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Gene L
25th August 2008, 07:06 PM
That's the question. I'm sure there is a formula, just as there is a formula for why bullets fired in a rain storm don't hit individual rain drops. (Don't ask me, I just know one exists. Something to do with both occupying the same space at the same time.)

Anyway, you see birds, which are pretty mysterious creatures, in a flock, all seeming to swerve at the same time. I've never seen a slo-mo of this, but they all seem to go the same way, with no one in particular in charge. I wonderd why they don't run into each other. Track and field runners, essentially going the same way, run into each other when competing. Birds asumptively aren't competing, but still, in HUGE flocks, they all seem to veer the same way at the same time.

Again, slo-mo films might reveal 1/10000 second difference (metric seconds, of course :)) in their wing movement. And it's a matter of occupying the same space at the same time, like a bullet in a rainstorm, but where missing is guided by intellegence and decision, if you can call a bird intelligent. You would think that a bird occasionally makes a wrong decision in the crowded skies, but apparently not.

Maybe this is a water-cooler topic with some, but I've never seen it addressed. Any ideas?

Brian-M
25th August 2008, 08:56 PM
That's the question. I'm sure there is a formula, just as there is a formula for why bullets fired in a rain storm don't hit individual rain drops. (Don't ask me, I just know one exists. Something to do with both occupying the same space at the same time.)

Somehow, I suspect this formula is Please don't try to bypass the autocensor.. Do you have a link to more info?

(I don't know if there are any formulas about flocks of birds, of schools of fish, but their behavior would be determined by instincts. You might need a different formula for each different species.)

AntiTelharsic
25th August 2008, 09:03 PM
http://en.wikipedia.org/wiki/Flocking_(behavior)
http://www.red3d.com/cwr/boids/

tsig
25th August 2008, 09:35 PM
That's the question. I'm sure there is a formula, just as there is a formula for why bullets fired in a rain storm don't hit individual rain drops. (Don't ask me, I just know one exists. Something to do with both occupying the same space at the same time.)

Anyway, you see birds, which are pretty mysterious creatures, in a flock, all seeming to swerve at the same time. I've never seen a slo-mo of this, but they all seem to go the same way, with no one in particular in charge. I wonderd why they don't run into each other. Track and field runners, essentially going the same way, run into each other when competing. Birds asumptively aren't competing, but still, in HUGE flocks, they all seem to veer the same way at the same time.

Again, slo-mo films might reveal 1/10000 second difference (metric seconds, of course :)) in their wing movement. And it's a matter of occupying the same space at the same time, like a bullet in a rainstorm, but where missing is guided by intellegence and decision, if you can call a bird intelligent. You would think that a bird occasionally makes a wrong decision in the crowded skies, but apparently not.

Maybe this is a water-cooler topic with some, but I've never seen it addressed. Any ideas?

When Birds Collide.

Amapola
25th August 2008, 09:36 PM
Umm... I think animals don't usually run into each other because they are paying close attention to tiny clues such as air or water currents, eye movements of flock mates and so on. They do occasionally collide. But I don't think it has anything to do with math.

Aepervius
25th August 2008, 10:11 PM
there is a formula for why bullets fired in a rain storm don't hit individual rain drops.

Considering the speed of rain, and the speed of the bullet , considering rain as an ideally constant repartition of water sphere in a volume, and considering the density of water in a rain storm, we can consider the bullet travelling a path described by a volume, and the relative speed make it so at first approximation all water sphere are fix in space. That would mean that there is a non zero probability that the bullet will have a few rain drop on its path.

The question is , do the shockwave before the bullet push away the raindrops before it hits the bullet ?

DevilsAdvocate
25th August 2008, 10:31 PM
I wonderd why they don't run into each other. Track and field runners, essentially going the same way, run into each other when competing.Track and field runners are confined to a certain space and are jockeying for position. Birds in a flock are not confined. The bigger the flock, the larger the space they take up. In a big flock flight there always seems to be considerable distance between birds so that they wonít collide. They simply spread out far enough from each other to have enough room.

They donít all change courses immediately. Some change course in reaction to an event. Others follow suit. It takes some time, but birds are pretty fast and one bird will be about as fast as the next bird. I think schools of fish move much faster.

DevilsAdvocate
25th August 2008, 10:34 PM
http://www.red3d.com/cwr/boids/That's very cool!

Jonnyclueless
26th August 2008, 12:39 AM
http://www.metacafe.com/watch/1489511/birds_collide/

rjh01
26th August 2008, 12:48 AM
With bullets it has a lot to do with weight. A bullet has a far bigger weight than a rain drop. So if a bullet came near a rain drop the bullet will not be deflected much.

A very small gun (air pistol?) may have a bullet that is lighter than a rain drop. However the probability that it will come near a rain drop is very low.

Edit. One of the above links does not work. Here is the right link.
http://en.wikipedia.org/wiki/Flocking_(behavior)

shadron
26th August 2008, 01:10 AM
Don't know about birds (or fish, for that matter), but the bullets in rain is simple. The bullet collides with them. The bullet is traveling faster than sound, so the raindrop isn't affected until it is rammed by the air being physically pushed out of the way by the bullet. The turbulence shatters the droplet and supplies heat enough to vaporize most of it. In it's small way, the raindrop absorbs some of the bullet's energy and slows it down with its demise. At long range, rain probably has a considerable cumulative effect.

Worm
26th August 2008, 05:03 AM
and that's why in the rain scene at the end of Lethal Weapon, both Martin Riggs and Sgt Murtagh had to shoot the baddy - one bullet just wouldn't have been enough.*





*Not necessarily true.

leon_heller
26th August 2008, 05:13 AM
Birds in a flock and fish in a shoal only need to keep tabs on their nearest neighbours. It's actually quite a simple strategy.

Leon

Gene L
26th August 2008, 06:16 AM
Don't know about birds (or fish, for that matter), but the bullets in rain is simple. The bullet collides with them. The bullet is traveling faster than sound, so the raindrop isn't affected until it is rammed by the air being physically pushed out of the way by the bullet. The turbulence shatters the droplet and supplies heat enough to vaporize most of it. In it's small way, the raindrop absorbs some of the bullet's energy and slows it down with its demise. At long range, rain probably has a considerable cumulative effect.


No, a bullet and a raindrop don't collide. A raindrop is pretty heavy, considering, and would explode a bullet traveling at high speed at most, or divert it severely at least. As I said earlier, there is a formula to explain why they don't collide, but I don't have it. As I recall, it has to do with the time a bullet is in the space that a raindrop is in, and the amount of raindrops in a that space at the same time, which is very, very small. It may be possible, but the odds are against it.

The size of a group of, say, 5 shots in rain will be the same as in clear weather. If one or two bullets hit a raindrop, the group would be widely and randomly dispereed, and that's not what happens.

What does happen with shooting groups in rain is the air is saturated with water (moisture, not raindrops) which makes it more dense. The bullets go low, but in the same size group as in dry weather.

Gene L
26th August 2008, 06:18 AM
Birds in a flock and fish in a shoal only need to keep tabs on their nearest neighbours. It's actually quite a simple strategy.

Leon


Humans collide. You see it in track events, fairly often. You see it on crowded streets as well. The differnce, I think, is birds and fish have 3 dimensions to avoid running in to other birds and fish, humans only have two.

Doubt
26th August 2008, 06:22 AM
No, a bullet and a raindrop don't collide. A raindrop is pretty heavy, considering, and would explode a bullet traveling at high speed at most, or divert it severely at least. As I said earlier, there is a formula to explain why they don't collide, but I don't have it. As I recall, it has to do with the time a bullet is in the space that a raindrop is in, and the amount of raindrops in a that space at the same time, which is very, very small. It may be possible, but the odds are against it.

The size of a group of, say, 5 shots in rain will be the same as in clear weather. If one or two bullets hit a raindrop, the group would be widely and randomly dispereed, and that's not what happens.

What does happen with shooting groups in rain is the air is saturated with water (moisture, not raindrops) which makes it more dense. The bullets go low, but in the same size group as in dry weather.

And your source for this info is?

Rob Lister
26th August 2008, 06:35 AM
And your source for this info is?

Is that just a nice way of saying, "Pfffft."

If so, I ditto.

I want to see the "formula" for why a bullet won't hit a raindrop. If I don't, I'm going to say, "Pfffft" outright.

quarky
26th August 2008, 06:40 AM
If you need to drive ten miles in a rainstorm, you will collide with less rain if you drive faster.

(officer)

Doubt
26th August 2008, 06:43 AM
Is that just a nice way of saying, "Pfffft."

If so, I ditto.

I want to see the "formula" for why a bullet won't hit a raindrop. If I don't, I'm going to say, "Pfffft" outright.

Nope. Pfffft would happen if there is no info to back up the claim.

Rob Lister
26th August 2008, 06:47 AM
If you need to drive ten miles in a rainstorm, you will collide with less rain if you drive faster.

(officer)

mythbusters did a show on this. I forgot the conclusion, but I think it was you would get "less" wet. that is not to say you wouldn't get wet. I think a bullet would have the same result.

Rob Lister
26th August 2008, 06:48 AM
Nope. Pfffft would happen if there is no info to back up the claim.

Well, so far....how much has been presented?

leon_heller
26th August 2008, 07:06 AM
I remember seeing an investigation on TV into whether it was best to run or walk through rain. I think that one got less wet if one walked, but I'm not certain. My mother always maintained that it was best to dart about at random to avoid raindrops, but I'm quite sure that was wrong.

Leon

Delvo
26th August 2008, 07:15 AM
Anyway, you see birds, which are pretty mysterious creatures, in a flock, all seeming to swerve at the same time.There's the problem. They don't swerve simultaneously. One swerves and then others nearby react by swerving, then others near them do so, then others near them do so. The swerving action spreads across the flock like a wave. (And it can be seen with just your eyes; no slow-motion camera needed.)

Ocelot
26th August 2008, 07:19 AM
http://www.snipershide.com/forum/ubbthreads.php?ubb=showflat&Number=637165

Doubt
26th August 2008, 07:35 AM
http://www.snipershide.com/forum/ubbthreads.php?ubb=showflat&Number=637165

I found that discussion too. But still no sources for the info.

Vaporization sounds plausible, but there are many kinds of bullets and guns and powder loads. Add in air resistance and you get a wide variety of velocities for bullets.

I doubt there is a single, simple answer for what happens when bullet meets water droplet.

GreyICE
26th August 2008, 07:58 AM
Considering the speed of rain, and the speed of the bullet , considering rain as an ideally constant repartition of water sphere in a volume, and considering the density of water in a rain storm, we can consider the bullet travelling a path described by a volume, and the relative speed make it so at first approximation all water sphere are fix in space. That would mean that there is a non zero probability that the bullet will have a few rain drop on its path.

The question is , do the shockwave before the bullet push away the raindrops before it hits the bullet ?
Of course not. The bullet is travelling faster than the speed of sound. The shockwaves, obviously, aren't. The bullet hits the raindrops.

Gene L
26th August 2008, 08:05 AM
And your source for this info is?


I'm glad you asked this, since it allows me to restate that I don't have the formula. I will continue to look, however.

But for the rest of it, the source is personal experience, and training as a LEO sniper at the Army Marksmanship Training Unit. The course was two weeks, and it was called the PC "Counter-Sniper Instructor's Training." They made a point that bullets don't hit raindrops. I'm not sure if they gave the formula, but gave a synopsis of why. I've read the raindrops/bullet thing several times and seen reference to the formula, but wouldn't be able to repeat it even if I saw it. I believe it from shooting rifles, and sure enough of the claim that I would be willing to bet you couldn't hit a falling raindrop with a bullet on demand.

Basically, relying on memory, the idea is to think of 1 bullet and one raindrop, both falling at the speed of gravity, with the bullet also moving at a forward velocity, and the infintesimally small time each is exposed to the other. For them to hit, they would both have to occupy the same space in that small time, and it ain't likely.

As for shooting groups in damp air, I know from experience that the groups shoot lower, since resistance is greater. A headwind will also make bullets hit lower, more air over the bullet (decreased velocity.) In dry air, groups will rise a bit becuse there's less friction. Groups at higher elevations, where the air is thinner, will rise a bit over sea-level groups. This is pretty well known among benchrest shooters. None the less, group size remains pretty much the same, impact just changes a bit.

Doubt
26th August 2008, 08:24 AM
I'm glad you asked this, since it allows me to restate that I don't have the formula. I will continue to look, however.

But for the rest of it, the source is personal experience, and training as a LEO sniper at the Army Marksmanship Training Unit. The course was two weeks, and it was called the PC "Counter-Sniper Instructor's Training." They made a point that bullets don't hit raindrops. I'm not sure if they gave the formula, but gave a synopsis of why. I've read the raindrops/bullet thing several times and seen reference to the formula, but wouldn't be able to repeat it even if I saw it. I believe it from shooting rifles, and sure enough of the claim that I would be willing to bet you couldn't hit a falling raindrop with a bullet on demand.


I can see where the probability of striking a rain drop may be low. But that is like saying nobody ever wins the lottery because the odds are so low.

Having been through a few military training courses, I do know they don't always have their facts straight. Last time I looked at it the land navigation manual it still said the reason the North Pole existed was that their were large iron deposits in Northern Canada.

Ocelot
26th August 2008, 08:25 AM
Basically, relying on memory, the idea is to think of 1 bullet and one raindrop, both falling at the speed of gravity, with the bullet also moving at a forward velocity, and the infintesimally small time each is exposed to the other. For them to hit, they would both have to occupy the same space in that small time, and it ain't likely.

That simply makes it unlikley that a bullets would interact with a particular raindrop, not impossible. Multiply that probability by the number of raindrops and the the liklihood increases. Essentially it is the liklihood of a raindrop being on the bullet's path at the time of firing. In heavy rain that seems quite likely indeed.

The question then becomes what happens when raindrop and bullet do interact. Does the pressure wave push the raindrop out of the way? Is the raindrop splattered with little effect on the bullet? Does the bullet explode?

Well I can answer that last one. Bullets do not explode when they hit water. In fact firing bullets into a hydroballistics tank is an everyday mundane experience for some.

Cuddles
26th August 2008, 08:30 AM
the idea is to think of 1 bullet and one raindrop, both falling at the speed of gravity

Speed of gravity?:rolleyes:

with the bullet also moving at a forward velocity, and the infintesimally small time each is exposed to the other. For them to hit, they would both have to occupy the same space in that small time, and it ain't likely.

Saying that any particular bullet is unlikely to hit a raindrop is not the same thing as saying that bullets simply don't hit raindrops. The former may be true, the latter certainly isn't.

GreyICE
26th August 2008, 09:04 AM
That simply makes it unlikley that a bullets would interact with a particular raindrop, not impossible. Multiply that probability by the number of raindrops and the the liklihood increases. Essentially it is the liklihood of a raindrop being on the bullet's path at the time of firing. In heavy rain that seems quite likely indeed.

The question then becomes what happens when raindrop and bullet do interact. Does the pressure wave push the raindrop out of the way? Is the raindrop splattered with little effect on the bullet? Does the bullet explode?

Well I can answer that last one. Bullets do not explode when they hit water. In fact firing bullets into a hydroballistics tank is an everyday mundane experience for some.
The pressure wave cannot build up in front of the bullet because the bullet is moving faster than the speed of sound. Pressure waves preceding a moving object (like a car or train) is a sub-sonic phenomena.

There is no pressure wave in front of a super sonic object. This is extremely obvious. I wish people would stop saying it exists.

AndyD
26th August 2008, 09:15 AM
I saw a short documentary years ago which included a POV of a fly on a breakfast table. The argument was that the fly thinks much, much faster than we do so, to it, we are moving in slow motion. Because of this, the fly escaped when someone tried to swat it with a newspaper.

I think there was a general correlation between metabolism, lifespan and the speed of thought.

If any of this still holds water (or ever did) then perhaps the same principle applies to birds (ever tried to catch one?). This might also explain why two birds can chase each other through the branches of a tree, seemingly without hitting a twig. Maybe, to them, it's really not that amazing?

Vorticity
26th August 2008, 09:27 AM
It's relatively straightforward to figure out whether a bullet in a rainstorm will strike a raindrop during its flight by calculating the mean free path of the bullet between successive collisions with raindrops. Let's do it:

Assuming a uniform number density of spherical raindrops in the space through which the bullet is flying, the bullet's mean free path will be*:


$\lambda = \frac{1}{\sigma\, n}\, ,$


where

\sigma = \pi {(R_b + R_r)}^2

is the cross section of bullet-raindrop collisions, R_b is the bullet's radius, R_r is the average raindrop radius, and n is the average number of raindrops per unit volume in the rain storm.

Let's find some typical rough numbers for these quantities, to get an idea of the order-of-magnitude of \lambda:

Suppose, for example, that we are using a 9mm bullet. Then

R_b \approx 4.5\times {10}^{-3}\, \text{meters}.

This (http://www.atmos.uiuc.edu/courses/atmos410-fa04/Z%20to%20other%20met%20quantities.ppt) PowerPoint presentation I found at the University of Illinois website mentions a distribution of 1000 1-mm raindrops per cubic meter. Supposing this is roughly typical, we then have:

R_r \approx 0.5\times {10}^{-3}\, \text{meters}

and

n \approx 1000\, {\text{m}}^{-3}.

Plugging these numbers into the formula above for the mean free path yields:

\lambda \approx 12\, \text{meters}.

Thus: 12 meters (on average) between bullet-raindrop collisions. Of course, the numbers I used were rough, so this may be off by a factor of 2 or 3 or whatever, but it does seem to suggest that a bullet will probably strike a handful of raindrops on its journey, unless you are shooting at a target only a few meters away.


*: This expression can be derived by considering the cylinder swept out by the combined bullet-raindrop radius moving a distance x. The number of raindrops in this cylinder is distributed according to the Poisson distribution. By looking at the probability that the bullet will have NOT hit a raindrop after a distance x, you can get this expression for the average distance between collisions. (I've left out most of the steps, but it's straightforward.)

By the way: The fact that the rain is falling makes little to no difference to this expression, since the bullet's speed and the raindrop's falling speed are so different.

leon_heller
26th August 2008, 09:57 AM
I saw a short documentary years ago which included a POV of a fly on a breakfast table. The argument was that the fly thinks much, much faster than we do so, to it, we are moving in slow motion. Because of this, the fly escaped when someone tried to swat it with a newspaper.

I think there was a general correlation between metabolism, lifespan and the speed of thought.



I always thought it was because flies detect the air movement caused by the newspaper long before it gets anywhere near them.

Leon

JoeTheJuggler
26th August 2008, 10:02 AM
No, a bullet and a raindrop don't collide. A raindrop is pretty heavy, considering, and would explode a bullet traveling at high speed at most, or divert it severely at least. As I said earlier, there is a formula to explain why they don't collide, but I don't have it. As I recall, it has to do with the time a bullet is in the space that a raindrop is in, and the amount of raindrops in a that space at the same time, which is very, very small. It may be possible, but the odds are against it.

The size of a group of, say, 5 shots in rain will be the same as in clear weather. If one or two bullets hit a raindrop, the group would be widely and randomly dispereed, and that's not what happens.

Sorry, but I don't buy this explanation at all.

As someone else pointed out, compared to a bullet, you could consider the falling drops of rain to be holding still. At any moment, in a nearly straight line (actually a segment of a very large parabola), there are bound to be rain drops in the line of fire. The question is what happens when they hit.

The energy is the mass and the velocity. A bullet has a lot*. Raindrops have very little. The effect of the raindrop on the bullet is pretty small (probably negligible in most cases). There would probably be a noticeable effect in heavy rain at a greater distance (hitting more drops). On the other hand, when shooting greater distances, I imagine people usually use a higher-powered gun (bigger bullets fired at a higher velocity, I guess?).

*This is why a bullet, even with relatively little mass can be fatal but killing someone with one blow of a 2 x 4 is pretty difficult. Better yet, killing someone by throwing a 2 x 4 at them from any distance is darn near impossible.

GreyICE
26th August 2008, 10:21 AM
Joe - two by fours tend to have a ton of kinetic energy, probably at least the same as a bullet. They don't kill because they apply their force over a larger surface.

As an example, knives and kevlar. Kevlar isn't great at stopping the force of the knife (especially if you put your body weight behind it).

shadron
26th August 2008, 10:29 AM
The pressure wave cannot build up in front of the bullet because the bullet is moving faster than the speed of sound. Pressure waves preceding a moving object (like a car or train) is a sub-sonic phenomena.

There is no pressure wave in front of a super sonic object. This is extremely obvious. I wish people would stop saying it exists.

Yes you are absolutely right. What you do have is a thin layer of air molecules that are being physically compressed by the bullet or by other molecules being pressed out of the way as it is making its way through them. They cannot extend that compression outward to the front but they do extend their disturbed behavior to the side at the speed of sound, creating the wakes of sound that appear to trail the object.

At or above the speed of sound one cannot propagate energy by compression, shear or transverse wave phenomena traveling through the medium, but that certainly doesn't mean you cannot accelerate particles of that medium as high as you need to (in fact, supersonically) in order to move them out of the way.

Rob Lister
26th August 2008, 10:30 AM
Ok. Back to the theory that birds don't collide. I don't buy it but failing me finding a video of birds colliding, I certainly can't prove it. I suspect they collide as often as cars on I-95. Not often, but it happens...and with less litigious results.

Opps! guess what...they do.

http://www.ebaumsworld.com/video/watch/750719/

all of somebody's base are belong to us.

erm...no lawsuit is pending, so far as I know.

GreyICE
26th August 2008, 10:50 AM
Yes you are absolutely right. What you do have is a thin layer of air molecules that are being physically compressed by the bullet or by other molecules being pressed out of the way as it is making its way through them. They cannot extend that compression outward to the front but they do extend their disturbed behavior to the side at the speed of sound, creating the wakes of sound that appear to trail the object.

At or above the speed of sound one cannot propagate energy by compression, shear or transverse wave phenomena traveling through the medium, but that certainly doesn't mean you cannot accelerate particles of that medium as high as you need to (in fact, supersonically) in order to move them out of the way.

Certainly true, but saying a layer a few molecules thick could possibly deflect anything is absurd. It's certainly nothing like the large and powerful pressure fronts that build up in front of near-sonic objects.

Those 'wakes' are in fact the shockwaves mentioned earlier, which is why shockwaves never deflect anything from the projectile (if they're occurring, the projectile is outrunning them, unless it suddenly goes subsonic).

Rob Lister
26th August 2008, 10:58 AM
Certainly true, but saying a layer a few molecules thick could possibly deflect anything is absurd.

Stop there and reflect upon your assertion. It is not only NOT absurd, it is a fact beyond refutation.

Let's at least be accurate in our debunking.

shadron
26th August 2008, 12:17 PM
No, a bullet and a raindrop don't collide. A raindrop is pretty heavy, considering, and would explode a bullet traveling at high speed at most, or divert it severely at least. As I said earlier, there is a formula to explain why they don't collide, but I don't have it. As I recall, it has to do with the time a bullet is in the space that a raindrop is in, and the amount of raindrops in a that space at the same time, which is very, very small. It may be possible, but the odds are against it.

The size of a group of, say, 5 shots in rain will be the same as in clear weather. If one or two bullets hit a raindrop, the group would be widely and randomly dispereed, and that's not what happens.

What does happen with shooting groups in rain is the air is saturated with water (moisture, not raindrops) which makes it more dense. The bullets go low, but in the same size group as in dry weather.

I'll reserve decision on this until you produce the math, but I really have doubts that it is saying what you think it does and if so, it is correct; it is much more likely some kind of professional rifleman's "urban legend". I've never heard of such a theory or law or statement to that effect in physics. The energy required to vaporize a droplet, and thus render it no different from any of the other gasses the bullet is passing through is such a small part of it's kinetic energy as to be undetectable. That plus the bullet's gyroscopic stability (the spin provided by the rifling) should keep it from deflecting noticeably. I can think of no principle that excuses a bullet from attempting to occupy the same space that a droplet may be in regardless of their relative velocities except the tunnel effect, which works for electrons over nanometers, not bullets over millimeters.

At rest, the probability that a drop and a bullet encounter is the probability that a drop occupies any volume equal to the bullet, so it depends on the volume of the bullet, the average volume of the drops and the "drop density" in the atmosphere. The effect of the bullet's velocity would seem to be to extend the effective length of the bullet, and therefore its volume; at infinite speed, it would occupy it's entire path volume (length of travel times its cross sectional area). So the bullet's velocity has the effect of increasing the probability of a collision, not decreasing it, up to a limit.

Skeptical Greg
26th August 2008, 12:43 PM
Joe - two by fours tend to have a ton of kinetic energy, probably at least the same as a bullet. ....

That has got to be so wrong, but someone who has more of this in their head will have to explain it.

I will start by saying, I do know kinetic energy increases with the square of the velocity; so you need to define the mass of the 2 x 4 , say a pound or two , then explain how you would accelerate it to a velocity that would give it the kinetic energy of a 8 gram bullet traveling at 2,500 fps or more ..


P.S

That raindrop, which you say will destroy a bullet, has so little kinetic energy, you might as well call it zero ..

calebprime
26th August 2008, 01:05 PM
That's very cool!

I'd like this to be my screen saver...

Gene L
26th August 2008, 01:17 PM
I maintain if you can't hit a falling raindrop on purpose, it's unlikely you'll hit it by accident. :)

A 168 gr. bullet traveling at 2800 fps (muzzle velocity of a .30-06) has something over a ton of kinetic energy. I've never figured how much fast you'd have to swing a 2 x 4 to get that, but it would have to be pretty fast.

I'm still looking for the formula, but doubt I'll find it, since it is irrelevant anyway.

We weren't taught anything about iron masses in Canada, but we were taught that True North and Magnetic North are not the same, and if you rely on a compass to get you somewhere on a map without figuring in delcination, you're pretty much screwed. Here in Georgia, it's only a sligth amount, but increases the farther East/West you go. The Angle of Declination is marked on maps.

ben m
26th August 2008, 01:19 PM
That has got to be so wrong, but someone who has more of this in their head will have to explain it.


A professional baseball player can swing a 20oz baseball bat at 60 mph. That's about 225 J of kinetic energy.

A .22 caliber long rifle fires a 2.5 gram bullet at 380 m/s, for 180 J of kinetic energy.

sanguine
26th August 2008, 01:26 PM
A professional baseball player can swing a 20oz baseball bat at 60 mph. That's about 225 J of kinetic energy.

A .22 caliber long rifle fires a 2.5 gram bullet at 380 m/s, for 180 J of kinetic energy.

Notably, I can swing a baseball bat in the rain without noticeable deflection or "vaporization," just like with bullets. :) I am confident the baseball bat also is, in fact, striking raindrops.

Per the OP's post, birds usually don't collide with each other in flocks because it's a bad idea. That's why they're all watching where the other birds are. That said, I've seen birds collide, so I can refute by specific example the basic premise of the question.

Fortunately, birds are light, and tend to bounce off each other and recover.

Skeptical Greg
26th August 2008, 01:31 PM
A professional baseball player can swing a 20oz baseball bat at 60 mph. That's about 225 J of kinetic energy.

A .22 caliber long rifle fires a 2.5 gram bullet at 380 m/s, for 180 J of kinetic energy.

I thought we were talking about a sniper bullet - maybe 8 grams or more at 2500 feet per second or more ( 4000 fps is not unrealistic ) -- At 2500 fps I get about 2322 j

A 2 pound ( .9kg ) 2 x 4 would have to go about 72 meters/sec ( 161mph ) to get 2351 j

http://www.csgnetwork.com/kineticenergycalc.html

sol invictus
26th August 2008, 01:39 PM
I'll reserve decision on this until you produce the math, but I really have doubts that it is saying what you think it does and if so, it is correct; it is much more likely some kind of professional rifleman's "urban legend".

Number density of raindrops during medium-heavy rain ~ 50/m^3. Crosssection of bullet times trajectory length of 500m ~ .01 m. Probability of collision ~ 50%.

Disclaimer: I guessed on all those numbers, particularly the raindrop density.

Skeptical Greg
26th August 2008, 01:40 PM
I maintain if you can't hit a falling raindrop on purpose, it's unlikely you'll hit it by accident. :)

.......
I'm glad that brings you comfort, but it has no basis in reality ..

That said, you need to get from " won't hit it " to " will destroy the bullet if you do " ....


P.S.
:)

jmercer
26th August 2008, 01:50 PM
A raindrop is pretty heavy, considering, and would explode a bullet traveling at high speed at most, or divert it severely at least.

No. Bullet in a barrel, unharmed:

a7vAXZIkhHo&NR=1

Bullet through gelatin (much denser than water), not diverted at all:

tTHo0K2Sc0g

I love youtube. :D

GreyICE
26th August 2008, 02:05 PM
I thought we were talking about a sniper bullet - maybe 8 grams or more at 2500 feet per second or more ( 4000 fps is not unrealistic ) -- At 2500 fps I get about 2322 j

A 2 pound ( .9kg ) 2 x 4 would have to go about 72 meters/sec ( 161mph ) to get 2351 j

http://www.csgnetwork.com/kineticenergycalc.html

If we're talking about high velocity sniper rifles (and their muzzle velocities), no. Humans can't match the force of impact of a sniper rifle fired at point blank range.

However for Mythbusters Grant was able to get a kN of force with his sword swings, and swords aren't even kinetic energy based. An actual mace-like weapon, like a weighted two-by-four, has the equivalent kinetic energy of a bullet impact, for a typical range of bullet speeds (and range of weapon weights and range of human swings, etc). The bullet just hits in an area that's smaller.

Can high-powered sniper rifles can exceed the force a human can deliver? Sure. They'll also punch through any sort of body armor.

Gagglegnash
26th August 2008, 02:27 PM
Hi

First: If a flying object generates a pressure wave of sufficient density to deflect a falling water drop, it's of sufficient density for the water drop deflects the pressure wave, and hence the object causing it, proportionally. As such, being hit by the pressure wave and being hit by the bullet are ballistically identical. (Isn't it interesting that supersonic jets have windshield wipers, but they stop being effective around 300 knots....)

See: Bullet Hits Water Drop (http://photo.net/photodb/photo?photo_id=3782128)

Second: You don't usually get too much deflection from expected trajectory in the rain because the bullet/water-drop impacts are at random points on the bullet, and resulting in random directional changes, creating a pretty much self-canceling effect on it. In an atmosphere, we see the same things as molecules of air are struck and pushed aside by the bullet. We call it the ballistic coefficient. Same thing with water drops, just bigger drag.

Any arguments?

...and if hitting a water drop would, "explode," a bullet, why don't they explode when they hit a chunk of ballistic gelatin? ...except .17-'06 hollow points - they WILL disintegrate, sometimes.

Gene L
26th August 2008, 02:43 PM
Any arguments?

...and if hitting a water drop would, "explode," a bullet, why don't they explode when they hit a chunk of ballistic gelatin? ...except .17-'06 hollow points - they WILL disintegrate, sometimes.

Yes, because most rifle bullets DO explode in ballistic gel, unless they're designed not to expand. A soft-point rifle bullet even at 3200 fps will go to peices, unless it's a full metal jacket, then it might still break in two at the crimping groove. A ballistic tip round (plastic-tip) will momick a block of ballistic gel and you'll have to collect bits, with a reasonable velocity and bullet weight. They're designed to NOT penetrate.

Forensic testing is usually to determine scoring on the outside of the bullet, so they use bullets that don't come apart as easily. Unless you have a "bonded" bullet, the jacket will frequently separate in gel, or "other media."

The latest ballistic testing medium is not gel, btw, it's another product that can be used again. It's green.

The guy in the reference above who apparently did have a bullet hit a raindrop (or more than one) didn't reach the target with his bullets at 100 yards.

Also, knives and arrows will penetrate kevlar vests. They're designed to resist bullets, and a sharp kinfe will cut the fibers. A ROTC student fatally stabbed another on a challenge the dagger wouldn't penetrate at North Georgia Colleget several years ago.

Skeptical Greg
26th August 2008, 02:46 PM
If we're talking about high velocity sniper rifles (and their muzzle velocities), no. Humans can't match the force of impact of a sniper rifle fired at point blank range.

However for Mythbusters Grant was able to get a kN of force with his sword swings, and swords aren't even kinetic energy based. An actual mace-like weapon, like a weighted two-by-four, has the equivalent kinetic energy of a bullet impact, for a typical range of bullet speeds (and range of weapon weights and range of human swings, etc). The bullet just hits in an area that's smaller.

Can high-powered sniper rifles can exceed the force a human can deliver? Sure. They'll also punch through any sort of body armor.
I don't have a problem with any of that..

I was responding to your claim that:

....two by fours tend to have a ton of kinetic energy, probably at least the same as a bullet. ....

... And ben m

A professional baseball player can swing a 20oz baseball bat at 60 mph. That's about 225 J of kinetic energy.

A .22 caliber long rifle fires a 2.5 gram bullet at 380 m/s, for 180 J of kinetic energy.

We can start with the former, by pointing out that kinetic energy is not measured in tons, and the rest of the claim goes downhill from there without assigning any mass or velocity to the two by four or the bullet ...

Then ben m manages to give an example where a piece of wood happens to have more kinetic energy than a bullet ..

I'm sure I can come up with a scenario where cheese beats a tomato ..


When all is said and done, bullets = 1, raindrops = 0 ... ( bottom of the fourth, and raindrops haven't won a game yet )

Brian-M
26th August 2008, 02:47 PM
If we're talking about high velocity sniper rifles (and their muzzle velocities), no. Humans can't match the force of impact of a sniper rifle fired at point blank range.

I'd have thought the kinetic force of a bullet would be identical to the kinetic force imparted to the person firing the gun (recoil), as every action has an equal and opposite re-action.


Somehow, I suspect this formula is Please don't try to bypass the autocensor.. Do you have a link to more info?
Sorry about that. :o

GreyICE
26th August 2008, 02:48 PM
I'd have thought the kinetic force of a bullet would be identical to the kinetic force imparted to the person firing the gun (recoil), as every action has an equal and opposite re-action.
You would, would you? Well, assuming you mean kinetic energy by kinetic force...

You'd be wrong.

Sorry about that ol' chap.

It's a simple mistake, but a mistake nevertheless. The bullet has far more kinetic energy than the gun.

Gene L
26th August 2008, 02:58 PM
I don't have a problem with any of that..

I was responding to Gene L's claim ....


We can start with the former, by pointing out that kinetic energy is not measured in tons, and the rest of the claim goes downhill from there without assigning any mass or velocity to the two by four or the bullet ...

Then ben m manages to give an example where a piece of wood happens to have more kinetic energy than a bullet ..

I'm sure I can come up with a scenario where cheese beats a tomato ..


When all is said and done, bullets = 1, raindrops = 0 ... ( bottom of the fourth, and raindrops haven't won a game yet )


First, it wasn't me who said what you attributed to me.

Second, kinetic energy relating to ammunition is measured in ft/lbs, and 2000 ft/lbs is known as a "ton" in shooter parlance, rightly or wrongly. It may ALSO be measured in jules, or whatever which have nothing to do with shooting a gun, and a jule is unknown and irrelvant to a bullet's energy. Miles can be measured in meters, energy can be measured in ft/lbs. I took the figures of a ton (it's actually 2521 ft/lbs) from an RCBS Ballistics Program.

But I'm sure an ammo website has energy figures for some of their loads.

Skeptical Greg
26th August 2008, 03:04 PM
Wow Gene L, I sincerely apologize..:o

I was able to correct that post ..

Toke
26th August 2008, 03:16 PM
I'd have thought the kinetic force of a bullet would be identical to the kinetic force imparted to the person firing the gun (recoil), as every action has an equal and opposite re-action.

Yes, bullet and gas weight times velocity, divided by weapon weight determines recoil.

The kinectic energi of the bullet is mass times velocity squared.

So an shotgun with a slow heavy bullet can have more recoil than a rifle, but less energy in the bullet.

GreyICE
26th August 2008, 03:19 PM
We can start with the former, by pointing out that kinetic energy is not measured in tons, and the rest of the claim goes downhill from there without assigning any mass or velocity to the two by four or the bullet ...

Out of curiosity, you're either still in school, or not American. Which?

We get nailed in the face by the damn English system all the time in the real world, and it's really frikkin annoying. I think the ONLY good thing to come out of the English system was the BTU, all the rest is broken garbage.

Very ironically, you can also use tons for air conditioning...

blobru
26th August 2008, 03:36 PM
I remember seeing an investigation on TV into whether it was best to run or walk through rain. I think that one got less wet if one walked, but I'm not certain. My mother always maintained that it was best to dart about at random to avoid raindrops, but I'm quite sure that was wrong.

Leon


In terms of getting someplace to get out of the rain, I think you're right: you should get less wet the faster you run. There are raindrops falling on your head, and raindrops you're walking into. The raindrops you're walking into are like a diffuse pool you have to wade through, so a constant total whether you walk or run. However, the raindrops that are falling on your head, they keep fallin' (according to prof's David and Bacharach), so the faster you get out of the rain, the drier your head. OTOH, if you're not heading for shelter, moving around is a bad idea (assuming the rain's coming straight down, not blown at a slant).

DavidS
26th August 2008, 06:10 PM
In terms of getting someplace to get out of the rain, I think you're right: you should get less wet the faster you run. There are raindrops falling on your head, and raindrops you're walking into. The raindrops you're walking into are like a diffuse pool you have to wade through, so a constant total whether you walk or run. However, the raindrops that are falling on your head, they keep fallin' (according to prof's David and Bacharach), so the faster you get out of the rain, the drier your head. OTOH, if you're not heading for shelter, moving around is a bad idea (assuming the rain's coming straight down, not blown at a slant).
The math is straightforward though generalization can be tedious. This was actually a homework problem for my collegiate "Problem Solving and Recreational Mathematics" course (a pass-fail geek amusement class) many (too many) years ago.

Basically, instead of stationary coordinates consider your path between two points in "rainspace" that falls with the raindrops. The length of that path corresponding to two points in stationary "stillspace" depends on your velocity and the velocity -- including direction -- of the raindrops. Because "rainspace" moves with the rain, no drops enter or leave the volume your body sweeps along the rainspace path, so that volume represents wetness, the essence of moisture. Of course, that volume depends on the area you project in that rainspace direction, which is also a function of your velocity and the rain's (not so much for me, I'm a bit more spherical that I'd wish).

At least for bodies of "reasonably round" aspect, only if the rain is blowing at your back is there is a speed which minimizes how wet you get. Whether terrestrial meteorlogical conditions ever place that speed within the range of human ability is left as an exercise for the reader.

If the rain falls perpendicularly or into your face path, faster is drier.

Gene L
26th August 2008, 06:29 PM
So how about a raindrop hitting a bullet traveling at, say, 3000 fps (a pretty generic velocity.) The bullet is a .22 caliber, and is .500 inch long and .224 inches in diameter. The rain is falling straight down.

sol invictus
26th August 2008, 08:18 PM
So how about a raindrop hitting a bullet traveling at, say, 3000 fps (a pretty generic velocity.) The bullet is a .22 caliber, and is .500 inch long and .224 inches in diameter. The rain is falling straight down.

Doing it in my head and assuming the collision is inelastic, I get that the bullet will be deflected down through an angle of about 1/40 of a degree. It will also slow down and heat up, very slightly.

Skeptical Greg
26th August 2008, 08:25 PM
How do you get the deflection without the mass of each object and the velocity of the raindrop?

Wouldn't the raindrop heat up and the bullet actually cool slightly ?

Gagglegnash
26th August 2008, 09:01 PM
Hi

Yes, because most rifle bullets DO explode in ballistic gel, unless they're designed not to expand. A soft-point rifle bullet even at 3200 fps will go to peices, unless it's a full metal jacket, then it might still break in two at the crimping groove. A ballistic tip round (plastic-tip) will momick a block of ballistic gel and you'll have to collect bits, with a reasonable velocity and bullet weight. They're designed to NOT penetrate.


Different definitions of, "explode."

I've flown lots of bullets into lots of... ummm... media - none have done what I would call, "explode," except the .17-'06 hollow-point, which once disintegrated so close to a groundhog-analogous medium that it knocked the thing out.

Forensic testing is usually to determine scoring on the outside of the bullet, so they use bullets that don't come apart as easily. Unless you have a "bonded" bullet, the jacket will frequently separate in gel, or "other media."

The latest ballistic testing medium is not gel, btw, it's another product that can be used again. It's green.


I had to revise that. I didn't want to enumerate the, "other media," into which I've fired projectiles.

Again, my definition of, "explode," may be more... ummm... explosive than yours. (Look up Army M.O.S. 55D, historically, 89D, currently, if you're interested.)

The guy in the reference above who apparently did have a bullet hit a raindrop (or more than one) didn't reach the target with his bullets at 100 yards.


Statistical clustering. Occasionally, weird results happen because of the granularity of the media. Raindrop impacts line up more on one side, and your bullet makes tracks for parts unknown.

Also, knives and arrows will penetrate kevlar vests. They're designed to resist bullets, and a sharp kinfe will cut the fibers. A ROTC student fatally stabbed another on a challenge the dagger wouldn't penetrate at North Georgia Colleget several years ago.


No argument here.

Gagglegnash
26th August 2008, 09:20 PM
Hi

I'd have thought the kinetic force of a bullet would be identical to the kinetic force imparted to the person firing the gun (recoil), as every action has an equal and opposite re-action.


It is (well... pretty much), but recoil occurs as acceleration over a distance of 20 to 24 inches, spread out over the area of contact between the shoulder and the butt-plate with a lot of rifle in between the two, while impact is deceleration over about six inches, spread out over a 1/3rd inch circle with nothing to speak of between the two.

Measure penetration as foot pounds per second per square inch, and the bullet wins.

DevilsAdvocate
26th August 2008, 10:38 PM
Now we just need to calculate the probability of a baseball pitched by Randy Johnson colliding with a bird (http://ballhype.com/video/youtube_randy_johnson_hits_bird_with_95mph_fastbal l_1/)

:D

X
26th August 2008, 11:03 PM
Haven't read the thread, so I apologize if this has been said:

Birds can, and do, collide.

I've seen it myself. When the Canada Geese start getting ready to fly South for winter. They sometimes seem a little, er, rusty at formation flying, after walking around eating and pooping all summer long. I've seen geese collide in midair a few times. It's not at all common, but it does happen. They snap at each other a bit and continue onwards.

blobru
27th August 2008, 01:39 AM
The math is straightforward though generalization can be tedious. This was actually a homework problem for my collegiate "Problem Solving and Recreational Mathematics" course (a pass-fail geek amusement class) many (too many) years ago.

Basically, instead of stationary coordinates consider your path between two points in "rainspace" that falls with the raindrops. The length of that path corresponding to two points in stationary "stillspace" depends on your velocity and the velocity -- including direction -- of the raindrops. Because "rainspace" moves with the rain, no drops enter or leave the volume your body sweeps along the rainspace path, so that volume represents wetness, the essence of moisture. Of course, that volume depends on the area you project in that rainspace direction, which is also a function of your velocity and the rain's (not so much for me, I'm a bit more spherical that I'd wish).

At least for bodies of "reasonably round" aspect, only if the rain is blowing at your back is there is a speed which minimizes how wet you get. Whether terrestrial meteorlogical conditions ever place that speed within the range of human ability is left as an exercise for the reader.

If the rain falls perpendicularly or into your face path, faster is drier.

Thx for the better answer, David. I think that makes sense (been awhile since PHYS111 for me too). If you idealize yourself as a sphere ("ideal" for physics -- not phys-ed ;)), so your top's as exposed as your front, moving with the rain at your back (up to the same speed as the wind I would think, so the rainfall is straight down relative to you) beats standing still. Otherwise, just get out of the rain... now, dummy! :scarper:

;3980474']Haven't read the thread, so I apologize if this has been said:

Birds can, and do, collide.

I've seen it myself. When the Canada Geese start getting ready to fly South for winter. They sometimes seem a little, er, rusty at formation flying, after walking around eating and pooping all summer long. I've seen geese collide in midair a few times. It's not at all common, but it does happen. They snap at each other a bit and continue onwards.

:nope: Geez, what a flock of hosers, eh?

Ocelot
27th August 2008, 02:58 AM
The pressure wave cannot build up in front of the bullet because the bullet is moving faster than the speed of sound. Pressure waves preceding a moving object (like a car or train) is a sub-sonic phenomena.

Sorry I did read that the first time you wrote it. I even clicked reply and then thought to myself. "He's taking the mickey - don't fall for it you sap"

Now I don't know if the pressure wave can or can't move a drop out of the way. But I do know that a bullet compresses the air in front of it. It is that patch of compressed air in front of the bullet that causes the pressure wave.

The wave trails off from this point in all directions and does so at the speed of sound, but that patch must keep pace with the bullet. It's not that the patch of compressed air causes a wave that keeps pace with the bullet, it's that the bullet is constantly regenerating this point.

Now big a "point" is this? How far in front of the bullet does it reach? There's no point in me waffling on if we're just talking a few atoms widths.

Well the average velocity of molecules in the air is faster than the speed of sound. 440ms-1 for room temperature and typical air pressure if I remember correctly. It has to be or a sound wave (330ms-1) couldn't be communicated by them. The actual average speed is a function of air pressure and temperature and so is the speed of sound. Higher temperature and higher density (pressure) leads to faster sound. Molecules that have just colided with a supersonic bullet tend to be going much faster than the average for "still air" Some of them made a head on collision with the bullet and bounced back along the bullet's path, for the moment travelling faster than the bullet they just bounced off. They will be slowed when they collide with another air molecule then we'll have two molecules travelling faster than average. This extra speed will be shared out amongst the other molecules they collide with until the average is back to normal. So we've got the molecules in front of the bullet that are there because they've just bounced off the front of the moving bullet and we've got the molecules that would be there anyway. That's more molecules than normal - higher density so higher pressure. We've also got molecuels travelling much faster than normal. That's higher temperature and so higher pressure again. Effectively we've got a higher speed of sound in front of the bullet.

This all depends on what proportion of the molecules hit the tip head on (or close enough) the finer and pojtier the tip the more molecule will spill off the side. That's why high velocity bullets are so aerodynamic and pointy.

"There is no pressure wave in front of a super sonic object. This is extremely obvious. I wish people would stop saying it exists.

I'm afraid your wish will not be granted whilst there are clear Schlieren photographs (http://en.wikipedia.org/wiki/Schlieren_photography) that show the pressure wave in front of supersonic objects.

http://web.mit.edu/Edgerton/www/schlieren.html

Sorry but you're wrong.

jmercer
27th August 2008, 03:42 AM
Doing it in my head and assuming the collision is inelastic, I get that the bullet will be deflected down through an angle of about 1/40 of a degree. It will also slow down and heat up, very slightly.

Depending on the distance, you'd get more of a drop from gravity. When I was in the USMC, we used M16A1 rifles with .223 ammo. (It's changed to the NATO standard now). We were prohibited from firing at the range on windy days because bullets would occasionally loft and loop back to the firing line. Prior to my advent, they had to actually reduce the muzzle velocity by adjusting the powder (either chemically or load, not sure which) in the cartridges for jungle fighting. The bullets were powdering when hitting leaves and even clothing or flesh. Once the velocity was lowered, it worked fine. (Lots of guys before the A1 refused to give up their M14's due to the jamming/fouling problem and the powdering issue.) So in one sense, I can see a bullet of a specific type and velocity being vulnerable to rain, I guess.

However, I never heard anything about rain or water being an issue re:deformation, and I know for a fact that guys would use their weapons to shoot fish successfully. Given that the .223 was a high-velocity and relatively light round, I'd say that's pretty solid evidence rain isn't a factor for either.

Inelastic or not - as far as hitting raindrops go (or the shock wave pushing drops out of the way), is actual contact necessary for the impact of mass to matter? What I'm asking is ... if there's a bow-wave effect and a drop is deflected, it would seem to me that some of the kinetic energy used to create that bow-wave would be lost in the deflection, and that should affect either the momentum or angle of the bullet. (or both) Be kind of a violation of the laws if it didn't, right?

Just askin', I don't have the math for this. :)

Worm
27th August 2008, 04:11 AM
Another handy picture.

Bullet shot through falling water drop (http://photo.net/photodb/photo?photo_id=3768150)

Note the lack of exploding (although that looks like a fairly small bullet to me)

I suppse the question is whether is the actual bullet that hits the water, or some kind of shockwave/pressure wave thingy (also note my stunningly exact use of scientific terminolgy)

sol invictus
27th August 2008, 05:26 AM
How do you get the deflection without the mass of each object and the velocity of the raindrop?

I guessed them. I'm sure I was off a bit (and I think there's probably a large range of raindrop sizes and speeds in a storm) but it's probaly about right.


Wouldn't the raindrop heat up and the bullet actually cool slightly ?

The collision will generate heat in both. If the bullet was hot to begin with the rain could cool it off, but there's very little time for heat transfer to take place, so my guess is that the first effect is more important. Just a guess, though.

Doubt
27th August 2008, 05:27 AM
We were prohibited from firing at the range on windy days because bullets would occasionally loft and loop back to the firing line.

Are you sure they would loop back? Does not make sense to me at all.

Back in the stone age. (The early '80s) I was stationed at Ft. Ord. Most of our small arms ranges were down at the beach and our ultimate backstop was Monterey bay. We were firing right into winds coming in from the Pacific ocean.

I have never, ever, heard of a bullet looping back from the wind.

sol invictus
27th August 2008, 05:28 AM
What I'm asking is ... if there's a bow-wave effect and a drop is deflected, it would seem to me that some of the kinetic energy used to create that bow-wave would be lost in the deflection, and that should affect either the momentum or angle of the bullet. (or both) Be kind of a violation of the laws if it didn't, right?

Right. But the range of that effect should be pretty small, so at the level of rigor I was working at it could probably be taken into account by pretending the bullet is a little larger than it is.

GreyICE
27th August 2008, 06:47 AM
Sorry I did read that the first time you wrote it. I even clicked reply and then thought to myself. "He's taking the mickey - don't fall for it you sap"

Now I don't know if the pressure wave can or can't move a drop out of the way. But I do know that a bullet compresses the air in front of it. It is that patch of compressed air in front of the bullet that causes the pressure wave.

The wave trails off from this point in all directions and does so at the speed of sound, but that patch must keep pace with the bullet. It's not that the patch of compressed air causes a wave that keeps pace with the bullet, it's that the bullet is constantly regenerating this point.

Now big a "point" is this? How far in front of the bullet does it reach? There's no point in me waffling on if we're just talking a few atoms widths.

Well the average velocity of molecules in the air is faster than the speed of sound. 440ms-1 for room temperature and typical air pressure if I remember correctly. It has to be or a sound wave (330ms-1) couldn't be communicated by them. The actual average speed is a function of air pressure and temperature and so is the speed of sound. Higher temperature and higher density (pressure) leads to faster sound. Molecules that have just colided with a supersonic bullet tend to be going much faster than the average for "still air" Some of them made a head on collision with the bullet and bounced back along the bullet's path, for the moment travelling faster than the bullet they just bounced off. They will be slowed when they collide with another air molecule then we'll have two molecules travelling faster than average. This extra speed will be shared out amongst the other molecules they collide with until the average is back to normal. So we've got the molecules in front of the bullet that are there because they've just bounced off the front of the moving bullet and we've got the molecules that would be there anyway. That's more molecules than normal - higher density so higher pressure. We've also got molecuels travelling much faster than normal. That's higher temperature and so higher pressure again. Effectively we've got a higher speed of sound in front of the bullet.

This all depends on what proportion of the molecules hit the tip head on (or close enough) the finer and pojtier the tip the more molecule will spill off the side. That's why high velocity bullets are so aerodynamic and pointy.



I'm afraid your wish will not be granted whilst there are clear Schlieren photographs (http://en.wikipedia.org/wiki/Schlieren_photography) that show the pressure wave in front of supersonic objects.

http://web.mit.edu/Edgerton/www/schlieren.html

Sorry but you're wrong.

I know about Schlieren Photography, I built a Schlieren Camera for pete's sake. Look at the photos again. The wave radiates from the very tip of the object.

You have a significant amount of nonsense math going on here. 440 m/s for air molecules because otherwise the sound couldn't be transmitted? Given the speed is a function of the temperature and that the fact kinetic energy is mv^2/2, water (speed of sound 1500 m/s, give or take a few meters) seems to be in a whole lot of trouble.

I'm not going to dissect exactly where you're going wrong here. I can't really figure it out, but we're at one conclusion absolutely contradicted by evidence (the speed of sound in water is too fast) so something went wrong.


I will give you that a bullet can push molecules. However, once two collide, they share the kinetic energy. Since a bullet couldn't accelerate a molecule to significantly faster than itself, once they share, they're slower, and they trail behind it.

This would suggest to me that the layer would be at most the thickness of the mean free path in atmosphere. That's 68 nanometers. I'm perfectly content with my description of that as 'negligible.'

Cuddles
27th August 2008, 07:06 AM
Out of curiosity, you're either still in school, or not American. Which?

We get nailed in the face by the damn English system all the time in the real world, and it's really frikkin annoying. I think the ONLY good thing to come out of the English system was the BTU, all the rest is broken garbage.

Very ironically, you can also use tons for air conditioning...

Tons and the other units you're using are the English units. You're trying to complain about the metric system. If you're going to be racist, at least try to get it right.:rolleyes:

II will give you that a bullet can push molecules. However, once two collide, they share the kinetic energy. Since a bullet couldn't accelerate a molecule to significantly faster than itself, once they share, they're slower, and they trail behind it.

Nope. What makes you think a bullet can't accelerate molecules to faster than its own velocity? In an elastic collision, a massive object hitting a much lighter, stationary object will result in the lighter object travelling many times faster than the massive one, while the massive one's velocity will remain pretty much unchanged.

I agree with you that since the mean free path is so small, there will only be a very thin layer in front of the bullet, but you should really not complain about "nonsense math" from others when your own physics is so horribly wrong.

GreyICE
27th August 2008, 08:28 AM
Tons and the other units you're using are the English units. You're trying to complain about the metric system. If you're going to be racist, at least try to get it right.:rolleyes:
I agree they're all English units.


Nope. What makes you think a bullet can't accelerate molecules to faster than its own velocity? In an elastic collision, a massive object hitting a much lighter, stationary object will result in the lighter object travelling many times faster than the massive one, while the massive one's velocity will remain pretty much unchanged.

I agree with you that since the mean free path is so small, there will only be a very thin layer in front of the bullet, but you should really not complain about "nonsense math" from others when your own physics is so horribly wrong. Argh. Exactly right. I fail.

I still doubt you'd have much more than one or two mean free paths in front of the bullet simply because too many collisions are going to knock fast atoms right out of the way of the bullet (there's just no way that it's going to statistically keep following the bullet's path).

Also, as far as I can tell from my back of the envelope, sound waves in water are handily outrunning the speed of the individual molecules. So either water and other densely packed materials act significantly differently than air (always a possibility) or there's something wrong with that theory in general. I'm inclined to think water acts differently, because it doesn't make any sense for a wavefront to be outrunning individual molecules, but I've been wrong before.

JoeTheJuggler
27th August 2008, 08:33 AM
Joe - two by fours tend to have a ton of kinetic energy, probably at least the same as a bullet. They don't kill because they apply their force over a larger surface.
If I throw a 2 x 4 at you, and it hits you in the arm with just one corner (a surface area about the size of a bullet) do you think it will go right through you?

No way does a thrown 2 x 4 have the same kinetic energy as a bullet fired out of a gun.

Toke
27th August 2008, 08:43 AM
I'm inclined to think water acts differently, because it doesn't make any sense for a wavefront to be outrunning individual molecules, but I've been wrong before.

Electric current move alot faster than individual electrons.

Hydralic valves on the seabottom* are controlled through very long hoses. Instead of waiting for preassure to build up, you can use pulses in the oil. Each valve actuator can be build with a different resonance frequency.
*under oilrigs

GreyICE
27th August 2008, 08:56 AM
If I throw a 2 x 4 at you, and it hits you in the arm with just one corner (a surface area about the size of a bullet) do you think it will go right through you?

Yes, (http://en.wikipedia.org/wiki/Javelin) I (http://en.wikipedia.org/wiki/Bec_de_Corbin) do. (http://en.wikipedia.org/wiki/Spear)
No way does a thrown 2 x 4 have the same kinetic energy as a bullet fired out of a gun.Okay...

I'll trust my math over your faith.

alfaniner
27th August 2008, 08:59 AM
If I throw a 2 x 4 at you, and it hits you in the arm with just one corner (a surface area about the size of a bullet) do you think it will go right through you?

No way does a thrown 2 x 4 have the same kinetic energy as a bullet fired out of a gun.

Well, the corner is backed up by the rest of the mass which is much wider.

If you imagine a bullet attached to the edge of a 2X4 with a short but impossibly rigid rod (or instead of a bullet, a nail), you would probably get some good penetration, although it would almost certainly lack the explosive power and damage potential of a bullet fired from a weapon.

GreyICE
27th August 2008, 09:07 AM
Well, the corner is backed up by the rest of the mass which is much wider.

If you imagine a bullet attached to the edge of a 2X4 with a short but impossibly rigid rod (or instead of a bullet, a nail), you would probably get some good penetration, although it would almost certainly lack the explosive power and damage potential of a bullet fired from a weapon.

Well yeah. A lot of the damage a bullet does is in the fact that it flattens upon impact, creating the huge exit wounds. Spears and other weapons don't exhibit that characteristic.

If you consider steel-jacketed slugs, which don't expand upon impact, and are much less deadly and much less damaging than lead slugs (one of the reasons the military uses them actually), the resemblance suddenly becomes much more apparent.

Ocelot
27th August 2008, 09:09 AM
I know about Schlieren Photography, I built a Schlieren Camera for pete's sake. Look at the photos again. The wave radiates from the very tip of the object.

You're joking right. You did look at the photo's before saying this?

http://web.mit.edu/Edgerton/www/images/06.jpg

See my annotated version attached

This would suggest to me that the layer would be at most the thickness of the mean free path in atmosphere. That's 68 nanometers. I'm perfectly content with my description of that as 'negligible.'

At most? Don't you mean at least?

GreyICE
27th August 2008, 09:20 AM
You're joking right. You did look at the photo's before saying this?

http://web.mit.edu/Edgerton/www/images/06.jpg

See my annotated version attached
Oh for pete's sake. It's near sonic and the shockwave is maybe half the width of a candle flame in front of it.
http://images.google.com/imgres?imgurl=http://www.nasa.gov/images/content/185591main_f-516.jpg&imgrefurl=http://www.nasa.gov/mission_pages/galex/20070815/f.html&h=357&w=516&sz=60&hl=en&start=8&sig2=sHRjcSQJrp9ZG7y36kU4Cw&um=1&usg=__OUEsAvGgk52QhJ7TBWug03NDhik=&tbnid=iln-gVByT7VqHM:&tbnh=91&tbnw=131&ei=TH61SIzHIqHyeYq60ZkI&prev=/images%3Fq%3Dshock%2Bwave%26um%3D1%26hl%3Den%26cli ent%3Dfirefox-a%26rls%3Dorg.mozilla:en-US:official%26hs%3DBsN%26sa%3DN

http://www.nasa.gov/images/content/185591main_f-516.jpg

It's a small fraction of the length of a BULLET. And it's deflecting objects right and left? I call bull.

At most? Don't you mean at least? Sure, a couple. Seriously, how far does this 'wall of hypersonic air' go?

sol invictus
27th August 2008, 09:23 AM
Electric current move alot faster than individual electrons.


The speed of sound has essentially nothing to do with the average speed of the molecules making up the substance the sound propagates through. For example, materials with a rigid crystal lattice still propagate sound.

Imagine a long line of lead weights connected by springs, sitting on a slippery table. Wiggle one weight and a disturbance will propagate down the chain at some speed. That speed depends only on the mass of each weight and the stiffness of the springs - it has nothing to do with some random motions (for example, from thermal effects) of those weights.

Gagglegnash
27th August 2008, 09:24 AM
Hi

Depending on the distance, you'd get more of a drop from gravity. When I was in the USMC, we used M16A1 rifles with .223 ammo. (It's changed to the NATO standard now). We were prohibited from firing at the range on windy days because bullets would occasionally loft and loop back to the firing line. Prior to my advent, they had to actually reduce the muzzle velocity by adjusting the powder (either chemically or load, not sure which) in the cartridges for jungle fighting. The bullets were powdering when hitting leaves and even clothing or flesh. Once the velocity was lowered, it worked fine. (Lots of guys before the A1 refused to give up their M14's due to the jamming/fouling problem and the powdering issue.) So in one sense, I can see a bullet of a specific type and velocity being vulnerable to rain, I guess.


When was this?

I was at Ft. Leonard Will It Ever Stop Snowing and Raining Wood in '70 and my Basic Training Batallion was the first to have the M16A1 all the way through Basic. We didn't even have a Manual of Arms for it yet!

We shot rain or shine, wind or hard vacuum.

....

Ok - not so much hard vacuum, but you get the idea. :D

Maybe it's a Marine thing - afraid you guys might melt. :D

A guy I know calls coyote with a .223 loaded as hot as he can get it without endangering his hands and eyesight, and the bullet doesn't powder. On the other hand, a stand of river cane on his farm counts as hard cover for the coyote.

...and we always used the 5.56mm NATO round.

However, I never heard anything about rain or water being an issue re:deformation, and I know for a fact that guys would use their weapons to shoot fish successfully. Given that the .223 was a high-velocity and relatively light round, I'd say that's pretty solid evidence rain isn't a factor for either.


I'd heard about that, too. I was pretty surprised when Mythbusters couldn't get the 5.56mm to penetrate much water in their test.

Inelastic or not - as far as hitting raindrops go (or the shock wave pushing drops out of the way), is actual contact necessary for the impact of mass to matter? What I'm asking is ... if there's a bow-wave effect and a drop is deflected, it would seem to me that some of the kinetic energy used to create that bow-wave would be lost in the deflection, and that should affect either the momentum or angle of the bullet. (or both) Be kind of a violation of the laws if it didn't, right?

Just askin', I don't have the math for this. :)


Right. I think.

logical muse
27th August 2008, 09:27 AM
I saw a short documentary years ago which included a POV of a fly on a breakfast table. The argument was that the fly thinks much, much faster than we do so, to it, we are moving in slow motion. Because of this, the fly escaped when someone tried to swat it with a newspaper.
If a fly is sitting on the table and you want to catch or kill it, just clap your hands together a few centimetres directly above it. It will fly straight up and into your colliding palms.

eta: sorry 'bout the derail

Gene L
27th August 2008, 09:57 AM
I'll trust my math over your faith.

I'm not math guy, but I think you guys are comparing kinetic energy to momentum.

GreyICE
27th August 2008, 09:59 AM
I'd heard about that, too. I was pretty surprised when Mythbusters couldn't get the 5.56mm to penetrate much water in their test.
Just a point on this - in an incompressible medium, the bullet doesn't have to HIT the fish in order to hurt it. Were the fish riddled with bullet holes, or did they lack bullet wounds?

That's why I think the Mythbusters might have dropped the ball on that one. While they did demonstrate the bullet doesn't get very far, they did not demonstrate that the bullet would have no effect on a person in the water - a rather different issue altogether.

Gene L
27th August 2008, 10:14 AM
Well yeah. A lot of the damage a bullet does is in the fact that it flattens upon impact, creating the huge exit wounds. Spears and other weapons don't exhibit that characteristic.

If you consider steel-jacketed slugs, which don't expand upon impact, and are much less deadly and much less damaging than lead slugs (one of the reasons the military uses them actually), the resemblance suddenly becomes much more apparent.


A couple of minor things: our army doesn't use "steel jacketed" slugs, as they would quickly wear out a barrel. The jacketing material is copper alloy. LIkewise, no one uses lead slugs, as lead is too soft and would melt in the bore, causing massive pressure problems.

Hard lead (an alloy of lead with tin or antimony) won't expand, but the less the percentage of other material, the softer the bullet becomes. The point where expansion must take a back seat to leading velocity is pretty narrow.

On large animals, cape buffalo and the such, a hard, heavy, non-expanding bullet is needed. These are called "solids" and may be jacketed with the copper alloy to save the bore, I don't know. Expanding bullets on this potentially dangerous game will expend their energy too soon. Bullet expansion is not a 100% controllable science. So, the bullets kill by penetrating vital organs rather than expanding and using hydrostatic shock to destroy vital organs. Cape buffalo have been taken with hard lead.

Personally, I think it impossible that a bullet would do a loop. Never heard that, and I've fired in winds that were so stiff I had to hold 36" to the right to hit the target, and still, I hit the target.

To clarify, I used "explosive" to describe what a bullet would do when it hits ballistic gel. It's purely a non-scientific explanation, and I took it for granted that people would apply common sense and realize no "explosion" took place in the scientific definition, whatever that might be. Fragmentation is what I meant. And that only happens with bullets designed to be frangible. Which includes hollow points (some, not all) lead tipped soft points, ballistic tips, and no doubt, some others.

As a clarification of my statement above that the army doesn't use steel-jacketed ammo, I mean the US Army. The former Soviet Union used a very mild ferrous material for their jackets, and used steel for the cases. It can be attracted by a magnet, but it's soft enough so it doesn't damage the bore of their rifles. Maybe other countries do as well, but the US does not. We use brass for our cases.

Gene L
27th August 2008, 10:17 AM
Just a point on this - in an incompressible medium, the bullet doesn't have to HIT the fish in order to hurt it. Were the fish riddled with bullet holes, or did they lack bullet wounds?

That's why I think the Mythbusters might have dropped the ball on that one. While they did demonstrate the bullet doesn't get very far, they did not demonstrate that the bullet would have no effect on a person in the water - a rather different issue altogether.


As I recall, the MBs got a lot more penetration as the angle of the shot decreased. Shooting straight down, they got excellent penetration of the water.

But in all, the MythBusters aren't very scientific and work some huge assumptions into their "tests." I remember one where they sank a small rowboat to see if it created a funnel to suck people down. Unsurprisingly, the small rowboat did not.

One more thing, almost all rifle bullets are supersonic, while most pistol bullets are not ("rounds" not bullets, but you know what I mean.) The .22 LR is about 1300 fps, slightly less, but I don't have the exacts. That's for the most common, "high velocity" .22 rounds.

This presents a problem for target shooters, because a supersonic bullet, when it goes subsonic, starts to wiggle at the back end. Such a small case as the .22s can't maintain supersonic velocity out to 50 yards, typcial target range. So, .22 LR target rounds start off sub-sonic and remain that all the way to their terminal. Standard velocity which you don't see much of, are also subsonic, but not as carefully made as match bullets.

With centerfire targets, bullet designers try to maintain supersonic speed as long as possible. One way they did this was with "boat-tail" bullets, whose rear-ends were angled inward, hence the name. Combined with a sptizer point (sharp) this added enormously to the range the bullet would travel by reducing air drag. This was discovered in WW I or thereabouts, for machine-guns to get long range, and now is pretty much standard. Apparently, they settle down in flight at distance.

But the boat-tail bullets don't leave the muzzle as steady as flat-based bullets. So, Bench REst shooters who measure their groups of 5 shots in thousands of an inch use flat-based bullets at 100 yards.

GreyICE
27th August 2008, 10:18 AM
*sigh*

Full Metal Jacket. There we go, now no one can pick nits, even if no one knows what I'm talking about.

Gravy
27th August 2008, 10:51 AM
Here's an article about how starlings flock without colliding:
http://www.ctv.ca/servlet/ArticleNews/story/CTVNews/20080203/starlings_secret_080203/20080203?hub=TopStories

Gagglegnash
27th August 2008, 10:52 AM
Hi

*sigh*

Full Metal Jacket. There we go, now no one can pick nits, even if no one knows what I'm talking about.


The movie?? :D

JoeTheJuggler
27th August 2008, 11:06 AM
Yes, (http://en.wikipedia.org/wiki/Javelin) I (http://en.wikipedia.org/wiki/Bec_de_Corbin) do. (http://en.wikipedia.org/wiki/Spear)


You surely must have put the wrong links up because you linked to articles on the javelin, some kind of pike and a spear.

I certainly can't throw a 2 x 4 hard enough to penetrate your arm even if it hits you on a surface area the size of a bullet. (The points in the weapons you described are significantly smaller, and can be thrown a helluva lot harder than a 2 x 4.)

At any rate, you choose one of those weapons and I'll choose a gun. Who wins the fight?

Back to the premise in the OP that a bullet never hits a raindrop, the issue is comparing the kinetic energy of a bullet to that of a falling drop of rain. Also, in a heavy rain, I doubt you could find any line of fire that doesn't intersect a raindrop, so how could they possibly miss? (I'd believe what several people said that the shock wave in front of the bullet might vaporize or disperse the drop so the bullet isn't technically colliding with the rain drop, but that's not what the OP meant, right?)
_______
I just got caught up on some of the intervening posts about 2 x 4s, baseball bats and bullets.

I wanted to add something, though---with the 2 x 4, I was treating it as a projectile (thrown) and not a lever (swung).

In the calculations of the force of a swung baseball bat, did you guys use the entire mass of the bat? At least part of that mass isn't traveling very fast (the part the batter is holding). Granted, a baseball bat is shaped so that most of its mass is out there at the fast end, but it's still not the same as throwing it.

GreyICE
27th August 2008, 02:30 PM
[QUOTE=JoeTheJuggler;3982020]You surely must have put the wrong links up because you linked to articles on the javelin, some kind of pike and a spear.

I certainly can't throw a 2 x 4 hard enough to penetrate your arm even if it hits you on a surface area the size of a bullet. (The points in the weapons you described are significantly smaller, and can be thrown a helluva lot harder than a 2 x 4.) I believe we were talking about hitting people with two-by-fours. Not throwing them around. Two by fours are not throwing weapons, for the same reason you don't smack people across the face with bullets.

At any rate, you choose one of those weapons and I'll choose a gun. Who wins the fight? If the argument is "which wins in a fight determines maximum kinetic energy" then I choose genetically tailored plagues. I won't even have to get out of bed. Genetically tailored plagues obviously have the maximum kinetic energy.

Back to the premise in the OP that a bullet never hits a raindrop, the issue is comparing the kinetic energy of a bullet to that of a falling drop of rain. Also, in a heavy rain, I doubt you could find any line of fire that doesn't intersect a raindrop, so how could they possibly miss? (I'd believe what several people said that the shock wave in front of the bullet might vaporize or disperse the drop so the bullet isn't technically colliding with the rain drop, but that's not what the OP meant, right?)
_______
I just got caught up on some of the intervening posts about 2 x 4s, baseball bats and bullets.

I wanted to add something, though---with the 2 x 4, I was treating it as a projectile (thrown) and not a lever (swung).

In the calculations of the force of a swung baseball bat, did you guys use the entire mass of the bat? At least part of that mass isn't traveling very fast (the part the batter is holding). Granted, a baseball bat is shaped so that most of its mass is out there at the fast end, but it's still not the same as throwing it. Well baseball bats are not particularly designed to maximize kinetic energy as much as they are to maximize baseball velocity upon impact.

Using something like a warhammer, yes, you get very similar kinetic energies between a hammer swing and many bullet impacts.

As far as thrown weapons, I can't imagine the human body throwing any object fast enough to possibly get anywhere near the kinetic energy of a bullet. If it was a thrown two-by-four I understand the problem. Even at 100 mph baseballs are nowhere close to the kinetic energy of bullets.

JoeTheJuggler
27th August 2008, 04:29 PM
I believe we were talking about hitting people with two-by-fours. Not throwing them around.

I'm the one who brought up 2 x 4s, and I specifically mentioned throwing them. (That way you can compare a projectile with a projectile.)

My point was that I can throw a 2 x 4 in the rain, and the rain won't deflect it. Yet a thrown 2 x 4 has nowhere near the kinetic energy of a fired bullet.

The bullet will hit rain drops (depending on your definition of "hit" with that business about the shockwave). The raindrops will not significantly deflect a bullet.


Well baseball bats are not particularly designed to maximize kinetic energy as much as they are to maximize baseball velocity upon impact.
How do you calculate kinetic energy if it's not dependent on the velocity? I suspect a swung baseball bat will have more energy than a swung 2 x 4 (for the reason I mentioned earlier--more of the mass of the bat is traveling at the higher speed whereas a lot of the mass of the 2 x 4 isn't).

At any rate, you can swing a bat through the rain and the rain won't deflect the bat noticeably (or cause it to break).

As far as thrown weapons, I can't imagine the human body throwing any object fast enough to possibly get anywhere near the kinetic energy of a bullet. If it was a thrown two-by-four I understand the problem. Even at 100 mph baseballs are nowhere close to the kinetic energy of bullets.
OK--that's what I was talking about. And a thrown baseball will hit rain drops and won't noticeably be deflected by them.

I don't buy the claim in the OP that a fired bullet won't hit rain drops, and that if it did the bullet would explode or deflect severely. In fact, it would cause the raindrop to "explode" and not affect the bullet much at all.

sol invictus
27th August 2008, 04:38 PM
The bullet will hit rain drops (depending on your definition of "hit" with that business about the shockwave). The raindrops will not significantly deflect a bullet.
<snip>
I don't buy the claim in the OP that a fired bullet won't hit rain drops, and that if it did the bullet would explode or deflect severely. In fact, it would cause the raindrop to "explode" and not affect the bullet much at all.

Did you see my posts on that? A bullet traveling 500m has about a 50% chance of hitting a raindrop, based on my estimate of the raindrop density during a medium rainfall. I also estimated how much deflection that would cause, and it's pretty negligible.

Brian-M
27th August 2008, 05:00 PM
It is (well... pretty much), but recoil occurs as acceleration over a distance of 20 to 24 inches, spread out over the area of contact between the shoulder and the butt-plate with a lot of rifle in between the two, while impact is deceleration over about six inches, spread out over a 1/3rd inch circle with nothing to speak of between the two.


Yes, I understand the difference the contact area and the duration in which the impact is absorbed makes. What now has me confused is this...

You would, would you? Well, assuming you mean kinetic energy by kinetic force...

You'd be wrong.

Sorry about that ol' chap.

It's a simple mistake, but a mistake nevertheless. The bullet has far more kinetic energy than the gun.


Why not?

Assuming hypothetically, for ease of understanding, that the gun was fired by remote control and attached to a frictionless rail that allows it to move freely (but not spin around), why would it have less kinetic energy? If the gun had 100 times the mass of the bullet, it should slide backwards at one tenth the speed. (Kinetic energy being proportional to mass times velocity squared?)

Or do you just mean the kinetic energy is absorbed by the person firing it over a longer period of time (several milliseconds) and a larger surface area than the person being hit by the bullet?


Tons and the other units you're using are the English units. You're trying to complain about the metric system. If you're going to be racist, at least try to get it right.:rolleyes:

I agree, don't blame the English. The French are to be congratulated for everything from the Metric system to the Statue of Liberty. :)

AntiTelharsic
27th August 2008, 05:10 PM
As far as thrown weapons, I can't imagine the human body throwing any object fast enough to possibly get anywhere near the kinetic energy of a bullet. If it was a thrown two-by-four I understand the problem. Even at 100 mph baseballs are nowhere close to the kinetic energy of bullets.

Edit: I screwed up converting miles per hour to meters per second, so the baseball calculation below is incorrect. See sol's correction in the next post.

Kinetic energy E = 1/2 m v^2

For a 100 mph baseball, m = 0.142 kg, v = 160 m/s
E = 1/2 * 0.142 * 160^2 = 1817.6 J

http://en.wikipedia.org/wiki/Baseball_(ball)

For a Wather PPK, .380 ACP, m = 0.0058 kg, v = 244 m/s
E = 1/2 * 0.0058 * 244^2 = 172.6544 J

http://en.wikipedia.org/wiki/Walther_PPK
http://en.wikipedia.org/wiki/.380_ACP

The baseball has ten times the kinetic energy of the bullet.

sol invictus
27th August 2008, 05:37 PM
Kinetic energy E = 1/2 m v^2

For a 100 mph baseball, m = 0.142 kg, v = 160 m/s
E = 1/2 * 0.142 * 160^2 = 1817.6 J

100 mph is 45 m/s, not 160. So the energy is about 142 J, or slightly less than what you got for the bullet.

GreyICE
27th August 2008, 06:36 PM
Why not?

Assuming hypothetically, for ease of understanding, that the gun was fired by remote control and attached to a frictionless rail that allows it to move freely (but not spin around), why would it have less kinetic energy? If the gun had 100 times the mass of the bullet, it should slide backwards at one tenth the speed. (Kinetic energy being proportional to mass times velocity squared?)

Or do you just mean the kinetic energy is absorbed by the person firing it over a longer period of time (several milliseconds) and a larger surface area than the person being hit by the bullet?

No, it would not. It would slide back at 1/100th of the speed. Momentum is conserved. Momentum, however, is equal to mass times velocity (as a vector). The gun would have 1/100th the kinetic energy of the bullet.

Note that it actually has significantly more than that, because the gun, in addition to firing bullets, is also firing rather large quantities of hot gasses downrange at significant speed, and has to take the momentum for all of those (which is where muzzle brakes on fast-firing guns come into play, but that's a whole different ballpark).

This is why head on collisions between two cars at 30 MPH are usually not fatal, whereas headon collisions at 60 are a complete mess, and headon collisions at 90 all the safety technology we have becomes mostly worthless. It's also why modern cars accordian when they are hit - they can't dissipate any momentum, but by slowing down the collision they reduce the amount of kinetic energy that you're taking in the face. Hence all the jokes about modern cars being made out of plastic and folding up in simple fender benders. That's actually by design (steel cars can come out of collisions that kill the driver virtually undamaged).

Kinetic Energy is most manifestly not conserved. By example, consider two objects moving at equal speeds that collide inelastically (say, two balls of silly putty) and stop dead. Momentum was conserved. Kinetic Energy is now zero.

ENERGY is always conserved, however.

AntiTelharsic
27th August 2008, 06:38 PM
100 mph is 45 m/s, not 160. So the energy is about 142 J, or slightly less than what you got for the bullet.

I can't believe I did that. Thanks :)

JoeTheJuggler
27th August 2008, 07:14 PM
Did you see my posts on that? A bullet traveling 500m has about a 50% chance of hitting a raindrop, based on my estimate of the raindrop density during a medium rainfall. I also estimated how much deflection that would cause, and it's pretty negligible.

Thanks, sol. That matches up pretty well (based on different givens as far as raindrop density and bullet size and speed) with the calculations vorticity did (post #33) of a bullet hitting a raindrop on average every 12 meters.

My non-mathematical guess was that a bullet fired in the rain would probably hit at least one raindrop. Even if it were far less than that, it would still have had to have happened many times. After all, lots of raindrops fall, and lots of bullets get fired, so even a one in a million event should have happened many times. Yet, I've never heard of a bullet exploding on impact with a raindrop or ricocheting off of a raindrop. Not even once! :)

GreyICE
27th August 2008, 07:16 PM
Thanks, sol. That matches up pretty well (based on different givens as far as raindrop density and bullet size and speed) with the calculations vorticity did (post #33) of a bullet hitting a raindrop on average every 12 meters.

My non-mathematical guess was that a bullet fired in the rain would probably hit at least one raindrop. Even if it were far less than that, it would still have had to have happened many times. After all, lots of raindrops fall, and lots of bullets get fired, so even a one in a million event should have happened many times. Yet, I've never heard of a bullet exploding on impact with a raindrop or ricocheting off of a raindrop. Not even once! :)

Nor will you ever. It's completely true that from the bullet's perspective, it just got hit by a very small drop of water moving at 600 m/s, but the amount of energy transmitted to the bullet by that collision is completely negligible. Raindrops, honestly, just don't weigh that much. Nor do they have good impact properties (they tend to, well, splash).

A rain of ball bearings would probably be totally different.

Brian-M
27th August 2008, 07:29 PM
No, it would not. It would slide back at 1/100th of the speed. Momentum is conserved. Momentum, however, is equal to mass times velocity (as a vector). The gun would have 1/100th the kinetic energy of the bullet.

Ah! :)

It's momentum and not energy that's equal and opposite? That must be where I've been getting mixed up.

GreyICE
27th August 2008, 08:22 PM
Ah! :)

It's momentum and not energy that's equal and opposite? That must be where I've been getting mixed up.
Yup! Also note that momentum is a vector quantity here, which means it has direction. So two objects with the same weight moving at eachother at the same speed have the same kinetic energy, but opposite momentums (which is how they can cancel out).

Note energy is also conserved (the objects get hotter when they stop moving). It changes form though (and if you're wondering where it comes from in the first place, that's the gunpowder ;) ).

Ocelot
28th August 2008, 03:44 AM
Oh for goodness sake. Kinetic Energy of a high speed bullet.

Lets take this (http://en.wikipedia.org/wiki/Barrett_M82) as a rough example, the Barret M82

Muzzle velocity: 853 m/s [2850 f/s] with 660 grain, 42.8 g projectile

IN SI Units.
velocity, v = 853 m/s
mass, m = 0.0428 kg

Kinetic Energy = (mv2)/2 = 31141.6652 J (Calculator)
~= 31.1 kJ (3 Significant Digits)

That's a heck of a lot of kinetic energy.

Now for the lumber where we have to make a few assumptions. I'm going to start by severely highballing it.

2 metres of 2 by 4 has a volume of 0.01 m3 Pine has a density (http://www.simetric.co.uk/si_wood.htm) of roughly 500 kg/m3 so we're talking about 5kg of lumber.

Here's where we really highball it. I'm going to assume that someone is throwing the 5kg of lumber as fast as the fastest recorded baseball pitch (http://www.baseball-almanac.com/articles/fastest-pitcher-in-baseball.shtml). Roughly 100 mph or 45 m/s

so mass, m = 5
velocity, v = 45
Kinetic Energy = (mv2)/2 = 5062.5 J (Calculator)
~= 5 kJ (to 1 Significant Digit)

Even with this immense highballing the lumber has a fraction of the kinetic energy of a sniper bullet.

What about a less hefty rifle. Lets take this (http://en.wikipedia.org/wiki/.22_Long_Rifle) for example. A .22 Long Rifle


40 grain (http://en.wikipedia.org/wiki/Grain_(measure)) (2.6 g) lead: 1082 ft/s (330 m/s) .22 LR Subsonic


So mass, m = 0.0025 kg
Velocity, v = 330 m/s
Kinetic Energy = (mv2)/2 = 136.125 J ~= 140 J (2 Significant Digits)

So that would be the same as (and I'll leave the workings out for the interested parties to do themselves) our 2 metre length of 4 x 2 being thrown at a much more sedate 16.5 miles per hour. Something closer to what I'd estimate is achievable without a woomera like launching device. It's about half the speed of a shot put of similar mass being thrown by a professional athlete (http://people.brunel.ac.uk/~spstnpl/LearningResources/ShotPutLab.pdf).

I can assure you that even a length of 2x2 hitting edge on can break the skin when swung playfully (we were playing Star Wars) at a little sister and hope I don't have to tell you not to try this experiment at home.

ETA I see I've been beaten to this calculation. Please ignore the repetition.

GreyICE
28th August 2008, 06:21 AM
Edit: I screwed up converting miles per hour to meters per second, so the baseball calculation below is incorrect. See sol's correction in the next post.

Kinetic energy E = 1/2 m v^2

For a 100 mph baseball, m = 0.142 kg, v = 160 m/s
E = 1/2 * 0.142 * 160^2 = 1817.6 J

http://en.wikipedia.org/wiki/Baseball_(ball)

For a Wather PPK, .380 ACP, m = 0.0058 kg, v = 244 m/s
E = 1/2 * 0.0058 * 244^2 = 172.6544 J

http://en.wikipedia.org/wiki/Walther_PPK
http://en.wikipedia.org/wiki/.380_ACP

The baseball has ten times the kinetic energy of the bullet.

Err, you do know that 100 mph represents the baseball pitch of a professional pitcher, while the Walter PPK is a gun deliberately designed to minimize kinetic energy.

Lets use at least a Glock 17, mmkay?

Also I'm finding your 6 gram number a little questionable.
http://www.winchester.com/products/catalog/components/handgunbullets.aspx

I'm really not seeing how you can defend this assertion.

Edit: Well, I'm thoroughly beaten to it.

Skeptical Greg
28th August 2008, 11:52 AM
Actually we did the math on bullet vs 2 x 4 back on page 2 ..

I'm still waiting on tomato vs cheese...

AntiTelharsic
28th August 2008, 12:19 PM
I'm really not seeing how you can defend this assertion.

Well, I didn't really -- you even quoted my edit where I said I was wrong :)

homer
28th August 2008, 01:13 PM
We had house martins nesting this year and two of them did collide , no doubt those that do this too often will die and fail to reproduce . Hence natural selection favours birds that don't collide . Evolution at work before my eyes .

Gene L
28th August 2008, 05:58 PM
Err, you do know that 100 mph represents the baseball pitch of a professional pitcher, while the Walter PPK is a gun deliberately designed to minimize kinetic energy.



How so to minimize? Walther didn't design the round, Colt did (.380 Colt Automatic Pistol) and Walther wasn't the first .380 out there.

What Walther did with the PP (and PPK) was design a double action pistol around what was then the most powerful round that could utilize the blow-back design.

But as for energy and the effect of bullets against hard targets, Safiriland did a series of tests about 25 years ago, where they had their ballistic ceramic armor plate in a vest and a guy allowed himself to be shot with it...Chuck Taylor, ( worked for Glock for a while) from a .22 Short to a .308. The bullets did not move him hardly at all, even the .308 barley rocked him.

This monumentally dumb feat was filmed, and Chuck paid a great deal of money. So, Hollywood aside, a bullet does not knock someone off his feat by monentum.

GreyICE
28th August 2008, 06:24 PM
How so to minimize? Walther didn't design the round, Colt did (.380 Colt Automatic Pistol) and Walther wasn't the first .380 out there.

What Walther did with the PP (and PPK) was design a double action pistol around what was then the most powerful round that could utilize the blow-back design.

But as for energy and the effect of bullets against hard targets, Safiriland did a series of tests about 25 years ago, where they had their ballistic ceramic armor plate in a vest and a guy allowed himself to be shot with it...Chuck Taylor, ( worked for Glock for a while) from a .22 Short to a .308. The bullets did not move him hardly at all, even the .308 barley rocked him.

This monumentally dumb feat was filmed, and Chuck paid a great deal of money. So, Hollywood aside, a bullet does not knock someone off his feat by monentum.

The PPK is designed to minimize recoil, so you can fire bullets quite quickly. Obviously minus muzzle brakes, the best way to do that is reduce velocity.

The 0.357 SIG exceeds the speed of sound, as does the 9 mm. They're both more powerful cartredges, even in pistols.

The PPK is also quieter than the Glock, because there's no sound barrier (hence it being a Bond favorite - Flemming actually sometimes cared about these things). Breaking the sound barrier is LOUD.

Gene L
28th August 2008, 08:25 PM
The PPK is designed to minimize recoil, so you can fire bullets quite quickly. Obviously minus muzzle brakes, the best way to do that is reduce velocity.

The 0.357 SIG exceeds the speed of sound, as does the 9 mm. They're both more powerful cartredges, even in pistols.

The PPK is also quieter than the Glock, because there's no sound barrier (hence it being a Bond favorite - Flemming actually sometimes cared about these things). Breaking the sound barrier is LOUD.

For the short range a pistol is shot at, the sound barrier doesn't add much sound. A .380 is quieter because it doesn't have as much powder.

I did a test with a 300 Winchester Magnum that was suppressed. The suppressor quieted the muzzle blast so low you didn't have to wear ear plugs if you were behind the gun. If you stood to one side you could hear the bullet crack quite loudly. It also acted as a muzzle brake. I'm getting one for my rifle because of the reduced recoil (WAY reduced).

Bond carried a .32, by the way..."like a brick hitting a plate glass window" as M described it. Here, a .32 is considered sub-par, and a .380 is on the low end of defense. And there are plenty of 9mm pistols smaller, lighter, and easier to carry than a PPK.

But they were the first to build a DA pistol, and the world owes them that. I have a PPK-S. Good gun for its time.

CaveDave
28th August 2008, 10:43 PM
That has got to be so wrong, but someone who has more of this in their head will have to explain it.

I will start by saying, I do know kinetic energy increases with the square of the velocity; so you need to define the mass of the 2 x 4 , say a pound or two , then explain how you would accelerate it to a velocity that would give it the kinetic energy of a 8 gram bullet traveling at 2,500 fps or more ..


P.S

That raindrop, which you say will destroy a bullet, has so little kinetic energy, you might as well call it zero ..

You may find this PDF (http://www.wind.ttu.edu/Research/DebrisImpact/Reports/DIF_reports.pdf) instructive.:)

I saw a very early prototype (that they kept in the access tunnels under the structural test lab of the civil engineering dept.) fired through a concrete block wall. Impressive even at that small scale.:D

Dave

jmercer
29th August 2008, 05:39 AM
Are you sure they would loop back? Does not make sense to me at all.

Back in the stone age. (The early '80s) I was stationed at Ft. Ord. Most of our small arms ranges were down at the beach and our ultimate backstop was Monterey bay. We were firing right into winds coming in from the Pacific ocean.

I have never, ever, heard of a bullet looping back from the wind.

What size round? As I understand it, it was only with the .224 and not the NATO ball round, which is what the military switched to in the late 70's/early 80's. (I was in from '73-76)

Gene L
29th August 2008, 07:19 AM
When you were in, it was the 5.56, 55 gr. bullet round. Later, the military switched to a heavier round, and eventually when it became NATO accepted, it got a designation as 5.56 NATO. The first changes were to a 62 gr. bullet, fairly briefly, and they changed the twist on the M 16 bore from 1 turn in 12" (1:12) to 1:10 and then to 1:8. Now, it's 1:7 and the bullet is 69 grain, teel-core "penetrator" bullet. It was developed to be able to penetrate a helmet at 500 yards. It's not very accurate, according to what I heard from a well placed source. All 5.56 rounds are boat-tailed spitzer types.

I still don't think a bullet could loop, or defy gravity by going upward.

GreyICE
29th August 2008, 07:49 AM
When you were in, it was the 5.56, 55 gr. bullet round. Later, the military switched to a heavier round, and eventually when it became NATO accepted, it got a designation as 5.56 NATO. The first changes were to a 62 gr. bullet, fairly briefly, and they changed the twist on the M 16 bore from 1 turn in 12" (1:12) to 1:10 and then to 1:8. Now, it's 1:7 and the bullet is 69 grain, teel-core "penetrator" bullet. It was developed to be able to penetrate a helmet at 500 yards. It's not very accurate, according to what I heard from a well placed source. All 5.56 rounds are boat-tailed spitzer types.

I still don't think a bullet could loop, or defy gravity by going upward.

I still don't get why the 5.56 gets the nod over the 7.62 NATO. Seriously, can you clear that one up for me?

Is it just the weight issue? I know that's a serious problem, but now that we're not humping it in a jungle, doesn't mechanized infantry kind of clean that up?

Gene L
29th August 2008, 09:33 AM
That's a good question. There are two problems with the 7.62 round IMO. First is the recoil, and that's WAY important. The big problem with the 7.62 in an automatic rifle is you can't control it. I trained on the M 14, but took the M 16 to war.

If you fire an M 14 on full auto, the first round would be on target, the second round about 3 feet high, the third round 12 feet high, as I recall from tests. All M 14s had the capability of full auto fire, but the selection of auto-riflemen was with the squad leader. He had a wrench, or a key, to turn certain rifles into full....either semi, or full, no operator selector.

Second is indeed weight. I was a Recon Platoon leader in Viet Nam, and our basic load was 440 rounds of 5.56, and most people carried more. In addition to that, we each carried 100 rounds of 7.62 for the machine guns, 3 frags and one smoke grenade, 5 quarts of water, food, sleeping gear, etc. I don't think a soldier can hump 400 rounds of 7.62 and still be effective as moving quickly. Basic load for a M 14 is 120 rounds.

As for accuracy, the M 16 has it all over the M 14 to the point no serious shooters are using them in Service Rifle matches any more. The M 14 uses WW II technology, basically, and it weighs a couple of pounds more than the M 16, and ammo weighs at least twice as much.

I wouldn't want an M 14 in the desert. The US government is emptying their warehouses of these rifles and sending them to whoever wants them (usually those who don't have to hump the ammo). There are no parts available; last parts made around 1967 or so, and only one magazine per rifle.

Mechanized infantry becomes foot infantry once contact is made, and vehicles get the hell out of danger, taking away spare ammo.

The current penetrator round is not a good round, as it isn't living up to the accuracy potential of the 5.56 round. They're looking at it, and I personally wish they'd go back to a higher-velocity round like the 62 gr. round with the twist in today's barrels. Match 5.56 is 69 grain, and that would be a great choice.

I'm a LEO firearms instructor, by the way, and live pretty close to Ft. Benning, the home of the Marksmanship Unit, where they test all the rounds for use. The then-head of that unit spoke before our annual conference two years ago, and he's a world-class shooter, spec ops guy and now deployed, I understand. He's the one who gave the information above about the no-parts, one-mag, but he wouldn't denigrate the M 14, said he'd fight with what he was issued. He was asked about stopping power and said if you hit a target right, it'll go down.

From additional sources, the 55 gr. bullet will break at the crimping cannulure at 5.56 speeds and open up a wound channel and ancillary damage. The steel penetrator core on the current round prevents this.

Anyway, most casualties are inflicted at 75 yards on in today, and the 5.56 allows you to shoot faster and more accurately.

Rocko
30th August 2008, 08:20 AM
If a fly is sitting on the table and you want to catch or kill it, just clap your hands together a few centimetres directly above it. It will fly straight up and into your colliding palms.

eta: sorry 'bout the derail

Sorry to continue the derail, but there's a - typically scant ondetails - story on the BBC about that today:

http://news.bbc.co.uk/1/hi/sci/tech/7586868.stm