THE LIBERTARIAN ENTERPRISE
Number 756, February 2, 2014
EVERY SINGLE AMERICAN CITIZEN HAS THE
RIGHT TO CARRY A CONCEALED WEAPON
Is Far Wind More Important Than Near Wind?
Special to L. Neil Smith's The Libertarian Enterprise
And now for something completely different -- a ballistics discussion.
I have gotten the impression that some top shooters have a wrong idea about which wind (that near the target or that near the shooter) has more effect on the drift of the bullet; they think the wind at the target is more important. For example, this individual writes,
I suspect, since he says "the accepted wisdom" is that the far wind is more important, he's not the only one in error on this point! He's not the only shooter I've heard say it.
Now, I have no doubt these guys could shoot rings around me; but they need to learn just a little bit of physics to get this point correct.
The following proof can be checked out on any ballistics calculator. I used Art Pejsa's equations in a spreadsheet, and in fact this proof was first encountered in his ballistics books, which I do recommend.
We will perform a thought experiment with two scenarios, both having to do with shooting at a target 600 yards distant. In scenario "far wind", there is no wind from the shooter to 300 yards; from that point to the target there is a constant, full-value 10mph wind. In scenario "near wind" it is the reverse; that is, the 10mph crosswind is from the shooter to 300 yards, and from that point to the target there is no wind.
Using a load I developed with a 175gr Matchking, I have a muzzle velocity of 2492 fps. Consulting the ballistics program, at 300 yards the velocity should be 2052fps and the time of flight of that segment 0.398 seconds. At 600 yards the velocity should be 1647fps and the time of flight of that segment (from 300 yds to 600 yards) is .489 seconds.
So the writers are correct in that instance; the time of flight in the second segment is definitely longer than that in the first segment, due to the reduced velocity of the bullet farther down range. And yes, the wind out there has more time to operate on the bullet. But that is certainly not the end of the story!
In scenario "far wind" there is of course no drift up to 300 yards. The drift at 600 yards is found by entering in the ballistics calculator the reduced velocity at 300 yards, 2052fps, and seeing how much drift you would get in 300 yards with a 10mph crosswind. That drift is 9.6 inches. That's all the drift there is in scenario "far wind".
In scenario "near wind", we put the muzzle velocity in the calculator and figure the drift for going 300 yards with a 10mph crosswind. That drift is 7.9 inches at the 300 yard mark, and yes that is a little bit less.
But that ain't the end of it! People might think that the bullet stops drifting, but why would it? As Newton noted, things in motion tend to stay in motion. For the first 300 yards there was a (relatively) constant force pushing the bullet sideways, which means a (relatively) constant acceleration according to Newton's equation F=ma. That means a linearly increasing velocity. This is why the illustrations of the wind's effect on a bullet show a curved path. But when the wind has stopped, the bullet still has the same sideways velocity it had at 300 yards, all the rest of the way to the target. The velocity is now constant, so the path is straight, not curved, but it is still diverging from a no-wind bullet strike. I suspect this is the effect our shooters are forgetting.
So, how far does it continue to drift in the no-wind segment of scenario "near wind"? D=vt when velocity "v" is constant; but it is not constant in the first segment of this scenario. Since we have (relatively) constant acceleration we can use the average sideways velocity, which is one-half the sums of the sideways velocity at the muzzle (zero) and our unknown velocity at 300 yards.
We have the time, .398 seconds as mentioned earlier. We know how far it drifted in the first 300 yards, 7.9 inches. So v(average)=7.9 inches / .398 seconds and that is 19.8 inches per second. But since the end velocity is twice the average velocity, the sideways velocity at the 300 yard point is 39.7 inches per second.
Now that we have the sideways velocity, and the remaining time of flight, it is simple to calculate the further displacement with this constant velocity: D=vt or 39.7 * .489, which is 19.4 inches. That drift in the second segment added to the drift in the first segment gives the total drift, 27.3 inches for the "near wind" scenario. That is nearly 3 times as much drift as the "far wind" scenario! The wind at the shooter is most important. In fact, the wind right at the target has no effect at all.
Well, we have made an approximation here that should be noted. The constant 10mph wind does not actually provide a constant force on the bullet; the faster the bullet is pushed sideways, the less pressure that 10mph wind provides. If the bullet ever got up to 10mph sideways, the pressure would drop to zero in that 10mph crosswind, and it would stop accelerating sideways. That has to be so otherwise the bullet could go faster than the wind is pushing it!
Our sideways velocity at 300 yards is 39.7 inches per second, which translates to 2.3 mph, an appreciable percentage of the 10mph sidewind, so the side pressure is not quite constant. I believe the ballistics programs account for this; however there is also the similar point that in passing 300 yards our bullet starts slowing down again (because it is dead air, not air going 2.3 mph sideways) and by the time it hits the target it is also no longer going 2.3 mph sideways. How much has it slowed down? In other words, how much less drift has occurred in the "near wind" scenario than what I predicted? Tell me what the ballistic coefficient of a 175 grain Sierra Matchking is, going through the air sideways in the (way) subsonic region, and I'll be sure to let you know! One thing we do know though, is that in the subsonic region, bullets take a long time to slow down...
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