Z-Man

A hundred percent. You will have to work your way out of the gravity well, but that's only about ten times harder than getting out of the Sun's gravity well, so trivial compared to the other problems. And, of course, even at substantial FTL speed, it would still take ages to get anywhere.

Umm, your math is wonky. short bursts of radiation are not less lethal than longer periods of the same total radiation. If 1 Sv/h kills you in 6-8 hours, and an hour has 60 minutes, then 60 Sv/h kill you in 6-8 minutes, and 3600 Sv/h in 6-8 seconds. 300 Sv/h kills you in a minute. (Plus, expect a 10% chance to die soonish at 1-2 Sv already. 6-10 Sv is only "needed" if you want to be sure. So don't linger around.)

No, that would be a non-local interaction. Instead, it's probably best to think of the momentum being "borrowed". Remember how the warp drive requires negative energy densities, so negative mass? Well, negative mass going in the same direction as the spaceship will have momentum pointing to the other direction. That would cancel out the spaceship's momentum, so at all times, the total momentum of spaceship plus warp bubble is zero. (The full picture is more complicated than that.) I don't quite understand the question. If you mean the gravity generated by the bubble itself, the AD geometry is as flat (gravity free, especially tidal forces free) as you like inside the bubble. It's a "wishful thinking" kind of spacetime specifically constructed with that property. If you mean gravity from external sources, that still influences the bubble and the spaceship.

Read your second source again. 160 (or rather 166) is the total number of accidents in 2013, including those where nobody was hurt. The source lists 146 total fatalities to date. Still higher than your nuclear death toll, but considering the wind fatality list also contains deaths from traffic accidents during transport of material and the like and your nuclear death toll nothing of the sort (on top of the other significant omission)... well, there is a certain bias visible. (You are of course right in one point: simply plonking down wind turbines is not going to give us enough energy.) On the actual topic, it's a bold idea. Forget about the LEDs, though. And do durability tests under realistic conditions. The scratch tests from thunderf00t's video hardly count (tires normally do not scratch), but it's probably going to be the main issue. And when that is solved, do a cost comparison to the obvious alternative of placing the panels next to the road. Or on roofs. In short, worth investigating, probably not worth doing.

Neglecting air resistance, the minimal dv requirement can be calculated in reverse: start at 200 km, fall down, calculate the speed as you hit the ground. The formula for that is v = sqrt(2gh) and gives 2 km/s. If you don't want to get shot out of a cannon and prefer to be gently accelerated, say feeling 2g, for the first 100 km and then free fall for the second 100 km, that number needs to be multiplied with sqrt(2). Air resistance can be neglected for these kinds of napkin calculations, but if you must, make that a factor of 1.5, which yields 3 km/s. Sooo... Deathsoul097's intuition is 100% correct.

Don't despair. You probably just don't remember enough specifics to find it again.

Actually, that he is using the group velocity for the momentum calculation is the most sensible part of it all. He is not saying they are equal, just using that they are proportional. That is what you have to do... if you want to abstract away the reason why the group velocity is different from the vacuum speed of light. Here, of course, the reason for the change are the side walls, and ignoring them is a mistake. What baffles me most about the paper is that he is using Special Relativity, so I assume he considers it valid, yet the results are manifestly not covariant with thrust direction depending on the current absolute velocity. As for the Chinese experiments, they are feeding the resonator with a wave guide and the actual microwave source is fixed away from the resonator. A lot can happen there that has nothing to do with propellant-less propulsion.

No, radiation has momentum, too.

Yes and yes. Your assumption is correct: Electrons have mass and therefore momentum, and to conserve that, the gun and surface have to provide/absorb it. Like visible light on a mirror? A contained magnetic field perpendicular to the boundary, such as the one inside a very long coil, does that. The Lorentz force makes the electron turn around and for symmetry reasons the angle it leaves the field at is the same as the angle it enters at. You have to make the field stronger to reflect higher energy electrons, of course. Nothing much different from when you fire it at a regular conductor: it gets absorbed, the object gains a small net charge and, if grounded, will get rid of another electron somehow to compensate. Superconductivity will not break down because of a single electron, or even a stream of electrons, and it won't do anything funny to the new electron either. As you suspect, the material charges up (positively) as it sheds electrons. If the electrons are not replaced (they usually are), after a long while, the field induced by that charge would stop further electrons from escaping the material. They would still be released, but quickly fall back.Photons of the energies usually used to trigger the photoelectric effect are not strong enough to knock away protons.

RIC specifically asked about an exothermic reaction, which you had implied (that's what "burn" means). Reactions that require energy are not.

The same thing that prevents mud from getting dirty That it already is as oxidized as it can get without adding another substance, in both senses of the word.Definition 1: Having reacted with Oxygen. Definition 2: Gave away one or several electrons at least partially to another atom. Since all the atoms are the same, there is no incentive for the electrons to move anywhere between them. It could form Ozone, but that is a stronger oxidizer than O2 and less tightly bound, and thus a) reduced relative to O2 and the reaction does not happen on its own.

While I don't know the answer, early spy satellites predating Apollo used film for their images and that was returned with small reentry capsules with an appropriately dimensioned heat shield and parachute. I'd imagine a similar setup was used here.

Yep, "said to be". I guess they come close enough.

Well, with a non-fixed conjunction longitude, they are not in a 1:2:4 resonance. They only would look like that in a frame co-rotating with the conjunction. But apart from that nitpick, yes.

Another simple test you can run with the material you already have: Put the device on the cart so it should produce backward thrust. If it really does produce thrust, it needs to stop now after about 2 feet max. Even if it does, it still would not be conclusive. Friction is horrifyingly complicated in detail. Your device has a motor, it will vibrate. Vibration modifies friction, and different orientations will produce different vibration modes, influencing friction differently. sgt_flyer is correct. The best test you can perform without a vacuum chamber and microgravity is the pendulum test, with your device in a box to rule out air flow as source of the thrust.

That is correct.

About four years ago: http://en.wikipedia.org/wiki/Active_Denial_System

Gee, attitude. Assuming perfectly circular orbits (*), it's the equation that holds iff every time the outer and middle moon are in conjunction, the inner moon is in the same relative position to them. And overtakes the outer moon three times in between two such events. So yes, it is the equation that determines resonance of that type. More generally, (AB-1)/(B-1) needs to be a rational number, in this case 3. Massage it a bit and you get your form. *: They aren't, of course, which I guess is why A and B are close to 2.

Yep: "Even in this extreme situation the second law holds as seen by taking into account the information that could be gathered by a photon detector, realizing the original scheme of Szilard." They don't claim the device IS Maxwell's Demon, they say it is similar, which is true.

I learned to program on this thing: http://de.wikipedia.org/wiki/Kosmos_CP1 (Could not find an English description) 1 kBit of usable memory! 1 Register! 8 Bits! Code as readable as assembler and as fast as an interpreted language! Good times.

It's not a one way divider, just a selective one. I don't know whether such a thing exists. It still does not work as Maxwell's Demon, of course. What you put in is low entropy laser light, and what you neglect is going out is the high entropy radiation emitted by the atoms when they fall back into the ground state. So it's a fancy roundabout laser cooler, basically. (Ignoring the other problems with the description. Like how an ideal gas would not cool down if you simply let it expand by removing the barrier, what you'd do instead is let it perform work by pushing the barrier first. That the excitation puts the gas out of thermal equilibrium, making temperature undefined and so the claim that temperature does not rise makes no sense.)

That. I don't mind the paging all that much, but you can't really tell the part size from the picture or name. I will always, always pick the wrong size reaction wheel, probe core, battery, RCS tank or decoupler first.

For a given central body and a small orbiter, the orbital period only depends on the semi-major axis and none of the other parameters you mention. So just take a circular synchronous orbit and make the sum of the apoapsis and periapsis height match two times the apo/periapsis height of that circular orbit.

Just posting to agree with this. It is quite possible to stay on one side of the center line 90% of the time without producing thrust. And if you want to be really, really precise, it also is possible to create a slight deviation of the average position if you shift a mass up and down a lot in sync with the pendulum motion. But the effect is very small and I don't think it can contribute here. Sadly, MEGA does not like me very much and I can't watch the videos. Here's how I would deal with the averaging of the dead time at the end: Take K^2's suggestion to use the dead time at the end for center calibration. For the actual measurement, average over individual periods, that is from one time where the dot passes the center line from left to right to the next. That will give you a nice series of numbers you can analyze statistically. Do they sum up to something statistically significantly different from zero? There are bound to be deviations from zero for each individual period due to your device not being in resonance with the pendulum frequency (presumably), it is important that they enter the analysis, and they do that way.

No, it is not. For lifter engines for launches from kerbin, use I_sp*(1-2/TWR). Rationale: Assume you build a huge asparagus. You already have the N upper stages, total mass M, capable of lifting themselves once state N+1 detaches. Let's design stage N+1 (using negative numbering). It is well know that for the ascend phase, a total TWR of 2 is optimal. The upper stages have that, so all we need to do is make stage N+1 have total TWR of 2 and we're set. Assume the engine has mass m and fuel tanks have zero dry weight (for simplicity). We put one engine on the stage (add symmetry later), how much fuel can we take? Since the engine has a TWR of, umm, TWR, and the target TWR is 2, it can lift a total weight of m_t = TWR*m/2, the fuel to take will be m_t-m. How much dv do we get out of that stage, assuming all other engines are also the same or have the same ISP? The rocket equation tells us it's dv = I_sp * ln((M+m_t)/(M+m)) = I_sp * ln(1 + (m_t-m)/(M+m)) If we assume M >> m_t (a large asparagus), that is approximated via ln(1+x) ≈ x by dv = I_sp * ln((M+m_t)/(M+m)) = I_sp * (m_t-m)/M That's what we get out of a stage with that engine. What's the price? Clearly, a good metric is the logarithmic mass increase: price = ln((M+m_t)/M) ≈ m_t/M The metric for the engine to use would be dv/price: Metric = dv/price = I_sp * (m_t-m)/m_t = I_sp * (TWR*m/2 - m)/(TWR*m/2) = I_sp * (1-2/TWR). Voila. Of course, you can make other assumptions. These are merely the ones that lead to a concise result that includes both I_sp and TWR and nothing else.