Physics Question


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Blofeld
September 20, 2008, 01:11 AM
You take a 1911 and launch it into space. There is zero gravity. You have a remote primer ignition system. The gun is stable and floating, anchored to nothing. You have a full magazine. You remotely fire the gun.

Does the muzzle flip? Does the gun travel straight back? Since there is no resistance, do both the gun and the bullet travel equal distances? Does the returning slide push the gun forward?

Things Red Bull evokes.:scrutiny:

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VARifleman
September 20, 2008, 01:17 AM
The muzzle flips because the force is off of the center of gravity. This is true everywhere. Then when the slide chambers the new round the flip should be slightly reduced, but I imagine that it will keep going end over end until it hits something.

kingjoey
September 20, 2008, 01:18 AM
Although it has no 'weight' it still has mass. The gun would do everything it does on Earth except fall to the ground. Due to the mass differential between the slide area and the grip, there would be a slight amount of rotation imparted to it, but not much. What causes muzzle flip is the slide trying to move rearward while the grip's movement is resisted by the hand of the shooter. The return of the slide will have little effect on the movement of the gun, the net result will be:

m(bullet) x a(bullet)=m(pistol) x a(pistol)

Because the pistol weighs more than the bullet, its rearward acceleration will be a lot less than that of the bullet.

RyanM
September 20, 2008, 01:19 AM
Since there is no resistance, do both the gun and the bullet travel equal distances?

If infinity = infinity, then yes.

If not, then I don't know.

JesseL
September 20, 2008, 01:23 AM
It seems likely that the slide wouldn't cycle fully. Without a supporting hand, arm, and body, the frame and loaded magazine alone probably wouldn't provide enough inertia to resist the recoil spring.

It would be like the worst possible case of limp wristing.

jakk280rem
September 20, 2008, 01:31 AM
stovepipe.

RecoilRob
September 20, 2008, 01:57 AM
The pistol would spin backwards like a top, with a sideways motion imparted by the rifling torquing from the bullet mass, and whether it fully cycled would depend on the strength of the ammo vs. the timing of the gun and recoil spring strength.

kingjoey
September 20, 2008, 02:11 AM
with a sideways motion imparted by the rifling torquing from the bullet mass

Ah yes, I forgot to take that into consideration.

Old Grump
September 20, 2008, 03:18 AM
I think you are in the realm of calculus because of so many different factors involved. The muzzle flips because the center of mass will be below the barrel. It will spin because even though the pistol has more mass it will still spin the opposite of the bullet spin. With each shot the pistol will gain speed as each shot accelerates the gun a little more. You have the gun spinning on a vertical axis and a lateral axis and depending what the attitude of the gun is when the next shot is fired you won't know the direction of travel, we need to know how often the gun is fired, exact time intervals. Each shot will affect the gun differently because it will have less mass with the ejection of the bullet, powder and brass. We need to know where the nearest gravity wells are and distance traveled by the projectile and the pistol will be dependent on that. I expect you are going to need a pretty good computer system to handle the trajectory of the pistol with pitch and yaw. The bullets will be slightly easier if you know the exact direction the gun was pointed at when fired. I almost forgot the mass of the falling hammer will affect the motion of the pistol too. Interesting question.

Loosedhorse
September 20, 2008, 11:08 AM
Even though the gun is now spinning around its Z-axis (from muzzle flip) and its X-axis (from rifling), the three points of gun, point of ignition, and bullet will always form a straight line (Euclidian straight if there are no other gravitational bodies, or Einsteinian time-space curved if there are), as long as the brass stove-pipes (I think it would, although with a light or no recoil spring, it obviously wouldn't).

If the brass is ejected from the gun, the gun would now get a different trajectory (no longer on the same line as the bullet and point of ignition), as well as slightly change its spin. All objects would now move in straight lines, and the gun, brass, and point of ejection would always form a straight line. Therefore, the bullet, gun, brass, point of ejection and point of ignition would all remain in the same Euclidian plane (discounting gravity).

Since space is not a perfect vacuum (solar wind, etc.), it is likely the objects would eventually fall off their origninal lines and plane, and the brass would eventually stop, then the gun, then the bullet. However, there is too little gravitational attraction between the gun, brass and bullet to ever cause them to be attracted back to each other.

And they'd most likely eventually crash onto another body, or burn up approaching it.

See what happens when you ask a question?

Loosedhorse
September 20, 2008, 11:21 AM
As the bullet was launched during muzzle flip, its axis of rotation (from the rifling) will be angeled away from the trajectory, with bullet nose "down" (toward the 6 o'clock position on the muzzle, at the time of launch).

If there is not perfect bullet symmetry and perfect rifling, the bullet will also precess (wobble) as it spins.

Sean Dempsey
September 20, 2008, 11:22 AM
Put a gun on it's side on a very smooth table in a bullet proof chamber.

Same effect.

Blofeld
September 20, 2008, 11:26 AM
Does the placement of the ejection port come into play, would the trajectory of the components change if the port was on top of the gun?

And what if it was Extreme Shock ammo? (couldn't resist, sorry):D

Loosedhorse
September 20, 2008, 11:26 AM
The table will not allow x- or y- axis rotation from rifling and brass ejection.

Clarification: muzzle flip is not cause by slide action, but by the bore being above the center of gravity of the pistol. :)

benEzra
September 20, 2008, 11:27 AM
It would behave pretty much the same as if you threw it out a window here on earth and fired it remotely as it was falling. The round in the chamber would fire normally, the round would exit the barrel in the direction the pistol was pointing, the pistol would be propelled backward in the opposite direction (not terribly fast, maybe 10 mph or so?), it would be spinning, and the next round probably wouldn't chamber. It would still be falling, of course, but all else would be as it would be in space until it started falling fast enough for aerodynamic forces to come into play.

Jst1mr
September 20, 2008, 02:25 PM
No way would the next round chamber, the force required to operate the slide would be transferred into backward movement of the gun, the exaggerated version of "weak" gripping a semi-auto and causing FTF. In a perfect vaccuum, the gun would be propelled backwards with the same energy(E)as was expended by the bullet mass( M) times the square of the velocity. With the mass of the gun differing significantly from the bullet, it would not react with the same velocity as the bullet.

brickeyee
September 20, 2008, 04:34 PM
"the three points of gun, point of ignition, and bullet will always form a straight line (Euclidian straight if there are no other gravitational bodies, or Einsteinian time-space curved if there are)"

All the bodies are going to move about their respective centers of gravity (since it is also the center of mass).
The CGs will describe a straight line at short distances until other masses (and their gravity if they are large) come into play.

Friendly, Don't Fire!
September 20, 2008, 04:43 PM
This is fun!

Any more hypothetical questions?

Dr. Tad Hussein Winslow
September 20, 2008, 05:10 PM
Does the muzzle flip?

Yes. It will flip end over end backward, and continue to do so until colliding with some matter.

Does the gun travel straight back? Not sure. In addition to the spinning movement, it will also travel backward - whether it's perfectly horizontal - parallel to the bullet line of travel, I'm not sure - it may go mostly back but slightly down, or mostly back but slightly up. IOW, it will go back but probably at a small angle one way or another. Also, due to the rifling twist, in addition to the end over end movement and backward movement, it will also take on a right-to-left spinning movement.

Since there is no resistance, do both the gun and the bullet travel equal distances?

Yes, which is to say, an infinite distance, or until they collide with matter.

Does the returning slide push the gun forward?

It will, but probably does not have nearly enough magnitude to overcome the recoil forces which have first acted upon it - it may stop or greatly decrease the backward spin due to muzzle flip.

Those are my somewhat-educated guesses.

texfed
September 20, 2008, 05:20 PM
I don't know, But I'm going to wack the guy that own's the 1911 with my light saber, before he figures it out!

Claymore1500
September 20, 2008, 06:34 PM
OK, Next question, When the gun in space goes bang, and no one is there, Does it really go BANG?

JesseL
September 20, 2008, 06:45 PM
OK, Next question, When the gun in space goes bang, and no one is there, Does it really go BANG?

Zen riddles aside:
Although there would obviously be no atmosphere to convey sound, if you were in a spacesuit close enough to the muzzle when it discharged, you would probably hear some of the propellant gas impacting your suit.

Shung
September 20, 2008, 07:13 PM
Dont fire your guns into space ;) gave me headaches..

GRB
September 20, 2008, 07:19 PM
The muzzle flips because the force is off of the center of gravity. The center of exactly what gravity in a weightless environment?

physics
September 20, 2008, 07:27 PM
I'm an idiot, with a degree in physics. ;) What I originally said was incorrect upon further review.

Loomis
September 20, 2008, 07:37 PM
Sound waves do not propagate in a vacuum.

Therefore, the gun does not go "bang" regardless wether anyone is present or not. So, to make it perfectly clear...there are no sound waves in space, therefore there is no sound in space.

As for the first question...When the gun fires and the slide begins it's rearward motion, the spring will compress, and then the compression of the spring will impart a force on the frame and cause the frame to move backwards to "catch up" to the slide

The "opposite and equal reaction" (as described by newton) from the bullet is going to be seen on the breech face, and thus the slide...NOT THE FRAME. Therefore, the CG of the reacting mass(slide) is not lower than the bore centerline, it is HIGHER than the bore centerline. Initially, in the first fraction of a microsecond, the muzzle flip will be down, not up. Once the spring compression begins to cause movement of the frame, the muzzle flip will reverse back to what we are accustomed to.

Because the spent gasses also have mass and will continue to produce "thrust" on the gun long after the bullet leaves the gun and the gun starts moving, the direction of the gun and the direction of the bullet WILL NOT FORM A STRAIGHT LINE. The deviation from a straight line will be very small, however.

brickeyee
September 20, 2008, 07:43 PM
The center of exactly what gravity in a weightless environment?

While gravity may not be present mass still is.

The 'center of gravity' is the same as the center of mass, and that is the point motion will be about.

On earth we have gravity. so locationg the center is not hard.

Loomis
September 20, 2008, 07:45 PM
There's no such thing as "absence of gravity". All matter has it's own gravity.

But that is besides the point. A force applied off-center to a mass will cause the mass to rotate due to inertal forces resisting the applied force create a couple...THATs TORQUE. Earth's gravitational pull has NOTHING to do with it.

Ok, mr physics, why are you using that screen name since you just proved that you are physics impaired?

JesseL
September 20, 2008, 08:08 PM
Sound waves do not propagate in a vacuum.

Therefore, the gun does not go "bang" regardless wether anyone is present or not. So, to make it perfectly clear...there are no sound waves in space, therefore there is no sound in space.

But in the immediate vicinity of the gunshot in space is not a vacuum. There is a quantity of rapidly expanding gas, roughly equivalent in mass to the original powder charge.

Therefore a gun in space does go bang, if you're close enough to hear it.

Claymore1500
September 20, 2008, 08:14 PM
Zen riddles
aside:

Sound waves do not propagate in a vacuum.

Therefore, the gun does not go "bang" regardless wether anyone is present or not. So, to make it perfectly clear...there are no sound waves in space, therefore there is no sound in space.

Ok, I didn't REALLY expect anyone to try to seriously answer my "zen riddle", but since you did, I would like to point out that "sound waves" are nothing more than air movement, Having said that, I submit that, IF, no reciever (ears, microphone, or what ever) is present to translate the moving air to sound, There is no sound wether in space or the middle of Rhode island.

Prove me wrong, I dare you!

NOTEIt's all in fun guys

rcnixon
September 20, 2008, 08:26 PM
There is some confusion, as someone pointed out, about the center-of-gravity and the center-of-mass. Also, I believe that the gun won't cycle at all. The falling hammer will impart a small force moving the muzzle downwards, very slowly, rotating about the center of mass. As the powder burns, the expanding gasses force the bullet down the barrel. This (and the gasses escaping the muzzle behind the bullet) will cause the muzzle to rise, again rotating about the center of mass. Someone else mentioned the ultimate limp-wristing; I don't conceive of any resistance to the backward, rotating motion that would allow the barrel and slide to unlock. There will be a rotational force imparted from the rifling, again as some else pointed out, causing a rotational movement. Now, would enough initial forward velocity allow the action to cycle? The math of this gets into differential equations and is beyond my worn-out skills from long ago. Anyone out there got mathlab or a handy engineering or physics student they can bribe into looking at this?

Russ, who will post another physics quiz in a moment

Owen
September 20, 2008, 08:30 PM
the pistol will spin around its center of mass.

In common parlance, center of mass = center of gravity.

rcnixon
September 20, 2008, 08:31 PM
BTW, Claymore, there most certainly is sound, if in an atmosphere. The physical disturbance is there whether perceived or not. In a vacuum, no. Ponder this as well; time, as we know it, does not exist, t is an artificial constraint placed on the universe by humans so we can understand it better.

the foot
September 20, 2008, 08:32 PM
You would absolutely have a remotely-fired limp-wristed jam on the first attempt. That being said, the action/reaction/blah blah blah would result in the weapon doing cartwheels.

Loomis
September 20, 2008, 08:46 PM
Wrong again, claymore:

Sound is energy, not the perception of energy. Therefore, no reciever is necessary for the presence of sound.

Loomis
September 20, 2008, 08:49 PM
rcnixon:

The falling hammer will cause muzzle to rise, not fall. As the hammer spring pushes hammer forward, it also, at the same time, pushes gun towards hammer. When the hammer strikes the gun(or gun strikes the hammer, depending on how you look at it) the relative motions of the two cease.

Claymore1500
September 20, 2008, 08:50 PM
BTW, Claymore, there most certainly is sound, if in an atmosphere. The physical disturbance is there whether perceived or not. In a vacuum, no. Ponder this as well; time, as we know it, does not exist, t is an artificial constraint placed on the universe by humans so we can understand it better.

Just a couple of points, Agreed the physical disturbance is there, But it does need to be translated into sound by SOMETHING or it is simply moving air, Secondly, Space is NOT a vacuum, the only way to achieve a "vacuum" is to draw all of the air out of a sealed enclosure, Space is lack of/ or possably "negative air pressure".

I must admit, I agree with the time thing though.

This should be a sticky, it's fun and nobody is getting bashed!

J Star
September 20, 2008, 08:57 PM
You're all forgetting the one thing the gun needs to fire at all: Oxygen. It's never been proven that a gun will go off in a vacuum, despite the presence of a built in oxidizer.

I believe this has been discussed to some length on this board, as well as on various physics boards.

jonmerritt
September 20, 2008, 09:16 PM
O2 is not needed for a gun to fire in a non o2 enviroment, that IS why there is an oxidizer! science class, a bullet will fire in a vacume. space is NOT a vacume. And if you did this experiment, you would lose a nice hand gun

Loosedhorse
September 20, 2008, 09:18 PM
Initially, in the first fraction of a microsecond, the muzzle flip will be down, not up.

No. This might be right if the slide traveled without friction. In the first fraction of a microsecond, friction would prevent the slide from moving at all and the recoil spring would initially resist compression (just like the surface of water initially feels much harder to a cannon-baller than it "really is"), and the slide and frame will move as one. So muzzle flip will be "up," as soon as the bullet (or any gases that escape ahead of it) starts to move.

I would expect the gun to spin from the bullet being fired above the center of mass, but absent gravity, no.

No. The muzzle will flip: the pistol retains its center of mass in a weightless environment. (And as has been said by Loomis, as long as there is mass there is gravitational force.)

When the gun in space goes bang, and no one is there, Does it really go BANG?

Definitional question: if by "go BANG", you mean ignites its charge or agitates molecules, then of course it does. If by "goes bang" you mean produces a subjective sensation of sound--well that would depend on an auditorily aware organism being present, and you have specified that there are none.

The falling hammer will impart a small force moving the muzzle downwards, very slowly

No. As the hammer spring forces the hammer "down," so also would it force the gun "up" to meet the hammer. But...

There is NO falling hammer: OP specified "remote primer ignition."

"At the end of the class notes, you will find 20 questions. At the end of the 20 questions, you will find 20 answers. The answers are correct. If you get any other answer, YOU ARE WRONG."--Jack Holladay, MD, MSEE, FACS

RecoilRob
September 20, 2008, 09:21 PM
It certainly HAS been proven, time and again, that a firearm will fire in a no-oxygen environment.

Put your 1911 in the swimming pool and pull the trigger....report back what happens!

rcnixon
September 20, 2008, 09:25 PM
Loomis, I do believe you are right about the hammer and the rest of the gun. The masses and reletive velocities are something to think about.

Sorry, J Star, you are wrong. What about guns (and for that matter, explosives) that fire underwater? There's not much air at "full fathom five". As a matter of fact, solid (and liquid) rockets fire quite well in a (near) vacuum. It's how Yuri Gagarin de-orbited.

Loomis
September 20, 2008, 09:26 PM
Loosedhorse:

You are wrong about friction. Theoretically, there is no friction force without a force normal to the friction face. Where is your force normal to the slide and frame in the first fraction of a microsecond?

Loosedhorse
September 20, 2008, 09:36 PM
I may be wrong--or perhaps just unclear--about what the first "fraction of a microsecond" is, but not about friction: are you saying that the slide initially travels with no friction? If that is so, I was unaware that that was possible--I had always assumed that as soon as a body begins to move (unless suspended in theorectically ideal vacuum and not in contact another body), it encounters friction.

If friction does begin at first slide movement, that friction will impart spin (as well as translational movement) on the frame the moment the slide begins its movement. Additionally, if the spring initially resists compression, then the slide movement will be inhibited as the bullet begins to move, and muzzle flip (upward) and gun translation will begin even before the slide begins to move relative to the frame.

If not, I stand corrected.

Blakenzy
September 20, 2008, 09:38 PM
What would be the velocity of that 230gr. bullet fired in outerspace?

Loomis
September 20, 2008, 09:45 PM
Let me put it this way...

With no "clamping force" between slide and frame, there is no friction. On earth, gravity provides the initial "clamping force". After firing, the muzzle flip provides the "clamping force"...slide is pushed towards frame at the rear, and pulled away from the frame at the front of gun. But even with the friction force present, it is very very small compared to other forces acting on gun.

In space, there is no initial "clamping force, therefore, no initial friction force.

Loomis
September 20, 2008, 09:50 PM
Initial muzzle velocity of bullet relative to gun will be slightly higher in space than on earth. This is because there is no air pressure in front of the bullet to resist bullet forward motion. This difference is very small, however.

But...notice I said "relative to gun". Since the gun accelerates away from the bullet when fired in space, the velocity of the bullet relative to a fixed point will be equal to the value on earth minus the velocity of the gun in space, plus the small bonus from no air pressure in front of the bullet.

Loosedhorse
September 20, 2008, 09:58 PM
Loomis--I disagree: simple metal-on-metal fit (or the viscosity of gun oil--let's assume the 1911 is well cared for, as it ought to be!) provides your missing clamping force that "connects" slide to frame.

(I mean, I know some 1911s have a loose frame to slide fit, but you're proposing so loose a fit that there is no contact at all between slide and frame unless gravity pushes the slide down? Don't think I've encountered that yet!)

:)

But let me sign off for a while, drink some scotch, and more thoroughly consider your argument.

Until later, my best regards, and thanks.

GarandOwner
September 20, 2008, 11:08 PM
It is always entertaining to see what people answer when physics questions are asked (and I found humor in the fact that someone with the name "physics" doesnt understand basic dynamics) Ill try to explain, but physical descriptions are sometimes hard in words without any visual aids anyway, here it goes:


Does the muzzle flip?

yes, the force of the bullet accelerating away from the gun will create a moment, which in turn enacts an angular acceleration on the gun about its center of mass.

Does the gun travel straight back?

No, as the objects rotates, it will "precess" (think of a top or a gyroscope) but the path will be mostly linear

Since there is no resistance, do both the gun and the bullet travel equal distances?

Each will travel until it encounters another body, is slowed by particles in space, or enters the gravity of another object


Does the returning slide push the gun forward?


possibly, but not enough to stop the gun from moving, Here you have two frames of reference, an inertial frame (fixed on the gun) and a universal frame (fixed at some point in space that does not move with the gun) we think of how we perceive things in the universal frame, however trying to do the calculations in this frame will be extremely complicated and far to "messy" that is why we have easy conversions between frames (hurray orbital mechanics class) vector calculations and frame conversions to calculate how much of an effect the moving slide will cause. More than I would want to do from a hypothetical situation with no data

Think about it. When you exert a force off center of mass, what causes it to spin? Gravity. You exert a force in the x, gravity in the y, so the object spins. In space, you exert a force in x, and there is no force in y, so the object just accelerates in x.

This is incorrect, the moment created by the force acting at a certian distance away from the center of mass causes an object to spin. A moment creates an angular acceleration, which would cause the object to spin if it is not constrained. What you are talking about is forces, not moments. While a force in the x direction will cause the gun to move opposite that, if it is not on the center of mass, then it will cause a moment (force x distance from CM) which will cause an angular accleration. This is basic dynamics.

Sum of moments:

Force x distance = I x alpha

Where alpha is the angular acceleration and I is the moment of inertia.

Gravity does not cause rotation, because the force has no moment arm, it acts at the center of mass in most applications.

Theoretically, there is no friction force without a force normal to the friction face

contact force, however slight IS, a force normal to the surface. (normal means perpendicular for anyone who was unclear about that term)

I wont even touch the hammer question, WAAAAAY two many variables with 3 frames of reference which you have to convert between.

saspic
September 21, 2008, 01:09 AM
I wont even touch the hammer question, WAAAAAY two many variables with 3 frames of reference which you have to convert between.
Don't forget the firing pin. If it was a normal 1911, not a "remote primer ignition system"(?), the firing pin would, after ignition, be pushed backwards by the firing pin spring.

Also you will have to stake the grip safety.:neener:

Axctal
September 21, 2008, 05:01 AM
... could not resist to answer some misconceptions ...

1. "zero gravity" - FALSE
Objects in space do affected by gravity. There are plenty of massive objects still around - planets, stars ... even if we are FAR away from everything, we still going to orbit at least the galaxy.
Now, there is no -weight-, but still -mass-. Those two quantities are DIFFERENT physics things, describing completely different things.
Mass is a measure of matter, it is a coefficient in a Newton's famous equation: F=ma. This quantity does not depent on where you are. The units of "m" is kilograms [kg].
Now -weight- is a FORCE which object produces on a support, measured in newtons [N]. This quantity does depend on how you move. When you are standing still, your weight = your mass X gravity constant of the place you are standing. NOTE: scales are measuring WEIGHT but shows it to you in kilos or pounds. Spring scales shows your mass correctly only when they are standing still. Now imagine you and the scales are jumping from a building ... you and scales are accelerating with the same g and flying the same way. You not going to exert any force on the scale, and they will show zero.
Now back to space ... spacecraft, astronaughts and ALL objects inside are orbiting the planet, affected by the same gravity acceleration and in never-ending free-fall, thus the -weight- of all objects are zero, but they still have mass.

2. "Does the muzzle flip?"
It depends on the design. On a typical handgun - yes, it will spin it.
The reason is that the line of linear momentum (not force as people before write) are not lined up with -center of mass- (not gravity)
The off-center linear momentum means we have angular momentum which will result in angular acceleration. Note that a shot has finite time of interaction with a gun, and thus resulting with finite final angular speed of the gun (it will be rotating). Every additional shot will increase gun's angular speed.

3. "Does the gun travel straight back?"
The gun will travel straight back while rotating. If you manage to fire second shot at the moment gun will point back compared to first shot (and assuming all ammo is perfectly identical) ... and NO YOU GUESSED WRONG ... the gun will not stop, because at the point of 2nd shot (compared to 1st) the gun is lighter by the mass of one cartridge - so it will recoil with slightly faster speed.
(of course, here I am not considering the momentum of the cartridge expelled to the side - we are talking only about gun and a bullet as they are the main contributors)

3. "do both the gun and the bullet travel equal distances?"
The gun and the bullet acquire equal MOMENTUM (magnitude same, opposite directions), but since they have different masses, they will have different speeds. Considering same amount of time they will travel different distances.
Is it infinite? - It depends. The gun and the bullet are on elliptical orbits, you would need to see if those orbits are intersecting the surface of the planet. If yes - they will perform a fiery reentry, if not - they will orbit the Earth like other cosmic junk.

4. Gravity or not - your slide/bullet still under 3rd Newton's law - it will operate same way anywhere. Now on the practical side, if your gun jams if you holding it light or "not holding it at all" on the planet - it will jam same way in space. But you your gun can fire and not jam while being "free" on the Earth, it will do same in space.

5. "no resistance" - FALSE, space have plenty of resistence. There is still molecules and ions, solar radiation etc. In fact "space vacuum" is considered pretty dirty and filled with stuff compared to laboratory and even microelectronic production vacuum chambers.



So ...
Gun will fire as usual and fly back while rotating.
Initially gun is on one elliptical orbit. The act of firing will send a bullet on an elliptical orbit and gun will change its orbit to a different one. Momentums are same and in opposite directions, M_gun > m_bullet --> gun travel back slower than a bullet. Bu conservation of momentum, the center of mass of gun+bullet is travelling on the original orbit (like if never fired).

Axctal
September 21, 2008, 05:04 AM
... deleted my dup post ... and posting some add-ons

About hammer - Any internal part which stay inside (and not fly out) will NOT affect any motion of the gun due to conservation of linear and angular momentums. (This is the same thing as to why you can't pull yourself from a swamp by you own hair - all objects are within the system)

Rotation of the gun due to the rotation of the bullet (rifling) is around unstable inertial axis and thus will end up in rotations about two stable axises. Try spin and toss a book - there is three different ways you can spin it - you will quickly find that there is two axises (minimum and maximum ones) that are stable and the middle one will result in "wobble" - sum of rotations around two stable axises.

moooose102
September 21, 2008, 07:57 AM
so, if the muzzle was pointed at earth when the gun fires, would the bullet "burn up" during re-entry? also, would it make a difference if it was plain lead, or a fmj?

Friendly, Don't Fire!
September 21, 2008, 09:14 AM
Yes, the bullet would most likely burn up upon re-entry into the atmosphere. It would most likely melt into a blob of whatever metal it is made of. At the speed it will be traveling, it will most likely vaporize into very minute particles.

Blofeld
September 21, 2008, 11:22 AM
If the bullet was HP, and travelled an infinite distance without impacting on anything, would it expand based solely on its speed an friction with all those cosmic particles?

Eric F
September 21, 2008, 12:10 PM
Every single movement would have an effect.

The trigger being moved the hammer falling

the discharge

bullet rotation slide movement

everything comes into play no matter how small it is.

brickeyee
September 21, 2008, 12:13 PM
While space is not a perfect vacuum, it is still pretty darn good.

"Cosmic particles" are mostly electrons, protons, and individual atomic nuclei.

The nuclei are mostly in the lower mass part of the periodic chart.

Heavier elements are present, but at very decreased rates compared to lighter elements.

Think many orders of magnitude like ~10^-8 (or more, tables are at work) from hydrogen to even iron.

We still worry about them hitting satellites though.

There are charts with spans of 16 orders of magnitude covering atomic distribution in cosmic rays.

Jim Watson
September 21, 2008, 12:15 PM
Do we have any empiricists here who would like to suspend a pistol by a thread and touch it off with a cable release? Probably have to do it several times hanging in different orientations to average out surface gravity. Or with some little accelerometers stuck to it.

Otherwise, Axtal has the physics down best.

And just remember, if you have a space vessel equipped with a Bergenholm, your velocity will immediately assume the value at which the thrust of the conventional drive is equaled by the frictional resistance of the medium being transited.

The Bushmaster
September 21, 2008, 12:38 PM
And it wouldn't even make a sound...

GarandOwner
September 22, 2008, 12:35 PM
so it will recoil with slightly faster speed.

If you assume that all the cartridges are identical, then it will recoil at the same speed, the speed of cycling is independent of how many rounds are in the magazine. If is base on the spring constant of the recoil spring, and the force of the shot (to determine how much time it will take for the spring to oscillate one cycle (compression and decompression).....unless you are counting the little change in friction of the round beind stripped since there is a little less compression in the magazine spring

About hammer - Any internal part which stay inside (and not fly out) will NOT affect any motion of the gun due to conservation of linear and angular momentums. (This is the same thing as to why you can't pull yourself from a swamp by you own hair - all objects are within the system)

Sorry but this is not true, the hammer CAN give the gun an angular acceleration, although it will be extremely small. Try this, if you are in a computer chair that swivels, hold your arm out, then quickly move it to the side, you will spin, because you give yourself an angular acceleration. Not to mention the movement of any part changes the center of mass of the gun.

Friendly, Don't Fire!
September 22, 2008, 05:53 PM
Are we still shooting that bullet in outer space? :rolleyes:

eyebrows
September 22, 2008, 06:15 PM
Brings all sorts of questions to mind.

What effect would the vacuum and extreme temperatures of space have on a steel 1911?
Would being in a vacuum increase the stress on the barrel?




With all that high pressure gas INSIDE and negative pressure OUTSIDE, with the 1911 either brittle cold or blazing hot, I think you'd possibly have a ka-boom and ruin your gun. I could be wrong, somebody should try it.

Friendly, Don't Fire!
September 22, 2008, 06:22 PM
Are there any engineers from NASA lurking in the shadows?:cool:

JesseL
September 22, 2008, 07:34 PM
With all that high pressure gas INSIDE and negative pressure OUTSIDE, with the 1911 either brittle cold or blazing hot, I think you'd possibly have a ka-boom and ruin your gun. I could be wrong, somebody should try it.

There's no such thing as negative pressure. The vacuum of space is roughly 0 psi. Compared with the roughly 14 psi of sea level on Earth and the 21,000 psi chamber pressure, you'd see about a 0.06% increase in relative pressure when firing in space. Probably significantly less than the typical shot to shot variation.

gvnwst
September 22, 2008, 09:52 PM
Star Wars!!!!!:D (sorry i had to do that) I would say teh question has been awnsered and that this post is utterly useless except for this. MAKE IT A STICKY PLEASE!!!

:):):)

Loomis
September 22, 2008, 10:04 PM
Negative pressure means less than atmospheric pressure. Therefore, one could say that a perfect vacuum is -14.7psi.

Loosedhorse
September 22, 2008, 10:29 PM
the hammer CAN give the gun an angular acceleration, although it will be extremely small.

No. Think of the hammer as a separate body, with its own center of mass, and the gun with a larger mass, and the two connected by a spring under tension.

The spring is released: the smaller mass speeds toward the larger mass--but the larger mass also moves toward the smaller, more slowly, but with equal momentum. The two collide (inelastically--like two balls of stickum with a stretched rubber band between them), and the momentum of one cancels out the other, so the stuck-together balls no longer move in either direction. And the center of gravity of the two-objects-considered-together never moved throughout the event.

This is easier to envision linearly, but the same effect holds with angular moments. The reason the swivel chair thing "works" is because the effect of friction allows your chair to not move at all when you start your arm moving (so that your body does NOT begin to move in the opposite direction) but when it decelerates abruptly (as your forearm hits your opposite shoulder) now your chair budges a bit in the direction of arm movement (and stops) since there is no counter-movement of your body to counter it.

In a frictionless chair, the chair would move (in the opposite direction) when the arm moved, and stop when the arm stopped. Anyone who has attempted to walk on wet, smooth ice knows it's very difficult in a frictionless environment to acclerate your center of mass in the desired direction, no matter your arm (or leg) movements--until gravity kindly solves the problem for you!

(The position of the gun's center of mass within the gun silhouette WILL change, slightly, as the postition of the hammer changes, and to this extent the position of the gun silhouette in reference to space will change ever so slightly during hammer fall. But the position of the gun's center of mass in reference to space will not change as a result of hammer fall.)

JesseL
September 23, 2008, 12:06 AM
Negative pressure means less than atmospheric pressure. Therefore, one could say that a perfect vacuum is -14.7psi.

If one said that, they would be wrong. A perfect vacuum is 0 psi and you can't get less than 0 psi.

You can talk about relative pressures if you like, but then you really should be specifying what they are relative to.

It would be awfully silly to say that the air I'm breathing here at 5200ft altitude is -2.47 psi, wouldn't it?

Stevie-Ray
September 23, 2008, 12:31 AM
O2 is not needed for a gun to fire in a non o2 enviroment, that IS why there is an oxidizer! science class, a bullet will fire in a vacume. space is NOT a vacume. And if you did this experiment, you would lose a nice hand gunMeh. We could use a Jennings......

Flyboy
September 23, 2008, 12:48 AM
The spring is released: the smaller mass speeds toward the larger mass--but the larger mass also moves toward the smaller, more slowly, but with equal momentum and energy. The two collide (inelastically--like two balls of stickum with a stretched rubber band between them), and the energy of one cancels out the other, so the stuck-together balls no longer move in either direction. And the center of gravity of the two-objects-considered-together never moved throughout the event.
Close, but no cigar.

Momentum and energy are almost never equal.

Momentum = mass * velocity
Energy = 1/2 * mass * velocity^2

For the two to be equal, mass * velocity = 1/2 * mass * velocity * velocity

This can be satisfied, but is very rare. In most (nearly all) cases, it won't be. This is an inelastic collision (http://en.wikipedia.org/wiki/Inelastic_collision). Since inelastic collisions are governed by conservation of momentum, the action-reaction pair will result in a net motion (equations for final velocity are available at the link provided; I'd include them here, but mathematical markup isn't supported in vBulletin).



Incidentally, thank you very much for making me remember my long-repressed high school physics. I'm going to go have a drink. :neener:

Axctal
September 23, 2008, 01:53 AM
Loosedhorse - Yes (about friction{less}) ! In fact if you keep thinking in that direction, you will realize that the only force that propell you forward is a friction force! Thinking more you will realize that this frinction is a simple static one. So if we have a friction coefficient [mu] between two surfaces, and the mass of the object (think your car) is M, than maximum allowed friction force (before we will go into slipping mode) will be [mu]Mg. Sticking this into Newton's F=Ma will give you your maximum acceleration:
a_max = [mu]g
( here we assumed flat road )
So if you want to build a cool race car and thus maximize your acceleration, you need to either increase planet's gravity (realistically impossible) or maximize friction coefficient - get wider tires with better rubber.
Note that if you go into slipping, you will be under mercy of kinetic friction coefficients which is typically smaller than the static one.
Another note: Can your engine exert enough torque to match maximum-optimal force? Well ... depends on engine.
Another note (practical): you are having max acceleration if you are just on the brink of going into "peeling" but not entering it.

Flyboy - You just got big fat "F" for equating apples and oranges - energy and momentum

Guys, once again:
ANY moving part which STAYS with a gun (slide/hammer/whatever) will NOT affect the gun's final state of motion in any way. Yes, DURING the time of motion of said parts they have effect, but after the motion is done the gun moves like if that action had never happen.
This comes from conservation of linear and angular momentums.

So considering that I dont really care how slide moves unless my task is to describe the motion DURING the shot. If my task is to find final gun's motion AFTER the shot, I do not need any information on moving parts which stays with the gun.

So, during the 1st shot, my gun's mass is (M) and I have n bullets each of mass m. Let's say bullet exits with speed v. Then by conservation of linear momentum (in center-of-mass coordinates):

0 = [ M + (n-1)m ]V1 + mv, so the speed of the gun after 1st shot will be
V1 = -{ m / [ M + (n-1)m ] }v
(Negative sign shows that they are in opposite directions)

Note that i purposfully here writtem the answer as V1 = {something unitless (kg/kg) } times v

Let's say second shot is directed against the gun's motion (as to "stop" it).
The gun is moving with above-found speed V. Applying same principle (but remember now we have one less bullet). Here we are taking just the value of the V1 and the signs are as follows: V1 is to the right (+), bullet exits also to the right (+), final gun's V2 is unknown, so keep it with (+) - if it comes out as (+) then gun flies right, if negative, then gun flies left.

[ M + (n-1)m ]V1 = [ M + (n-2)m ]V2 + m( v+V1 ), so
V2 = { [ M + (n-1)m ]V1 - m( v+V1 ) } / { [ M + (n-2)m ] }; substituting V1 from above:

V2 = { [ M + (n-1)m ]{ m / [ M + (n-1)m ] }v - m( v + { m / [ M + (n-1)m ] }v) } / { [ M + (n-2)m ] }

After simple algebra we will find the velocity of the gun after 2nd shot:

V2 = - { { m^2 } / { [ M + (n-1)m ][ M + (n-2)m ] } } v

WHICH IS ALWAYS NEGATIVE no matter what the (physycally real*) values of those variables are.

(Physically real means all your masses are positive, n>=2 to be able to produce two shots, v>0 )

So if 1st shot was pointing left and sent the gun flying to the right, the second shot aimed to the right will not stop the gun but send it (quite slower) to the left.


Uff ... formulas on paper (compared to text) is so easier so see ... hope you got through that.

Loosedhorse
September 23, 2008, 12:53 PM
Flyboy--

Inelastic collisions (in effect, all real collisions) result in a loss of kinetic energy AFTER impact: the two moving bodies had more energy (summed) before the collision than after it.

What I was speaking of was a spring simultaneously acting (with the same force) on two bodies, one heavier than the other, before collision. The force will act for the same amount of time on each body (starts when hammer is released, ends when it impacts).

So, agreed, the two bodies will head toward each other with equal momentums, but not with equal energies.

Previous post amended.

Old Grump
September 23, 2008, 04:14 PM
Hot diggety dog, this is a fun thread. I have lain awake a couple of nights just thinking about all the factors that changes the equations after each shot, like the reduced mass after each shot, the effect of lubrication on the moving parts after it warmed up, The effect of the nearest or strongest gravity well in relationship to the gun and on and on.

Kentak
September 23, 2008, 04:43 PM
A few tidbits, which may already have been mentioned since I didn't read every reply yet.

1) "Space" is not free from gravity. When astronauts are orbiting in a Shuttle, both the shuttle and everything in it are free falling fully within the effects of Earth's gravity. If you think of a typical schoolroom globe, a shuttle orbits at perhaps 225 miles or so above the surface, which would be only about maybe a half inch or so above the globe, keeping the same scale.

2) There is no need to go into space to perform this experiment unless you had to reduce the minor effects of air resistance to a bare minimum. Merely drop the gun from something at altitude, like a weather balloon, high bridge, etc. Track the gun with a camera with a very high power lens if you want to examine the motion in detail.

3) Upon remote firing, the reaction forces on the gun will cause it to change it's motion relative to itself in several ways. Assuming the axis of the forces are above the gun's center of gravity, the gun will rotate around the center of gravity and there will be a slight sideways rotation imparted by the rifling of the barrel acting against the inertia of the bullet as it travels through the bore. The *exact* motion of the gun is based on the complex interactions of all the factors in play, barrel length, gun design, spring weights, time of the bullet in the bore, slide mass, all kinds of things. A revolver's motion would likely be easier for a really good physicist to calculate than a semi-auto pistol due to the fewer moving parts in play.

4) If you did do this experiment in space by having an astronaut simply place the gun in orbit along side the Shuttle, the basic results would be the same, with a little more velocity due to the lack of air resistance. The gun and bullet, however, would, in this case, not travel in straight lines, being captured in Earth orbit, and would eventually fall to Earth as the orbits decayed.

K

Claymore1500
September 23, 2008, 06:34 PM
There's no such thing as negative pressure.

While I can not truthfully say that space is "Negative air pressure" ('cause I've never been there), Negative air pressure is for real, Think about flight, What gives an airplane it's lift, (or at least what do they call it) also, If you talk about carburetors on an internal combustion engine, The fuel is drawn out of the float bowl by,.......wait for it........Negative air pressure.

Like I said, While I can't say for sure that space has neg. air pressure, you cannot say it doesn't exist.

JesseL
September 23, 2008, 06:46 PM
Nice try but no cigar.

An airplane gets lift from positive pressure on the underside of the wing. The lower pressure on top of the wing simply allows the higher pressure on the bottom to exert a net lifting force.

Same with a carburetor. The fuel is pushed out of the float bowl by positive atmospheric pressure from the vents. The venturi simply lowers the pressure of the incoming air so the the pressure differential is enough to force fuel through the jets.

This is a subtle but important distinction. Air pressure never pulls on anything. Pressure is by definition a pushing force. When the pressure on all sides of something are equal, it cancels out. When the pressure on one side is reduced, the pressure on the opposite side is what pushes the object towards the lower pressure side.

brickeyee
September 23, 2008, 08:06 PM
Negative air pressure is for real ...
You are confusing absolute pressure with relative pressure.

'Gauge' pressure is measured relative to whatever ambient is present.
A manometer shows gauge pressure.
It has two ports.


Absolute pressure is relative to a 'perfect' vacuum.
A barometer is a decent example, the main source of error being the vapor pressure of the mercury above the column of liquid mercury.
Water or oil barometers are typically less accurate since the vapor pressure above the liquid of the column is more than mercury.
Notice there is only one port on a barometer.

Barometers using a vessel evacuated are not as accurate.
They are comparing the pressure inside the evacuated bellows with the air pressure outside.

There are semi-conductor pressure transducers that approach absolute since the vacuum trapped under the silicon strain gauge is very good, easily approaching that of free space.

Claymore1500
September 23, 2008, 10:19 PM
Yes but, The American society of automotive enginiers,"A.S.E." refers to the carburator principal as negitive air pressure in the venturies allowing the atmospheric pressure in the float bowl to push the fuel out.

At one time I was certified by the A.S.E. in tune-up, and argued that point with one of the instructors, ( I lost) But, in the process I had the very detailed explination drilled home, I fully understand that it is simply a term used to discribe a condition, You should have noticed in the post where it was first mentioned I stated "Lack of, or poss. negitive air pressure", Essentially, I think it may be more appropriate than calling space a vacuum, which it is not.

Owen
September 23, 2008, 10:43 PM
its only negative relative to gage pressure. its positive relative to absolute pressure.

Stevie-Ray
September 23, 2008, 11:49 PM
Wouldn't jet wash be considered negative air pressure?

Kerf
September 24, 2008, 12:52 AM
is whether the gun would function or cycle completely in space. I know it's not part of the original post, but several posts have mentioned that the gun wouldn't completely cycle in outer space. I think we need to nail down this one loose end, so we can all get a good nights sleep.

I held off, hoping that 1911 Tuner or Old Fluff would step in with their much-respected expertise in this matter and enlighten us mortals and physicist/rocket scientists, alike. If my memory serves me correctly, one of my above mentioned experts, stated, “if a gun runs, then there is no such thing as limp-wristing”. Limp-wristing is what you tell your wife is the reason her gun has jammed, (it’s not the crummy reloads or the cheap Wally World ammo you provided her, because she can’t hit anything, anyway); or, the new shooter, even though he usually does a hundred chin-ups a day. I’ve never been a believer in limp wristing as a cause for a malfunction in an auto loading pistol, just as a matter of experience.

So, begging the issue to be resolved, would that Glock/1911 cycle normally in space? I think it would. What say you?

Kerf

Kentak
September 24, 2008, 09:32 AM
Claymore--

I totally understand how you are using the term "negative" air pressure. In a way you are right, and so are those saying there is no such thing. IF you define ambient air pressure as zero, then you could describe a lesser pressure with negative values. You could say, "the air pressure in this room is zero, and the air pressure in this vessel is -5 psi." When the manual you mentioned spoke of negative air pressure, it was using it to mean negative *relative* to ambient air pressure.

But, it's more accurate technically to speak of space as being a vacuum or partial vacuum or having "little or no" air pressure. A total vacuum exerts 0 psi air pressure. Can't go lower than that.

Hope this helps.

K

Kentak
September 24, 2008, 09:43 AM
Just a little bit about the bullet burning up upon reentry. Keep in mind that the intense heat generated by an orbiting object reentering the atmosphere is due mostly to its *forward* velocity relative to the Earth's surface, not it's downward velocity. If one could merely place a bullet in space at an orbital altitude, with no forward velocity, and drop it, it's not clear to me what would happen to it as it fell back to Earth. Certainly, it's going to accelerate rapidly due to the pull of gravity, and that can be calculated, but it's also going to start to encounter ever increasing air resistance as it nears the surface. Whether it will ever reach a velocity high enough to generate enough friction to melt or vaporize is above my pay grade. Anyone want to take a stab at that?

K

brickeyee
September 24, 2008, 10:00 AM
If you are in orbit you have a high velocity.
There is no such thing as "an orbital altitude, with no forward velocity."

Even if you just fired it from the surface of the earth it would have a velocity component from the surface.
At the equator you are moving at over 1000 mph since the earth spins every 24 hours.

Blofeld
September 24, 2008, 10:11 AM
Just wait for my next thread, "If you fired a 1911 gangsta style on Mars." Thanks for the well reasoned and incredibly astute responses. I knew gun people were intelligent, but some of you guys are amazing.:)

Kentak
September 24, 2008, 12:34 PM
If you are in orbit you have a high velocity.
There is no such thing as "an orbital altitude, with no forward velocity."

I didn't say, "in orbit," I said, "at an orbital altitude." By that I meant a distance above the Earth at which objects are put in orbit, which just for this hypothetical here let's say is 200 miles, or so. So, if the term orbital altitude bothers you, reword my post to say, "...place a bullet in space at an altitude of 200 miles, with no forward velocity..."

K

Old Grump
September 24, 2008, 02:17 PM
I think it would still burn up even if fired straight down to the center of the gravity well. A falling bullet fired straight up from the earth will come back down at less than 300fps and will be tumbling, laying on its side or come down base first. We are launching a spin stabilized bullet nose first at 850 fps assuming a standard hard ball round and it is spin stabilized. Add in the acceleration from gravity and the resistance encountered as it enters the atmosphere. Its horizontal velocity will have to slow down in reference to the surface of the planet it had been aimed at and will now be coming down at an ever increasing angle, assuming we are over the equator the earth's surface is moving at 1467 fps and the atmosphere discounting winds is moving at the same rate the bullet would have the effect of its downward and horizontal speed adding to the amount of resistance its going to have from the increasingly denser mass of molecules in the atmosphere. I'm no physicist but I would think that the initial velocity + acceleration plus the ever increasing affect of the speeding atmosphere on the bullet would have to increase the temperature of the bullet from friction. I don't know how hot a 45 bullet is but a 30 caliber bullet fired at a sedate 1900 fps will have a base temp of around 500 C, a tip temp of 170 C and a groove temp of 320 C. I expect the temp of the 45 would be lower with the low pressure and short barrel but still considerable and would only get warmer with friction. Even if the jacket of the bullet didn't melt I would think the core would pretty much have to vaporize and the low pressure behind the bullet would suck it right out of the core. I admit I am only speculating but it seems logical to me. However I am the first to admit I will never be a rocket scientist, I would be the guy in Crews quarters who would have to fix the control panels after the Klingons got done making them spark and smoke again for the third time this week. I wouldn't be as fast as Spock or Scotty either because I don't have pointy ears or a Scottish accent.

harrygunner
September 24, 2008, 05:22 PM
Good to see people interested in physics.

One of my many life experiences was being a professor of physics at a major university.

The principles are basic. Students learn them around the time they are told total momentum, the vector sum of translational and angular momentum is conserved.

An object's moment of inertia is a tensor and if represented as a matrix, a handgun will have significant non-zero off diagonal elements.

On my Sig 229 with a full magazine, the center of mass is in the frame, in front of the decocker lever.

If I erect a coordinate system fixed to the gun at the center of mass, it is clear that there is no transformation that would diagonalize the moment of inertia matrix (where the eigenvector lies in real space). Nothing special in principle or mathematically, but that's why people are struggling with what the motion would look like.

I've learned not to discuss physics on gun forums, but as I said, good to see the interest.

Old Grump
September 25, 2008, 01:59 PM
Just looked up eigenvector

Understood a teeny tiny portion of it. It's official, I definitely will not become a rocket scientist.

harrygunner
September 25, 2008, 06:03 PM
Be careful, physics can be fascinating. You'll be wanting to design rockets before you know it. :)

The mechanical theories associated with firearms were resolved hundreds of years ago, so it's difficult to get someone interested in creating simulations in response to these kinds of questions.

A junior mechanical engineer could get an animation going pretty fast. I'm sure it would be a hit on gun forums.

Loomis
September 26, 2008, 07:50 PM
"It would be awfully silly to say that the air I'm breathing here at 5200ft altitude is -2.47 psi, wouldn't it?"

It would be silly for YOU to say that, but not for ME. Your gauge should be set to read zero at your atmospheric pressure. My gauge is set to zero for my atmospheric pressure.

halfbreed808
September 27, 2008, 03:15 AM
Huh? I thought you need oxygen to cause the primer and powder to burn?!:neener:

brickeyee
September 27, 2008, 12:12 PM
So, if the term orbital altitude bothers you, reword my post to say, "...place a bullet in space at an altitude of 200 miles, with no forward velocity..."


The prpblem is that as you move above the earth you become a free body, and all of a sudden you have a whole lot of velocity relative to the earth just by climbing to space.
The surface of the earth has a speed of over 1000 miles per hour at the equator (circumference divided by the 24 hours it takes to rotate once).

While we do not see this speed on the surface, or even in an air plane, but satellites see it (long range artillery sees it also).

If you fire a shell straight up relative to the local surface it will not return to the same spot since the earth rotates under the projectile.

Orbital mechanics is a strange world.
You move faster and slower to change altitude, and cannot change speed without also changing altitude, they are fundamentally linked by orbital mechanics.
Kepler's laws prevent you from trying to have zero velocity under orbital conditions.

benEzra
September 27, 2008, 08:47 PM
Just a little bit about the bullet burning up upon reentry. Keep in mind that the intense heat generated by an orbiting object reentering the atmosphere is due mostly to its *forward* velocity relative to the Earth's surface, not it's downward velocity. If one could merely place a bullet in space at an orbital altitude, with no forward velocity, and drop it, it's not clear to me what would happen to it as it fell back to Earth. Certainly, it's going to accelerate rapidly due to the pull of gravity, and that can be calculated, but it's also going to start to encounter ever increasing air resistance as it nears the surface. Whether it will ever reach a velocity high enough to generate enough friction to melt or vaporize is above my pay grade. Anyone want to take a stab at that?
Not sure, but Burt Rutan's Space Ship One fell back to earth from an altitude of 100 km/62 miles, and its speed was around 2170 mph, or 3182 ft/sec, which is about the same velocity as a 55-grain .223 at the muzzle. The heat pulse may be longer, so it is possible that you might have some core melting, but it won't burn up.

Dropped from 150 or 200 miles or so, your bullet would likely exceed the melting point of copper, and if you do then the bullet would indeed "burn up" (melt/ablate and resolidify as small droplets or dust).

RecoilRob
September 27, 2008, 09:54 PM
If you could somehow place a bullet 200miles up in space....with no forward velocity (which is needed to orbit), it would fall directly back to earth.

And, seeing as the gravitational force working at 200miles is still about 90% of surface gravity, it would pick up speed very close to the theoretical 32 ft/sec squared.

Without significant atmospheric resistance for the first 130 miles or so...it would be moving pretty darn fast when it came in. I haven't the ability to calculate it, but I'm betting that it would melt the slug for sure.

Claude Clay
September 28, 2008, 01:13 AM
ask NASA to try it on the Vomit Comet. weightlessness is maintained for almost 2 minutes. though the air in the cabin is 8psi, it is sure to drop after the first bullet passes through the fuselage. the govt loves to share; after all it is our $ making all this possible. i have even had real time on the Hubble. present them with an idea and a outline for how to do it. don't know if you don't try.

the difference between achieving orbital velocity and just say, taking an elevator there is great. and well explained by Arther C. Clark in his novel The Fountains Of Paradise

Kind of Blued
September 28, 2008, 01:30 AM
Has anyone mentioned TEMPERATURE yet?

From what I've heard all external parts exposed to the sun might be fine, but the internal parts would become VERY cold being shadowed.

Drusagas
September 28, 2008, 01:35 AM
And what if it was Extreme Shock ammo? (couldn't resist, sorry)


Then it would destroy whatever it hit first completely and with no exit wounds :evil::neener::banghead:

Kentak
September 28, 2008, 04:33 AM
The prpblem is that as you move above the earth you become a free body, and all of a sudden you have a whole lot of velocity relative to the earth just by climbing to space.
The surface of the earth has a speed of over 1000 miles per hour at the equator (circumference divided by the 24 hours it takes to rotate once).

Guys, you're making my hypothetical more complex than it has to be. The question I posed was whether or not a bullet (not in orbit) falling from an altitude of 200 miles would gather enough velocity to burn up at some point during it's reentry into the atmosphere (yes, I know there's a very thin atmosphere, even at 200 miles). I noted that a satellite reentering from orbit has more "forward" velocity than vertical velocity, which is the source of most of the heat-producing friction. I also noted that the answer is complicated by the fact that the bullet will be encountering ever increasing denser air as it falls, which will reduce it's acceleration from the theoretical 32 ft/sec^2.

If it helps to clarify, assume the bullet is dropped from a 200 mile high tower built directly in line with the Earth axis at the North Pole, hence, no motion relative to the surface except downward. Also, assume a nose first stabilized bullet, just to avoid the complications of a tumbling bullet.

I don't really expect an answer, it was just one of those interesting hypotheticals that my warped mind dwells on.

K

Claude Clay
September 28, 2008, 09:43 AM
would the bullet burn up--YES if released from 200 miles up
or it could be argueed--NO. if released very much lower.

Old Grump
September 28, 2008, 03:34 PM
If dropped I doubt if it would burn up, if fired I believe it probably would. One thing is for certain, there had better be a large landing area for that bullet because it will not come down at the base of the tower. Between winds and the movement of the earth before it could drop 200 miles it could shift miles away from the target area.

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