NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/sci/military/11248

Text File | 1992-11-20 | 3.5 KB | 76 lines

Newsgroups: sci.military
Path: sparky!uunet!zaphod.mps.ohio-state.edu!rpi!psinntp!psinntp!ncrlnk!ciss!law7!military
From: ultra![e-mail] (Norm Finn)
Subject: Re: RPG Combat Rules & Reality
Message-ID: <[e-mail]>
Sender: [e-mail] (Sci.Military Login)
Organization: Ultra Network Technologies
References: <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]>
Date: Thu, 19 Nov 1992 14:02:33 GMT
Approved: [e-mail]
Lines: 63


From ultra![e-mail] (Norm Finn)

In article <[e-mail]> [e-mail] (MILLER, JIMMY A.) writes:
>
>> >Modern jackets/vests are fairly effective, but one thing most games ignore is
>> >kinetic energy. Sure, so you got DAMN lucky and the .44 didn't punch through
>>
>> I've always wondered about this.  The person *shooting* the .44 had to
>> accept just as much momentum when shooting as the bullet imparts to the
>> recipient.
>>
>> It seems like if the vest distributes this over a wide enough area you
>> could void any damage at all.
>
>  Well, yes but this is pretty tricky, I would think.  And in answer to the
>question concerning about the shooter receiving momentum, consider that the
>shooter has shock absorbtion available (stocks, gas bleeds, the motion of the
>body or hands backwards, momentum absorbed by the mass of the weapon itself,
>etc.).  The target has to absorb all that kinetic force on a spot somewhat
>smaller than a dime...

Let's get back to basic physics, guys.  The explosion of the powder
imparts equal momentum to 1) the bullet and blow-by gasses in one
direction, and 2) the gun in the other direction.  The bullet gets most
of the momentum in its direction.  Now, momentum (p) = mass (m) *
velocity (v).  If the bullet is 1/50th the weight of the gun, it gets 50
times the velocity.

What about energy?  Kinetic energy (e) = 1/2 * m * (v squared).  So if
the bullet gets 50 times the velocity, it gets 2500 times the energy.
That is, most all the energy of the power explosion is converted
into the kinetic energy of the bullet.  (OK, OK, there is some angular
momentum and energy, too, because the bullet is spun up by the rifling
grooves, and a big *bang*, etc., but let's stick to the main factors.)

We'll ignore the energy and momentum the bullet loses to the air during
its passage to the target, and worry about the target.  The inelastic
collision at the target transfers the momentum of the bullet to the
momentum of the bullet+target mass.  It is equal to the momentum given
the gun when it was fired.  It doesn't amount to a hill of beans.

The energy, however, is very important to the target.  The kinetic
energy of the bullet is dissipated upon the bullet and the target by
being transformed into heat, shock waves, ripping, tearing, and general
mayhem.  If the target is kevlar armor, the energy is dissipated into
the ripping and heating of the kevlar, shock waves through the vest and
person, etc.

You can think of a gun as a reasonably efficient energy transmission
device.  It transmits the energy of an explosion to a target.  It also
transmits momentum, but that's not important.

(Does this come up often enough for the FAQ file?  I've seen similar
questions often enough.)

[mod note -- It might be in the rec.guns FAQ ]
--
Norm Finn                       domain:   [e-mail]
Ultra Network Technologies      Internet: ultra![e-mail]
101 Daggett Dr.                 uucp:     ...ames!ultra!norm
San Jose, CA 95134