K^2

I take it you aren't familiar with concept of dirt bikes and quads. If it goes blazing fast over terrain that isn't safe to even walk over, you can bet your rear that there will be people screaming to get in line for it. If it makes a hell of a racket doing so, it's a bonus.

I could not find anything in regs that clarifies whether ground effect is flight or not. In fact, FAA doesn't seem to define "flight", which is rather silly. So it's hard to tell whether ground effect bike would require flight certification. On the other hand, I can totally see ground effect bike qualifying for FAR Part 103: Ultralight. It should be possible to get that thing under 254lb. And power-off stall speed is basically zero, since a ground effect rotorcraft with a shot-out engine simple hovers to the ground. Which leads us to 103.7. In other words, you wouldn't need so much as a motorcycle license. But there are other rules that limit where and when you'd be able to ride one. The short version, it'd have to be in unpopulated area, during daytime hours only, and with no passengers. P.S. Rules in other countries differ. I'm not sure if JAA has an ultralight. One might need a Light Sport Aircraft certificate for one of these in Europe. And while that's a little easier to get than a full PPL, it's still a pilot's license, and it still takes a lot of flight hours.

I mean expending energy to move heat from colder surface to a hotter one by means of Peltier Effect or mechanical heat pump. Refrigeration would be an example. Thermoelectric cooler, as I've mentioned, another. Simply helping heat flow down a temperature gradient is still passive cooling in my books.

Sure, but external heat sinks on yeolde water-cooled systems had the same limitation. They were pretty crap, so you had to have a significant sized heat sink, usually sitting on the side of your expansion tank.

The later. If your external heat sink has the same parameters as your CPU sink, then liquid cooling will make things worse, since you now effectively have two TIM layers. The whole point is that you still have an external heat sink with much greater area and better ventilation than the heat sink you could possibly place inside. This is why we started talking about water cooling still being air cooling in the end. It just lets you be much more efficient about it. As for TIM being a bottleneck, if that's the case, the only way to combat it is active cooling. You can increase the heat flow through TIM by increasing temperature gradient. Since you want CPU to stay cool, you need to make external temperature positively frigid. You can either use thermoelectric cooler between CPU and liquid cooler, or you can put your liquid through a refrigerator. Both are done occasionally for extreme overclocking.

That's the problem, you "toyed" with the idea. As a person who actually studied Quantum Field Theory and knows what virtual particles actually are, how to make computations with them, and which rules for ordinary matter they do and do not obey, I can tell you that the idea of using virtual particles as propellant is complete and utter nonsense. You cannot transfer momentum into quantum vacuum. This is a fundamental violation of conserved currents. I have written pages in this thread about how many things would be completely broken if this was possible, and I have only scratched the surface. Idea that you can generate thrust from quantum foam is just as absurd as the idea that you can make a wheel generate infinite energy if you arrange magnets on it just right. And for many of the same reasons. We are simply dealing with more sophisticated fields. The only thing that virtual particles can do is act as very short range intermediaries. They must transfer momentum to something else almost instantly. If no propellant is involved, then the net efficiency of a Q-thruster is that of a photon drive. EMDrive demonstrates efficiency at least an order of magnitude better. That means that a massive, on-shell (not virtual) propellant is involved. And if there is massive propellant, then all you have is a fancy ion thruster. Which is still neat, but it is going to be limited by the same delta-V cap as any other ion thruster. That's all there is to it.

Precisely. And the most common reason to use water cooling is because heat-sink-to-air is the bottleneck. In which case you have to either use a much larger external heat sink and water cooling, or you can go with active cooling and force heat into heat sink.

That's not how pricing works. At high end, demand for components becomes fairly inellastic. So very small changes in quality result in huge changes in equilibrium prices. This is particularly true because Intel has very little competition on the high end. Still, that $800 gap is there for a reason. Any CPU. You will need to install a special heatsink instead of the stock one. They do make water-cooling kits for GPUs. But it tends to be more work to install it. There are also water-cooling ready graphics cards out there. Just hook up the hoses and you're set. Water cooling only transfers heat. It doesn't remove it. In fact, it's impossible to just get rid of the heat. It has to go somewhere. In air-cooled system, heat goes directly from CPU heatsink to air. Which limits cooling rate by surface area of the CPU heatsink and air flow. With water-cooling, water transfers heat from CPU heat sink to a larger external heatsink, which usually can be significantly larger and allows for better airflow. But yes, eventually, all that heat still has to go into air.

Ah, I missed the part where 254/256 work. That is, indeed, extra weird. But it's hard to say why these two are special without knowing some details on the algorithm used. I was picturing this with a^b = 2^(b ln2(a)). This makes the exponent really fast to compute using FPU arithmetic. The ln2 can also be computed very efficiently by using FPU to get integer part, and iterating from there. But if that was the case, ln2 wourld return strictly 1.0, which would allow the rest of the algorithm to procede exactly the same for 2^254 and 2^256. I suppose, there could be some sort of a check in there that fails for some odd reason, but it's also possible that it's actually a completely different algorithm. Edit: What would the result of 2^253 be? Does it still give you correct value?

Here is some insider perspective for you. Most modern PC games are optimized under assumption that you'll have four cores. Console games tend to be built around 6 cores, but CPUs on modern consoles are such fail, that if it's a straight up port, you won't be bottlenecking on CPU anyways. Having more cores won't hurt, of course, so if you can afford an 8 core Intel, great! But if you have budgetting constraints, and are considering AMD CPUs, I wouldn't. Get yourself a quad-core Intel instead, and you'll still have better gaming performance.

Well, obviously, Wikipedia isn't in jurisdiction they have to worry about it. In US, at least for a while, the state of these laws was very odd. You could only get into trouble if you cross an international border in posession of such an algorithm. Hopefully, this silliness is on decline.

Several. For starters, as I've indicated, a bunch of calculators use double precision. That has a larger range of exponents, -1023 to +1023. Next, there could be hardware difference. Particular chip might respond to over/underflow differences. Finally, there are usually flags that specify what should happen on floating point error. Typical responses are to do nothing, to replace result with Inf/NaN as appropriate, or to throw exceptions. It's possible that this is normal "do nothing" behavior for this calculator, while others tend to treat it as an error.

Please, read the thread. I have explained exactly what happens. Sign IS the first bit, but it's the exponent that's overflowing. It does not overflow the same way as it does in integer math. And by the way, all errors on over/underflows are always optional, even on PC. By default, neither integer nor floating point over/underflows generate an error. You have to set some flags to get them to trigger an exception. No, it is not. In two's complement, zero is zero. In IEEE exponent, zero is 127. It has the same modulus property as two's complement, but the number is shifted by a fixed offset.

No, this is precisely an overflow error, as I have explained.

The format stores a number as sign * 2^exponent * (1 + mantissa). The sign is stored as a single bit. Exponent is stored as a number between -127 and +127. To achieve that, it adds 127 to the actual exponent and stores that. So number 1.0 is 2^0 * (1 + 0). And so the exponent it's actually going to store is 0 + 127. Since it's positive, sign bit is zero, hence 0x3F800000. Likewise, 0x3F000000 = 0.5, because now the exponent stored is just 126. 126 - 127 = -1, and 2^(-1) = 0.5. I don't see why it would work differently with 2^254. You should get 0.25 as an answer. I can imagine 2^256 not working because of some other limitation. But it's hard to tell. If it works the same, the answer should be 1.0. In general, with this simple exponentiation algorithm, instead of 2^n, you should get 2^(((n + 127)%256) - 127), where % denotes modulus operator. Edit: Many systems use double-precision instead. So the limit is 2^1023 instead of 2^127. But I haven't been able to find one that would give the same mistake on 2^2047. Everything I've tried so far has correctly identified an overflow and displayed result as +Inf. (Impressively enough, Ti-89 managed to actually compute a value even for 2^2048)

That's because it uses IEE floating point. 255 + 127 = 382 = 0x17E. But it only has room for 2 hex digits. So the 1 is lost. It becomes just 0x7E. As an actual floating point number, it's 0x3F000000 = 0.5. Easy.

Well, they were all secret at some point. There have been accidents that USSR did not report. Most notably, death of Valentin Bondarenko. But these are not secrets anymore. They have not been for a long time.

It's a radio source, and it's quasi-stellar. I don't see why not.

Exhaust travels at several km/s. If your rocket is twisting fast enough to hit its own exhaust, it's already ripping itself into pieces due to extreme centrifugal stress. Otherwise, this is a problem with any multi-engine configuration, regardless of where the engines are placed. Of course, if your engines are in the front, you are forced into multi-engine. So there's that.

Neither have you, and I asked first. Then, as a courtesy, I've answered yours. Now, until you give me a derivation of some relevant physical quantity, I see no reason why I should keep answering your questions. Science is about models and predictions. Until you can demonstrate that you can show some sort of a derivation, all of your blabbering is pointless.

That's all you've been doing. I actually posted a derivation. Mathematics. Care to try? Or were you homeschooled, and never learned how to do fractions?

Consider a particle located at space-time coordinate x, with two detectors both in un-triggered states labeled as -. The initial state |x--> then needs to propagate to the final location z, passing in transit via one of the two slits at locations y1 and y2. Each of the slits is equipped with a detector that triggers if the particle passes through it. Lagrangian for interaction with detectors is given by: L = S(z,y)[δ(y-y1)|y1+-><y1--| + δ(y-y2)|y1-+><y1--|]S(y,x) The target state is labeled as |z>, which is to be understood as particle at z with any combination of possible detector states. We then consider transition amplitudes, given by the following expression (to within phase factors). <z|exp(∫L dy)|x--> = <z|x--> + <z|∫L dy|x--> + (1/2)<z|(∫L dy)²|x--> + ... Only first order contributes. Expanding the integral and the propagators: <z|z>(<y1|y1+-><y1--|y1> + <y2|y2-+><y2--|y2>)<x|x--> The two non-zero components of this amplitude are as follows. <z+-|z+->(<y1+-|y1+-><y1--|y1--><x--|x--> and <z-+|z-+>(<y2-+|y2-+><y2--|y2--><x--|x--> Which leaves us with the final detector state |+-> + |-+>. This is a mundane entangled pair. So the particle has passed through both slits, however, the detectors are entangled so that if one of them is triggered, the other is not. This is the full explanation of the double-slit experiment and results, 100% consistent with observed data and derived using actual theory. Now it is your turn. I want you to derive an actual physical quantity using your model of the universe. And enough with the excuses. It's getting pathetic. I want to see a derivation, or feel free to get out.

Your orbital-fu is weak, grasshopper. The further out the asteroid, the better. You need almost no dV to get from edge of the Solar system to the inner system. Then you can aero-capture to rendezvous with a ship in LEO. Or Mars orbit. Or a cycler traveling between the two. Best part, even what little dV you need to launch cargo into a transfer orbit - doesn't have to come from fuel. You can mag-rail stuff directly off a large asteroid. Then you just need a few puffs along the way for adjustments. These can be performed with ion thrusters if you're really worried about wasting fuel along the way. So outer Sol System mining can make a difference between needing 9k dV or a few hundred m/s. And later can come from ion thrusters in this scheme. So again, effectively free. By far, the hardest part about space travel is leaving a planet with atmosphere. Once you are in interplanetary space, you can go anywhere you want, combining fly-bys and aero-braking maneuvers. Provided you have time for all of that, of course. But presumably, we don't care if our fuel has to spend a decade in transit, so long as we have a steady stream.

Or you can, you know, mine just about any icy asteroid/planetoid for fuel and get it to your orbital ship for practically nothing. You can also mine rocky asteroids for everything you need to build ships at your orbital shipyard.

You are still not posting any computations. Go. Do math. Come back with results.