Question regarding full value wind on elevation

zenbiker

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Watched a History Channel special on Snipers; the one that has GSGT Hathcock in it. There was also a sniper who served in Fallujah who made a 1350 meter triangle/center mass shot on enemy personnel. The shot was complicated by a full value 9 o'clock wind from an incoming sandstorm that required him to hold his Barret 50BMG loaded with 750 AMAX 2mils to the left AND 2 mils high.

My question is this: why did the sniper have to hold high? The narrator said it was because of the wind, which in this case was left to right. Was there another factor not mentioned such as compass point of aim, i.e. a westerly direction of the shot, or would the full value wind slow down a 750 AMAX enough to require 2 mils of elevation holdoff? What am I missing, so to speak?


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I would think wind represents resistance which represents drag which causes drop.

I've heard people dispute that theory.


But who knows if the HC even got the story right.
 
I'm not experienced enough to have seen it in the field nor have I actually done it in practice but I think I remember reading something in litz's book about projectiles behavior in the cross winds where for a bullet leaving a right twist barrel a left to right cross wind will cause the bullet to drift right and down and a right to left cross wind will cause the bullet to drift left and up slightly...

THe vertical component will reverse for left twist barrel... Someone with my knowledge and experience chime in please :)
 
More likely he was out of travel in the scope and using the reticle to add the extra in for distance. It's already been said, its the history channel, its like asking a woman what's wrong with her car, you don't get an accurate picture
 
There is a an elevation component to wind, but it's nowhere near that large.

Some of the tests they reference and use, is from WWII shooting M2s out of bombers. Except those things are going 450MPH. They "observed" dissimilar wind drift with an elevation component to it when shooting their .50s. They are also shooting non-precision machine guns and crappy ammo, so take that observation with a grain of salt.

Taken to the smallest possible point, David Tubb's DTR reticle has a slight, (and I mean slight) offset for dissimilar wind drift and he uses a different twist barrel depending on the direction of the wind. If it's coming from the right he uses a left hand twist barrel to counter the offsets.

If you cant shoot 1/4 MOA at 1000 yards, you'll never see it.
 
OK, thanks guys. I don't pretend to know as much about ELR as I intend to, but that clip had me thinking I had missed something of monumental importance along the way, not to mention questioning the results of personal experience at those ranges.


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I believe this to be an excerpt from one of bryan's writings:

"Aerodynamic jump is what causes groups to slant when shot in varying wind conditions. Basically, when the bullet exits the muzzle into a cross wind, the bullet tries to yaw slightly to align itself with the airflow. When the bullet yaws to the side, gyroscopic action causes it to nose up or down by a small amount depending on the wind direction. This initial yaw has an effect on the trajectory, and is known as aerodynamic jump. The more severe the cross wind, the more pitch the bullet ends up with. Flying to the target at a pitch angle will result in an elevation error that's proportional to crosswind."



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I believe this to be an excerpt from one of bryan's writings:

"Aerodynamic jump is what causes groups to slant when shot in varying wind conditions. Basically, when the bullet exits the muzzle into a cross wind, the bullet tries to yaw slightly to align itself with the airflow. When the bullet yaws to the side, gyroscopic action causes it to nose up or down by a small amount depending on the wind direction. This initial yaw has an effect on the trajectory, and is known as aerodynamic jump. The more severe the cross wind, the more pitch the bullet ends up with. Flying to the target at a pitch angle will result in an elevation error that's proportional to crosswind."



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Makes sense; like another poster mentioned, the direction of the wind also affects elevation or declination of POI, although I don't know why. Thanks for posting that quote.

Wes


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You have to also understand, Terrain in any given area can effective this with higher winds.

If the terrain is hilly or has reliefs that guide the wind up, and if the wind is strong enough, you get a ski jump like effect where the wind is moving UP and not laterally.

This updraft can move the POI up depending where you are shooting. It would have to be really moving to give you a giant shift, but it can happen.
 
You have to also understand, Terrain in any given area can effective this with higher winds.

If the terrain is hilly or has reliefs that guide the wind up, and if the wind is strong enough, you get a ski jump like effect where the wind is moving UP and not laterally.

This updraft can move the POI up depending where you are shooting. It would have to be really moving to give you a giant shift, but it can happen.

Last month I took a shot with the 338 at 1000 yards and it was pretty windy that day, but you can see in the camera footage It went right through a small twister about 700 yards out and my shot went high about a foot. was wondering what happened until my friend said I broke the shot right through a twister. Didn't think it would move it that much. Could of been me though.
 
You have to also understand, Terrain in any given area can effective this with higher winds.

If the terrain is hilly or has reliefs that guide the wind up, and if the wind is strong enough, you get a ski jump like effect where the wind is moving UP and not laterally.

This updraft can move the POI up depending where you are shooting. It would have to be really moving to give you a giant shift, but it can happen.

Copy that. Used to shoot at a 1000 yd range in NC that was essentially a bowl between FFP and targets. God help the shooter who drew the leftmost firing point with a full value L to R wind; as the wind came up over the ridge and down into the bowl, deflection could change from elevation to declination with no visible evidence in the mirage.


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A full value cross wind at any range with a 2.0+ bullet gyroscopic stabilization will move the bullet impact up and or down (depending on direction) 3/8 of a MOA per 10 MPH full value wind.

Example - you can shoot 1/2 MOA at 200 yards with your rifle.
You go to a prairie dog town in the Texas Panhandle where I have been working them over. The only target you will see all day are their heads poking out of the their holes. We will give them a 2" head diameter for this illustration. All of the shots in this example are at 200 yards on a flat level plane.

You CAN kill them all when in a calm condition....
10 AM the wind comes full value from 3 oclock at 10 MPH. (right wind and right hand twist rifle will make you shoot high)
Your rifle's point of impact now shifts upwards 3/8 MOA or .75" at 200 yards. Your rifle's true zero is now 1/4" under the top edge of the Prairie Dog's head and the rifle shoots only 1/2 MOA.
This means your % of hits will fall slightly to slightly over 50%. The Right hand wind will cause you to shoot over the PD's 2" diameter target.

You shoot till lunch with a much lower kill %. You probably will dismiss your misses to Sunlight direction /Temp Sensitivity of you ammo/ change in position ETC...

When you finish lunch the Panhandle wind has now increase to it's afternoon full value of 15 MPH and you POI has shifted again upwards another 3/16 MOA or .375 MOA. Your total move is now .75" and .375" for a move of 1.125" (1 1/8"). You will now shoot the rest of the afternoon and only score a hit when your shot impact is in the lower edge of your 1/2 Moa group. The bottom 3/8" to be specific.

Your hit probability now approaches 25%..
If the wind increases to 20 MPH YOUR HIT PROBABILTY WILL GOTO ZERO.

Let us also imagine that when you sighted your rifle in at the 200 YD range there was a 5MPH wind from the right.. You already went to the dog town with your elevation zero off 3/8" high to start with.
Little stuff does add up..

DTubb
 
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Wow! Thanks David. Really appreciate some feedback on practical experiences. No wonder the BR guys choose a twist that will JUST stabilize the bullet they want to shoot.

Regards
Fritz


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A full value cross wind at any range with a 2.0+ bullet gyroscopic stabilization will move the bullet impact up and or down (depending on direction) 3/8 of a MOA per 10 MPH full value wind.

Example - you can shoot 1/2 MOA at 200 yards with your rifle.
You go to a prairie dog town in the Texas Panhandle where I have been working them over. The only target you will see all day are their heads poking out of the their holes. We will give them a 2" head diameter for this illustration. All of the shots in this example are at 200 yards on a flat level plane.

You CAN kill them all when in a calm condition....
10 AM the wind comes full value from 3 oclock at 10 MPH. (right wind and right hand twist rifle will make you shoot high)
Your rifle's point of impact now shifts upwards 3/8 MOA or .75" at 200 yards. Your rifle's true zero is now 1/4" under the top edge of the Prairie Dog's head and the rifle shoots only 1/2 MOA.
This means your % of hits will fall slightly to slightly over 50%. The Right hand wind will cause you to shoot over the PD's 2" diameter target.

You shoot till lunch with a much lower kill %. You probably will dismiss your misses to Sunlight direction /Temp Sensitivity of you ammo/ change in position ETC...

When you finish lunch the Panhandle wind has now increase to it's afternoon full value of 15 MPH and you POI has shifted again upwards another 3/16 MOA or .375 MOA. Your total move is now .75" and .375" for a move of 1.125" (1 1/8"). You will now shoot the rest of the afternoon and only score a hit when your shot impact is in the lower edge of your 1/2 Moa group. The bottom 3/8" to be specific.

Your hit probability now approaches 25%..
If the wind increases to 20 MPH YOUR HIT PROBABILTY WILL GOTO ZERO.

Let us also imagine that when you sighted your rifle in at the 200 YD range there was a 5MPH wind from the right.. You already went to the dog town with your elevation zero off 3/8" high to start with.
Little stuff does add up..

DTubb

Thanks a lot David, this is a very well know issue at short ranges. For people interested in a technical perspective, see chapter 10 of "rifle accuracy facts" by H.R.Vaughn.

The vertical drift will vary with the BC and velocity, as well as the gyroscopic stability, etc.

But now, for LONG RANGE, what is the effect?

For example, for a load that has a 9.6 MOA wind drift with a full value 10 mph blowing from 3:00, with a RH barrel and 2.0 GS, witll the POI move up 3.6 MOA?
 
But now, for LONG RANGE, what is the effect?

For example, for a load that has a 9.6 MOA wind drift with a full value 10 mph blowing from 3:00, with a RH barrel and 2.0 GS, witll the POI move up 3.6 MOA?
I seriously doubt it. I think the effect is counter intuitive, being more pronounced at shorter ranges?



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Copied from FM 1-140 Aerial Gunnery

FWIW

(3) Projectile jump (vertical plane gyroscopic effect).

(a) When a crew fires a weapon from a helicopter in flight and the weapon's muzzle is pointing in any direction other than into the helicopter's relative wind, the projectile will experience projectile jump. Projectile jump begins when the projectile experiences an initial yaw as it leaves the muzzle. The yaw is in the same direction as the projectile's direction of rotation. The jump occurs because of the precession (change in axis of rotation) induced by crosswind.

(b) The amount a projectile jumps is proportional to its initial yaw. Firing to the right produces a downward jump; firing to the left produces an upward jump. To compensate the gunner must aim slightly above a target on the right of a helicopter and slightly below a target on the left. The amount of compensation required increases as helicopter speed and angular deflection of the weapon increase. Compensation for projectile jump is not required when firing from a hover.
 
Tiro,

A 3/8 MOA per 10 MPH vertical component would be present at all ranges. So a 600 yard target with 10 MPH true crosswind will effect your elevation 2.25 inches. This of course is not taking into account the coriolis effect which would also have a slight influence (about the size of a dime at 600 yd).

Dtubb
 
Copied from FM 1-140 Aerial Gunnery

FWIW

(3) Projectile jump (vertical plane gyroscopic effect).

(a) When a crew fires a weapon from a helicopter in flight and the weapon's muzzle is pointing in any direction other than into the helicopter's relative wind, the projectile will experience projectile jump. Projectile jump begins when the projectile experiences an initial yaw as it leaves the muzzle. The yaw is in the same direction as the projectile's direction of rotation. The jump occurs because of the precession (change in axis of rotation) induced by crosswind.

(b) The amount a projectile jumps is proportional to its initial yaw. Firing to the right produces a downward jump; firing to the left produces an upward jump. To compensate the gunner must aim slightly above a target on the right of a helicopter and slightly below a target on the left. The amount of compensation required increases as helicopter speed and angular deflection of the weapon increase. Compensation for projectile jump is not required when firing from a hover.

Very nice! Honestly was not aware of the effect on helicopter fire. But makes perfect sense now.


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Question regarding full value wind on elevation

Tiro,

A 3/8 MOA per 10 MPH vertical component would be present at all ranges. So a 600 yard target with 10 MPH true crosswind will effect your elevation 2.25 inches. This of course is not taking into account the coriolis effect which would also have a slight influence (about the size of a dime at 600 yd).

Dtubb

I was wrong. It does actually maintain the 3/8MOA angle of deflection throughout.


Y = 0.01SG - 0.0024L + 0.032

Where Y = vertical deflection in MOA/mph full value crosswind
SG = gyroscopic stability factor
L = bullet length in calibres

Taken from applied ballistics by bryan litz.


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It's due to a phenomenon known as aerodynamic jump. If you shoot a string of shots in a varying full value wind without changing the the point of aim, you will not see a perfectly horizontal line of impacts, but rather one tilted by about 20-30 degrees. The angle of the tilt depends on the twist rate: more twist = more elevation due to wind (yet another reason not to overspin bullets). You won't need 2 up for 2 right, though. More like 1/2 up for 2 right. Practically speaking the effect is hard to see because you need a dramatic wind shift and an accurate rifle/hold to notice, but it is both real and significant. A lot of the time, you might write it off as just a horrible shot because in order to see it clearly you've already whiffed by a significant amount. Pay attention next time you call a perfect shot that ends up wide due to an unnoticed large wind shift. Chances are the elevation is a little off too.