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Joined: Jan 2002
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The traditional wind drift formula is the delay formula. The amount the bullet is slowed is mulitplied by the wind velocity at a lineal rate. Thus the amount of drift is X at 10 mph and 2X at 20 mph as I understand it.

My question is that the force or energy of wind increases with the square of the difference. Thus the energy of wind at 20 mph is four times as great as wind at 10 mph. Then why are the formulas lineal? I would think a stronger wind would have a effect relevant to it's energy and not lineal velocity!

The link is just a discussion of wind drift www.rifleshootermag.com/shooting_tips/determining_wind_values.htm

At that site under "Shooters Tips" there is a good discussion of wind drift showing the lineal values.

GB1

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Think this through more carefully. Mentally list and examine the factors involved.

Wind deflection is a function of the bullet's time of flight. During that interval, it's moving laterally, driven by the speed of the cross-wind vector that's blowing it off-course. The time of the bullet's exposure to that lateral vector and the speed of the wind produce deflection. Double the speed, and you double the deflection. Increase the time of flight, and you increase the deflection. I can not imagine how four times as much force laterally applied could move the bullet laterally twice as fast as the cross-wind speed. How would you explain the bullet being deflected at twice the speed of the cross-wind?

I do not know how to factor-in the gyroscopic stability of the spinning bullet. You may have a point, but I don't see it yet.


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If you stick your hand out of the window of your car at 30 it will not feel like much. At 60 mph you will really feel it as it's four times the force.

The force on a boats sails are far greater at 20 knots than at 10. There is an enormous difference. Ten knots is a nice day. Small craft advisories go up at 18!

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I fail to see the parallel.


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E-mail your PO address to me (kenhowell1931earthlink.net), and I'll send you some material to study that'll do more for you than I can.

For one thing, this isn't a subject that I have the time (right now) to dig deep into -- which I'd have to do, to do your question justice. It's not something that I know much about, so I'd have to study it long and hard. Since you're more deeply and immediately interested in pursuing it than I am, I'll leave the studying to you until I can get into it later.


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Always take your responsibilities seriously but never yourself.



















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Wind drift bothered me for years. It makes no sense that it should be a function of "delay time".
Harold Vaughn in his book "accuracy Facts" finally explained it. ( A most excellent book)
The spinning bullet aligns itself with the vector result of it's velocity, and the wind vector velocity.
( zero angle of attact). The drag is now acting with a vector down wind, which is why drift is a function of
Ballistic coefficient. The fact that B.C. is defined in terms of velocity loss explains why
"drift varies with delay time".
It is not the wind blowing against the bullet as that would be a direct function of flight time.
And no Virginia, boat tail bullets do not " drift more due to increased length".
Take Care!

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"Reasoning" by analogy is one of the weakest forms of "logic" -- especially when the cited "analogy" isn't really analogous.

The bullet in flight is more nearly (but not quite) analagous to the boat driven sideward by water currents (not the wind) or the aircraft flying east in a north wind. Neither the boat nor the aircraft has the insistent "I gotta go this way" resistance to the cross-wind that the gyroscopic effect of its spin gives the bullet. Both the boat and the aircraft go sideward at exactly the speed of the body of water and air that they're traveling in. Neither has a connection to the ground, so the sideward speed of each, relative to the ground, is solely a function of the speed of the medium surrounding it.

When the bullet flies across the prevailing wind, how much of the wind blows past it? Ah, there's the rub! (This question doesn't even apply to a boat crossing a river or an aircraft flying perpendicular to the direction of the wind.)

An object tethered in some way to the ground (or a body of water) and secondarily subject to the wind simply isn't analagous to the bullet, the boat, or the aircraft, relative to the way that the wind affects it.


"Good enough" isn't.

Always take your responsibilities seriously but never yourself.



















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irv Offline
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WOW! I made no analogy between a bullet and a boat. I did refer to the common error that the wind blows on the
side of the bullet forcing to "drift". If thet were so longer bullets (as in boattail) would present a larger cross
. section and drift furthur. If fact due to their increased Ballistic Coefficient they drift less.
I'll try once more; The gyroscopic action of the bullet aligns it in such a manner that the airflow is head on.
That is to say it points into the wind at a vector resultant of the wind, and the bullet velocity.
With the bullet turned into the wind there is a component of drag down wind which causes the drift.
Take Care!

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Since Dr. Howell hasn't the time to really dig into the subject, may I suggest you ask your question over on the Long Range forum. Those guys can very quickly tell you the relationship that wind has to bullet drift. They deal with it all the time. E

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"WOW!" all you want, Irv -- I wasn't responding or referring to you or your post.


"Good enough" isn't.

Always take your responsibilities seriously but never yourself.



















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It is helpful to look at the top of each post to see to whom the poster is responding. Irv didn't do this, so he mistook Dr. Howell's comments to be in reference to his own post. It clearly was not. <img src="/ubbthreads/images/graemlins/wink.gif" alt="" />


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