Gurney equations, Explosively formed penetrators and kinetic energy

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Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-17 10:52am

Hello!

I'm having trouble with a set of equations meant to describe the velocity of an Explosively Formed penetrator, called the Gurney equations. Specifically, the kinetic energy delivered by the explosive charge, and the kinetic energy contained in the EFP, do not add up.

Here is the set-up:

Image

M is the metal plate, or the 'flyer'. C is the explosive charge. N is the tamper or backplate.
In our scenario, M is the projectile, C is the beryllium filler and N is the nuclear warhead of mass about 10 times greater than explosive charge. It allows us to consider the metal plate as infinitely tamped.

Here is the equation:

Image

Vm: Velocity of the metal plate
E: yield energy converted into thermal energy within the filler. (2E)^0.5 is the specific velocity of our device.
M: mass of metal plate
C: mass of filler

Let's use a 1 kiloton yield warhead, massing 100kg. It is configured like a pulse propulsion unit for an Orion driver, delivering about 85% of its energy into heating a beryllium filler. This is 3.56 TJ.

The beryllium masses 10kg.

The metal plate is 10kg.

Using the equation, we get an EFP flying out at 2311km/s.

The problem:

10kg at 2311km/s contains 26TJ of kinetic energy. This is higher than the energy delivered by the warhead.

What could be the problem?
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby LaCroix » 2017-05-17 11:48am

I'm not qualified to say if your scenario with the switched components would work the way you imagine it would, but could you please write down your equations and how you solved it to get these numbers.
coming up with a number that is 10 times of a number that seems sensible (2.6TJ kinetic for 3.56 TJ input) means that it could be that you just accidentally used one zero too much at one point.
A minute's thought suggests that the very idea of this is stupid. A more detailed examination raises the possibility that it might be an answer to the question "how could the Germans win the war after the US gets involved?" - Captain Seafort, in a thread proposing a 1942 'D-Day' in Quiberon Bay

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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-17 01:44pm

LaCroix wrote:I'm not qualified to say if your scenario with the switched components would work the way you imagine it would, but could you please write down your equations and how you solved it to get these numbers.
coming up with a number that is 10 times of a number that seems sensible (2.6TJ kinetic for 3.56 TJ input) means that it could be that you just accidentally used one zero too much at one point.


Okay then.
The calculations in order.

1 kiloton yield is 4.184*10^12 Joules.
85% efficiency is 4.184 * 0.85 = 3.5564 TJ
(2E)^0.5
(2*3.5564*10^12)^0.5 = 2666983
(M/C +1/3)^-0.5
(10/10 +1/3)^-0.5 = 1.3333^-0.5 = 0.866025

Vm = 2666983 * 0.866025 = 2309675.3

Kinetic energy in plate:
0.5 * 10 * 2309675.3^2 = 5 * 5334599991430.09 = 26672999957150.45

Kinetic energy / warhead energy = 26672999957150.45/3556400000000 = 7.4999

Either only 13.33% of the plate gets ejected at that velocity, which shouldn't be the case with a flat plate, or we follow the kinetic energy limits and the plate only gets ejected to 843.4km/s.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby jwl » 2017-05-17 04:32pm

Looked up the equation because it looks fishy due to failing dimensional analysis. It turns out E does not stand for energy, it stands for the Gurney constant.

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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-17 05:02pm

jwl wrote:Looked up the equation because it looks fishy due to failing dimensional analysis. It turns out E does not stand for energy, it stands for the Gurney constant.


I see. I thought it was a function of energy density.

How would I go about determining gurney constants of explosive fillers containing X joules at Y density? The only numbers I have found so far are empirical studies on conventional explosives.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby jwl » 2017-05-18 04:38am

matterbeam wrote:
jwl wrote:Looked up the equation because it looks fishy due to failing dimensional analysis. It turns out E does not stand for energy, it stands for the Gurney constant.


I see. I thought it was a function of energy density.

How would I go about determining gurney constants of explosive fillers containing X joules at Y density? The only numbers I have found so far are empirical studies on conventional explosives.

I don't know what it is, I hadn't heard of the equation before looking it up. But if E is energy density you didn't put energy density in your equation, you put energy in.

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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Sea Skimmer » 2017-05-18 06:51pm

1) A flyer plate warhead isn't an EFP, no deformation of the plate is required for it to function correctly.
2) Equation is not really valid for the purpose you are attempting in general; among other problems the equations are never valid for extreme situations in charge to liner energy-mass. That said you might get a more useful result out of it if you assumed a more realistic efficiency then 85%, which is basically implausible, in fact any claim of over 50% for any shaped charge is dubious. It could be oh, like a half percent instead, if your expecting to couple a 1kt nuke to only 10kg of material. I know people have claimed nuke shaped charges might transmit that much heat into the material, based off about nothing solid I can tell, but that's far different then converting that heat into KE projected in a useful direction in the first place.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Elheru Aran » 2017-05-19 12:48pm

Totally irrelevant, but I keep looking at this and thinking it's about James Gurney and Dinotopia. 'Explosively formed penetrators' don't really fit that mental picture, alas...
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-19 01:58pm

Sea Skimmer wrote:1) A flyer plate warhead isn't an EFP, no deformation of the plate is required for it to function correctly.
2) Equation is not really valid for the purpose you are attempting in general; among other problems the equations are never valid for extreme situations in charge to liner energy-mass. That said you might get a more useful result out of it if you assumed a more realistic efficiency then 85%, which is basically implausible, in fact any claim of over 50% for any shaped charge is dubious. It could be oh, like a half percent instead, if your expecting to couple a 1kt nuke to only 10kg of material. I know people have claimed nuke shaped charges might transmit that much heat into the material, based off about nothing solid I can tell, but that's far different then converting that heat into KE projected in a useful direction in the first place.


2) The 85% figure is from a NASA study on nuclear pulse propulsion units. It is the 'directivity' of the warhead, ie, how much energy is pumped into the beryllium filler.
Experimental results suggest 5% nuclear-to-kinetic coupling. Unless the first experiments have already unlocked the full potential of the technology, I doubt 5% is a realistic figure either. The truth is somewhere in between.

I have been looking into gaseous models for the explosive filler. The problem is that I cannot find any estimates of heat capacities that go into the tens or hundreds of thousands of kelvins. This is necessary to determine the temperature and therefore the root mean square velocity.

But, if it reaches temperatures of 10 million K, then beryllium gas travels at 166km/s.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Simon_Jester » 2017-05-20 12:29am

Can you link to that study?

Also, the reason you can't find estimates of heat capacities at those temperatures is that plasmas behave very differently from ordinary states of matter and don't necessarily fit neatly into the equations you're accustomed to. I confess I've forgotten pretty much all my plasma physics, but it is not a simple or trivial thing to predict how a plasma will behave.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Imperial528 » 2017-05-20 09:58am

The equations for explosively formed projectiles are not applicable here. EFP warheads do not vaporize the liner, whereas a casaba howitzer or Orion charge unit does vaporize the entire assembly including projectile liner. Standard shaped charges are also too dissimilar due to the energies involved.

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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-20 01:53pm

Simon_Jester wrote:Can you link to that study?

Also, the reason you can't find estimates of heat capacities at those temperatures is that plasmas behave very differently from ordinary states of matter and don't necessarily fit neatly into the equations you're accustomed to. I confess I've forgotten pretty much all my plasma physics, but it is not a simple or trivial thing to predict how a plasma will behave.


85% thermal efficiency is derived from Atomic Rocket's entry on Orion Pulse Propulsion, itself citing from GA-5009 vol III "Nuclear Pulse Space Vehicle Study - Conceptual Vehicle Design" by General Atomics (1964) (https://web.archive.org/web/20070704104944/http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770085619_1977085619.pdf) and further studies.

Considering that a nuclear explosion delivers its energy over the course of 10 nanoseconds, would a simple monoatomic gas model be sufficient to get a rough temperature since no expansion of electromagnetic effects have time to affect the plasma?

Using the ideal gas constant, 3 degrees of freedom and a molar mass of 9g/mol for beryllium, I get a heat capacity of 1385J/kg/K. This is close to the 1825J/kg/K listed here (http://www.engineersedge.com/materials/specific_heat_capacity_of_metals_13259.htm). Dumping 85% of a kiloton yield into 10kg of beryllium results in a temperature of about 256 million K.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-20 01:54pm

Imperial528 wrote:The equations for explosively formed projectiles are not applicable here. EFP warheads do not vaporize the liner, whereas a casaba howitzer or Orion charge unit does vaporize the entire assembly including projectile liner. Standard shaped charges are also too dissimilar due to the energies involved.


So what can be used?
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Simon_Jester » 2017-05-20 04:47pm

matterbeam wrote:85% thermal efficiency is derived from Atomic Rocket's entry on Orion Pulse Propulsion, itself citing from GA-5009 vol III "Nuclear Pulse Space Vehicle Study - Conceptual Vehicle Design" by General Atomics (1964) (https://web.archive.org/web/20070704104944/http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770085619_1977085619.pdf) and further studies.

Considering that a nuclear explosion delivers its energy over the course of 10 nanoseconds, would a simple monoatomic gas model be sufficient to get a rough temperature since no expansion of electromagnetic effects have time to affect the plasma?
No, it would not. Understanding the physics of how matter behaves at millions of degrees is not something you can conveniently simplify using formulas from a high school science textbook and expect reasonably accurate results.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-20 05:04pm

Simon_Jester wrote:
matterbeam wrote:85% thermal efficiency is derived from Atomic Rocket's entry on Orion Pulse Propulsion, itself citing from GA-5009 vol III "Nuclear Pulse Space Vehicle Study - Conceptual Vehicle Design" by General Atomics (1964) (https://web.archive.org/web/20070704104944/http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770085619_1977085619.pdf) and further studies.

Considering that a nuclear explosion delivers its energy over the course of 10 nanoseconds, would a simple monoatomic gas model be sufficient to get a rough temperature since no expansion of electromagnetic effects have time to affect the plasma?
No, it would not. Understanding the physics of how matter behaves at millions of degrees is not something you can conveniently simplify using formulas from a high school science textbook and expect reasonably accurate results.


What would you use?
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Simon_Jester » 2017-05-21 03:10pm

Personally? I'd ask a real rocket scientist or plasma physicist.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Sea Skimmer » 2017-05-21 08:48pm

You need a hydrocode program to do this right, which do exist as open source but a lot of them still run in Fortran because that's what the original math was validated in.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Imperial528 » 2017-05-21 10:25pm

Simon_Jester wrote:Personally? I'd ask a real rocket scientist or plasma physicist.


I would ask a nuclear physicist as well, just to be on the safe side of things. Especially since the first stage of major energy loss is in the efficiency of the actual reflection/focusing of radiation into the filler via the casing.

Frankly matterbeam if you can't find the equations necessary in the design study or elsewhere online, then the documentation for that work is probably still classified or was possibly destroyed. You or someone with the knowledge to do so would have to do the calculations more or less from the ground up.

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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby LaCroix » 2017-05-22 05:16am

I don't think that anybody ever did any work on how to form a kinetic penetrator by setting off a nuke. Matterbeam is just running amok with a known formula, and substituting "explosive" with the biggest bang he can find, probably for some sci-fi story reason.

Biggest problem with this setup is that in the currently stated configuration, he'd lose about 80-90% of the energy due to the fact that there is no tamper plate, as he switched them around. Most likely, this is not going to work any other way other than making a nice fireball and a partly vaporized piece of metal flying off due to the normal blast wave.
A minute's thought suggests that the very idea of this is stupid. A more detailed examination raises the possibility that it might be an answer to the question "how could the Germans win the war after the US gets involved?" - Captain Seafort, in a thread proposing a 1942 'D-Day' in Quiberon Bay

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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-22 05:35am

LaCroix wrote:I don't think that anybody ever did any work on how to form a kinetic penetrator by setting off a nuke.


http://scienceandglobalsecurity.org/archive/sgs01fenstermacher.pdf
Pages 204-206

Matterbeam is just running amok with a known formula, and substituting "explosive" with the biggest bang he can find, probably for some sci-fi story reason.


Do you have a better model?

Biggest problem with this setup is that in the currently stated configuration, he'd lose about 80-90% of the energy due to the fact that there is no tamper plate, as he switched them around. Most likely, this is not going to work any other way other than making a nice fireball and a partly vaporized piece of metal flying off due to the normal blast wave.


https://en.wikipedia.org/wiki/Gurney_equations#Open-faced_sandwich
A 1:1 filler:plate ratio allows for a flyer velocity of 55% the gurney constant.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-22 05:36am

Simon_Jester wrote:Personally? I'd ask a real rocket scientist or plasma physicist.


They seem to be in very short supply these days.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-22 05:38am

Imperial528 wrote:
Simon_Jester wrote:Personally? I'd ask a real rocket scientist or plasma physicist.


I would ask a nuclear physicist as well, just to be on the safe side of things. Especially since the first stage of major energy loss is in the efficiency of the actual reflection/focusing of radiation into the filler via the casing.

Frankly matterbeam if you can't find the equations necessary in the design study or elsewhere online, then the documentation for that work is probably still classified or was possibly destroyed. You or someone with the knowledge to do so would have to do the calculations more or less from the ground up.


The energy losses in the nuclear shaped charge itself is not in question - I directly used figures from the propulsive units designed for the Orion drive. I take 85% of the effective yield of the nuke and work from there.

A model for the acceleration of a flyer by an arbitrarily energetic plasma should exist, since it is being used in fusion ignition research.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby LaCroix » 2017-05-22 07:51am

That document is about nuclear shaped charges, and the concept of shaped charges is a bit different from epf. Enough to matter. Sea Skimmer is the guy who knows about it, he briefly discussed this once here on the board. Also - it usually means that there is some back plate to direct the charge.

And it states on page 204 that
up to 5 percent of the energy of a small nuclear device reportedly can be converted into kinetic energy of a plate, presumably by employing some combination of explosive wave-shaping and "gun-barrel" design, and produce velocities of 100 kilometers per second and beam angles of 10-3 radians.


There you have your better model. The footnote contains the maths, too, but it is a value that seems to be based on tests.
So, no 85%. Far from it.

Also, they think of a kind of shotgun - the plate getting fractured into millions of fragments (molten, most likely), that will produce a spread of 1 particle per square meter at 2000km range.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby matterbeam » 2017-05-22 09:16am

LaCroix wrote:That document is about nuclear shaped charges, and the concept of shaped charges is a bit different from epf. Enough to matter. Sea Skimmer is the guy who knows about it, he briefly discussed this once here on the board. Also - it usually means that there is some back plate to direct the charge.


The difference is the metal liner. Modern lab testing uses shaped charges without the liner, to find the detonation velocity, and then with the liner to test EFPs.

The 'back plate' is the tamper and is not strictly necessary.

And it states on page 204 that
up to 5 percent of the energy of a small nuclear device reportedly can be converted into kinetic energy of a plate, presumably by employing some combination of explosive wave-shaping and "gun-barrel" design, and produce velocities of 100 kilometers per second and beam angles of 10-3 radians.


Is this a physical limit? Can we go faster? Is this efficiency the result of lab tests or theoretical limits? Even 5% of a megaton yield can push a 1 ton projectile to 647km/s.

So, no 85%. Far from it.


"I take 85% of the effective yield of the nuke and work from there."

Also, they think of a kind of shotgun - the plate getting fractured into millions of fragments (molten, most likely), that will produce a spread of 1 particle per square meter at 2000km range.


Useful for striking down clusters of nuclear re-entry vehicles in low orbit using untested aiming technology, but certainly not a generalization we must rely upon.
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Re: Gurney equations, Explosively formed penetrators and kinetic energy

Postby Imperial528 » 2017-05-22 09:38am

I think I see where the confusion over the 85% is coming from. That's the efficiency of the Orion pulse units, and the Casaba charges described in this paper have a different design from the Orion units. It looks like the design described in the paper is attempting to shape the actual burn wave to converge on the projectile liner as in a conventional shaped charge, whereas Orion pulse units get very high efficiencies by focusing the radiation products of the detonation instead.

If you were to use an Orion charge with the right propellant material you can get a beam with a 5.7 degree cone-angle, much better than the 22.5 degree cone of the standard charge but not as focused as the design described in the paper, so you would lose in range for gains in overall efficiency.

Matterbeam, the section from the paper that appears to be the most helpful to you would be this footnote:

THE EFFECTS OF NUCLEAR TEST-BAN REGIMES ON THIRD-GENERATION-WEAPON INNOVATION wrote:‡ The energy fluence per beam, E in J/m2, is approximately ηY/(NbR2θ2), where η is the fraction of overall yield transferred to the pellets, Y is the bomb yield (1 kiloton is equivalent to 4.2 × 1012 joules), Nb is the number of individual beams being driven by one bomb, R is the distance to the target, and θ is the individual full-beam divergence angle. A maneuvering target could accelerate out of the path of the beam if amR/vf2 > θ, where am is the magnitude of the target's average acceleration, vf is the particle velocity, and τ = R/vf is the particle fly-out time. (For comparison, the average acceleration of ICBMs is about 40 m/s2.) To deliver this energy requires a total mass per beam of Mb = 2E(Rθ)2/vf2.


I don't know how accurate this would be to apply to an Orion-derived device, though, since the two designs rely on very different operating mechanisms.


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