Wind shooting

Discussion in 'Long Range Hunting & Shooting' started by Brian Rybicky, Feb 15, 2004.

  1. Brian Rybicky

    Brian Rybicky Well-Known Member

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    This weekend i got a chance to shoot my 6.5/284 for the first time out to 788 yards, where i was the shooing the wind was very non-constant, gusting between 5/20mph,i had wind flags set out at 10,500,and 788 yards, and they were never in the same position for any of my shots, and guessing wasn't working to well because i only hit the soda bottle once, if it was a 3ltr there might have been a few more holes [​IMG]. Is there any way to accuratly shoot in these conditions?

    Thanks,
    Brian
     
  2. Brian Rybicky

    Brian Rybicky Well-Known Member

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    what i should have asked was how do you guys accuratly judge the windage when its blowing hard at the bench but not at all/very little at the target or vis versa. I know from shooting that day that bullets were most effected by wind at the muzzle rather than at the longer range.

    Thanks,
    Brian

    [ 02-16-2004: Message edited by: Brian Rybicky ]
     
  3. Dave King

    Dave King Well-Known Member

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    Brian

    I've been toying with a graph to show you how wind works based on start, stop and overall trajectory distance effected but my brain broke!

    Think on this for a bit. (I shoot a 308 Win with 175 Sierra at 2660 fps) If you're shooting 1000 yards and there is a 10 mph full value wind for the first 100 yards and no wind for the remaining distance the amount of correction you'll need is about 1 MOA, same as if you're only shooting 100 yards in a 10 mph full value wind.

    If you think you have a grasp on that and the reason answer back this question; the amount of correction you'd need if the 10 mph wind started at the 500 yard mark and stopped at the 600 yard mark with no wind anywhere else on the range with the target still at 1000 yards.
     
  4. winmagman

    winmagman Well-Known Member

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    OH MAN , That hurt, now I'm gonna have a headache for the rest of the night.
    Chris

    [ 02-16-2004: Message edited by: winmagman ]
     
  5. Brian Rybicky

    Brian Rybicky Well-Known Member

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    Dave,

    Is the answer 1/5 moa adjustment, if it is then say 0/200 would be 2moa or say 800/900 it would be 1/8moa, from what i saw shooting this weekend more, wind at my position caused more drift and no wind at my position and more wind at target the bullet drifted alot less, so these answers make alot of sense. Thanks alot for clearing this up for me.

    Good shootin,
    Brian

    [ 02-16-2004: Message edited by: Brian Rybicky ]
     
  6. JBM

    JBM Well-Known Member

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    The reason is that the wind provides an angular deflection. The earlier the angle change, the more it changes the point of impact. I've done this with my MPM program. It allows you to define a "wind field" which is velocity as various points, <x,y,z>. It then interpolates over the points to find wind at any point.
     
  7. Dave King

    Dave King Well-Known Member

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    Brian

    I'll get back to you on your addition in a little bit, we still need some work I think.

    For the 10 mph wind at 500 and stopping at 600 and the target at 1000. You'll need .5 MOA correction as I see it. Here's why, the 100 yards of 10 mph wind will cause 1 MOA of deflection from 500 to 600 yards. This deflection is caused by the projectile adopting an additional sideways motion caused by the wind. This adopted motion does not cease at the 600 yards line when the wind stops, it continues on the same vector, so we have a projectile with a 1 MOA deflection, this equates to 1.047 inches per 100 yards of travel. Add 1.047 inches of deflection for the 600 through 1000 yardsages and your get 5.235 inches of Point Of Impact (POI) to Point Of Aim (POA) error. To figure the amount of correct the shooter needs to correct this ~5" miss we need to turn it into MOA. 5" of error / 10 (for the 1000 yards because MOA is used per 100 yard increments herein)= .5 MOA of correction.

    Hopefully one of our resident "egg heads" will double check this and chime in if I'm all wet.
     
  8. Brian Rybicky

    Brian Rybicky Well-Known Member

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    Dave,

    That answer makes alot of sense, but im going to also ask would that only work if the bullet was moving at a constant speed, by the time the bullet gets out there its moveing alot slower and the wind has more time to act on it causing more than 1/2moa drift [​IMG].
     
  9. JBM

    JBM Well-Known Member

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    I don't really think it has all that much more time. There are really two things we're talking about here. First, how much angular deflection you get for a given wind and second, how much that angular deflection moves the bullet. Of course if the deflection is farther down range then it causes less wind drift on the target.

    I've never been good at remembering numbers so I ran a few of test cases with a point mass program. They are all 10 mile/hour winds. One is a constant wind from 0 to 1000 yards. The next is from 0 to 100 yards and the third is 500 to 600 yards. All have trajectories to 1000 yards.

    Here are the URLs:

    wind_10_000_1000.txt
    wind_10_000_100.txt
    wind_10_500_600.txt

    Please note the version number of the program running these (<1) so I'm covering my butt.
     
  10. JBM

    JBM Well-Known Member

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    Also I might add that the deflection at the near range (0 to 100 yards) is about .6 inches and at the 500-600 yard range is about an inch. The difference at 1000 yards ends up being a few inches.

    Here's another file with the same 10 mile an hour wind from 900 to 1000 yards.

    wind_10_900_1000.txt

    Note that the deflection is at little more than twice (1.5") about the 0 to 100 yard wind deflection which is what we would expect since the velocity has fallen to less than half the muzzle velocity. (See the standard wind formula and you'll see that it has the velocity in the denominator...)
     
  11. Dave King

    Dave King Well-Known Member

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    JBM

    Thanks for the reply and most welcome info. This is the first time I've seen this type of output from a ballistics program and I'm excited.

    Would you be so kind as to rerun this same info with a muzzle velocity of 2660, BC of .470 @ STP. This I believe is about the standard for the "older" standard military "sniper" rounds data that I've somewhat committed to memory.

    Thanks.
     
  12. Brian Rybicky

    Brian Rybicky Well-Known Member

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    JBM,

    Thanks for posting that, I looks that wind effect the bullet at the same angle thru out its trajectory, it just depends on where the wind start/stops inorder to determine the amount of drift.

    Thanks,
    Brian
     
  13. Dave King

    Dave King Well-Known Member

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    Brian

    That's what I read from JBM's posts and have understood to some degree. Constant wind over the entire flight is additive to previous wind caused deflection. Singlular (short wind period(s)) are more difficult but there is no more cumulative induced deflection after the initial deflection, just continuation of the angular error. Confused myself on that one!

    [ 02-17-2004: Message edited by: Dave King ]
     
  14. JBM

    JBM Well-Known Member

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    <BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>Singlular (short wind period(s)) are more difficult but there is no more cumulative induced deflection after the initial deflection<HR></BLOCKQUOTE>

    I think it's interesting to see the angular change stop after a while. Take a look at the file test_470_10_000_100.txt (link above) and you can see that the bullet is accelerated off course, and then the velocity in the cross wind direction goes to a constant after about 600 yards and the windage MOA stays at 1.4 MOA all the way to 1000 yards (constant anglular deflection).

    [ 02-17-2004: Message edited by: JBM ]