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 hed 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 bullets trajectory by studying the movement of heat haze and dust across the arid desert landscape.
I was concentrating so hard that I didnt 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 ...
Assume that the shot is 2600 feet. Assume that the shooter sites his weapon at a point 56 feet to the left of the target. Assume also, for the sake of strictly conservative calculations, that the bullet's speed is essentially constant at 2600 feet per second. This gives it a travel time of 1.0 second from the muzzle to the target.
Firing the weapon into the wind at the angle as described above effectively steals some of the straight-downrange velocity and gives the bullet a leftward component of 56 feet per second. (You can confirm this by a simple vector analysis which is made even simpler by the fact that the target is 2600 feet away and the travel time is 1.0 second.)
It is important to grasp the idea and magnitude of this leftward component of the bullet's intitial velocity, because it is this rather enormous leftward component which the wind has to arrest and reverse and push back (i.e., in a rightward direction) to the target.
In other words, if we are in a helicopter looking straight down at the shooting range, the bullet will be seen to follow a parabolic path. In the first half of the bullet's travel, the bullet will be moving to the left. In the last half, the bullet will be moving to the right.
Arguing from symmetry, the bullet will have a rightward velocity component of 56 fps when it impacts the target at roughly 2600 fps total velocity.
Now, it also follows from this analysis, that when the bullet reaches the halfway mark, the wind has necessarily and completely arrested the lateral velocity of the bullet--by a deceleration effect of friction acting sideways on the bullet--and must now accelerate it sideways from zero lateral velocity to 56 fps rightward velocity.
The problem is, this velocity change has to happen in one-half second. In other words, the crosswind has to exert a force capable of accelerating the bullet sideways at 112 feet per second per second. (This is the same as 56 fps per half-second.) This represents an enormous sideways force, and I frankly don't see how a crosswind can exert that much force on a bullet!
To illustrate, I would point out that a 100-mph crosswind acting on a suspended human body only exerts a force capable of an acceleration of 32 feet per second per second. (The reason why I say this is because the terminal velocity of a spreadeagled skydiver is roughly 100 mph, I believe--and terminal velocity is defined as the point at which the friction force of the relative airflow offsets the acceleration due to gravity [which is 32 feet/sec/sec].)
If a shockwave-and-bullet arrangement has "terminal velocity characteristics" similar to those of, say, a human body, then a 100-mph crosswind would offer a decelerating force effect of only 32 feet per second per second. This would not be enough to overcome the initial vector component at the muzzle, much less bend the bullet back to the target and even reaccelerate it to a 56-fps rightward component speed by the time of impact.
I conclude that if you sight 56 feet to the left of the target to account for a "mere" 100-mph crosswind, you will miss the target by about 50 feet!
The only potential flaw in the analysis is the possibility that a shockwave-encased and spinning bullet behaves in a completely different way than we see in "regular" fluid mechanics. I will cheerfully concede the fact that the shockwave may mess up my worst case assessments of the friction which a crosswind imposes on a traveling bullet. But I am still profoundly suspicious that the reporter has simply gotten the facts wrong.
If the wind has that much effect on a bullet, then I don't see how snipers make the 1600-meter shots in even a light wind.
The fact is, experienced marksmen take 500-meter shots pretty often, and they make deliberate but relatively small corrections for stiff crosswinds even at 500 meters. (My calculations, following the logic I have presented above, confirm that the correction needs to be only about a foot or so at 500 meters. Why, then, should we believe that the shooter suddenly has to add a sighting correction of 50 feet when he goes out another 300 meters or so? And how on earth could he hit the broad side of a barn if he is sniping at 1600 meters in even a light wind?)
Please do not send me reams of vector calculus- my head hurts today already. On your main point, though, I also assumed about a one-second travel time for the bullet- and the figure of 56 feet of lateral movement just looked impossibly large. The cross-sectional density of the bullet, as well as gyroscopic effects, surely would not allow it to attain a velocity of 56 ft/sec in only 1 second. A piece of balsa wood might, or a ballon.
You have to be right. That is a very hard shot, but he would have to be using a crossbow to get 56 feet of windage.
So9
So9
How did you get your lateral drift figure?
Either way, if the story is true, that was one lucky shot. Combining wind drift and drop into a one shot kill at that range takes alot of skill, and a whole lot more luck.
Don't want to start another whole argument, but the vertical drop of the bullet is dependent ONLY on time-of-flight- which in our example is just about 1.2 seconds, and the gravitational attraction of the Earth (9.8 meters/sec/sec).
Therefore, the vertical drop could not be much more than 35 feet or so- regardless of any other complicating factors.
Check out #41, #61 and #91 among others.
Trajectory for Sierra .308 dia. 168 gr. HPBT MatchKing at 2600 Feet per Second
At an Elevation Angle of: 0 degrees
Ballistic Coefficients of: 0.462 0.447 0.424 0.405 0.405
Velocity Boundaries (Feet per Second) of: 2600 2100 1600 1600
Wind Direction is: 0.0 o'clock and a Wind Velocity of: 0.0 Miles per hour
Wind Components are (Miles per Hour): DownRange: 0.0 Cross Range: 0.0 Vertical: 0.0
Altitude: 0 Feet with a Standard Atmospheric Model.
Temperature: 59 F
Data Printed in English Units
Range (yds) |
Velocity (ft/sec) |
Energy (Ft/Sec) |
Drop (in) |
Bullet path (in) |
Time of flight (sec) |
0 | 2600 | 2521 | 0 |
-2.5 |
0 |
100 |
2405 | 2159 | -2.71 |
34.6 |
0.12 |
200 |
2219 | 1838 |
-11.4 |
65.7 |
0.25 |
300 |
2038 | 1550 |
-27.2 |
89.7 |
0.391 |
400 |
1861 | 1292 |
-51.4 |
105.4 |
0.545 |
500 |
1694 |
1070 |
-85.7 |
111 |
0.714 |
600 |
1536 |
881 |
-132 |
104 |
0.9 |
700 |
1390 |
721 |
-193 |
83 |
1.11 |
800 |
1264 |
596 |
-272 |
43.6 |
1.33 |
900 |
1159 |
501 |
-373 |
-17.6 |
1.58 |
I would imagine there is a certain amount of leadership dropping going on in downtown Baghdad as well.
I'd believe that for a .223 -- that's a fairly light bullet with a lot of powder behind it -- but not for a 7.62. I don't recall the muzzle velocity of that offhand (I haven't fired that in 30 years), but half your 2600 fps sounds more like the right ballpark.
Plus, the supersonic drag on the bullet is considerable. It's going to lose some velocity over that 900 yards.
Finally, you're treating the crosswind force as a constant force, like gravity, which it is not. True it's not the simple vector calculation (like the old "man trying to row across a river" problem) that some are making it out to be, but because of the changing velocity of the bullet (which is never as fast as your assumption), the leftmost point of the parabola will be closer to the target than the halfway point.
(Furthermore, that "56 foot curve" may mean both the left and right components, not a shot with a 56 foot aim offset.)
Heavier bullets will of course be less deflected by crosswind (higher mass per cross-sectional area, square/cube law and all that), so .50 cal will have an easier time of it. (May have higher muzzle velocity, too, and drag would affect it less, again because of square/cube.)
Range (yds) |
30 mph (in) |
30 mph (ft) |
40 mph (in) |
40 mph (ft) |
0 | 0 |
0 | 0 | 0 |
100 |
2.41 |
0.2 | 3.2 | 0.27 |
200 |
10.0 |
0.84 | 13.4 | 1.16 |
300 |
23.5 |
1.96 | 31.4 | 2.62 |
400 |
43.9 |
3.7 | 58.6 | 4.88 |
500 |
72.3 |
6.0 | 96.3 | 8.03 |
600 |
110 |
9.1 | 146 | 12.2 |
700 |
157 |
13.1 | 209 | 17.4 |
800 |
216 |
18 | 288 | 24.0 |
900 |
286 |
23.8 | 381 | 31.8 |
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