mwezell
Mike Ezell
Isn't it nice to just pull the trigger............get your eye back in the scope and watch the impact in relation to the clay bird you are sighting on??? or see the results on the sighter target? Beats all of the above mumbo jumbo
Rich De
Yep! All that matters is what's on the target.--Mike
Actually Boyd, I think I got that part OK, it was something else where I'm (probably) wrong. Let's suppose you are in space -- no friction. If you give your craft a small burn that moves you 1 degree off the original course, you pick that up, and of course, accelerate along that vector. If you now give a burn along the original axis (zero degrees), you haven't canceled the small sideways vector. To do that, you'd need a burn of -1 degrees. Only if it were timed to cancel out how long you'd been at 1 degree would you be back on the original course.
To put that back into the world of targets, if, for the first 100 yards you had a 20 mph right crosswind, and for the next 100 yards you had a 20 mph left crosswind (and ignoring the variations in BC), you'd be back on the x ring.
One thing I'd like to point out to the original question -- which flag counts more -- is that in short range BR, a fairly common pattern is the two far flags (at 100) point one way, and the near flag points the other. This happens when there is a hole on one side of the range out near the targets, and fairly solid tree lines or berms on the sides. The wind comes in the hole & bounces off the tree line. Once you figure it out, it's a joy to shoot, but you can't assume the first flag counts for more.
To put that back into the world of targets, if, for the first 100 yards you had a 20 mph right crosswind, and for the next 100 yards you had a 20 mph left crosswind (and ignoring the variations in BC), you'd be back on the x ring.
One thing I'd like to point out to the original question -- which flag counts more -- is that in short range BR, a fairly common pattern is the two far flags (at 100) point one way, and the near flag points the other. This happens when there is a hole on one side of the range out near the targets, and fairly solid tree lines or berms on the sides. The wind comes in the hole & bounces off the tree line. Once you figure it out, it's a joy to shoot, but you can't assume the first flag counts for more.
Boyd Allen
Active member
Charles, the second hundred yards would be started at the velocity that the first hundred ended at, and there would be some distance required for the opposite wind to bring the cumulative lateral momentum from the first hundred yards back to zero, before the bullet could start to move with the downrange wind, so you wouldn't make it back to the X . In space, your second small burn would put you at an angle to your second heading and if it was the right size and angle it could put you on a course that was parallel to the original heading. To put it on a course that would intersect the original one at some distance would require a longer burn the second time, and a precisely calibrated third correction would be required to put it back on the same line that it started on, I have seen some brutal conditions at ranges, but so far, I have not had to shoot in a vacuum, and I have yet to see a bullet with maneuvering rockets.
Boyd
Rich : I would agree that the utility of this for actual shooting, is about zero.
Boyd
Rich : I would agree that the utility of this for actual shooting, is about zero.
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Wind drift is directly tied to LAG TIME
It is crosswind speed times lag time
Wd = Ws * Tlag
Lag time is Time of flight(which encompasses many factors) minus time of flight in a vacuum
Tlag = TOF - TOFvac
This can be applied AT a target of distance, and/or any increment of travel beforehand, with any vectors applied along the way, and it's VERY accurate.
THIS is what experienced long-range HUNTERS know first hand.. They gross adjust, and fine hold-off for it, based on this math (best they can).
Now with a bullet in-flight, the only variable above is TOF which changes with drag applied from one velocity to the next throughout a given displacement. There is no direction assigned to TOF, only to the crosswind component sign.
So there is nothing that would change the deflected path other than another crosswind, applied to ever changing deceleration/TOF rates.
Deceleration is highest nearest the muzzle, so Tlag is highest here, therefore wind drift -deflection angle is highest here.
Bullets are not wings propelled. They don't FLY, they FALL.
And this wind drift math applies to bullets pointy or not, spinning or not.
I suggest we all putt a few golf balls, watch this model with each undulation and break of greens, and agree on something more fun!
It is crosswind speed times lag time
Wd = Ws * Tlag
Lag time is Time of flight(which encompasses many factors) minus time of flight in a vacuum
Tlag = TOF - TOFvac
This can be applied AT a target of distance, and/or any increment of travel beforehand, with any vectors applied along the way, and it's VERY accurate.
THIS is what experienced long-range HUNTERS know first hand.. They gross adjust, and fine hold-off for it, based on this math (best they can).
Now with a bullet in-flight, the only variable above is TOF which changes with drag applied from one velocity to the next throughout a given displacement. There is no direction assigned to TOF, only to the crosswind component sign.
So there is nothing that would change the deflected path other than another crosswind, applied to ever changing deceleration/TOF rates.
Deceleration is highest nearest the muzzle, so Tlag is highest here, therefore wind drift -deflection angle is highest here.
Bullets are not wings propelled. They don't FLY, they FALL.
And this wind drift math applies to bullets pointy or not, spinning or not.
I suggest we all putt a few golf balls, watch this model with each undulation and break of greens, and agree on something more fun!
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Boyd Allen
Active member
Looking at an example from Quick Target of a 68 gr. Berger with a MV of 3,269 and a 20 MPH crosswind, the drift from 0 to 100 yd. is 2.04 in. and from 0 to 200 yd. is 8.21 in. which means that the drift from 100 to 200 was 6.17. The only thing that I don't have a way to know is how much distance would be used up cancelling out the initial drift momentum. Playing with the interactive graph, it looks like it would take till about 140 yd. to cancel out the drift from the first wind with its opposite, which still looks like about a net drift from the line to the target, in the direction of the second cross wind of a little over 4", which would take us past center by around 2". I think....
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4
4Mesh
Guest
I did not read the rest of this thread cuase it's growning faster than I can read!
Charles, I think your supposition here is incorrect and I'll explain why. This is not like the situation MikeW below describes with the feather. This bullet has mass that gives it a higher BC than that of the feather. Furthermore, the very fact that the feather has weight and keeps falling is evidence that it would also continue drifting after it was accelerated laterally.
Now, your statement above makes one assumption that I do not think applies. You say there is wind to begin, then there is still air. Whatever has happened to the bullet in the moving air is not going to get undone when it gets to still air. It will only get undone if it gets to opposite moving air. That supposition implies that somehow the bullet remembers it's original path. It has no memory, it simply knows what is ahead. Still air does not alter the bullet, moving air does. The vast majority of the bullets movement is due to inertia. Only a very subtle part of that movement is due to drag/air/whatever. The bullet may not get displaced more, but it still has been displaced, and there is no taking back that displacement by removing all acting force.
Hope that makes sense...
Hope that's correct!
I wasn't going to reply anymore, because it's obvious there are others who know far more than me.
But consider this: If a bullet orients itself to meet the center of pressure, So, when it encounters wind, it reorients itself accordingly, and there is a small change in the angle of attack. Should the wind stop, the bullet will reorient itself, that is, take on a new angle of attack. And since drag is a force in this model, this is a "new" force being applied, not the absence of an old one.
Whatever. I still have trouble with the model, because I can't see how a round musket ball will "reorient" itself. It has no nose. From time to time people explain this to me & I think I get it, but then time passes, and I again have trouble with the round ball/angle of attack/drag as an applied force model.
Don't bother to explain yet again
Yes, usually it's more fun to shoot. Yesterday, it was not.
D
Damon
Guest
What's the difference between a stable round ball and an unstable round ball anyhow? I know they like to spin them a bit.
Boyd Allen
Active member
Charles,
OMO that you have trouble with the round ball speaks well of you. The problem is that correlation is not proof of causation. Just because two things happen at the same time does not mean that one causes the other. A half degree turn (that Sierra says is in the wrong direction for the popular explanation of drift to work) does not mean that the side of a bullet does not "see" see a 20 MPH crosswind, and vectors that are the result of a graphic solution to a sum of forces problem does not mean that there is an actual applied force. If you put a cup in front of you at the edge of the desk, and touch it with your index finger tips on both sides 45 degrees each way from the line of the edge of the desk, and apply equal force along the line of each finger (as it points to the center of the cup) the cup will move at right angles to the edge of the desk, even though no force is being applied at that angle. If the forces applied are drawn as vectors an a graphic solution is done, there will be a resultant vector that shows the magnitude and direction of the sum of the forces. This should not be confused with a third hand and finger. Over and over again, posters have looked at a schematic drawing of a bullet, overlaid with a schematic vector addition problem. Looking at this they start to imagine how bullets are moved, and make up stories that remind me of the cargo cult in New Guinea, and when I ask for details of how it works, they say something about how many books they have read, and who wrote them, or simply that it's physics. Most of what we ague is one persons unsubstantiated imagining vs. another's. Keep asking questions, and when a gust of wind blows a ping pong ball across the yard, keep repeating that it was not blown by the wind. BTW just what the heck do these fellows think blown means? To me it means moved as a result of the force of the wind.
Boyd
OMO that you have trouble with the round ball speaks well of you. The problem is that correlation is not proof of causation. Just because two things happen at the same time does not mean that one causes the other. A half degree turn (that Sierra says is in the wrong direction for the popular explanation of drift to work) does not mean that the side of a bullet does not "see" see a 20 MPH crosswind, and vectors that are the result of a graphic solution to a sum of forces problem does not mean that there is an actual applied force. If you put a cup in front of you at the edge of the desk, and touch it with your index finger tips on both sides 45 degrees each way from the line of the edge of the desk, and apply equal force along the line of each finger (as it points to the center of the cup) the cup will move at right angles to the edge of the desk, even though no force is being applied at that angle. If the forces applied are drawn as vectors an a graphic solution is done, there will be a resultant vector that shows the magnitude and direction of the sum of the forces. This should not be confused with a third hand and finger. Over and over again, posters have looked at a schematic drawing of a bullet, overlaid with a schematic vector addition problem. Looking at this they start to imagine how bullets are moved, and make up stories that remind me of the cargo cult in New Guinea, and when I ask for details of how it works, they say something about how many books they have read, and who wrote them, or simply that it's physics. Most of what we ague is one persons unsubstantiated imagining vs. another's. Keep asking questions, and when a gust of wind blows a ping pong ball across the yard, keep repeating that it was not blown by the wind. BTW just what the heck do these fellows think blown means? To me it means moved as a result of the force of the wind.
Boyd
Knuckleball! And it's time for baseball again.
Boyd, thanks for the (sort of) explanation. Leaving the mirror of reality alone for a moment, do any of the theories actually predict what happens? It was my impression most did. And the, desire for mirror still strong, are any of the book-models acceptable beyond prediction? (For those of you who think that's nonsense, consider this: we *think* we could get the aether theory to predict everything in physics as well as Einstein's relativity theory predicts, and as accurately. But the math would be horrible to even contemplate. To resolve this, the philosophy of science (metascience, if you will) holds that if two theories predict equally well, the simpler one is correct. Great. Now try to define "simpler."
And that, I think, is a partial answer for the original poster. Which flags predict, and the weight you assign to them, depends on the range, and sometimes the day. We almost never get the simple situations. There is a reason for practice before a match, and unless you know the range very well, it's for more than determining a load.
R.G. Robinett
"That'll never work."
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.
Ray
I hear ya, Ray!
Boyd Allen
Active member
Charles,
I believe that if you look up the formula for calculating wind drift, that there is no part of it that requires knowledge of how the wind "gets a hold of" a bullet, which would seem to make discussions of that topic irrelevant to the task of predicting where bullets go when the wind blows. If you scroll down on this page http://stevespages.com/page8e.htm you will come to the formula for computing wind drift. If you come up with a different one, I would be interested.
Boyd
I believe that if you look up the formula for calculating wind drift, that there is no part of it that requires knowledge of how the wind "gets a hold of" a bullet, which would seem to make discussions of that topic irrelevant to the task of predicting where bullets go when the wind blows. If you scroll down on this page http://stevespages.com/page8e.htm you will come to the formula for computing wind drift. If you come up with a different one, I would be interested.
Boyd
Jerry H
Different Drummer
My experiment with a bank of fans in a warehouse was an utter failure. However based on experiments with my "Smart Flag" my current opinion (changes quite often) is the following relationship from the bench to the target with equally spaced segments and a flag centered in each. 8.5-8.25-7.75-7.25-6.5-5.75-4.75-3.75-2.5-1.25 in any combination of winds between the muzzel and the target. The accuracy level with this relationship is .15 MOA. Based on this, in my opinion, the first flag has dramatically more effect on the impact at the target than the last flag for the same wind force.
4
4Mesh
Guest
There are two things that make a bullet different than most things we observe.
One is that it is spinning.
The other is that it is supersonic.
We can replicate one of those and see what happens in real time. Take a gyroscope. When it is spinning, it is stable for the most part. Drag causes it to precess. A really really nice gyroscope, (like the one I built) has very little drag, and lots of inertia. If I push on that gyroscope to alter its position, while it does fight that motion a lot, if it is moved, it does not attempt to return when I let go. It does go on its way at the pace it prefers, but it does not ever come back.
That would be the model I would apply to the wind stopping somewhere downrange. The damage is done. The original path is no more. Therefore, the bullet continues to diverge from the line of sight, though now at a constant rate, not at a rate that is accelerating.
One is that it is spinning.
The other is that it is supersonic.
We can replicate one of those and see what happens in real time. Take a gyroscope. When it is spinning, it is stable for the most part. Drag causes it to precess. A really really nice gyroscope, (like the one I built) has very little drag, and lots of inertia. If I push on that gyroscope to alter its position, while it does fight that motion a lot, if it is moved, it does not attempt to return when I let go. It does go on its way at the pace it prefers, but it does not ever come back.
That would be the model I would apply to the wind stopping somewhere downrange. The damage is done. The original path is no more. Therefore, the bullet continues to diverge from the line of sight, though now at a constant rate, not at a rate that is accelerating.
D
Damon
Guest
The happy fact in all of this is that the spin allows the bullet to orient itself into the oncoming air, which is simply the air flow relative to the bullet. Think about that for a moment.
We have a terrible time defining drag for a bullet facing head on into the wind. We resort to goofy drag functions and BC's. Then we get all riled and demand that G1 is no longer sacred and the new idol is G7. But all we're doing is talking about what the drag on a bullet is given that it is stable and pointing with a tiny yaw (again relative to the ariflow - the ground is irrelevant).
If not for this happy fact (that bullets point into the airflow), we'd need to know how the drag function looked at various yaw angles. The mathematical and data requirements would mushroom (into something resembling the 6DOF model), and we, as sportsmen, would have no idea where the bullets would hit (except by experimenting).
But since bullets do turn into the flow, we know that drag is drag and the simple drag functions we use are useful in figuring out motion in all three directions. It is drag that causes wind deflection, drag that modifies drop, and drag that slows down the bullet as it heads down range. The same, simple drag that we calculate with a G7 and a BC.
So, to the extent that bullet impact is well predicted by point mass solvers is the extent to which you should believe that bullets point into the airflow. Because if they did not turn, the drag profile would be very different and dependent on wind speed. And, since I've written a point mass solver, I can guarantee you that there is but one and only one drag function used to calculate wind drift, and it's the same one used to calculate bullet deceleration and drop.
Yes, there are some neglected effects here, but I'd say that overall, the results are not too bad.
Now if someone could tell me why a round ball is more accurate when spun up, my head might stop hurting.
We have a terrible time defining drag for a bullet facing head on into the wind. We resort to goofy drag functions and BC's. Then we get all riled and demand that G1 is no longer sacred and the new idol is G7. But all we're doing is talking about what the drag on a bullet is given that it is stable and pointing with a tiny yaw (again relative to the ariflow - the ground is irrelevant).
If not for this happy fact (that bullets point into the airflow), we'd need to know how the drag function looked at various yaw angles. The mathematical and data requirements would mushroom (into something resembling the 6DOF model), and we, as sportsmen, would have no idea where the bullets would hit (except by experimenting).
But since bullets do turn into the flow, we know that drag is drag and the simple drag functions we use are useful in figuring out motion in all three directions. It is drag that causes wind deflection, drag that modifies drop, and drag that slows down the bullet as it heads down range. The same, simple drag that we calculate with a G7 and a BC.
So, to the extent that bullet impact is well predicted by point mass solvers is the extent to which you should believe that bullets point into the airflow. Because if they did not turn, the drag profile would be very different and dependent on wind speed. And, since I've written a point mass solver, I can guarantee you that there is but one and only one drag function used to calculate wind drift, and it's the same one used to calculate bullet deceleration and drop.
Yes, there are some neglected effects here, but I'd say that overall, the results are not too bad.
Now if someone could tell me why a round ball is more accurate when spun up, my head might stop hurting.
Boyd Allen
Active member
Hint: Just because a ball is round does not mean that when shot from a rifled bore, that it does not have a spin axis. I would guess that round balls are hardly ever perfectly round, particularly if made of pure lead and subjected to the peak pressure developed as they are propelled down the bore. If they are not symmetrical, as they leave the muzzle, they will probably tumble and thus present random angles of attack as they do so, flying off in one direction and then another. I gather that this is what makes a well thrown knuckle ball hard to hit. On the other hand, when spun, with sufficient force a ball will present itself in a uniform manner. Granted, this is all just speculation. on the other hand, it pleases me greatly that the projectile shape that has no real front to turn into the wind, is most effected by it. Drag does determine time of flight which determines how long gravity and the wind will act on a projectile. This does not however change the fact that a cross wind will always be at close to 90 degrees to the spin axis of a bullet, an that the wind velocity will be greater than the lateral acceleration that it imparts to a bullet.
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All right, I have to chime in on this one, first off I am sure you have all watched the vapor trail, or wash, from the bullet, when a shooter is shooting from 1000 yards. The gun has to pointed into the wind, to compensate for the push the wind is giving the bullet. Otherwise the bullet isn't even going to get the right start, you're bucking the wind to start with. And how about the Pa. club having a beaver pond to shoot over, so what does the bullet do if there's an updraft? I've watched bullets in 30 mph cross winds do tricks you wouldn't believe and still hit the targets where they were supposed to.
Joe Salt
Joe Salt
OK, so let's say the bullet hits a branch, or a raindrop, or a pool ball...........or a blip of wind.
It's deflected, NEW direction of travel.
What could possibly be the force or moment ("momentum") trying to bring it back to it's original course? It's on a NEW course..... it'll STAY there until acted upon by an outside force.
al
and BTW Joe Salt..... I too have watched some vapor trails, and even a few bullets. Those buggers go all over creation and then HOOK right into the target.
too cool
al
It's deflected, NEW direction of travel.
What could possibly be the force or moment ("momentum") trying to bring it back to it's original course? It's on a NEW course..... it'll STAY there until acted upon by an outside force.
al
and BTW Joe Salt..... I too have watched some vapor trails, and even a few bullets. Those buggers go all over creation and then HOOK right into the target.
too cool
al