Physics of shooting a rifle

[JohnKSa]

Ultimately we are talking about deflection--the specific mechanism and magnitude may be complex to characterize, but it all comes down to something that starts off on one path ending up not traveling the original path. So it had to be deflected at some point and in some manner.

Given an isolated occurrence that generates deflection (say a gust of wind), the earlier in the trajectory that occurs, the greater the magnitude of the result on target. If the bullet is "knocked off" its initial course very early in the trajectory, it will follow the new course for a longer distance resulting in being farther off target than if it were given a new course just before it hits the target.

Given a constant input that generates deflection (constant wind), the longer that happens the greater the magnitude of the result on target.

It does not--the air mass it is entrained in has a sideways movement relative to the fixed objects on the ground.

It's worth noting that even if one takes the "bullet in the air mass" interpretation, the bullet can't instantly acquire the speed of the air mass--in fact there might not be time for it to fully do that before reaching the target.

 

[tangolima]

It's worth noting that even if one takes the "bullet in the air mass" interpretation, the bullet can't instantly acquire the speed of the air mass--in fact there might not be time for it to fully do that before reaching the target.

Agreed. There are 2 states in physical dynamics; transient state and steady state. Flight of bullet is mostly transient state. Time is never long enough for the projectile to settle and become part of the air mass. Air mass model is handy for stead state situations; aerial / maritime navigation for instance, where time is long enough for the vessel to settle with its surroundings. The pilot's input often ensure such settings, keeping constant airspeed/heading for example. In such cases the error incurred by ignoring transient part of the motion is insignificant.

-TL

Sent from my SM-N960U using Tapatalk

 

[davidsog]

Flight of bullet is mostly transient state.

The Sierra Engineers answered this in the report I first posted. The transient phase as you call it is almost instantaneous.

I also think it difficult to grasp how quickly the bullet achieves equilibrium on 3 axis like a glider. Sierra Engineers relate that the stabilization occurs in the heaviest, most powerful bullets in 1/10th of second with most bullets achieving it much quicker. The order of these effects are very small.

Most bullets achieve it on the order of 1000th of a second so unless you are shooting .577 Tyrannosaur there just isn't a "transient state" of any consequence and even then Sierra says it is extremely small.

To quote Sierra:

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.

A cyclical process has a specific definition in physics.

Cyclic process - work
A process that eventually returns a system to its initial state is called a cyclic process.

In other words, it does not affect the outcome as it returns to its initial state.

Given a constant input that generates deflection (constant wind), the longer that happens the greater the magnitude of the result on target.

The wind no more exerts a constant force on the bullet than your train seat beats, shoves, and pummels you all the way to your destination.

It is an interesting discussion.

 

[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 noise 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

 

[JohnKSa]

A process that eventually returns a system to its initial state is called a cyclic process.

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. What they are saying is that the bullet exhibits some transient cylical motion due to the forces acting on it and then that motion settles down with the bullet now pointing in whatever direction the forces acting on it changed it to. I think they are mostly talking about what happens to the bullet immediately after it exits the muzzle and is kicked around by the exiting gases.

The wind no more exerts a constant force on the bullet than your train seat beats, shoves, and pummels you all the way to your destination.

Let's say there's no wind from 0 yards to 150 yards, then a constant full value wind the rest of the way to the target at 300 yards.

You're saying that the effect on the target will be exactly the same as if there is a constant full value wind all the way from the muzzle to the target?

I think that would be very easy to prove or disprove.

 

[stagpanther]

Let's say there's no wind from 0 yards to 150 yards, then a constant full value wind the rest of the way to the target at 300 yards.

You're saying that the effect on the target will be exactly the same as if there is a constant full value wind all the way from the muzzle to the target?

I'm getting confused here--but isn't that essentially what Tangolima is saying by asserting that the bullet "aquires" a sideways inertial velocity upon exiting an airmass?

 

[tangolima]

Guess we still need to stay in the old rabbit hole a bit longer.

It is impossible for an object to change its energy (potential + kinetic) instantaneously. It always takes finite amount of time, or infinite power is required.

A projectile in flight can change its posture (yaw / pitch) in response to disturbance in air mass relatively quicker, but still it takes time. Changing posture doesn't mean it catches up with the cross wind or has become part of the air mass. It is just the first step. Theoretically the bullet's lateral speed will approach the cross wind speed, meaning it can get close but can never be the same, given enough time. I will try to find out the order of magnitude of that time, but most likely it is in terms of seconds.

In the original question, the bullet has acquired certain lateral speed when it crosses the 200yd line. It can't possibly snap to zero speed instantly. Most certainly it won't by the time it reaches 300yd, unless it encounters a very strong cross wind in the opposite direction.

-TL

Sent from my SM-N960U using Tapatalk

 

[stagpanther]

As mentioned--I'm not a fizzics guy (in other words I be igorant)--but I do know air behaves like a fluid. True laminar flow with a "wall" boundary almost never happens in nature, especially at or near the surface where convective and advective disturbance is almost always present. I'm guessing there probably is some kind complex intermixing transition between the air masses that alters the ground-relative trajectory to a very small degree.

 

[JohnKSa]

It's less to do with the concept of the existence of a very clearly defined line between the two conditions and more to do with how long it takes the bullet to achieve a steady state in the new condition.

But that doesn't happen in an instant, in fact, it happens quite slowly in the overall scheme of things.

https://www.sierrabullets.com/exterior-ballistics/5-3-2-crosswinds/

"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."

As long as the bullet's sideways velocity due to crosswind never matches the crosswind's velocity, then a crosswind (relative to the bullet) still exists and the bullet continues to be turned in the direction of the crosswind. Said more simply, until the bullet's crossrange velocity matches the crosswind velocity, the crosswind continues to exert a force on the bullet and that force will continue to work on the bullet to change the impact point on the target.

All of that to say that my comment: "It's worth noting that even if one takes the "bullet in the air mass" interpretation, the bullet can't instantly acquire the speed of the air mass--in fact there might not be time for it to fully do that before reaching the target. " is consistent with the Sierra document. In fact, they go farther and say that it never fully acquires the speed of the air mass.

 

[tangolima]

Hornady's 4DOF calculator. 6mm 103gr eldx bullet is arbitrarily chosen. MV set to 3000fps. 10mph cross wind. Here are the results.

Range, wind deflection, bullet speed, TOF, bullet's lateral speed

0yd, 0", 3000fps, 0ms, 0mph
100yd, 0.5", 2835fps, 103ms, 0.5mph
500yd, 14.6", 2200fps, 582ms, 2.8mph
1000yd, 72.3", 1488fps, 1411ms, 5mph
2000yd, 402.3", 859fps, 4285ms, 7.32mph
3000yd, 948.9", 647fps, 8381ms, 7.99mph
3500yd, 1310", 577fps, 10917ms, 8.2mph

The calculator runs out of range.

Even after 10 full seconds, the lateral speed of the bullet is still only 82% of the cross wind speed. The bullet never reaches steady state.

While we are at it, let's plug the 1000yd figures in the "obselete" wind deflection equation.

W=(TOF-D/Vo)*Vw=72.3"

It still works!

-TL

Sent from my SM-N960U using Tapatalk

 

[JohnKSa]

The lateral speed looks like it should be of the general form:

Crosswind* [1 - exp( -x/Constant) ]

Where exp is the exponential function--the constant 'e' (~2.718281828) raised to the power of the expression in the parentheses.

'Constant' would be determined by the interaction of the bullet and the crosswind that turns the bullet.

'Crosswind' is the full crosswind value in mph.

'x' is the independent variable. Traditionally it would be time or distance.

It does come close to fitting, but doesn't quite work whether I make the independent variable 'x' distance or time of flight.

I suspect that 'Constant' isn't really a constant but probably varies based on bullet velocity and maybe other parameters.

 

[tangolima]

The lateral speed looks like it should be of the general form:



Crosswind* [1 - exp( -x/Constant) ]



Where exp is the exponential function--the constant 'e' (~2.718281828) raised to the power of the expression in the parentheses.



'Constant' would be determined by the interaction of the bullet and the crosswind that turns the bullet.



'Crosswind' is the full crosswind value in mph.



'x' is the independent variable. Traditionally it would be time or distance.



It does come close to fitting, but doesn't quite work whether I make the independent variable 'x' distance or time of flight.



I suspect that 'Constant' isn't really a constant but probably varies based on bullet velocity and maybe other parameters.

I gave it a try on excel. It fits pretty ok with R^2=0.98. Constant = 4167.

V=10*(1-exp(-D/4167))

In order for the lateral speed to get within 95% of the cross wind speed, it will take 3*4167=12,500yd. Of course this is extrapolation. The point is the bullet probably never has enough time during its meaning flight to settle with the air mass, at least so for this particular bullet.

-TL

Sent from my SM-N960U using Tapatalk

 

[JohnKSa]

Yup, the general rule of thumb for that equation is that when the independent variable is 3x the Constant, the equation output is at 95% of full value. At 5x, it's over 99% of full value.

Are you sure about 4167 though? Maybe I've got a glitch somewhere but I'm getting a lot closer to 1950.

That also seems to fit with your data table better since it's showing 82% of full value at 3500. Normally you'd expect that to happen at nearly 2x the Constant (1x is 63%, 2x is 87%). So you'd expect that with a Constant of 4167, 82% would happen between 4167yards and 8334yards but much closer to the latter.

Made some changes. Final version post at about 2:48

 

[tangolima]

Yup, the general rule of thumb for that equation is that when the independent variable is 3x the Constant, the equation output is at 95% of full value. At 5x, it's over 99% of full value.



Are you sure about 4167 though? Maybe I've got a glitch somewhere but I'm getting a lot closer to 1950.



That also seems to fit with your data table better since it's showing 82% of full value at 3500. Normally you'd expect that to happen at nearly 2x the Constant (1x is 63%, 2x is 87%). So you'd expect that with a Constant of 4167, 82% would happen between 4167yards and 8334yards but much closer to the latter.



Made some changes. Final version post at about 2:48

Oops. I mixed up LN() with LOG(). I was using libre office, not the real excel. Yes, constant = 1808.

V=10*(1-exp(-D/1808))

3*1808=5425yd

The conclusion still holds though.

Thanks for catching the error. I should have done a sanity check.

-TL

PS. R^2 =0.999 (much better fit) if 2nd order polynomial is used. But the equation becomes a bit more complicated. 1st order will do.

Sent from my SM-N960U using Tapatalk

 

[JohnKSa]

The conclusion still holds though.

Correct. For the bullet in question, it will probably be around 3 miles downrange before its crossrange velocity gets within 95% of the crosswind velocity. That's even with a constant crosswind velocity from the muzzle to as far as the bullet can stay in the air.

I would say that for practical purposes, it's safe to assume that the bullet's crossrange velocity is always significantly different from the crosswind velocity.

I suppose there's always the coincidental case. Let's say there's a high crosswind at the muzzle and downrange, when the bullet is moving crossrange at maybe 10% of the crosswind value, the crosswind changes to only 10% of its original value. Or the trivial case where the crosswind value is very small or zero.

 

[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/blogs/the-apex-files/dampers-part-5-critical-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.