Sniper tells of long-shot success(1/2 mile shot in the chest)

http://www.timesonline.co.uk ^ | 4-06-03 | OXENinFLA

[the_doc]

I still don't buy the scenario.

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?)

[RANGERAIRBORNE]

In an entirely uncharacteristic gesture, I will defer to SPODEFLY on this. He has the ballistics software, and I don't even have a reloading manual with me today! However, I am not really comfortable with your assumption that the lateral forces on the bullet must necessarily bring it to a stop(in the plane at 90 degrees to it's direction of travel), and then move it the exact displacement in the opposite direction. This is not only counter-intuitive, it cannot be correct- or I would never even hit my 50-Meter targets with my .45 in a light crosswind. 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.

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.

[Servant of the Nine]

56 feet at 900 yards? The must mean "inches".

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

[RANGERAIRBORNE]

For a thought experiment, I have set up the rapidly spinning, small, smooth and heavy bullet lying at rest (other than it's rotational motiuon) and applied a wind force to the side of the bullet. I cannot see it being 56 feet to my left at the end of one second. (I think that distances cancel out, and only time of flight matters here).

[Servant of the Nine]

Not to contradict myself, but with a time of flight of around 2 seconds (say 3,000 feet at an average 1500ft/sec.) that could be the actual vertical drop of the bullet.
Gravity at 32ft/sec/sec gives 16 feet in the first second around 40 in the second.

So9

[RANGERAIRBORNE]

I used this calculator, but got no answer for lateral correction (not surprising, since there no way I could find to enter the wind direction- only the wind speed. Perhaps that refers only to the component of windspeed at right angles to the direction of bullet travel?)

How did you get your lateral drift figure?

[Double Tap]

Definitely not inches. And 56 feet is wrong too. The drop at that range should be no more than approximately 20 feet. Must be an error on the reporters part.

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.

[RANGERAIRBORNE]

"The bullet's drop from the muzzle if fired level with the ground would be 85.1 Ft."

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.

[RANGERAIRBORNE]

Unless I have the time-of-flight wrong...

[RANGERAIRBORNE]

No, the "average" is not going to be 1500 ft/sec. The bullet DOES slow to zero velocity on impact, but JUST PRIOR to impact it was still moving at good clip- perhaps half of it's muzzle velocity, or a little more. So the time-of-flight is not obtainable by a simple arithmetic average.

[maccraze]

One shot one kill. The British are bad to the bone. America could not ask for a better partner!

[F-117A]

The rest of the story

Check out #41, #61 and #91 among others.

[Calamari]

Listed in the column titled "Windage in" next to the "TOF" column (time of flight)in the table that comes up at the bottom of the screen.

Most ballistics tables assume a 90 degree to line of sight cross wind. More sophisticated programs allow for variable input of wind direction. Also the muzzle velocity and ballistic coefficient of the bullet have an effect on the calculations.

To calculate the total "drop" from the muzzle enter a very close zero range such a 1 or 5 yards. Also the zero range has a large effect on the sight correction needed from the zero range and the path of the bullet from line of sight.


Enter a "range interval" of 100 and a "max range" of 1200 to get a table output. I used a muzzle velocity of 2755 fps, Ballistic Coefficient of .422, bullet weight 150 grains,wind speed 50, altitude 300 zero range 500.

[Calamari]

Took the output from the column titled bullet path using a zero range of 5 yards. Shows path of 1022 inches. The more I change zero range inputs this column appears to be calculating path above and below line of sight not drop. So my statement of 85.1 feet is incorrect.

[spunkets]

The following illustrates the shot pretty good, except for wind...later. Shooters like this use well prepared loadings, with FMJBTs. They are excellent at ranging and environmental conditions judgement. The shooter would have to judge quite well the range, 'cause between 800-900 yds the beullet drops 5'. The highest the bullet traveled in the arc was 9.3'(not 56 as the story claims).

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

[BOBTHENAILER]

Thanks for the Ping. What an incredible shot. Friend of mine's nephew is now there as a sniper as well. He's very good, as they all are.

I would imagine there is a certain amount of leadership dropping going on in downtown Baghdad as well.

[algol]

Assume also, for the sake of strictly conservative calculations, that the bullet's speed is essentially constant at 2600 feet per second.

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.)

[spunkets]

The art. just says big wind, so here's the effects at 30 & 40 mph.

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

[algol]

Okay, I take some of that back. 2600 fps is indeed a reasonable muzzle velocity for 7.62mm. A quick search turns up weapons of that caliber with about 850 m/sec velocity, closer to 2800 fps.

Sorry 'bout that.

[spunkets]

Here's an interesting thread.