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i thought the whole concept was patently absurd.

but, it is actually a nice looking tartan

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speaking of solar, here is a vimeo video demonstrating just what could be done with all that heat..

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Rezwan wrote: Hey, I didn’t know Bananas were radioactive!

yes, because they’re rich in potassium. K-40 is a beta emitter with half-life of 1.21 billion years and natural abundance of 0.0117%

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may i suggest…. fission? chicken kiev

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according to wikipedia, “Under tests Starlite withstands attack by a laser beam producing a temperature of 10,000 degrees Celsius.” But the material is proprietary, and its inventor is asking for a huge stake from its commercialisation, which has really hindered its uptake. — http://en.wikipedia.org/wiki/Starlite

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a few fundamental things will differ, with different fusion approaches: among them the reaction rate, and the energy conversion efficiency.

for fusion to occur, atoms either have to strike each other hard enough (high temperature), or be held in close proximity for long enough (high pressure); and these cost energy. heat can leak out of the reaction chamber, materials and fields can only be so strong, and plasmas are extremely corrosive. these facts affect the size and shape of the reaction chamber.

many fusion approaches attempt to scale the reactor based on a steady, sustained reaction. whereas DPF does fusion in short pulses.

then, after fusion occurs, how do you extract energy? perhaps with a heat engine, or by transforming electromagnetic fields.

efficiency of a heat engine depends on the difference in temperature between the hot chamber and the cool one. this is limited by the materials you choose, to make the chamber.

efficiency of an electric transformer depends on voltage, current, pulse width and shape, and period of firing, and likewise is also limited by the materials chosen.

a single proton-boron reaction produces 8.7 MeV of energy, from a collision in the hundreds of keV range. while this is a gain of more than ten to one, a single reaction is only about a trillionth of a joule. obviously, you need to get many of them going at once, before you can make the reactor do useful work.

all these approaches introduce variables that differ, affecting break-even.

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has anyone shown any residual radioactivity after shutting down the Rossi reactor?

if it’s doing fusion for real, there should be some positron emitters created by it.

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dennisp wrote: But the question will be: Is it cheaper for others to try to steal our resources, or to go to space and get their own?

The solar system has millions of times the resources of the Earth. It’ll be a long time before fighting over them will be cheaper than going after the stuff that’s still free for the taking.

hopefully, once you have fusion, it will be cheaper to recycle resources than to steal ours

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You can measure the build-up of geo tectonic stress, and say roughly how much energy is stored, but nothing i know of will help pinpoint the location, magnitude, and timing of an earthquake.

try this experiment: drop grains of sand onto a mound, and notice they build up and collapse. now try to predict the timing and magnitude of the collapses. you will see that it is impossible.

(if you find it is not impossible, then you could use your method to predict earthquakes as well.)

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zapkitty wrote:
Talked about FF as better for robotic missions without mention of electrode core changeout…

(that would mean a disposable stage or a tug that returns for reuse…. … or multiple staged cores? contact mad scientist central 8) )

Talked about FF as power supply for bases, Mars base talk.

(what about in situ reforming of electrodes?)

That’s where the general-purpose orbital factory comes in.

Here are some numbers that Lerner-hakase tossed out during his Space Show appearance and my patella-twitching responses to a few of them are in parantheses:

50 kj pulse = 500 hz

(isn’t this supposed to be about 10 kj now? 50 kj at 500 hz gives us 25 megawtfs… )

nominal 5 to 20 MW has been talked about for some time.. with possible impact on cooling and anode wear.

5 mw genset @ 3 tons and 2 meters wide

(Unshielded mass? He uses 3 tons as current mass of LPPX-1 but then uses this # for mass for FFs in general use… and they will need shielding )

output 1/3 x-rays – caught by PV with thousands layers of metal
2/3 beam !

(this needs confirmed please)

current LPPX-1 has shielding. maybe the 3 tons includes that?

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Ivy Matt wrote:

what is the parts list and cost?

The patent application describes the basic device. The interview linked to above mentions a price of $2000 per kW for the device. However, there are apparently some catalysts needed for the device to work whose nature has not been disclosed by Ing. Rossi.

…

zapkitty wrote: They claim they are awaiting patent protection before proceeding.

If Ing. Rossi wants patent protection he will have to disclose more than he has disclosed so far.

yes. failure to disclose could invalidate the patent. on the other hand, it is a common tactic used by scammers, who seek to bankroll a “patent pending” technology.

by the way, there was a technical difficulty downloading from the above link (503 Service Unavailable); but, here is an alternate
http://www.freepatentsonline.com/pdfb/documents/wipo/patent_pdf/2009/125/WO2009125444A1/pdf/WO2009125444A1.pdf

curiously, their description seems to require elevated pressure (2 to 20 bars preferred), and ß+ decay of unstable Cu isotopes, with consequent annihilation with an electron.

Looking at cost and yield of reactions in terms of mass excess, in keV, I can see that the lowest cost reaction would be
61-Ni + p +1422.9 keV –> 62-Cu; and the best yield reaction among these would be
60-Cu –> 60-Ni + ß+ +6128 keV

The minimum cost of 1422.9 keV (did i get that value right?) would seem to suggest that the fusions are rather endothermic and will occur at low probability, even at high pressures and temperatures.

At 500°C, what’s the fraction of protons with > 1423 keV energy?

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departing from all hype and conspiracy theory,

what’s the smallest of these reactors that produces greater than break-even?
can an individual buy one today?
what is the parts list and cost?
physical size and mass?

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the water moderates neutrons, so even the odd fast-neutron will be slowed through it; the boron captures neutrons; the lead captures gammas.
all this brings it down to background level.

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“One prototype cable is 6.5 millimeters (mm) in outer diameter and carries a current of 1,200 amperes; a second cable is 7.5 mm in diameter and carries a current as high as 2,800 amperes. They are roughly one-tenth the diameter of typical HTS cables used in the power grid.”

these are very nice, but the range required by FoFu is around 2.5 million amps.. about a thousand times these; and it works out to requiring a pair of flat, conductive sheets, of opposite polarity, separated by ~50 mm.

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i edited the numbers, to bring it up a bit. but, no, this is based on no use of on-board propellant in atmosphere.

the 600 kN is based on 80 kg/s air inflow at 36 km altitude and mach-2.4; inflow can be a lot higher at full atmosphere, and the 5 MW anodes could maybe run at 20 MW each for the first minute.

Ive tried to keep things as simple as possible; hybrid propulsion gets really complicated; the only thing to manage here is the amount to choke the input flow in the engine. even so, it is quite challenging to find an operating niche.

Perhaps I should revise that to say propellant cost, in terms of mass, is king.
for launches beyond LEO, you could spend 99.9% of your lift mass using conventional rockets. I’m seeking an end to that as a fundamental barrier

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