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Comment: Overstating the case like mad (Score 1) 140

by Phanatic1a (#45090443) Attached to: Two-Laser Boron Fusion Lights the Way To Radiation-Free Energy

This doesn't light the way to radiation-free energy.

http://dspace.mit.edu/handle/1721.1/11412

"Although there have been a few proposals for fusion reactors employing plasmas far out of thermodynamic equilibrium (such as migma and inertial-electrostatic confinement), there has never been a broad, systematic study of the entire possible range of such devices. This research fills that gap by deriving fundamental power limitations which apply to virtually any possible type of fusion reactor that uses a grossly nonequilibrium plasma. Two main categories of nonequilibrium plasmas are considered: (1) systems in which the electrons and/or fuel ions possess a significantly non-Maxwellian velocity distribution, and (2) systems in which at least two particle species, such as electrons and ions or two different species of fuel ions, are at radically different mean energies. These types of plasmas would be of particular interest for overcoming bremsstrahlung radiation losses from advanced aneutronic fuels (e.g. ^3He-^3He, p-^{11}B, and p- ^6Li) or for reducing the number of D-D side reactions in D-^3He plasmas. Analytical Fokker-Planck calculations are used to determine accurately the minimum recirculating power that must be extracted from undesirable regions of the plasma's phase space and reinjected into the proper regions of the phase space in order to counteract the effects of collisional scattering events and keep the plasma out of equilibrium. In virtually all cases, this minimum recirculating power is substantially larger than the fusion power, so barring the discovery of methods for recirculating the power at exceedingly high efficiencies, reactors employing plasmas not in thermodynamic equilibrium will not be able to produce net power. Consequently, the advanced aneutronic fuels cannot generate net power in any foreseeable reactor operating either in or out of equilibrium."

You're shooting a beam of protons through a gas of fuel, this is about as far away from thermal equilibrium as you can get. Only a small proportion of protons will actually wind up fusing, the power it takes to generate those protons and shoot them into the fuel (or the power to take the ones that miss fuel ions and recirculate them to give them another pass through the fuel) will dwarf the power you get from the fusion reactions. In other words: big fat hairy deal. Fusion is easy. It's the extracting useful amounts of energy from it that's hard, and this process can't do that.

Comment: Re:Three levels of break-even (Score 3, Informative) 429

by Phanatic1a (#45068225) Attached to: Fusion Reactor Breaks Even

You're not delusional. JT-60 in Japan sort of reached breakeven, but with one hell of a caveat: JT-60 only uses D-D fuel, but it achieved conditions in the plasma such that if the D-D fuel was replaced with D-T fuel, it would have achieved Q=1.25.

What's delusional is the notion that ICF can ever be a commercial source of fusion power. Even after you squint and wave your hands and say "We reached break-even, if you count only the energy absorbed by the fuel," you need to realize the huge inefficiencies at every step along the chain. Conversion of electricity into laser energy is really inefficient. The IR lasers are frequency-converted into UV beams, a process which is only 50% efficient. And only about 10% of *that* actually goes into compressing the fuel.

And that fuel is frozen D-T contained within a copper-doped beryllium capsule that needs to be spherical to micron tolerances, and the surfaces of that sphere need to be smooth to *nanometer* tolerances. The beryllium must be precisely 150 microns thick, and a 5-micron hole is laser-drilled through it. The capsule in turns rests within an equally-precisely made hohlraum comprised of a gold/uranium alloy. Each one of these precision assemblies costs tens of thousands of dollars to make, assembly of the various parts also must be done to micron tolerances. And out of this, if fusion works perfectly and every bit of the fuel is used, you can expect a maximum possible energy output of 45 megajoules. That's 12.5 kilowatt-hours of energy; if you can manage the miraculous feat of 100% efficiently converting that back into electricity, you could sell that electricity for about $1.25.

For commercial fusion, they'll need to burn 15 of these targets per second, every second, indefinitely. Which means that in addition to needing a fusion gain factor of about *60* (compared to 20 for a tokamak, which will also probably never produce commercial fusion power), they'll need to get the fuel cost down to like 10 cents per target.

Meanwhile, fission just works. Figure out how many LFTRs we could build for the cost of the NIF and weep. ICF is a jobs program for engineers who got scared as hell when the cold-war ended and started pimping their bomb-research machines to environmentalists who don't understand physics or economics.

Comment: Re:And the story is...? (Score 1) 453

by Phanatic1a (#44328217) Attached to: TSA Orders Searches of Valet Parked Car At Airport

"If you're going to park a car full of explosives, you can either create a small crater in a car park, or you will go for the airport - so cars that are left outside are checked."

Not quite. What you probably meant to say was that "...cars that are left outside are checked by some minimum-wage monkey whose chief marketable skill is being able to maneuver a car into a parking space."

If bombs in parked cars were an actual concern, they'd be using trained and skilled personnel in order to search them. You think a parking lot attendant is going to check underneath the car, or under the hood, or in the trunk, or in the wheel wells, or behind the dash, or any of the other places in a car where substantial quantities of explosive can be concealed from view?

  No. Of course not. At best he's going to check under the seats and steal your loose change. This should let intelligent people know that there are two possibilities here:

1. Bombs in parked cars at short-term valet-parked lots are not a serious security threat.
2. The TSA is criminally incompetent and that useless titsuck of a government agency should be sequestered with extreme prejudice.

Note: These are not mutually exclusive possibilities.

Comment: Re:Uranium means it is not a silver bullet (Score 1) 293

by Phanatic1a (#43987525) Attached to: <em>Pandora's Promise</em> and the Problem of "Solutionism"

I qualified, "With breeders and/or sane fuel cycles/reactor designs." The fissionable U235 goes into reactors because we designed the present-day fuel cycle around the military's interest in bomb production. If you designed a fuel cycle around electrical production, there are reactor designs that will either breed U238 into fissile plutonium (all conventional reactors do this to an extent), or which will use 238 as fuel directly (U238 will not sustain a fission chain reaction but it will fission and release energy when bombarded by fast neutrons).

But the underlying point is that whever someone says "There are only [x] years of uranium|oil|coal|lithium reserves left!" that means "at current prices," because that's how "reserves" *are defined.*. With lithium at $y/lb, there are x years of lithium reserves left to mine. But as the price of lithium increases because those reserves start running low, it becomes economical to mine sources of lithium that previously weren't worth mining. So at price $y+1/lb, there are now x+z years of lithium reserves left. Uranium, unlike coal or oil, is so ridiculously energy-dense that it would have to become ridiculously expensive for it not to be worth using to produce energy.

Comment: Re:Uranium means it is not a silver bullet (Score 3, Interesting) 293

by Phanatic1a (#43985623) Attached to: <em>Pandora's Promise</em> and the Problem of "Solutionism"

"something that is very limited on this planet."

Bullshit. Uranium is ridiculously abundant. There's more uranium than silver, or tin, or cadmium, or antimony.

"If Fukashima has not occurred, we would be currently looking at a global uranium shortage in the next 5 years as existing major sources (re-purposing from old warheads) dry up and are not replaced with new mines."

Utter and total nonsense. Old warheads are not major sources of uranium, because warheads are fabricated from *plutonium*, not uranium, which is produced in reactors specifically so we can build warheads out of it. There are billions of tons of uranium dissolved in seawater, with another 32000 tons being carried into the oceans by rivers every year. With breeders and/or sane fuel cycles/reactor designs, there's enough uranium to provide our present electrical demands for, literally, millions of years.

And there's three times as much thorium as there is uranium.

"Whenever production of power plants comes back on track, we will once again be facing such a shortage."

Only someone who completely fails to understand what constitutes ore reserves would say such a thing. As uranium prices rise, ore reserves increase, because a higher price for uranium means other sources become economical to exploit. There will only ever be a shortage of uranium *at a given price*, and once that price gets high enough to make extraction from seawater economical, supplies become effectively limitless. And since nuclear fuel is so energy dense, orders of magnitude moreso than chemical fuels, the raw price of ore contributes very little to the cost of electricity coming out of the plant.

Comment: Density (Score 1, Interesting) 198

by Phanatic1a (#43270379) Attached to: Graphene Aerogel Takes World's Lightest Material Crown

" the ultra-light aerogel has a density of just 0.16 mg/cm3, which is lower than that of helium and just twice that of hydrogen."

Picture in the article shows a chunk of the stuff being supported by a blade of grass. If the density's lower than that of helium, why isn't it floating away instead of sitting there like a thing that's denser than the atmosphere around it?

Comment: Re:Maybe they do. Maybe they don't. (Score 1) 861

by Phanatic1a (#42043055) Attached to: Israel's Iron Dome Missile Defense Shield Actually Works

Israel actually has a free press. If missiles were falling into populated areas and killing a bunch of Israelis, we'd hear about it.

And beyond that, I'd expect to hear about it *from the government*. On one hand, there's embarassment from deploying a defense system that doesn't do what it's supposed to do, and a morale cost from the failure to meet the expectations of the citizenry. On the other hand, there's a bit of a propaganda benefit from showing the world a bunch of people who died because Palestinians or Egyptians fired rockets at their homes. I suspect the math there works out in favor of not concealing the bodies.

Comment: NG, what happened to you (Score 5, Informative) 77

It's sad to see that even National Geographic now has to tart up the very real risks of this attempt with dramatic bullshit like the first sentence: "the atmosphere above 12 miles, or 63,000 feet (19,200 meters)—known as the Armstrong line (named for Harry George Armstrong, who founded the U.S. Air Force's Department of Space Medicine in 1947)—is so thin that, if not protected, human blood will literally boil. To prevent that, Baumgartner's airtight suit and the capsule around him will be continuously pressurized to create a personal atmosphere that isolates him from the void surrounding him."

Nonsense. Even if you're in an environment of pure vacuum, your circulatory system is *pressurized*. This is called "blood pressure." Your blood will not boil in space. It will outgas, as dissolved gases in it come out of solution, but that's not boiling; Scuba divers who ascend too rapidly get the bends as N2 leaves solution, but their blood doesn't boil, they don't die. Fluids exposed to atmosphere, like the water on the surface of the eyes and lining the mucous membranes will boil, but not the blood.

"The smallest crack in this protective layer would cause almost immediate death."

Again, why tart this up? The guy who holds the current record and who's helping with this jump, Joe Kittinger? He suffered a "crack" in his "protective layer," in one of his gloves. His hand swelled up like a balloon, and it hurt, and he had some bruising/soft tissue damage, but he continued with the mission and his hand returned to normal size when he descended and healed normally.

Sad to see National Geographic turning into Discover.

Comment: Re:Old news... (Score 5, Interesting) 321

by Phanatic1a (#41518765) Attached to: The US Navy's Railgun Program

"They won't be willing to wait and it's not exactly a simple thing to change the power output of a Rankine cycle nuclear power plant at a whim."

Actually, it is, it's called a throttle. When you're in a nuclear submarine puttering along at 5 knots and someone drops a torpedo on you, and you want to get up to 30+ knots as fast as you can, you do it. You take more heat out of the coolant, which cools down the water in the reactor, which increases the reaction rate, which produces more power, this relationship is very tight and the changes can happen very rapidly. Way more rapidly than shoveling in more coal.

The power source is a non-issue. Gas turbine, nuclear, whatever, there's plenty of available power. A single destroyer carries 4 gas turbine engines that are each capable of 40,000+ shaft horsepower. It's generation capacity that's more of an issue, but even that just means "wait for a longer period of time between shots."

The means of delivering electrical power to the projectile without arcing destroying the rails is an issue. Ideally you want all the current in the world at as low a voltage as you can manage it, so capacitors aren't as good as a magnetohomopolar generator. But getting the power to put into the capacitors of MHG is not a complex problem.

Comment: Argh science journalism. (Score 5, Insightful) 155

by Phanatic1a (#41273423) Attached to: Violation of Heisenberg's Uncertainty Principle

This article is horrible.

"The Heisenberg uncertainty principle is in part an embodiment of the idea that in the quantum world, the mere act of observing an event changes it."

That's not the Heisenberg uncertainty principle. That's just the observer effect, and it's not something peculiar to quantum mechanics. You want to measure the temperature of a system, so you stick a thermometer in there. Okay, the mercury in the thermometer absorbs a bit of heat from the system, providing you with a temperature measurement at the same time it changes the temperature of the system. If you want to measure the parameters of a particle, you stick a bubble chamber in the way, and as the particle flies through the chamber it smacks into hydrogen molecules, showing you what it's doing but also taking a different path than it would have if none of those hydrogen molecules were in the way. Big fat hairy deal.

The HUP doesn't just say that you can't simultaneously measure the position and momentum of a particle, it says that a particle *does not simultaneously possess* a well-defined position and momentum. If the particle's doing something in a system and is interacting in such a way that you can define its position to arbitrary precision, then it *does not have* a well-defined momentum for you to measure, and vice versa. Position and momentum are what are called quantum conjugate variables, and the HUP says that when you have a pair of those variables, then the product of their uncertainties is greater than or equal to a constant. There is *no state* in which that particle is even *allowed* to exist in which it possesses both a well-defined position and well-defined momentum.

A signal processing analogy, for any analog people. A particle's wavefunction carries information about its position and its momentum. Where the wave exists is where the particle actually is, and the wavelength is the particle's momentum. Take a particle whose momentum you know to the utmost precision, and graph that. Range of momentums on the x axis, probability of the particle having that momentum on the y axis. You'll get a graph that looks like a Dirac function, a value of 0 everywhere except for a single spike corresponding to the particle momentum, area under the curve of 1.

Now switch domains, change from the momentum to the position domain, this is mathmatically the same thing as changing from a time domain to a frequency domain, which means you can use your old friend the Fourier Transform.

What do you get when you do an FT of a Dirac function? You get a constant value everywhere, from -infinity to +infinity. If you know exactly where that particle is, you have no idea *where* it is, and it's not because you disturbed it in measuring it, it's because *it* has no idea where it is, a well-defined position does not exist; since the uncertainty in the momentum measurement approaches zero than the uncertainty in the position measurement has to approach infinity so that the product of those uncertainties remains greater than a constant.

The "you change the system by measuring it" is an analogy, and it's one that Heisenberg himself used to explain the HUP, but *that is not what it says*. The HUP is not a statement about the process of measuring things, it is a statement about the nature of the universe, and finding a way to improve a measuring system to reduce the disturbance it creates in the system it's measuring has nothing to do with the HUP.

Comment: Re:pump it into the air (Score 1) 347

by Phanatic1a (#40946013) Attached to: US Freezes Nuclear Power Plant Permits Because of Waste Issues

This is a very silly comparison. 1700 PBq of the Chernobyl release was in the form of I-131, which has a half-life of 8 days. Which means that 3 months after the disaster, it was effectively gone. Thousands more Pbq of Xenon-133 were released, but Xe133 has a half-life of 5 days. So after 2 months, that was effectively gone, 99.98% of it had decayed to stable cesium.

The only radioisotopes released from Chernobyl that are still exist in significant amounts, 26 years after the release, are Sr90 and Cs137, with half-lives of about 30 years. Total release of those isotopes was 100 Pbq. So about equal to the total radioactive release from burning coal for 100 years. But that stuff from burning coal? That's going to last for many thousands of years. (And that's just the radioactive release, the arsenic, mercury, etc? That stuff's forever.)

Meanwhile, 300,000 people a year die to air pollution. That beats Chernobyl's total by a factor of 75.

Comment: Re:My little sister picked my BB gun's trigger loc (Score 1) 646

by Phanatic1a (#40797763) Attached to: How a 3-Year-Old Can Open a Gun Safe

"Same thing applies to guns in the home. Even if your kids are perfectly safe around the guns, you need to be cognisant that their friends may not have the same education."

There's a flip-side to that coin.

Even if you don't have a pool in the backyard, teaching your kids to swim is a good idea. Likewise, even if you don't keep guns in your house, you should teach your kids about them and about gun safety. Imagine how you'd feel if your kid kills himself or another kid with a gun he found in a neighbor's house.

Comment: Re:It probably makes sense. (Score 2) 403

by Phanatic1a (#39699935) Attached to: Sixty Years On, B-52s Are Still Going Strong

"But as for the airframe ... as long as they can confirm that the fuselage is sound and in good shape, there's no reason why they can't continue to fly"

The life-limiting factor on the B-52 isn't the fuselage, it's the upper wing, which has a maximum life of 37,500 flight hours.

Given how many flight hours are on the airframes (at *most* 21,000) and the rate of accumulation, the mid-2040s is when we can't maintain the required numbers.

Comment: Re:Hmm (Score 2) 195

by Phanatic1a (#39542407) Attached to: Navy Planning To Build Laser Cannon In Four Years

I wish people would cut this out.

Have you ever seen a high-energy mirror? It's not something you pick up at Bed, Bath & Beyond. They are expensive, they are fragile, they must be kept completely clean. The reflective surface has to be on the *front* of the mirror, not the rear, because there aren't materials transparent enough to pass high-energy laser light through without absorbing enough of it to react unpleasantly and spoil the reflection. So if there's something like a fingerprint, or a dust speck, on the reflective surface, that bit of crud absorbs the incident light, heats up/explodes, and damages the mirror coating. Which means it's not reflective anymore, which means that area of mirror coating now heats up/explodes and damages adjacent areas, leading to catastrophic failure of the mirror.

You are not going to put mirrors on your greasy *boats* that go bouncing around the surface of the *ocean* and have them remain clean enough to offer protection against a multi-kilowatt laser beam.

I am not now, nor have I ever been, a member of the demigodic party. -- Dennis Ritchie

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