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Message no. 1
From: The Deb Decker <RJR96326@****.UTULSA.EDU>
Subject: Ballistics
Date: Thu, 30 Sep 1993 17:27:30 -0500
Ok, several items to respond too:

Justin, are you sure that AP rounds work by tossing hot molten metal? I
know that's how HEAT rounds work: They use a funnel to focus the molten
explosion against the armor-but AP rounds are usually made of tungsten or
depleted uranium, which are dense, and what carries them through is the
incredible amount of force afforded by their velocity and mass-no molten
stuf involved. I may be wrong but your explanation of tank AP rounds
contradicts everything I've ever read or viddied regarding them-which is
a lot.

I also don't see why dikote would destroy the groove of a barrel, provided
the process was fine enough.

J Roberson
"You, you. . .you YOU you!"
Unimaginative Gangster
Message no. 2
From: Justin Kim <jlkim@******.COM>
Subject: Re: Ballistics
Date: Thu, 30 Sep 1993 15:49:03 -0800
>Ok, several items to respond too:
>
>Justin, are you sure that AP rounds work by tossing hot molten metal? I
>know that's how HEAT rounds work: They use a funnel to focus the molten
>explosion against the armor-but AP rounds are usually made of tungsten or
>depleted uranium, which are dense, and what carries them through is the
>incredible amount of force afforded by their velocity and mass-no molten
>stuf involved. I may be wrong but your explanation of tank AP rounds
>contradicts everything I've ever read or viddied regarding them-which is
>a lot.

Pretty sure, at least for the DU rounds. The mass of the DU *is*
important, too. I'm fairly sure that one of the selling points of the DU
round is that it arrives in a semi-molten state. However, I'm *not* an
expert on AT rounds. I'm fairly sure (70-75%) of my comments, though.
Your points are very good, though-I'll have to check my data.

Your comments on the rifling of the barrel of a cannon are well
taken, but I can't imagine that the rifling would steal so much velocity
from the round that it would end up making much of a difference.

Work is about to become rather busy and is likely to remain so for
about 3 weeks (working on the weekend kind of busy), but I'll do some
reading and see if I can back up my assertions. There are also a fair
amount of ex-military men around here (though not many Army) that I might
be able to ask.

Justin

--------
Justin Kim jlkim@****.edu Defense Analyst
Science Applications International Corporation San Diego, California USA
I'm paid to have opinions. If SAIC wants them, they had better pay me for them!
"The perimeter sensors are picking up subspace ocilations. What the hell
does that mean?" -Major Kira Nerys, DS9
Message no. 3
From: Stainless Steel Rat <ratinox@***.NEU.EDU>
Subject: Re: Ballistics
Date: Thu, 30 Sep 1993 19:38:45 EDT
>>>>> "JK" == Justin Kim <jlkim@******.com> writes:

JK> Pretty sure, at least for the DU rounds. The mass of the
JK> DU *is* important, too. I'm fairly sure that one of the selling
JK> points of the DU round is that it arrives in a semi-molten state.

This is incorrect. When DU impacts the target, it is definitely in a
solid, unmolten state. What happens is that the energy of impact
causes both the DU and steel (assuming a tank) to start to melt. This
molten metal forms an alloy which is incredibly soft (like mixing lead
and tin to get solder). The remaining DU projectile slices through
this alloy and reacts with the remaining steel, until there's either
no steel or no DU left.

Can you say 'hot knife through butter'? I knew you could.

Rat <ratinox@***.neu.edu> Northeastern's Stainless Steel Rat
ask about rat-pgp.el v1.63 PGP Public Key Block available upon request
||| | | | | | | | | | | | | | | | | | | | | | | |||
I'd rather be a pig than a fascist. --Porco Roso (The Crimson Pig)
Message no. 4
From: "Stephen R. Wilcoxon" <wilcoxon@****.EDU>
Subject: Re: Ballistics
Date: Sat, 2 Oct 1993 01:40:06 -0400
[sorry for taking a few days to respond to this, but I'm still in the
process of moving (so that I don't have a 2 hour commute every day (each
way)...]

> Pretty sure, at least for the DU rounds. The mass of the DU *is*
> important, too. I'm fairly sure that one of the selling points of the DU
> round is that it arrives in a semi-molten state. However, I'm *not* an
> expert on AT rounds. I'm fairly sure (70-75%) of my comments, though.
> Your points are very good, though-I'll have to check my data.

I believe (though *I* could be wrong) that the whole point of the DU round
is its mass/density. It's a kinetic energy round, and from all the
descriptions I've read, I believe it works by simply hitting the target
tank REALLY hard (since DU is one of the densest materials they can use (I
believe - though I could easily be wrong on this part - that the other
"common" core for the kinetic energy round is Tungsten)).


Twilight

The Crystal Wind is the Storm, and the Storm is Data, and the Data is Life.
-- The Player's Litany
Message no. 5
From: rogue@*****.fr (Sebastien Andrivet)
Subject: Re : Ballistics
Date: Wed, 14 Aug 1996 15:03:44 GMT
Azh wrote :
> Tumble rounds are extremely ineffective. They don't have any stopping power
> and are inaccurate at long ranges.

By tumble round, do you refer to 5,56mm (.223 for you Americans) ?
I was under the impression it was a pretty effective ammo.

Paul Adam wrote :
[widely spaced entry and exit points from tumble bullets]
> It's a very persistent myth, mind you, I've heard tales of it going back
> at least as far as World War 2.

Then I'm guilty of spreading an urban legend in my former combat group. Oh
well, greens *need* to be scared... :-)

> The "shock of being hit" has some effect, and where you're hit matters,
> but there are just too many variables for most "vary damage by hit
> location" systems to work well unless you get into Rolemaster-like

1) Is combat experience likely to decrease the "shock of being hit" effect
? If so, should experienced soldiers be more resistant to bullet wounds
than green ones ? This effect is present in Shadowrun, with the Combat
Pool.

2) Since determining wound trauma takes too many factors into account, does
rolling Xdice to determine bullet damage (as in GURPS and other games) does
a good job ?

> For a long time "shock" was a goal, and high-velocity lightweight
> bullets proliferated. The infamous 1986 Miami shootout (made into a


Yeah, but the USArmy muast have traded their .45 for a 9mm for a reason.
Was it simply due to prevailing theories ?

> Anyone still interested is referred to the work of Dr Martin Fackler. I
> have some International Defence Review articles of his, and will
> summarise them if there's interest.

Am strongly interested.

BTW, is it me, or does Shadowrun strongly underestimated the effect of
rifle bullets ? (A 5,56mm is 8M, a 10mm is 9M... sounds wrong to me).

Sebastien Andrivet
rogue@*****.fr
France / Europe
"I'm not gonna try to hit him. I'm gonna try to hit myself. Since my skill
with a staff is so low, I have a good chance of achieving critical failure
and hitting the wrong target, and there's only me and him around. So I
attack him by trying to hit myself".
- Fred
Message no. 6
From: "Paul J. Adam" <shadowrn@********.demon.co.uk>
Subject: Re: Re : Ballistics
Date: Mon, 19 Aug 1996 17:50:01 +0100
In message <v0153050bae379d77ea2d@[194.250.185.56]>, Sebastien Andrivet
<rogue@*****.fr> writes
>Paul Adam wrote :
>> The "shock of being hit" has some effect, and where you're hit matters,
>> but there are just too many variables for most "vary damage by hit
>> location" systems to work well unless you get into Rolemaster-like
>
>1) Is combat experience likely to decrease the "shock of being hit" effect
>? If so, should experienced soldiers be more resistant to bullet wounds
>than green ones ? This effect is present in Shadowrun, with the Combat
>Pool.

Never having been shot myself, I wouldn't know :) I would say that
experience helps a great deal when under fire, even experience of
training: even near-misses can make people freeze up (one person locked
up solid during a live-fire exercise, terrified by the realisation that
bullets were passing near him).

>2) Since determining wound trauma takes too many factors into account, does
>rolling Xdice to determine bullet damage (as in GURPS and other games) does
>a good job ?

As good a system as any, as long as most firearms are able to kill
almost at once: rules systems where a PC can survive a max-damage hit
from a .44 Magnum always annoy me :)

>> For a long time "shock" was a goal, and high-velocity lightweight
>> bullets proliferated. The infamous 1986 Miami shootout (made into a
>
>Yeah, but the USArmy muast have traded their .45 for a 9mm for a reason.
>Was it simply due to prevailing theories ?

Nope. NATO ammo standardisation (everyone else uses 9mm) and the need to
replace the increasingly elderly stock of M1911A1s, many of which were
Second World War vintage. 9mm is a pretty effective round, it was just
the particular type of ammunition used by the FBI that proved to be less
than totally effective.

>> Anyone still interested is referred to the work of Dr Martin Fackler. I
>> have some International Defence Review articles of his, and will
>> summarise them if there's interest.
>
> Am strongly interested.
>
> BTW, is it me, or does Shadowrun strongly underestimated the effect of
>rifle bullets ? (A 5,56mm is 8M, a 10mm is 9M... sounds wrong to me).

Grossly so, but necessary in the interests of playability. Making rifles
realistic would make weapons like the G3 or L1A1 firing APDS far too
deadly for game balance: scaling down pistols would make them way too
weak.

I don't think the damage system, or its scaling applied to weapons, is
particularly realistic, but it is playable and it does make most weapon
types (rifle, SMG, shotgun, even pistol) competitive so a PC has to
choose carefully: so I didn't mess with it.

--
"There are four kinds of homicide: felonious, excusable, justifiable and
praiseworthy."
Ambrose Bierce, "The Devil's Dictionary"
Paul J. Adam paul@********.demon.co.uk
Message no. 7
From: rhoded01@******.STCLOUD.MSUS.EDU (Ahzmandius)
Subject: Re: Re : Ballistics
Date: Mon, 19 Aug 1996 18:18:57 -0600 (CST)
>Azh wrote :
>> Tumble rounds are extremely ineffective. They don't have any stopping power
>> and are inaccurate at long ranges.
>
> By tumble round, do you refer to 5,56mm (.223 for you Americans) ?
> I was under the impression it was a pretty effective ammo.

Myth for profit. .223 cal is a pathetic excuse for a round. 7.62x52 has both
killing power and accuracy. They changed for economic purposes.

>Yeah, but the USArmy muast have traded their .45 for a 9mm for a reason.
>Was it simply due to prevailing theories ?
.45 carries 7 rounds, 9mm carries 15 + 1 in the chamber....any questions?


> BTW, is it me, or does Shadowrun strongly underestimated the effect of
>rifle bullets ? (A 5,56mm is 8M, a 10mm is 9M... sounds wrong to me).
Drastically.


Ahz
Message no. 8
From: "Gurth" <gurth@******.nl>
Subject: Re : Ballistics
Date: Tue, 20 Aug 1996 13:39:00 +0100
Sebastien Andrivet said on 15:03/14 Aug 96...

> 1) Is combat experience likely to decrease the "shock of being hit" effect
> ? If so, should experienced soldiers be more resistant to bullet wounds
> than green ones ? This effect is present in Shadowrun, with the Combat
> Pool.

I don't know for sure, but I don't think combat experience accounts for
much here. OTOH, experience with getting hit might.

> Yeah, but the USArmy muast have traded their .45 for a 9mm for a reason.
> Was it simply due to prevailing theories ?

Maybe because 9 mm rounds are smaller, and so more can be held in a
certain size weapon? Or because everybody else was/is using 9 mm too?

> BTW, is it me, or does Shadowrun strongly underestimated the effect of
> rifle bullets ? (A 5,56mm is 8M, a 10mm is 9M... sounds wrong to me).

I think nearly everybody here has reached that conclusion. The best
solution is to go with the flow and not pay attention to real-world
data, IMHO.

--
Gurth@******.nl - http://www.xs4all.nl/~gurth/index.html
I get a little bit nervous
-> NERPS Project Leader & Unofficial Shadowrun Guru <-
-> The Plastic Warriors Page: http://www.xs4all.nl/~gurth/plastic.html <-

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Message no. 9
From: Richard M Conroy <Richard_M_Conroy@***.ir.intel.com>
Subject: Re : Ballistics
Date: Tue, 20 Aug 96 10:01:00 PDT
Sebastien wrote:
[big snip]
: BTW, is it me, or does Shadowrun strongly underestimated the effect of
:rifle bullets ? (A 5,56mm is 8M, a 10mm is 9M... sounds wrong to me).

I wouldn't call those Heavy Pistols 10mm's, more likely they're at least
.45's possibly .50's (big bullets). What's more daft is the fact that
they've ignored the difference between 5.56's and 7.62's, all rifles
seem to be uniformly 8M.

Richard.
O--------------------------------------------------------------------O
\Food for thought lies in the\Richard_M_Conroy@\Roadkill on the Info \
\depth of an inedible brick. \ccm.ir.intel.com \-rmation SuperHighway\
O-------------------------------------------------------------------O
Message no. 10
From: "Paul J. Adam" <shadowrn@********.demon.co.uk>
Subject: Re: Re : Ballistics
Date: Tue, 20 Aug 1996 15:49:20 +0100
In message <01I8GWTPL8MM0000IA@******.STCLOUD.MSUS.EDU>, Ahzmandius
<rhoded01@******.STCLOUD.MSUS.EDU> writes
>Myth for profit. .223 cal is a pathetic excuse for a round. 7.62x52 has both
>killing power and accuracy. They changed for economic purposes.

Actually... (wicked grin) the SS109 ball (US equivalent M855) has better
penetration and very similar wounding effects to 7.62mm x 51. The
earlier M193 was questionable at medium or long ranges, though, and its
penetration was poor.

--
"There are four kinds of homicide: felonious, excusable, justifiable and
praiseworthy."
Ambrose Bierce, "The Devil's Dictionary"
Paul J. Adam paul@********.demon.co.uk
Message no. 11
From: "Paul J. Adam" <shadowrn@********.demon.co.uk>
Subject: Re: Re : Ballistics
Date: Tue, 20 Aug 1996 17:58:19 +0100
In message <Tue@?>, Richard M Conroy <Richard_M_Conroy@***.ir.intel.com>
writes
>I wouldn't call those Heavy Pistols 10mm's, more likely they're at least
>.45's possibly .50's (big bullets).

I'd hestitate before saying that. Having shot a Desert Eagle, it's a
non-trivial handful: and I'm 6' 2". My wife fired one shot and went back
to .38 Specials :)

These weapons are pretty widely used by your average Strength 3 Joe. I'd
have to say that heavy pistols represent the 9mms and the .40cals of the
world, not the .454s and the .50AEs.

>What's more daft is the fact that
>they've ignored the difference between 5.56's and 7.62's, all rifles
>seem to be uniformly 8M.

When one of my PCs acquired a L1A1 SLR, we made it a 9S and gave it
Sporting Rifle ranges.
--
"There are four kinds of homicide: felonious, excusable, justifiable and
praiseworthy."
Ambrose Bierce, "The Devil's Dictionary"
Paul J. Adam paul@********.demon.co.uk
Message no. 12
From: "Sambo" <polan881@******.edu>
Subject: Re: Re : Ballistics
Date: Tue, 20 Aug 1996 20:45:11 +0000
> In message <01I8GWTPL8MM0000IA@******.STCLOUD.MSUS.EDU>, Ahzmandius
> <rhoded01@******.STCLOUD.MSUS.EDU> writes
> >Myth for profit. .223 cal is a pathetic excuse for a round. 7.62x52 has both
> >killing power and accuracy. They changed for economic purposes.
>
> Actually... (wicked grin) the SS109 ball (US equivalent M855) has better
> penetration and very similar wounding effects to 7.62mm x 51. The
> earlier M193 was questionable at medium or long ranges, though, and its
> penetration was poor.

I think we are confusing our calibers here. The 7.62x51mm is used by
medium machine guns and sniper rifles. It's in a totally different
class than the .223. However the 7.62x39mm, your common Russian
small arms round,
has been compared to the .223 for a long time. Both have their
advantages and disadvanteges. At close range (<150m) the .223 is beaten
hands down for damage. However, at long ranges the .223 outperforms
the 7.62x39mm by a wide margin.

Pop quiz: What is the max effective range of a M16A2 HBAR?

600m

There isn't a weapon chambered for 7.62x39mm that will come close to
this. BTW, I used to shoot at ground squirrels for target practice
with a .223 - at 200+ meters.
***Sambo***
Message no. 13
From: "Gurth" <gurth@******.nl>
Subject: Re : Ballistics
Date: Wed, 21 Aug 1996 12:26:16 +0100
Richard M Conroy said on 10:01/20 Aug 96...

> I wouldn't call those Heavy Pistols 10mm's, more likely they're at least
> .45's possibly .50's (big bullets). What's more daft is the fact that
> they've ignored the difference between 5.56's and 7.62's, all rifles
> seem to be uniformly 8M.

Technically, a full-power 7.62mm rifle firing on full-auto (FAL, early
M14, G3, etc.) is not an assault rifle, at least not according to the US
Army. Lower-powered weapons and/or smaller-caliber ones are:
AK-47/AKM/dozens of other variants (7.62x39mm), M16s (5.56mm), AK-74
(5.45mm) are all assault rifles. They have comparable damages if you work
it out using an RPG weapon system like 3G3 (but don't trust the real-world
bullet data in that -- it's not completely accurate).

I tend to look on sporting rifles as being 7.62mm in SR, with the 7S
damage that fits more or less nicely IMO.

--
Gurth@******.nl - http://www.xs4all.nl/~gurth/index.html
See the amazing tourist-eating dolphins!
-> NERPS Project Leader & Unofficial Shadowrun Guru <-
-> The Plastic Warriors Page: http://www.xs4all.nl/~gurth/plastic.html <-

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Message no. 14
From: rhoded01@******.STCLOUD.MSUS.EDU (Ahzmandius)
Subject: Re: Re : Ballistics
Date: Wed, 21 Aug 1996 10:31:26 -0600 (CST)
>I think we are confusing our calibers here. The 7.62x51mm is used by
>medium machine guns and sniper rifles. It's in a totally different
>class than the .223. However the 7.62x39mm, your common Russian
>small arms round,
>has been compared to the .223 for a long time. Both have their
>advantages and disadvanteges. At close range (<150m) the .223 is beaten
>hands down for damage. However, at long ranges the .223 outperforms
>the 7.62x39mm by a wide margin.

7.62x51mm is also used in the M-14 carbine (still in use in the USN)
Message no. 15
From: "Paul J. Adam" <shadowrn@********.demon.co.uk>
Subject: Re: Re : Ballistics
Date: Wed, 21 Aug 1996 21:02:27 +0100
In message <199608210340.UAA00406@****.csrv.uidaho.edu>, Sambo
<polan881@******.edu> writes
<I said...>
>> Actually... (wicked grin) the SS109 ball (US equivalent M855) has better
>> penetration and very similar wounding effects to 7.62mm x 51. The
>> earlier M193 was questionable at medium or long ranges, though, and its
>> penetration was poor.
>
>I think we are confusing our calibers here. The 7.62x51mm is used by
>medium machine guns and sniper rifles.

Not to mention my beloved old L1A1 SLR, before they took it away and
gave us SA80s instead.

>It's in a totally different
>class than the .223.

Actually, it isn't.

SS109/M855 ball penetrates about 20% better than 7.62mm ball ammo, and
achieves equivalent (for European manufacture) or better (than US
rounds) wounding effects.

Out to 300-400 metres, there's no noticeable difference in accuracy,
except the lighter weight of the SA80 made hitting Figure 11s at 300
metres easier at the end of a long day.

>However the 7.62x39mm, your common Russian
>small arms round,

Not any more, they've gone to 5.45mm for some time now.

>has been compared to the .223 for a long time. Both have their
>advantages and disadvanteges. At close range (<150m) the .223 is beaten
>hands down for damage. However, at long ranges the .223 outperforms
>the 7.62x39mm by a wide margin.

Actually, .223 outperforms it for damage (to people) across the board.

>Pop quiz: What is the max effective range of a M16A2 HBAR?

About four feet. (You didn't say it was loaded...) :)

--
"There are four kinds of homicide: felonious, excusable, justifiable and
praiseworthy."
Ambrose Bierce, "The Devil's Dictionary"
Paul J. Adam paul@********.demon.co.uk
Message no. 16
From: "Paul J. Adam" <shadowrn@********.demon.co.uk>
Subject: Re: Re : Ballistics
Date: Wed, 21 Aug 1996 22:28:06 +0100
In message <199608210340.UAA00406@****.csrv.uidaho.edu>, Sambo
<polan881@******.edu> writes
>I think we are confusing our calibers here. The 7.62x51mm is used by
>medium machine guns and sniper rifles. It's in a totally different
>class than the .223. However the 7.62x39mm, your common Russian
>small arms round,
>has been compared to the .223 for a long time. Both have their
>advantages and disadvanteges. At close range (<150m) the .223 is beaten
>hands down for damage. However, at long ranges the .223 outperforms
>the 7.62x39mm by a wide margin.

Apologies for the length: I snipped several of the less relevant
weapons, but they're available for anyone who still wants them.

Extract from "Wounding Patterns of military rifle bullets" by Martin L.
Fackler MD

Reproduced without permission from International Defence Review, January
1989.

Current Rifle Bullets
Soviet 7.62mm x 39 - The Soviet AK-47 Kalashnikov fires a full-metal-
jacketed, boat-tailed bullet that has a copper-plated steel jacket, a
large steel core, and some lead between the two. In tissue, this
projectile typically travels about 26cm point-forward before beginning
any significant yaw. This author observed, on many occasions, the damage
pattern shown in Figure 2 while treating battle casualties in Da Nang,
Vietnam (1968). The typical path through the abdomen caused minimal
tissue disruption: holes in organs were similar to those caused by a
non-hollowpoint handgun bullet. The average uncomplicated thigh wound
was about what one would expect from a low-powered handgun: a small,
punctate entrance and exit wound with minimal intervening muscle
disruption.

US M193 5.56mm x 45 - This bullet is fired from the US armed forces'
first-generation smaller-calibre rifle, the M16A1. The large permanent
cavity it produces, shown in the wound profile (Fig. 6), was observed by
surgeons who served in Vietnam, but the tissue disruption mechanism
responsible was not clear until the importance of bullet fragmentation
as a cause of tissue disruption was worked out and described. As shown
on the wound profile, this full-metal-jacketed bullet travels point-
forward in tissue for about 12cm, after which it yaws to 90 degrees,
flattens, and breaks at the cannelure (groove around the bullet
midsection into which the cartridge neck is crimped). The bullet point
flattens and remains in one piece, retaining about 60 per cent of the
original bullet weight. The rear portion breaks into many fragments that
penetrate up to 7cm radially from the bullet path. The temporary cavity
stretch, its effect increased by perforation and weakening of the tissue
by fragments, then causes a much larger _permanent cavity_ by detaching
tissue pieces. The degree of bullet fragmentation decreases with
increased shooting distance (as striking velocity decreases) as shown in
Figure 7. At a shooting distance over about 100m the bullet breaks at
the cannelure, forming two large fragments and, at over 200m, it no
longer breaks, although it continues to flatten somewhat, until 400m.
This consistent change in deformation/fragmentation pattern has an
important forensic application. It can be used to estimate shooting
distances if the bullet is recovered in the body and has only penetrated
soft tissue.

The effects of this bullet in the abdomen shot will show the temporary
cavity effects as described for the Yugoslav AK-47 and, in addition,
there will be an increased tissue disruption from the synergistic
effect of temporary cavitation acting on tissue that has been weakened
by bullet fragmentation. Instead of finding a hole consistent with the
size of the bullet in hollow organs such as the intestine, we typically
find a hole left by missing tissue of up to 7cm in diameter (see
permanent cavity in Fig 6). The thigh entrance wound will be small and
punctate. The first part of the tissue path will show minimal
disruption. The exit will vary from the small punctate hole described
for the Soviet AK-47 to the stellate exit described for the Yugoslav AK-
47, depending on how thick the thigh is where the bullet perforates
it. In a sufficiently thick thigh, the M193 bullet fragmentation is also
likely to cause a significant loss of tissue and possibly one or more
small exit wounds near the large stellate one.

NATO M855/SS109 5.56mm x 45 - The slightly heavier and longer American
M855 bullet shot from the M16A2 assault rifle is replacing the M193
bullet shot from the M16A1 as the standard bullet of the US armed
forces. Fn Herstal originally developed this bullet type (which has a
steel penetrator as the forward part of its core) designating its bullet
the SS109. The wound profile (Fig. 8) is very similar to that produced
by the M193 bullet. Although the SS109 and the M855 are _not_ the same
bullet, their differences are small and one almost needs a magnifying
glass and a side-by-side comparison to differentiate the two. There is
little difference in their performance in tissue.

The abdominal and the thigh wound produced by the M855 or the SS109
bullets would be essentially the same as those described above for the
M193 bullet.

The longer 5.56mm bullets (M855, SS109) need a higher rotational
velocity to maintain stabilisation in air. FN claimed that this faster
rotation also causes the SS109 to have a significantly longer path in
tissue before marked yaw occurs, thus producing wounds of less severity.
This is simply untrue (compare Fig. 6 with Fig. 8). Additional rotation
beyond that needed to keep the bullet straight in air appears to have
little or no effect on the projectile's behaviour in tissue. However,
there is a situation regarding rotational rates whereby these longer
5.56mm bullets can cause _increased_ wound severity. Shooting the SS109
or M855 bullet in the older M16A1 rifle (they are not intended for use
in this 1-in-12 twist barrel, but in the newer M16A2's 1-in-7 twist)
produces a bullet spin rate insufficient to stabilise the longer
bullets. Such a bullet will yaw up to seventy degrees in its path
through air. Striking at this high yaw angle (essentially travelling
sideways) these bullets break on contact and the marked fragmentation,
acting in synergy with the temporary cavity stretch, creates a large
(over 15cm) stellate wound with the loss of considerable tissue.

NATO 7.62mm x 51 (US version) - the wound profile of this full-metal-
jacketed military bullet (Fig. 9) shows the characteristic behaviour in
tissue observed in all non-deforming pointed bullets. It yaws first
through 90 degrees and then, after reaching the base-first position,
continues the rest of its path with little or no yaw.

The uncomplicated thigh wound might show very minimal tissue disruption
since the streamlined bullet tends to travel point-forward during the
first 16cm of its tissue path. The abdominal wound, with a sufficiently
long path so that the bullet will yaw, causing the large temporary
cavity that is seen at depths of 20 to 35cm, would be expected to be
very disruptive. If the bullet path is such that this temporary cavity
occurs in the liver, this amount of tissue disruption is likely to make
survival improbable.


--
"There are four kinds of homicide: felonious, excusable, justifiable and
praiseworthy."
Ambrose Bierce, "The Devil's Dictionary"
Paul J. Adam paul@********.demon.co.uk
Message no. 17
From: Rand Ratinac docwagon101@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 17:56:14 -0800 (PST)
Okay, smartasses, got a question for you.

How much would you have to raise the muzzled of a gun
(a shotgun, to be precise) to change a chest shot inot
a head shot at a range of 10 metres?

What about at 15 metres?

====Doc'
(aka Mr. Freaky Big, Super-Dynamic Troll of Tomorrow, aka Doc'booner, aka Doc' Vader)

.sig Sauer

Can you SMELL what THE DOC' is COOKIN'!!!

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Message no. 18
From: Matthew Bond mgb@*****.swinternet.co.uk
Subject: Ballistics
Date: Tue, 7 Nov 2000 02:35:38 -0000
----- Original Message -----
From: "Rand Ratinac" <docwagon101@*****.com>
To: <shadowrn@*********.com>
Sent: Tuesday, November 07, 2000 1:56 AM
Subject: Ballistics


> Okay, smartasses, got a question for you.
>
> How much would you have to raise the muzzled of a gun
> (a shotgun, to be precise) to change a chest shot inot
> a head shot at a range of 10 metres?
>
> What about at 15 metres?

Ignoring spread (which at those ranges will only be a few inches at most
in the real world), you need to raise the centre of the impact point by
about 40cm to go from the centre of the chest to between the eyes...

For a given angle (a)in a right angled triangle, the tangent is the
length of the opposite side (O)divided by the length of the adjacent
(non hypotenuse) side (A).

Tan a = O/A

Tan a = 0.4/10 = 0.04

So a = ~2.3 degrees for 10m or ~1.5 degrees for 15m.

Assuming the shotgun is about 1m in total length, you need to raise the
muzzle

1m * Tan 2.3 = 0.04m or 4cm at 10m

or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m

HTH

Matt
Message no. 19
From: Wagemage wagemage@**.rr.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 23:27:59 -0500
>Tan a = O/A
>
>Tan a = 0.4/10 = 0.04
>
>So a = ~2.3 degrees for 10m or ~1.5 degrees for 15m.
>
>Assuming the shotgun is about 1m in total length, you need to raise the
>muzzle
>
>1m * Tan 2.3 = 0.04m or 4cm at 10m
>
>or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m

Sick, sick, sick.
Message no. 20
From: Paul Collins paulcollins@*******.com
Subject: Ballistics
Date: Tue, 7 Nov 2000 15:38:47 +1100
----- Original Message -----
From: "Matthew Bond" <mgb@*****.swinternet.co.uk>
To: <shadowrn@*********.com>
Sent: Tuesday, November 07, 2000 1:35 PM
Subject: Re: Ballistics


>
> ----- Original Message -----
> From: "Rand Ratinac" <docwagon101@*****.com>
> To: <shadowrn@*********.com>
> Sent: Tuesday, November 07, 2000 1:56 AM
> Subject: Ballistics
>
>
> > Okay, smartasses, got a question for you.
> >
> > How much would you have to raise the muzzled of a gun
> > (a shotgun, to be precise) to change a chest shot inot
> > a head shot at a range of 10 metres?
> >
> > What about at 15 metres?
>
> Ignoring spread (which at those ranges will only be a few inches at most
> in the real world), you need to raise the centre of the impact point by
> about 40cm to go from the centre of the chest to between the eyes...
>
> For a given angle (a)in a right angled triangle, the tangent is the
> length of the opposite side (O)divided by the length of the adjacent
> (non hypotenuse) side (A).
>
> Tan a = O/A
>
> Tan a = 0.4/10 = 0.04
>
> So a = ~2.3 degrees for 10m or ~1.5 degrees for 15m.
>
> Assuming the shotgun is about 1m in total length, you need to raise the
> muzzle
>
> 1m * Tan 2.3 = 0.04m or 4cm at 10m
>
> or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
>
> HTH
>
> Matt
>
Unless of course, we are talking about a troll :o)

Annachie
Message no. 21
From: Rand Ratinac docwagon101@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 22:50:25 -0800 (PST)
> Okay, smartasses, got a question for you.
>
> How much would you have to raise the muzzled of a
gun (a shotgun, to be precise) to change a chest shot
inot a head shot at a range of 10 metres?
>
> What about at 15 metres?

Crud, that spelling is SHOCKING. I need a break...or a
brain transplant...

*Doc' goes to see a typing therapist...*

====Doc'
(aka Mr. Freaky Big, Super-Dynamic Troll of Tomorrow, aka Doc'booner, aka Doc' Vader)

.sig Sauer

Can you SMELL what THE DOC' is COOKIN'!!!

__________________________________________________
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Message no. 22
From: Rand Ratinac docwagon101@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 23:01:03 -0800 (PST)
> > How much would you have to raise the muzzled of a
gun (a shotgun, to be precise) to change a chest shot
inot a head shot at a range of 10 metres?
> >
> > What about at 15 metres?
>
> Ignoring spread (which at those ranges will only be
a few inches at most in the real world),

Hey, yeah, forgot about that! Thanks, Matt! That
solves ANOTHER problem I had. :)

> you need to raise the centre of the impact point by
about 40cm to go from the centre of the chest to
between the eyes...
>
> For a given angle (a)in a right angled triangle, the
tangent is the length of the opposite side (O)divided
by the length of the adjacent (non hypotenuse) side
(A).
>
> Tan a = O/A
>
> Tan a = 0.4/10 = 0.04
>
> So a = ~2.3 degrees for 10m or ~1.5 degrees for 15m.
>
> Assuming the shotgun is about 1m in total length,
you need to raise the muzzle
>
> 1m * Tan 2.3 = 0.04m or 4cm at 10m
>
> or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
>
> HTH
>
> Matt

Thanks, Matt! Very useful!

====Doc'
(aka Mr. Freaky Big, Super-Dynamic Troll of Tomorrow, aka Doc'booner, aka Doc' Vader)

.sig Sauer

Can you SMELL what THE DOC' is COOKIN'!!!

__________________________________________________
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Message no. 23
From: Rand Ratinac docwagon101@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 22:35:30 -0800 (PST)
<snipt!(TM)>
> >Assuming the shotgun is about 1m in total length,
you need to raise the muzzle
> >
> >1m * Tan 2.3 = 0.04m or 4cm at 10m
> >
> >or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
>
> Sick, sick, sick.

Don't be like that, WM...

Matt's a GOOD boy.

At least, I'm glad SOMEONE remembers - or still uses -
their high school maths. :)

====Doc'
(aka Mr. Freaky Big, Super-Dynamic Troll of Tomorrow, aka Doc'booner, aka Doc' Vader)

.sig Sauer

Can you SMELL what THE DOC' is COOKIN'!!!

__________________________________________________
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Message no. 24
From: Rand Ratinac docwagon101@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 22:40:06 -0800 (PST)
<snipt!(TM)>
> > Assuming the shotgun is about 1m in total length,
you need to raise the muzzle
> >
> > 1m * Tan 2.3 = 0.04m or 4cm at 10m
> >
> > or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
> > Matt
> >
> Unless of course, we are talking about a troll :o)
> Annachie

*lol*

Good point. I would have mentioned something like
that, though. :)

====Doc'
(aka Mr. Freaky Big, Super-Dynamic Troll of Tomorrow, aka Doc'booner, aka Doc' Vader)

.sig Sauer

Can you SMELL what THE DOC' is COOKIN'!!!

__________________________________________________
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Message no. 25
From: Sebastian Wiers m0ng005e@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 22:43:47 -0600
:> Okay, smartasses, got a question for you.
:>
:> How much would you have to raise the muzzled of a gun
:> (a shotgun, to be precise) to change a chest shot inot
:> a head shot at a range of 10 metres?
:>
:> What about at 15 metres?
:
:Ignoring spread (which at those ranges will only be a few inches at most
:in the real world), you need to raise the centre of the impact point by
:about 40cm to go from the centre of the chest to between the eyes...
:
:For a given angle (a)in a right angled triangle, the tangent is the
:length of the opposite side (O)divided by the length of the adjacent
:(non hypotenuse) side (A).
:
:Tan a = O/A
:
:Tan a = 0.4/10 = 0.04
:
:So a = ~2.3 degrees for 10m or ~1.5 degrees for 15m.
:
:Assuming the shotgun is about 1m in total length, you need to raise the
:muzzle
:
:1m * Tan 2.3 = 0.04m or 4cm at 10m
:
:or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
:
:HTH
:
:Matt

HUH? I just assumed he meant the whole gun was kept on the same angle the
whole time. In that case, you just lift the whole thing vertically the
exact distance by which you want to raise the point of impact.

Its the same as with a hose- if you want to spray a point 12 inches higher,
raising the hose 12 inches will do exactly that, presuming you keep the hose
at the same angle relative to level.

Looks like it was a trick question to me...

-Seb
Message no. 26
From: Rand Ratinac docwagon101@*****.com
Subject: Ballistics
Date: Mon, 6 Nov 2000 23:38:51 -0800 (PST)
> :Ignoring spread (which at those ranges will only be
> a few inches at most
> :in the real world), you need to raise the centre of
> the impact point by
> :about 40cm to go from the centre of the chest to
> between the eyes...
> :
> :For a given angle (a)in a right angled triangle,
> the tangent is the
<snipt!(TM)>
> HUH? I just assumed he meant the whole gun was kept
on the same angle the whole time. In that case, you
just lift the whole thing vertically the exact
distance by which you want to raise the point of
impact.
>
> Its the same as with a hose- if you want to spray a
point 12 inches higher, raising the hose 12 inches
will do exactly that, presuming you keep the hose at
the same angle relative to level.
>
> Looks like it was a trick question to me...
> -Seb

*lol*

No, Seb - that was just me not explaining things
fully. As Matt surmised, I meant how much would you
have to TILT the muzzle upwards. So your hands stay in
the same place (more or less), but instead of shooting
him in the chest, you shoot him in the head.

====Doc'
(aka Mr. Freaky Big, Super-Dynamic Troll of Tomorrow, aka Doc'booner, aka Doc' Vader)

.sig Sauer

Can you SMELL what THE DOC' is COOKIN'!!!

__________________________________________________
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Message no. 27
From: Matt Bond MBOND@******.demon.co.uk
Subject: Ballistics
Date: Tue, 7 Nov 2000 10:09:56 -0000
> -----Original Message-----
> From: Sebastian Wiers [mailto:m0ng005e@*****.com]

<snip my *brilliant* analysis ;)>

> :Assuming the shotgun is about 1m in total length, you need
> to raise the
> :muzzle
> :
> :1m * Tan 2.3 = 0.04m or 4cm at 10m
> :
> :or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
> :
> :HTH
> :
> :Matt
>
> HUH? I just assumed he meant the whole gun was kept on the
> same angle the
> whole time. In that case, you just lift the whole thing
> vertically the
> exact distance by which you want to raise the point of impact.
>
> Its the same as with a hose- if you want to spray a point 12
> inches higher,
> raising the hose 12 inches will do exactly that, presuming
> you keep the hose
> at the same angle relative to level.
>
> Looks like it was a trick question to me...
>
> -Seb

Well, assuming the shotgun in question is being fired from the shoulder
position, to raise the whole gun 40cm will entail standing on a box...

Alternatively, tilting the muzzle up a few cm does exactly the same
job...

Matt
Message no. 28
From: Gurth gurth@******.nl
Subject: Ballistics
Date: Tue, 7 Nov 2000 12:27:17 +0100
According to Rand Ratinac, on Tue, 07 Nov 2000 the word on the street was...

> Okay, smartasses, got a question for you.
>
> How much would you have to raise the muzzled of a gun
> (a shotgun, to be precise) to change a chest shot inot
> a head shot at a range of 10 metres?
>
> What about at 15 metres?

This is a rather complicated question due to the fact that bullets actually
rise up into the air for a while after firing even if the weapon was held
level. But if you're satisfied with a simple answer, just assume that at
these ranges the bullet goes in a straight line so you can do some basic
math.

Distance between chest and head *grabs ruler* is, say, 30 cm (that's from
my nose to the middle of my chest; to my forehead is maybe 10 cm more).
Assuming the weapon is held level for a shot at the chest, you need to
angle it up by 1.7 degrees at 10 meters -- divide the chest-head distance by
the range, and take the inverse tangent. At 15 meters, it would be 1.1
degrees. If you take the distance between chest and head as 40 cm, BTW,
these values become 2.3 degrees and 1.5 degrees, respectively.

But like I said, this is the very simple way of calculating this by
assuming the bullet does not drop. If you want to include that factor as
well, things become a lot more complicated.

--
Gurth@******.nl - http://www.xs4all.nl/~gurth/index.html
"20% of all Americans are going to vote; the remaining 80% can't
fit into the polling booths." --Have I Got News For You
-> NAGEE Editor * ShadowRN GridSec * Triangle Virtuoso <-
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Incubated into the First Church of the Sqooshy Ball, 21-05-1998
Message no. 29
From: Matt Bond MBOND@******.demon.co.uk
Subject: Ballistics
Date: Tue, 7 Nov 2000 12:34:02 -0000
> -----Original Message-----
> From: Gurth [mailto:gurth@******.nl]

<snip>

> Assuming the weapon is held level for a shot at the chest, you need to
> angle it up by 1.7 degrees at 10 meters -- divide the
> chest-head distance by
> the range, and take the inverse tangent. At 15 meters, it would be 1.1
> degrees. If you take the distance between chest and head as
> 40 cm, BTW,
> these values become 2.3 degrees and 1.5 degrees, respectively.
>
> But like I said, this is the very simple way of calculating this by
> assuming the bullet does not drop. If you want to include
> that factor as
> well, things become a lot more complicated.

Not really, as we are assuming that the 'chest hit' situation has
already accounted for that. The bullet drop is the same in both the
chest and head hits (well, to within fractions of a millimetre at those
ranges...). All we are doing is adjusting one angle to covert a
successful chest hit to a successful head hit... all other factors are
the same, and can thus be ignored for this calculation.

Matt
Message no. 30
From: Wage Mage wagemage@**.rr.com
Subject: Ballistics
Date: Tue, 07 Nov 2000 09:57:58 -0500
> > >Assuming the shotgun is about 1m in total length,
>you need to raise the muzzle
> > >
> > >1m * Tan 2.3 = 0.04m or 4cm at 10m
> > >
> > >or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
> >
> > Sick, sick, sick.
>
>Don't be like that, WM...
>
>Matt's a GOOD boy.
>
>At least, I'm glad SOMEONE remembers - or still uses -
>their high school maths. :)

I know. I'm just jealous. I always sucked at math.
Message no. 31
From: Matt Bond MBOND@******.demon.co.uk
Subject: Ballistics
Date: Tue, 7 Nov 2000 15:36:14 -0000
> -----Original Message-----
> From: Wage Mage [mailto:wagemage@**.rr.com]
> Sent: 07 November 2000 14:58
> To: shadowrn@*********.com
> Subject: Re: Ballistics
>
>
>
> > > >Assuming the shotgun is about 1m in total length,
> >you need to raise the muzzle
> > > >
> > > >1m * Tan 2.3 = 0.04m or 4cm at 10m
> > > >
> > > >or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m
> > >
> > > Sick, sick, sick.
> >
> >Don't be like that, WM...
> >
> >Matt's a GOOD boy.
> >
> >At least, I'm glad SOMEONE remembers - or still uses -
> >their high school maths. :)
>
> I know. I'm just jealous. I always sucked at math.


It's quite easy... For trigonometry just remember "Tommy On A Ship Of
His Caught A Herring" (For an angle 'a' then Tan a = Opposite/Adjacent,
Sin a = Opposite/Hypotenuse, Cos a = Adjacent/Hypotenuse). ArcTan,
ArcSin and ArcCos are the inverses.

(Bad ascii art example)
|_
| -_ a=angle
O| -_H O=Opposite
|_ -_ H=Hypotenuse
|_|_ _(a-_ A­jacent
A

If you have a length and an angle you can work out the other two
lengths.

Say we know a=5 degrees and A is 10m. O = 10 * tan 5 = 0.87m; H = 10/cos
a = 10.04m.

If you have two lengths, you can work out the angle and the other
length.

if O=1m and H%m then sin a = 1/25 (0.04) so ArcSin 0.04 = a = 2.29
degrees; A% cos 2.29 = 24.98m

See, simple!

:)

Matt
Message no. 32
From: James Dening james.dening@****.co.uk
Subject: Ballistics
Date: Tue, 7 Nov 2000 16:06:04 -0000
<snip geometry>

>>Assuming the shotgun is about 1m in total length, you need to raise the
>>muzzle

>>1m * Tan 2.3 = 0.04m or 4cm at 10m

>>or 1m * Tan 1.5 = 0.026m or 2.6cm at 15m

Actually, there's a much easier way of doing this. Assuming the pivot point
of the shotgun (i.e. where the angle is measured from) is 1 metre back from
the end of the barrel, then you can simply think of triangle ratios.

Think of the gun as a triangle, one metre on its base side, and say,
4cm vertically high (i.e. you lift the muzzle 4cm up).

Now, expand the whole thing by a factor of 10. You now have a triangle
10 metres on its base side, with a height of 40cm. This is where the shotgun
round will end up.

So, with a 1 metre shotgun, and the target 10 metres away, the muzzle
movement is 1/10th of the vertical displacement of the aiming point change -

around 40cm in this case, from torso to head.

At 15 metres, you'd only need to move the muzzle 1/15th of 40cm ~ 26 cm.

There, wasn't that simple? And all without getting a tan. ;-)

J.
Message no. 33
From: Nexx nexx@********.net
Subject: Ballistics
Date: Tue, 7 Nov 2000 12:19:12 -0600
----- Original Message -----
From: "Matt Bond" <MBOND@******.demon.co.uk>

> It's quite easy... For trigonometry just remember "Tommy On A Ship Of
> His Caught A Herring" (For an angle 'a' then Tan a = Opposite/Adjacent,
> Sin a = Opposite/Hypotenuse, Cos a = Adjacent/Hypotenuse). ArcTan,
> ArcSin and ArcCos are the inverses.

No wonder the math majors I deal with can never write a coherent sentence.
;-)

Personally, I like the way we remembered the proper order for biological
Taxonomy... Kinky People Came Over For Group Sex.

Further Reading

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