Upgrades in modern combat armor?

[imported_Darthan]

everman, it was actually UC Berkely that developed the BLEEX (Berkeley Lower Extremity Exoskeleton) system (http://www.berkeley.edu/news/m.../2004/03/03_exo.shtml).

Gamingphreek, I suggest that Gannon has a valid point. He isn't suggesting stopping military developement, just limiting it. If we limit our military development to projects that, well, work (you know, like canceling the missile defense program) we would have billions and billions of dollars to spend other items important to national security (like energy independence) or maybe reduce the deficit/national debt.

Also, no, we do not have force fields.

 

[CycloWizard]

Originally posted by: Darthan
everman, it was actually UC Berkely that developed the BLEEX (Berkeley Lower Extremity Exoskeleton) system (http://www.berkeley.edu/news/m.../2004/03/03_exo.shtml).

Gamingphreek, I suggest that Gannon has a valid point. He isn't suggesting stopping military developement, just limiting it. If we limit our military development to projects that, well, work (you know, like canceling the missile defense program) we would have billions and billions of dollars to spend other items important to national security (like energy independence) or maybe reduce the deficit/national debt.

Also, no, we do not have force fields.

If we don't spend the money on military, it won't get spent at all. Besides, the military is typically the driving force for major advancements in areas such as fuel cells, et cetera. Taking away all the research funding would result in a dramatic drop in overall research output from the entire country, if not the world.

 

[natenut]

YES!!!!! we do have force fields. Have you ever seen an electrically conductive material droped through a high tesla magnetic field? it slows to a crawl due to the induced current in the conductor. Bullets are usually copper jacketed lead. If there was a way to make a large enough magnetic field any projectile would be slowed to a crawl. the faster the projectile the greater the opposing force. Please correcet me if i'm wrong but i draw my theory from dropping pennies on to superconducting magnets. it defys logic when you see it happen and i'm not talking about the messinger (sp?) effect.

-natenut

edit: and by any projectile i mean electrically conductive projectile.

 

[LethalWolfe]

Originally posted by: CycloWizard

Originally posted by: Darthan
everman, it was actually UC Berkely that developed the BLEEX (Berkeley Lower Extremity Exoskeleton) system (http://www.berkeley.edu/news/m.../2004/03/03_exo.shtml).

Gamingphreek, I suggest that Gannon has a valid point. He isn't suggesting stopping military developement, just limiting it. If we limit our military development to projects that, well, work (you know, like canceling the missile defense program) we would have billions and billions of dollars to spend other items important to national security (like energy independence) or maybe reduce the deficit/national debt.

Also, no, we do not have force fields.

If we don't spend the money on military, it won't get spent at all. Besides, the military is typically the driving force for major advancements in areas such as fuel cells, et cetera. Taking away all the research funding would result in a dramatic drop in overall research output from the entire country, if not the world.

Just to add a few more to the list:
Radio, internet, computers, jet engines, interstate highways in the US, GPS... All things that, at one point or another, the military helped develope.


Lethal

 

[f95toli]

That is in part because of how the research funding works in the US.
Projects that would be supported by civilian agencies in other countries are often funder through for example DoD or directly through the military in the US (for example the US Navy which supports a lot of materials research, even if the research has no obvious military application).
There are also a few of well-known research facilties which do both secret military research and "civilian" basic research, Los Alamos is one example.

 

[Atomicus]

any ideas on the use of magnetic fields (or force fields) which generates a countering force proportional to the KE of the object entering the field. Not exactly feasible with our current technology, but the principle is the same as pushing a copper disk through a U-magnet.

 

[f95toli]

*It would be VERY hard to "shape" a magnetic field so that you can use it for protection (in all directions)

*Where do you get the power? You need a LOT of power.

*There is a limit to how high fields you can create even with a superconducing magnet, I think the maximum (continous) field is around 20T or so, and that is in a very small space (about 1x1x1 cm).

*Your enemy would just stop making bullets out of conducting materials: Alloys, ceramics etc could be used to make bullets.

 

[drpootums]

Originally posted by: skyking

Originally posted by: Matthias99

Originally posted by: TuxDave
Why focus on body armor? With some gene therapy perhaps we can start churning out fleets of super soldiers with redundant organs and a very high pain threshold.

Well, sure, but that would take at least 30-40 years at our current rate of biotechnology R&D. Plus, without some serious breakthroughs in decreasing development time, you're looking at years (if not decades) to replace one of them. Ideally, you'd have super-soldiers with good armor anyway.

*Stopping* bullets isn't that hard. Ceramic composite strike plates are fairly light (up to about 6 pounds for 10x12" plates) and will prevent penetration from just about any pistol or rifle round (at least a few of them; ceramics are brittle, and will fracture after repeated impacts). However, the kinetic energy of a ballistic projectice is still considerable. At close range, you're talking 1500-3000 ft-lbs of energy (about 2000-4000 newton-meters) for an assault rifle. That's enough energy to accelerate a 100kg mass (like, say, a soldier) to 6-9 m/s (~15-20mph). Admittedly, not all of the energy is turned back into KE (some is wasted in fracturing the bullet and/or the strike plate, dispersed as heat and sound, etc.), and I have no idea how big that fraction is. But that's a *lot* of impact force, even spread over your whole torso. Low-caliber bullets (which have a much lower velocity) wouldn't be that bad, but to absorb an impact like that and not be (at the very least) knocked down, you'd have to be wearing significantly more massive armor.

Are you sure about your energy figures? Where does the energy come from?
I figure that energy has to act against the fellow holding the gun, and even moreso. The bullet is at it's maximum velocity as it exits the barrel.

Those energy figues are correct. The energy from a 30-06 is about 3000-3100 ft-lbs, and the energy from the .223 (m-16) and the 7.62x39mm (ak-47) are about 1500-1800 ft-lbs. The person shooting the gun isnt going to get that kind of energy put on him just because the energy is escaping (both out the barrel and through the reciever when the slide opens).

Energy comes from a equation (forgot it now) of the weight of the bullet and the velocity it has.

 

[drpootums]

Also, during one period in time Germany had a tank with balistic plates on the sides of it. When a heavy enough shell got very close to it the plate would just blow straight off, making the shell much MUCH less powerful.

 

[Velk]

Originally posted by: drpootums

Originally posted by: skyking

Originally posted by: Matthias99

Originally posted by: TuxDave
<SNIP>
However, the kinetic energy of a ballistic projectice is still considerable. At close range, you're talking 1500-3000 ft-lbs of energy (about 2000-4000 newton-meters) for an assault rifle. That's enough energy to accelerate a 100kg mass (like, say, a soldier) to 6-9 m/s (~15-20mph). Admittedly, not all of the energy is turned back into KE (some is wasted in fracturing the bullet and/or the strike plate, dispersed as heat and sound, etc.), and I have no idea how big that fraction is. But that's a *lot* of impact force, even spread over your whole torso. Low-caliber bullets (which have a much lower velocity) wouldn't be that bad, but to absorb an impact like that and not be (at the very least) knocked down, you'd have to be wearing significantly more massive armor.

Are you sure about your energy figures? Where does the energy come from?
I figure that energy has to act against the fellow holding the gun, and even moreso. The bullet is at it's maximum velocity as it exits the barrel.

Those energy figues are correct. The energy from a 30-06 is about 3000-3100 ft-lbs, and the energy from the .223 (m-16) and the 7.62x39mm (ak-47) are about 1500-1800 ft-lbs. The person shooting the gun isnt going to get that kind of energy put on him just because the energy is escaping (both out the barrel and through the reciever when the slide opens).

I am not sure about how you are figuring the energy effects, but I know the conclusion is absurd, so I suspect you have dropped a couple of zeros somewhere.

If you disregard force being dissipated into fracturing armor, heat etc, then it becomes an extremely basic exercise in conservation of momentum.

A bullet weighing 10g travelling at 850m/s hitting a 100kg man who is at rest will end up with both of them together moving at a rate that is determined by :

Vf =(M1V1+M2V2) / (M1+M2)

Or in this case (0.01kg x 850m/s + 100kg x 0m/s ) / (0.01kg + 100kg ) = 0.08 m/s, so the original poster is off by a factor of roughly 100, which I believe he acknowledged a couple of posts further down...

Of course, a typical soldier isn't an inelestatic body, however the above is a *best case* scenario given that the deformation of the armor, bullet, and/or soldier decrease the final velocity.

 

[Matthias99]

Originally posted by: Velk
I am not sure about how you are figuring the energy effects, but I know the conclusion is absurd, so I suspect you have dropped a couple of zeros somewhere.

I thought maybe I had screwed up going from ft-lbs to Newton-meters, but it turns out I actually computed the output velocity assuming all the KE of the round was being transferred into KE of the target (ie, such that .5 * m_bullet * (v_bullet^2) = .5 * m_target * (v_target^2) ). However, doing that violates conservation of momentum (such that m_bullet * v_bullet + m_target * v_target must always remain constant). The calculation you gave is correct (and much more sensible).

Assuming an inelastic collision, the "extra" KE that doesn't get turned into motion must be dissipated by deforming the armor, and/or producing sound/heat energy. If the collision is elastic, some fraction of the KE will remain with the ricochet, and some fraction will be lost as friction in the collision. It's been too long since I had to do any of this stuff for a class; I majored in CS so that I would have a computer to do this crap for me.

I still stand by the conclusion that, even if you have armor that will stop the bullet dead, that's a *lot* of force acting on your body in a *very* short time frame (probably well under a millisecond). If you are accelerated from 0 to 0.8m/s in .001 seconds, that's 800 m/s^2 -- or about 80 Gs of acceleration. It's only for a brief instant, but that's not gonna be pleasant, and a blow like that could easily cause a concussion, dislocate a joint, or cause internal bruising, depending on where it was delivered.

 

[CycloWizard]

Originally posted by: Matthias99
I still stand by the conclusion that, even if you have armor that will stop the bullet dead, that's a *lot* of force acting on your body in a *very* short time frame (probably well under a millisecond). If you are accelerated from 0 to 0.8m/s in .001 seconds, that's 800 m/s^2 -- or about 80 Gs of acceleration. It's only for a brief instant, but that's not gonna be pleasant, and a blow like that could easily cause a concussion, dislocate a joint, or cause internal bruising, depending on where it was delivered.

That's why armor is designed the way it is: it dissipates this force over as large an area as possible. Whatever isn't absorbed by the armor during fracture will be passed on to you, but hopefully spread over a larger surface area than it would be had you not been wearing armor. Kevlar is fiber-based, so it would only spread the impact over a relatively small area, which usually resulted in internal bleeding and such. Newer armors hope to overcome this problem by distributing the load.

 

[Matthias99]

Originally posted by: CycloWizard

Originally posted by: Matthias99
I still stand by the conclusion that, even if you have armor that will stop the bullet dead, that's a *lot* of force acting on your body in a *very* short time frame (probably well under a millisecond). If you are accelerated from 0 to 0.8m/s in .001 seconds, that's 800 m/s^2 -- or about 80 Gs of acceleration. It's only for a brief instant, but that's not gonna be pleasant, and a blow like that could easily cause a concussion, dislocate a joint, or cause internal bruising, depending on where it was delivered.

That's why armor is designed the way it is: it dissipates this force over as large an area as possible. Whatever isn't absorbed by the armor during fracture will be passed on to you, but hopefully spread over a larger surface area than it would be had you not been wearing armor. Kevlar is fiber-based, so it would only spread the impact over a relatively small area, which usually resulted in internal bleeding and such. Newer armors hope to overcome this problem by distributing the load.

Right -- but 80 Gs is already the acceleration when the blow is distributed across all the mass in your body (assuming you weigh around 100kg). The force involved to do that would be (100kg * 800m/s^2 = 80,000N, or about the equivalent weight of four tons). Distributed over a 10x16" strike plate (as in modern combat armor), that's ~8000 lbs / (10*16 inches square) ~= 50 pounds per square inch over 160 square inches of your torso. Not pleasant. And that's the best-case here, not the worst -- flexible armor like Kevlar would produce a MUCH higher impact force over a small area.

 

[Gibsons]

But the plate isn't perfectly coupled to the torso... Assume a plate has a mass of say, 90 grams and has to travel a very small distance (1mm will do) before it starts really pushing (transferring energy to) on the body. You go from a 10g 800m/sec bullet to 100g (bullet plus plate) at 80m/sec. The kinetic energy of the two is vastly different. Thats not counting all the energy that will be wasted by shattering/deforming the bullet.

 

[Matthias99]

Originally posted by: Gibsons
But the plate isn't perfectly coupled to the torso... Assume a plate has a mass of say, 90 grams and has to travel a very small distance (1mm will do) before it starts really pushing (transferring energy to) on the body. You go from a 10g 800m/sec bullet to 100g (bullet plus plate) at 80m/sec. The kinetic energy of the two is vastly different. Thats not counting all the energy that will be wasted by shattering/deforming the bullet.

It's been a while since I took physics, but I'm *pretty* sure it doesn't work like that. It's the momentum that's the problem here, *not* the KE (directly). Getting hit by a 100g (not accurate, BTW; strike plates are more like 3kg) object at 80m/sec is not any better than getting hit by a 10g object at 800m/sec in terms of momentum transfer and the ultimate acceleration that's going to be inflicted on you. The final result calculates out to the same amount of net impact force.

 

[Gibsons]

Kinetic energy is the capacity to do work. I think it's a much more important factor than momentum. If momentum was the story, then the back end of the gun would be about as dangerous as the front end, no? By definition it has to have the same momentum as the bulletl. It's all because kinetic energy is determined by the square of the velocity (1/2m(v)^2) as opposed to momentum (mv).

 

[Matthias99]

Originally posted by: Gibsons
Kinetic energy is the capacity to do work. I think it's a much more important factor than momentum. If momentum was the story, then the back end of the gun would be about as dangerous as the front end, no? By definition it has to have the same momentum as the bulletl. It's all because kinetic energy is determined by the square of the velocity (1/2m(v)^2) as opposed to momentum (mv).

Actually, the gun *has* to have the same momentum as the bullet (rather, m_bullet * v_bullet + m_gun * v_gun must equal 0, assuming the gun-bullet system was at rest when it was fired). It's just that the gun weighs several thousand times more than the bullet (usually a few kilos versus a few grams for a high-powered rifle), so its velocity (and thereby KE) is much, much lower.

Read the posts a few above this one, where the math is gone over in detail. Conservation of momentum is what leads to the force calculations, not conservation of energy. The 'excess' energy has to be dissipated by deformation of the armor (assuming an inelastic collision).

 

[Gamingphreek]

Just out of curiosity would this work.

You know in a car how if you slam on the brakes and come to a stop like that it jerks you. However if you slame on them and as you are slowing down release some of that energy by letting up on the break a little bit. What if they made thick armor like they did but they put minor air pockets all around to "dissipate" the momentum built up from the sudden stop.

-Kevin

 

[Gibsons]

Actually, the gun *has* to have the same momentum as the bullet (rather, m_bullet * v_bullet + m_gun * v_gun must equal 0 assuming the gun-bullet system was at rest when it was fired). It's just that the gun weighs several thousand times more than the bullet (usually a few kilos versus a few grams for a high-powered rifle), so its velocity (and thereby KE) is much, much lower.

My point exactly. The same process works when the bullet has to move a ceramic plate before it can impart force to the wearer. The weights aren't the same, but it's the same idea: for a given amount of momentum, the available ke decreases dramatically as the weight increases/velocity decreases.

Read the posts a few above this one, where the math is gone over in detail.

You might be talking about my post (?).

 

[Velk]

Originally posted by: Gibsons
Kinetic energy is the capacity to do work. I think it's a much more important factor than momentum. If momentum was the story, then the back end of the gun would be about as dangerous as the front end, no? By definition it has to have the same momentum as the bulletl. It's all because kinetic energy is determined by the square of the velocity (1/2m(v)^2) as opposed to momentum (mv).

Actually energy is irrelevant with regards to calculating what the velocity of someone struck by a bullet is, mostly because conversation of energy only applies for perfectly elastic collisions.

The reason that it isn't much help is because energy, unlike velocity, does not have a *direction*.

In the case of a perfectly elastic collision between the solider and the bullet, kinetic energy would be conserved, but it would be conversed by knocking the soldier back slightly and sending the bullet richochetting in the opposite direction.

The dangerous part is actually the force required to bring the bullet to a standstill in a given distance - on the rifle this force can be much smaller because it is applied over a longer distance and over a much larger surface area. While you can use, say, 50cm of barrel to accelerate the bullet from the gun, if you use even 5cm of your body to slow it down it's already punched a hole in you.

Were it possible to have a situation where a bullet was fired and at the other end it passed into the barrel of an identical gun then experienced the same acceleration force as being fired, there would be no difference in impact between the shooter and the person being shot at.

 

[TerribleTerryTate]

Disclaimer: This will probably be incredibly stupid.

Would there be any way to make a flexible (wearable) armor that works using a viscous fluid and valves like a shock absorber in a vehicle? Suppose a kevlar vest was made with several compartments that are separated very securely by some means I cannot think of. Each section is filled with a viscous fluid and the sections are connected by small openings. When a bullet hits the vest in a particular section, the struck section receives the impulse, compressing the fluid until it is forced through the rest of the vest. This transfers some of the momentum/energy/ying/yang from the bullet into the motion and temperature of the fluid. Of course, the exterior fabric would have to be thick enough to prevent the bullet from penetrating each fluid cell.

And if it doesn't work, it could at least be used as a make-shift waterbed...

 

[BujinZero]

What would you, as the party being faced with super-armor wearing soldiers, do to combat them? Especially when it's your duty to defend those you love from the invaders?

Projectile weapons are just part of the arsenal, and super armor would make war far more terrible. If I encountered soldiers that couldn't be killed with bullets, I'd be quick to adopt other tactics. Booby traps, attacks on supply convoys, stealthy infiltrations and guerilla war, anything. Chemical weapons, biological weapons, and nuclear weapons, whatever it takes. When there's a will, there tends to be a way.

I find the war robot interesting. It's an excellent sniper and anti-armor tool, but it would need troops to support it, and would be terrible if threatened. A decent counter sniper would probably make quick work of it. Hell, you could walk up right behind it, it's only dangerous in one direction.

 

[thecrecarc]

i didn't read the entire thread so i don't know id this was mentioned before but i think there is a tpe of armor whose molecle is purposly deformed so that when a bullet hits it it turns hard and stops the bullet. then there are sientists who are tring to make armour of a of spider's silk because that is like harder that steel

 

[jjzelinski]

Originally posted by: klaviernista

Originally posted by: ReiAyanami
Raytheon was already working on something of an anti-bullet system. basically a laser mounted tank shoots and vaporizes incoming bullets, the problem so far is energy requirements (MASSIVE). to top it off acoustics analysis can tell the soldier almost exactly where the bullets are being fired from so they can respond.

Couple thoughts on this:
1) The targeting and tracking system for a tank based anti-ballistic system would have to be ridiculous to be effective. The targeting/tracking system for this is well well beyond most if not all current systems (the system would have to track hundred of tiny fast moving objects with sufficient precision to direct the laser directly onto them). This is obviously a big problem. This could be offset by making the beamwidth of the laser large (i.e. feet), but then you run into problems with #2

2) Energy delivery: Not only must the targeting and tracking system be sufficient to track incoming bullets and direct a laser onto them, it must do so with sufficient time for the laser to impart enough energy to either vaporize the incoming round or partially melt it, thereby changing its trajectory by changing the aerodynamics of the round. Seeing as how bullets fly quite fast, the radiant power of the laser would have to be huge. Further, if you expand the beamwidth of the laser to help out with the tracking sytem, then energy in the laser beam is (generally) more diffuse, meaning less energy is imparted to a target by the laser. I.e. imagiune this O is representive of the beamwidth of a laser. This . is be a bullet. Obviously, if a large beamwidth laser is used only a portion of the beam actually shines on the bullet, thus, less energy is imparted to the bullet.

ONly thing I heard similar to this is the airborne laser missile defense system. The chemical laser used in that system required a 747 to carry it and could only be fired once or twice before the plane had to land to replenish the laser system. Check it out HERE

As for the acoustic positioning system, that is much more plausible I think. At least for current or near future applications.

Haven't bothered to read on and see if anyone's responded to this, but I'd like to offer that it is totally feasible. You're right in assuming a mechanical steering apparatus would be far to slow to take down a bullet, but if you instead use electrical steering you would be working in an element of time that makes bullets slow by comparison. I could see one of two methods being used:

1.) Using a phased array (radar term) that would either hit one predetermined coordinate using radar tracking, or simply transmit all beams at once in a "region."

2.) Use beam steering like you would use on a cathode ray tube. Now Imp starting to second guess myself on this one since I don't think light behaves the same way as electrons, so it may not be possible to use electromagnetism on the beam like you would with an electron gun.

Okay one more...

If a mechanical approach were necessary should the physics in my first two theories prove to be crap Then why not employ small, highly maneuverable mirrors to reflect the beam. This would eliminate much of the mass of a laser that would make it otherwise to unwieldy to track a bullet.

 

[jjzelinski]

Originally posted by: natenut
YES!!!!! we do have force fields. Have you ever seen an electrically conductive material droped through a high tesla magnetic field? it slows to a crawl due to the induced current in the conductor. Bullets are usually copper jacketed lead. If there was a way to make a large enough magnetic field any projectile would be slowed to a crawl. the faster the projectile the greater the opposing force. Please correcet me if i'm wrong but i draw my theory from dropping pennies on to superconducting magnets. it defys logic when you see it happen and i'm not talking about the messinger (sp?) effect.

-natenut

edit: and by any projectile i mean electrically conductive projectile.

You know that got me thinking, what if extrmely high powered electrmagnets were placed around an armored vehicle to attract bullets to less dangerous parts (away from the crew.) I can think of many caveats to this but it still intrigues me.