For you ballisticians.

...because this isn't the case.

With respect - no.

...it's probably simpler to use conditions that avoid transonic flight...

The exception I'm aware of ... has to do with increased drag in the transonic range ...

Right.

Which is exactly the range (850 fps vs 1700 fps) that I was talking about. The exception doesn't stop (ie, wind drift doesn't get back to 850 fps levels) until velocity increases to about 2300 fps.

I'm a pistol silhouette shooter. This is the range in which I live. BPCR shooters have to deal with it, too.

When I first stumbled on this it seemed quite counter-intuitive so I wrote Sierra and asked for some explanation why their ballistics software was showing less wind drift for the same bullet over the same distance in the same conditions when it was launched at 900 fps than when it left the barrel at 1400. Here's a part of the response I got from Dr. Bill McDonald:

You have discovered a phenomenon which is real and which affects both handguns and rifles firing bullets at transonic and subsonic velocities. We first ran into the same effect several years ago when we were working with bullets from a .45-70 rifle fired at velocities ranging from about 1200 to 2000 fps. I could not believe the wind deflection numbers either until I performed a considerable amount of analysis of the physics involved. I analyzed the situation using the closed form ballistic solutions that we published in the second and later editions of Sierra’s Reloading Manuals. To shorten a very long story, the apparent anomaly in wind deflection occurs because the air drag on a bullet decreases so radically when velocity falls through the lower transonic and upper subsonic velocity regions.

At muzzle velocities above about 1500 fps, wind deflection decreases as muzzle velocity rises because lag time (difference between actual time of flight and the time of flight the bullet would have in a vacuum) decreases as the bullet travels faster between muzzle and target. So, as a result maximum wind deflection occurs for a muzzle velocity somewhere in the 1300 to 1500 fps region. At lower muzzle velocities the air drag on the bullet decreases very rapidly, so that the bullet is not slowed down nearly so much by drag. Then, the actual time of flight begins to approach the vacuum time of flight, and wind deflection then decreases at these lower muzzle velocities.

He then went on to talk about still lower velocities as seen in air guns and to spend quite a bit of time taking me to task for some very sloppy arithmetic I had included in my original email to Sierra. :rolleyes:

So, yeah, a bullet launched at 900 fps will drift less than one launched at 1700.

I see, after doing some reading, that this does NOT contradict the Didion approximation you referenced.

Learning stuff is fun. Seriously.
 
Sure, that makes sense for loads that are transonic part of the time. I didn't realize that anybody other than rimfire shooters dealt with transonic velocities on a regular basis.

Everything I shoot competitively stays well above the speed of sound from muzzle to target, so for me more muzzle velocity (with the same bullet) always produces less wind deflection.

Toby Bradshaw
baywingdb@comcast.net
 
A couple of things to keep in mind...Your model should work with a round ball. If it does not, you might want to take another look. As to airplanes and rockets, they are not spinning like a bullet, so there are significant limitations to the usefulness of comparisons between them. Yea, I know...stirring more stuff:D Oh and one more thing, just because things happen at the same time, does not mean that there is a causal relationship, or that all of them are important to a particular outcome. For example, if turning into the wind is important to wind drift, how does that work with a round ball?
 
A couple of things to keep in mind...Your model should work with a round ball.

Boyd, what makes you think the model doesn't work with a spin-stabilized round ball?

For example, if turning into the wind is important to wind drift, how does that work with a round ball?

Turning into the wind isn't important for wind deflection, it's just that a spin- or fin-stabilized projectile always points its nose into the relative wind because of the relationship between the center of pressure, the center of mass, and, in the case of a spin-stabilized projectile, gyroscopic forces.

DRAG is what produces wind deflection. A round ball is symmetrical, and so doesn't have a "nose" to point into the relative wind, but in a crosswind a round bullet is pushed off the line of bullet departure by drag just like a pointed bullet is. A supersonic round ball (in typical sporting arms calibers) has a very low BC -- around 0.07 -- so drag is very high, deceleration is therefore rapid, and wind deflection is large when compared to a pointed bullet of the same weight.

Plug the BC and muzzle velocity of a round ball into a ballistics program (which uses the same algorithms we have been discussing) and it will output the correct wind deflections.

Toby Bradshaw
baywinddb@comcast.net
 
I'm gonna' try this, once... ;)

Any of y'all who want to point out "I already said that" .... or not... feel free. This is MY shot. There will be repetition. :D


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

The references to 'lag time' are generally referring to the difference between a given bullet's flight in a vacuum VS reality. The reality is that bullets must pummel through AIR not vacuum, and different bullets have different lag times due to their shape. Different bullets slow down at different rates.

It's reasonable that a short/fat bullet shaped like a soup can will slow down 'more' or 'faster' than a long pointy bullet shaped like a javelin. AND, the bullet which slows down more will always show more "winddrift.".

Here's why...

Take this from the bullet's perspective......

Let's say that a bullet leaves the bore at about 3000fps in a 20mph crosswind....... you are riding the bullet. Your flight will be between one and two seconds, a short ride. Here's what you'll see/feel:


From the bullet's view
you see a 2000mph headwind and a 20mph crosswind. The 20mph crosswind is negligible, you don't even notice it. If you were sitting still you wouldn't even be significantly "blown over" in the second or two of flight.

Really.... if you DROP a bullet off the barn roof say 40-50ft it won't blow way over will it???

So why does the wind "blow" that 22-250 over 15ft at 1000yds? In less than a second?

Here's why:

From the bullet's view You're feeling 2000mph from the front and 20mph from the side. The "resultant vector" is angled a little off your flight path.

From the bullet's view that 2000mph wind is pushing you BACKWARDS......

And from the bullet's view that 20mph wind is adding a little sideward motion.

A bullet which is very streamlined (high BC) will slow down LESS, the 2000mph wind will not get enough grab to send you BACKWARDS as fast.. "Flight time" has really nothing to do with it. The key is how much does the bullet slow down? That slow down is actually backing up when viewed from the bullet.

You're riding the bullet remember??? Your cheeks'r flappin' in that 2000mph headwind. And you're backing up, being BLOWN BACKWARDS by a 2000mph wind........

And just a little sideways by the 20mph wind.

You duck down on the bullet..... lay on your belly so's you don't catch much wind..... and the backing up is minimized. (High BC bullet)

Sit up and WHAMMM!!!! you're being blown backwards like CRAZY! (Low BC bullet)

So now it becomes obvious that if you back up just a little bit that sideward component isn't all that big a deal..... but if you back up A LOT then even that little sideways becomes a big deal.

If your car is sitting crooked and you back up for a second s l o w l y you won't get very far off line......... but if you're a little sideways and you TROMP IT!! you're gonna' get perty sideways perty quick.......


A high BC bullet which bucks wind really well gets the same sidewise push as the low BC bullet......... just a slight "angle"........ but the low BC bullet, the 'soup can', backs up 'wayyy further!

Consequently it drives further sideways off the path. The path of its original trajectory.

Hey, it's a try! :) The engineering types 'er gonna' HATE IT because it's pure-dee laymanese except for one use of "resultant vector", :D but it's essentially accurate....

LOL

al
 
Al,
The force that the wind exerts on an object can be expressed in PSI, and the area that it acts upon is the silhouette that you would see when looking at the bullet parallel to and in the direction of the wind. The mass of the bullet is also a factor in determining its rate of wind induced lateral acceleration, as is time. If you drop a bullet 50 ft. in a 20 mph crosswind, it will be blown sideways. Bullets don't back up after they leave the muzzle, from any perspective, or relative to anything terrestrial, they decelerate until they stop. When a sail boat is capsized by the action of the wind on its sails, we say that it has been blown over, not pulled or sucked over. This is a simple convention of language.

How has the shooting been going? Discovered anything new lately?
Boyd
 
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The force that the wind exerts on an object can be expressed in PSI,

Strictly speaking, the force should be expressed in pounds or Newtons or some other unit of force. PSI is force/area.

and the area that it acts upon is the silhouette that you would see when looking at the bullet parallel to and in the direction of the wind.

The drag force that acts on the bullet can't be calculated from PSI and frontal area alone -- the form factor (coefficient of drag) also matters.

I'm not clear on your description of the bullet's silhouette -- do you mean that the bullet is pointed straight into the relative wind, or that one side of the bullet has more air molecules hitting it than the other?

The mass of the bullet is also a factor in determining its rate of wind induced lateral acceleration, as is time. If you drop a bullet 50 ft. in a 20 mph crosswind, it will be blown sideways.

To a first approximation, the only things affecting the acceleration of the bullet are its mass and the force acting on it.

If you are saying that the drag force changes with time, then I understand where you're coming from.

Bryan posted (#249) a linearized version of the physics involved that I thought gave the simplest perspective on the situation.

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

Toby Bradshaw
baywingdb@comcast.net
 
I think that force per square inch is appropriate.
For example, in a 90 degree crosswind, the force of the wind on the bullet can be seen as the profile (side view) in square inches, times the unit force in PSI. As the wind varies from 90 degrees, its influence is diminished in proportion to the decrease in the area of the bullet's profile from the angle that the bullet is striking it. At the extremes, the area is the full profile of the bullet or a circle. This is why a 90 degree crosswind, of a given speed, has the largest influence on a bullets flight, larger sail area.

Apparent wind is a construction used to solve a problem. If you push on a coffee cup with two fingers, at 90 degrees to each other, each finger pushing with equal force, in the direction that it is pointed, the cup will move along a line that bisects the angle between them. This is a practical demonstration of the sort of vector analysis that has caused some to speak of an apparent wind. If you believe that the actual wind has changed, you may want to look for the apparent finger that is pushing the coffee cup. Saying that a thing moves according to the sum of the forces acting on it is quite different than saying that a new force is somehow created.

I said lateral acceleration to make a distinction from the bullets primary acceleration, since wind drift is not the bullets primary acceleration. Drift is dependent on air density, wind velocity and time that it is acting on the bullet. The time that it is acting on the bullet is a function of BC initial bullet velocity, and distance to target. I was attempting to address wind drift.

My earlier mentioning of the round ball referred to a previous discussion where much was made of an elongated bullet turning so that it is pointing directly into the "apparent wind". We agree. I was poking at some who offered explanations in that discussion.

Are you saying that a bullet will not be blown sideways by the wind as it falls from a height of 50'?
 
Ride it Al...ride it

So I said "tuck in Al" and he said "shut up and take the picture"....


20qmcs01.jpg
 
For example, in a 90 degree crosswind, the force of the wind on the bullet can be seen as the profile (side view) in square inches, times the unit force in PSI.

I guess we'll have to agree to disagree on this one. Except for a very brief interval after leaving the muzzle, while the pressure from the crosswind is equilibrating on the spinning bullet, the bullet is pointed with its nose straight into the relative wind. A spin- or fin-stabilized projectile can't fly (stably) any other way. Seen from the shooter's point of view, the bullet will be canted (into the crosswind) relative to its angle of departure from the bore, and relative to its downrange trajectory.

As the wind varies from 90 degrees, its influence is diminished in proportion to the decrease in the area of the bullet's profile from the angle that the bullet is striking it. At the extremes, the area is the full profile of the bullet or a circle. This is why a 90 degree crosswind, of a given speed, has the largest influence on a bullets flight, larger sail area.

Nope. The reason the 90 degree crosswind produces the largest deflection is because it has the largest component of drag in the direction perpendicular to the bullet's flight.

Either mental image -- wind blowing on the side of a bullet, or drag along the long axis of the bullet with the bullet's nose canted into the wind -- produces the same expected effect -- wind deflection.

But only one of these mental images is physically correct, and a spin- or fin-stabilized projectile isn't stable with more pressure on one side than the other. It has to fly point-first into the only wind it "feels."

Further, if the wind really did impinge on the side of the bullet, we WOULD need a separate "frontal BC" and "side BC", which is how this thread started. But one BC will do (with all its problems) because the bullet always presents the same profile (nose first) to the drag produced by the air.

Are you saying that a bullet will not be blown sideways by the wind as it falls from a height of 50'?

Sure it will. If you spin it fast enough to stabilize it, then drop it, it will also point its nose into the relative wind (produced by the crosswind and the increasing velocity of air "rushing up at it" from the ground.

If you drop a spinning bullet from 50 feet (or even 1000 feet), at what attitude will it hit the ground?

How about an arrow?

Toby Bradshaw
baywingdb@comcast.net
 
But only one of these mental images is physically correct, and a spin- or fin-stabilized projectile isn't stable with more pressure on one side than the other. It has to fly point-first into the only wind it "feels."

Further, if the wind really did impinge on the side of the bullet, we WOULD need a separate "frontal BC" and "side BC", which is how this thread started. But one BC will do (with all its problems) because the bullet always presents the same profile (nose first) to the drag produced by the air.


Toby Bradshaw
baywingdb@comcast.net

Not quite a true "point first" into the wind it feels", the yaw of repose would either have to be deducted or added depending on wind direction and barrel twist. Actually, when we consider all the external factors and forces on a bullet in flight, then the word "Stable" is a bit of a misnomer when discussing external ballistics. Don't we always have a little more pressure on one side???I can't wait until you guys get to the Poisson effect...I am all ears!

But no matter what, I will never fire another round without thinking of little Al riding on my bullet! :D Hold on tight Al, I got those 105's up to 3212 last night with no pressure signs...it's going to be a short ride! ;)
 
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?
 
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Are they...or do you just think they are round??? Round like or perfectly round???

"If the doors of perception were cleansed every thing would appear to man as it is: infinite. For man has closed himself up, till he sees all things thro' narrow chinks of his cavern." - William Blake

Or, you could be shooting a round ball! ;)
 
Not quite a true "point first" into the wind it feels", the yaw of repose would either have to be deducted or added depending on wind direction and barrel twist. Actually, when we consider all the external factors and forces on a bullet in flight, then the word "Stable" is a bit of a misnomer when discussing external ballistics. Don't we always have a little more pressure on one side???I can't wait until you guys get to the Poisson effect...I am all ears!

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

Toby Bradshaw
baywingdb@comcast.net
 
An excerpt from Sierra that has the bullet pointing with the wind, not into it.
"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."
Evidently, experts disagree on this point.
 
An excerpt from Sierra that has the bullet pointing with the wind, not into it.
"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."
Evidently, experts disagree on this point.


Ahhhh, this is where it gets fun :D

Generally I've found Robert Rinker and the Sierra techs to be in agreement.

Boatright, McCoy, Vaughn and Litz generally agree and The Nennstiel Rupprecht site seems to parrot whoever uses the coolest sounding phrases......

At least the way I read them ;)

LOL


al


BTW, that last sentence, I read it as "left + downward"....... agreed? and agree?
 
When "experts" disagree and there are no pictures to prove either side correct, the resulting discussions remind me more of theological discussions, or perhaps differing schools of economics, or possibly string theory, or the cargo cult.:D
 
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