A lot of words are spewed forth in the War on Drugs, and when the drug war is combined with trouble at the border, the result is quite predictable. Every once in a while, though, I see things that push credulity beyond what I consider acceptable.
I saw a recent example in this AP story, headlined “Drug war sends bullets whizzing across the border“:
…one bullet came across the Rio Grande, crashed through a window and lodged in an office door frame at the University of Texas at El Paso. Police are also investigating reports that another errant round shattered a window in a passing car. Witnesses at a nearby charity said at least one bullet hit their building, too.
El Paso police spokesman Darrel Petry said authorities have only confirmed the single bullet found at the university, but it is possible that several other shots flew across the border.
“As a local municipality, we are doing everything we can,” Petry said. “Looking where we’re at, the community we live in, that’s all we’ve got. It’s the reality of life here in El Paso for right now.”
Officers say the types of bullets used in the drug war can travel more than a mile before falling to the ground.
There’s no question that bullets being fired across the border constitute a growing problem, but I have a serious problem with that last sentence, because it implies that there’s something uniquely awful about “bullets used in the drug war” as opposed to what I guess the AP thinks are “normal bullets.”
There are no normal bullets that won’t travel more than a mile before falling to the ground. The most typical, garden variety rounds (ranging from .22 to .44) all travel to maximum distances in excess of a mile. The following chart displays the range of ordinary handgun bullets of the sort sold today.
Notice that there is no bullet available which if it is fired straight will drop to the ground in has a range limited to less than a mile.
So, while it is technically correct to say that “the types of bullets used in the drug war can travel more than a mile before falling to the ground,” it is very misleading. The .22 (often thought of as the smallest commercially available gun) can kill from a mile away. And while I doubt the .22 is the weapon of choice for the Mexican drug gangs, it will do as an illustration.
Of either AP ignorance or anti-gun bias.
UPDATE: The correction came from commenter “Lt York,” who pointed out that “all listed bullets, fired straight, (level, parallel to the ground, I assume you mean…) would hit the ground in less than a mile,” which means that the maximum ranges listed have to factor in elevation. So I shouldn’t have said “if it is fired straight.” As to how many bullets are fired exactly straight, who knows?
OTOH, what is “straight”? Does the word necessarily mean “level”? And is it really possible under the laws of physics to fire a bullet in a mathematically absolute straight trajectory? (At some point both gravity as well as the curvature of the earth would come into play.)
No matter how I look at it, it seems that “straight” was a poor word choice.
MORE: Not to digress, but because of my poor word choice I found myself wondering whether it is theoretically possible for any bullet to ever be fired in a mathematically truly straight trajectory. There’s an interesting theoretical discussion here about bullets fired in space. Yes, they can be fired, but the consensus seems to be that if they are within the earth’s gravitational field, they will orbit the earth for a very long time; otherwise they will head for the nearest influential planet
If a bullet is fired outside the Earth’s gravitational field, it will be acted on by the nearest large mass; The Sun, The Moon, one of the planets, the Milky Way, etc. The bullet will turn toward that large mass and what happens then depends on the parameters. Since no gun has ever been outside the Earth’s gravity, it’s kind of moot up until now. The effect of a vacuum on the bullet would probably give it a slightly higher muzzle velocity since the bullet doesn’t have to expend energy pushing the air out of it’s way as it leaves the barrel. You would have to measure the aerodynamic drag of the bullet in the atmosphere and add that velocity to the muzzle velocity. It won’t be much, but it might be measurable withing the repeatability of loads.
At some point, it seems that a bullet in zero gravity might be going “straight.” But if pull (however slight) is always being exerted by one planet or another, is there really such a thing as zero gravity in the mathematical sense?
Might there be true zero gravity beyond that area we circumscribe as delimited by the Big Bang, commonly called “the Universe”?
Comments
37 responses to ““bullets used in the drug war””
Actually, (and it is a technical argument) due to bullet drop, all listed bullets, fired straight, (level, parallel to the ground, I assume you mean…) would hit the ground in less than a mile.
However, any upward angle transferred at firing would greatly increase range, thereby making the gist of your argument correct.
From Mike on the Rifle Team.
Actually, (and it is a technical argument) due to bullet drop, all listed bullets, fired straight, (level, parallel to the ground, I assume you mean…) would hit the ground in less than a mile.
However, any upward angle transferred at firing would greatly increase range, thereby making the gist of your argument correct.
From Mike on the Rifle Team.
Thanks for the correction. I guess “maximum range” is not measured on a straight line.
You have to define your terms. Do you mean straight in the abstract, mathematical ideal? Or straight as best we can measure? Do you mean truly zero gravity, or measurably zero?
The ideal is, simply, unobtainable in the real universe. Every object of even the smallest mass has some gravitational field, including the shooter and the gun; but it takes a pretty large mass before that gravity is measurable, so we don’t usually notice. (Personally, I’m convinced that my couch has an abnormally large gravitational field, explaining why I’m stuck here so often.) If you imagine a shooter out in deep intergalactic space where the nearest large objects are trillions of miles away, those objects still have some non-zero impact. Furthermore, unless the shooter were aligned precisely parallel with the angle of the gun, his center of mass would lie off the line of the bullet, and would exert some non-zero pull on it that would deflect it.
Could you measure these deflections? Well, I suppose if you have the technology to fly out to deep interstellar space, you might have instruments sufficiently advanced to measure them. Using today’s technology? Probably not. But you asked about a mathematically truly straight line, and that’s never going to happen.
The closest we get to travel in a straight line is light; but even light bends in a gravitational field. One way of interpreting the equations of gravity is that space itself bends in a gravitational field.
That should’ve been “trillions of light years away”. Trillions of miles in space is just a long way down the road to the chemist’s.
And is it really possible under the laws of physics to fire a bullet in a mathematically absolute straight trajectory?
If you fire one bullet exactly parallel to the ground and drop an identical one at exactly the same time, they will both hit the ground at the same time. The one dropping goes straight, the one fired curves.
I would say that you could theoretically do the math and figure out at exactly what angle to use to fire your bullet and it could go straight for a short period of time.
Very short.
Although….. maybe you could shoot straight at something with its own gravity. It would have to not have a rotation.
Or you shoot a steel bullet at a gigantic magnet.
That seems the most promising.
So there really is no such thing as a straight shooter?
Another illusion shattered!
Wow, that got existential, quick.
Ok, First off:
“If you fire one bullet exactly parallel to the ground and drop an identical one at exactly the same time, they will both hit the ground at the same time. The one dropping goes straight, the one fired curves.”
No. Travel time is significantly longer [by a factor of as much a 5, depending on the round] for the fired bullet.
2. A 30-06* Springfield round is a rifle round. A rifle is a long gun has a grooved/rifled barrel that fires single bullets. A handgun is a firearm designed to be held and operated by one hand, with the other hand optionally supporting the shooting hand.
3. The whole “what is straight” debate is not really applicable to the scenario I envisioned. The scenario is a man firing in standing, kneeling or prone position, over relatively flat ground, at a target a mile away at his same elevation. For our purposes, we can quantify straight, parallel or level as the vector an Army or Marine sniper would use.
4. Yes, in attempting to make a kill at long ranges, the curvature of the earth [along with a host of other factors including, windage, temperature and humidity] must be calculated.
That is all from the Rifle Range today…
“the vector an Army or Marine sniper would use,… if he was instructed to fire straight at the target, rather than attempt to hit the target, which would require elevating the gun.
“Might there be true zero gravity beyond that area we circumscribe as delimited by the Big Bang, commonly called “the Universe”?”
Even if you got outside the expanding Universe to a place where there was nothing, there would still be the Universe to exert a gravitational pull.
Sorry Lt York- two bullets of the same mass one dropped and one fired level with the ground at the same height, will hit the ground at the same time. Well, there could be minute differences due to air resistance but it would require very precise measurements to see the difference.
“Sorry Lt York- two bullets of the same mass one dropped and one fired level with the ground at the same height, will hit the ground at the same time. Well, there could be minute differences due to air resistance but it would require very precise measurements to see the difference.”
You are grossly incorrect.
What is your expertise?
Mine is as a graduate of U. S. Army sniper training at Kemper Military College.
How many Camp Perry matches have you shot in?
The travel time on a 30-06 alone, depending on the charge, can be over 3 seconds.
A bullet dropped takes less than 1 second using the standard formula of Time = the square root of 2(Distance)/g.
.78 seconds dropped from 3 ft. to be precise.
The travel time for a fired round, particularly from higher velocity weapons, is significantly more.
Take for example, the 30-06. The force applied actually makes the bullet hit above the aiming point after 100 yards:
From Guns & Ammo Magazine’s website:
* 100 yards: 1.7 inch rise above the muzzle for a 180-grain bronze tip round.
The force applied adds travel time.
Or perhaps I have it wrong…and missed all those time on target calculations…
LT York, yes, a bullet fired will hit the ground at exactly the same time as one dropped.
Discounting air resistance.
The vector perpendicular to gravity does not affect the gravity vector.
In other words, side to side vectors do not affect up and down ones.
That’s physics.
I’ve done the experiments in lab settings and the math works every time.
Now, if you’re firing at an angle, well, that changes everything because there will be a component going up as well as going side to side.
You don’t fire your .30-’06 directly at a target that’s 300 yards away, you aim a little up.
For instance, look at the sights on a Garand, they have notches at every 100 yards, only at 100 yards do you fire straight, at 600 yards, you’re firing upward at quite an angle.
They figured those notches out using the equations I’m talking about.
Why don’t you come out to the range?
I’ll show you.
You simply don’t know what you are talking about.
I’ll avoid arguing with someone who has no experience except inside a lab.
Drop a paper ball from 5 feet, and throw it at varying velocities from the same height.
And, pray tell, what does an orbiting body use to avoid crashing into the earth?
If velocity didn’t matter, then orbital physics would be invalid.
Dude, what was wrong with my Garand proof?
At only 200 yards you need to elevate the sights to hit your target, you don’t shoot straight at it. At 500 yards it’s quite a dramatic angle.
A .30-’06 has a muzzle velocity of around 2700fps, 200 yards is 600 feet, that’s under a quarter second and you’re already aiming up a little.
I have a couple Mausers, at 600 yards I’m aiming way over the target.
How long does it take for an 8mm bullet to travel 500 yards when fired by a Mauser?
It’s around 2800 fps muzzle velocity. 500 yards is 1,500 feet, so it travels there in just over a half second and the muzzle is pointed way over the target. Because the bullet is falling because of gravity, which fall is not affected whatsoever by its horizontal vector, the fall is only affected by whatever upward vector you put on the barrel.
A bullet fired horizontally falls at exactly the same speed as one dropped. That’s basic physics.
I’ve done time on target equations, notice that the artillery tubes are not parallel to the ground, they’re pointed at an angle.
If you’ve really done them, then you know that the artillery tubes are pointed at an angle, not straight there.
Ask any physicist or engineer you trust if horizontal acceleration has any affect whatsoever on how a projectile reacts to gravity and he will tell you exactly what I just said.
I said, “Lab”, because that’s where I saw it proven. I didn’t believe it at first either, then I did the math and it worked each and every time.
We used zip guns to fire a projectile at one angle, 10 times to average it out, measured where it landed and then predicted where it would land when fired at a different angle. We broke its vectors from the first firing up into horizontal and vertical and then computed where it would land using different horizontal and vertical because of the different angle.
It just kept hitting the box.
The math works.
Thank God I’m not relying on you for cover fire…
I see you’re done, so am I.
Ask an engineer you know, it’s physics 101.
Good sir:
You are obviously an intelligent man.
And so I offer this humbly and respectfully:
I don’t need to see an engineer: I’ve done this for a living.
Orbital physics is a subject we must master to understand bullet trajectory.
However, for the purposes of comparison, we can make this easy.
Think of the space station or shuttle.
Ff you stopped the velocity on either, they would drop to the ground, or burn up in a matter of hours. [a guess about the duration of the drop.]
The velocity overcomes gravity, and therefore keeps them in orbit.
The same applies to a bullet, does it not?
Theoretically, if you could input enough force, and design a bullet that would not melt, and find a clear path, you could orbit the earth at there feet, could you not?
Respectfully…
Perpendicular vectors do not affect each other.
If you’ve ever done vector analysis, you would know that.
That’s physics 101. Ask any engineer.
http://www.math.hmc.edu/calculus/tutorials/vectoranalysis/
If you want to measure the result of two vectors (in this case, they’re the acceleration due to gravity and that due to the gunpowder), you put them nose to tail and measure them.
So the vector for gravity is constant, the vector for horizontal acceleration is variable (one we fire, the other we drop), the horizontal vector does not affect the vertical.
That means that the drop is the same not matter how big the horizontal vector is.
That’s for a bullet fired parallel to the ground.
In real life you have other factors, drag due to air resistance mostly, but perpendicular vectors do not affect each other.
Again, that’s physics 101.
So when I load more grains, the bullet does not go further, and spend more time in the air?
Sir, I understand your calculations, but they seem not to apply in this case.
The more force I apply to a bullet, the farther it will travel, and the longer it will stay in the air. There are no two ways about this question…
I understand what you are trying to say, but it does not apply here.
Fire a baseball out of a pitching machine. With a small amount of force applied, the ball travels only a few feet. More force = more distance. Enough force and no obstructions and you’d blast it out of the atmosphere…
I’d like to have the time to do the research and offer your a link, but I just don’t have the time.
Again, you say one vector does not affect the other, but you ignored my shuttle example.
I wish I had the time to sort through this, but let me assure you that if you applied enough force to a bullet that wouldn’t melt, you’d be able to overcome gravity and orbit, or leave the orbital path entirely.
Anyway, I respectfully disagree, but don’t have the time for the research, and therefore, as one gentleman to another, must concede the argument.
Have a good weekend, Sir.
Mike
To speak of the “kind of bullets used in the drug wars” sounds like the “kind of cars used in the drug wars.” Those cars will kill someone if they hit him hard enough! Must be special cars.
Your shuttle example is not relevant, orbital mechanics is different.
So when I load more grains, the bullet does not go further, and spend more time in the air?
Assuming you mean “grains of powder” not “grains of projectile”,
No/yes.
More force=more distance because the bullet travels farther in the same amount of time.
If I drive my car at 50MPH for 1 hour I go 50 miles.
If I drive my car at 20MPH for 1 hour, I go 20 miles.
Why the difference? I’m driving for one hour, right?
It’s because the object is going faster so it travels more distance in the same amount of time.
That’s also physics 101.
As I said, you add the vectors nose to tail.
Your downward vector is constant, more and less horizontal force means the projectile will travel more or less distance in the same time period.
The bullet is dropping at exactly the same speed no matter what happens perpendicular to gravity.
Physics 101.
Now I’m really done.
Ask an engineer you trust since you obviously don’t trust this one.
I have a degree in Mechanical Engineering with a minor in math.
I tutored physics, math and engineering in college.
The bullet has to push more air below it out of the way, to fall, the faster it goes.
So a bullet falls slower if it goes forward faster.
If you want to experience the effect personally, ride a bike in a heavy straight crosswind and notice you can’t go very fast at all. You’re pushing a lot more air up to the speed of the bicycle when there is air going by sideways.
Yes, but any slowing is not from gravity, it’s from air resistance. If the wind is up or down that would affect travel because it introduces another vector.
And no, it doesn’t fall slower the faster it goes.
If it is fired exactly horizontally, it falls at exactly the same rate no matter how fast you fire it.
If you point it up, for a while it falls slower because there is an upward vector, not just down for gravity and horizontal from the powder.
That’s why high velocity rounds are more accurate, they travel farther in the same amount of time so they get there before too much passes for them to fall.
Bullet A takes .5 seconds to fall to the ground.
If I fire it horizontally at 2800ft/s, it will travel about 1400 ft and hit the ground.
If I fire it at 5000ft/s, it will travel about 2500 ft before hitting the ground.
(about because the bullet is slowing but it doesn’t have much time to slow in .5 seconds).
That’s why you aim higher the farther your target.
I’ve done the equations, they work in the real world every single time.
That’s why I like physics, you can’t bullshit it.
Your pellet either hits the box or not.
I should also say about the bullets being pushed up by air below.
Yes, buy they’re round, and spinning if you’re firing out of a rifled barrel, so they’re being pushed down by the air above equally.
Wings are shaped to take advantage of lift and that’s a whole nother set of equations and vectors.
The Navier Stokes equations are nonlinear, which is how the vertical air drag comes to depend on horizontal velocity.
Vertical air drag is different, so the fall rate is different.
Try the bicycle experiment: the forward velocity of the bicycle corresponds to fall rate; the sideways air movement corresponds to the bullet’s horizontal velocity.
You can’t pedal very fast even though the two vectors are perpendicular.
What do you mean by “sideways air movement”? A cross wind?
What does a windy day with people pedaling have to do with perpendicular vectors?
Navier-Stokes is fluid motion, which motion has nothing to do with perpendicular vectors as there are vectors in any direction in fluid flow.
We’re talking solid objects acted on by gravity and a force due an chemical reaction.
As to “vertical air drag”, you’re talking about introducing another vector.
Perpendicular vectors do not affect each other.
I’m going to have to assume you’re just messing around now, trying to look smart by trying to get me into some stupid troll loop.
Right, take bicycling in a straight crosswind. So the crosswind is perpendicular to the bicycle’s motion.
If it’s a very strong crosswind, your speed is severely reduced.
Why? You bring the air in front of you up to the speed of the bicycle. If there’s a strong crosswind, you bring a lot of air up to the speed of the bicycle, because new air is being put in front of you by the crosswind all the time.
Even though the crosswind is not setting the air against you with a headwind component, it’s merely putting new (stopped) air before you to deal with.
You’re thinking in effect that the bicycle deals only with the “headwind” from its own motion working against it; whereas the major contributor to drag in this case comes from a crosswind perpendicular to the motion of the bicycle.
The analogy in the bullet case is that the motion of the bicycle is the fall to the ground of the bullet; and the perpendicular crosswind is the horizontal motion of the bullet through the air.
The drag on the bicycle that keeps it from going very fast is “lift” that slows the bullet’s fall to the ground, caused by that perpendicular vector.
Looked at another way, suppose the bullet falls unimpeded as you say, regardless of horizontal speed.
So at some fixed time t, the velocity downwards is vd, regardless of launch speed.
The total drag on the bullet is in the direction of its motion, and has magnitude proportional to its speed squared (there’s that Navier-Stokes nonlinearity!).
So at that time t, the bullet’s velocity is v = vh + vd (vectors)
and drag is
drag = Const*(vh^2+vd^2)*(vh + vd) (vectors)
So the vertical drag is
dragv = Const*(vh^2+vd^2)*vd
The vertical drag depends on vh! In particular if vh, launch speed, is bigger, the vertical drag dragv is bigger.
If vertical drag dragv is bigger, the assumption must be wrong that vd is independent of the launch velocity.
In particular, the bullet must fall more slowly downwards the faster it is launched horizontally, because the vertical drag opposing the fall is bigger than the assumption of independence says.
Sorry, drag should be multiplied componentwise by vd/sqrt(vd^2+vh^2) and vh/sqrt(vd^2+vh^2)
leaving
dragv=Const*sqrt(vh^2+vd^2)*vd
Were the drag proportional not to v^2 but to v, then dragv would have depended only on vd, and your assumption would have been right.
But alas, if you’re going twice as fast, you hit twice as much air twice as hard, and that multiplies the drag by four, not two.
I was right, trying to make yourself look smarter by getting me involved in irrelevencies.
Yes or no rhhardin
Do orthogonal vectors affect each other?
Certainly orthogonal vectors affect each other.
Suppose you always move at exactly 30mph.
Your velocity component northwards is affected by your velocity component eastwards.
Which is pretty close to what happens with bullets, except it’s drag (in particular the component of drag that’s “lift”) that’s the focus of attention.
Certainly orthogonal vectors affect each other.
Bzzt. Wrong answer. Johnny, tell him about his parting gifts.
Suppose you always move at exactly 30mph.
Okay, I supposed.
A speed is not a vector.
Your velocity component northwards is affected by your velocity component eastwards
Buh? So if I run directly east that affects my northward running how? I’m not running north, I’m running east. So I’m not going north.
Or are you saying that if I leave Los Angeles and run directly east I’ll end up in Nova Scotia?
Or are you trying to say I’m running both north and east at the same time?
In which case, I would be running northeast and there would not be orthogonal vectors.
You know, except you could break your northeast vector up into a “north” and “east” vector.
Which, if you put them nose to tail, would end up at the same place as your northeast vector.
Hmmmmmm, I’ve seen you here before and I don’t think you’re just a troll, that’s why I kept responding, but now I’m really thinking you think you’re being funny.
And since you’re all excited about drag, which is pretty insignificant considering how much time the bullet is in the air, I’ll cut and paste from my second comment on the subject.
quote
LT York, yes, a bullet fired will hit the ground at exactly the same time as one dropped.
Discounting air resistance.
The bullet travels for less than a second, that means drag might affect it by a whole .0000005 seconds.
If you’re shooting a bullet 80 miles, drag would become significant. At less than 1 mile, well, it’s just not a factor.
But all that doesn’t matter because you got the first answer wrong.
So no credit, no points. A solid B+.
http://www.nydailynews.com/news/politics/2009/12/14/2009-12-14_b_for_me_bam_sez_first_couple_hosts_oprah_on_abc.html
Your velocity northward is affected by your velocity eastward because both velocities are in the same equation, namely the equation that says you’re going exactly 30mph.
Just in the same way, the downward velocity of the bullet is in the same equation as the horizontal velocity of the bullet. Unless the equation is linear, and it is not linear, the one velocity affects the other, by way of horizontal and vertical drag each depending on the other.
No, you are introducing another force.
And what the hell north and east equations are you talking about?
I have to admit, I don’t understand some people’s motivations.
Are you one of the ones who thinks it’s funny to hassle people who try to treat you like an adult?
The equation of motion for the bullet is the equation in question. The horizontal and vertical vectors couple because of that equation, namely square law drag.
Just as, analogy, the equation always-going-30mph couples the north and the east directions in that equation.
If you think drag is a small force for a bullet, you’re wildly underestimating it.
Look at it this way.
The air just below the bullet is brought to the downward speed of the bullet.
If you just drop the bullet, that’s about 4 feet of air.
If you fire the bullet, that’s about a mile of air.
A mile of air weighs a lot more than 4 feet of air.
That reduces the downward speed of the bullet.
You know, you’re right.
I was hit right in the face with a mile of air once.
That hurt.