Are the winds strong enough to turn a bullet around and launch it back towards me? I'm (incorrectly) assuming there's no debris to bounce off of; I just want to know if the winds are strong enough.

  • $\begingroup$ This is equivalent to asking whether a hurricane can fire a bullet from rest. My guess is probably not, it's too small to experience a significant push from the air. $\endgroup$
    – Javier
    Sep 9, 2017 at 23:11

3 Answers 3


A 150 mph (241 kph) hurricane is still only blowing at 250 feet per second (76.2 m/s). A .270 caliber (6.8 mm) rifle has a muzzle velocity of about 3,100 feet per second (944.8 m/s) and a max range of about 2.72 miles (4.377416 km). At 500 yards (457.2 m) the bullet is still moving at about 1,800 feet per second (548.64 m/s). Assuming the bullet is fired directly into a head wind with the rifle barrel elevated +30 degrees (+30 degrees), the hurricane is too slow to have much impact on the bullet. It'll be too far downrange for a 150 mph (241 kph) wind to reverse its course and bring it back to you. What will happen is that the bullet impacts the earth a little less than 2.72 miles (4.377 km) downrange.

What happens in a 150 mph (67 m/s) crosswind? Using my .270 Winchester, at a distance of 1000 yards (914 m) I get 44 yards (40.2 m) of lateral drift. But my bullet is spin stabilized, so it acts like a gyroscope. This causes it to rise or fall depending on wind direction relative to spin direction. A rule of thumb puts the ratio between drift and rise or fall at 10:1 (for every 10 units of drift I get 1 unit of rise or fall). That means my bullet either rises or drops 53.4" (135.6 cm). Wind shear (vertical wind) is just a crosswind in a vertical plane. Thus at 1000 yards (914 m) I'll rise or fall 44 yards (40.2 m) and drift laterally 53.4" (135.6 cm).

Thus, no matter what the wind direction is, at 1000 yards (914 m), hurricane force winds of 150 mph (67 m/s) don't even come close to reversing bullet direction. The maximum ideal range for this projectile is only 2.72 miles (4377 meters) and at 1000 yards we've already consumed more than half the projectile's initial velocity. At this turn rate, we run out of energy before we reverse the bullet's direction.

The bullet used in this example is a 130 grain (8.42 gram) Hornady boat tail soft point.

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    $\begingroup$ Any chance you can add SI equivalents of those numbers so they make sense to people from countries on the post-18th-century side of metrology? (otherwise it looks good, though) $\endgroup$ Sep 10, 2017 at 1:51
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    $\begingroup$ If the you shoot a subsonic 45 pistol caliber from a distance, the billet may slow down enough to be picked up be the wind. If there happens to be a category F5 tornado in the storm, it may propell the bullet back at up to 300 mph. This is not too far from the speed of a bow arrow. Not likely to kill, but can, if hits in the wrong spot. Imagine a car passing you by at 300 mph and the driver throwing a marble at you. Extremely unlikely, but not impossible. $\endgroup$
    – safesphere
    Sep 10, 2017 at 7:37
  • $\begingroup$ It's important to remember that a bullet in flight is being tugged earthward by gravity. In my .270 example, at 500 yards the bullet drops about 4.5 feet, and at 1000 yards it drops more than 23 feet. Relative to horizontal, it's traveling downhill. This is why a +30 degree barrel elevation only produces a max range of 2.72 miles. $\endgroup$
    – Dave
    Sep 10, 2017 at 14:59
  • $\begingroup$ Given a bullet max range of 2.72 miles, consider that a 9-mile hurricane eye has a circumference of about 28 miles. A bullet with a range of less than 3 miles can't make the trip around the eye--it hits dirt long before it completes the trip. Thus, traveling around the eye is no more feasible than blowing the bullet backwards. $\endgroup$
    – Dave
    Sep 10, 2017 at 15:19
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    $\begingroup$ An F5 tornado "may" produce a velocity of 300 mph (440 fps) but think of that as muzzle velocity. The moment the bullet leaves the tornado, wind drag rapidly robs it of velocity. It's probably tumbling at this point, which produces even more drag. Velocity very quickly falls to that of the prevailing winds outside the tornado. Gravity wins this one. Now, if one were close enough to an F5 tornado to be hit by the bullet, the rain of airborne coconuts, cows, and cars would be a much bigger concern. $\endgroup$
    – Dave
    Sep 10, 2017 at 15:30

@Dave 's answer is reasonable. However, let us imagine for a moment that there is no gravity. Then a bullet flying directly into a headwind will eventually stop and then fly back with the wind. Therefore, if there is not just a headwind, but also a significant upward component of the wind,I don't see why the bullet cannot return back in the presence of gravity, as the upward component of the wind counteracts gravity.

Another situation when a bullet can return back in the wind is when it is fired in the direction that is close to vertical, so the horizontal component of the velocity of the bullet is comparable to the velocity of wind.

  • $\begingroup$ I'd say upward angle is everything, with two likely outcomes: (1) The upward angle is so steep that the bullet falls out of a ballistic arc, tumbles, and returns to earth at terminal velocity (approx. 300 feet per second for a .30 caliber, 180 grain projectile). (2) The upward angle is shallow enough to permit the bullet to retain its ballistic arc and stability. In this case, the bullet's forward velocity exceeds wind velocity and it continues downrange without reversing course. This is the more deadly scenario because bullet velocity is higher and the projectile strikes point-first. $\endgroup$
    – Dave
    Sep 11, 2017 at 2:53
  • $\begingroup$ @Dave: Again, if the upward angle is shallow, but there is upward component of wind, I don't see why the bullet cannot reverse course. $\endgroup$
    – akhmeteli
    Sep 11, 2017 at 2:59
  • $\begingroup$ Let's define what "reverse course" means. To me it means the forward end remains pointed in the direction of travel and the bullet remains stable. Getting my example bullet to turn 180 degrees means causing it to turn more than it travels downrange. I can expect hurricane-force winds to displace the bullet laterally about 40.2 meters per 914 meters of travel. To turn 402 meters, the bullet has to travel 9,140 meters. If it's turning horizontally, it hit the ground at 4,377 meters or less. In a vertical turn it reaches max altitude, stops, and tumbles without nosing over. $\endgroup$
    – Dave
    Sep 11, 2017 at 3:29
  • $\begingroup$ @Dave : Your definition may be reasonable, but it is not the only possible definition, moreover, I fail to see anything in the question that would require such a definition. In my book, "reverse course" means that the bullet starts to fly in the direction that is opposite to the initial direction (with respect to the earth). I do not consider lateral movement. I consider a case where the bullet has a shallow angle and does not fall on earth because of upward component of the wind. The headwind will eventually reverse the course of the bullet. $\endgroup$
    – akhmeteli
    Sep 11, 2017 at 3:52
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    $\begingroup$ @FilipHaglund: The smallest known Atlantic hurricane eye was Hurricane Wilma, at about 2 miles diameter (about 3.2 km). That produces a circumference of 6.2 miles, which is quite a bit longer than the max range on my .270 Win. I'd say you're safe, but someone about 1/3 the way around is gonna need a flak vest and helmet. $\endgroup$
    – Dave
    Sep 11, 2017 at 4:37

I think it could be possible but extremely unlikely. Even if it did come back and hit you or someone else it would definitely be tumbling instead of acting like a normal bullet. If you look at how a bullet shoots under water you have to realize air is doing almost the same thing with way less resistance. The wind is always gonna change the course of the bullet nomatter what calibre. @dave @akhmeteli I think your both right in your own way. Unless you own a rail gun even just 1-5 km of wind will change the course. You also have to factor in if there's updraft, down draft or other equations. Technically nobody here is right until we put an RFID chip in a bullet and test it in different scenarios

A sniper rifle would be the most likely to make it through unless you're shooting from a distance of about 2 miles or 3.2 km aiming for the strongest part of the storm. It'd most likely go through with minimal change in direction.. however if you're using a small calibre. Pistol that'd be more than likely to turn into added flying debree or end up falling before getting even close to back to you.. try it in a wind tunnel that's circular and change the statistics every time, then do the math. I'd actually like to know lol. I know if you shoot a bullet straight up, it comes down in a completely different area, id like to know what happens in different scenarios. Example, with an updraft, would it continue following the course of the storm as it turns into a projectile, or is the wind too weak and will it just go through or fall to the ground..

P.s. ever wonder why a bunch of dead fish show up after big storms... Cause they gettin shot at from ppl who don't understand physics xD


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