Bullet energy
ED ROGERS
velocity squared divided by 450,436.What factors make-up the 450,436
figure?
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Clark Magnuson
grains.
energy is 1/2 m v squared
so energy = 1/2 [# grains] [# of feet per second squared] in units of
grain feet squared per second squared.
The desired units are foot pounds
7000 grains per pound in weight, but 7000 grains = 1 pound/32 ft/ sec
squared in mass.
So 1 grain = pound of mass/[[7000][32 ft/ sec sec]]
Putting it all together:
E = .5 [ # of grains / [[7000 grains per pound][32 ft/ sec sec]] [# of
feet / sec][# of feet / sec]
Reducing:
E = [# of grains of bullet] [Velocity of bullet in feet per sec]/44800
all in units of foot pounds of energy
Richard I Landon
On Tue, 18 Mar 2003 22:09:00 +0000 (UTC), SAY...@webtv.net (ED ROGERS) wrote:
#
#velocity squared divided by 450,436.What factors make-up the 450,436
#figure?
Using accepted methodology prove all four of the following: That the universe is infinite; that truth is beauty; that there is not a little person who turns off the light in the refrigerator when you close the door, and that you are the person taking this exam. Now disprove all of the above. Be specific; show all work.
Bill Smith
wrote:
#The formula for calculating bullet energy is bullet weight times
#velocity squared divided by 450,436.What factors make-up the 450,436
#figure?
#
The bullet's weight is in grains, and velocity is in feet per second.
The energy is in ft lbs. It all just a conversion factor.
Richard E. Karlsen
least part of the answer.
Bullets are generally weighed in grains (about 7000 to the pound?)
Bullet velocity is measured in "feet per second".
Energy is expressed in foot/pounds, (actually correctly stated as
"pounds/feet" in physics)
I believe that the figure 450,436 was calculated to be a conversion factor
to convert grains to pounds, and "feet per second" to "feet per minute" in
the formula.
I hope there are some math majors out there. Trying to remember this stuff
is making my head hurt!!!
"ED ROGERS" <SAY...@webtv.net> wrote in message
news:b585ds$555$1...@grapevine.wam.umd.edu...
> ...
Ralph Wardlaw
: velocity squared divided by 450,436.What factors make-up the 450,436
: figure?
:
Well, you'll get a lot of replies on this one.
The equation for energy is E=1/2 mv*2. In the fps system of uinits, v is in
feet per second and mass (m) in in the lowly slug. If the bullet weight is
in grains, you must divide by 7000 to get pounds and then divide by 32.1740
to get slugs. And you still have that 1/2 to take care of, so you end up
dividing by 2 x 7000 x 32.1740= 450,436. You end up with energy in foot
pounds.
RW
Thomas
# The formula for calculating bullet energy is bullet weight times
# velocity squared divided by 450,436.What factors make-up the 450,436
# figure?
Here is an interesting site that refers your number:
http://home.sprynet.com/~frfrog/qalist.htm
Although I was getting close, I did not get a definate answer. I was
as far
as Gravitational Constant with the number when I gave up. Gravity is
the key.
Thomas
Tony Williams
# velocity squared divided by 450,436.What factors make-up the 450,436
# figure?
This guide to basic ballistics explains it:
http://www.quarry.nildram.co.uk/ballistics.htm
Tony Williams
Military gun and ammunition website: http://www.quarry.nildram.co.uk
Discussion forum at: http://forums.delphiforums.com/autogun/messages/
Clark Magnuson
I USED to calculate bullet energy for self defense rounds.
Now I don't.
What stopped me was reading rec.guns a year ago:
http://groups.google.com/groups?selm=a0hnni%24oa6%241%40xring.cs.umd.edu&output=gplain
> Nichol Simpson's head was nearly cutoff, and she bled to death nearly
instantly
> with hardly any kinetic energy whatsoever.
>
I could immediately imagine a deadly knife stroke with 10 to 100 foot
pounds of energy.
I could see that it is not an energy problem. It is more of a cut nerves
and blood vessels problem.
Then there is the phenomena of rifle bullets that have so much velocity
that they make a shock wave of destruction in regions the bullet does
not touch. If the bullet hits a bone, the bone turns into Shrapnel with
an even bigger cone of destruction.
I still calculate energy for rifles.
Clark
Ralph Wardlaw
"Thomas" <cano...@yahoo.com> wrote in message
news:b5cd5c$pmv$1...@grapevine.wam.umd.edu...
: SAY...@webtv.net (ED ROGERS) wrote in message
news:<b585ds$555$1...@grapevine.wam.umd.edu>...
: # The formula for calculating bullet energy is bullet weight times
: # velocity squared divided by 450,436.What factors make-up the 450,436
: # figure?
:
: Here is an interesting site that refers your number:
: http://home.sprynet.com/~frfrog/qalist.htm
:
: Although I was getting close, I did not get a definate answer. I was
: as far
: as Gravitational Constant with the number when I gave up. Gravity is
: the key.
: Thomas
As indicated in my post above, a gravitational acceleration of 32.1740
ft/sec**2 is the accepted value and gives the exact number the original
poster inquired about. It is the average value at 45 degrees latitude.
RW
ED ROGERS
pointed me to the right sources.
Gene Paul
in message news:b5f7kf$ss9$1...@grapevine.wam.umd.edu...
# "Thomas" <cano...@yahoo.com> wrote
# : # The formula for calculating bullet energy is bullet weight times
# : # velocity squared divided by 450,436.What factors make-up
# : # the 450,436 figure?
#
# As indicated in my post above, a gravitational acceleration of
# 32.1740 ft/sec**2 is the accepted value and gives the exact
# number the original poster inquired about. It is the average value
# at 45 degrees latitude.
# RW
Actually, the "standatd acceleration of gravity" is an international
standard used for "metrology", the science of weights and
measures.
The reason this gets used with systems of units is that some
systems are "coherent", based on F=ma with kinetic energy
given by .5mv^2, while some systems are "gravitational",
based on F=ma/g, with k.e. = .5mv^2/g. Here "g" means the
dimensionless constant (g sub c, g_c) equivalent to the
standard acceleration of gravity (g sub n, g_n.)
The g_c value must be used in converting certain units to others,
e.g. the pound(mass) to/from the slug, or the pound(force)
to/from the poundal.
Of course, the pound force (lbf) is the force that will accelerate
a pound mass (lb) at the rate g_n. When we use "foot-pound"
as a unit of energy, it always means ft-lbf
So what is the "Standard Acceleration of Gravity" (which is
merely an agreed-upon value, and not an actual measured
physical constant)?
Modern usage now uses the equivalent of the internationally
accepted standard for SI (the modern metric system):
g_n = 9.80665 m/s^2 (exactly!, by definition).
The equivalent "English" system constant is
g_n = 32.1740485... f/s^2 (again, exactly...)
and is used because any other value would give inconsistent
results when converting English system units to SI units.
This gives the divisor in the formula noted as 450436.67...
I prefer to use a variant on the mass and velocity units when
dealing with bullet (smallarms) energy and momentum :-) :
With mass/weight in grains and velocity in kilofeet/sec:
momentum = mv = IPSC/IDPA power factor
ke = 2.22mv^2 (foot-pounds)
The 2.22 is accurate to better than 1 part in 30000.
(reciprocal of .45043667 )
A link to the NIST site showing the Standard Acceleration of Gravity:
http://physics.nist.gov/cgi-bin/cuu/Value?gn
Regards,
Gene