# Physics of automatic weapons, mass of 5.56?

**URL:** <https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220>\
**Category:** Factual Questions\
**Created:** [September 21, 2004, 12:59am UTC](https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220 "2004-09-21T00:59:42Z")\
**Posts on this page:** 5\
**Page:** 2

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**Author:** ![Dag\_Otto](https://avatars.discourse-cdn.com/v4/letter/d/439d5e/32.png) [@Dag\_Otto](https://boards.straightdope.com/u/Dag_Otto)\
**Post date:** [September 21, 2004, 11:13pm UTC](https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220/21 "2004-09-21T23:13:35Z")

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> [@Squink](#):
>
> If you’re going to sweat casing weight, you also need to take wind resistance into account. This’ll make things much more complicated.

I don’t think it will make it that much more complicated, but then again, **MikeS** is doing the heavy lifting here.

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**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\
**Post date:** [September 22, 2004, 1:23am UTC](https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220/22 "2004-09-22T01:23:28Z")

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> [@Dag Otto](#):
>
> I don’t think it will make it that much more complicated, but then again, MikeS is doing the heavy lifting here.

Great, the pressure’s on…

First, wind resistance may very well be a factor. Unfortunately, I’m going to have to ignore it, since it’s impossible to know what the magnitude of its effects would be. My rough estimate is that it might change things by a factor of two or three, but that’s really a WAG. If anyone wants to give me a rough analytic form to work with for the wind resistance (force as a function of velocity), then I might be able to modify things.

Here’s the result I get when I work through through this. We’ll denote q[sub]1[/sub] the mass loss due to shell casings, q[sub]2[/sub] the mass loss rate due to combustion products, and q[sub]3[/sub] the rate due to the bullets. Let u[sub]1[/sub], u[sub]2[/sub], u[sub]3[/sub] be the downward ejection velocity of each item, respectively. Then we end up with a very similar result to last time:

v(t) = -gt - (P/q) ln(1 - q t/m[sub]0[/sub])

where I’ve redefined some of my notation: P = q[sub]1[/sub] u[sub]1[/sub] + q[sub]2[/sub] u[sub]2[/sub] + q[sub]3[/sub] u[sub]3[/sub], and q = q[sub]1[/sub] + q[sub]2[/sub] + q[sub]3[/sub].

Note that the form of this motion is similar to the previous derivation, and thus the basic conclusion above still holds: the guy will be unable to stop himself unless his mass is below a certain value. In this case, M \< P / g.

So what are the numbers? Well, we already know q[sub]3[/sub] and u[sub]3[/sub] from below. q[sub]2[/sub] is about 25 grains per round per gun, or about 40 gms/sec for both guns assuming the fire rate we’ve established below. u[sub]2[/sub] is unknown, but let’s assume that it averages out to about half of u[sub]3[/sub]. q[sub]1[/sub] is about 170 gms/sec from both guns, and we’ll assume that u[sub]1[/sub] = 0, i.e. the casings are not ejected with an appreciable vertical velocity (compared to the bullet and the combustion products, at least.)

Putting it all together, we find that :::taps away furiously on his calculator::: M \< 14.8 kg, or about 33 pounds. I’m afraid that our intrepid hero is out of luck.

:::goes to massage his temples:::

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**Author:** ![Squink](https://avatars.discourse-cdn.com/v4/letter/s/b5e925/32.png) [@Squink](https://boards.straightdope.com/u/Squink)\
**Post date:** [September 22, 2004, 1:52am UTC](https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220/23 "2004-09-22T01:52:33Z")

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> [@MikeS](#):
>
> Great, the pressure’s on…

Hee! 🙂

> [@](#):
>
> First, wind resistance may very well be a factor.

A factor that varies at roughly the cube of velocity, and limits a falling human to ~60 meters per second.  
Humans have such a complicated shape that I think you’d actually have to measure things to find out how wind affects their acceleration.

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**Author:** ![DougC](https://avatars.discourse-cdn.com/v4/letter/d/7feea3/32.png) [@DougC](https://boards.straightdope.com/u/DougC)\
**Post date:** [September 22, 2004, 5:18am UTC](https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220/24 "2004-09-22T05:18:33Z")

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- 
  - 
    - Hmmm, gonna need a couple M249s here. Maybe miniguns…  
~

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**Author:** ![emarkp](https://avatars.discourse-cdn.com/v4/letter/e/3be4f8/32.png) [@emarkp](https://boards.straightdope.com/u/emarkp)\
**Post date:** [September 22, 2004, 8:13am UTC](https://boards.straightdope.com/t/physics-of-automatic-weapons-mass-of-5-56/265220/25 "2004-09-22T08:13:49Z")

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> [@Squink](#):
>
> If you ignore gravity…  
> 72,500,000/2,200,000 = 33  
> It’d take 33 bullets to stop the man.

Since it isn’t an elastic collision KE isn’t conserved. But p always is. Note that this answer would mean that a single bullet would impart ~3mph to the man. Enough to knock you over with a single bullet.

If you conserve momentum, it comes out as more like 686 bullets (if the other numbers are right and I got my math right).

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