Physics of shooting a rifle

[davidsog]

JohnKsa says:

A cyclical transient state means that there is oscillation that degrades to zero. it doesn't mean that the bullet will necessarily end up in its initial state.

While that maybe correct in other applications, in physics it is just wrong.

What is a Cyclic Process?
The process in which the initial and final states are the same is known as a cyclic process. It is a sequence of processes that leave the system in the same state in which it started.

https://byjus.com/physics/cyclic-process/

[JohnKSa]

A cyclical process does leave a system in the same state that it started in--assuming that it stops the cycle at the same point where it started.

However it is possible to have processes that are transient but exhibit cyclical motion, or oscillations. Those processes need not end up in the same state they start in, in fact, depending on the input "stimulus" they often don't.

Here's a longer version of the quote from the Sierra PDF you uploaded in post 218.

Clearly if the motion results in the bullet pointing the same direction that it did when it exited the muzzle, the rest of the passage would be meaningless and the first highlighted portion would contradict the ones that come later on in the text.

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. Only a tiny lift force and a tiny side force are maintained to generate moments of torque which cause the nose of the bullet to turn. We will describe these tiny effects a little later. First, let us consider the “average” angular orientation.

When the bullet exits the muzzle, its nose turns upwind. This is the necessary direction to align the longitudinal axis with Vbullet relative to the air . If the bullet did not turn in this direction, the crosswind would cause a strong side force on the bullet which would ultimately destabilize it. 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.

[davidsog]

Clearly if the motion results in the bullet pointing the same direction that it did when it exited the muzzle

There is your misconception. Nobody ever said it points the same direction as the muzzle. It picks an angle to equalize the forces acting upon it. That equalization process involves a period of oscillation.

That oscillation is small, cyclical, and transient.

It is also dampened.

Only a tiny lift force and a tiny side force are maintained to generate moments of torque which cause the nose of the bullet to turn.

This is also not a positional change in the bullet but a explains why the bullet rotates around the CG to change orientation during its flight. Once more, the torque moment is what equalizes the tiny side force and lift force.....

[JohnKSa]

Nobody ever said it points the same direction as the muzzle.

I believe it was asserted that after the small, cyclical and transient motions the bullet goes through after exiting the muzzle it would be left in the same state it was due to the nature of cyclical processes.

Clearly the bullet is pointing in the same direction as the muzzle as it exits, but the quoted source states that after the small, cyclical and transient motions, if there is a crosswind, it will not be pointing in the same direction as it was at exit. The crosswind will result in the bullet turning and aligning.

Which, of course, means that the bullet does not end up in the same state that it was after the small, cyclical and transient motions complete upon exiting the muzzle.

It is also dampened.

Correct, that's why it is transient. If it were undamped, it would continue oscillating forever--indefinite cyclical behavior.

When a stimulus is applied to a system (a spin stabilized bullet, in this case) there are three general categories of damping in the response. Well, actually four if you include the absence of damping (undamped) where the system continues oscillating/cycling forever without any tendency for that cyclical behavior to die out.

Overdamped in which the system responds by slowly moving back to the original state or to the new state.

Critically damped in which the system responds by moving back to the original state or to the new state in minimal time but without any oscillation or overshoot.

The third is called Under Damped in which the system responds by moving back to the original state or to the new state but with an oscillatory/cyclical behavior before the response settles. It overshoots, then undershoots, then overshoots, but with each under/overshoot amount being less than the one before it due to the damping effect.

That is what the document is talking about when it says there are small, cyclical, transient motions. It's basically an underdamped system where there is some oscillatory/cyclical behavior before the system either returns to the original state or achieves a new state.

Here's a decent discussion of the classes of damping in system responses.

https://en.wikipedia.org/wiki/Damping

Here's one that talks about it in a mechanical system. Figure 6 plots a number of system responses some of which show transient cyclical/oscillatory behavior.

https://www.racecompengineering.com/...itical-damping

[davidsog]

I believe it was asserted that after the small, cyclical and transient motions the bullet goes through after exiting the muzzle it would be left in the same state it was due to the nature of cyclical processes.

It almost instantly picks an angle that achieves equilibrium in 3 axis that will move it as a part of that airmass at the same velocity. The movement along that angle it picks is small, cyclical, and transient.

[davidsog]

Correct, that's why it is transient.

Are you trying to explain the concept of dampening to me or just throwing it out there? It just means the amplitude is not constant. We aren't designing a shock absorber.

It is not an under or over dampened system either. You realize it would not be on the same path if was in either of those conditions, right? It would never achieve zero velocity relative to the air as the Sierra Engineers tell you.

[JohnKSa]

We aren't designing a shock absorber.

The same general principles apply in many types of systems whether they are mechanical, electrical or aerodynamic, whether natural or designed intentionally.

It almost instantly picks an angle that achieves equilibrium in 3 axis that will move it as a part of that airmass at the same velocity.

Correct. That means the angle changes. The result of the small, cyclical transient motions is a new angle.

Are you trying to explain the concept of dampening to me or just throwing it out there? It just means the amplitude is not constant.

I was explaining the concept of damping to explain what they meant by small transient cyclical motion and demonstrate that it can result in a new state when it fades--just as in this case. The 'cyclical' component need not result in only the original state.

It is not an under or over dampened system either.

It is an underdamped system, otherwise it would not demonstrate any oscillatory behavior.

The figure is slightly misleading. The bullet does not ever achieve zero velocity with respect to a crosswind moving air mass in practice because it takes too long for that to happen. What it does achieve relatively quickly is a new orientation that will help it begin to move in the same direction as an air mass that is moving crosswind.

[davidsog]

The bullet does not ever achieve zero velocity with respect to a crosswind moving air mass in practice because it takes too long for that to happen.

Again, not according to Sierra Engineers. You keep saying this but it is contrary to what their measured results show.

These angular motions are small, cyclical, and transient.

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.

100 yards for the worst BC and less for average BC and even less for a good BC.

The Angle is small....

This angle is small. For example, for a rifle firing the 308 Winchester cartridge with Sierra’s 168 grain MatchKing bullet at 2650 fps muzzle speed in a 25 mph crosswind, θmuzzle = 0.793 degree = 47.6 minutes of angle. For lower wind speeds higher muzzle velocities values of θmuzzle are even smaller.

Here is the average monthly wind speeds for 2020:

https://www.ncei.noaa.gov/pub/data/c...a/wndspd20.dat

Notice NONE of them are 25mph or even close.

Think about that. The Body Angle of the bullet is 47.6 MOA. That is 47.6 inches OFF the target LOS at 100 meters.

Running a ballistics calculator shows us a 168 grain SMK .308 round with a G1 BC of .462 yields a wind drift of 1.8 inches.

Time of Flight in our bullet is .11 seconds at 100 yards. Explain to me how our bullet covers the 47.6 inches plus the 1.8 inches wind drift....that is 49.4 inches of movement without picking up the crosswind velocity and picking it up very quickly.

Think about that and after your answer, I will run the math for you.

[JohnKSa]

Again, not according to Sierra Engineers. You keep saying this but it is contrary to what their measured results show.

It is consistent with their measured results.

1. The small transient cyclical motions adjust the bullet's ANGLE after muzzle exit. That happens fairly quickly. They do not impart the crosswind velocity to the bullet.

2. The bullet then acquires the crosswind velocity gradually because the angle of the bullet (acquired in step 1) "steers" it in the same direction as the crosswind. But the angle is small and air is not very solid so this takes time. A lot of time, in terms of bullet flight.

100 yards for the worst BC and less for average BC and even less for a good BC.

Yes, that is when the cyclical oscillations die out--how long it takes for the bullet to achieve its new angle--but it takes much longer for the bullet to reach the same crossrange velocity that the crosswind has. Maybe 100yds for the transients to settle into a new bullet angle, but miles before the bullet actually gets to the same crossrange velocity as the crosswind.

The Angle is small....

Yes, that's why it takes it a while to "catch up" to the crosswind's velocity.

Explain to me how our bullet covers the 47.6 inches plus the 1.8 inches wind drift.

It doesn't cover 47.6inches. It covers 1.8"

The bullet is pointed 47.6 minutes (about 3/4 of a degree) off the target after the transients settle. If it were on a rail and it achieved that angle immediately after exiting the muzzle, it would go 47.6 minutes off the target. But it's not on a rail, instead it's traveling through a relatively insubstantial medium (air) with only a very small angle (about 3/4 of a degree) steering it to one side. So it only goes about 1.8 inches off the point of aim.

1.8" over 0.11 seconds works out to an average velocity of 0.93mph. If the bullet had achieved 25mph of crossrange velocity by the time it hit the target one would expect the average velocity to be much higher than that.

The ballistics calculator results in tangolima's post (#270) on this thread show how the crossrange velocity of a bullet changes with time and distance. In that case, a 10mph crosswind resulted in only about 8mph of crossrange velocity on the bullet after 3500 yards and over 10 seconds of flight.

[davidsog]

It doesn't cover 47.6inches. It covers 1.8"

No, John. That is where it lands but it not how it gets there.

308 Winchester cartridge with Sierra’s 168 grain MatchKing bullet at 2650 fps muzzle speed in a 25 mph crosswind, θmuzzle = 0.793 degree = 47.6 minutes of angle.

One degree = 1/60 of a MOA

1 MOA = 1 inch at 100 yards.

The bullet angle is pointing 47.6 inches off the target the moment it exits the muzzle.

3/4 of a degree

Is the 47.6 inches. You are on the right track with the Body angle vs path. However our bullet is not on rails and it must obey the laws of physics. While it is in the air, the physics of aerodynamics must be obeyed.

How much distance in inches will an object moving at 25 mph move in .11 seconds?

Distance = Rate * Time

Units are critical. MPH must be converted to Feet per Second.

25mph = 36.6 fps

D = 36.6 fps * .11 seconds = 4.026 feet

Seconds cancel leaving us with at total distance in .11 seconds of 4.026 feet or 48.3 inches.

48.3 inches minus the 46.7 inches to bring us back to LOS = .71 inches downwind of the target.

That represents the distance the bullet travels outside the muzzle while it seeks the body angle it requires to equalize the forces.

1.8 inches - .71 = 1.09 inches

1.09 inches represents the both the short oscillation period and the flight path angle.

That conforms with all laws of Aerodynamics and Physics. The bullet very quickly picks up the velocity of the air mass it is moving in and moves with that air mass.

Sierra Engineers agree:





The other theory seems to be some magical force fights the bullets ability to obey the laws of aerodynamics to stay on some path the forces are not pushing it onto.

[JohnKSa]

The bullet angle is pointing 47.6 inches off the target the moment it exits the muzzle.

The bullet is pointing straight ahead when it exits the muzzle. After some small, transient, cyclical motions, it is re-oriented to an angle of 47.6 minutes by the crosswind from its original angle.

That angle is what steers it through the air so that it lands 1.8" away from the point of aim.

Is the 47.6 inches.

Minutes of angle, not inches.

How much distance in inches will an object moving at 25 mph move in .11 seconds?

It is not moving 25mph. It will never move 25mph. Look at the post I linked to and you can see that the bullet in that example never reached the crosswind velocity even after 3500 yards.

Is the 47.6 inches. You are on the right track with...
...
...and the flight path angle.

That conforms with all laws of Aerodynamics and Physics. The bullet very quickly picks up the velocity of the air mass it is moving in and moves with that air mass.

The bullet only hits 1.8" from the point of aim. This 4 feet you are coming up with doesn't make any sense.

I can't tell if you are trying to say the bullet moves 4 feet going upwind to cancel the wind (which would mean the bullet never takes the velocity of the air mass) or 4 feet going downwind and then back so it only ends up 1.8" from the point of aim which would mean that the bullet would go significantly off windage trajectory on its way to the target which is clearly inconsistent with reality. Either way, the argument can't work.

The bullet very quickly picks up the velocity of the air mass it is moving in and moves with that air mass.

Based on the POI alone, that obviously can't be true.

If the bullet takes the velocity of the air mass very quickly then it would hit 4 feet downwind. You did the math yourself. 25mph for 0.11 seconds is over 4 feet.

How does an object traveling crossrange at 25mph for 0.11 seconds hit within 2" of where the barrel was aimed?

[davidsog]

Minutes of angle, not inches.

1 MOA = 1 Inch at 100 yards.

The Relationship between MOA and Distance
MOA is not dependent on distance. If you thought there was a relation, well it’s a bit complicated here. We use MOA as an angular measurement as opposed to linear. That’s what we adjust a long range target scope on a rifle using the elevation turret. So, how do we translate MOA to a linear measurement?

Apply this rule: 1 minute (or 1 MOA) is one inch at 100 yards.

https://americanshootingjournal.com/...nute-of-angle/

This 4 feet you are coming up with doesn't make any sense.

308 Winchester cartridge with Sierra’s 168 grain MatchKing bullet at 2650 fps muzzle speed in a 25 mph crosswind, θmuzzle = 0.793 degree = 47.6 minutes of angle.

Explain how the bullet starts out 47.6 MOA off the target to land 1.8 inches to the other side of the target without violating physics.

I did just that. What's your explanation?

If the bullet takes the velocity of the air mass very quickly then it would hit 4 feet downwind.

Not if it requires a 47.6 MOA body angle to equalize the aerodynamic forces and then moves as a part of the air mass. It works out rather nicely in fact without violating any laws of physics.

[JohnKSa]

I know what a minute of angle is. There are 60 minutes in a degree, 60 seconds in a minute.

At 100 yards, a minute of angle is approximately 1.047"

Explain how the bullet starts out 47.6 MOA off the target to land 1.8 inches to the other side of the target without violating physics.

You didn't explain how it does that AND takes the velocity of the air mass. The bottom line is that if it takes the velocity of the air mass very quickly after exiting the muzzle, it would hit 4 feet downwind.

Not if it requires a 47.6 MOA body angle to equalize the aerodynamic forces and then moves as a part of the air mass.

If it is effectively cancelling out the crosswind drift using aerodynamic forces, then it never moves sideways at 25mph and therefore, by any reasonable definition of the words involved, it never takes the velocity of the air mass.

In post 269, I quoted Sierra stating that:

"The crossrange bullet motion is accelerated relatively slowly, and in fact the crossrange component of the bullet’s velocity never does grow to equal the crosswind velocity."

And also provided a link.

[davidsog]

it would hit 4 feet downwind.

You are struggling with Frame of Reference.

This would be completely wrong in a ground frame of reference. The ground is not moving in relation to the bullet.

The air mass is moving for 4 feet downwind in .11 seconds in relation to the ground.

In order for an aerial object to strike a specific point on the ground, it must angle into the wind enough to account for that airmass movement. It does just that by equalizing the forces acting upon it and in this case our .308 168 grain SMK bullet takes up an angle 47.6 MOA thru the air to strike that object on the ground 1.8 inches downwind from the target while moving 25mph across the ground with the airmass.

If it is effectively cancelling out the crosswind drift using aerodynamic forces, then it never moves sideways at 25mph and therefore, by any reasonable definition of the words involved, it never takes the velocity of the air mass.

This is just wrong. Equalizing the forces means it moves with the airmass and just like the Sierra Engineers relate...it takes on the velocity of that airmass. It MUST take on the velocity of the airmass.

It works exactly the opposite of what you are claiming in terms of this specific assertion.

[davidsog]

In post 269, I quoted Sierra stating that:

"The crossrange bullet motion is accelerated relatively slowly, and in fact the crossrange component of the bullet’s velocity never does grow to equal the crosswind velocity."

And I will quote Sierra from the PDF I posted about your earlier webpage:

Section 4.3 of the Exterior Ballistics chapter of the
Sierra Rifle and Handgun Reloading Manual, Edition V, is not correct. This author will
eat some humble pie and correct that section at the next printing of the Sierra Manual.

https://thefiringline.com/forums/att...1&d=1706067709

[JohnKSa]

You are struggling with Frame of Reference.

Crossrange is specifically referenced to the shooter and the target, by definition. It is completely unambiguous in meaning and frame of reference.

The only way what you are saying makes sense is if crossrange velocity means one thing for a crosswind and something completely different for a bullet.

In order for an aerial object to strike a specific point on the ground, it must angle into the wind enough to account for that airmass movement. It does just that by equalizing the forces acting upon it and in this case our .308 168 grain SMK bullet takes up an angle 47.6 MOA thru the air to strike that object on the ground 1.8 inches downwind from the target while moving 25mph across the ground with the airmass.

Which is to say that it isn't moving at 25mph across the ground, since it's cancelling most of that motion out by steering into it.

The forces are not completely equalized, because the bullet is still accelerated in the crosswind direction, gaining crossrange velocity slowly.

The bottom line is that the bullet never acquires the crossrange velocity of a crosswind because it effectively steers into it. It is accelerated very slowly in the direction of the crosswind by the same force that causes the steering effect and does gain crossrange velocity slowly. So slowly, in fact, that it will never acquire the full value of the crosswind in the time it takes to reach the target--even for very long range shooting. As stated in the quote I have twice provided by Sierra. As clearly demonstrated by the actual difference between the POA and POI. As shown by the ballistic calculator results posted in 270.

Section 4.3 of the Exterior Ballistics chapter of the Sierra Rifle and Handgun Reloading Manual, Edition V, is not correct. This author will eat some humble pie and correct that section at the next printing of the Sierra Manual.

The quote I provided was not from Section 4.3.

[davidsog]

Which is to say that it isn't moving at 25mph across the ground

Nobody ever said it did, that is your misconception because your struggling with the different frames of reference.

Which is to say that it isn't moving at 25mph across the ground, since it's cancelling most of that motion out by steering into it.

Here is where you confuse frames of reference or are unable to seperate them.

Just like an airplane matches the crosswind velocity of the airmass to arrive at a specific point on the ground that is the runway....

The bullet does the same thing. It achieves equilibrium and that angle that it achieves it moves at the same velocity as the air mass it is now apart.

In the frame of reference to the ground, it has canceled that airmass velocity to stay on track with a point on the ground.

To then claim it does not acquire the velocity of the air mass when it absolutely MUST acquire that air mass velocity to even cancel it out is conceptually and factually wrong.

[JohnKSa]

Nobody ever said it did...

From the post directly above mine.

Originally Posted by davidsog

In order for an aerial object to strike a specific point on the ground, it must angle into the wind enough to account for that airmass movement. It does just that by equalizing the forces acting upon it and in this case our .308 168 grain SMK bullet takes up an angle 47.6 MOA thru the air to strike that object on the ground 1.8 inches downwind from the target while moving 25mph across the ground with the airmass.

I can't tell if you're screwing with me...

In the frame of reference to the ground, it has canceled that airmass velocity to stay on track with a point on the ground.

See, here's the thing. If it has canceled that airmass velocity then it isn't moving at the same velocity as the air mass.

The air mass is moving, but the bullet is moving differently. We know this because a 25mph crosswind moves 4 feet in the time it takes the bullet to get to the target but the bullet moves less than 2". Clearly they are not moving the same.

Sierra says they are not moving at the same velocity. The ballistic calculator shows they aren't.

Yes, the bullet is moving within the air mass and is affected by it, but it is not moving with the air mass. It is in fact canceling out much of the movement of the air mass.

Look, if you want to say that crossrange velocity means something different for a bullet than it means for wind, then I guess that's ok as long as you explain adequately and don't complain when people don't use your definitions. But when it is plain to see that the bullet's crossrange velocity is very different from the crosswind's velocity, it doesn't make sense to tell people they are wrong for noting that fact.

And yes, you have been doing exactly that. Even trying to contradict the quote from the Sierra manual which states explicitly that they are different.

Again, from the Sierra manual, which you brought up as a source.

"The crossrange bullet motion is accelerated relatively slowly, and in fact the crossrange component of the bullet’s velocity never does grow to equal the crosswind velocity."

It's all about communication. If you say that a person is on a train and moving with the train but then it turns out that the train moves 48 miles and the person only moves 2, that's going to generate a lot of confusion.

[davidsog]

I can't tell if you're screwing with me...

I am not. I can tell you are a very intelligent guy and it bothers you tremendously the concepts. You are not alone and it's one of the reason's we just don't see many people running around with Aeronautical Engineering degrees. It is called "Rocket Science" for a reason. It is hard and there is no shame in questioning in the search for knowledge.

If it has canceled that airmass velocity then it isn't moving at the same velocity as the air mass.

Why does it have to move at the same velocity?? Nothing says that it does move at the same velocity but it must equalize those forces and therefore it will take on the velocity of the air mass it is traveling in. That is principle of aerodynamics. Even the Large Transport Category Aircraft in the Airlines must obey it.

For any vector v,

v + - v = 0

and thus an additive inverse exists for every vector.

Sierra's analysis appears to be body angle and not the aerodynamic angle to the Relative Wind. Sierra states that is impossible to determine because of the angle of repose cannot be measured with the equipment they have at hand.

We know this because a 25mph crosswind moves 4 feet in the time it takes the bullet to get to the target but the bullet moves less than 2"

No, the bullet moves the full distance of 4 feet THRU THE AIR and not in relation to the ground. That is how it makes up the 49.4 inches it needs to travel to land 1.8 inches downwind of the POA.

Think of it like running on a treadmill going 25 mph. If you hop on that treadmill, you are going to have to match speed with that treadmill or be thrown off. That does not mean you are moving a single inch in the room the treadmill occupies.

Sierra says they are not moving at the same velocity.

Sierra is contradictory on this point but they are not contradictory on acknowledging the error in 4.3. There are basic errors in some of their published assumption on bullet behavior specifically to how the bullet turns. That will compound errors in predicting behaviors and the magnitude of those effects.

4.3 TURNING OF A BULLET TO FOLLOW A CROSSWIND AND RESULTING DEFLECTIONS

https://www.sierrabullets.com/exteri...g-deflections/



In fact,

Sierra says:

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. Only a tiny lift force and a tiny side force are maintained to generate moments of torque which cause the nose of the bullet to turn. We will describe these tiny effects a little later. First, let us consider the “average” angular orientation.
When the bullet exits the muzzle, its nose turns upwind. This is the necessary direction to align the longitudinal axis with Vbullet relative to the air. If the bullet did not turn in this direction, the crosswind would cause a strong side force on the bullet which would ultimately destabilize it. 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.

So Sierra agrees that the bullet picks the aerodynamic angle it requires to equalize the forces and that includes side force of the crosswind.

At that point our vector math says, v + - v = 0

That is how our bullet strikes the target in the ground frame of reference by flying thru the air pointed 46.7 inches off the target. It needs to move that 4 feet thru the air to stay in the same relative ground reference as the target.

[JohnKSa]

I'm going to try again, against my better judgement.

Sierra is contradictory on this point but they are not contradictory on acknowledging the error in 4.3.

The quote I provided is not from 4.3. It is from 5.3.2 If you believe you have a point, it's fair to raise it. Once it is categorically addressed, please don't raise it again as if it's a new point.

The error in 4.3 is not relevant to a quote from 5.3.2. So we're done with that, right? Furthermore, you then go on to quote from 4.3 after saying twice that 4.3 is in error. How does that make sense? Why wouldn't you quote, instead, from the article that claims to be correct?

There are a number of equations in that article that involve the difference (Vcrosswind - Vcrossrange), that is the difference in the crosswind velocity and the crossrange velocity of the bullet. That expression reduces to zero if the two are the same. Clearly they are not the same or all those equations would be fancy ways of writing 'zero'. Clearly the bullet is not moving at the same crossrange velocity as the crosswind velocity.

As far as contradictory goes, let's address this.

You say: "...our .308 168 grain SMK bullet takes up an angle 47.6 MOA thru the air to strike that object on the ground 1.8 inches downwind from the target while moving 25mph across the ground with the airmass."

I say: "...Which is to say that <the bullet> isn't moving at 25mph across the ground"

You respond with: "Nobody ever said it did..."

Somebody obviously did say that--in fact, it was you in the very quote I was addressing. If you say something, then don't deny saying it and then move on as if nothing happened.

Next.

Why does it have to move at the same velocity??

Because you said it did when you stated very clearly: "our .308 168 grain SMK bullet takes up an angle 47.6 MOA thru the air to strike that object on the ground 1.8 inches downwind from the target while moving 25mph across the ground with the airmass".

You say that the bullet is moving 25mph across the ground with the air mass which you stated was also moving 25mph earlier.

In fact, I was the one saying it was not moving at the same velocity while you were saying it was, but you ask me why it has to be moving at the same velocity as if that were an assertion I had made. If you don't mean that they are moving at the same velocity, then don't say it. And if someone calls you on it, admit it, don't pretend instead that they are the one who made the claim that you made.

Ok. Let's move on.

Nothing says that it does move at the same velocity but it must equalize those forces and therefore it will take on the velocity of the air mass it is traveling in.

Well, sorta.

1. The crosswind force acts on the bullet but the bullet counteracts it by "steering" into the wind and cancelling out much of the crosswind velocity. By doing so, it effectively "flies" in the upwind direction in the air mass instead of being carried along at the same velocity as the wind. This is easily verified by comparing the crossrange movement of the bullet during TOF to the crossrange movement of a hypothetical object moving at the crosswind velocity during the same timeframe and noting that the bullet will travel only a small fraction of the distance crossrange as the hypothetical object moving at crosswind velocity does.

2. Because of the nature of the interaction, a bullet will not ever achieve the full crosswind velocity because TOF is limited by practicality. Even at distances far exceeding a mile the crossrange velocity of a bullet will never take on the full velocity of the crosswind. That is what Sierra says and that is what Hornady's ballistic calculator says.

[stagpanther]

I appreciate the info from everyone.

[tangolima]

Originally Posted by tangolima

Found another rabbit hole nearby to explore.

Although over simplified, it is quite true to say all stabilized (not tumbling) projectiles have ability to turn into the wind, weathervane or weather helm.

It is pretty easy to visualize in fin stabilized projectiles, such as arrow, pellet, etc. Center of effort (CE) is behind center of gravity (CG).

It is rather opposite in bullet, where CE is ahead of CG. It requires spin / gyroscopic stabilization. But exactly how does it work? What makes a bullet turn its nose into the wind?

Here are some facts to consider. Assuming right-hand riflings and pure cross winds (no up/down drafts), cross wind from the right causes high-left POI, and cross wind from the left causes low-right POI.

-TL

Sent from my SM-N960U using Tapatalk

Let's try again with this new rabbit hole. All are welcome. No special degree/qualification/credential/title is required. . Here we are all amateurs.
-TL

Sent from my SM-N960U using Tapatalk

[georgehwbush]

arrow dynamic jump. it's a real (THANG) believe it or not.

[davidsog]

arrow dynamic jump. it's a real (THANG) believe it or not.

Yep

[stagpanther]

If the twist in the bore is reversed (left instead of right)--is the jump/fall of the aerodynamic jump also reversed? Reminds me of climb vs conventional cut in milling.