A British Army sniper has set a new sharpshooting distance record by killing two Taliban machinegunners in Afghanistan from more than a mile away.
Craig Harrison, a member of the Household Cavalry, killed the insurgents with consecutive shots — even though they were 3,000ft beyond the most effective range of his rifle.
The distance to Harrison’s two targets was measured by a GPS system at 8,120ft, or 1.54 miles. The previous record for a sniper kill is 7,972ft, set by a Canadian soldier who shot dead an Al-Qaeda gunman in March 2002.
Full Story – HERE
Very impressive kills. I couldn’t even imagine the adjustments you would have to make at that range, and even the mirage that he must have seen looking through the scope.
Tom Irwin, a director of Accuracy International, the British manufacturer of the L115A3 rifle, said: “It is still fairly accurate beyond 4,921ft, but at that distance luck plays as much of a part as anything.”
If he took shot after shot and EVENTUALLY hit the insurgents I would call it luck. According to the article though, Harrison fired 3 shots. The first 2 hit and killed the insurgents, then the 3rd and final shot he used to disable the machine gun they were using. If that is luck, and not skill, then Harrison is pretty damn lucky.
Anyone know why it took so long for this story to reach the news? According to the article, the record breaking kills happened in November 2009.
I think it’s interesting that the article refers to the caliber of the L115A3 as 8.59 mm , and the caliber of the previous record holder, Canadian Sniper Rob Furlong’s rifle as 12.7 mm. I suppose most of the rest of the world is on the metric system, and they might be familiar with those numbers, but referring to them as .338 Lapua Magnum and .50 BMG in addition would likely have avoided some confusion (here in the U.S. anyway).
More information on the L115A3 – HERE
Accuracy International Website (one of the worst websites I have ever seen) – HERE
Comments
24 responses to “British Army Sniper Craig Harrison Sets New Record”
I would say that they were probably pretty lucky to get the shot dialed in correctly in the first place at such a long distance. Once the shot was set up correctly they probably had another luck factor in that the conditions held steady for all three shots. I’m sure the slightest shift in the wind would have massive effects on drift at that range.
I was curious about what the trajectory would be like on something like this. I used the Federal 2010 Ammunition Catalog program I have on my computer to pull information about .338 Lapua Mag (load #GM338LM) trajectory. You can adjust parameters in the program, and I set the zero distance at 500 yards, and was able to pull trajectory for 0 to 1,000 yards at 100 yard increments (it won’t go beyond 1,000). Putting that data into a statistical analysis program and running a regression analysis gave me a very statistically accurate equation for the trajectory relative to the range. Now, it’s not exact, but it should be in the right ballpark. According to those calculations, the bullet would strike a target 8,120 feet (2,707 yards) away at 3,890 inches below point of aim (unadjusted for 500 yard zero). You’d be aiming at a point 324 feet above the target to lob it in there!
(This should be close, the equation was an EXTREMELY close fit for the known trajectory from 0 to 1,000 yards.)
WOW that is amazing, thanks for doing the math Josh.
That doesn’t sound right. The other analysis I’ve seen on this shot put the number closer to 6 feet.
Also, it took so long to get the word out on this because he was still in theater on tour. Announcing his name to the world while he was still on deployment is considered a no no.
I would love to see the link for that. The information from the Federal ballistics table puts the shot, when zeroed at 500 yards, -185.4 inches at 1,000 yards. So at a thousand yards you’re already 15 feet below point of aim.
Seriously though, if you can find the link to whatever you read, I’d like to read it. If I’m wrong, I don’t see how.
By the way, 6 feet at 8,120 feet is about .04233 degrees of angle, 2.54 MOA. Your assertion is that the bullet deviates from zero by less than 3 MOA at 1.5 miles???
I don’t think the constraints of the physical world would allow for any bullet to have such a trajectory. I would think the forces required to create such a flat shot would cause the bullet to disintegrate, either from the force required to accelerate it to a sufficient speed, or the friction from the air in flight.
324 feet elevation at 8,120 feet is approximately 2.3 degrees. So, what I’m saying is that the shooter had to increase the elevation of the barrel by 2.3 degrees to make this shot (with a weapon zeroed at 500 yards). That sounds pretty plausible, doesn’t it?
Hiya,
In some other article, it says that it took nearly 3 seconds for the bullets to hit their targets. If the bullets were free-falling, they would drop about 45m (150ft) in those 3 seconds. Not sure the ‘3 seconds’ is accurate, but it can’t be too far off. The free-fall assumption is probably not accurate for ballistics, but it can’t be too far off either. So, in any case, it means that your 384ft calculation isn’t unrealistic at all, and that we’re talking about hundreds of feet as opposed to tend. Pretty amazing shots in any case, especially with the guy scoring 3 hits with 3 rounds!
I cannot say as though I know a tremendous deal about shooting anything outside of a good shouting distance, but at those ranges, damned near everything comes into play – the line from Shooter comes to mind, with him listing off every damned variable that would need to be accounted for, times two for the distance.
Stand by for a brief dip in the stocks of .338 Lapua and anything Accuracy International sells to the public, though, as fanboys around the world convince themselves it is the hardware, and not the skill and ability of the man behind the trigger. Hell, this dude broke both his arms, and was back in the fight in six weeks – he might even give Chuck Norris a run for his money!
I do love the folding stock on that rig, though… Throw in “removable” and that is damned near perfect for travel and suchlike.
Definitely an awesome rifle and skilled shooter.
I laughed when I read that 2 broken arms part of the story… that’s hardcore!
Linoge is right. It’s like the kid in Quigley Down Under after Quigley put three rounds through a wooden bucket, offhand at about 600 yards: “If I had that gun, I could do that!”
About 457 feet of additional drop from the 500 yard zero mark.
Travel time for bullet to cover this distance is about 6.5 seconds.
Bullet arrives with about the same energy as a 9mm bullet at the muzzle.
http://www.jbmballistics.com/cgi-bin/jbmtraj_simp-5.1.cgi
Bullet weight: 250 grains
Bullet diameter: .338
BC: .587
Muzzle velocity: 2950 FPS
Chart from 0 to 2700 yards:
http://img191.imageshack.us/img191/7572/traj02.jpg
Chart out to 500 yards, with entry data:
http://img687.imageshack.us/img687/5382/traj01.jpg
All estimates, of course. Temp and elevation will make HUGE differences at this range.
After posting earlier, I realized I had a ballistics calculator spreadsheet on my computer. I got similar results to the calculator you linked to (looks like less than 1% difference at any of the distances up to 1,000 yards anyway) , although if I adjusted elevation to 5,000 feet and temperature to 80 degrees Fahrenheit, the bullet drop reduced to about the results from the regression analysis. At 1,000 feet elevation and 70 degrees, my spreadsheet says approx 410 ft.
Regardless, the bullet drop is far more than 6 feet. Even at twice the initial velocity, the bullet drop would still be nearly 100 feet.
Might be worth noting also, I just looked at a table of ballistic coefficients for .338 bullets, and .587 was the most aerodynamic on the list.
Interesting stuff DJ.. thanks!
Bullet drop whether from a pellet gun or 50 cal is the same. Drop one from your hand. Bullets from a gun drop at the same rate. Not a lot of math involved once the time to target is known. Distance = 16 Ft. X the square of the time to target.
If the bullet takes 6.5 sec. to reach the target, which sounds more or less correct, the total bullet drop from line of sight would be 676 FEET!
That would be true, EXCEPT, the bullet is not being fired perpendicular to the ground. It is being fired at an angle with an initial upward velocity, based on the angle of the barrel and the muzzle velocity.
There is actually quite a lot of math involved in the ballistic computations that are used in the online calculators, or the spreadsheet I used.
The sniper might have fired off three rounds before the first one hit hoping to vertically bracket the target and the second guy placed himself right in the path when he tried to man the MG. I’m guessing he went high medium low and with no sidewind nailed both guys plus the MG.
I’m interested to know what the target looked like with 25X power at that distance. You aren’t going to be able to read their license plate due to mirage and distortion from the air itself. Even if your spotter is watching through 60 power, all he is doing is compressing the water vapor/mirage/dust/haze between himself and the target. More magnification doesn’t help. Add to that the needed hold-over-300+feet….and it gets a little shaky. You certainly could NOT see where you hit someone. The crosshair would cover them even if you were on the moon. A 2.5 MOA mil dot would cover 25 inches at 1000 yards, 50 inches at 2000 yards and more at the advertised distance.
I’m thinking their GPS numbers are off and he was inside 1000 yards. Still a hell of a shot.
I’ve taught army shooting and coached folks fresh out of Army sniper school. Most of their 7.62 M24s carry 10 power scopes. The 300+ foot drop would have been beyond their capacity.
It certainly seems like an amazing (and lucky) shot to be able to make once, and to be able to make it twice seems nearly impossible. And it seems like the kind of thing that would be nearly impossible to either prove, or disprove, although there is no burden of disproof. As you said, it doesn’t seem like you’d be able to make out ANYTHING at that distance. That 2.5 MOA at that distance would be (tan(2.5/60)*8120 = 5.905 feet. So, large enough to cover up an entire person standing up (most people, anyway).
I just looked up the specs on that 5-25×56 Schmidt & Bender scope. It has a maximum elevation adjustment of 26 Mrads. That would be 0.026 radians, or (0.026*(180/π)) = 1.4897 degrees. At 8,120 feet, that would equal tan(1.4897)*8,120 = 211 feet.
How would you even aim at a target? Clearly the bullet drop is well beyond 211 feet at that distance.
The “n” in that equation is supposed to be the symbol for “pi.” Apparently this format doesn’t support it.
So basically, a kill at 7,972ft is perfectly plausible yet 8,120ft is a bullshit figure?
One shot is plausible, because it is quite possibly due to luck/perfect circumstances, whatever you want to call it. Three is amazingly less plausible. But feel free to explain it, in light of the mathematics surrounding it. You did so well with the math on that last post of mine you responded to on a similar topic. How do you successfully aim at a target not once, but multiple times, that would be completely obscured by the reticle at that distance? How do you make three successful shots when your scope only allows you to aim approximately 200 feet above the target, and the bullet drop is multiple times that at that distance?
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A shot of this magnitude is possible with the shooter and his spotter taking into effect the curvature of the earth and the kills could be acknowledged through the use of satellite or even a spotting scope as I have glassed deer at further distances to see whether or not they were worth stalking and have had no trouble with mirages or anything of that nature.
its all complete bullshit Household Cavalry!!!!!! he spends all fucking year on a horse!!!