# Celebratory Gunfire Danger

**URL:** <https://boards.straightdope.com/t/celebratory-gunfire-danger/608206>\
**Category:** Factual Questions\
**Created:** [January 3, 2012, 12:50am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206 "2012-01-03T00:50:40Z")\
**Posts on this page:** 20\
**Page:** 3

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**Author:** ![quenda](https://avatars.discourse-cdn.com/v4/letter/q/b5e925/32.png) [@quenda](https://boards.straightdope.com/u/quenda)\
**Post date:** [January 4, 2012, 1:11am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/41 "2012-01-04T01:11:07Z")

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> [@Shodan](#):
>
> As far as wind resistance is concerned, wouldn’t a larger bullet present a greater surface area, and therefore accelerate slower?

The opposite. Larger surface area would cause greater accel, both positive from the detonation, and negative from air resistence. But again you’d have to make the silly assumption that all else is equal.

> [@](#):
>
> Actually, if I remember my physics, …

Another bad assumption there 🙂

This is all about terminal velocity and energy - a function of front area, drag coefficient, and mass.  
Muzzle velocity is irrelevant, as is angle of fire, unless it is so low the bullet still has substantial horizontal velocity on impact.

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**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\
**Post date:** [January 4, 2012, 3:33am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/42 "2012-01-04T03:33:52Z")

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> [@Shodan](#):
>
> I assume you fire two bullets directly upward. One is a .22; the other a .44. You loaded your own ammunition, and put the same charge into each bullet. I would expect the .22 to go higher, but the .44 would fall faster. Is that correct?

Let’s take this to a logical extreme: instead of a .44, let’s assume the given charge tries to loft a thirty-pound lead weight. That hefty lead weight won’t have much muzzle velocity at all, and won’t get very high; it also won’t come down very fast. Even if it achieves 99% of its original muzzle velocity during the freefall, that’s still not much. I’d bet the .22 bullet will come down with more speed, but a lot less energy.

> [@](#):
>
> Now assume two more shots. Same charge, but both are .44 caliber. One, however, is hollow so that it contains half the mass of the solid slug. Would the lighter one still go higher? I assume the answer is Yes, because the lesser mass is easier to accelerate for a given input of energy.

Again, take it to a logical extreme, and assume a featherweight .44 slug, one that’s only 0.1 percent of the mass of a solid slug. Like a badminton shuttlecock, it’s going to leave the barrel with lots of velocity, but decelerate extremely rapidly. In my scenario, the heavy one goes higher.

I’m not sure there are any absolutes here. There may be crossover points where a greater powder charge or a larger mass or a different drag profile causes a change in which bullet wins.

> [@](#):
>
> Would the solid, heavier one still fall faster? I am guessing No, because wind resistance is the same for both.

Wind resistance is the same for both, but gravity is pulling harder on the heavier one; the heavier one will fall faster.

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**Author:** ![RickJay](https://avatars.discourse-cdn.com/v4/letter/r/bb73d2/32.png) [@RickJay](https://boards.straightdope.com/u/RickJay)\
**Post date:** [January 4, 2012, 3:50am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/43 "2012-01-04T03:50:41Z")

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> [@engineer\_comp\_geek](#):
>
> If you fire a bullet straight up, it will go up until it stops, and then it will freefall back down, increasing in speed until it reaches its terminal velocity. A bigger heavier bullet will fall faster and will cause more damage when it hits. Handgun rounds are pretty light and probably wouldn’t hurt you all that much. Rifle rounds like the 7mm are heavier and would be a bit more painful.

Handgun ROUNDS may be lighter than rifle ROUNDS, but handgun BULLETS certainly are not. There is perhaps some misunderstanding here; what comes out of a firearm when you pull the trigger is a bullet, not a round. A round is a bullet combined with its shell and propellant.

In fact, most common handgun bullets are just as heavy as most common rifle bullets. A typical 7.62mm rifle bullet weighs about the same as a 9mm bullet; the rifle bullet is longer, but, obviously, thinner. The 5.56mm bullet used by most Western militaries is only half the weight of a 9mm pistol bullet; it derives its lethality from its speed, not its weight.

What would make a rifle bullet fired straight into the air more dangerous is not that it is heavier, because it isn’t heavier. It’s that it has vastly more speed and so will get up higher and have more time to accelerate before it hits you in the noggin.

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**Author:** ![Elendil\_s\_Heir](https://avatars.discourse-cdn.com/v4/letter/e/7cd45c/32.png) [@Elendil\_s\_Heir](https://boards.straightdope.com/u/Elendil_s_Heir)\
**Post date:** [January 4, 2012, 5:11am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/44 "2012-01-04T05:11:44Z")

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> [@Bloodless\_Turnip](#):
>
> A local kid got [hit by one this year](http://www.tampabay.com/news/publicsafety/crime/ruskin-boy-hit-by-celebratory-new-years-gunfire-remains-critical/1208808).

See post #16.

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**Author:** ![MichaelEmouse](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/michaelemouse/32/101_2.png) [@MichaelEmouse](https://boards.straightdope.com/u/MichaelEmouse)\
**Post date:** [January 4, 2012, 7:15am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/45 "2012-01-04T07:15:16Z")

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> [@RickJay](#):
>
> Handgun ROUNDS may be lighter than rifle ROUNDS, but handgun BULLETS certainly are not. There is perhaps some misunderstanding here; what comes out of a firearm when you pull the trigger is a bullet, not a round. A round is a bullet combined with its shell and propellant.
> 
> In fact, most common handgun bullets are just as heavy as most common rifle bullets. A typical 7.62mm rifle bullet weighs about the same as a 9mm bullet; the rifle bullet is longer, but, obviously, thinner. The 5.56mm bullet used by most Western militaries is only half the weight of a 9mm pistol bullet; it derives its lethality from its speed, not its weight.
> 
> What would make a rifle bullet fired straight into the air more dangerous is not that it is heavier, because it isn’t heavier. It’s that it has vastly more speed and so will get up higher and have more time to accelerate before it hits you in the noggin.

I agree, but the weight still matters, though not in a straight comparison.

Even though the weight of a 62\* grains 5.56 bullet is lower than the weight of a 125\* grains 9mm bullet, the 62 grains 5.56mm bullet has a higher sectional density than the 125 grains 9mm bullet. A higher sectional density means that it will lose less speed to air resistance and penetrate more deeply.

The 5.56mm will also be more aerodynamic than the 9mm. That sort of calculation is above my knowledge and skills.

- To take typical weights as examples.

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**Author:** ![Mangetout](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mangetout/32/19_2.png) [@Mangetout](https://boards.straightdope.com/u/Mangetout)\
**Post date:** [January 4, 2012, 8:09am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/46 "2012-01-04T08:09:25Z")

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> [@RickJay](#):
>
> What would make a rifle bullet fired straight into the air more dangerous is not that it is heavier, because it isn’t heavier. It’s that it has vastly more speed and so will get up higher and have more time to accelerate before it hits you in the noggin.

Surely this is moot, isn’t it? The velocity of launch is greater than the terminal velocity the falling bullet can attain - obviously there’s wind resistance on the way up, but the vertically-fired bullet must still be ascending to a sufficient height that it will achieve terminal velocity on the way back down, must it not?

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**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\
**Post date:** [January 4, 2012, 7:00pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/47 "2012-01-04T19:00:33Z")

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> [@Machine\_Elf](#):
>
> Let’s take this to a logical extreme: instead of a .44, let’s assume the given charge tries to loft a thirty-pound lead weight. That hefty lead weight won’t have much muzzle velocity at all, and won’t get very high; it also won’t come down very fast. Even if it achieves 99% of its original muzzle velocity during the freefall, that’s still not much. I’d bet the .22 bullet will come down with more speed, but a lot less energy.

But energy varies directly with the square of the velocity.

take 2 masses M1 and M2 and M1 \> M2. Given the same force F over the same time T they are given accelerations of A1 and A2.  
We know the velocities V1 = F(t/M1) and V2 = F(t/M2). By symmetry and no air, the bullet hit the ground (or your head) at the same velocity yeilding the kinetic energy  
K1 = (M1/2)(t/M1)^2 = (t/2)(1/M1^2)  
K2 = (M2/2)(t/M2)^2 = (t/2)(1/M2^2)  
Since M1 \> M2, K1 \< K2

> [@Machine\_Elf](#):
>
> Wind resistance is the same for both, but gravity is pulling harder on the heavier one; the heavier one will fall faster.

After reading the works of Aristotle, you may want to read Galileo’s response.

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**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\
**Post date:** [January 4, 2012, 8:43pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/48 "2012-01-04T20:43:34Z")

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> [@Saint\_Cad](#):
>
> take 2 masses M1 and M2 and M1 \> M2. Given the same force F over the same time T they are given accelerations of A1 and A2.

If they have different accelerations, then they will not have the same barrel residence time T; instead, T2 \< T1.

That’s kind of beside the point, though. The assumption implicit in Shodan’s and my analyses is that both bullets leave the barrel with the same kinetic energy.

> [@](#):
>
> We know the velocities V1 = F(t/M1) and V2 = F(t/M2). By symmetry and no air

I stopped reading when you said no air. Aerodynamic drag is a critically important part of this entire thread; you cannot ignore it.

> [@Saint Cad](#):
>
> > [@Machine Elf](#):
> >
> > Wind resistance is the same for both, but gravity is pulling harder on the heavier one; the heavier one will fall faster.
> 
> After reading the works of Aristotle, you may want to read Galileo’s response.

I worded that poorly. What I should have said was that the drag vs. speed function is the same for both bullets, but gravity is pulling harder on the heavier one; therefore the heavier one will achieve a higher terminal velocity.

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**Author:** ![UltraVires](https://avatars.discourse-cdn.com/v4/letter/u/ecccb3/32.png) [@UltraVires](https://boards.straightdope.com/u/UltraVires)\
**Post date:** [January 4, 2012, 10:03pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/49 "2012-01-04T22:03:58Z")

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> [@engineer\_comp\_geek](#):
>
> Fire the rifle an an angle though and the bullet follows a parabolic arc and can easily have a fatal velocity when it returns to earth.

This. For all of the physics talk about terminal velocity in this thread, drunken revelers rarely get out their protractors and design a mechanism for firing at a 90 degree angle to the ground.

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**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\
**Post date:** [January 5, 2012, 4:01pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/50 "2012-01-05T16:01:27Z")

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> [@Machine\_Elf](#):
>
> If they have different accelerations, then they will not have the same barrel residence time T; instead, T2 \< T1.
> 
> That’s kind of beside the point, though. The assumption implicit in Shodan’s and my analyses is that both bullets leave the barrel with the same kinetic energy.
> 
> I stopped reading when you said no air. Aerodynamic drag is a critically important part of this entire thread; you cannot ignore it.

I realize my calculations were extremely simplified. I presented it simple to address your statement

> [@](#):
>
> I’d bet the .22 bullet will come down with more speed, but a lot less energy

. and demonstrate that speed is actually MORE of a factor in kinetic energy than mass.  
How about this: KE = (mv^2)/2 so doubling a mass doubles the energy but doubling the speed quadruples the energy.

Also I was addressing your thought experiment of the 30 lbs weight vs. a bullet. RickJay had already addressed the issue of two bullets. In your thought experiment you said “given charge” so I assumed that the force on both objects would be the same over the same time e.g. the propellent produces 100 lbs of force over 0.1 sec.

> [@Machine\_Elf](#):
>
> I worded that poorly. What I should have said was that the drag vs. speed function is the same for both bullets, but gravity is pulling harder on the heavier one; therefore the heavier one will achieve a higher terminal velocity.

I understand your concept but gravity does not “pull harder” on heavier objects. A simple way to make your point is that terminal velocity in freefall is when air resistance equals its weight. All else being equal, terminal velocity varies with the square root of the object’s weight so if A is twice as heavy as B, A’s terminal velocity is about 1.4 times that of B.

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**Author:** ![Gray\_Ghost](https://avatars.discourse-cdn.com/v4/letter/g/c0e974/32.png) [@Gray\_Ghost](https://boards.straightdope.com/u/Gray_Ghost)\
**Post date:** [January 5, 2012, 5:08pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/51 "2012-01-05T17:08:15Z")

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[Some expansion](http://www.loadammo.com/Topics/March01.htm) of **silenus** ’s early post, for those without access to the 1920 BuOrd study.

Don’t know if it’s been mentioned already, but the researchers cited above noted that the bullets came down base first, presumably due to their rotational inertia. As such, the ballistic coefficients for those bullets are going to be quite a bit different than the listed, point-forward values. As for lethality, there’s always the eggshell plaintiff issue. Consider also that the [.25 ACP](http://en.wikipedia.org/wiki/.25_ACP) handgun bullet in many of its common loadings, has about 60-65 ft-lbs of energy. I don’t have the stats in front of me, but I would guess that while 95%+ of people shot once with a .25 eventually recover, it’s had to have killed someone over its history.

Further expanding on **engineer** ’s point, the cite has this quote about terminal velocities for large artillery shells:

> [@](#):
>
> Major Julian Hatcher in his book Hatcher’s Notebook estimates that a 12 inch shell weighing 1000 pounds and fired straight up would return with a speed of 1,300 to 1,400 feet per second and over 28 million foot pounds of striking energy.

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<div class="post-metadata">

**Author:** ![JBDivmstr](https://avatars.discourse-cdn.com/v4/letter/j/b4bc9f/32.png) [@JBDivmstr](https://boards.straightdope.com/u/JBDivmstr)\
**Post date:** [January 5, 2012, 5:51pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/52 "2012-01-05T17:51:22Z")

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> [@Gray\_Ghost](#):
>
> [Some expansion](http://www.loadammo.com/Topics/March01.htm) of **silenus** ’s early post, for those without access to the 1920 BuOrd study.
> 
> Don’t know if it’s been mentioned already, but the researchers cited above noted that the bullets came down base first, presumably due to their rotational inertia. As such, the ballistic coefficients for those bullets are going to be quite a bit different than the listed, point-forward values. As for lethality, there’s always the eggshell plaintiff issue. Consider also that the [.25 ACP](http://en.wikipedia.org/wiki/.25_ACP) handgun bullet in many of its common loadings, has about 60-65 ft-lbs of energy. I don’t have the stats in front of me, but I would guess that while 95%+ of people shot once with a .25 eventually recover, it’s had to have killed someone over its history.
> 
> Further expanding on **engineer** ’s point, the cite has this quote about terminal velocities for large artillery shells:

> [@](#):
>
> Major Julian Hatcher in his book Hatcher’s Notebook estimates that a 12 inch shell weighing 1000 pounds and fired straight up would return with a _ **speed of 1,300 to 1,400 feet per second and over 28 million foot pounds of striking energy.** _

(bolding mine)

Well… _ **that’s** _ probably gonna leave a mark. 😃

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<div class="post-metadata">

**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\
**Post date:** [January 5, 2012, 5:53pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/53 "2012-01-05T17:53:22Z")

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> [@Machine\_Elf](#):
>
> If they have different accelerations, then they will not have the same barrel residence time T; instead, T2 \< T1.
> 
> That’s kind of beside the point, though. The assumption implicit in Shodan’s and my analyses is that both bullets leave the barrel with the same kinetic energy.

Which isn’t necessarily a true assumption.

As you note, there are extremes here. An extremely light bullet doesn’t get much kinetic energy. It’s similar to trying to impart energy to a ball of paper by throwing it. The mass is low, and your velocity is capped by arm speed. Similarly, something like a bowling ball would be just as difficult because you can’t generate hardly any velocity. The best is something like a baseball. Enough mass to have a high kinetic energy, but low enough so that you can impart a high velocity.

Guns are similar. Each type of gun is going to have a sweet spot based on barrel length, caliber, bullet weight, and propellant type. So there is no real way to say what bullet would be worse without testing it, or doing extensive analysis.

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**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [January 5, 2012, 6:15pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/54 "2012-01-05T18:15:23Z")

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> [@Saint\_Cad](#):
>
> … gravity does not “pull harder” on heavier objects.

It doesn’t??

When you put a 100-lb object on a scale, gravity causes it to exert a force on the scale that’s twice that of a 50-lb object. Isn’t that “pulling harder”?

It is true that in a vacuum, both objects when dropped will accelerate at the same rate - but not because the force is equal. The heavier one experiences a greater force, but also has greater inertia - so its change in velocity matches that of the 50-lb object.

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<div class="post-metadata">

**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\
**Post date:** [January 5, 2012, 6:21pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/55 "2012-01-05T18:21:17Z")

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There is some discussion here about a bullet in freefall vs. one on a ballistic trajectory.

Isn’t there some point at which air resistance will slow a bullet to terminal velocity even if it was fired on a ballistic parabola? Or does the propelling force overwhelm that in any practical scenario?

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**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\
**Post date:** [January 5, 2012, 7:44pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/56 "2012-01-05T19:44:04Z")

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> [@CookingWithGas](#):
>
> There is some discussion here about a bullet in freefall vs. one on a ballistic trajectory.
> 
> Isn’t there some point at which air resistance will slow a bullet to terminal velocity even if it was fired on a ballistic parabola? Or does the propelling force overwhelm that in any practical scenario?

There is some critical angle where it slows to terminal velocity. A bullet fired at 89.9999 degrees will reach terminal velocity coming down. One fired at 0.0001 degrees never will. Somewhere in between there it hits exactly terminal velocity when it reaches the ground.

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<div class="post-metadata">

**Author:** ![Patch](https://avatars.discourse-cdn.com/v4/letter/p/a183cd/32.png) [@Patch](https://boards.straightdope.com/u/Patch)\
**Post date:** [January 6, 2012, 6:15am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/57 "2012-01-06T06:15:51Z")

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> [@RickJay](#):
>
> In fact, most common handgun bullets are just as heavy as most common rifle bullets. A typical 7.62mm rifle bullet weighs about the same as a 9mm bullet; the rifle bullet is longer, but, obviously, thinner. The 5.56mm bullet used by most Western militaries is only half the weight of a 9mm pistol bullet; it derives its lethality from its speed, not its weight.
> 
> What would make a rifle bullet fired straight into the air more dangerous is not that it is heavier, because it isn’t heavier. It’s that it has vastly more speed and so will get up higher and have more time to accelerate before it hits you in the noggin.

A 124gr 9mm FMJ bullet has a terminal velocity about 219 fps, assuming a base-first return. A 5.56mm 55gr FMJ bullet has a terminal velocity of 244 fps if returning base-first, 141 fps if tumbling. Data from the source I cited earlier.

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<div class="post-metadata">

**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\
**Post date:** [January 6, 2012, 4:08pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/58 "2012-01-06T16:08:43Z")

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> [@Xema](#):
>
> It doesn’t??
> 
> When you put a 100-lb object on a scale, gravity causes it to exert a force on the scale that’s twice that of a 50-lb object. Isn’t that “pulling harder”?
> 
> It is true that in a vacuum, both objects when dropped will accelerate at the same rate - but not because the force is equal. The heavier one experiences a greater force, but also has greater inertia - so its change in velocity matches that of the 50-lb object.

Yes and no.  
The gravitational force between two objects is proportional to the mass and so I guess that in that way it is pulling harder.  
And yes, all else being equal a body in freefall with a higher mass will have a higher terminal velocity because of the greater weight.

What I am trying to get at with Machine Elf is his Aristotilian way of phrasing things. ME neglects that the higher terminal velocity occurs when air resistance matched weight. His post comes across as heavier objects fall faster than lighter objects because there is “more gravity” on a heavier object which was disproved by Galileo.

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<div class="post-metadata">

**Author:** ![Malthus](https://avatars.discourse-cdn.com/v4/letter/m/85f322/32.png) [@Malthus](https://boards.straightdope.com/u/Malthus)\
**Post date:** [January 6, 2012, 4:21pm UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/59 "2012-01-06T16:21:07Z")

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> [@CookingWithGas](#):
>
> There is some discussion here about a bullet in freefall vs. one on a ballistic trajectory.
> 
> Isn’t there some point at which air resistance will slow a bullet to terminal velocity even if it was fired on a ballistic parabola? Or does the propelling force overwhelm that in any practical scenario?

My understanding is that for all practical purposes one can assume that a rifle bullet “fired in the air” in celebration will in fact be in a ballistic parabola - for one, celebratory firers are unlikely to be too particular about ensuring accurate angle of fire. 😉

Hence the talk of freefalling bullets is probably of more theoretical interest than real.

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<div class="post-metadata">

**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [January 7, 2012, 2:20am UTC](https://boards.straightdope.com/t/celebratory-gunfire-danger/608206/60 "2012-01-07T02:20:28Z")

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> [@Malthus](#):
>
> … a rifle bullet “fired in the air” in celebration will in fact be in a ballistic parabola

Until the parabola is distorted by drag - which happens in the first half second.

> [@](#):
>
> Hence the talk of freefalling bullets is probably of more theoretical interest than real.

As Mangetout and others have noted, the concept of terminal velocity can apply even if the shot is far from vertical.

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