For you ballisticians.

Boyd Allen said:
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.
Click to expand...

Boyd, we all agree that, qualitatively, the predictions of the "nose-into-the wind" and "sail area" models are the same -- the bullet moves downwind. I might note that, qualitatively, Rosie O'Donnell and Catherine Zeta-Jones are the same. :)

If we can't distinguish between the models qualitatively, we're going to have to be quantitative.

I've already given one example of different quantitative predictions of the two models. Wind deflection under "sail area" is proportional to time-of-flight, but proportional to lag time under "nose-into-the-wind."

Here's another quantitative difference. The "sail area" model predicts that wind deflection will increase as the square of crosswind speed, because drag ("side drag," in this case) increases as the square of velocity. So, a 20mph crosswind should move the bullet 4 times as far as a 10mph crosswind at any given range.

On the other hand, the "nose-into-the-wind" model predicts that total drag on the bullet increases as:

bullet velocity^2 + wind velocity^2

which is very, very nearly the same as bullet velocity^2 for typical small arms projectiles and terrestrial winds.

Further, the angle of the drag vector relative to the line of bullet departure is given by:

arctan(crosswind velocity/bullet velocity)

which is essentially LINEAR with changes in crosswind velocity as long as the crosswind velocity is much less than bullet velocity. Thus, the "nose-into-the-wind" model predicts a linear increase in wind deflection with crosswind velocity -- twice as much at 20mph compared to 10mph.

Now, to me, the difference between twice as much wind deflection ("nose-into-the-wind") and four times as much ("sail area") is a pretty big deal, even though both models have qualitatively predicted that the bullet will deflect more in a crosswind of 20mph than one at 10mph.

Of course, these different predictions are subject to experimental test, and, as you might imagine, they have been tested extensively. I don't need to tell you which model is supported by the evidence, do I? :)

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.
Click to expand...

I know the differences seem significant to you, but many aspects of the physics of flight don't depend on the method of stabilization, the projectile density, or the method of propulsion.

Toby Bradshaw
baywingdb@comcast.net
 
Last edited by a moderator:
Hang in there guys,

,,this thread is a real jewel! :D

The reason we have difficulty letting go of the "wind blowing on the side of the bullet" idea is because since birth, we have been attached to the earth's surface. A couple of flying lessons, conducted at low altitude by a competent flight instructor would dispell this myth once and for all and open your eyes to how bullets really drift in the wind.

Oh,,, by the way; you can remove a couple of stumbling blocks to learning if you will dismiss forever the ideas of shooting in a vacuum and the "delay theory." It's hogwash. :rolleyes: Once our bullets leave the muzzle, they fly through the earth's atmosphere experiencing tremendous drag which constantly slows them down and it is this decelleration that causes drift.

My .02 worth

Gene Beggs
 
Last edited:
"I've already given one example of different quantitative predictions of the two models. Wind deflection under "sail area" is proportional to time-of-flight, but proportional to lag time under "nose-into-the-wind."

Why must sail area be tied to time of flight? Why not use lag time? It's correlation has been demonstrated.
 
Boyd Allen said:
Why must sail area be tied to time of flight? Why not use lag time? It's correlation has been demonstrated.
Click to expand...

Lag time does correlate with wind deflection, but "sail area" predicts (incorrectly, as it turns out) that wind deflection is a function of time-of-flight only, not lag time.

Let me see if I can be clearer (and briefer!) this time.

Force = mass x acceleration

The "sail area" model assumes a "side drag" produced by the air molecules hitting the side of the bullet. Drag is a force. If the crosswind velocity is constant, and the bullet doesn't change its orientation relative to the crosswind, the bullet receives a constant "side drag" force. Since the mass of the bullet constant, the constant (drag) force produces a constant acceleration. Under constant acceleration, the distance traveled is:

1/2 x acceleration x time^2

So, in the example I gave before, I concluded that the distance traveled sideways by the bullet (aka wind deflection) must be proportional to the time-of-flight squared. NOT TO LAG TIME. Under the "sail area" model, the bullet is simply accelerating sideways at a constant rate, and how far it goes sideways depends only on how long it spends flying between the muzzle and the target (time-of-flight).

Luckily, nobody took me to task for making the "side drag" constant (even though, for most practical purposes, it can be considered constant over the typical flight time of a bullet). In reality, even though the crosswind velocity is constant, as the bullet picks up speed in the downwind direction it experiences an ever-decreasing "push" from the crosswind (just as if you push a person on a bicycle the force you are able to apply declines steadily as the bicycle approaches your maximum running speed).

Eventually, the bullet is moving sideways at the same velocity as the moving air mass within which it is flying and has no net "side drag" force on it at all. So, if I had done a more competent job I would have said that the sideways force is declining with time as the bullet approaches the velocity of the crosswind.

If, instead of a constant force, we say that the "side drag" declines with time-of-flight (that is, force is proportional to 1/time), I would have concluded that the wind deflection is proportional to:

1/2 x acceleration x time^2 x 1/time

which means that wind deflection is proportional to time-of-flight rather than the square of time-of-flight.

But, either way, the "sail area" model predicts wind deflection as a simple function of time-of-flight, and can't account for the dependence of wind deflection on time lag.

Boyd, you are absolutely right that any reasonably good physical model of the bullet's flight must account for the mathematical correlation between lag time and wind deflection. "Sail area" can't do that, at least if my description of the sail area physics is approximately correct.

But you will find that, as a rule, mathematical models are derived from physical models (or hypotheses), rather than the other way 'round. This is the case for wind deflection, too.

See: http://saxtech.eu/Ballistik/Didion/Didion.htm

Note that Peake's mathematical modeling is based on a physical model which assumes that the bullet's nose points into the relative wind (this can be seen in the expression for "combined airflow" -- re-read Bryan's post #249 in the other thread if needed). Under that assumption, Didion's approximation (with its prediction of proportionality between lag time and wind deflection) follows.

Well, I guess I missed the mark on "briefer." I probably didn't do any better on "clearer," either.:(

Toby Bradshaw
baywingdb@comcast.net
 
Toby, it's a difficult thing to explain; isn't it? But you're doing a much better job than I did.

Oh,, if only I could take each shooter for a thirty minute flight in a small aircraft. There's nothing like experiencing something first hand.

Gene Beggs
 
tobybradshaw said:
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.




Toby Bradshaw
baywingdb@comcast.net
Click to expand...

True, theoretically, but I feel it must be noted that this is where the bullet/plane analogy really falls apart :)

A bullet will be displaced much further/faster than can be accounted for by acceleration due to the sidewind. Unlike a boat or a plane or a sail.

al
 
alinwa said:
True, theoretically, but I feel it must be noted that this is where the bullet/plane analogy really falls apart :)

A bullet will be displaced much further/faster than can be accounted for by acceleration due to the sidewind. Unlike a boat or a plane or a sail.

al
Click to expand...


Al, I don't understand what you mean by,

"A bullet will be displaced much further/faster than can be accounted for by acceleration due to the sidewind."

Other than the crosswind (you call it 'sidewind') what would be causing the bullet to be displaced "further/faster" as you say?

Gene Beggs
 
What Al said got me to thinking. So a bullet, plane and boat all leave the 0 yard line to take a 1000 yard trip. Between 500 and 600 yard they have a 3 oclock 10mph wind...all other ranges there is 0 wind. So we have discussed how all three will act in that wind but what happens when they get to 601 yards? We know the plane and boat under their own power can then make corrections (take a new bearing) and the bullet will be affected by MOA rules but since it is already in motion (from the side push) would it stay in motion or would the spin forces overcome that motion?

I don't know about you guys but the plane, boat, sail, & Al on a bullet just makes this harder for me. Hard enough to get my pea brain wrapped around one thing let alone three or four. Now I have all my books out like I am writing a term paper just trying to keep up!:eek:
 
Don't worry,,

,,it doesn't matter whether or not we shooters understand exactly 'HOW' wind drift works, the end result is the same. Assuming a right hand twist, a crosswind from left to right drifts our bullets right and down; crosswind from right to left pushes them left and up. How much? Go to the sighter and check it out. :)

It's okay if you want to believe the wind "Blows on the side" of the bullet; although, that is not exactly how it works, and it's okay if you want to believe Magnus effect not gyroscopic forces cause the left/up, right/down thing.

These discussions are wonderful and I learn something from every one of them. Most shooters have a pretty good understanding of crosswinds and how they work; how about headwinds and tailwinds? Ask ten shooters and you'll get ten different answers. Got a question for you. ;)

Let's say we are shooting in a direct headwind of 10 mph; will the bullets strike higher or lower than in a dead calm? :eek: How 'bout a direct tailwind?

Later,

Gene Beggs
 
Just wondering

to make the appropriate comparison between an airplane and a bullet, wouldn't the airplane need to decelerate at the same rate as a bullet does once fired? How could this be done?

Pete
 
Gene Beggs said:
Al, I don't understand what you mean by,

"A bullet will be displaced much further/faster than can be accounted for by acceleration due to the sidewind."

Other than the crosswind (you call it 'sidewind') what would be causing the bullet to be displaced "further/faster" as you say?

Gene Beggs
Click to expand...



Well, first of all does it? :) DOES the crosswind adequately account for displacement?


Put a bullet into the airstream for two seconds.... drop it..... and it will not 'blow over' like a bullet which has been fired from a rifle. Mann built a machine to test this. He came up with a ratio of 640:1 drop VS drift.

THROW a bullet through a 2 second trajectory and it won't 'blow over' as if you'd shot it.

Shoot two bullets over the same distance and time and the faster bullet will drift (be dragged) off course more... Lob a lead slug 1500 yds using an initial velocity of 2000fps. Send a copper jacketed hyper-velocity .22 bullet over the same course but start it at 4000fps. SAME distance, SAME time ....................... different drift.

Two bullets, SAME time, SAME distance, one is blown over much more.... WHY?



This discovery back in the day is what prompted people to ask "WHY" does a bullet act differently...... Franklin W Mann couldn't explain this. In fact in his his last paragraph on the subject he states that no attempt is being made to explain the bulk of side drift. It's now been explained but is certainly not intuitive.

I don't know how to structure an airplane analogy such that two planes cover the same course in the same time but drift differently and this is the root of the OP...... he asks WHY??? How can a lighter bullet drift less or a faster bullet drift more??


How can a bullet that's in the airstream longer drift less???

GOOD question model14! :)






The answer to Mann's dilemma can be found in deceleration. Deceleration with a small sideward component accounts neatly for drag-induced "wind drift.".


Again, maybe this whole thing is just terminal terminology block but I don't think so :confused: as long as planes and sails are being used to illustrated bullet flight I feel that there's still a failure to account for the above example of 2 bullets, same time and distance etc...


al
 
Ohhh, and BTW, in the example of a large slow bullet VS a small fast bullet it can be noted that the large slow bullet presents a larger sail....... more surface for the wind to blow against.

but it drifts less...

al
 
A round ball turns into the wind, too

Boyd Allen said:
... if turning into the wind is important to wind drift, how does that work with a round ball?
Click to expand...

Boyd,
The center of pressure on a spinning round ball is offset from the center of mass by a crosswind, same as for a pointed bullet, thus its spin axis also turns into the wind. The round ball does not turn if it isn't spinning, because even though the pressure vector moves, it still passes through the center of mass, thus there is no torque created to turn the ball.

Hang in there, you will see the light. I'm sure other readers have the same questions as you, and hopefully are learning from this thread.

Cheers,
Keith
 
mks said:
Boyd,
The center of pressure on a spinning round ball is offset from the center of mass by a crosswind, same as for a pointed bullet, thus its spin axis also turns into the wind. The round ball does not turn if it isn't spinning, because even though the pressure vector moves, it still passes through the center of mass, thus there is no torque created to turn the ball.

Hang in there, you will see the light. I'm sure other readers have the same questions as you, and hopefully are learning from this thread.

Cheers,
Keith
Click to expand...

mks,

this one makes me wonder...... :) not about the drift, a round ball DOES act exactly like a pointed bullet re drift....... but about the "pointing into the wind."

Without a moment arm there is nothing to produce a torque, I'm not sure that the round ball does adjust it's spin axis. Irrelevant to the question but interesting....

al
 
You must log in or register to reply here.
Back
Top