Physics of shooting a rifle

[tangolima]

Gas exiting muzzle over takes the bullet momentarily. It is indeed a critical moment, even with perfect crown.

99.9% of a projectile's flight is with a dominant head wind, and therefore it is an indispensable component in stabilization. With head wind suddenly gone and reversed to tail wind, even momentarily, something can easily go wrong.

I have experimented seating bullet backwards. The bullet tumbles every time. Without going into details, a spin stabilized projectile requires center of pressure (CP) be ahead of center of gravity (CG). A tail wind reverses that. The bullet becomes unstable for fraction of a millisecond. Any minute imbalance of air flow, a cross wind, a less than perfect crown, can impart yaw and pitch on the bullet that requires some iterations in bullet's posture in flight to correct.

Not much can be done to avoid this I am afraid. A good crown is of course essential. Some suggested flash cone / can to shield the bullet from cross wind. Its efficacy is questionable unless the cross wind is super strong, such as waist guns on a B17. How about taking away the gas before the bullet exits muzzle? Muzzle device won't work, ported barrel may.

This is for spin stabilized projectile. Situation is more hopeless in fin stabilized airgun pellet. That's why the barrel is rifled even though a pellet does need that for stabilization.

-TL


Sent from my SM-N960U using Tapatalk

[davidsog]

I’ve seen where the bullet is doing some pitching and yawing before it’s stabilized, some much more than others. Pretty interesting.

It is interesting. The concept of those pitch and yaw movements having an overall effect of zero is weird. It is counterintuitive but if you understand vector math then the vectors cancel each other out. 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.

I have experimented seating bullet backwards.

Current Ballistic math is only approximates behavior in just a few basic shapes. The most common, G2 and G7 refers to just two basic shapes.

https://kestrelmeters.com/pages/g1-g...the-difference

Reversing your bullet is not one of those shapes and represents being a bullet test pilot so I am not surprised if you observe unusual behaviors. In fact, I would think your bullet would tumble and be unstable as you have changed the design relationship of the CP to CG.

[Vinootz]

On page 1 of this thread some were calculating physics formulas. Why not just point and shoot a rifle you’re comfortable with. And if democrats catch wind of that, they may start drafting gun control legislation that requires people to prove they understand physics formulas.

[davidsog]

On page 1 of this thread some were calculating physics formulas. Why not just point and shoot a rifle you’re comfortable with. And if democrats catch wind of that, they may start drafting gun control legislation that requires people to prove they understand physics formulas.

Just use "The Spontaneous Generation Theory" of Ballistics....

https://youtu.be/AngYmfS7pFI?si=K7O_6FWGI2Biqdso

[tangolima]

Another rabbit hole on wind deflection.

Target at 300yd. Wind flag at firing line and every 100yd. The one at 300yd is in scope's FOV so it is handy. The other ones I have to scan with left eye. Wind is varying at different distances. Which flag should I pay most attention to and why?

Say I got 2" wind deflection at 200yd. Same wind up to 200yd, but no wind from 200yd to 300yd. The wind deflection at 300yd is still 2"?

-TL

Sent from my SM-N960U using Tapatalk

[tangolima]

Nobody home apparently. Here are my inputs.

A sideways force is asserted on the bullet as result of cross wind. That force accelerates (velocity keeps increasing) the bullet sideways. More time the bullet is in cross wind, more it is deflected by the wind.

Following this mechanism, it is the wind at the firing line that matters the most, and the wind at the target has little or no effects.

Varying wind down the range is difficult. Different shooter has his own method. I use weighted average. Wind at firing line as weighting of 4, mid range 2, target 1. The weighted average is used for initial dope. Fire and adjust.

The wind accelerates the bullet sideways. Even if the wind stops, the bullet still retains that sideways speed; it just doesn't increase. So the poi shift at 300yd will still be more than 200yd. Certainly it is would be more if the wind is there. Halfway is probably a good enough estimate.

In the example, the wind deflection up to 200yd is 0.5moa / 100yd. Should the wind be there from 200yd to 300yd, the poi would be off by 3*0.5*3=4.5". Without the wind, it would halfway between 2" and 4.5", or 3.25".

-TL

Sent from my SM-N960U using Tapatalk

[stagpanther]

I'll take a stab at it--per the previous discussion of object entrained in an air mass (of which it appears that I'm one of only two people who ascribe to this theory on this thread )--I do not believe this is correct--the object does not "aquire" a sideways velocity vector upon exiting an air mass.

[tangolima]

Originally Posted by stagpanther

I'll take a stab at it--per the previous discussion of object entrained in an air mass (of which it appears that I'm one of only two people who ascribe to this theory on this thread )--I do not believe this is correct--the object does not "aquire" a sideways velocity vector upon exiting an air mass.

The bullet has sideways speed before crossing the 200yd line, correct? It cannot go to zero speed immediately unless a force of infinite magnitude is applied to it in the opposite direction.

-TL

PS. Notice that wind deflection is characterized as x moa / 100yd, instead of x inches / 100yd. That means the bullet is accelerating sideways in cross wind.

Sent from my SM-N960U using Tapatalk

[stagpanther]

The bullet has sideways speed before crossing the 200yd line, correct? It cannot go to zero speed immediately unless a force of infinite magnitude is applied to it in the opposite direction.

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

[tangolima]

Originally Posted by stagpanther

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

Sorry I meant the bullet's speed relative to ground, to which the target belong.

Say the cross wind up to 200yd is 10mph. If the bullet is indeed moving with the air mass, its sideways ground speed is also 10mph. When it cross the 200yd line, it's ground speed cannot drop instantly to 0mph, unless some infinite "brake force" is applied.

-TL

Sent from my SM-N960U using Tapatalk

[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]

Originally Posted by JohnKSa

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/exteri...-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]

Originally Posted by 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.

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]

Originally Posted by 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

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.