Longest "confirmed kill" sniper shot

[MrGenius]

There's some math involved that is way beyond my knowledge to give an accurate depiction of the event. What I do "know" is that the bullet wasn't traveling faster than the speed of sound during its entire flight. Exactly for how long or how far along its flight path it was supersonic is something I can only gather from the previously linked chart. Which only provides a rough idea. Since I didn't know some of the values to enter to make a more accurate calculation. And I can't verify the accuracy of results even if I did. As I'm no ballistics expert.

Anyhow, the chart says that the bullet slowed below the speed of sound(or 1116 ft/s) at a distance of ~2550 yds., and/or after ~4.52 seconds of flight time. And for the last ~1216 yds., and/or ~3.76 seconds, of its trajectory it was subsonic. There's a discrepancy in the 10 second flight time reported and the ~8.28 second flight time the chart states. I don't know who's right there. At any rate, we can estimate that the bullet was supersonic for roughly around the first half of the time, or two thirds of the distance, it took to reach the target. And for roughly the last half of the time, or last third of the distance, it was subsonic. But its subsonic speed was still quite fast. In fact it never, at any point, slowed below Mach .75, or ¾ the speed of sound, or 837 ft/s. Its slowest speed was ~895 ft/s when it reached the target.

So did the sound of the rifle firing ever catch up with the bullet? Good question. I don't know. My best guess is no, it did not. Another question is could it be easily heard at that distance even if it did? I can answer that with a better guess. Yes, it most likely could. Since I've heard how loud a .50 BMG being fired is first hand. It's LOUD! VERY LOUD!! Sounds like a small cannon being fired(which I've also heard first hand). Makes a .30-06 sound like a small firecracker in comparison. I wouldn't be surprised if you could easily hear it 5 miles away or more.

 

[ZenZimZaliben]

I suppose the sound could surpass the shell eventually but like you said sound dissipates. Another aspect is the angle of the muzzle...the sound was not directed at the target it was elevated way above to adjust for drop..so my point there is even less sound would be in the direction of the target. I guess if sound didn't dissipate and was not directional it could happen.

 

[agent_x007]

Another aspect is the angle of the muzzle...the sound was not directed at the target it was elevated way above to adjust for drop..so my point there is even less sound would be in the direction of the target. I guess if sound didn't dissipate and was not directional it could happen.

Aren't all soundwaves generated by explosion/detonation onmidirectional by default (+ can bounce off the ground) ?

 

[ZenZimZaliben]

sure, except this is a controlled explosion with a directional force. Stand behind a rifle and shoot, then stand on the side. Which is louder? Try not to test this by standing in front.

 

[peche]

the laws of physics prevent this.

if the bullet was travelling at mach 2 and took 10 seconds to arrive then the sound would take 20 seconds

so

the isis guy would have been shagging virgins for a good 10 seconds before the sound arrived

" From a place you dont see, comes a sound you wont heard... to take your life..."
Learned that in a place i want to come some day back !

Regards,

 

[agent_x007]

sure, except this is a controlled explosion with a directional force.

But side and rear don't matter in this case because "Target" is ALWAYS in front of the rifle/gun.
Also, I highly doubt sniper that was shooting used bullet drop to loop over buldings or rocks or stuff like that (ie. sound should have a direct path to target as well).
^If he indeed loop over stuff to get a kill... thats some BF3 jet sniper kill from air stuff right here (God-mode engaged ).

 

[ZenZimZaliben]

But side and rear don't matter in this case because "Target" is ALWAYS in front of the rifle/gun.
Also, I highly doubt sniper that was shooting used bullet drop to loop over buldings or rocks or stuff like that (ie. sound should have a direct path to target as well).
^If he indeed loop over stuff to get a kill... thats some BF3 jet sniper kill from air stuff right here

No, I am just saying the main cone of sound is not being aimed directly at the target. Like a trumpet. A trumpet is louder when you are directly in front of it because most of the sound is being directed in a cone in front. The bullet also didn't have a linear path to the target..huge elevation was needed which points the "trumpet" up in the air. Just having fun here it is a bit of a brain exercise.

 

[agent_x007]

I agree it's not direct, but in this case that effect doesn't matter.
Here's a basic "paint job" :

^This is based on .50 BMG but it's a good example : Target is only 14,5% under direct line.
So, in case of trumpet example :
How good can you hear it, if it's 3 feet away and your ears are located ~5 inches below direct line to it ?
I hope you understand my point

 

[Bones]

the isis guy would have been shagging virgins for a good 10 seconds before the sound arrived

Could be they were shagging him - I don't recall them ever being described beyond just being "Virgins".... And 72 of them?
Talk about a line.....
Not to mention the imagination running wild over it too.....

 

[EarthDog]

BOOM HEADSHOT!!!

That is all.

 

[MrGenius]

I screwed up. Let me show you what, how, and why it matters.

What?

2.14 miles = 3766.4 yards
M1022 Long-Range Sniper Ammunition = .50 caliber 650-grain projectile, initial/muzzle velocity 2750 fps
Estimated drop in elevation(distance the bullet drops from barrel level to target) = ~9409 inches/~784 feet/~261 yards/~239 meters
Estimated energy(ft.lbs. force retained by the bullet at target) = ~1155 ft.lbf.
Estimated velocity(speed in Feet Per Second retained by the bullet at target) = ~895 ft/s

Estimations based on the following chart :
http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=1da2e6db

I used default .50 BMG values for calculation(except for bullet weight and initial/muzzle velocity, which I guessed would be that of the M1022). And without any correction for atmosphere.

So...the energy of the bullet related to its potential to kill a human at that range is roughly equivalent to dropping a half inch diameter pointy object that weighs ~1155 lbs. a distance of 1 foot.

How?

Because for the zero range I used the default value of 100 yds. for the calculation. When I should have changed that to 3766 yds(for a better approximation). Since, due to the way a sniper typically aims at a target, the zero range effectively(more or less) becomes the distance from the barrel(or rather the muzzle) that the target is located. Either through adjusting the optics to be zeroed on the target(+ or - in elevation from a known zero range/distance), and/or by holding over or under(aiming directly above or below) the target by a certain amount. The correct hold over or under amount can be determined by using a MIL-Dot reticle scope and knowing the precise distance to the target. Meaning...if you know how far away the target is, and you know what distance the scope is zeroed for, you can then adjust the scope so that the target is directly centered in the reticle(requiring no hold over/under). And/or if you know the distance the scope has been zeroed to(the zero range), and you now how much farther or closer the target is to that distance, you can then calculate how much hold over or under is needed, and determine(by understanding how the MIL-Dot system works) which dot in the reticle will give you the correct zero when centered on the target. Which is oversimplifying things a little. Since, depending on the distance to the target, and the limits of how fine of an adjustment can be made to the scope, you may need to center the target(more or less) between two dots(or the vertical post and a dot) to get a good zero. There's also windage adjustments...but that's a whole other story.

Why does that matter?

Because it gives a significantly different trajectory. Which has a direct effect on several of the numbers derived from the calculations. And that's been bugging me. So I'm going to jump in here one more time and fix it. Alright...here goes.

2.14 miles = 3766.4 yards
M1022 Long-Range Sniper Ammunition = .50 caliber 650-grain projectile, initial/muzzle velocity 2750 fps
Estimated climb in elevation(distance the bullet climbs upward from muzzle level before dropping downward towards the target) = ~3640 inches/~303 feet/~101 yards/~92 meters
Estimated energy(ft.lbs. force retained by the bullet at target) = ~1116 ft.lbf.
Estimated velocity(speed in Feet Per Second retained by the bullet at target) = ~879 ft/s

Estimations based on the following chart:
http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=08306eb8

Again, I used default .50 BMG values for calculation(except for bullet weight and initial/muzzle velocity, which I guessed would be that of the M1022). And without any correction for atmosphere.

So...the energy of the bullet related to its potential to kill a human at that range is roughly equivalent to dropping a half inch diameter pointy object that weighs ~1116 lbs. a distance of 1 foot.

As you can see, that changes the calculations/estimations considerably.

What does the new trajectory with a zero range of 3766 yds. look like when compared to the previous trajectory with a zero range of 100 yds.?

Zero range of 3766 yds.

Zero range of 100 yds.

As you can also plainly see, it looks quite different too.

Since the bullet flight time is only increased by ~.05 seconds(from ~8.28 to ~8.33), the distance at which the bullet slows to subsonic speed is only ~37 feet less(from ~2550 to ~2513), the estimated speed of the bullet at the target is only ~16 ft/s less(from ~895 to ~879), and the difference between ~1155 ft.lbf. and ~1116 ft.lbf. is you're dead and you're still dead, I don't really need to correct my previous statements on any of those. Since it's pretty much irrelevant. So my job is done here.

Cool...I feel much better know.