this is the guy according to the Sun newspaper...
Posted on 04/06/2003 10:05:41 AM PDT by OXENinFLA
Sniper tells of long-shot success From Nick Parker with the Royal Marines in southern Iraq
A BRITISH sniper told yesterday how he killed an Iraqi gunman from more than half a mile. Corporal Matt Hughes, 28, a Royal Marines marksman, was ordered to “take out” an Iraqi holding back a vital advance during a fierce gale. He pulled off an incredible feat of marksmanship by gauging perfectly the wind speed to bend the bullet to its target. The 7.62 calibre round from his L96 sniper rifle curved 56ft in the air before striking its target in the chest, killing him instantly.
Next to him, another Royal Marines sniper killed a second Iraqi at exactly the same moment.
Corporal Hughes, of the Marines’ spearhead brigade patrol troop, said yesterday: “It was a bit like David Beckham taking a free kick. I knew I only had one shot and had to get the angle exactly right. It was hot and the wind was blowing strongly from left to right as we crept up to a vantage point about 860 metres from the target.
“I saw I had a clear shot at my man — he was in what he thought was a secure position, but his head and chest were exposed. He was still wearing his green Iraqi uniform and was holding the rifle he’d been using to shoot at Marines.
“My training then took over and I got myself quickly, but calmly, into the perfect sniping position. We follow a set pattern, placing parts of our bodies in the optimum position, starting with the left hand followed by the elbow, legs, right hand and cheek.”
His sniping partner fixed a separate sight on the target, then the pair calculated the bullet’s trajectory by studying the movement of heat haze and dust across the arid desert landscape.
“I was concentrating so hard that I didn’t have time to think about him as a person or the fact that I was about to kill him. He was just a distant shape magnified ten times in a telescopic lens. He was a target — the enemy.”
(Excerpt) Read more at timesonline.co.uk ...
this is the guy according to the Sun newspaper...
An observation on the photo in #8. That sniper is using an Anti-Reflective Device on the objective. That's outstanding.
To your question, his scope may be equipped with a mil-dot reticle (mil = milliradian),which would allow him to estimate distance. The Marine Corp really did good with this innovative idea in the 70's? and the other services have caught on and now use them.
Mil-dots are either etched on glass or comprised of wire. They used to be made for use at 10X , but now companies are providing mil-dots that are usable and allow the mil spacing to remain constant throughout the range of a variable power scope.
Army spec mil-dots have posts that are ~40% narrower than the posts used by the Marines. The common 3/4 minute mil-dot is just that - the dot mass is 3/4 minute of angle. Actually for the oval mil-dot, the dot is slightly longer and and narrower that 3/4 minute, BUT the dot length is 1/4 mil.
Mil-dots are used on both weapon scopes and also spotting scopes. You can buy products (e.g. Mil-Dot Master) that allows you to compute a firing solution very quickly. It's basically a slide card that allows you to move an inside card based on data you collect (considerations of knowing the size of various objects (e.g. road signs, standard piece of plywood, etc and also how many mil-dots does the object take up in your scope . Knowing that 1 minute of angle (MOA) at 100 yards is 1.047" (or ~1") and also 1 mil = 3.438 MOA allows these types of devices to be easily used. Of course, they offer the calculations in meters as well.
You should also know the ballistics of the particular round you are shooting. For example, several companies that do mil-dot modifications have ballistic info on the Federal Gold Medal Match .308 168 and also 175 grain round. The military also has the ballistics info for the round they use.
That shot is very, very good. He was probably contending with wind velocities and directions that varied quite a bit from his position downrange to the target. If you wanted a 1 MOA dot on the target at 800 feet, that would give you around ~8" to work with - not much. This guy's shot was at ~850 meters - that's good shooting.
I propose we rename it WV windage.
You are correct.
Checking the table I have gives a 900 yard velocity of 1300fps so the time of flight is somewhere between one and two seconds. That makes bullet drop between 16 and 64 feet as long as gravity behaves.
The table shows a wind drift for the 7.62 national Match round of 12.4 MOA / 10MPH of crosswind. That is 118 in / 10 MPH, or 9.83 feet / 10 MPH at 900 yards.
That gives a 56 foot drift for a 57 MPH crosswind at 900 yards. The alternative 56 in drift would correspond to a 5 mph crosswind.
So9
(The total distance would be closer to 39 ft... Pythagorean theorem, 26ft for the base, 30ft for the altitude, 39.5 would be the hypotenuse.)
All of my simplifying assumptions aside, I think you may have misinterpreted the point of physics which I was making. A simpler illustration is as follows:
If you brisky roll a beachball toward a target in a room with a steady left-to-right crosswind, you have to aim it upwind, i.e., slightly to the left. The beachball will follow a curve of some sort. At some point, it will cease to have any leftward component to its velocity and will at that instant begin to acquire an ever-increasing rightward component until it hits the target on the other side of the room.
The precise point at which it ceases to be moving in a leftward direction and starts moving back the other direction--still continuing forward across the room at all times, of course--is a point of exactly zero lateral velocity. This point is by no means a point of zero lateral acceleration, because the wind is continuing to act on it at all times. The wind forces merely slow down its leftward movement--i.e, in a way of acceleration (what the layman might prefer to call a deceleration component, since the ball is moving left, but accelerating to the right). The same rightward-accelerating forces will continue to accelerate the beachball to the right until it hits the target.
This is not only counter-intuitive, it cannot be correct...
But the whole idea of letting the wind bend the rolling beachball toward the target does entail the wind's conversion of a left-to-right velocity component to a right-to-left component. This is why you aim upwind, why you have to aim slightly away from the target in order to hit the target.
The point is, you are actually counting on the wind to change the course of the beachball. Hopefully, you are correct in the amount of arc correction which you deliberately "install" by your sighting compensation to make the wind work for you rather than against you.
...or I would never even hit my 50-Meter targets with my .45 in a light crosswind.
But this is actually the point which I am trying to make. I don't think the wind effects are negligible, but I don't think that they are as enormous as the article has suggested. My point here is that I just don't see how a crosswind can accelerate a bullet at 112 feet/sec/sec under any real-world sniping circumstances whatsoever. I would think it would take about a 400-mph crosswind to exert that much friction force on a bullet aimed 56 feet to the left of the target.
Another way to appreciate this is to realize that the beachball model is basically correct in the major issues it raises, but that a bullet--even a 7.62 mm NATO slug--is not affected by the wind as much as a beachball is.
Since a rifle slug doesn't experience as much frictional effect from a crosswind as the beachball does, we ought to look at the scenario in the article as telling us that we have misaimed by 56 feet feet and that the bullet is screaming toward the wrong point and that the wind has a mere second to correct our aiming mistake. It's gonna take a big wind indeed to correct a "mistake" of this magnitude.
As I said in my earlier post, a bullet encased in a shockwave may be affected in ways that I am not accounting for, but I know nothing to tell me that supersonic rounds are especially susceptible to winds even if they slow down after, say, 500 meters. And until I read the account at the top of the thread, I had never heard of anyone shooting 56 feet to the left of the target at 800 meters. I think the reporter got the facts wrong. (The .308 was a pretty popular sniper rifle in Vietnam, and 800-meter kills were pretty common. Is it that different from NATO's 7.62mm standard? If it required almost 60 feet of sighting correction for crosswinds, I don't see how it would have been useful.)
It seems to me that the trajectory you are describing is not a parabola, but a sigmoid curve- which cannot occur with constant forces acting on the bullet.
This gets into the question of the simplifying assumptions I have made. If the crosswind friction effect is completely constant, and if the bullet's forward velocity component is constant, then the path is, indeed, paraboic.
Of course, we know that the bullet will slow down. But I think that this is more significant for the matter of the amount of the drop of the bullet versus distance traveled. (And the only way I can see the path being sigmoid, as viewed from above, is for the bullet to speed up at the end or for the crosswind effect to decrease at the end.)
Aside to algol: One of the other threads gave me the impression that the aiming offset itself was 56 feet. I can't really confirm this, though.
I am still uncomfortable with that 56-foot figure, although the ballistics programs do seem to give pretty similar figures. I think that it may be more useful to think in terms of the subtended angles, rather than actual distances. If you look at the aiming point as being just a few minutes of arc to the left at 900 meters, rather than "56 feet to the left", the whole thing looks much more reasonable. After all, 56 feet is only about .018% of the range (say, 3000 feet). So the point of aim adjustment is relatively small, in angular terms.
It would take a 71mph xwind to move the bullet 56', see the table in #78. The effect of the wind only depends on the bullets ballistic coefficient. That's the ratio of the drag of a std. projectile/actual bullet. In this case it's ~0.46, see #75. That means, for any given bullet with constant B.C.(ballistic coeff.), the effect of the wind is linear with wind velocity. A bullets B.C. may change slightly throughout a velocity range, but the effect is normally small. In this case the shooter had to consider those changes to hit his target.
The only num. that seems close to 56, is the 60" drop between 800 and 900 yards, assuming he's sighted in exactly on target. The reporter probably got lost in the explaination.
Looks like I am late to the party!
Yep - when I read this, the paragraph mentioning Beckham was not yet visible, and that was the first thing I thought of! Good analogy. GOAL!
GREAT work, guys. Hats off and thank you.
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