G
gt40
Guest
I have read that the wind closer to the muzzle has more influence on pushing the bullet then down range. I am thinking long range like 1,000 yds. What do you think?
gt40
gt40
C
Cheechako
Guest
gt40
This has ben discussed more times than moly-coating bullets. Do a search.
Wind does not push a bullet. The bullet turns to follow the wind. It's called deflection.
A 10 mph wind at 50 yards has just as much influence as a 10 mph wind at 950 yards. However, since the wind at 50 yards will influence the the bullet over a much longer distance and time, it will result in much more deflection on the target at 1000 yards.
Ray
This has ben discussed more times than moly-coating bullets. Do a search.
Wind does not push a bullet. The bullet turns to follow the wind. It's called deflection.
A 10 mph wind at 50 yards has just as much influence as a 10 mph wind at 950 yards. However, since the wind at 50 yards will influence the the bullet over a much longer distance and time, it will result in much more deflection on the target at 1000 yards.
Ray
G
gt40
Guest
Thanks Ray I've never had it explained like that. That makes more sense then I have ever read.
gt40
gt40
Just a little mind game from never-never land.
Suppose there is a 10 mph wind blowing from left to right for the first 100 yards. At 100 yards, miraculously, the wind completely stops. Is the bullet any farther right at 200 yards than it is at 100 yards?
More of the same. As above, but at 900 yards, there is a right-to-left wind of 10 mph. Where does the bullet strike the target (ignoring, for the moment, the difference in drag the bullet has at the lower velocity)?
In never-never land, the winds are pure 90-degree crosswinds, and are constant for the yardages mentioned.
Suppose there is a 10 mph wind blowing from left to right for the first 100 yards. At 100 yards, miraculously, the wind completely stops. Is the bullet any farther right at 200 yards than it is at 100 yards?
More of the same. As above, but at 900 yards, there is a right-to-left wind of 10 mph. Where does the bullet strike the target (ignoring, for the moment, the difference in drag the bullet has at the lower velocity)?
In never-never land, the winds are pure 90-degree crosswinds, and are constant for the yardages mentioned.
G
Greg Culpepper
Guest
Here we go again
Yes, the bullet would be farther to the right at 200 than 100, and as it progressed down range past 200 it would continue to acquire right displacement compared to line of sight and at an increasing rate compared to distance.
The bullet stills strikes to the right. Even if it moved back to the left the last 100 yds travel exactly as much as it traveled to the right (compared to line of sight) the first 100 yds. (it won't except in your version of "never-never land), it would still begin its reversal from a position that was farther to the right at 900 than its position was to the right after the first 100 yds. traveled.
Greg
Yes, the bullet would be farther to the right at 200 than 100, and as it progressed down range past 200 it would continue to acquire right displacement compared to line of sight and at an increasing rate compared to distance.
The bullet stills strikes to the right. Even if it moved back to the left the last 100 yds travel exactly as much as it traveled to the right (compared to line of sight) the first 100 yds. (it won't except in your version of "never-never land), it would still begin its reversal from a position that was farther to the right at 900 than its position was to the right after the first 100 yds. traveled.
Greg
Last edited by a moderator:
I disagree with the 'deflection' notion, and contend that bullets 'drift'.
The drift is tied to decceleration, defined as T-LAG, which is the basis of wind drift calculations.
Bullets deccelerate at their highest, earliest in flight(in fall), because drag is highest there. So wind drift in the first 100yds only would be higher in both MOA, and actual displacement, than produced from wind in the last 100yds only. This, even with all the extra TOF from 900 to 1000, and with very little TOF from muzzle to 100.
It's because it's not about TOF, but T-LAG and the drag that caused it.
Just throwin another perspective out there..
The drift is tied to decceleration, defined as T-LAG, which is the basis of wind drift calculations.
Bullets deccelerate at their highest, earliest in flight(in fall), because drag is highest there. So wind drift in the first 100yds only would be higher in both MOA, and actual displacement, than produced from wind in the last 100yds only. This, even with all the extra TOF from 900 to 1000, and with very little TOF from muzzle to 100.
It's because it's not about TOF, but T-LAG and the drag that caused it.
Just throwin another perspective out there..
C
Cheechako
Guest
Charles
That's a good example of why you need to think and talk in MOA. The answer is no, and yes. It is the same MOA to the right but twice the inches to the right.
Mike & Al
You say tomato, I say tomahto. Deflection, Drift. - As my old friend Ole Andersen would say, same ting only different.
I sure hope this thread doesn't run into 20 pages like it did the last time.
Ray
Last edited by a moderator:
I don't care to drag it out either, and align with Al, but this is assumptive in that MOA is considered here to be w/resp to half the range, or every 100yds.
In otherwords, you're implying that 1/2" of 'deflection' due to wind drift, apparent at 100yds, will impact ~1" off at 200yds.
But much of that deflection occurred somewhere BETWEEN the muzzle and 100yds, which means that 1/2" of deflection by 100yds(due to wind) is MORE than 1/2moa. Hell that could impact 2-3" off by 200yds.
This is why you wouldn't see 1/2moa of near wind drift as ~1/2" by 100yds. It would show more like 1/10th" at 100yds, but a full 5.235" by 1000yds (without further wind beyond the 100yd point).
This applies as well to the far wind scenario(900-1k), and there would be very little apparent deflection from it.
Not meaning to nit-pick, or create chaos. But we can't understand wind drift without attention to the details of it.
C
Cheechako
Guest
Well, this has quickly drifted (pun intended) from gt40's question, so I'll get out of the way and let you guys give him an answer.
Ray
Ray
4
4Mesh
Guest
Well, I agree with Greg n Ray...
R
Roy Allain
Guest
Hey, wil you guys stop it. You're making my teeth hurt. Dang.
Roy
Roy
I think, yes.
D
Damon
Guest
The interesting thing is what happens when the wind stops so quickly. I'd wager you'd see a vertical deflection after that point one way or another.
why?
D
Damon
Guest
This is getting a little on the silly side, so bear with me. If you had a a step function in wind (that is, going from 10 mph to zero instantaneously), perhaps that would cause an outsized yaw right at the transition. I'd think that situation is very similar to the one where a side wind at the muzzle causes slanted groups.
Boyd Allen
Active member
GT40,
Getting back to your original question, what you heard is correct. I will add that, IMO the best way to learn about the effects of wind is with a rifle that is known to be accurate, shooting over flags, observing the correlation between, what the flags show, and where the bullet goes.
Boyd
Getting back to your original question, what you heard is correct. I will add that, IMO the best way to learn about the effects of wind is with a rifle that is known to be accurate, shooting over flags, observing the correlation between, what the flags show, and where the bullet goes.
Boyd
I didn't (and still don't) know the answer to the question I posed. I do know why I asked it (evidence of at least a modicum of sanity). From time to time, I like to understand the physical situation, not just the model. Wind drift has always confused me.
O.K., Harold Vaughn probably did the best job of explaining wind drift for the layman. From p. 195 of Rifle Accuracy Facts:
OK, that's all background, and does not address the original question of this thread. But what I see is that almost everyone believes that the drift caused by the (drag) force initially applied continues after the force itself is removed. After all, it would take additional work to remove that vector. So, as I read most of the replies, if there were a wind that caused 1 inch deflection at 100 yards, then the wind quit, total deflection at 20 yards would NOT be 1 inch.
I think that's wrong, or at least, that's where I get lost; additional work is done. When the wind quits, the bullet realigns itself to the center of pressure (resulting in 0 angle of attack with respect to the flight path), and the drag force is so applied. In short guys, you have a new force, if vector is included in "force."
Now that does relate to the original question of this thread. The subtleties Damon points out would explain why you wouldn't have 0, 1", 1" for bullet position in the first model (right wind for 100 yards, then zero wind), or 0" 1", "0" in the second (where there is an opposite wind for the second 100 yards). What I can't see is that we're talking about an initial force that continues because it isn't countered.
So that says in never-never land, all your flag placements are equally important with respect to "how much."
The only guy I know of who might have empirically tested this is Jerry Hensler. I *think* he did try a test, and wasn't able to quite pull it off. Maybe he'll get back to it, or maybe he did get the answer. Jerry Hensler is one of those people it pays to listen to. He doesn't post much, and usually on the CF (short range) forums, but if you see that name, read it.
OK back to our discussion. If I'm wrong, where's the error?
O.K., Harold Vaughn probably did the best job of explaining wind drift for the layman. From p. 195 of Rifle Accuracy Facts:
OK, that's all background, and does not address the original question of this thread. But what I see is that almost everyone believes that the drift caused by the (drag) force initially applied continues after the force itself is removed. After all, it would take additional work to remove that vector. So, as I read most of the replies, if there were a wind that caused 1 inch deflection at 100 yards, then the wind quit, total deflection at 20 yards would NOT be 1 inch.
I think that's wrong, or at least, that's where I get lost; additional work is done. When the wind quits, the bullet realigns itself to the center of pressure (resulting in 0 angle of attack with respect to the flight path), and the drag force is so applied. In short guys, you have a new force, if vector is included in "force."
Now that does relate to the original question of this thread. The subtleties Damon points out would explain why you wouldn't have 0, 1", 1" for bullet position in the first model (right wind for 100 yards, then zero wind), or 0" 1", "0" in the second (where there is an opposite wind for the second 100 yards). What I can't see is that we're talking about an initial force that continues because it isn't countered.
So that says in never-never land, all your flag placements are equally important with respect to "how much."
The only guy I know of who might have empirically tested this is Jerry Hensler. I *think* he did try a test, and wasn't able to quite pull it off. Maybe he'll get back to it, or maybe he did get the answer. Jerry Hensler is one of those people it pays to listen to. He doesn't post much, and usually on the CF (short range) forums, but if you see that name, read it.
OK back to our discussion. If I'm wrong, where's the error?
Last edited:
mwezell
Mike Ezell
Charles, I agree. That's why I made the feather analogy. Imagine, if you will, holding a feather out in front of your face at arms length. Now drop the feather. When it passes in front of your face, blow on it. It will be blown off course but after the puff of air stops it will continue straight down on it's new path with gravity being the only force(in never neverland) left to act upon it. Now, I'm no engineer and I haven't written any books on the subject, but what am I missing here? All that being said, all that really matters is what the target says.--Mike Ezell
Last edited: