how much does a bullet drop?
donald girod
My box of .22 long rifle cartridges says
muzzle velocity 1250 fps
100 yds--1000 fps
drop at 100 yds: 3.6 inches.
Ok, I don't know what is going on here. Ordinary physics says that the
bullet should drop 3.6" = .3 ft in .137 sec, and it can't possibly travel
100 yds in that time--you'd get an average velocity of 2190 fps. So what's
up? Either "drop" means something other than the the distance of the bullet
below the line through the bore at 100 yds, or else the bullet "flies" in
some sense, the way an arrow does. I don't think bullets fly in this way,
but I guess I don't really know much about ballistics in the presence of air
resistance.
And how can you apply this information to estimate how far the bullet would
drop in 200 yds? That is, if the gun is sighted in at 100 yds, how far
above the target would you have to hold at 200 yds? I realize that the
right way to attack this is to shoot the gun at a 200 yd target, see what
happens, and use this empirical data instead of speculating. But still, you
should be able to figure it out roughly anyway.
When I did this empirically for 100 yds, I found that with the gun sighted
at 100 ft, it shot about 6-7" low at 100 yds. This suggests that the bullet
is really dropping about 8 or 9 inches in 300 yds, so I don't understand the
3.6 inch figure at all.
-------------------------------------------------------------------------
Please find out about rec.guns at http://doubletap.cs.umd.edu/rec.guns
WARRIOR HUNTER
is only 300 feet the bullet covers 1250 feet in a second even with the
resistance it would still be moving quite quickly IMHO a 22lr fireing
gun wouldnt be any where near accurate for firing at 200 yards with the
light relitavely slow bullet
fl...@alaska.net
# My box of .22 long rifle cartridges says...drop at 100 yds: 3.6 inches.
#
# Ok, I don't know what is going on here. Ordinary physics says that the
# bullet should drop 3.6" = .3 ft in .137 sec... I found that with the gun sighted
# at 100 ft, it shot about 6-7" low at 100 yds. This suggests that the bullet
# is really dropping about 8 or 9 inches in 300 yds, so I don't understand the
# 3.6 inch figure at all.
While I don't have the poster's box in hand, Remington's published
ballistics for their .22LR HiVel loads list a mid range trajectory of
3.5" for 100 yards. This is the trajectory for a bullet sighted at 100
yards. I 'suspect' that this may be the 3.6" value referenced above.
[ Note that this value changes with differing height of the sights
above the bore centerline. ]
Ordinary physics is complicated here, since the bullet is constantly
slowing down; simple calculations like the above don't work. There
are plenty of ballistic programs available which do the tedious math
for us shooters.
Jay T
Seafin 41
You are right and the box is wrong. It should read "Mid-range trajectory at
100 yards: 3.6 inches.
If you sight the gun 3.6 inches high at 50 yards it will be on target at 100
yards. Actually this is not quite correct, because theoretical mid-range
trajectory is calculated from the center of the bore, not from the line of
sight. So you must subtract 1/2 the distance from the line of sight to the
center of the bore to get the actual mid-range trajectory.
If your gun is scope-sighted and the center of the scope is 1.2 inches above
the center of the bore, then half of 1.2 inches is .6 inches. Subtract that
from the theoretical mid-range trajectory of 3.6 inches and you get the actual
mid-range trajectory of 3.0 inches. Now, assuming your gun gives the velocity
listed on the box, if you sight 3.0 inches high at 50 yards you will be on
target at 100.
But you must shoot to be sure. There are many variables which can cause actual
results to differ from the mathematical model.
Regards,
Phil
Regards,
Phil
Gandalf
> ...
The only real way to get accurate data about bullet trajectory is
empirically. Even the best computer model are often off by quite a bit.
The computer models are great to give you an idea what to expect, and
can save significant amounts of ammunition in sighting a rifle in, but
if you want to hit what you're aiming at, you have to sight your gun,
with the load you are going to be shooting, under conditions as close to
the ones you will be under when you are hunting. Cold weather means less
energy from the powder, and the air is denser, but with less molecular
energy and motion ( the physics of being cold...), all of which can make
a big difference shooting at 75F compared to 20 or 30F. Reloading
manuals, at least the better ones, have extensive tables of data on
bullet trajectory and drop, derived from actual bench shooting, but the
variables found between firearms is enough to make them only about as
accurate as the computer software that generates trajectories. Again,
these tables will almost certainly put you on the paper at 200 or 300
yards, but if your rifle has a 4" shorter barrel than the test gun, with
a slightly different rate of twist, this can affect the impact point to
a greater or lesser degree. This is why bench rest competitors weigh
every powder charge, every slug, check for concentricity of each slug,
make the case flash holes uniform, turn the case necks, weigh the empty
cases, etc. etc. etc. Lots of variable that can't be 'predicted' by any
computer model or table, only a fairly accurate approximation. Good
observation and question on your part, to be sure.
Dan Varner
trajectory figures, which is the distance above line of sight the bullet
must be at what ever mid range distance they chose to hit the target at
the given range.
There is probably a flaw in your average velocity required for the
bullet to get to the 100 yards. The bullet in question is traveling at
an average velociy of 1125 feet per second and thus will take .27
seconds which should drop about 14 (32*.27/2*.27*12) first part gets
average drop velocity for the time of the drop then multiply that
velocity for the time of drop to determine feet droped then times 12 to
convert feet to inches below the line of bore. Just for drill I checked
my Sierra Reloading manual's ballistic table the closest approximation
of velocities I could find for what we are discussing is 115 g .355
(9mm)
muzzle velocity 1200 fps velocity at 100 yards 1011 fps drop 13.76 inchs
so the 14 is in the ball park.
The same table shows if zeroed for 100 yards that bullet is 3.28 inchs
above the line of sight at 50 yards so I would tend to believe you have
a mid range trajectory fiqure rather than a true drop figure.
donald girod wrote:
> ...
Sir Onan De Schmeckie
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LMAO! And all of that for a .22. You must be one serious shooter.
Your forgetting mass, Bullet coefficiency, temperature, and humidity.
Oops, I almost forgot altitude, and pollen count for atmospheric
density.
Is the bullseye, on the same level, as your Muzzle?
--WebTV-Mail-27082-1310
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--WebTV-Mail-27082-1310--
Dan Bollinger
J. Freeman
100 yards = 300 feet. So if a bullet starting at 1250fps and traveling at
1000fps at 100 yards it should take it a little less than .3 seconds to make
it to the 100 yard mark.
Most .22 rimfire rifles are sighted in at 50 yards so that would account for
the drop of 3.6 inches at 100 yards.
Richard R. Manasek
drop". A 36 gr .22LR with 1280 fps muzzle velocity when sigthed in
(far zero) at 100yds will rise 3.5 in. over the straight line (sight
line) between the muzzle and the target. Note, that the bullet will
cross the sight line in two places - the near zero and the far zero
which generally is the sight in distance. I personally like to sight
my .22LR rifles in at 50 yds. Assuming a scope centerline to bore
distance of 1.5 in. the near zero will be at approx. 25 yds, the far
zero (sight in) 50yds. The MPT will be approx. +.25 in.at 35 yds.
The bullet will be -2.5 in at 80 yds and -5.7 at 100 yds
Regards...Dad.
On 15 Apr 2000 10:40:00 -0400, "donald girod"
<d.g...@worldnet.att.net> wrote:
> ...
vZc
http://www.cybermesa.com/~jbm/ballistics/traj_basic/traj_basic.html
WVanhou237
<d.g...@worldnet.att.net> writes:
#
#muzzle velocity 1250 fps
#100 yds--1000 fps
#drop at 100 yds: 3.6 inches.
#
#Ok, I don't know what is going on here. Ordinary physics says that the
#bullet should drop 3.6" = .3 ft in .137 sec, and it can't possibly travel
#100 yds in that time--you'd get an average velocity of 2190 fps. So what's
#up? Either "drop" means something other than the the distance of the bullet
#below the line through the bore at 100 yds, or else the bullet "flies" in
#some sense, the way an arrow does. I don't think bullets fly in this way,
#but I guess I don't really know much about ballistics in the presence of air
#resistance.
#
For the given trajectory. The rifle would have a scope with the center ~1.5"
above the bore. Zeroed to hit point of aim @ 50 yds. That would give you ~
4" below POA at 100 yds.
Bill Van Houten (USA Ret)
"No matter how hard you try, you can't throw a potato chip very far."
"Linus"
Jeff Bissette
resistance (like a bullet, in this case), should drop at approximately 9.8
meters per second per second (acceleration) - that equals 32
feet/second/second. That means that, for every second your bullet is
traveling horizontally, it is also falling 9.8 meters faster (so, after one
second, it's moving 9.8 meters per second toward the ground; after two, it's
moving 19.6 meters per second toward the ground, etc.). Just do the math...
In article <38f91221...@news.knoxnews.infi.net>, rr...@knoxnews.infi.net
(Richard R. Manasek) wrote:
> ...
fl...@alaska.net
# ...That means that, for every second your bullet is
# traveling horizontally, it is also falling 9.8 meters faster (so, after one
# second, it's moving 9.8 meters per second toward the ground; after two, it's
# moving 19.6 meters per second toward the ground, etc.). Just do the math...
Unfortunately, the "math" is complicated by the vector of the bullet's
velocity, among other things. With the barrel's bore parallel to the
ground the 'math' is relatively easy, but all firearms of which I'm
familiar use sights to elevate the bullet departure vector at a
positive angle to the ground; this so they can actually hit something.
Now the bullet does not fall at a right angle to the departure vector,
and without knowing that departure angle, the average bullet velocity
over the range of interest, or the lift factor of the bullet ( which
can slow its actual drop rate to less than 9.8 mps/s ), the bullet
drop cannot be accurately calculated by simple math. That's why there
are so many ballistic programs to do all that work for us - and even
they are not perfect.
Jay T
J. Melton
starting at 1250 fps and at 300 feet it has the velocity of 1000 fps = 1125 fps
average over the distance
to go 100 yds = 300 ft. = 300 / 1125 = .266 seconds
distance the bullet will if fired horizontally = 1/2( 32)(time)(time)
or said. 1/2 times downward acceleration times time squared
.5 X 32 X (.26)(.26) = 1.08 feet = 12.97 inches
if sited in for 50 yards, then that would leave the following 50 yards to travel
average velocity over the distance would be 1062 fps
150 / 1062 = .141 seconds
distance of drom from the sited in 50 yard point = 1/2 (32) (.14)(.14) = .31
feet = 3.76 inches
Rambling Wreck and a hell of
donald girod wrote:
# I know I could find the answers in a library, but it is 50 miles away....
#
# My box of .22 long rifle cartridges says
#
# muzzle velocity 1250 fps
# 100 yds--1000 fps
# drop at 100 yds: 3.6 inches.
David Steuber
--
David Steuber | Hi! My name is David Steuber, and I am
NRA Member | a hoploholic.
I've found my niche. If you're wondering why I'm not there, there was
this little hole in the bottom ...
-- John Croll
prestonj
The bullet will drop all the way from the muzzle to the ground!
How fast it falls depends on bullet weight and speed.
See, I did make the smart ass comment.
wedgew
> ...
I believe that how fast any bullet falls is dictated by the acceleration due
to gravity and independent of bullet weight and speed.
--
OK!
So whose smart ideA was it to put the
CAPSLOCK kEy rIGHT nExT TO tHE sHiFt kEy
(Seek not to confound me as I am dim witted and slow of study)
Just remember...if the world didn't suck, we'd all fall off.
~Wedge
The Polymath (Jerry Hollombe)
#
# prestonj wrote:
#
# > ...
#
# I believe that how fast any bullet falls is dictated by the acceleration due
# to gravity and independent of bullet weight and speed.
32 ft/sec^2 modulo air resistance.
--
The Polymath (aka: Jerry Hollombe)
http://www.babcom.com/polymath/
http://www.babcom.com/gla-mensa/
Query pgpkeys.mit.edu for PGP public key.
Richard Molay
totally wrong.
A bullet, and every other object on this planet, drops at the rate of 32
feet per second/per second. It does not matter how much it weighs or how
fast it comes out of the barrel. It still drops at exactly the same rate.
One second after a bullet leaves the muzzle, it has dropped 32 feet. Yes,
32 feet. I know it sounds crazy, but consider... Most targets are pretty
close, and if the bullet is moving at 1000 feet per second, it'll get to a
target at 50 feet before it has much time to drop.
There's a name for this bullet-drop business: Rectilinear motion. It is the
same kind of figuring that goes into aiming angles for big artillery guns.
Why else would they be aimed so far up in the air? And if you've ever seen
the ramp sight on a rifle intended for very long-range service, you'd become
a believer.
I am a total novice when it comes to most gun subjects. I have a couple and
I like to shoot them at a range. I read this list day after day, year after
year, and admire the knowledge of people who engage in back-and-forth. I
almost never have anything worth contributing. But when it comes to the
physics of a projectile, well, I did my homework and I passed the course.
Gravity asserts itself on fast moving objects as well as weights dropped
from the Leaning Tower of Pisa.
- Richard
=========
I guess I'll make the smart ass comment.
The bullet will drop all the way from the muzzle to the ground!
How fast it falls depends on bullet weight and speed.
See, I did make the smart ass comment.
================
alf sauve
> ...
Mass, area and shape all play a factor.
Plus we could reheat the debate on possible aerodynamic lift afforded
bullets due the to forward momentum, initial upward angle of attack and
gyroscopic stability. [nah---we've been there before]
Alf
gruhn
#from the Leaning Tower of Pisa.
And feathers and bricks and air planes.
Jeffrey C. Dege
#Ok, I don't know what is going on here. Ordinary physics says that the
#bullet should drop 3.6" = .3 ft in .137 sec, and it can't possibly travel
#100 yds in that time--you'd get an average velocity of 2190 fps. So what's
#up? Either "drop" means something other than the the distance of the bullet
#below the line through the bore at 100 yds, or else the bullet "flies" in
#some sense, the way an arrow does. I don't think bullets fly in this way,
#but I guess I don't really know much about ballistics in the presence of air
#resistance.
One thing to keep in mind is that if the sights are parallel to
the ground, the barrell isn't. The bullet will rise to cross the
line-of-sight, peak, and fall to cross the line-of-sight a second, time.
On an M16A1, the crossing points were something like 21m and 225m from
the muzzle, meaning that you could fire at a human-sized target without
worrying about ballistics out to approximately 250m.
--
The Windows API has done more to retard skill development
than anything since COBOL maintenance.
--Larry O'Brien
J. Freeman
bullet in the hand and the one in the cartridge in the gun weighing the
same. Hold the barrel of the gun level with the ground.
Drop the bullet in you hand and fire the gun at the same time. Both bullets
will hit the ground at the same time!
The only difference in the 2 bullets is the bullet fired from the gun will
be a far away from you and the one you dropped will be at your feet.
bill horne
#
# I'm not sure I'd characterize your answer as smart-ass, as much as it is
# totally wrong.
# A bullet, and every other object on this planet, drops at the rate of 32
# feet per second/per second. It does not matter how much it weighs or how
# fast it comes out of the barrel. It still drops at exactly the same rate.
# One second after a bullet leaves the muzzle, it has dropped 32 feet. Yes,
# 32 feet. I know it sounds crazy,
It not only sounds crazy, it's wrong. When a bullet leaves the muzzle,
it is not dropping at 32 ft per sec - it is at zero fps and beginning to
accelerate at 32 ft per sec per sec. In the first second, it has an
average drop speed of 16 ft per sec. So, in the first second after
leaving the muzzle, it drops about 16 feet - not 32. Sierra ballistics
tables confirm this.
but consider... Most targets are pretty
# close, and if the bullet is moving at 1000 feet per second, it'll get to a
# target at 50 feet before it has much time to drop.
#
# There's a name for this bullet-drop business: Rectilinear motion. It is the
# same kind of figuring that goes into aiming angles for big artillery guns.
# Why else would they be aimed so far up in the air? And if you've ever seen
# the ramp sight on a rifle intended for very long-range service, you'd become
# a believer.
#
# I am a total novice when it comes to most gun subjects. I have a couple and
# I like to shoot them at a range. I read this list day after day, year after
# year, and admire the knowledge of people who engage in back-and-forth. I
# almost never have anything worth contributing. But when it comes to the
# physics of a projectile, well, I did my homework and I passed the course.
I think you should redo your homework.
# Gravity asserts itself on fast moving objects as well as weights dropped
# from the Leaning Tower of Pisa.
#
# - Richard
--
bill
Theory don't mean squat if it don't work.
JMartin957
#feet per second/per second. It does not matter how much it weighs or how
#fast it comes out of the barrel. It still drops at exactly the same rate.
#One second after a bullet leaves the muzzle, it has dropped 32 feet. Yes,
#32 feet.
Not quite. Acceleration due to gravity is 32 feet per second per second. So,
at theend of one second it is travelling 32 feet per second, 64 fps after 2
seconds, 96 fps after 3 seconds, etc. But an object accelerating from 0 - 32
fps in one second averages only 16 fps for that second, and travels only 16
feet. 64 feet in 2 seconds, 144 feet in 3 seconds.
John Martin
alf sauve
You've got the math wrong.
If you ignore air resistance, then objects fall at the rate of 32ft/sec/sec.
So at the END of the first second, the object is moving at a RATE of
32ft/sec, however it only fell 16 feet during that first second because it's
initial velocity was 0.
And remember, AIR. AIR resistance throws this all off. We can't ignore
it. Mass, area and shape all play important factors in how fast an object
falls. Not only do objects not really fall at 32ft/sec/sec, but at some
point they reach terminal velocity where acceleration becomes 0.
Oh, and the 32ft/sec/sec rate, not only does that ignore air resistance,but
it's only good at sea level. The further away from the core of the earth
the lower this number would be since the pull of gravity is less (inverse of
the square of the distance).
Alf
Lee DeRaud
wrote:
#I'm not sure I'd characterize your answer as smart-ass, as much as it is
#totally wrong.
#A bullet, and every other object on this planet, drops at the rate of 32
#feet per second/per second. It does not matter how much it weighs or how
#fast it comes out of the barrel. It still drops at exactly the same rate.
#One second after a bullet leaves the muzzle, it has dropped 32 feet. Yes,
#32 feet. I know it sounds crazy, but consider... [snip]
Neither crazy nor smart-ass, but WRONG by a factor of two. Downward
*acceleration* due to gravity is a constant 32 ft/sec/sec, which makes
downward velocity as a function of time:
v = a * t
Integrate velocity to get displacement (drop) as a function of time:
d = 1/2 * a * t**2
with a = 32, t in seconds, v in ft/sec, d in feet. So after one
second, drop is 16 feet, not 32.
All usual caveats concerning air resistance etc apply.
YMMV, but not enough to really matter :-)
Lee
vZc
Brian Liedtke
: I'm not sure I'd characterize your answer as smart-ass, as much as it is
: totally wrong.
: A bullet, and every other object on this planet, drops at the rate of 32
: feet per second/per second. It does not matter how much it weighs or how
: fast it comes out of the barrel. It still drops at exactly the same rate.
: One second after a bullet leaves the muzzle, it has dropped 32 feet. Yes,
: 32 feet. I know it sounds crazy, but consider... Most targets are pretty
: close, and if the bullet is moving at 1000 feet per second, it'll get to a
: target at 50 feet before it has much time to drop.
Nope. Wrong. acceleration due to gravity is 32 ft/sec*sec. But this
is not the rate of drop (velocity) This is the rate in change of velocity.
After one second, an object dropped from rest, has not travelled
32 feet, it travels 16 feet (X = .5 * g * t * t).
: There's a name for this bullet-drop business: Rectilinear motion. It is the
: same kind of figuring that goes into aiming angles for big artillery guns.
: Why else would they be aimed so far up in the air? And if you've ever seen
: the ramp sight on a rifle intended for very long-range service, you'd become
: a believer.
: I am a total novice when it comes to most gun subjects. I have a couple and
: I like to shoot them at a range. I read this list day after day, year after
: year, and admire the knowledge of people who engage in back-and-forth. I
: almost never have anything worth contributing. But when it comes to the
: physics of a projectile, well, I did my homework and I passed the course.
No, I'm sorry. You need to go back to school and retake Physics 101.
Brian
: Gravity asserts itself on fast moving objects as well as weights dropped
: from the Leaning Tower of Pisa.
: - Richard