Physics Genius Please!!!

[Jake M]

We are having a heated discussion in our office today about bullet trajectory, one person believes that when a bullet is fired on a completeley flat surface with no wind conditions and normal earth atmospheric conditions that the bullet will rise slightly before falling, does anyone know if this is true, if not true please provide the argument or maybe some facts to back it, he believes that there is a small rise right after leaving the barrell of the gun.

[vintage1]

Now that depends entirely on how the gun is sighted.

What I find is that if I use a scope on top of a rifle, and zero in on a shortish range target,because the scope is higher then the target the scope has to point down slightly.

Then what happens is that you get two points at which the bullet hits the sight center exactly, one at close range where the bullet is still travelling upwards, and another at longer range when the bullet has dropped back to the line of the sights.

In between the gun shoots high, and beyond it shoots low.

I suspect this is what is being said, but its not the physics of travel, its the parallax bteween the bore and the gunsight.

A similar thing happens with multiple gun installatins on warbirds: The bullets are set to converge at a certain distance. In front of and behind they are more spread out.



From pure physics, assuming no backspin on the bullet. it flies in more or less a parabolic path, and will start to drop immediately if fired horiozontally, unless it gets fired at escape velocity, in which case its drop is matched by the curvature of the earth :D

[Jake M]

so you believe that it is more of a visual trick than the actual upward travel of the bullet?

[pimmer]

If the bullet is fired fast enough, then there is the possibility that the curvature of the earth will come into play, but then the bullet would have to leave the muzzle at > 18,000 mph, which far exceeds all high power rifles, and would tend to vaporize the bullet before it got very far.

The bullet will begin to drop the millisecond it leaves the barrel. The reason why these guys are saying that is because of the way we set up scopes. The rifle is suspended about 2 inches below the scope. If the scope is set up to point at the target, the the rifle barrel is actually pointing slightly upwards so that the bullet trajectory will intersect this line of sight. If the scope is level, then the bullet trajectory will be pointed slightly upward. The fired shot will intersect the line of sight of the scope at a close range, rise slightly above the line of the scope, and then back down to the line of sight of the scope, making the scope alignment accurate at two different ranges. One is close, the other is farther away. This gives the impression that the bullet rises when fired flat, but that is not the case.

[Jake M]

Thanks if anyone else has any more arguments please post.

[robert harik]

Its like water out of a hose , its only going to rise if you aim the end of the hose above horizontal.

[Ledbetter]

I agree with robert harik. Obviously he has aimed a few hoses in his day.

Bullets have no inherent lift. Given a "perfectly flat surface" (of which there are virtually none* naturally occurring on a spherical planet) and a horizontally aimed bore, a rifle bullet fired out of that bore's muzzle will fall at the same rate as one dropped alongside it at the same instant.

Most rifles are sighted to compensate for this; maybe that's what your friend means. The bullet's trajectory is normally adjusted upwards to intersect (50, 100, 200, 500 or more yards away, depending on caliber) the scope's centerline, an inch or so above the muzzle.

_______________________
*none large enough to serve as a rifle range.

[mwraight]

Bullets are ballistic...they don't develop lift...and they don't rise.

If you want to drive people nuts at the office then use this little fun fact on them:
Let's say you had two bullets. One is in your hand and you're holding it at shoulder level. The other bullet is in a gun that you are holding at shoulder level aimed horizontally. At the same time you fire the gun you drop the bullet held in your other hand. Both bullets will hit the ground at the same time!

That'll be sure to drive them nuts, stump them and cause feverish debate...

[Sparky Paul]

We are having a heated discussion in our office today about bullet trajectory, one person believes that when a bullet is fired on a completeley flat surface with no wind conditions and normal earth atmospheric conditions that the bullet will rise slightly before falling, does anyone know if this is true, if not true please provide the argument or maybe some facts to back it, he believes that there is a small rise right after leaving the barrell of the gun.
.
Ask the guy what creates this "...small rise after leaving the barrel of the gun".
It's always amusing to see where these things get to. :)

[Ham-fisted-oaff]

Bullets are ballistic...they don't develop lift...and they don't rise.

If you want to drive people nuts at the office then use this little fun fact on them:
Let's say you had two bullets. One is in your hand and you're holding it at shoulder level. The other bullet is in a gun that you are holding at shoulder level aimed horizontally. At the same time you fire the gun you drop the bullet held in your other hand. Both bullets will hit the ground at the same time!

That'll be sure to drive them nuts, stump them and cause feverish debate...

Shot yourself in the foot on this one (pun intended:D ) , and I don't need a lot of math to prove it (although if I didn't have WORK to do I could produce).

Practical counter-argument:

For both bullets to hit the ground simultaneously, the bullet from the gun would have to fall to Earth THE INSTANT it left the barrel. That wouldn't be much use as a projectile now, would it?

Mathematical (kinematics) argument:

The basic explanation is that the "free-fall" bullet only has gravitationally derived POTENTIAL ENERGY, whereas the projectile bullet has both KINETIC and POTENTIAL ENERGY.

Remember Newton #1 - A body will remain at rest or at a constant velocity (and remember velocity is not the same as speed) unless acted upon by an external force or system of forces).

The forces acting on the projectile are drag and gravity:

Drag force Fd = 0.5*Cd*A*U^^2

Cd = drag coefficient
A = projected area (the cross sectional area of the projectile)
U = the projectile''s instantenous velocity
^^2 = squared

Gravity force Fg = m*g*z

m = projectile mass
g = acceleration due to gravity
z = altitude

The resulting accelerations (remember F = M*a) are:

Ad = (0.5*Cd*A*U^^2)/m and Ag = g*z

Now draw the free body diagrams of the projectile and free-fall bullets and use calculus to determine the time taken to hit the ground, remembering that U and z are not constant. For argument's sake I'll let you assume drag in the freefall case is negligible (basically true in air, not so in water I assure you) and that Cd is constant with U (it'll actually vary with the Reynolds' Number, but that's for the advanced class).

Firs correct answer wins a night in the Baghdad Hilton (to be taken between now and the end of October 2004), second prize is TWO nights at the same venue. Certain restrictions apply, not valid where legally binding as an offer of contract. All successful winners must be accompanied by all eight great-grandparents, etc. etc.

[Jake M]

Shot yourself in the foot on this one (pun intended:D ) , and I don't need a lot of math to prove it (although if I didn't have WORK to do I could produce).

Practical counter-argument:

For both bullets to hit the ground simultaneously, the bullet from the gun would have to fall to Earth THE INSTANT it left the barrel. That wouldn't be much use as a projectile now, would it?

Mathematical (kinematics) argument:

The basic explanation is that the "free-fall" bullet only has gravitationally derived POTENTIAL ENERGY, whereas the projectile bullet has both KINETIC and POTENTIAL ENERGY.

Remember Newton #1 - A body will remain at rest or at a constant velocity (and remember velocity is not the same as speed) unless acted upon by an external force or system of forces).

The forces acting on the projectile are drag and gravity:

Drag force Fd = 0.5*Cd*A*U^^2

Cd = drag coefficient
A = projected area (the cross sectional area of the projectile)
U = the projectile''s instantenous velocity
^^2 = squared

Gravity force Fg = m*g*z

m = projectile mass
g = acceleration due to gravity
z = altitude

The resulting accelerations (remember F = M*a) are:

Ad = (0.5*Cd*A*U^^2)/m and Ag = g*z

Now draw the free body diagrams of the projectile and free-fall bullets and use calculus to determine the time taken to hit the ground, remembering that U and z are not constant. For argument's sake I'll let you assume drag in the freefall case is negligible (basically true in air, not so in water I assure you) and that Cd is constant with U (it'll actually vary with the Reynolds' Number, but that's for the advanced class).

Firs correct answer wins a night in the Baghdad Hilton (to be taken between now and the end of October 2004), second prize is TWO nights at the same venue. Certain restrictions apply, not valid where legally binding as an offer of contract. All successful winners must be accompanied by all eight great-grandparents, etc. etc.


Are you saying that the bullet does rise or just travels farther.

[Rodney]

mwraight is correct, what the following discussion fails to consider is the direction of the forces, drag in the horizontal direction has no effect on the rate of fall, both bullets hit the ground at the same time.

[RGinCanada]

I am assuming that the barrel is perfectly parallel to the flat surface and the bottom edge of the muzzle is inline with the surface. Is this the scenario?

If so, your co-workers may be arguing that there will be a ground effect between the projectile and the plate. (Possibly higher pressure under the projectile imparting upward momentum to the bullet?) I wouldn't bank on it, but I can see the reasoning.

Aside from any interaction of this nature, gravity rules, and there will be no lift.

Regards,
Raymond

[vintage1]

I am sure that the co-worker is a practical rifle shooter, and has noted that with his sights set for eg 100 yards, the gun fires high at say 150 yards.

[Grejen]

For both bullets to hit the ground simultaneously, the bullet from the gun would have to fall to Earth THE INSTANT it left the barrel.

The bullets will hit the "ground" at the same TIME not the same place - unless the height of the barrel above the ground is zero. The only thing relevant is gravity which acts on both bullets.

btw - while we're "driving people nuts at the office" how bout this: The bullets are traveling in a straight line. The parabolic 'curve' is an illusion caused by gravity which is curving space .... lol . :p

One could also say that the earth is rising to meet them. Can't you feel that constant acceleration (9.8M sec sec) pushing you up? No?... step off a ledge. ;) :D

[RyanPSU21]

They hit the ground at the same time. Do the free body diagram. The only vertical acting force on both bullets is gravity and negilible drag in the vertical direction from falling. The bullet fired from the gun will instantly start to fall the moment it leaves the gun barrel

[Mike Taylor]

NB: Bullets don't follow a true parabolic path. Wind resistance slows them down as they fly and the rate of drop per distance traveled increases. A bullet's true trajectory is more of a french curve. Vintage's image is correct except that the third target is closer to the middle target than the first target is. The higher the muzzle velocity, the less the drop.

[vintage1]

heck, I didn't want to complicate matters unduly.

Thers such a range of muzzle velocities out there that its almost impossible to predfict anything - thats why me mate with the real guns spends a whole afternoon getting sights lined up and marking off drop points at various ranges, and when we go bunny shooting, we have marker posts all along the field to tell us how far the rabbit is.

(he shoots silenced with subsonic ammo, so as not to scare the rest)

[Sparky Paul]

I spotted this same question elsewhere..
Note to Jake:
The laws of physics are not subject to popular opinion or majority vote.
They apply to all participants, whether they're known or not.
Ballistics are well documented, and what happens to a bullet is quite well established.
As with everything else in the universe, that object which isn't supported falls down.
A bullet fired precisely horizontally -must- descend as soon as it leaves the barrel of the weapon.
Isaac said it, I believe it, that settles it! :D

[macr0t0r]

Heh, I'll assume we are leaving the curvature of the earth out of the discussion with the two bullets, right? :D

In the fictional flat world, they will hit the ground at the same time. While the forward velocity DOES increase the aerodynamic pressure surrounding the bullet, one has to remember that the pressure increases on both the top and bottom of the bullet, canceling out any change in vertical acceleration.

[Kaos116]

When talking about a bullet, we are assuming a projectile that has a spinning axis parallel to the direction of travel. I know for a fact that I get projectiles to rise when shot from a specially designed barrel. In easier terms... Paintball. Just as a basball pitcher puts spin on the ball to get it to curve, sink or rise. I can do the same with a .68 caliber projectile.

The question, as asked, needs to be more specific to be answered with out assumptions.

but...

If we 'assume' that we are talking about a spear point boat tail bullet. It will not rise in relation to the barrel in which it is shot from. Because a bullet is not a lifting body it begins to fall the moment it leaves the barrel. The rotation of the bullet is to increase accuracy (gryoscopic effect) not create lift.

As Vintage was saying, for a rifle that is sighted at 100M, the bullet will be shot at a trajectory and appear to rise, but in relation to the barrel that shot it, it will start to fall the moment it leaves the barrel. It won't fall in relation to ground beneath it but to the angle of the barrel that shot it.

Clear as mud? :-D

[Jake M]

I spotted this same question elsewhere..
Note to Jake:
The laws of physics are not subject to popular opinion or majority vote.
They apply to all participants, whether they're known or not.
Ballistics are well documented, and what happens to a bullet is quite well established.
As with everything else in the universe, that object which isn't supported falls down.
A bullet fired precisely horizontally -must- descend as soon as it leaves the barrel of the weapon.
Isaac said it, I believe it, that settles it! :D

I wasn't looking for a popular vote or majority thought, this is just to provide more proof to this person to prove him wrong. I also want to thank everyone who has provided good information on this topic.

[olmod]

hows this http://www.wcsscience.com/bullet/trajectory.html

[banktoturn]

mwraight is correct, what the following discussion fails to consider is the direction of the forces, drag in the horizontal direction has no effect on the rate of fall, both bullets hit the ground at the same time.

Actually, I don't think mwraight is correct. First, because the bullet which is dropped will likely tumble, while the bullet which is fired will not, so the two bullets would not have the same coefficient of drag. Secondly, it is not valid to decompose the velocity into horizontal and vertical components, and calculate drag independently in the two directions. As a thought experiment, consider an object other than a bullet, such as a sailplane. If you launched one sailplane at moderate speed forward, and dropped an identical sailplane at the same time, would you expect the two to hit the ground at the same time? If you think the situation is different for bullets than sailplanes, then you need to explain why.

banktoturn

[Sparky Paul]

I wasn't looking for a popular vote or majority thought, this is just to provide more proof to this person to prove him wrong. I also want to thank everyone who has provided good information on this topic.
.
It's not all that difficult to see bullets in flight. With the sun behind you, a large black target and a spotting scope more or less aligned with the bullet path, even the faster bullets can be seen on their way to the target, both due to the sun on the bullet, and the disturbed air they leave behind them.
I have some videos of bullet flight in these conditions.
BB's or pellet can be seen without optical aid.
Tests:.
Set up a safe range, where BBs or pellets can be fired. In the path of the projectile set up some paper targets. With the gun barrel leveled, fire a few times and measure the height of the gun barrel and the position of the holes in the paper as the paper is moved downrange.
Or;
With a bolt-action rifle, remove the bolt and center the target in the barrel, looking from back to front, and then do the same test.
The BB/bullet will never pierce the paper above the level of the gun barrel, and will never hit the spot bore-sighted, with the paper at any distance from the gun, but will always be below it.
Vintage's diagram shows the typcial condition with the sight neccessarily being above the rifle bore. To get the bullet to hit the spot sighted at, the barrel must be tilted up, and the bullet will rise, but only because it has an initial positive angle when leaving the barrel. And it will also pass back thru the line of sight on the way down from the peak of the trajectory.
MOF, some weapons can be "sighted in" without having to use a long distance range, with this principle. Adjust the sights on an M-16 say, to have the bullet go thru the point of aim at 25 feet, and the weapon will be sighted for 200 meters, with the bullet never more than 2 inches above or below the line of sight for this distance. (The "point blank" range.)
A popular military rifle with a fixed sight was deadly to 300 meters, if the shooter held his aimpoint at the target's belt buckle. At any shorter distance the bullet was never more than 16" above the line of sight. And above the belt buckle is where the neat stuff to hit is.

[Sparky Paul]

Actually, I don't think mwraight is correct. First, because the bullet which is dropped will likely tumble, while the bullet which is fired will not, so the two bullets would not have the same coefficient of drag. Secondly, it is not valid to decompose the velocity into horizontal and vertical components, and calculate drag independently in the two directions. As a thought experiment, consider an object other than a bullet, such as a sailplane. If you launched one sailplane at moderate speed forward, and dropped an identical sailplane at the same time, would you expect the two to hit the ground at the same time? If you think the situation is different for bullets than sailplanes, then you need to explain why.

banktoturn
A bullet is a symmetrical airfoil at zero angle of attack. The spin keeps it there.
A sailplane is -intended- to fly using lift.
On the moon, propelling the sailplane and dropping its twin would be a useful analogy, but not in an atmosphere.
The horizontal component of velocity and the vertical component -are- considered seperately, as they are independent.
With no air drag, the horizontal velocity doesn't change, while the vertical velocity is affected by gravity.
In an atmosphere, drag will have an -additional- effect to this first principle effect, on both velocities.

[banktoturn]

A bullet is a symmetrical airfoil at zero angle of attack. The spin keeps it there.
A sailplane is -intended- to fly using lift.
On the moon, propelling the sailplane and dropping its twin would be a useful analogy, but not in an atmosphere.
The horizontal component of velocity and the vertical component -are- considered seperately, as they are independent.
With no air drag, the horizontal velocity doesn't change, while the vertical velocity is affected by gravity.
In an atmosphere, drag will have an -additional- effect to this first principle effect, on both velocities.

The angle of attack of the fired bullet is not zero. If it is falling toward the earth, it has a small positive angle of attack.

More importantly, it is not valid to consider the horizontal and vertical components of velocity separately. That is the intuitive view that we tend to take, and the one which leads us to the conclusion that the two bullets hit the ground at the same time, but it is not correct. The reason I used the sailplane in my example is to make this point. The bullet or sailplane which is dropped will have some value of CD which affects its downward velocity. The bullet or sailplane which is launched forward will have some value of CL which affects is downward velocity. If you claim that it is OK to consider the two components of velocity separately, you are claiming that it is OK to substitute a body's CD in one direction for its CL in the perpendicular direction. This is quite clearly not true, even though it intuitively seems like the right thing to do.

banktoturn

[Mike Taylor]

For a discussion with this wide a range of positions to remain so polite is no small feat. Congratulations, all.

This is actually a very tough problem to solve intellectually. It took gunners and philosophers hundreds of years of study to get even close to the problem's solution. There are so many independant variables that come into play to spoil the theorectical results.

[vintage1]

To quote my friends quote of a top target shooter

" I spend a day sighting it in at the target range, then rest easy, take my time., and loose off 5 rounds and let the wind take em wandering around the center spot"

Sure bullets are affected by everythig including the wind.

I suspect that early gunners simply bracketed anyway - certainly up to WWI that was the way to do it.

Prior to cartridged shot, a few more ounces of black powder in a field piece might vary the impact point 20-30 meters.

I think some guns had - maybe still have - the ability to lob shells across the English channel. Some 25 miles.

Aircraft changed that somewhat.

[Ollie]

"Sure bullets are affected by everythig including the wind."

Some thermals strongly of vertical ~200 MPH (a horiz. altitude of few thousands of feet).

"- provide the argument or maybe some facts to back it."

Ha, Ha.

[macr0t0r]

I think this is where the Engineer differs from the Mathematician. Theoretically, we can't even measure the physics of aerodynamics well enough to come up with a solution of infinite accuracy. For most purposes, it is safe to say the fired and dropped bullets will land at the same time. We might be off by a nanosecond or two, but does it really matter, and could the offset be reliably repeated anyway?

Most equations involving air-friction on a moving body are actually estimated "fits" based on sampled data. Number-crunching to 5 decimals of precision is pointless.

[raptor22]

ASSUMING a perfect situation for theory (vacuum, flat) they will hit at the same time.

The angle of attack of the fired bullet is not zero. If it is falling toward the earth, it has a small positive angle of attack.

More importantly, it is not valid to consider the horizontal and vertical components of velocity separately. That is the intuitive view that we tend to take, and the one which leads us to the conclusion that the two bullets hit the ground at the same time, but it is not correct. The reason I used the sailplane in my example is to make this point. The bullet or sailplane which is dropped will have some value of CD which affects its downward velocity. The bullet or sailplane which is launched forward will have some value of CL which affects is downward velocity. If you claim that it is OK to consider the two components of velocity separately, you are claiming that it is OK to substitute a body's CD in one direction for its CL in the perpendicular direction. This is quite clearly not true, even though it intuitively seems like the right thing to do.

banktoturn

Looking at my calculus book, ballistic calulations ARE to be made by seperating the horizontal and vertical components by seperating them into vectors. The sum of these vectors is to equal their net forces. Therefore, it is perfecly plausible to seperate these forces.

As for the nose-high aoa, that is true but to a VERY LITTLE amount. At supersonic speeds, a nose high aoa of a body of revolution produces an amazingly strong destabilizing action. So, if it was very high it would most likely overcome gyroscopic forces and tumble.

To the original qeustion: any lift produced would require the nose-high aoa cited by banktoturn, and that is only created by it falling in the first place. So, it does not rise.

--Alex

[Sparky Paul]

If a rotating bullet "rose" due to lift, the very successful introduction of the MkII gyro stabilized gunsight in the Spitfire (and others) would not have made aces of otherwise mediocre shots particularly in "curving flight".
The gyro computed the rate of turn of the pursuer, with the presumption the target was being held in the middle of the gunsight combining glass during the chase.
The computations in the analog computer attached to the sight used normal ballistics equations, and predicted the lead needed to strike the target at whatever range the target was. The pursuing pilot would have set the wingspan of the target into the computer, and with a thumb control on the throttle, adjust the graticule in the combining glass to match the wingspan of the target as it manuvered. This sight extended the kill range of the Spitfire to 600 yards, with as much as a 50° deflection.*
.
A rotating bullet developing "lift" would precess off the straight line trajectory, making accuracy impossible at any range.
.
*ref: "Late Marque Spitfire Aces 1942-45, 4 Major Improvements" Osprey Aircraft of the Aces .5

[robert harik]

I think this is where the Engineer differs from the Mathematician. Theoretically, we can't even measure the physics of aerodynamics well enough to come up with a solution of infinite accuracy. For most purposes, it is safe to say the fired and dropped bullets will land at the same time. We might be off by a nanosecond or two, but does it really matter, and could the offset be reliably repeated anyway?

Most equations involving air-friction on a moving body are actually estimated "fits" based on sampled data. Number-crunching to 5 decimals of precision is pointless.

Yes!
For all practical purposes they hit the ground at the same time.
If you dont believe it, go try it.

[raptor22]

Yes!
For all practical purposes they hit the ground at the same time.
If you dont believe it, go try it.

That's be pretty hard to do, since seeing the bullet at that kind of speed and range accurately would be quite difficult.

--Alex

[mwraight]

The dropped bullet and the fired bullet are both subject to the law of gravity the instant they leave the object supporting them - the hand on one side and the barrel of the gun on the other.

If only I had the ability to draw in a program I could show you a simple way to perform this experiment in independence of motion in your home (without firing a gun off!).
I'll have to see if I can come up with something.

I knew this would drive everyone nuts! It's just one of those things. I didn't believe it until my good friend (a physics major in college) proved it!

[robert harik]

That's be pretty hard to do, since seeing the bullet at that kind of speed and range accurately would be quite difficult.

--Alex

You must not do alot of shooting, you dont have to see the bullet fly, just hit.
Make a rest for your rifle at the height you want, tape a level to the barrel, and shoot away. Have someone drop a bullet when you pull the trigger, when your bullets hit,( if your too blind use a spotting scope) you each call out. We ended up using steel targets( one down range, and one at our feet) and the "Clinks" were right on( at close range).
I have done it at various heights and it works as advertised.

[mwraight]

Hopefully you'll all be able to figure out the basics of the experiment from my rudimentary drawing. Computer illustration is not exactly one of my finer abilities. Anyhow, you build this simple thing to shoot one penny forward while the tray is pulled out from underneath the other penny. This is the same as shooting one bullet and dropping the other...although at a slightly reduced velocity!

Oh yeah, and the pennies get placed on the tray...they don't magically hover out in front of it as the illustration shows. I didn't want to spend the time placing the pennies on the tray. I'm lazy...so shoot me!

I also found the following website http://www.crocodile-clips.com/absorb/AP5/sample/010105.html It explains everything we're discussing a little better and shows the same experiment - although electrified. Very cool. It has little interactive things you can do - firing a cannon at certain angles and velocities, driving a motorcycle off a cliff...

[raptor22]

You must not do alot of shooting, you dont have to see the bullet fly, just hit.
Make a rest for your rifle at the height you want, tape a level to the barrel, and shoot away. Have someone drop a bullet when you pull the trigger, when your bullets hit,( if your too blind use a spotting scope) you each call out. We ended up using steel targets( one down range, and one at our feet) and the "Clinks" were right on( at close range).
I have done it at various heights and it works as advertised.

I see. That's smart to time the noises. I've been proven wrong about that.

Used to shoot occasionally with friends, but none of us have a scoped rifle I have used (and I only have a pelet gun that doesn't count). I have to admit that usually the attenion is given to the semi-automatic rifle with the quick-change clip :D :D :D.

I have tried it for arrows, though, and it works (I like archery).

--Alex

[banktoturn]

I think this is where the Engineer differs from the Mathematician. Theoretically, we can't even measure the physics of aerodynamics well enough to come up with a solution of infinite accuracy. For most purposes, it is safe to say the fired and dropped bullets will land at the same time. We might be off by a nanosecond or two, but does it really matter, and could the offset be reliably repeated anyway?

Most equations involving air-friction on a moving body are actually estimated "fits" based on sampled data. Number-crunching to 5 decimals of precision is pointless.

You're right; the bullets would hit the ground at very nearly the same time, probably. If it were a practical question, I would be fine with that answer. It wasn't. It was a theoretical, head-scratching kind of question.

banktoturn

[banktoturn]

ASSUMING a perfect situation for theory (vacuum, flat) they will hit at the same time.



Looking at my calculus book, ballistic calulations ARE to be made by seperating the horizontal and vertical components by seperating them into vectors. The sum of these vectors is to equal their net forces. Therefore, it is perfecly plausible to seperate these forces.

As for the nose-high aoa, that is true but to a VERY LITTLE amount. At supersonic speeds, a nose high aoa of a body of revolution produces an amazingly strong destabilizing action. So, if it was very high it would most likely overcome gyroscopic forces and tumble.

To the original qeustion: any lift produced would require the nose-high aoa cited by banktoturn, and that is only created by it falling in the first place. So, it does not rise.

--Alex

Alex,

The forces can be legitimately decomposed into horizontal and vertical components, but the components of velocity used to calculate those forces cannot be.

banktoturn

[Sparky Paul]

Alex,

The forces can be legitimately decomposed into horizontal and vertical components, but the components of velocity used to calculate those forces cannot be.

banktoturn
.
???
And why is this, and how must the velocity diagram look?

[raptor22]

I don't see any reason why not. The velocity can easily be composed into two vetors. In fact, by definition, velocity is expressed by vectors and can be decomposed into any compnents as long as the net result is the same.

--Alex

[banktoturn]

.
???
And why is this, and how must the velocity diagram look?

I'm assuming your question is regarding my assertion that it is not valid to decompose the components of velocity in order to determine aerodynamic forces on a body. Consider a body, say a cylindrical one with a rounded nose and a blunt tail. We could measure the CL and CD of this body from all angles, and all relevant speeds. If that body is in motion relative to the air at some attitude, would it be correct for me to decompose the relative velocity of the air into perpendicular components of my choice, and use the CD and CL values for those angles to determine the net aerodynamic force? The answer is no, lift and drag simply do not behave that way. A math guy would say that this problem is "nonlinear", so that kind of "superposition" is not valid. Once I know all the forces acting on a body, it is perfectly OK to add them up, and you will get the correct net force. For the bullet problem, you can say that the both bullets have the same downward component of velocity, and conclude that they experience the same upward component of drag. The lift and drag acting on a body are the result of the overall flow of air relative to the body, and you can't calculate components of lift and drag resulting from components of the flow and add them up. It would be great, but it doesn't work.

banktoturn

[banktoturn]

I don't see any reason why not. The velocity can easily be composed into two vetors. In fact, by definition, velocity is expressed by vectors and can be decomposed into any compnents as long as the net result is the same.

--Alex

Alex,

You can decompose the velocity into any components you like. But if you calculate lift and drag on the body for each of the components into which you decomposed the velocity, and add them up, they won't add up to the lift and drag that result from the overall flow. It seems like they should, but they don't, because that's the way the universe was put together.

banktoturn

[Sparky Paul]

... For the bullet problem, you can say that the both bullets have the same downward component of velocity, and conclude that they experience the same upward component of drag. The lift and drag acting on a body are the result of the overall flow of air relative to the body, and you can't calculate components of lift and drag resulting from components of the flow and add them up. It would be great, but it doesn't work.

banktoturn
.
The component of drag is the same because the down velocity is the same.
A bullet dropping 4 feet at 200 meters takes the same time going the 4 feet down as a bullet dropping the same 4 feet vertically at the rifle muzzle.
Lift is zero. It's a symmetrical body, spinning around its zero alpha axis. If it does anything else and develops a side-force.. lift say, it wil depart from stable flight and yaw-tumble-precess off the trajectory.

[Viper Pilot]

. . . . . . Gravity force Fg = m*g*z

m = projectile mass
g = acceleration due to gravity
z = altitude

The resulting accelerations (remember F = M*a) are: . . . .


You got it right the second time, F=mA and the z=altitude is not in the equation.

Gravitational; force has nothing to do with the altitude.

[2000-redrider]

Here's something you guys might find interesting.

http://www.nennstiel-ruprecht.de/bullfly/faq.htm#Q4

[banktoturn]

.
The component of drag is the same because the down velocity is the same.
A bullet dropping 4 feet at 200 meters takes the same time going the 4 feet down as a bullet dropping the same 4 feet vertically at the rifle muzzle.
Lift is zero. It's a symmetrical body, spinning around its zero alpha axis. If it does anything else and develops a side-force.. lift say, it wil depart from stable flight and yaw-tumble-precess off the trajectory.

The bullet which is dropped has its downward progress slowed by drag. This drag would be calculated by using the CD of the body and its velocity at each point in time. We can make the assumption that this bullet maintains the same attitude as the fired bullet, and we use the values of CD appropriate for that attitude (ie, the CD for flow hitting the bullet from the side).

Now let's look at the fired bullet. If it maintains a constant attitude, it does not have a zero AoA. The bullet is travelling at a very shallow angle toward the earth, giving it a small positive angle of attack. There are two components of aerodynamic force tending to slow the fall of this bullet. One is the upward component of the lift force. The overall lift force acts at a very slight angle from the vertical, since lift is, by definition, perpendicular to the direction of flight, which is just slightly downward from horizontal. The drag force, which by definition acts toward the rear of the bullet's motion relative to the air, is just slightly upward. Thus, the net upward aerodynamic force is the vertical component of the lift plus the vertical component of the drag. Because the trajectory is so nearly horizontal, this sum ends up being most of the lift plus a small fraction of the drag. It is simply not true that the sum of these components is equal to the drag force of the dropped bullet. This is a fundamental characteristic of aerodynamic forces. I understand how counterintuitive this is, but it's just the way it is.

banktoturn

[10cent]

Viper Pilot is right. The FORCE of gravity has nothing to do with distance from the earth. However, the potential NRG does vary with height and that is what hamm gave the eqn for.

Also, if you neglect atmospheric/drag forces then a bullet fired from a horizontal gun and one dropped from the same plane would hit the ground at the same time.

Common sense would say that the forces experienced by the bullet from the air are negligable in comparison to the Kinetic NRG the bullet has but there could theoretically be pressure or drag forces caused by the barrel to open atmosphere transition which could push the bullet slightly up. A word of causion though: The force would have to be a uniform force over the bullet. Think about it, if it was caused by drag on the angled front face of the bullet then bullet would have to tumble through the air right? i.e a point force would cause a moment about the center of mass...

I think :D

[vintage1]

.

Now let's look at the fired bullet. If it maintains a constant attitude, it does not have a zero AoA.
banktoturn

It doesn't maintain a constant attitude in the way you mean.

Its nose does indeed point down, and gyroscopic precession probaly does shift it slightly sideways towards the end of its travel where its changing attitide rather faster, but its so way beyond accurate range that no one really cares.


If shells 'maintained constant attitude' being as how the fuses are on the ends of their noses, and the elevation on a large gun may well be up to 45 degrees or more, the bloody things wouldn't go bang when they hit the target.

[vintage1]

Also, if you neglect atmospheric/drag forces then a bullet fired from a horizontal gun and one dropped from the same plane would hit the ground at the same time.

:D

Even if atmospheric drag is taken into account, it stll holds (almost) true, because the vertical compnent of drag is almost te same in both case. However teh bullet that is fired will ultimately hit the ground nose down, which the one dropped may not, although even there its open to question..and that will slightly change the vertical drag component.

Its intesting to note that a shotygun has an effective range of only about 30m not because the pellets don't go much further, but because they have slowed so much due to wind resistance that they lack killing power. Whereas a similar charge is a single rifle bullet is lethal up to a mile or more. I am not sure exactly why this is so, I can only cnjecture that a spherical projectile has infinitely moredrag than a bullet shaped one, or does size in fact matter - yes, must be so because one can lob a 12 " shell about 15 miles...

[Sparky Paul]

Vint, in the "Hornady Handbook of Cartridge Reloading", they present the way a "ballistic coefficient" is determined.
In essence, it's the drag of the projectile under test compared to a "standard" projectile.
Hornady uses a standard projectile based on the Krupp Test Firings of 1881. This bullet is 3 calibers long, with an ogival head of two calibers radius, and was used to develop the Mayevski drag function used in the Ingolls Ballistics Tables.
A ballistic coefficient of a bullet, or a desired b.c. can be used to compare the results between shapes and sizes of bullets.
The closer the b.c. is to 1.000, the better the bullet approaches the Krupp shape in performance.
In the Hornady tables for instance, sharply pointed bullets outperform the round nose style in terms of velocity for a given amount of powder, based on allowable chamber pressures. They brag on their "extremely high ballistic coefficients", which for example their 6.5 mm series ranges from .441, their 140 grain spire point to .251, their 160 grain round nose. The former drops 68 inches at 600 yards, the latter 126 inches at the same distance, corresponding velocities being 1911 fps and 1493 fps, fired at the same muzzle velocity.
I don't have any information on round ball projectiles, other than in the Hogdon book in the muzzle loader section, where the low muzzle velocities of shotguns are listed... This is due most probably to the much higher weight of a shotgun load, typically more than an ounce. Blasting that amount of mass with a reasonably sized shoulder weapon to rifle velocities probably can't happen, the chamber pressures being too high, to say nothing of the recoil.
.50 caliber rifle shooters can attest to this.
.
Anyway, I'm going to have to regard banktoturn's "counter intuitive" as another of saying "wrong", since none of the literature I've ever encountered considers lift on a seriously spinning body, except to mention the precessional effect a force normal to the direction of motion exerts on spinning bodies.
As we know this doesn't happen, the postulated lift force doesn't exist.

[robert harik]

Round shot has a really low ballistic coefficient compared to streamlined bullets.The rule of thumb is the BC of spheres is equal to their sectional density.
00 buck shot loses about half its muzzel vel. at 150 yards, compared to streamlined ( high BC) bullets that lose less than 15%.
Even at 150 yards 00 buck is still moving at over 600fps and will kill you dead.

[raptor22]

I'll add why this is:

at high transonic comressible speeds (mach 1.0-1.2) or supersonic speeds, a blunt nosed projectile, aifoil LE, or anything else will not produce a nice oblique shockwave. It will produce a region of somewhat stagnant super high pressure subsonic air with a normal shock in from of it. Normal shocks destroy supersonic flow, and are to be avoided if even close to acceptable supersonic performance is to be found.

--Alex

[djindivik]

Whats up ? The weather no good for flying ??? He he. Ray

[flieslikeabeagle]

...and one more reason why larger projectiles go further: the kinetic energy in a projectile is proportional to its mass, which in turn is proportional to its volume. The drag force, on the other hand, is proportional to its area of cross section.

If you scale up the size of a projectile, keeping its shape the same, the mass grows faster than the area of cross section. This is because the area of cross section is proportional to size squared, while the mass is proportional to size *cubed*. For instance, if you double all the dimensions of a bullet, its area of cross section quadruples, but its mass increases *eight* times. That massive bullet will experience four times the drag of its small counterpart, but it will pack eight times as much kinetic energy (assuming its fired at the same muzzle velocity), so it should go roughly twice as far before its energy is spent fighting the air drag.

Incidentally, this same scaling effect is the reason why an ant can fall ten stories, and walk away unhurt, while a man cannot. It's also the reason why large animals have disproportionately thick legs - think elephant vs gazelle.

-Flieslikeabeagle

[raptor22]

Good point. Also, the small RN of the tiny bullets gives them alot more drag to deal with.

--Alex

[jlk]

To really muddy the water, if we consider a clock on each bullet that starts the instant the fired bullet leaves the muzzel of the rifle with the second bullet dropped simultaneously and the clocks stop the instant the bullets each strike the ground. Will the clocks read the same time passage or will one be more or less than the other? If the clocks do not read the same did the bullets strike the ground at the same time?:confused:

Opinoins or answers? :D

Jordan

[raptor22]

How accurate are the clocks?

If they are REALLY accurate we can add in that times slow a very tiny bit at those speeds. So, they can hit the ground at the same time with the clocks reading differently!

--Alex

[jlk]

Yes, they hit at the same time with the clocks different by how and why can that be so?

Jordan

[Neil Stainton]

I agree with BankToTurn. It would be interesting to see a numerical estimation and see how much longer it takes the fired bulet to hit the ground.

However this does imply that there will be a side force on the bullet which would affect accuracy at high range, which should be documented. Does anyone know of this effect?

Neil.

[RyanPSU21]

Why would time passage be any different at the speed of a bullet then at stand still? It's not like we are shooting bullets near the speed of light.

If anything, the logic of banktoturn's theory that the drag force is going to make the fired bullet hit the ground slower would make the bullet hit the ground faster. Since that bullet has a much faster forward velocity when it's angle starts heading toward the ground the forward velocity which the drag is slowing down will make it hit the ground faster then the dropped bullet. If there wasn't forward velocity there wouldn't be drag opposing it. So therefore the bullet must have a velocity component heading down for their to be a drag component opposing it.

[2000-redrider]

Did anyone get a chance to look at the link I gave?

http://www.nennstiel-ruprecht.de/bullfly/faq.htm#Q4

How accurate is the info provided? Apparently this was from a presentation to the Association of Firearm and Tool Mark Examiners. The arrow with the line over it (all the way at the bottom) will get you to the top of the presentation.

[raptor22]

Why would time passage be any different at the speed of a bullet then at stand still? It's not like we are shooting bullets near the speed of light.

If anything, the logic of banktoturn's theory that the drag force is going to make the fired bullet hit the ground slower would make the bullet hit the ground faster. Since that bullet has a much faster forward velocity when it's angle starts heading toward the ground the forward velocity which the drag is slowing down will make it hit the ground faster then the dropped bullet. If there wasn't forward velocity there wouldn't be drag opposing it. So therefore the bullet must have a velocity component heading down for their to be a drag component opposing it.


I understand the logic. I just think that the conclusion that the fired bullet has a higher vertical drag component is wrong.

BTW, in theory, time slows in relation to us whenever an object is travelling at a greater velocity that ourselves. It just happens in a VERY, VERY tiny amount. Too tiny to notice for even the most precise mathemeticians at most speeds. So, I said that if the clock is unbeleivably accurate, it would read an alkmost infetesimally small difference. evenb if it strikes at the same moment.

--Alex

[Sparky Paul]

Did anyone get a chance to look at the link I gave?

http://www.nennstiel-ruprecht.de/bullfly/faq.htm#Q4

How accurate is the info provided? Apparently this was from a presentation to the Association of Firearm and Tool Mark Examiners. The arrow with the line over it (all the way at the bottom) will get you to the top of the presentation.
.
The stuff there makes the most sense I've seen.

[RyanPSU21]

I disagree with the link regarding which bullet hits first. I still would think that the fired one would get to the ground first. For the drag force to be effecting the sink rate the bullet has had to turn so it is pointing towards the ground. If it is pointing towards the ground some component of it's original forward velocity is now vertical giving it a faster sink rate. The only reason that the drag force is slowing it to the square of the velocity is that it has the velocity to begin with.

Someone correct my logic if I am missing something here.

Without a downward velocity component there is no drag proportional to the square of the velocity. The downward velocity of the intially stationary bullet will be increasing from the moment it is dropped. The downward velocity of the fired bullet increases at the same rate as it falls + forward velocity as the trajectory rotates downward so it has a net higher velocity downward then the dropped bullet. The only reason that the fired bullet has a higher drag force is because it is going faster downward. The only way that the drag force due to the velocities high rate of speed will slow the sink rate is if it is opposing a high component of sink velocity.

[banktoturn]

I disagree with the link regarding which bullet hits first. I still would think that the fired one would get to the ground first. For the drag force to be effecting the sink rate the bullet has had to turn so it is pointing towards the ground. If it is pointing towards the ground some component of it's original forward velocity is now vertical giving it a faster sink rate. The only reason that the drag force is slowing it to the square of the velocity is that it has the velocity to begin with.

Someone correct my logic if I am missing something here.

Without a downward velocity component there is no drag proportional to the square of the velocity. The downward velocity of the intially stationary bullet will be increasing from the moment it is dropped. The downward velocity of the fired bullet increases at the same rate as it falls + forward velocity as the trajectory rotates downward so it has a net higher velocity downward then the dropped bullet. The only reason that the fired bullet has a higher drag force is because it is going faster downward. The only way that the drag force due to the velocities high rate of speed will slow the sink rate is if it is opposing a high component of sink velocity.

You're missing several things. There is drag proportional to the square of velocity whether there is a downward component of velocity or not. The dropped bullet's drag acts vertically at all times, and grows as the bullet accelerates. The fired bullet's drag acts only horizontally at first, and gradually develops a vertical component as the downward component of velocity grows. Both bullets have a downward component of velocity. Forward velocity does not somehow "become" downward velocity, as you seem to be asserting.

I don't know which bullet would hit first. My assertion is simply that one cannot declare them to hit at the same time. The reason for this is that it is simply not valid to decompose the fired bullet's velocity into horizontal and vertical components, and then calculate a vertical component of drag using the vertical component of velocity. The incorrect assumption that this method would work is the basis for the assertion that the bullets would hit the ground at the same time.

banktoturn

[banktoturn]

I agree with BankToTurn. It would be interesting to see a numerical estimation and see how much longer it takes the fired bulet to hit the ground.

However this does imply that there will be a side force on the bullet which would affect accuracy at high range, which should be documented. Does anyone know of this effect?

Neil.

Neil,

I was curious about this as well, never having heard reference to this phenomenon in the area of shooting rifled weapons. It turns out that it does occur, and is documented. In sniper circles, you will hear it called "spin drift". More generally, the phenomenon of a spinning body generating lift perpendicular to the axis of the spin is called the Magnus effect. A Google search on either of these terms, perhaps with the word "bullet" included, will result in some relevant hits.

banktoturn

[RyanPSU21]

After saying I'm missing several things you then stated exactly what I just said.....

"The fired bullet's drag acts only horizontally at first, and gradually develops a vertical component as the downward component of velocity grows."

There is no increased drag preventing fall without first an increase in downward velocity so therefore fired bullet hits the ground first.

[banktoturn]

After saying I'm missing several things you then stated exactly what I just said.....

"The fired bullet's drag acts only horizontally at first, and gradually develops a vertical component as the downward component of velocity grows."

There is no increased drag preventing fall without first an increase in downward velocity so therefore fired bullet hits the ground first.

There is not necessarily an increase in downward velocity caused by the fired bullet's horizontal velocity. You are asserting that there is, as though the horizontal velocity gets "bent" downward. This is simply not true.

It doesn't seem to me that I said the same thing you did.

[Antonsoarer]

Interesting bullet discussion, if you visualise the bullet staying still and the gun going away from the bullet …what changes?… Nothing! Or another visualisation would be two bullets back to back being fired simultaneously while falling to earth. They would acquire a trajectory that would place them some distance from the predicted landing point and that’s it.

If some one stood on the back of a train doing 500mph and shot a horizontal gun from the back off the train (bullet speed = 500mph), a well placed bystander at the side of the track would see the bullet fall on the ground the same as though it had been dropped by him, it might be hot and spinning.

Some assumptions, it all happens in a vacuum so no turbulence or drag etc. Even not in a vacuum the last statement I believe would still hold true. That the gun is exactly horizontal and the barrel imparts no deflection as the bullet exits the barrel, that a spinning bullet would not walk sideways through the air or react like a gyro to turbulence or other external influences.

[Neil Stainton]

> if you visualise the bullet staying still and the gun going away from the bullet …what changes?… Nothing!

Yes it does the gun now has the big drag force acting on it, not the bullet.

Neil.

[Antonsoarer]

Neil, See assumptions!

Also the gun will describe a trajectory unique to the gun shape and weight etc. Shape will only be relevant if we are not working in a vacuum and or the barrel is not center of the gun mass and the firing event introduces gyroscopic considerations.


Tony.

[PHaze]

From the post that started this thread:
one person believes that when a bullet is fired on a completeley flat surface with no wind conditions and normal earth atmospheric conditions that the bullet will rise slightly before falling

What about ground effect? It seems that if the bullet were fired close to the ground, the flow of the air being displaced by the bullet would be blocked by the ground, thus creating higher pressure under the bullet than above it. I don't know if this lift would be strong enough to actually cause the bullet to rise before it started to fall, but given that the amount of lift generated would increase with increased speed, it seems that given a fast enough bullet, the lift would be strong enough to cause the bullet to rise initially.

[Neil Stainton]

> Neil, See assumptions!

Hang on! They weren't there when I made my post. You really shouldn't change the meaning of a post when someone has already made an answer based on its original meaning.

> Also the gun will describe a trajectory unique to the gun shape and weight etc.

Agreed, but we are talking about bullet trajectories in this discussion.

Neil.

[Neil Stainton]

> What about ground effect?

Isn't ground effect mostly a reduction in induced drag? I could be wrong here. As the bullet will have practally no induced drag this will be a third order effect.

Actually if the bullet was trajecting very close to a smooth flat surface wouldn't Bernoulli effect cause a suction between the bullet and the surface, which would increse its vertical velocity?

Neil.

[Sparky Paul]

"trajecting"?????????????? :)
.
We've introduced relativistic effects...inconsequential at bullet speeds,
ground effect, active for 1/2 the bullet's diameter at best, and now..
new words.. :D
The Magnus effect might have some bearing on this situation.

[raptor22]

Ground effect effectively increases the AR, by decreasing the upwash and downwash of the wing. Since the bullet is not, for any practical purposes, producing either of these we do not need to worry about it.

--Alex

[PHaze]

Isn't ground effect mostly a reduction in induced drag?
I just did some quick reading, and it looks like there are actually two different ground effects. There's "span dominated ground effect" which is a reduction in induced drag, and there's "chord dominated ground effect" or "ram effect" which I guess is the type I described. I found a page describing the two types of ground effect. (http://www.turkstra.com/k-designs/WIGTheory.htm)

We've introduced relativistic effects...inconsequential at bullet speeds,
ground effect, active for 1/2 the bullet's diameter at best, and now..
new words..
I realize my hypothesis will not make much difference in the flight of a bullet shot under normal circumstances, but this whole discussion is hypothetical anyway. So if a bullet *is* shot parallel and very close to the ground, will ground effect indeed cause it to go up before it goes down?

The Magnus effect might have some bearing on this situation.
Hehe, I thought you were joking here, but I googled it and learned that there is a Magnus effect. It's what makes spinning balls traject :p in a curved manner. From what I read, it would have an effect in this situation only if the bullet had a front or back spin instead of a side spin. Or maybe if you use special magnus effect bullets. That's where the .44 magnum gets its name you know. ;)

[Sparky Paul]

I use .44 Mag bullets mashed flat as ballast. A 240 grain bullet is almost a 1/2 oz.
Magnus probably interests golfers more than ballisticians worrying about trajecting their ball so that it hits the really large one it started from.

[raptor22]

I doubt magnus is a huge issue except at very large ranges. I bet the paris gun had to deal alot with it.

The post on "magnus effect" bullets reminded me of something in paintball; the flatline barrel. It is designed to put a backspin on the paintball, so it doesn't curve and the backspin prosudes lift. This greatly increases the range and accuracy.

--Alex

[Purdue Aero Man]

according to my old prof, NASA guy, real smart....developed a way to determine wind speed out on the ocean by using "useless" signals from GPS sats, the affect was first noticed during the napoleanic wars.