Terminal velocity of a bullet
Norman F. Johnson
# Assuming a bullet is falling, i.e. no longer ballistic, what
# terminal velocity would it achieve? For example, fired straight
# up, it would eventually reach a speed (vertical) of zero, and
# then begin falling and accelerating at 32 fps^2. What terminal
# velocity would we expect, assuming, say a bullet with an average
# bc?
This is from Hatcher's Notebook:
The Miami and Daytona tests showed many other interesting facts
about bullets fired straight up. For example, the flat based
150-grain Service bullet fired straight up at 2700 feet per second
took 49.2 seconds to return...
Calculations indicate that the 150-grain .30-caliber Service bullet
fired straight up at a muzzle velocity of 2700 feet per second will
rise 9000 feet, taking about 18 seconds to do it; and that it takes
31 seconds to return to earth, the last few thousand feet of the
fall being at a nearly constant speed of a bit over 300 feet per
second [=30 foot pounds where he previously stated the Army
had decided an average 60 foot pounds was required to produce
a disabling wound].
God Bless!
Norm
Brian Hagen
I'm borrowing my brother's computer again (yes, he knows).
In article <33c04o$r...@xring.cs.umd.edu>,
Richard A. De Castro <deca...@netcom.com> wrote:
#A coworker asked me a question I couldn't answer the other day.
#
#Assuming a bullet is falling, i.e. no longer ballistic, what terminal
#velocity would it achieve? For example, fired straight up, it would
#eventually reach a speed (vertical) of zero, and then begin falling and
#accelerating at 32 fps^2. What terminal velocity would we expect, assuming,
#say a bullet with an average bc?
#
#TIA
#
#
#
#--
#============================================================================
#deca...@netcom.com Warning: I am a trained professional. No, Really!
Assuming that the bullet has no forward velocity? There have been tests
done on this. I believe it was Thompson at the turn of the century who
mounted a .30cal MG straight up at some docks and fired 500 rounds into
the air in an attempt to measure this. They found 6. (so much for field
testing).
With computers, we can simply write the appropriate programs and
calculate it out. I've got a report on this very subject at home. The
velocities that I remember were around 250fps for pistol rounds. Rifles
were, I believe, comparible.
When fired straight up, the bullet will fall back down base first. If it
should start tumbling (for whatever reason), the terminal velocity is
about 170fps.
Email me privately and I can sent some data out, or if it's requested I
can get it posted in a few days.
Mike Hagen
mha...@trumpet.calpoly.edu
Eric at CU-Denver
:>fired straight up at a muzzle velocity of 2700 feet per second will
:>rise 9000 feet, taking about 18 seconds to do it; and that it takes
:>31 seconds to return to earth, the last few thousand feet of the
:>fall being at a nearly constant speed of a bit over 300 feet per
:>second [=30 foot pounds where he previously stated the Army
:>had decided an average 60 foot pounds was required to produce
:>a disabling wound].
Hmmm... According to yer basic College Physics course, the bullet
would have a constant 32 ft/s/s (de)acceleration applied to it, so
it would go up in the air, decelerate, stop, and accelerate toward
the ground again, taking the same amount of time for the trip down
as for the trip up, and (finally!) would hit the ground with the
same velocity as when it left (albeit, in the other direction :).
Of course, all this is disregarding aerodynamic drag. Would drag
alone account for the discrepancy noted in the quote above? What
other forces are at work here? (Besides the obvious "Clinton Effect"
which would make such experiments impossible and moot by way of
not having any gun to shoot the experimental bullet out of)
Thanks for the enlightenment,
Eric (parenthetical statement) Rinehart
--
A Eric Rinehart is aeri...@ouray.denver.colorado.edu
Eric in the Arts 194 Mac Lab
How many nets would a network work if a network could work nets?
michael zarlenga
: alone account for the discrepancy noted in the quote above?
Yes.
It's aerodynamic drag that limits a falling human to approximately 120mph.
It's also aerodynamic drag that makes parachutes work.
--
mike zarlenga
Robert Dellicker <rjd@pica.army.mil>
#:>Calculations indicate that the 150-grain .30-caliber Service bullet
#:>fired straight up at a muzzle velocity of 2700 feet per second will
#:>rise 9000 feet, taking about 18 seconds to do it; and that it takes
#:>31 seconds to return to earth, the last few thousand feet of the
#:>fall being at a nearly constant speed of a bit over 300 feet per
#:>second [=30 foot pounds where he previously stated the Army
#:>had decided an average 60 foot pounds was required to produce
#:>a disabling wound].
#
#Hmmm... According to yer basic College Physics course, the bullet
#would have a constant 32 ft/s/s (de)acceleration applied to it, so
#it would go up in the air, decelerate, stop, and accelerate toward
#the ground again, taking the same amount of time for the trip down
#as for the trip up, and (finally!) would hit the ground with the
#same velocity as when it left (albeit, in the other direction :).
#
# Of course, all this is disregarding aerodynamic drag. Would drag
#alone account for the discrepancy noted in the quote above? What
#other forces are at work here? (Besides the obvious "Clinton Effect"
#which would make such experiments impossible and moot by way of
#not having any gun to shoot the experimental bullet out of)
#
Yes, aerodynamic drag is what causes this discrepancy. This scenario is a
pain to model due to the changing air density with altitude and the
trans-sonic velocity region on the way up. Even when you get to second
year physics they make convienient, invalid assumtions like the coefficient
of drag and the air density are constant.
Since air density and temperature change with altitude in an unusual fashon
the speed of sound must be calculated for each bit of time. (most of the
strange effects may be above the altitude of interest.) The bullet
velocity devided by the speed of sound gives the Mach number which is the
drag controlling factor for most of the bullet flight. The CD also changes
with Mach number.
Bob Dellicker r...@pica.army.mil
My WWW Home Page, <http://ramcad2.pica.army.mil/~rjd/>
Cliff Alexander
#:>Calculations indicate that the 150-grain .30-caliber Service bullet
#:>fired straight up at a muzzle velocity of 2700 feet per second will
#:>rise 9000 feet, taking about 18 seconds to do it; and that it takes
#:>31 seconds to return to earth, the last few thousand feet of the
#:>fall being at a nearly constant speed of a bit over 300 feet per
#:>second [=30 foot pounds where he previously stated the Army
#:>had decided an average 60 foot pounds was required to produce
#:>a disabling wound].
#Hmmm... According to yer basic College Physics course, the bullet
#would have a constant 32 ft/s/s (de)acceleration applied to it, so
#it would go up in the air, decelerate, stop, and accelerate toward
#the ground again, taking the same amount of time for the trip down
#as for the trip up, and (finally!) would hit the ground with the
#same velocity as when it left (albeit, in the other direction :).
# Of course, all this is disregarding aerodynamic drag. Would drag
#alone account for the discrepancy noted in the quote above? What
#other forces are at work here? (Besides the obvious "Clinton Effect"
#which would make such experiments impossible and moot by way of
#not having any gun to shoot the experimental bullet out of)
The basic College Physics course does ignore drag (in fact, it predicts
a greater than 100 mile range for a bullet travelling at 3000 ft/sec!).
The simple college physics constant acceleration model does not apply
to real bullets in air! Whenever there is a terminal velocity, there
*has* to be a velocity dependent force. When this velocity dependent
force equals the weight, gravity will no longer accelerate the object
because drag = weight.
---Cliff
#A Eric Rinehart is aeri...@ouray.denver.colorado.edu
#Eric in the Arts 194 Mac Lab
#How many nets would a network work if a network could work nets?
Scot Heath
: Hmmm... According to yer basic College Physics course, the bullet
: would have a constant 32 ft/s/s (de)acceleration applied to it, so
: it would go up in the air, decelerate, stop, and accelerate toward
: the ground again, taking the same amount of time for the trip down
: as for the trip up, and (finally!) would hit the ground with the
: same velocity as when it left (albeit, in the other direction :).
Using this same reasoning, a bullet fired horizontally would travel with
the same horizontal component of speed until it hit the ground. Think of
it, no loss in energy as the bullet whizzes along, easy to calculate hold
points for any given range. That would be nice but that dang nab'd air
keeps getting in the way.
-Scot
--
DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER
Everything I write is my opinion only. Nobody else would want it.
DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER
Scot E. Heath "Nothing beats turning clay to dust."
Fort Collins, CO
sc...@hpfcla.fc.hp.com
Norman Yarvin
Deceleration, of course; and this is just after it leaves the muzzle.
Air resistance is not only non-negligible, it's by far the most
important factor.
(Those curious about the exact figures are urged to go buy Hatcher's
Notebook -- anyone that curious will be interested in half the book at
least.)
--
Norman Yarvin yar...@cs.yale.edu
"People from the financial side seem to be good at keeping score --
but we are yet to see one man who can play the game himself."
-- John DeLorean
John Fowler
In a previous article, sc...@hpfcla.fc.hp.com (Scot Heath) says:
#Eric at CU-Denver (aeri...@ouray.Denver.Colorado.EDU) wrote:
#
#: Hmmm... According to yer basic College Physics course, the bullet
#: would have a constant 32 ft/s/s (de)acceleration applied to it, so
#: it would go up in the air, decelerate, stop, and accelerate toward
#: the ground again, taking the same amount of time for the trip down
#: as for the trip up, and (finally!) would hit the ground with the
#: same velocity as when it left (albeit, in the other direction :).
#
#Using this same reasoning, a bullet fired horizontally would travel with
#the same horizontal component of speed until it hit the ground. Think of
#it, no loss in energy as the bullet whizzes along, easy to calculate hold
#points for any given range. That would be nice but that dang nab'd air
#keeps getting in the way.
#
#-Scot
#
#--
#DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER
#
#Everything I write is my opinion only. Nobody else would want it.
#
#DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER_DISCLAIMER
#
#Scot E. Heath "Nothing beats turning clay to dust."
#Fort Collins, CO
#sc...@hpfcla.fc.hp.com
#
#
Doesn't make sense to me - up, ok; down ????. The bullet is STARTING down
from 0fps - at what popint does fluid (air) friction defeat 32fps/ps?
--
carpe diem, quam minimum credula postero
michael zarlenga
: : would have a constant 32 ft/s/s (de)acceleration applied to it, so
: : it would go up in the air, decelerate, stop, and accelerate toward
: : the ground again, taking the same amount of time for the trip down
: : as for the trip up, and (finally!) would hit the ground with the
: : same velocity as when it left (albeit, in the other direction :).
: Using this same reasoning, a bullet fired horizontally would travel with
: the same horizontal component of speed until it hit the ground. Think of
: it, no loss in energy as the bullet whizzes along, easy to calculate hold
: points for any given range. That would be nice but that dang nab'd air
: keeps getting in the way.
Close, but not right.
Ignoring aerodynamic drag, a bullet would have the sme horizontal compo-
nent of acceleration, not speed.
It would be gaining velocity every moment while in flight.
--
mike zarlenga
Brian Hagen
where I replied. This is Mike Hagen, using my brother's account (with
his humble permission). Direct comments to mha...@trumpet.calpoly.edu
Looking over a report entitled _Falling Bullets: terminal velocities and
penetration studies_ by Lucien Haag and presented at the Wound Ballistics
Conference in Sacramento last April, we have the following.
Lucien used a compute to model the effects of airflow on a variety of
bullets, thenn determined their terminal velocity, max altitude, etc.
Note that a bullet fired straight up will come back down base first.
Here's some excerpts from their results:
Cartridge Bullet Muzzle vel Max alt time up terminal time down
type (fps) (ft) vel (fps)
.22LR 40 LRN 1255 3867 12.5 198-BF 23.5
142-TU 30
9MMP 124FMJ 1110 4415 13.3 219-BF 24.6
.44 MAG 240 JHP 1180 4519 13.6 249-BF 23.1
5.56mm 55FMJ-BT 3240 8024 17 244-BF 38
141-TU 60
7.62SOVIET 123FMJ-BT 2400 8556 19 246-BF 38
158-TU 57
.30-06 180JSP 2700 10103 20.6 323-BF 37.5
Of course, if there's a horizontal velocity associated with the bullet,
then these numbers will change. Contact me if you'd like a copy if the
report.
Mike 'Great confrence, but a lousy drive' Hagen
mha...@trumpet.calpoly.edu
Tony Cattermole
Notebook'. He found that the 'time-in-the-air' formed two distinct
groups, caused by some bullets retaining gyroscopic stability and hence
descending tail first, and others that tumbled. Terminal velocity would
by greater in the first group of course.
Can't check the details, my copy is out on loan to a friend.
Tony Cattermole (tc...@cix.compulink.co.uk)
James F Johnson
michael zarlenga <zarl...@world.std.com> wrote:
#Close, but not right.
#
#Ignoring aerodynamic drag, a bullet would have the sme horizontal compo-
#nent of acceleration, not speed.
#
#It would be gaining velocity every moment while in flight.
And what, pray tell, would be the force acting upon the bullet which would
cause it to accelerate in the horizontal direction ?
One must be careful with their "close, but not right"s .
Jim Johnson
------------------------------------------------------------------------------
HUD secretary Henry Cisneros on "protected" guns and "sporting purposes":
"I mean, a 30-06 deer rifle with a scope is the kind of weapon that's in
use [by gangs] and the kind of weapon that was confiscated the other night."
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
michael zarlenga
The force generated by igniting the powder in the cartridge.
The gun is horizontal, the trigger is pulled, the powder ignites, a
force is created.
... The bullet leaves the barrel, the force disappears. Ooops!
I plead temporary insanity regarding my earlier reply. Sorry ...
--
mike zarlenga
Doug Gwyn (ACISD/MCSB) <gwyn>
#Doesn't make sense to me - up, ok; down ????. The bullet is STARTING down
#from 0fps - at what popint does fluid (air) friction defeat 32fps/ps?
It depends on various parameters. There are two major forces acting on the
bullet, the (nearly constant) pull of gravity and the (speed-dependent) drag
due to air resistance. The latter increases roughly as the square of the
speed, for subsonic motion, so when the speed is high enough the drag force
is comparable to the force of gravity. That is a stable condition known as
"terminal velocity" and skydivers are quire familiar with it. They also
know that you can change the terminal speed considerably by changing the
drag, for example decreasing drag by making the moving object becoming more
"streamlined", which leads to a higher terminal speed.
A bullet's terminal speed depends on the shape of the bullet. The path of a
bullet fired upward at an angle is not parabolic, because of the effect of
drag. Since there is no countering force in the horizontal direction, unlike
gravity that acts in the vertical direction, the horizontal component of the
velocity "scrubs off" and the final downward part of the trajectory will be
more vertical than the launch phase. I will try to illustrate this using
ASCII characters:
------
/ slow \
-- |
/ \
- |
/ \
/ fast | moderate speed
/ |
BANG SPLAT
Doug Gwyn (ACISD/MCSB) <gwyn>
#Note that a bullet fired straight up will come back down base first.
That's not necessarily true -- if the bullet is aerodynamically stable (CP
substantially behind CG), it should flip over and come down nose first. It
is true that most bullets are not stable when their spin scrubs off and at
that point will start tumbling. The most likely result is that the falling
bullet is tumbling.
Doug Gwyn (ACISD/MCSB) <gwyn>
#: Using this same reasoning, a bullet fired horizontally would travel with
#: the same horizontal component of speed until it hit the ground. Think of
#: it, no loss in energy as the bullet whizzes along, easy to calculate hold
#: points for any given range. That would be nice but that dang nab'd air
#: keeps getting in the way.
#Close, but not right.
#Ignoring aerodynamic drag, a bullet would have the sme horizontal compo-
#nent of acceleration, not speed.
No, that's absurd. The cited poster was correct. If it weren't for drag
and very minor effects such as Coriolis, precession, and Earth's curvature,
the horizontal component of velocity would remain constant as there is ZERO
horizontal component of acceleration.
This stuff is high school science, guys..
Rob Boudrie
#
#nent of acceleration, not speed.
#
Only if a force continued to act on the bullet. In the absence of force,
the energy (1/2 mv^2)* would be constant. If velocity increased without
appliction of force, you hvae "free energy".
* - yeah, I know this equation neglects reletavistic effects (negligible
at bullet velocities)
#It would be gaining velocity every moment while in flight.
#--
#mike zarlenga
#
James F Johnson
michael zarlenga <zarl...@world.std.com> wrote:
#I plead temporary insanity regarding my earlier reply. Sorry ...
Happens to the best of us, excepting my Omniomni self, of course.
It would be kinda neat if your idea worked though. In a sufficiently
rarified atmosphere, you could simply point a bullet at the intended target
and drop it from the proper height, at which point it would accelerate
toward the target of its own volition. Whats really neat, is that the
farther away the target, the flatter your trajectory would become, and the
harder the impact would be ! Who was this Newton guy, and why did he get
to make the rules, anyway ?
Gary Coffman
#from 0fps - at what popint does fluid (air) friction defeat 32fps/ps?
At the point where the aerodynamic drag force exactly balances the
force of gravity of course. That's called terminal velocity. The
simplistic drag (force) formula (valid subsonic) is
FL=0.5*Q*Cw*A*v^2
Where Q is air density, Cw is the drag coefficient, A is bullet frontal
area, and v is velocity. Using English units, FL will be in pounds. So
all you need to do is set FL to the weight of the bullet in pounds,
substitute in for air density (unfortunately variable over the bullet's
flight), insert the appropriate Cw and A for the bullet (different depending
on whether the bullet falls base down, nose down, or tumbles), and solve
for v.
I'll leave as an exercise calculating the heights required from which
a dropped bullet will reach terminal velocity. Hint, it's not very high.
Gary
Gary
--
Gary Coffman KE4ZV | You make it, | gatech!wa4mei!ke4zv!gary
Destructive Testing Systems | we break it. | uunet!rsiatl!ke4zv!gary
534 Shannon Way | Guaranteed! | emory!kd4nc!ke4zv!gary
Lawrenceville, GA 30244 | | ga...@ke4zv.atl.ga.us
Dennis Hagarty
In article <940829192...@SMOKEY.ARL.MIL>, gw...@arl.mil (Doug Gwyn (ACISD/MCSB) <gwyn>) writes...
Indeed, and they were BOTH HALF right. There IS a continuing increase
in velocity, but NOT in the horizontal component of that velocity.
=======================================================================
Dennis Hagarty Digital Equipment Corp (Aust)
haga...@snoc02.enet.dec.com Sydney, Australia
Opinions are solely those of the author, not Digital
=======================================================================