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Speed of shell ejection for class project

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Fmintz

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Nov 11, 1999, 3:00:00 AM11/11/99
to
I need to find the speed of shell ejection for different rifles and shotguns
for a class project. Can anyone tell me where to look to locate this
information.

Brady

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EUGENE NEIGOFF

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Nov 11, 1999, 3:00:00 AM11/11/99
to
There is no way to get this data other than experimentation. The variables
involved are so complex that it is a mess. Ejection velocity is a first degree
differential equation which include type of gas action, weight of bolt and
moving parts, weight of case, velocity of bolt during extraction, spring
constants of the recoil spring, and lubricational effects.

If you want a real scientific discussion of this and other subjects in the
design of firearms I would suggest that you get a copy of:
BRASSEY'S ESSENTIAL GUIDE TO MILITARY SMALL ARMS
Design Principles and Operating Methods
D. Allsop & L.Popelinsky
ISBN 1-85753-107-8
This book is expensive and is one of the most complete discussion of the physics
of firearms. It costs $85.00 per copy.

Eugene N. Neigoff P.E.
..

Fmintz wrote:

> ...

Dave Bostock

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Nov 11, 1999, 3:00:00 AM11/11/99
to

Fmintz (fmi...@sccoast.net) wrote:
: I need to find the speed of shell ejection for different rifles and shotguns

: for a class project. Can anyone tell me where to look to locate this
: information.
: Brady

No. The speed of the ejected round also varies with the ammo being used
for semi-automatic weapons, and most likely also with the temperature of
the round when it was fired, the temperature of the round will vary with
the outside temperature and also how long it's chambered in a barrel hot
from previous firings. Some semi-automatic weapons also have a manual
adjustment to allow for variances in the ammo, this adjustment will also
change the ejection speed. For manualy operated guns the speed of
ejection is determined by how fast the shooter moves the bolt.

Forensic technicians may be interested in the range and direction of how
far a shell may or may not travel for a given set of circumstances which
they probably would establish by testing under specific conditions.

Competitive shooters may be interested in getting the shells to land in
a consistent manner for purposes of collecting them, and maybe psyching
others out.

Not blowing up the gun - internal ballistics
Where to aim down range - external ballistics
What happens to incapacitate, damage, or kill - terminal ballistis

Are areas of importance for which information may be found on the web
via search engines. Such information can also be found in a library.

--
Dave Bostock dea...@dc.seflin.org Florida, USA
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Clark Magnuson

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Nov 11, 1999, 3:00:00 AM11/11/99
to
I can explain an example of how to calculate shell speed.

1)For a semi automatic case fired 5 feet form the floor, the time to
reach the floor is
t= square root (distance / acceeration )= root (5'/32'/sec/sec)= .4
seconds

2)In a Colt .45 the ejector contacts the case when the slide is back
1.3". The farthest the slide can travel is 1.8" where it hits a stop. If
the spring is perfectly sized for the gun and the round, then the slide
will just run out of energy at the stop. Assume Vslide = 0 at 1.8".

3)Energy Slide at 1.3 inches = (force) (distance)= (16lb
spring)(1.8-1.3=.5")=8 foot pounds of kinetic energy left in slide when
it hits the case

4)Energy is also = 1/2 mass velocity squared = .5 (mass of slide= weight
of slide/ grav accel=12 oz/32 ft/sec/sec)(V squared)

5) Combining equations 3) and 4): Vslide at 1.3" = square
root(E/(.5mass)) = root(8 ft lb/((.5)(.023 lb sec sec /ft)) = 26
feet/sec

6)Center of gravity is .25" from extractor claw and ejector hits the
case at .35" from the extractor claw.
Velocity of case = (.25"/.35")velocity of slide at 1.3" = (.25/.35)26
feet/sec = 18.7 feet per sec [here's your answer] = 12.75 miles per hour

7)Combining 1) and 6): Distance case travels=
(Velocity)(time)=(18.7ft/sec)(.4 sec)=7.46 feet horizontally from the
gun

Also see FAQ at Wolff springs recomends sizing spring to have shells
land at from 3 to 6 feet
http://www.gunsprings.com/
Clark


Fmintz wrote:
> ...

Tom Kunz

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Nov 12, 1999, 3:00:00 AM11/12/99
to
Brandy,
I've read all the previous posts to this question, and they get a
little complicated. I'm a mechanical engineer, and I don't work in the
firearms industry, but if there's 1 thing I learned from engineering
school, it was to get the simplest form of the answer as quickly as
possible. Your instructor is just trying to get you to think and do a
little research and talking to other people about the problem, not
really make a super-scientific endeavor on the matter. Make sure you
enumerate *all* your assumptions, and place them clearly in your text.
That's one of the kickers.
First, there is *no* average ejection distance. I have had .45's eject
and fall at my feet. I've also had .45's eject and fly 20'. Same for
other calibers, large and small, there is no "average" ejection
distance. But suppose that someone would just assume the distance is
lets say 5'. I have tuned my 9mm reloads to eject between 4' and 6'.
Assume that the casings eject perfectly horizontally and not in an arc
upwards at all. That's a minor simplification that can be generalized
and dealt with later on.
The casing as it flies through the air is a free body, acted on only by
gravity and the initial velocity it was given. You probably already
know how to compute the time it takes to get from a specified height
(shoulder-height of the shooter, usually) to the ground. Other than for
obtaining the time of flight, you can ignore all other aspects of the
vertical component of the casings motion, if we make a reasonable
assumption that the horizontal velocity of the casing is generally
constant. So we can just use the equation d=r*t, distance=rate*time.
Distance is the assumed 5', rate is X and time is what you got by
calculating how long it takes to fall. Simple math, and I'll bet it
falls within 10% or less of whatever you'll get through
experimentation. The engineer's motto is "KISS" - "Keep it Simple,
Stupid!". Use the simple equations, fully enumerate your assumptions,
and then at the end of the paper put in a section on how to improve the
accuracy of the tests. My guess is that you'll be within 10% with these
assumptions.
Granted, the .45's I mention which ejected out to 20' probably slowed
considerably during their flight, but for short distances it can
probably be ignored.
Tom


Fmintz wrote:
> ...

--
Tom Kunz Tool Developer Software Consulting Services
PGP Key http://www.users.fast.net/~tkunz/pgp.html
1452 1F99 E2BB 632E 6EAE 2DF0 EF11 4DFC
DB62 7EBC 3BA0 6C40 88C0 C509 DA85 91B4 D5E9 EFD3

Dan Varner

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Nov 12, 1999, 3:00:00 AM11/12/99
to
Not disputing your calculations, but do they factor in the vertical
component, decleration to 0 on the vertical and the acceleration during the
fall? Is that something that kind of magically cancels out?

Clark Magnuson wrote:

> ...

Doug

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Nov 12, 1999, 3:00:00 AM11/12/99
to
EUGENE NEIGOFF wrote:
#
# There is no way to get this data other than experimentation. The variables
# involved are so complex that it is a mess. Ejection velocity is a first degree
# differential equation which include type of gas action, weight of bolt and
# moving parts, weight of case, velocity of bolt during extraction, spring
# constants of the recoil spring, and lubricational effects.
#
# Eugene N. Neigoff P.E.
#
I don't think it is that complex, the shell is just being pulled by the
extractor (part of the bolt) and then it gets an added vector from one
side being stopped by the ejector. An approximate bolt speed could be
fudged from the rate of fire.
Or if you could get movies of cases being ejected then you could look at
the trajectory and do some figuring, or even simpler look at distance
traveled per frame, that could even be done from still pictures with
multiple cases shown in the air if you know the lenght of exposure.

Good Luck,
Doug T

Gas action, weight of bolt, weight of case don't seem nessecary to me.

Clark Magnuson

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Nov 12, 1999, 3:00:00 AM11/12/99
to
Yes I factored in the vertical drop time, but I did not vector sum the
falling speed and the ejection speed. The speed I show is right at the
point of ejection.
I did approximate with center of gravity when I should have used moment
of inertia of the casing. [It was easier to balance a casing and measure
the balance point]
Clark

MROBINHOOD

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Nov 12, 1999, 3:00:00 AM11/12/99
to
couldnt some nice person just use his chronagragh to read the spped of a shell
casing ejecting out of a gun for this poor lad.??

Clark Magnuson

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Nov 12, 1999, 3:00:00 AM11/12/99
to
You are right. It should say

5) Combining equations 3) and 4): Vslide at 1.3" = square

root(E/(.5mass)) = root(.66 ft lb/((.5)(.023 lb sec sec /ft)) = 7.6
feet/sec

6)Center of gravity is .25" from extractor claw and ejector hits the
case at .35" from the extractor claw.

Velocity of case = (.25"/.35")velocity of slide at 1.3" = (.25/.35)7.6
feet/sec = 5.4 feet per sec [here's your answer] = 3.6 miles per hour

7)Combining 1) and 6): Distance case travels=

(Velocity)(time)=(5.4ft/sec)(.4 sec)=2.2 feet horizontally from the
gun
Clark

Eric Benson wrote:
#
# > 3)Energy Slide at 1.3 inches = (force) (distance)= (16lb
# > spring)(1.8-1.3=.5")=8 foot pounds of kinetic energy left in slide when
# > it hits the case
#
# Shouldn't that be 8 in-lb?
#
# -Eric

Dan Varner

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Nov 22, 1999, 3:00:00 AM11/22/99
to
Interesting point, may be somewhat problematic to get the brass to fly thru both
sky screens.

Figuring out the formula to predict the speed of ejection combining bullet weight,
spring strength, barrel length, weight of slide, power, powder charge, mass of the
brass, muzzle velocity, temperature, humidity, posibly even the weight and stregth
of the person holding the pistol might be involved.

MROBINHOOD wrote:

> ...

Clark Magnuson

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Nov 24, 1999, 3:00:00 AM11/24/99
to
1) For a semi-automatic case fired 5 feet form the floor, the time to
reach the floor is
t= square root ((2)distance / acceleration )= root (5'/32'/sec/sec)= .54
seconds

2) In a Colt .45 the ejector contacts the case when the slide is back


1.3". The farthest the slide can travel is 1.8" where it hits a stop. If
the spring is perfectly sized for the gun and the round, then the slide
will just run out of energy at the stop. Assume Vslide = 0 at 1.8".

3) Energy Slide at 1.3 inches = (force) (distance)= (16lb
spring)(1.8-1.3=.5")=.66 foot pounds of kinetic energy left in slide
when it hits the case

4) Energy is also = 1/2 mass velocity squared = .5 (mass of slide=


weight
of slide/ grav accel=12 oz/32 ft/sec/sec)(V squared)

5) Combining equations 3) and 4): Vslide at 1.3" = square


root(E/(.5mass)) = root(.66 ft lb/((.5)(.023 lb sec sec /ft)) = 7.6
feet/sec

6)Center of gravity [this should be moment of inertia, but that would be
work] is .25" from extractor claw and ejector hits the


case at .35" from the extractor claw.

Velocity of case = (.25"/.35") velocity of slide at 1.3" = (.25/.35)7.6


feet/sec = 5.4 feet per sec [here's your answer] = 3.6 miles per hour

7) Combining 1) and 6): Distance case travels=
(Velocity)(time)=(5.4ft/sec)(.54 sec)=2.9 feet horizontally from the
gun

And Wolff FAQ wants your empties to land from 3 to 6 feet. They must
want the slide to barely hit the stop.
http://www.gunsprings.com/1ndex.html
Clark

David Steuber

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Nov 26, 1999, 3:00:00 AM11/26/99
to
Clark Magnuson <cmag...@home.com> writes:

-> 1) For a semi-automatic case fired 5 feet form the floor, the time to
-> reach the floor is
-> t= square root ((2)distance / acceleration )= root (5'/32'/sec/sec)= .54
-> seconds

Ejected brass has a vertical component as well as a horizontal
component. This sort of makes the rest of your calculations moot.

I've had brass land anywhere from at my feet to nearly 20 feet away.
The brass tends to keep moving after hiting the ground, even in
grass.

It seems the best thing to do would be to actually measure the
velocity. As someone pointed out, a chrony won't work. However, a
camera will. Get a sound activated shutter control and set up the
camera and the microphone so that the picture will be snaped at a
known time after the shot is fired. The distance the brass has moved
from the pistol can be measured from the image. That should give a
fairly good measure.

--
David Steuber | Hi! My name is David Steuber, and I am
SAJ7580C2 | a hoploholic.

"O, reason not the need!"
-- King Lear

When a fly lands on the ceiling, does it do a half roll or a half
loop?

Bill Keiser

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Nov 26, 1999, 3:00:00 AM11/26/99
to
This would work only if you know exactly where the shell was when the timer
started, but if you set the shutter speed to a long enough exposure time, you
will catch the shell moving through part of an arc. If a yardstick is suspended
in the air nearby, the distance can be measured and speed calculated.
example: 1/30 sec, 2 inches, 60 per second.

David Steuber wrote:
#

# It seems the best thing to do would be to actually measure the
# velocity. As someone pointed out, a chrony won't work. However, a
# camera will. Get a sound activated shutter control and set up the
# camera and the microphone so that the picture will be snaped at a
# known time after the shot is fired. The distance the brass has moved
# from the pistol can be measured from the image. That should give a
# fairly good measure.

--

*** Friends don't let friends buy Macs! ***
*** William Keiser Brooksville FL ***
*** http://keiserb.tripod.com/ (updated 10/1) ***
*** icq#43286512 ***

Dave Bostock

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Nov 27, 1999, 3:00:00 AM11/27/99
to

Good grief, if you realy realy realy have to pursue such measurement for
some particular set of prescribed circumstances.
Place a video camera some distance behind the shooter.
Erect a verticle reference marked every 3" or 6" or 1' or so.
Lay down a horiz reference, same units & plane as ejection port.
Put a $5. quartz clock with a sweep second hand within frame.
Video tape, and note how far or behind the landing spot is from the horiz
& verticle references.
Use the pythagorean therom to find horiz length of flight. (same as horiz
ref if it lands on the ref, longer otherwise)
Play back the tape using frame advance on large TV. You can tape strings
to the TV for a grid that lines up with the references within the video.
You can determine the time length of frame to frame by inspection of the
sweep second hand in the video during playback. You can interpolate and
demonstrate exact position/time information for every 1/30 second or so
of case flight. From this you can calc. speed of case movement.

--
Dave Bostock dea...@dc.seflin.org Florida, USA
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

-------------------------------------------------------------------------

Mark Linden

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Nov 27, 1999, 3:00:00 AM11/27/99
to
On 26 Nov 1999 10:29:06 -0500, David Steuber <tras...@david-steuber.com> wrote:

#Clark Magnuson <cmag...@home.com> writes:
#
#-> 1) For a semi-automatic case fired 5 feet form the floor, the time to
#-> reach the floor is
#-> t= square root ((2)distance / acceleration )= root (5'/32'/sec/sec)= .54
#-> seconds
#
#Ejected brass has a vertical component as well as a horizontal
#component. This sort of makes the rest of your calculations moot.
#
#I've had brass land anywhere from at my feet to nearly 20 feet away.
#The brass tends to keep moving after hiting the ground, even in
#grass.
#
#It seems the best thing to do would be to actually measure the
#velocity. As someone pointed out, a chrony won't work. However, a
#camera will. Get a sound activated shutter control and set up the
#camera and the microphone so that the picture will be snaped at a
#known time after the shot is fired. The distance the brass has moved
#from the pistol can be measured from the image. That should give a
#fairly good measure.
#
Or better yet, have a flash set up to fire several times at known
intervals while the shutter is open. The distance the casing moves
between successive flashes will allow one to calculate both the
speed and the accelleration.

Doug t

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Nov 27, 1999, 3:00:00 AM11/27/99
to
Mark Linden wrote:
#

# Or better yet, have a flash set up to fire several times at known
# intervals while the shutter is open. The distance the casing moves
# between successive flashes will allow one to calculate both the
# speed and the accelleration.
#
Well the positive acceleration is in the down direction, and we already
know what that is (32'/s^2). The negative acceleration is probably
ignorable, I doubt the case is going to experience that much air drag
during it's short trip. If you use a video tape with frame advance then
you have your time measurement, all you have to do is get an accurate
distance measure frame to frame.

Doug T
P.S. Is to original poster still looking for an answer?

rob...@gmail.com

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Feb 11, 2014, 6:55:35 PM2/11/14
to
On Friday, November 26, 1999 12:00:00 AM UTC-8, David Steuber wrote:
# Clark Magnuson <cmag...@home.com> writes:
#=20
# -> 1) For a semi-automatic case fired 5 feet form the floor, the time to
# -> reach the floor is=20
# -> t=3D square root ((2)distance / acceleration )=3D root (5'/32'/sec/sec
)=3D .54
# -> seconds
#=20
# Ejected brass has a vertical component as well as a horizontal
# component. This sort of makes the rest of your calculations moot.
#=20
# I've had brass land anywhere from at my feet to nearly 20 feet away.
# The brass tends to keep moving after hiting the ground, even in
# grass.
#=20
# It seems the best thing to do would be to actually measure the
# velocity. As someone pointed out, a chrony won't work. However, a
# camera will. Get a sound activated shutter control and set up the
# camera and the microphone so that the picture will be snaped at a
# known time after the shot is fired. The distance the brass has moved
# from the pistol can be measured from the image. That should give a
# fairly good measure.
#=20
# --=20
# David Steuber | Hi! My name is David Steuber, and I am
# SAJ7580C2 | a hoploholic.
#=20
# "O, reason not the need!"
# -- King Lear
#=20
# When a fly lands on the ceiling, does it do a half roll or a half
# loop?

Mr. Magnuson, Pls. use unit analysis to check your formula for energy. You
say 1/2M V2=Energy. NOT according to unit analysis. Your "Missing" something
in there. HINT, try dividing previous formula's answer by "G" (Gravity).
It's miraculous, units work out, and numbers work out. Always check
your math by doing unit analysis, can save some "oops" moments.

Rubaiyat of Omar Bradley

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Feb 12, 2014, 2:37:45 PM2/12/14
to
On Tuesday, February 11, 2014 4:55:35 PM UTC-7, rob...@gmail.com wrote:
# Your "Missing" something in there.

You're missing something too - the thread you are posting to is from 1999

clarkm...@gmail.com

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Feb 13, 2014, 4:32:10 PM2/13/14
to
On Tuesday, February 11, 2014 3:55:35 PM UTC-8, rob...@gmail.com wrote:
# It's miraculous, units work out, and numbers work out. Always check
# your math by doing unit analysis, can save some "oops" moments.

I will look at the original post for mistakes caused by units for mass of the
slide. I only used the mass of the slide so I assume you are asking about that:
https://groups.google.com/forum/?hl=3Den#!msg/rec.guns/EdUhT6XPqQA/0ObE4c_JqRYJ

"3) Energy Slide at 1.3 inches =3D (force) (distance)=3D (16lb
spring)(1.8-1.3=3D.5")=3D.66 foot pounds of kinetic energy left in slide
when it hits the case

4) Energy is also =3D 1/2 mass velocity squared =3D .5 (mass of slide=3D


weight
of slide/ grav accel=3D12 oz/32 ft/sec/sec)(V squared)
5) Combining equations 3) and 4): Vslide at 1.3" =3D square


root(E/(.5mass)) =3D root(.66 ft lb/((.5)(.023 lb sec sec /ft)) =3D 7.6
feet/sec"

I think I am getting the right answer for the assumptions, but I am skipping
a lot of arithmetic steps. To get mass from ounces.

In 2003 I added to that calculation and showed some problems with the original.
https://groups.google.com/forum/?hl=3Den#!msg/rec.guns/zfECQTdlFok/vVdOeWJpVcwJ

If we start at the momentum of the projectile and momentum of the gas.
My father used Hayes Elements of Ordinance 1938 to design guns. I called him
for the 1999 post to ask how to calculate a falling object. Anyway..

That book says that the recoil momentum is equal to the bullet velocity times
the bullet mass plus the center of mass of the gas velocity times 4700 fps.

If I start with that to calculate how fast the slide and barrel mass will move,
we get a velocity so high we would need a much bigger recoil spring than we really
need. The reason is that the slide velocity relative to the frame velocity is
lowered by the velocity that the frame recoils.

This gets into "limp wristing" which may really be "Low mass handing".

Mike Fontenot

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Feb 13, 2014, 7:49:02 PM2/13/14
to
On 02/13/2014 02:32 PM, clarkm...@gmail.com wrote:
#
# I will look at the original post for mistakes caused by units for mass of the
# slide.

I don't think this is what the objector meant, but I think there may
indeed be an error in your choice of mass units. In the English system,
the basic unit of mass is NOT the lbm (aka "pound mass"), but rather is
the "slug". So, in any of the basic laws of physics, like E =
(m*v*v)/2, the mass "m" must be expressed in slugs, when the velocity is
expressed in ft/s, and when the energy is to be expressed in ft-lb. One
slug is 32.20 lbm. (The same thing is true for the law F = ma ... use
slugs, not lbm ... otherwise, you will need a constant coefficient in
the basic equations of physics).

The "slug" unit hasn't been used recently (or maybe ever) in popular
usage. The popularly used lbm unit ("pound mass") is unfortunately used
instead, probably because non-scientists and non-engineers have always
confused mass and force. The "pound mass" (lbm) is the mass of an
object that receives a force of one "pound" (lb, or sometimes written
lbf, for "pound force") from the earth's gravitational field at the
surface of the earth. The metric units have never suffered from this
confusion: when Europeans refer to their body "heft", they properly use
kilograms (the unit of mass), rather than the gravitational force at the
surface of the earth on that mass (newtons of force).
--
Mike Fontenot

rfr...@gmail.com

unread,
Feb 14, 2014, 12:31:58 PM2/14/14
to
On Thursday, November 11, 1999 1:00:00 AM UTC-7, Fmintz wrote:
# I need to find the speed of shell ejection for different rifles and shotguns
# for a class project. Can anyone tell me where to look to locate this
# information.
#
# Brady

Interesting read, from 1999 ?? I do wonder what Brady is doing these days. Career??

[MODERATOR: We're starting to see more responses to old posts, which I
see as evidence people still find value in interaction with our group.
Not all of these are as relevant as other posts, and in particular are
responses to "for sale" ads or the like. I generally err on the side
of inclusion. ]

clarkm...@gmail.com

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Feb 14, 2014, 6:31:50 PM2/14/14
to
# [MODERATOR: .. I generally err on the side
# of inclusion. ]

when you ruled that staple guns for putting up targets were indeed a gun topic... I remember
that better than most supreme court rulings:)

[MODERATOR: Regrettably, I am in no danger of ever being appointed as
one of the Supremes... ]



David R. Birch

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Feb 15, 2014, 2:58:23 AM2/15/14
to

# [MODERATOR: Regrettably, I am in no danger of ever being appointed as
# one of the Supremes... ]

Dunno about that, what's your singing voice like?

David


Gunner Asch

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Feb 15, 2014, 7:50:50 AM2/15/14
to
On Fri, 14 Feb 2014 23:31:50 +0000 (UTC), "clarkm...@gmail.com"
<clarkm...@gmail.com> wrote:

## [MODERATOR: .. I generally err on the side
## of inclusion. ]
#
#when you ruled that staple guns for putting up targets were indeed a gun topic... I remember
#that better than most supreme court rulings:)
#
# [MODERATOR: Regrettably, I am in no danger of ever being appointed as
# one of the Supremes... ]
#
#
What..you have common sense and ethics? Shame on you!!
(Grin)


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