagpanther. Just like a passenger on a train does not feel the movement of the train, the bullet does not feel the movement of the volume of air it is in.

[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/ai.../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]

Originally Posted by 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.

I concur.

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

-TL

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