What wind influences bullet drift most?

mwezell

Your analogy is incorrect for one very important reason. A bullet is spin stabilized whereas a feather, or a golf ball, or an airplane, etc, is not. It is spin stabilization that causes a bullet to follow its trajectory, and it is spin stabilization that causes a slight vertical deflection in a horizontal crosswind, and vice-versa, and it is spin stabilization that causes spin drift, and it is spin stabilization that cause a bullet to turn to follow a crosswind.

I will agree that the only important thing is where the bullet meets the paper.

Ray
 
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Cheechako said:
mwezell

Your analogy is incorrect for one very important reason. A bullet is spin stabilized whereas a feather, or a golf ball, or an airplane, etc, is not. It is the spin stabilization that causes a bullet to follow its trajectory and it is spin stabilization that causes a slight vertical deflection in a horizontal crosswind, and vice-versa, and it is spin stabilization that causes spin drift.

Ray
Click to expand...

Agreed...but drag is what causes wind drift, whether spinning or not. Either way, the bullet moves with the wind. When the wind stops..so does the deflection. I don't see how vertical deflection, spin stabilization, or spin drift pertain to the subject...at least not in the context of the OP's question. My point is that the bullets path will be altered but not arched. It'd be more like this...________/-----------
Mike
 
So if the axis of the bullet is canted because of a wind, and the the wind stops, does the bullet stay canted for the rest of its flight or does rotational forces & drag straighten it back to zero degrees of cant? :confused:
 
Mike,
Doesn't the bullet have mass, and therefore momentum? This would seem to argue against your "drawing". It looks to me like you are saying that when a force is removed there is no further motion in the direction in which it was applied. It seems to me that if this was the case, bullets wouldn't make it very far down range.
Boyd
 
mwezell said:
. . .I don't see how vertical deflection, spin stabilization, or spin drift pertain to the subject...at least not in the context of the OP's question. . .
Click to expand...

I meant my post to illustrate that spin stabilization isn't everything - it's the only thing.

JMHO

Ray
 
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Rflshootr said:
So if the axis of the bullet is canted because of a wind, and the the wind stops, does the bullet stay canted for the rest of its flight or does rotational forces & drag straighten it back to zero degrees of cant? :confused:
Click to expand...

I don't know about "zero", but it'd correct itself to a "no wind" angle of attack, and continue straight on it's new path to the target...out of the no wind group unless held for. My whole point is that the path will change latterally but will not curve. Deflection and/or drift will stop when the force causing it stops, and the bullet will continue on it's course, albeit a new one. None of this is accounting for anything spin related.--Mike Ezell
 
Boyd Allen said:
Mike,
Doesn't the bullet have mass, and therefore momentum? This would seem to argue against your "drawing". It looks to me like you are saying that when a force is removed there is no further motion in the direction in which it was applied. It seems to me that if this was the case, bullets wouldn't make it very far down range.
Boyd
Click to expand...

Because the far greater force of mass and momentum is forward. Again, think about my feather analogy where the only constant force is gravity and the momentary force of wind only acts upon it while it's blowing. With a bullet in flight the greatest force is the one forcing it downrange, but gravity will eventually win. Forget about spin and velocity for a minute. The only difference between the feather analogy IMO, is mass and direction. The feather is going down because of the force of gravity, and the bullet is going downrange because of a force being exerted that is greater than the gravity. Exert side drag on either, and it will act the same way IMO.
 
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mwezell said:
. . .None of this is accounting for anything spin related.--Mike Ezell
Click to expand...

But Mike, spin stabilization is exactly what causes the bullet to maintain it's vertical and horizontal trajectory. You're disregarding the most important factor of all. It's what seperates rocks (and feathers) from bullets.

Get a copy of the Sierra loading manual. Read the chapter on Exterior Ballistics by McDonald & Almgren. Sierra Infinity is based on their expertise.

Ray
 
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Rflshootr said:
So if the axis of the bullet is canted because of a wind, and the the wind stops, does the bullet stay canted for the rest of its flight or does rotational forces & drag straighten it back to zero degrees of cant? :confused:
Click to expand...
No, the bullet "reorients" itself within one revolution. Both the "cant" and the "straightening out" refer to a downrange orientation. BTW, I'm uncomfortable with using "rotational forces" and "drag" to describe where the bullet points. Where the bullet points comes from orienting itself to meet the center of pressure. When some small component of that pressure comes from the wind, it then "cants" (from a downrange perspective).

Boyd:

Doesn't the bullet have mass, and therefore momentum? This would seem to argue against your "drawing". It looks to me like you are saying that when a force is removed there is no further motion in the direction in which it was applied. It seems to me that if this was the case, bullets wouldn't make it very far down range.
Click to expand...
As I understand it, this isn't quite the situation. It is true that if a force is remove from an object in flight, it will continue in that direction -- law of inertia. But we aren't talking about simply removing a force. We're talking about adding a force -- if, somehow, the bullet didn't reorient itself when the wind stopped, then it's momentum would continue. But it does reorient itself, so we're adding a new force -- doing work.

Does anyone remember just how far Jerry Hensler got with his empirical testing -- he had big fans, right, and was actually testing this question.
 
Charles

Both McDonald/Almgren and Vaughn agree that a bullet turns to follow a wind. IOW the centerline of a bullet is canted or turned horizontally from its original flight path. If that wind disappears, it is spin stabilization that makes it continue on that new path. Neither straight nor curved because it's not quite that simple, and a wind doesn't miraculously disappear. A ballistic chart is a great thing but it cannot account for all factors. Well, it can account for all factors but we, the shooter, don't have the ability to enter all factors into the calculation. GIGO.

Ray
 
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Charles,
The last time that I checked Sierra's external ballistics resuouce, on line, I believe that the article that I was reading said that bullets orient so that their points are slightly down wind, not into the wind (see below) Also, what is the source of the force that cancels out the bullet's lateral inertia? I find nothing to support this "goes straight, jogs over in a band of wind, and then proceeds parallel to the original, heading" theory, certainly no supporting facts. You may find this to be of interest.

[FONT=arial,helvetica,verdana,sns-serif]If the crosswind blows in the opposite direction, that is, from the right to the left as the bullet flies, the bullet must turn to the left to follow the wind. This necessitates a torque vector directed horizontally and to the left as the bullet flies. Such a torque can be generated by a negative lift force (directed downward as the bullet flies), and this can happen with a small, negative angle of attack. The final result is a crossrange deflection of the bullet to the left, and a vertical deflection of the bullet downward.

It came from this page, http://www.exteriorballistics.com/ebexplained/5th/43.cfm
That is listed on this index http://www.exteriorballistics.com/ebexplained/index.cfm

Enjoy



[/FONT]Boyd
 
Ray, I said it was never-never land. The reason is most postings on this subject assume certain things -- like the wind doesn't suddenly stop in that 1.6 seconds it takes to go 1,000 yards. Which is true. The wind has mass and momentum, too.

But there are some odd real-world situation, too. Like the 50-foot gully at about 35-degrees to downrange at the old Hawks Ridge range, where a right-to-left around 10 mph gave more vertical than horizontal. Or the almost-bullet-high berms at the Charlotte range, that meant a headwind shooting at 300 yards was "challenging."

So it's better to say just that, if I'm right about the physics (& that's a big maybe). All too often, people take a practical answer, then try to fit it to an impractical situation, and get the explanation all wrong.
I did find a post by Jerry Hensler (user name Jerry H)

The Boyer and Vaughn books say closest to the muzzle has the greatest influence on the POI at the target. That is provided there are no terrestrial effects on the wind. Buildings, berms, trees, etc can direct the wind in different ways than on an open field. In my experiments with a "Smart Flag" the relationship for three sensors with a PPC is: first 100, second 65, and third 42. With a rimfire at 50 yards and 5 sensors, it is: 100, 82, 67, 55, 45. These 2 examples do not match because of the great difference in exterior ballistics. PPC bc .26 @ 3400 and RF bc .14 @ 1050.
Click to expand...

In the general thread


http://benchrest.com/showthread.php?72730-Wind-Flags

As you said, a topic that comes up all too often.

Edit:

Boyd, I did give a source, but there have been so many intervening posts since this morning:

Harold Vaughn, p. 195, Rifle Accuracy Facts

Y'all need to read that, because I have also said other things, and those may well not be in Vaughn -- I could have misconstrued, or extended what I *thought* I knew too far. Always a problem with this topic.
 
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Charles

What you described is exactly what experienced long-range shooters know first hand. It's why a ballistic chart is only a guide. Unforunately, in the Internet age, some new shooters look at the charts as gospel and they scratch their heads in disbelief when they miss the paper completely with their first scientifically charted shot. I, on the other hand, am amazed each time I hit the paper at all.

The same goes for chronographs.:cool:

Ray
 
Charles, if I get what you're saying, you're talking about the drag force merely changing direction slightly when the wind stops.

Of course when the wind stops, the bullet will have a sideways velocity component.

You then have a sideways velocity relative to a zero wind, the total of which is non-zero. That's the same thing, mathematically speaking, as having a non-zero wind and a zero sideways velocity (which is what you had coming out of the muzzle). The numbers are just different.

With a ballistics program that has suitable inputs, you can calculate all this out, changing wind in steps as small as you like. I was thinking about updating my point mass program to allow the user to enter a distance-varying wind profile just to help people learn about this sort of question. (by the way the first cut of the program is online at http://bisonballistics.com/point_mass_calulations/new )

The way a point mass calculator does all this is very simple - it's just F = ma, transformed into a useful form. It assumes gravity points down, and drag is opposite of velocity (relative to air) and that's that. Crazy yaws during dynamic events like huge fictional wind shifts, and real effects like yaw of repose, are of course, ignored. For that you'll need a 6-DOF program, and that's well out of my league.

Not that I can even tell within +/- 2mph what the wind is doing right in my face, let a lone down range.
 
Charles,
I had the pleasure of discussing some of the points that he covered in his book, on the phone, with Mr, Vaughn (but not this one). He was a very smart fellow, but (and I know that I will take a shellacking for this one) he confused the solution to a graphic analysis of forces acting on a bullet with the actual direction of the wind. When, as he stated in his example, you have a 20 mph crosswind, and the relative wind vector is a little over a half degree off of the line that the bullet started on, as it left the muzzle, even if you accept that the bullet has turned that small amount into the wind, there is still a 20 MPH wind at 90 degrees, and the lateral velocity of the bullet is not that of the wind, so the bullet's angle of attack is no where near being directly into the wind. This is what he said, "It takes less than one fast procession cycle for the bullet to align itself to the relative wind vector and reduce the attack to the wind to near zero." The drawing that he provides is obviously schematic. Unfortunately those who have looked at it and others like it may have not given sufficient attention to that fact. In fact, the relative wind vector and the actual wind are about 89.5 degrees apart, in the 90 degree crosswind that he used in his example.
Boyd
 
What I said, is the way I THINK it is. Like I said, I'm no expert in this area. I do respect the opinions of the people posting though. Keep it up...I'm listening. I still see drag,(all directions), gravity, and force(forward)/mass as being the only things that affect the flight. I see drift being a result of drag in a crosswind. The vertical component, I attribute to spin.--Mike
 
And keep your comments coming Mike. None of us here, especially me, is an expert on this stuff either, and we can be wrong. Making each other think is what Forums are all about.

Vertical winds cause the bullet to turn and follow the wind just as a horizontal wind does. So there is more to it than just spin. But there is a vertical component of horizontal deflection, due to spin, so you're right to a certain degree.

Ray
 
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