For you ballisticians.

[tobybradshaw]

Than why are my bullet holes in the target always round? How much is a bullet canted by a 20 mph 90 degree crosswind? How has this been verified?

See Bryan's post #249 in

http://benchrest.com/forums/showthread.php?t=43279&page=17

if you want to calculate the change in axis angle. You'll need to specify a bullet velocity in addition to the wind velocity and direction.

The angle will be less than 1 degree for typical projectile and wind velocities, which you won't be able to see from the shape of the bullet hole on a target.

But since you mention bullet hole shape, let's think of this another way. Imagine firing a bullet upward at a 20 degree angle. Further imagine that the bullet remains spin-stabilized throughout its flight. What shape does the hole in the (distant) target have as the bullet is on the downward part of its trajectory, does the bullet go through the target point up or point down, and why?

When the USS Missouri touches off one of those 16" guns at a 45 degree upward angle, do you reckon the projectile lands point first or base first, and why do you think so (i.e., what forces have produced any change in orientation of the projectile long axis relative to its departure angle from the bore)?

There has been plenty of observation of projectiles in wind tunnels, spark photographs, and at impact to verify that, as long as they remain stabilized, they point into the relative wind.

Toby Bradshaw
baywingdb@comcast.net

 

[tobybradshaw]

An excerpt from Sierra that has the bullet pointing with the wind, not into it.

The bullet's center of mass does move to the left in a right-to-left crosswind. But the center of pressure of a spin-stabilized projectile is forward of the center of mass (making it unstable when not spinning), so the nose of the bullet moves in the opposite direction from the center of mass (i.e., into the relative wind).

Throw a football in a crosswind and see for yourself!

Toby Bradshaw
baywingdb@comcast.net

 

[alinwa]

But since you mention bullet hole shape, let's think of this another way. Imagine firing a bullet upward at a 20 degree angle. Further imagine that the bullet remains spin-stabilized throughout its flight. What shape does the hole in the (distant) target have as the bullet is on the downward part of its trajectory, does the bullet go through the target point up or point down, and why?

When the USS Missouri touches off one of those 16" guns at a 45 degree upward angle, do you reckon the projectile lands point first or base first, and why do you think so (i.e., what forces have produced any change in orientation of the projectile long axis relative to its departure angle from the bore)?

Toby Bradshaw
baywingdb@comcast.net

I'll take a lash at it but'cha won't like it!

I'm a glutton for punishment..... and this is how we learn. I don't even mind being shown WRONG as long as the RIGHT answer is established.



in each case the bullet goes through the target on a steep descending angle but point first in line with its trajectory. It has to to stay "centered" on the wind it sees. "Centered" because it's yawing a little (yaw of repose) due to precession.

This is just one of the place where I disagree with Nennstiel Rupprecht.... they insist that an "overstabilized" bullet will go through the target with its nose up, its axis essentially unchanged from launch angle.


Now, to your question " (i.e., what forces have produced any change in orientation of the projectile long axis relative to its departure angle from the bore)?
"

"WHY" does the bullet nose over??? Because it's falling, the wind on its belly tips it over.

After a couple seconds falling through the air it's got a nearly 50mph wind blowing UP on it. Or will someone dispute/refute this? Why do so many people insist that the "wind" generated by gravity drop doesn't exist?

And yes this tipover has been demonstrated, clear back into the mid 19th century folks were firing bullets down the beach and examining the holes in the sand.

al

 

[tobybradshaw]

And now for a sporting proposition ...

Just as Kekule’s vision of a snake swallowing its own tail led him to solve the circular structure of the benzene molecule, it came to me in a dream.

Physicists, aerodynamicists, and ballisticians have agreed for more than a century that fin- and spin-stabilized projectiles point into the relative wind in flight, canted relative to the line of departure from the bore in a crosswind. Indeed, this behavior is what DEFINES stability. Predictions of wind deflection based on this “nose into the wind” model are quite accurate, as determined experimentally.

Many shooters, on the other hand, subscribe to the “sail area” model, where wind deflection is produced by crosswind blowing on the side of the bullet, with the bullet maintaining its long axis parallel to the line of departure from the bore.

Is there an experiment that the average shooter can carry out to distinguish between these competing hypotheses? According to my dream, there is.

LET’S SHOOT BULLETS BACKWARDS!

The “nose into the wind” proponents (you know, the eggheads who brought you powered flight, the moon landing, and nuclear weapons) will claim that a typical pointed bullet will have much more drag if shot backwards (base-first), and so will have more drop and more wind deflection if started at the same muzzle velocity as the bullet fired in the normal point-first orientation.

The “sail area” intuitivists must predict that, while drop will be greater because of the increased “frontal drag” of a backwards bullet, the wind deflection should be the same whether the bullet is pointed forward or backward, because the “side drag” from the crosswind is the same regardless of bullet orientation enroute to the target.

Now, some “sail area” intuitivists are going to balk at this experiment because they will claim that any additional wind deflection in the backwards bullets is due to the increased time of flight, and therefore the longer time the “sail area” is being pushed on by the crosswind.

Since even die-hard “sail area” proponents seem to trust ballisticians to produce reasonably accurate drop tables based on aerodynamic models, to satisfy the “time of flight x sail area” crowd we could raise the muzzle velocity of the backwards bullets until they produced a time of flight to the target equal to (or even a little shorter than) that of the forward-pointed bullets. This would be done by estimating BC using drop data or chronograph data, and plugging the numbers into a ballistics program to find the right (i.e., higher) muzzle velocity needed for the backwards bullets.

Just for grins, I used Art Pejsa's program

www.jacksonrifles.com/files/pejsa ballistics.xls

to see how this experiment might play out. Let's shoot a 6mm bullet point-first with a BC of 0.435 at 2000 fps at a 300yd target with a 10mph crosswind. Let's turn the same bullet around backwards and just take a wild guess that the base-first BC is 0.270.

MV(fps) BC drop(MOA) deflection(MOA) time of flight(sec)
2000 0.435 5.5 4.0 0.52
2000 0.270 6.9 7.2 0.58

What the heck. Let's blast that bass-ackwards bullet downrange at 2500fps while we're at it.

2500 0.270 4.5 5.0 0.45

So, the "nose into the wind" physicists are claiming that the backwards bullet, even when speeded up to produce a shorter time of flight to 300yd, will have a flatter trajectory (by 3in at 300yd) BUT MORE WIND DEFLECTION (by 3in at 300yd) than the same bullet shot point-first!

Surely the "sail area" proponents will predict that the faster backwards bullet will have the least wind deflection of all.

Now, before Al runs to his loading room to jam some bullets into cases upside down, there is the matter of the amount of the wager. I have $100 that says those ballisticians understand the physics of bullet flight pretty well, and that the forward-pointing bullets will have less wind deflection than the base-first bullets either 1) when shot at the same muzzle velocity; or, 2) shot with the same time of flight to the target.

Who wants to bet $100 on their intuition?

Toby Bradshaw
baywingdb@comcast.net

 

[tobybradshaw]

in each case the bullet goes through the target on a steep descending angle but point first in line with its trajectory. It has to to stay "centered" on the wind it sees.

Of course it does. If the descent angle is steep enough the hole in the (vertical) target will show the point down and the base up.

Or you could just watch a football game to see how an arcing spin-stabilized projectile behaves during its flight.

Toby Bradshaw
baywingdb@comcast.net

 

[alinwa]

"Indeed, this behavior is what DEFINES stability. "



Toby Bradshaw
baywingdb@comcast.net

well said.....






Now, as per the b'zackwards spin-stabilized sand socks experiment...... not a chance this guy

I've already done a bunch of experimentation with weight-forward and weight-stabilized designs..... tail waggin' b'zastards all....

I will note though A'gain, that TOF is of no real note. It's all about loss of velocity. A bullet can be in the wind stream for three or four seconds (Scheutzen rifle bullets often ARE!) and show less wind deflection than a 4000fps 22-250 over the same yardage.

al

 

[Vibe]

I'm a glutton for punishment.....

Truer words were never presented.

and this is how we learn. I don't even mind being shown WRONG as long as the RIGHT answer is established.

Right being defined as one he agrees with.

in each case the bullet goes through the target on a steep descending angle but point first in line with its trajectory. It has to to stay "centered" on the wind it sees. "Centered" because it's yawing a little (yaw of repose) due to precession.

Try it with a "light for caliber" bullet from a fast twist. They hit nose up at longer ranges...and are less accurate because of it.

This is just one of the place where I disagree with Nennstiel Rupprecht.... they insist that an "overstabilized" bullet will go through the target with its nose up, its axis essentially unchanged from launch angle.

Try it yourself. It's the reason why the results are vertical slashes as opposed to round holes in the target.

 

[Boyd Allen]

"The “sail area” intuitivists must predict that, while drop will be greater because of the increased “frontal drag” of a backwards bullet, the wind deflection should be the same whether the bullet is pointed forward or backward, because the “side drag” from the crosswind is the same regardless of bullet orientation enroute to the target."

Wrong...because of the decrease in BC (from being shot tail first) the time to target in increased (lag time as well, if that makes you happier), increasing the time that a crosswind is acting on the bullet, thereby increasing drift.

 

[tobybradshaw]

"The “sail area” intuitivists must predict that, while drop will be greater because of the increased “frontal drag” of a backwards bullet, the wind deflection should be the same whether the bullet is pointed forward or backward, because the “side drag” from the crosswind is the same regardless of bullet orientation enroute to the target."

Wrong...because of the decrease in BC (from being shot tail first) the time to target in increased (lag time as well, if that makes you happier), increasing the time that a crosswind is acting on the bullet, thereby increasing drift.

Boyd, I already addressed this argument:

Now, some “sail area” intuitivists are going to balk at this experiment because they will claim that any additional wind deflection in the backwards bullets is due to the increased time of flight, and therefore the longer time the “sail area” is being pushed on by the crosswind.

Since even die-hard “sail area” proponents seem to trust ballisticians to produce reasonably accurate drop tables based on aerodynamic models, to satisfy the “time of flight x sail area” crowd we could raise the muzzle velocity of the backwards bullets until they produced a time of flight to the target equal to (or even a little shorter than) that of the forward-pointed bullets. This would be done by estimating BC using drop data or chronograph data, and plugging the numbers into a ballistics program to find the right (i.e., higher) muzzle velocity needed for the backwards bullets."

So, are you in for $100, or what?

Toby Bradshaw
baywingdb@comcast.net

 

[Old Gunner]

Not quite a true "point first" into the wind it feels",

Is that a quote from the great Jedi Master Yoda?

Sorry but I couldn't let it pass.

Looks like theres a lot of reading to be done and much for me to learn, always happy when that happens.

I'm still trying to figure out how the plastic ballistic tip bullets can show less wind drift, when the very similar long nosed match bullets of the early 20th century ended up all over the place once the wind kicked up a bit.
This was put down to a lever effect , the long nose increasing the force the air could put on the heavier base.

Less ambitious light weight nose plug bullets seem to hold their accuracy quite well, the MkVII .303 for example, and the hollow cavity Russian poison bullet. Both are known for tumbling in flesh, but seem to remain stable on the way down range.

Length of bullet exposed to cross wind would probably make little difference, at least at supersonic speeds, the air being pushed away by the shockwave.

At the speeds and density of the medium I'd figure that bullet flight through air would be more analogous to a unguided torpedo moving at top speed through water. Instead of a pressure on the projectile the entire environment moves carrying the bullet with it, once intial enertia is overcome shortly after leaving the muzzle.
Thats just the way it would seem to me, I'm probably wrong.

Of course one must factor in "Bullet Drill", the projectile rolling to one side according to rifling twist.

 

[Boyd Allen]

Obviously, I did not read far enough. Still, they would not have the same lag time. (Vacuum time at higher velocity would be shorter than point first vacuum time at lower velocity.)

 

[Tony Shankle]

just for you Old Gunner

 

[tobybradshaw]

Obviously, I did not read far enough. Still, they would not have the same lag time. (Vacuum time at higher velocity would be shorter than point first vacuum time at lower velocity.)

Boyd, see if you agree with my conversion of your "sail area x time" model into a mostly verbal physical model. I don't see a place for lag time in "sail area x time," at least as I understand it. To wit:

The bullet leaves the muzzle pointed straight at the target. On its nose it "feels" a headwind equal to the downrange component of its velocity, and on its side it "feels" a sidewind equal to the horizontal component of the crosswind ("horizontal" meaning perpendicular to the long axis of the bullet and, therefore, also perpendicular to the straight line from the bore to the target).

During its entire flight, the nose of the bullet remains pointed straight downrange (allowing for a little nose-up on the way up and nose-down on the downward arc if you like -- this is only a degree or so in any usual case). During its entire flight, the bullet also experiences a constant force on its side from the crosswind. (You described this as PSI * area of the bullet's side, and pounds/in^2 x in^2 = pounds of force.)

Now, if the bullet experiences a constant "side force" perpendicular to the bullet's long axis, by the Newtonian definition of force (F=ma) the bullet must experience a constant acceleration. Under constant acceleration, the distance traveled (sideways = downwind, in this case) is given by (a x t^2)/2, where t=time-of-flight. So, in this "sail area * time" model, wind deflection is proportional to the square of time-of-flight. NB: Not proportional to lag time, but to the square of time-of-flight.

How -- exactly (i.e., mathematically, physically) -- do you account for the relevance of lag time (not just time-of-flight) for wind drift in the "sail area x time" model?

---------------------------------

The "nose into the wind" model accounts for the effect of lag time (instead of time-of-flight) on wind deflection this way:

Because the bullet is spin-stabilized, the force (drag) on the bullet is always (MUST always be) straight down the long axis of the bullet. The bullet flies nose-first into the relative wind. In a crosswind the relative wind is at an angle to the line of departure from the bore, so the bullet's nose turns into the crosswind until the aerodynamic forces on its sides are equal. Now the drag force has a crosswind component (in the direction of the wind) as well as an "uprange" component pointed back at the bore centerline. If the shooter could see the bullet, the bullet would be canted relative to its line of departure and to its downrange trajectory, nose into the wind. The bullet just "feels" a headwind, always (i.e., drag always pushes the bullet straight back, from the bullet's perspective).

Unlike the constant force on the side of the bullet in the "sail area x time" model, under the "nose into the wind" model THE (DRAG) FORCE ON THE BULLET IS NOT CONSTANT. In fact, the drag force is decreasing as the square of velocity (e.g., at 1500fps the drag force is only 1/4 as great as it was at 3000fps -- 1/2 the velocity, (1/2)^2 the drag force). So, what matters for wind deflection under the "nose into the wind" model is not the time-of-flight, but the rate at which velocity decays as a result of changes in drag force. This rate of decay can be expressed in terms of lag time as in the Didion approximation.

One more thought experiment to be posted separately.

Toby Bradshaw
baywingdb@comcast.net

 

[tobybradshaw]

suspension of disbelief and a bullet, plus a fan

Imagine if we had a magnetic bullet suspended above a large flat magnet of opposite polarity. Imagine that we get this bullet spinning fast enough to stabilize it, but with no translational velocity. It's just sitting there, a few inches above the flat plate, spinning but otherwise motionless.

Now we put a fan at the 1 o'clock position relative to the pointed end of the bullet, and turn on the fan. We look down on this experiment from above.

The "nose into the wind" model predicts that the bullet (specifically, its center of mass) will begin to move in a 1-to-7 o'clock direction as it responds to the drag force from the moving air. The acceleration will be fairly rapid at first because the right (upwind) side of the bullet is being hit by air molecules along its entire length (i.e., the projected area is large). The unbalanced aerodynamic force on the sides of the bullet produce a downward deflection of the bullet's nose, resulting in a torque moment to the right, putting the bullet's nose into the wind, and aligning the bullet's long axis on the 1-to-7 o'clock path.

The spinning bullet is stabilized nose-into-the-wind by constant small motions of the nose and resultant torque moments between the center of mass and the center of pressure. The bullet continues along its 1-to-7 o'clock path but the the acceleration is much less now, because drag (force) is minimized with the bullet's nose pointed directly into the wind. The bullet's acceleration declines continuously as the bullet approaches the velocity of the wind, and eventually the bullet is moving at a constant velocity (a=0) equal to that of the wind.

-----------------------------

Now, what do the "sail area" theorists predict will happen to the stationary spinning bullet when the fan is turned on?

Toby Bradshaw
baywingdb@comcast.net

 

[Tony Shankle]

All true, but there's no use worrying about the details until the big picture comes into focus.

Toby Bradshaw
baywingdb@comcast.net

Toby,

With due respect, it is part of my big picture. When my first round hit or miss counts at 1000 or 1200 yards on a .5 moa or 1 moa plate then that 9 inches of spin drift is every bit as important to me as the wind. Even 2 inches of Coriolis can cost me that hit.

If I get sighter shots from a bench and machine-gun rounds down range then I agree there are details you can ignore. When you shoot for 30 minutes or an hour at various ranges and every shot counts then the detail determines who gets the lumber.

With that said, I am really enjoying your (and everyones) post. I would just like to lure you real "Jedi's" into all the details so I can learn something about the whole picture in lieu of everything about one piece.

Thank you all for your post(s), this has turned out to be a good thread!

Tony

 

[tobybradshaw]

With due respect, it is part of my big picture.

Sure, Tony, I understand what you're saying.

I will resort to quoting George E. P. Box. "All models are wrong. Some are useful."

I'm just trying to get the "wrong-est" models out of the way as a first step, and not feeling especially successful.

Toby Bradshaw
baywingdb@comcast.net

 

[Boyd Allen]

I am under the impression that the correlation of wind drift to lag time has be verified by actual measurement. Am I wrong?

 

[Tony Shankle]

Imagine if we had a magnetic bullet suspended above a large flat magnet of opposite polarity. Imagine that we get this bullet spinning fast enough to stabilize it, but with no translational velocity. It's just sitting there, a few inches above the flat plate, spinning but otherwise motionless.

Now we put a fan at the 1 o'clock position relative to the pointed end of the bullet, and turn on the fan. We look down on this experiment from above.

The "nose into the wind" model predicts that the bullet (specifically, its center of mass) will begin to move in a 1-to-7 o'clock direction as it responds to the drag force from the moving air. The acceleration will be fairly rapid at first because the right (upwind) side of the bullet is being hit by air molecules along its entire length (i.e., the projected area is large). The unbalanced aerodynamic force on the sides of the bullet produce a downward deflection of the bullet's nose, resulting in a torque moment to the right, putting the bullet's nose into the wind, and aligning the bullet's long axis on the 1-to-7 o'clock path.

The spinning bullet is stabilized nose-into-the-wind by constant small motions of the nose and resultant torque moments between the center of mass and the center of pressure. The bullet continues along its 1-to-7 o'clock path but the the acceleration is much less now, because drag (force) is minimized with the bullet's nose pointed directly into the wind. The bullet's acceleration declines continuously as the bullet approaches the velocity of the wind, and eventually the bullet is moving at a constant velocity (a=0) equal to that of the wind.

-----------------------------

Now, what do the "sail area" theorists predict will happen to the stationary spinning bullet when the fan is turned on?

Toby Bradshaw
baywingdb@comcast.net

Great example!

Question, would the Magnus effect also make the entire bullet move down assuming a clockwise/right spin? Or is it just the nose?

 

[Gene Beggs]

Great Thread!

Toby Bradshaw should receive the "Atta' Boy of the year" award; not only for his complete understanding of how wind affects bullets in flight, but also for his writing and teaching skills. Listen carefully to what he says; Toby knows exactly what he is talking about!

After reading several of his posts I knew Toby was a pilot, and I guessed him to be an aeronautical engineer, but I was only half right. He is indeed a pilot, an aerobatic pilot no less, but his profession is teaching. And he's damn good at it; isn't he?

Keep up the good discussion; I'm enjoying this immensely.

Gene Beggs

 

[Boyd Allen]

It will be blown off of the magnet and fall to the table, traveling in the same direction that the wind is blowing. If the magnet was of infinite size, it would still be blown in the direction of the wind. BTW, the whole turning thing is not important to the sail model if we are talking about a bullet that has been fired from a rifle. The change in sail area that can be attributed to less than a degree of turn is insignificant. It may turn into, or with the wind. The difference is not important. Also, for the pilots, airplanes are not spin stabilized, very dense, or propelled by momentum, unless they are gliding. All of these differences are significant when making comparisons to a bullet in flight. I am not saying that pilots don't know about airplanes, just that airplanes have significant differences from bullets that limit the usefulness of comparisons. In what you have called the sail theory of wind drift the path of the bullet will still be determined by the forces applied to it from the wind and its decreasing (due to the cumulative effects of drag) momentum. Since the wind ( in our model) remains constant and forward velocity is decreasing, the influence of the wind increases over time. The bullet's slight turning is irrelevant, indeed if Sierra has it wrong, it has not had much effect on their ability to produce a functional ballistic program.