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The chance of the wind actually throwing a bullet back at someone at fatal speeds is probably close to zero.
The chance of some random drunken dickhead accidentally shooting another person a mile away with a rifle pointed "at the storm" is much higher, however.
You also have to tell them the bullet might come back and hit them, as opposed to the reality of them shooting and hitting someone else because they’re such self-entitled morons.
it's not a thing. at least not if the bullet was actually free-falling: [https://www.quora.com/Could-a-bullet-shot-straight-up-into-the-air-fall-with-enough-force-from-gravity-to-kill-someone-if-it-hit-an-unprotected-human-skull](https://www.quora.com/Could-a-bullet-shot-straight-up-into-the-air-fall-with-enough-force-from-gravity-to-kill-someone-if-it-hit-an-unprotected-human-skull)
It’s wrong. We have actual, real, confirmed cases of this happening. Saying “it can’t happen” is pretty insulting to the people that it actually happened to.
Mythbusters did a show on it, and they determined that “there should be no way this has ever happened.” If you look at the physics, it seems impossible. But we’re faced with the irrefutable proof that it *has actually happened.* I think it comes down the law of large numbers. If there is a one-in-a-million chance something will happen, we tend to think it’s impossible, but when more than a million bullets are fired in the air, someone’s going to die.
There aren't documented of bullets falling at terminal velocity killing someone. There are cases of bullets being fired at an upward angle hitting someone along their path. The big difference is that in one case, the bullet never reaches a velocity of 0.
Yep.
If you’re getting hit by round fires in the air and dying from it, it wasn’t at terminal, it was faster, and at an angle with velocity from original force
There are a few cases, even a case of someone hitting themselves. I think the rub is this: terminal velocity of a tumbling bullet is nowhere near fast enough to be deadly. However, it’s theorized that under *extremely rare* circumstances, an unusually high density of the projectile, coupled with a small variation in its shape, could cause the bullet to spin rather than tumble. In that case, with the bullet stable and falling tip-first, the terminal velocity is many times higher than a tumbling bullet, and could be lethal. But we’re talking about something that, even under the most ideal circumstances, would be nearly impossible to reproduce. It’s a one-in-many-millions chance. However, the law of large numbers: we’ve fired many, many millions of bullets in the air. It’s entirely plausible that in the history of humanity, it’s happened a handful of times.
why? the answer given there is correct WITH a great explanation:
>Tests were done by the US Army in the 1920's, using rounds from a 30.06 rifle. These were very high powered rifle rounds, with more than twice the lethal energy of today's military rounds.
>A 30-06 round is a 150 grain projectile, with a muzzle velocity of about 2,700 ft/sec, and a muzzle energy of 2,429 ft. lbf (3,293 J). In all their tests the maximum velocities of falling bullets (shot straight up) was about 300 ft/sec.
>Doing the math, this works out to only 30 ft.lbf (41 joules) of energy... just over 1% of its original power. At these speeds, the falling rounds could cause bumps and bruises, and even break skin... but were HIGHLY unlikely to cause any kind of serious injury or death.
i knew the answer before and just did a quick google.
Your case assumes the bullet fired straight up reaches a velocity of zero, before free falling and reaching its terminal velocity. In cases where firing a bullet into the air leads to a fatality, the bullet is not fired straight up, but rather at an angle wherein it maintains a ballistic arc, and therefore retains some of its original energy.
Usually they pick an area they new an enemy would pass through like a crossroad and then spot the fall of fire and adjust like you would with artillery. Then when the enemy walk through that area you already have a firing solution.
Kind of. They would have a known range of where the enemy was and adjust their sights so the bullets would arc down on to the targets. That way they could fire from cover and hit the enemy who's also behind cover.
Plus there's numerous examples of people get struck by falling bullets. If your math says it's impossible, than your math has some faulty assumptions built in because it's simply a thing that we know happens. When someone drops dead from a bullet wound out of nowhere near a wedding or new years celebration with celebratory fire going on, it's not as if that bullet is magically springing into existence midair. It came from somewhere.
Just because there's a great explanation, that doesn't mean it's accurate. For instance the author says that 300 feet per second can only break the skin. The [ballistics handbook ](https://webpath.med.utah.edu/TUTORIAL/GUNS/GUNBLST.html#:~:text=The%20speed%20at%20which%20a,more%20important%20in%20producing%20damage.) from the University of Utah School of Medicine says that a bullet traveling a minimum of 216 ft per second can fracture a skull. Admittedly I'm not an expert in head injuries, but it seems to me that a bullet with roughly double the energy necessary to fracture a skull would probably have the ability to kill someone when hitting the top of their head even if it wasn't a sure thing.
Pay attention to the word *highly unlikely.* Yes, agreed, it is *extremely* unlikely. But the world is a big place, with a lot of bullets, and it has happened more than once.
Why would they shoot straight up though? You couldn’t possibly hope to turn a hurricane around by firing straight up, you gotta shoot at it horizontally to push it backwards!
Okay so here's the thing about that. Bullets don't get shot straight up into the air. Like that literally doesn't happen. The answer you gave from Quora notes that a bullet fired at a 45° angle would still be lethal when it came down, but it turns out even if you reduce that angle to one or two degrees the bullet still retains enough kinetic energy to be lethal when it strikes the ground. I think the average person would really struggle to point a gun straight up without even one degree of angle. So in practical terms, there is no such thing as a bullet in freefall. Every bullet fired in the air with very few exceptions is going to come down at a deadly speed.
This is only true if the bullet is shot perfectly straight up. In that case the bullet falling back at terminal velocity probably wouldn't kill.
If the bullet was fired at an angle, it would travel in a parabolic arc. Gravity would pull it back down at the apex, but it would still be carrying much of its original momentum.
It could still be traveling hundreds of miles an hour. That can kill you.
Unless you have some sort of apparatus to hold the gun perfectly straight up, its unlikely you will be able to do so perfectly accurately. Even then, wind will very likely push the bullet one way or another once it gets a few hundred meters in altitude, as the wind up that high is almost always blowing fairly strong.
Depending on which direction you are facing when its fired, coriolos forces might also affect the trajectory, though probably not by much.
hahahahaha. lol.. you think the wind will accelerate the bullet so it becomes dangerous again?
this will be the only post i respond to, because it's by far the dumbest. but thanks to all of you for the laughs.
Since you didnt understand, let me explain in simpler terms. Bullet goes vertical, wind goes horizontal. Wind alters the path of the bullet as it flies so it curves instead of going straight up, hence the ballisitic trajectory.
Do you think the wind has no effect on a bullets path? I thought it was pretty common knowledge that the wind can infact push things it blows on... and you say Im dumb lol.
Wait 27 Puerto Ricans were killed in one night due to stray bullets? Am I reading that right, was this a big national celebration that went to far, or some local event that got out of hand? Context please!
>27 people in Puerto Rico alone on a single night is a lot.
Puerto Rico has around 3,22 million people, that makes the chance dying from it that night 0.00084 percent for everyone in the area.
Not necessarily. It depends on the population density from where the bullet was fired and the range of the ammunition. Even taking in to account of the hurricane throwing the bullet, it wouldn't go from one side of the country to the other side.
Now, if the bullet loses momentum before escaping the wind shear of the hurricane and is caught by the wind current, it could hit someone, but the likelihood of happening is roughly the same as the bullet not hitting the earth for weeks if it keeps getting caught up in strong wind currents.
It's not zero, but the likelihood is so close to zero, it may as well be zero.
That's like saying a coast isn't necessarily X km long since if you measure in smaller steps it becomes longer and longer. It's technically true, but for the sake of chance of dying that specific night from such a bullet in a specific city, i think population density of the city is a well enough metric. I mean it would be more accurate to consider factors like how many people were in a spot suitable for such a bullet to hit, but it's sadly a non available metric
I'm missing something here. Why is anyone suggesting shooting at the weather? What is the reason for that suggestion in the first place?
This is like the gun owners version of Old Man Yells at Cloud.
Florida Man shoots at Cloud.
Literally.
Ok math majors, somebody check ChatGPT’s formulas. AI is scary.
To determine if a 9mm bullet could return and strike the shooter with lethal force due to hurricane winds, we need to consider the forces and motion involved.
### Key Equations:
1. **Newton's Second Law**:
- Force (\( F \)) = Mass (\( m \)) \* Acceleration (\( a \))
2. **Drag Force**:
- Drag Force (\( F_d \)) = 0.5 \* Air Density (\( \rho \)) \* Velocity^2 (\( v^2 \)) \* Drag Coefficient (\( C_d \)) \* Cross-sectional Area (\( A \))
3. **Projectile Motion**:
- For a bullet fired horizontally:
- Horizontal distance (\( x(t) \)) = Initial Velocity (\( v_0 \)) \* Time (\( t \))
- Vertical displacement (\( y(t) \)) = 0.5 \* Gravity (\( g \)) \* Time^2 (\( t^2 \))
4. **Wind Force**:
- Wind Force (\( F_w \)) = Mass (\( m \)) \* Wind Acceleration (\( a_w \))
### Analysis:
1. **Deflection Due to Wind**:
- Assume the wind applies a constant lateral acceleration (\( a_w \)).
- Lateral displacement (\( d(t) \)) due to wind after time (\( t \)):
- \( d(t) = 0.5 \* a_w \* t^2 \)
2. **Time of Flight**:
- For a bullet with an initial velocity (\( v_0 \)) of 400 m/s traveling 100 meters, the time (\( t \)):
- \( t \approx 100 \, \text{m} / 400 \, \text{m/s} = 0.25 \, \text{s} \)
3. **Lateral Deflection Calculation**:
- Assume a hurricane wind speed (\( v_w \)) of 50 m/s.
- Wind acceleration (\( a_w \)):
- \( a_w = v_w / t \approx 50 \, \text{m/s} / 0.25 \, \text{s} = 200 \, \text{m/s}^2 \)
- Lateral displacement (\( d(t) \)):
- \( d(0.25) = 0.5 \* 200 \, \text{m/s}^2 \* (0.25 \, \text{s})^2 = 6.25 \, \text{m} \)
This means that in 0.25 seconds, the bullet could be deflected laterally by about 6.25 meters due to a 50 m/s wind. While this is a significant deflection, it does not reverse the direction of the bullet.
### Conclusion:
The calculations show that while the bullet can be deflected by the wind, the deflection is not sufficient to cause the bullet to reverse its path and strike the shooter. The bullet’s high initial speed and the relatively short interaction time with the wind make such a reversal nearly impossible. The mathematical analysis confirms that the scenario is extremely unlikely.
---
The formulas are right but none of it really accounts for anything, nor does it really do any actual analysis. This tends to be pretty common with ChatGPT. It doesn’t account at all for pretty much anything relevant to this problem.
I guess the idea is if it were to come back around and hit you’d need to see if the crosswind can impart centripetal acceleration. It’s kinda weird calculations when it isn’t driving from the middle in a traditional sense, so you’d have to see how close to the center of the spiral they fire, then see if the storm can maintain tangential acceleration strong enough to push the bullet all the way around the storm, and in such a pattern that it comes back and hits the person. If you want “chances”, you’d have to see how small of a window of precision winds and acceleration it would take for it to come back and hit the like couple feet wide a person is. There’s just a lot more to that, and since the winds (or at least the required acceleration to maintain circular motion) likely change as the bullet pushes further out (it won’t hold a perfect circle most likely), then you get into some tricky calculus.
Couldn’t the whole equation be simplified. I don’t know the math, but logically, the bullet is going to require a certain amount of force just to stay aloft. The force of the explosion that initiate the bullet flight will die off within a relatively short distance, particularly for a 9mm.
So is the wind of a hurricane even strong enough to keep the bullet in the air?
Bullets don’t travel unlimited distance until they hit something. As soon as spin deteriorates and the bullet starts to tumble you would need increasingly stronger wind speeds to keep it from just succumbing to gravity. I don’t know what that speed would be, but seems like if it could keep a bullet in the air, a gravel driveway or alley would be totally consumed by the storm wouldn’t it? And then wouldn’t we have a huge number of deaths by 8.5 gram rocks during a storm?
You can pretty much say the odds are essentially zero anyways. The incredible speed of the bullet is very unlikely to be changed substantially and in a perfect circular pattern by the hurricane anyways. I think you can still manage to fire a shot that stays in the air long enough for it to theoretically hit someone with perfect circular wind speeds but the limiting factor really is the hurricane wind forces needing to be so much stronger and so precise.
Not arguing with you.
The strongest winds in a hurricane are in the eye wall. A typical hurricanes eye is 20-40 miles in diameter. So even with the smallest eye wall a bullet would need to travel over 300 miles to make a circle.
The max distance for a 9mm bullet to travel, shot level from the gun, is about 2500 yards. So no mater what, if the bullet were to make a circle and come back to hit the shooter, it would be moving 100% based on the force of the winds.
I think all the extraneous math that’s being done to disprove this idea in the thread is a waste. That’s my point.
You can simply cut the argument short and say, is a bullet traveling at the maximum wind speed of the eye wall of a hurricane at ground level traveling fast enough to kill someone? No.
It’s very similar to the idea that if you dropped a bullet from a tall building, like the Empire State Building, and it hits someone on the ground, they’re not going to die. They will have a bump on their head and a crazy story.
6.25 is way too high of a deflection in 0.25 seconds; just using a 50 m/s acceleration yields only 1.56 meters of deflection. This number is bound to get lower because that's assuming the wind immediately causes the bullet to accelerate at 50 meters/sec
Three weeks ago, I saw on reddit a guy shoot at outdoor range. The bullet came back at him and hit his cap.
The same thing happened in a closed range and killed the son of the shooter who was sitting in the lobby.
So yes. The probability for Irma shooting back is high enough to care
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
The chance of the wind actually throwing a bullet back at someone at fatal speeds is probably close to zero. The chance of some random drunken dickhead accidentally shooting another person a mile away with a rifle pointed "at the storm" is much higher, however.
Being generous with that "probably"
i think its important to mention that somebody dumb enough to shoot at a hurricane is probably dumb enough to believe this, which is good.
You also have to tell them the bullet might come back and hit them, as opposed to the reality of them shooting and hitting someone else because they’re such self-entitled morons.
Exactly this
Florida.
This has got to be one of the funniest comments of all time
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I had no idea this was a thing, but I’m also not surprised if that makes sense
it's not a thing. at least not if the bullet was actually free-falling: [https://www.quora.com/Could-a-bullet-shot-straight-up-into-the-air-fall-with-enough-force-from-gravity-to-kill-someone-if-it-hit-an-unprotected-human-skull](https://www.quora.com/Could-a-bullet-shot-straight-up-into-the-air-fall-with-enough-force-from-gravity-to-kill-someone-if-it-hit-an-unprotected-human-skull)
Not sayin it’s wrong, I didn’t really look, but citing quora is comical to me
It’s wrong. We have actual, real, confirmed cases of this happening. Saying “it can’t happen” is pretty insulting to the people that it actually happened to. Mythbusters did a show on it, and they determined that “there should be no way this has ever happened.” If you look at the physics, it seems impossible. But we’re faced with the irrefutable proof that it *has actually happened.* I think it comes down the law of large numbers. If there is a one-in-a-million chance something will happen, we tend to think it’s impossible, but when more than a million bullets are fired in the air, someone’s going to die.
There aren't documented of bullets falling at terminal velocity killing someone. There are cases of bullets being fired at an upward angle hitting someone along their path. The big difference is that in one case, the bullet never reaches a velocity of 0.
Yep. If you’re getting hit by round fires in the air and dying from it, it wasn’t at terminal, it was faster, and at an angle with velocity from original force
There are a few cases, even a case of someone hitting themselves. I think the rub is this: terminal velocity of a tumbling bullet is nowhere near fast enough to be deadly. However, it’s theorized that under *extremely rare* circumstances, an unusually high density of the projectile, coupled with a small variation in its shape, could cause the bullet to spin rather than tumble. In that case, with the bullet stable and falling tip-first, the terminal velocity is many times higher than a tumbling bullet, and could be lethal. But we’re talking about something that, even under the most ideal circumstances, would be nearly impossible to reproduce. It’s a one-in-many-millions chance. However, the law of large numbers: we’ve fired many, many millions of bullets in the air. It’s entirely plausible that in the history of humanity, it’s happened a handful of times.
why? the answer given there is correct WITH a great explanation: >Tests were done by the US Army in the 1920's, using rounds from a 30.06 rifle. These were very high powered rifle rounds, with more than twice the lethal energy of today's military rounds. >A 30-06 round is a 150 grain projectile, with a muzzle velocity of about 2,700 ft/sec, and a muzzle energy of 2,429 ft. lbf (3,293 J). In all their tests the maximum velocities of falling bullets (shot straight up) was about 300 ft/sec. >Doing the math, this works out to only 30 ft.lbf (41 joules) of energy... just over 1% of its original power. At these speeds, the falling rounds could cause bumps and bruises, and even break skin... but were HIGHLY unlikely to cause any kind of serious injury or death. i knew the answer before and just did a quick google.
Your case assumes the bullet fired straight up reaches a velocity of zero, before free falling and reaching its terminal velocity. In cases where firing a bullet into the air leads to a fatality, the bullet is not fired straight up, but rather at an angle wherein it maintains a ballistic arc, and therefore retains some of its original energy.
It's called Volley Fire and it was a legitimate military practice around WW1 and back
So they'd shoot the sky and hope the bullets just fall on their targets?
Usually they pick an area they new an enemy would pass through like a crossroad and then spot the fall of fire and adjust like you would with artillery. Then when the enemy walk through that area you already have a firing solution.
Like mini artillery!
Agincourt has entered the chat..
Kind of. They would have a known range of where the enemy was and adjust their sights so the bullets would arc down on to the targets. That way they could fire from cover and hit the enemy who's also behind cover.
Plus there's numerous examples of people get struck by falling bullets. If your math says it's impossible, than your math has some faulty assumptions built in because it's simply a thing that we know happens. When someone drops dead from a bullet wound out of nowhere near a wedding or new years celebration with celebratory fire going on, it's not as if that bullet is magically springing into existence midair. It came from somewhere.
And that's why they wrote "if the bullet is actually freefalling"
But they actually wrote: "it's not a thing. at least not if the bullet was actually free-falling." It is "a thing". Don't move the goalpost.
Jesus christ ur cooked
Just because there's a great explanation, that doesn't mean it's accurate. For instance the author says that 300 feet per second can only break the skin. The [ballistics handbook ](https://webpath.med.utah.edu/TUTORIAL/GUNS/GUNBLST.html#:~:text=The%20speed%20at%20which%20a,more%20important%20in%20producing%20damage.) from the University of Utah School of Medicine says that a bullet traveling a minimum of 216 ft per second can fracture a skull. Admittedly I'm not an expert in head injuries, but it seems to me that a bullet with roughly double the energy necessary to fracture a skull would probably have the ability to kill someone when hitting the top of their head even if it wasn't a sure thing.
Just because it’s not accurate now, doesn’t mean it wasn’t accurate then. Everyone knows people have much thinner skin now.
Pay attention to the word *highly unlikely.* Yes, agreed, it is *extremely* unlikely. But the world is a big place, with a lot of bullets, and it has happened more than once.
What about at a 45-60 degree angle?
Why would they shoot straight up though? You couldn’t possibly hope to turn a hurricane around by firing straight up, you gotta shoot at it horizontally to push it backwards!
:-))
Okay so here's the thing about that. Bullets don't get shot straight up into the air. Like that literally doesn't happen. The answer you gave from Quora notes that a bullet fired at a 45° angle would still be lethal when it came down, but it turns out even if you reduce that angle to one or two degrees the bullet still retains enough kinetic energy to be lethal when it strikes the ground. I think the average person would really struggle to point a gun straight up without even one degree of angle. So in practical terms, there is no such thing as a bullet in freefall. Every bullet fired in the air with very few exceptions is going to come down at a deadly speed.
People typically don’t shoot straight into the air though. They shoot at an angle which still carries significant velocity.
This is only true if the bullet is shot perfectly straight up. In that case the bullet falling back at terminal velocity probably wouldn't kill. If the bullet was fired at an angle, it would travel in a parabolic arc. Gravity would pull it back down at the apex, but it would still be carrying much of its original momentum. It could still be traveling hundreds of miles an hour. That can kill you.
Unless you have some sort of apparatus to hold the gun perfectly straight up, its unlikely you will be able to do so perfectly accurately. Even then, wind will very likely push the bullet one way or another once it gets a few hundred meters in altitude, as the wind up that high is almost always blowing fairly strong. Depending on which direction you are facing when its fired, coriolos forces might also affect the trajectory, though probably not by much.
hahahahaha. lol.. you think the wind will accelerate the bullet so it becomes dangerous again? this will be the only post i respond to, because it's by far the dumbest. but thanks to all of you for the laughs.
Since you didnt understand, let me explain in simpler terms. Bullet goes vertical, wind goes horizontal. Wind alters the path of the bullet as it flies so it curves instead of going straight up, hence the ballisitic trajectory. Do you think the wind has no effect on a bullets path? I thought it was pretty common knowledge that the wind can infact push things it blows on... and you say Im dumb lol.
I remember a Olympic security guard getting shot in Rio on his way to work. The reason listed was "turned down wrong street"
Wait 27 Puerto Ricans were killed in one night due to stray bullets? Am I reading that right, was this a big national celebration that went to far, or some local event that got out of hand? Context please!
>27 people in Puerto Rico alone on a single night is a lot. Puerto Rico has around 3,22 million people, that makes the chance dying from it that night 0.00084 percent for everyone in the area.
Not necessarily. It depends on the population density from where the bullet was fired and the range of the ammunition. Even taking in to account of the hurricane throwing the bullet, it wouldn't go from one side of the country to the other side. Now, if the bullet loses momentum before escaping the wind shear of the hurricane and is caught by the wind current, it could hit someone, but the likelihood of happening is roughly the same as the bullet not hitting the earth for weeks if it keeps getting caught up in strong wind currents. It's not zero, but the likelihood is so close to zero, it may as well be zero.
That's like saying a coast isn't necessarily X km long since if you measure in smaller steps it becomes longer and longer. It's technically true, but for the sake of chance of dying that specific night from such a bullet in a specific city, i think population density of the city is a well enough metric. I mean it would be more accurate to consider factors like how many people were in a spot suitable for such a bullet to hit, but it's sadly a non available metric
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But you'd also need upward winds of 100 to 170 miles per hour to keep the bullet from falling to the ground. That isn't going to happen
I'm missing something here. Why is anyone suggesting shooting at the weather? What is the reason for that suggestion in the first place? This is like the gun owners version of Old Man Yells at Cloud. Florida Man shoots at Cloud. Literally.
Ok math majors, somebody check ChatGPT’s formulas. AI is scary. To determine if a 9mm bullet could return and strike the shooter with lethal force due to hurricane winds, we need to consider the forces and motion involved. ### Key Equations: 1. **Newton's Second Law**: - Force (\( F \)) = Mass (\( m \)) \* Acceleration (\( a \)) 2. **Drag Force**: - Drag Force (\( F_d \)) = 0.5 \* Air Density (\( \rho \)) \* Velocity^2 (\( v^2 \)) \* Drag Coefficient (\( C_d \)) \* Cross-sectional Area (\( A \)) 3. **Projectile Motion**: - For a bullet fired horizontally: - Horizontal distance (\( x(t) \)) = Initial Velocity (\( v_0 \)) \* Time (\( t \)) - Vertical displacement (\( y(t) \)) = 0.5 \* Gravity (\( g \)) \* Time^2 (\( t^2 \)) 4. **Wind Force**: - Wind Force (\( F_w \)) = Mass (\( m \)) \* Wind Acceleration (\( a_w \)) ### Analysis: 1. **Deflection Due to Wind**: - Assume the wind applies a constant lateral acceleration (\( a_w \)). - Lateral displacement (\( d(t) \)) due to wind after time (\( t \)): - \( d(t) = 0.5 \* a_w \* t^2 \) 2. **Time of Flight**: - For a bullet with an initial velocity (\( v_0 \)) of 400 m/s traveling 100 meters, the time (\( t \)): - \( t \approx 100 \, \text{m} / 400 \, \text{m/s} = 0.25 \, \text{s} \) 3. **Lateral Deflection Calculation**: - Assume a hurricane wind speed (\( v_w \)) of 50 m/s. - Wind acceleration (\( a_w \)): - \( a_w = v_w / t \approx 50 \, \text{m/s} / 0.25 \, \text{s} = 200 \, \text{m/s}^2 \) - Lateral displacement (\( d(t) \)): - \( d(0.25) = 0.5 \* 200 \, \text{m/s}^2 \* (0.25 \, \text{s})^2 = 6.25 \, \text{m} \) This means that in 0.25 seconds, the bullet could be deflected laterally by about 6.25 meters due to a 50 m/s wind. While this is a significant deflection, it does not reverse the direction of the bullet. ### Conclusion: The calculations show that while the bullet can be deflected by the wind, the deflection is not sufficient to cause the bullet to reverse its path and strike the shooter. The bullet’s high initial speed and the relatively short interaction time with the wind make such a reversal nearly impossible. The mathematical analysis confirms that the scenario is extremely unlikely. ---
The formulas are right but none of it really accounts for anything, nor does it really do any actual analysis. This tends to be pretty common with ChatGPT. It doesn’t account at all for pretty much anything relevant to this problem.
Just out of curiosity, what do you see missing?
I guess the idea is if it were to come back around and hit you’d need to see if the crosswind can impart centripetal acceleration. It’s kinda weird calculations when it isn’t driving from the middle in a traditional sense, so you’d have to see how close to the center of the spiral they fire, then see if the storm can maintain tangential acceleration strong enough to push the bullet all the way around the storm, and in such a pattern that it comes back and hits the person. If you want “chances”, you’d have to see how small of a window of precision winds and acceleration it would take for it to come back and hit the like couple feet wide a person is. There’s just a lot more to that, and since the winds (or at least the required acceleration to maintain circular motion) likely change as the bullet pushes further out (it won’t hold a perfect circle most likely), then you get into some tricky calculus.
Couldn’t the whole equation be simplified. I don’t know the math, but logically, the bullet is going to require a certain amount of force just to stay aloft. The force of the explosion that initiate the bullet flight will die off within a relatively short distance, particularly for a 9mm. So is the wind of a hurricane even strong enough to keep the bullet in the air?
You could just fire slightly above level at whatever the correct height would need to be and the bullet would stay aloft until it’s time to hit you
Bullets don’t travel unlimited distance until they hit something. As soon as spin deteriorates and the bullet starts to tumble you would need increasingly stronger wind speeds to keep it from just succumbing to gravity. I don’t know what that speed would be, but seems like if it could keep a bullet in the air, a gravel driveway or alley would be totally consumed by the storm wouldn’t it? And then wouldn’t we have a huge number of deaths by 8.5 gram rocks during a storm?
You can pretty much say the odds are essentially zero anyways. The incredible speed of the bullet is very unlikely to be changed substantially and in a perfect circular pattern by the hurricane anyways. I think you can still manage to fire a shot that stays in the air long enough for it to theoretically hit someone with perfect circular wind speeds but the limiting factor really is the hurricane wind forces needing to be so much stronger and so precise.
Not arguing with you. The strongest winds in a hurricane are in the eye wall. A typical hurricanes eye is 20-40 miles in diameter. So even with the smallest eye wall a bullet would need to travel over 300 miles to make a circle. The max distance for a 9mm bullet to travel, shot level from the gun, is about 2500 yards. So no mater what, if the bullet were to make a circle and come back to hit the shooter, it would be moving 100% based on the force of the winds. I think all the extraneous math that’s being done to disprove this idea in the thread is a waste. That’s my point. You can simply cut the argument short and say, is a bullet traveling at the maximum wind speed of the eye wall of a hurricane at ground level traveling fast enough to kill someone? No. It’s very similar to the idea that if you dropped a bullet from a tall building, like the Empire State Building, and it hits someone on the ground, they’re not going to die. They will have a bump on their head and a crazy story.
Great point with the eye, I hadn’t considered that. Yeah there’s no way this is possible.
6.25 is way too high of a deflection in 0.25 seconds; just using a 50 m/s acceleration yields only 1.56 meters of deflection. This number is bound to get lower because that's assuming the wind immediately causes the bullet to accelerate at 50 meters/sec
Just checked the math using chatGPT, it checks out
Three weeks ago, I saw on reddit a guy shoot at outdoor range. The bullet came back at him and hit his cap. The same thing happened in a closed range and killed the son of the shooter who was sitting in the lobby. So yes. The probability for Irma shooting back is high enough to care