If I fired an arrow into the sky.....


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moredes
September 21, 2004, 12:47 AM
Given this hypothetical data, how far could this bullet travel if it didn't hit anything til it hit the ground?

.45 ACP
850fps
185 grain bullet
assume a BC of .135
barometric pressure- 29.70
300' above sea level
temp 75 degrees, humidity 60%

I'm told the optimum angle is ~30 degrees; let's say I fired it 'eyes-high', say 5' 7" (doubt whether the launch height contributes much to the results, though, wouldn't you?)

I'm just looking for a ballpark number--are we talking about 1400 yards, 1800 yards, 2400yds, 2 miles? I don't care about eeking out the last +-50 - 100yd, just a rough guess.

Thank you.

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trapperjohn
September 21, 2004, 12:50 AM
what is the angle? how far above the ground is it fired from?
to be close we also need barametric pressure, temperature and humididty data.
btw, drag coefficient also changes with velocity.

crewchief
September 21, 2004, 01:01 AM
Talk about a good math question. Why is it professors never ask questions like that instead of the average car a vs. car b question.:D

chevrofreak
September 21, 2004, 01:15 AM
I read somewhere that a bullet fired at a 45 degree angle will be traveling faster when it reaches earth than it was when it was fired. Who knows if thats true or not.

jefnvk
September 21, 2004, 01:38 AM
Dude, thats some fun physics and calc right there!

Since I've not gotten to the physics yet, I'll wait until some other answers popup. But, I think I got a good guess ;)

moredes
September 21, 2004, 01:40 AM
My guess is the answer's probably closer to 12-1400yd than any other distance I listed.

ilcylic
September 21, 2004, 02:03 AM
I come up with 4.25 miles in a vacuum. Not the question asked, but unfortunately, I've forgotten enough physics that I can't figure out the calculations for drag anymore. I'm embarrassed. :(

-Ogre

cracked butt
September 21, 2004, 02:14 AM
Its an apples to oranges comparison, but when I was a kid, me and hte neighbor kid used to fire our compound bows into the air as we had about a mile of hay/corn fields behind our houses. The arrows probably weighed 400 gr, and probably at about 250 fps. they typically travelled 400-500 yards. Just a WASG, I would think the bullet would land around 1600 yards.



We had an old Navy manual dated 1946 or so that belonged to my great uncle in our house. IIRC it had a mention of a .45 fired at a 45 degree angle had a range around 1200 yards, but that was a long time ago that I read that, and I was probably only 10 year sold.

jefnvk
September 21, 2004, 02:50 AM
I came up with 3.75 mi in a vacumn :uhoh:

I'll post my work when I get a chance to look over it rested, and make sure I didn't make one of my signature mathematical screw-ups (ie, 2*3 = 5).

Maybe we just need a mathematician or physicysts (sp?) in here.

strambo
September 21, 2004, 03:35 AM
1586 yards.
http://www.eskimo.com/~jbm/ballistics/maxdist/maxdist.html

Firing Elevation: 32.0 degrees
Terminal Range: 1586.0 yards
Maximum Height: 1231.6 feet
Terminal Angle: 62.0 degrees
Terminal Velocity: 249.3 ft/sec
Terminal Energy: 25.5 ft-lbs
Time of Flight: 17.2 secs

Tharg
September 21, 2004, 05:11 AM
OMG - gotta love the web - the new calculator... <grin>

J/Tharg!

moredes
September 21, 2004, 10:22 AM
Neat!

Thank you, Strambo.

Onmilo
September 21, 2004, 10:26 AM
If I fire my arrow to the sky, it may come down and hit me eye!

jpIII
September 21, 2004, 10:31 AM
45 degrees will give you the most horizontal distance travelled for any given projectile at any given speed.


.45 ACP
850fps
185 grain bullet
assume a BC of .135
barometric pressure- 29.70
300' above sea level
temp 75 degrees, humidity 60%

the bullet calliber, barometric pressure, temp, etc don't really matter unless you want to take wind resistance into account.

Usually calculations such as this neglect wind resistance, and only factor in the initial velocity of the projectile (which is dependent on charge, weight, and barrel length) and the negative acceleration due to gravity.

at least this is what was taught to me in my college physics a few years ago.:D

vertical distance - Dv
horizontal distance - Dh
Vertical Velocity= Vv
Initial velocity - Vi
Acceleration due to gravity - Ag
Initial height = Hi
Total time in flight - T
time - t

first find Vv=0 (appex of flight)
Vv = 0 = Vi (sin(angle))-Ag*t
therefore at t = Vi(sin(angle))/Ag seconds into flight, the Vv is 0.

At this point I think you can assume that T is 2* the time to appex of flight (in this case T=2*(Vi(sin(angle))/Ag)

then to find the horizontal distance travelled Dh
Dh = Vi*cos(angle) *T

I know you said...
I'm just looking for a ballpark number

but I couldn't resist trying to recall this from memory. I hope someone else can confirm that I'm not too off base here.

This is probably more information than anyone wanted.:eek:

HankB
September 21, 2004, 10:45 AM
I read somewhere that a bullet fired at a 45 degree angle will be traveling faster when it reaches earth than it was when it was fired. Who knows if thats true or not. Not true. Think about it - when going up, the bullet's velocity is reduced by both gravity (the vertical component) and air resistance (both vertical and translational components.) When coming down, gravity accelerates it downward, but drag forces continue to oppose both translational and vertical motion. Terminal velocity will necessarily be less than initial velocity.45 degrees will give you the most horizontal distance travelled for any given projectile at any given speed. This is the case in a vacuum. Bullets generally will achieve maximum range in air if fired at an inclination somewhere between 30 and 35 degrees - this varies by bullet, velocity, ballistic coefficient, etc. Ballistic calculations in undergraduate college courses invariably neglect drag forces as they're not readily and tidily computed from first principles . . . physics professors don't LIKE inelegant things like numerical integration and "fudge factors" like ballistic coefficients.

If one is talking about long range artillery, things like the Earth's curvature and rotation also have a noticeable effect . . . IIRC this was first noticed in a serious way during the long range naval duels between the English navy and the German High Seas Fleet at the Battle of Jutland during WWI.

Werewolf
September 21, 2004, 11:46 AM
At this point I think you can assume that T is 2* the time to appex of flight (in this case T=2*(Vi(sin(angle))/Ag)True in a vacuum but not true on a planet with an atmosphere.

One must take into account the effect that drag has. At apex the projectile has lost some horizontal velocity and the relative effect of the gravity component over the velocity component is increased - which is why the impact angle is considerably greater than the launch angle. In addition due to drag the projectile will not regain any horizontal velocity on the way down.

The loss of horizontal velocity would - all other things being equal - imply a longer downward time. However, due to the loss of horizontal velocity and the increased effect of the gravity component intuitively one would be led to believe that post apex flight time would be shorter than pre-apex flight time. Since the impact angle is greater than the launch angle intuitively it would seem that the time of travel post apex would be less than the time of travel pre apex.

The shape of the trajectory curve due to air resistance is not the shape of a pure parabola. The shape would be skewed thus the standard equation for a parabolic curve cannot be used to calcultate distance or flight times.

Drawing the hypothetical trajectory curve with the skewed shape would also tend to indicate that the downward time would be less than the upward time though that's just an intuitive assumption. I'm not sure but I think that solving for time would require the integration of the trajectory curve and then comparing the area under the curve less than apex distance to the area under the curve greater than apex distance. The real problem is getting the formula for the trajectory curve (however I suspect that since artillery can be fired with such great precision that tables exist somewhere that make knowing the forumla for the curve for the bullet in question would not actually be necessary).

It's been way too long since I had to use mathematics capable of proving the above but it seems intuitive to me that total flight time would be less than time to apex times two.

Should be an easy exercise though for those with both physics and advanced mathematics training.

strambo
September 21, 2004, 12:30 PM
The shape of the trajectory curve due to air resistance is not the shape of a pure parabola.

Yes, notice the firing angle was 32 deg. and the terminal angle was 62 deg. not even close to a true parabola. Almost twice as steep on the return to Earth. Also the terminal velocity of only ~250 fps. This severe velocity decay due to gravity and drag is the reason for the trajectory not being symmetrical. At 250fps it certainly did not accelerate on the steep (62 deg) angle down either. This figure would ovbiously be higher for a low drag bullet...but still way lower than the start velocity.

Mikul
September 21, 2004, 12:50 PM
I do remember reading an article about idiots who shoot bullets into the air for celebration. The author's calculations showed that almost any bullet (rifle or pistol) will reach the ground at roughly 300fps. Air resistance was the key. The other interesting point was that the bullets go up with their point to the sky and come down with their point to the sky.

benEzra
September 21, 2004, 01:02 PM
A draggy bullet-type trajectory in an atmosphere is sometimes called an "impetus trajectory," IIRC, to differentiate it from a pure ballistic trajectory. Impetus trajectories give greatest range at low elevations. An extreme example is the trajectory of a golf ball when hit by a driver off the tee; the ball visibly "runs out of steam" in midair and falls much more steeply than the initial trajectory. A driver with a 45-degree face angle wouldn't hit the ball very far at all. (Golf isn't an exact analogy due to the role of aerodynamic Magnus-effect lift in the early part of the ball's flight, but it's the same idea.)

At the other end of the spectrum, really heavy artillery (e.g., railroad guns) produce maximum range at elevations of 60 to 70 degrees, IIRC, to get the projectile out of the dense lower atmosphere as quickly as possible, thereby greatly reducing the kinetic energy lost to the atmosphere.

trapperjohn
September 21, 2004, 01:27 PM
Usually calculations such as this neglect wind resistance, and only factor in the initial velocity of the projectile (which is dependent on charge, weight, and barrel length) and the negative acceleration due to gravity. at least this is what was taught to me in my college physics a few years ago

We tell you that when we teach college physics because the calculations to solve it with air resistance is a PITA. there is actually no good way to do it analyticaly (with an equation that you simply solve). You have to set up an iterative process and genearlly use a computer code.
The vacuum assumption is valid for heavy slow moving objects, but is way off base when dealing with small arms trajectory.

Best bet is to use the online calculator that someone posted the link to above.

JohnKSa
September 22, 2004, 12:18 AM
The author's calculations showed that almost any bullet (rifle or pistol) will reach the ground at roughly 300fps. True if you're shooting more or less straight up. In that case, all of the velocity is bled off by air resistance and gravity and the terminal velocity is a result of the balance between air resistance and the pull of gravity.

If you're shooting at a significant angle off of vertical, then you have to consider more than terminal velocity. The DOWNWARD velocity component (pointing straight into the ground) won't be more than 300fps (whatever terminal velocity works out to) but if the angle wasn't very steep and the bullet is very aerodynamic, it may retain a significant amount of FORWARD velocity (pointing on the horizontal in the direction of bullet travel) that could only be bled off by air resistance.

Jrob24
September 23, 2004, 01:11 PM
assuming fired verticaly I got rougly 1.5 miles

lbmii
September 23, 2004, 06:21 PM
At the other end of the spectrum, really heavy artillery (e.g., railroad guns) produce maximum range at elevations of 60 to 70 degrees, IIRC, to get the projectile out of the dense lower atmosphere as quickly as possible, thereby greatly reducing the kinetic energy lost to the atmosphere.

This is what I understood also. I would think that an angle above 45 degrees would also favor bullets as well. Due to the thinner air at the higher elevation.

strambo
September 23, 2004, 10:14 PM
This is what I understood also. I would think that an angle above 45 degrees would also favor bullets as well. Due to the thinner air at the higher elevation.

No, rifle bullets go about 11,000 feet or so straight up (obviously depends on caliber~10K for .223/~ 12K for .300WM) that isn't nearly high enough to realize any major air thinning effects, and that is straight up...so 45 deg. would = 5,500 feet maximum ordinate roughly.
http://www.eskimo.com/~jbm/ballistics/maxdist/maxdist.html this link provides straight up distance as well and you can check the rest of the sight for lots of other equations.

"Trajectories Basic" is the most usefull for making custom ballistic tables. I have found them to be very accurate for my rifle and handloads.

themic
September 25, 2004, 03:43 AM
in a vacuum (i.e. no drag), with a perfectly spehrical earth, a bullet fired at a 45 degree angle, hypotheically, would land going faster than it was shot.

why? curvature of the earth. means it will have to fall just a bit farther to get to the same elevation relative to the earth's surface.

cracked butt
September 25, 2004, 03:53 AM
why? curvature of the earth. means it will have to fall just a bit farther to get to the same elevation relative to the earth's surface.

I don't think so... the curvature of the earth works both ways.

If anything it would hit the ground slightly faster than when its fired because the gun was 5 1/2' off the ground to start with.:cool:

c_yeager
September 25, 2004, 03:57 AM
Talk about a good math question. Why is it professors never ask questions like that instead of the average car a vs. car b question.

I once had an instructor that had us figure out the velocity of a 22lr prejectile by firing it through a pair of paper plates spinning at a known rate of speed about 10 inches apart. All i remember is that you have to measure the distance (laterally) between the two holes and then do some math and you get the answer :D

I still cant figure out how he managed to bring a rifle into a lecture hall at the university and fire it into a bullet trap (i do believe that he used a CB cap though).

BluesBear
September 25, 2004, 05:48 AM
I shot an arrow,
Into the Sky.
It fell to Earth,
OWWWW... MY EYE!

horge
September 25, 2004, 09:21 PM
Bluesbear!
:) :) :) :) :) :)

Piney
September 26, 2004, 07:12 AM
Since mathematics hadn’t been invented when I was in school I’ll have to claim ignorance. Anything more than 3 rocks and we were lost.

benEzra
September 26, 2004, 03:38 PM
Actually, in a vacuum, it would hit the earth's surface with the same velocity it started with (if it was fired from a point even with the surface).

Yes, the bullet has to "fall" a bit farther to get to the surface due to the earth's curvature. But the gravity vector also moves "behind" the bullet a little, so that the force is ALWAYS normal to the surface.

Another way to look at it is that on a perfectly spherical earth with uniform mass distribution, any two points the same distance from the center define identical gravitational potential energy. So a projectile gaining energy from gravity by going up and then coming back down to a point of the same elevation relative to the earth's center would violate the first law of thermodynamics.

Werewolf
September 27, 2004, 08:37 PM
What about the spin of the body from whence the projectile is fired?

In a vacuum if fired in a direction exactly opposite the direction of spin wouldn't that mean the relative impact velocity would be greater than the firing velocity kind of like two cars moving at 30 MPH hitting head on is the same as one car moving at 60 MPH hitting a stationary wall. Wouldn't the opposite be true?

JohnKSa
September 28, 2004, 12:18 AM
Only if you shoot someone on a neighboring planet.

When you toss a ball straight into the air, is it whipped away from you at the velocity of the earth's rotation?

Werewolf
September 28, 2004, 01:38 PM
When you toss a ball straight into the air, is it whipped away from you at the velocity of the earth's rotation?NO...

But that's because the ball's motion already has the rotational velocity vector in it even while you are holding it.

In a vacuum that would also be true. So if in a vacuum a ball was tossed upward would it return to the same spot from which it was tossed? It seems likely.

Still my gut tells me that the spin of the body from which a projectile is launched must have some affect on the projectile's terminal velocity and point of impact.

OK all you physicists out there! Time to chime in...

STW
September 28, 2004, 06:29 PM
I learned this as a child to apply in these situations:

I shot an arrow into the air
It fell to earth I knew not where
Until the next day, with rage profound
The man it fell on came around
And now I do not greatly care
To shoot more arrows into the air. :what:

(I, not really learning this lesson, played many a game of archery golf where distance was a necessity. Some "holes" were a 1/2 mile long.) :D

JohnKSa
September 28, 2004, 11:32 PM
Werewolf,

If you watched a shot from a point off the earth (not spinning) you would see a north/south shot curve in relation to YOU. However, if you looked at the launch point and angle versus the impact point, you would find that the shot DIDN'T curve. Although the shot DID curve in relation to you, the earth moved under the shot so that it landed in a straight line and at the proper angle from the starting point.

Just as the ball has the rotational velocity already applied when it is at rest in your hand, the bullet has the rotational velocity already applied when it is in the launcher.

Jonathan
September 29, 2004, 01:14 PM
Still my gut tells me that the spin of the body from which a projectile is launched must have some affect on the projectile's terminal velocity and point of impact.

OK all you physicists out there! Time to chime in...


The system needs to be defined better.

For our purposes, we probably care most about what the impact looks like with relation to the target. Likewise, the shot will probably be defined with respect to the shooter.

In this situation, assuming a vacuum, the rotation matters greatly.

However, in a simpler situation, if we assume no rotation, then you can make accurate approximations by looking at the potential energy. Moving away from and towards the earth, the projectile is going to experience the same deceleration and acceleration: it will return to an equivalent starting point. Since the distance to the center of mass of the earth will be the same for both start and finish, and no loss of energy via drag etc in flight, the the initial speed will be equal to the final speed.

The exact path travelled is irrelevant.

benEzra
September 29, 2004, 02:09 PM
So if in a vacuum a ball was tossed upward would it return to the same spot from which it was tossed? It seems likely.
If you were moving linearly, yes. The earth's rotation can be approximated as linear movement over short distances, but it's really rotational, so you get into the Coriolis effect there. If you shoot due north, for example, the ground you're standing on is moving faster to the east than the ground where the bullet lands, so there's an apparent velocity gain. The bullet didn't gain any velocity, though, it just carried some of the extra equatorial velocity with it to the north (hence the apparent bending to the northeast).

Tinker
September 29, 2004, 02:41 PM
Mikul,

"I do remember reading an article about idiots who shoot bullets into the air for celebration."

Guess we have that thing going on too.

Once, when I got to work, I found a spent bullet in the parking lot at work. It was in amazinly good condition. Looked to be about .30 cal. Riflling grooves on the side and only the nose was slightly bent were it hit pavement.

JohnKSa
September 29, 2004, 11:00 PM
BenEzra,

I was going to try to ignore the Coriolis effect and hope no one noticed... ;)

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