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It's tricky, but entirely possible to build an air tight seal that still allows for mechanical rotation. Usually, these involve oils with very low vapor pressure, similar to these used in vacuum pumps, or even ionic liquids to create the seal. You'll have losses, but even a conventional airlock will have some losses. So long as these are sufficiently low, it's not a problem.

You start out with excited electromagnetic field, which carries energy, momentum, and angular momentum. It transfers these quantities to an atom. So an atom is now in excited state, and electromagnetic field went down to ground state. Nothing actually stops being. The photon state is still there. It just lost excitation. The field is there. All of the conserved quantities are there. There are two self-consistent ways of interpreting what is a photon here. You can call it a collection of all relevant quantum numbers. These get transferred to the atom. I know it's a bit weird to think of it this way, but it's logically consistent, and sometimes useful, to consider atom as "holding" and extra photon. But a more conventional interpretation is that photon is just the state labeled with its quantum numbers, including the four-momentum. And that state remains intact regardless of whether it is excited or not. It all comes back to the fact that "particles" are really just a result of us quantizing the fields. It's purely a mathematical trick. We have the electromagnetic field, and instead of dealing with all possible configurations of it, we introduce particle operators which "add a particle" or "remove a particle" from the field. But all they do is change field configuration. Which particle? A proton? You seriously can't find a proton where you are? And yes, it is the proton you're looking for, because they're all the same. Are you seriously asking me to prove that other protons exist? Are you studying Relativistic Quantum Field Theory and doing research in Particle Theory? Because I simplify it a lot for students in lower level QM courses. I know professors do to, because it's impossible to talk about all of this in precise terms until you start doing field quantization.

You do understand the concept of indistinguishable particles, don't you? As soon as you explain to me how you want to tell the difference between not being able to detect a particle because it stopped existing and between being unable to detect it because it tunneled out of the ion trap, we can continue discussing this particular line of reasoning. And again, there is no such thing as two protons colliding and you being able to say that they were disintegrated. A collision between two protons can happen without a single valence quark being struck. With exactly the same shower produced. More importantly, that's always part of the process. Heck, the very distinction between valence and non-valence quarks gets really hazy at this point. This is precisely because composition of the proton doesn't matter. A proton is 99% just vacuum fluctuations. What's important about the proton is that it has a +1 baryon number, +1 electric charge, spin 1/2, and whatever energy and momentum it happened to carry. That's what really makes a proton. And all of these numbers are conserved. They might end up being distributed among different excitations of various fields, but what exactly stops existing that existed in the first place?

Fine. Can you tell me the exact time at which a particle stopped existing? No? How about an exact time after which you are certain it no longer exists. Also no? Ok, how about this. We have a positively charged pion. Its up quark can annihilate with the d-bar to produce a W+ which then interacts with an anti-neutrino it pulled out of vacuum to make a positron. That's pretty straight forward. But there are also a bunch of quark and anti-quark fluctuations in vacuum. So say, instead of the valence d-bar, the up quark pulled a d-bar from vacuum, leaving an extra down quark to continue on with a valence d-bar. So it still spits out a positron, but now it continues on as a neutral pion until that decays. This is absolutely equivalent to a process where the up quark simply decides to spit out the W+, and pion becomes neutral. All quarks of the same flavor are indistinguishable, after all. So did the original pion decay, leaving a neutral pion as a product, or did the pion simply let go of its charge? It's a Ship of Theseus question certainly, and you'll say, it's because pion is a composite particle. But a photon behaves exactly the same way. It propagates as a an electron-positron pair. And the electrons/positrons keep getting swapped out with ones from vacuum constantly. So at which point, exactly, should the photon stop existing? Or did it even exist to begin with, under your definitions?

Particles never disappear. Quantum Mechanics just doesn't work that way. There are a whole bunch of conserved quantities which have to keep going. You can have a neutrino encounter a W- boson and continue on as an electron. Or a photon splitting off into a positron and an electron. But none of them just disappear. That's a very misleading thing to say.

Have you tried to actually understand the physics of flight, rather than quote simplified rules of thumb written for pilots? Especially a book for naval pilots, when we are discussing airliners? Which is perfectly true, so long as L/D(max) occurs at the same place for any air speed. Which, of course, doesn't happen. I don't have 747 polars across a sufficient range, but lets look at 787 for sake of example. Even the text you quote is in agreement that if I load the plane heavier, it will have to glide faster to do best glide. So suppose, I load a 787 to have best glide at mach .87. Looking at the nice blue curve for that speed we see that L/D(max) is at CL .45 and the L/D is about 19.5. Now I load the plane more, so that at the same CL it has to glide at mach .88. Suddenly, we're on the orange curve, and the L/D is now just over 17. And isn't L/D(max). So the planes with different amount of load will have to glide at different CL to achieve L/D(max). But again, any time you'd like to bring out charts that show different, feel free. "If there is no change in L/D(max)." Bam. Right on the first sentence. There is a huge change in L/D(max) with altitude. See that 37k' on the polars? It's there for a reason. Polars for sea level are completely different. Now, I've been trying to explain why there is a change across several posts. I've shown via computation that 747 has L/D(max) a little over 5 at sea level, and well over 10 in cruise. Yet you keep insisting that L/D(max) doesn't depend on altitude with absolutely no argument, reference, or drag polars to back it up. So again, if you claim you have actual charts for 747, go ahead and post them.

Autorotation is done with positive pitch, and it works perfectly fine with vertical descent. I know, it can be difficult to accept that the rotor with positive pitch does not stall, but it does not. I strongly suggest you look up some explanations of that on-line. Wiki page on autorotation or autogyros might be a good place to start.

Maybe you'd like to remind us all what the definition of Reynolds number for an airfoil is? Or for a general flow? Or precisely in what ways does atmosphere of Mars differ from that on Earth, other than density? Dynamic viscosity of both is essentially the same. The fact that actual ascent profile of an airliner happens to follow more or less constant Reynolds number is besides the point here. Cruise speeds are nowhere near best glide. I'm pretty sure I covered that. Inertial flow cannot generate lift, but I'm sure that's not important at all for an airplane. I've managed to describe its physics, which tends to be more important than whatever hand-waving explanations that lead you to that level of confusion on Reynolds numbers and importance of viscous forces in aerodynamics. And this kind of confusion, too. Which aircraft has more drag, in proportion to weight, due to wingtip vortices, 747 or a glider? And why do you think that is? And as the final question, based on that, for which is extra weight going to make a larger difference? Yes, I would not doubt that advantages of bringing ballast on a plane with glide ratio over 60 significantly outweigh (sorry) the losses to wingtip vortices. But if you think that applies across the board, you understand nothing about the induced drag. Even the most casual glance at the polars, both the foils and the aircraft, show that glide ratio is very soft near best glide. The difference between ref+20 and ref+80 is not going to be dramatic. But hey, if you have specific references that show a completely different glide ratio at sea level, go ahead and share. If you can show me data that gives glide at sea level at anything like 10+ you get in cruise*, I'll be genuinely surprised and admit my mistakes. * I was actually surprised to find that glide ratio in nominal cruise of 747-400 is 12, after I found some polars. I was expecting a slightly lower number, but it's in the ballpark.

Tarantulae, I'm not sure what your point is. There was never a question that you can trade altitude for air speed. The question was if this is absolutely required for a rotor kite to work as a landing system fora capsule. I'm perfectly aware of how this works for a helicopter.

That's the idea, yes.

Simple logic dictates that we will either have colonies off-Earth, or we will go extinct. Human beings are very good at surviving. But we are also very good at cutting the branch we sit on. I honestly have no idea which one is more likely. There are no fundamental problems with starting off-Earth colonies. There are some engineering problems. There are lots and lots of socioeconomic ones. There would definitely have to be some changes, either in technology, politics, or environment, but I don't think anyone could remotely reliably predict these.

It's not so much about harvesting it, as it is about not wasting it. It's mass that's already in orbit, so it's free. You don't need to waste fuel to launch it. For example, a pair of tethered satellites can raise/maintain orbit by de-orbiting material without using any other propellant. If you have a good supply of co-orbital junk that you can collect without problem, it's a way to do station keeping while also clearing orbit of debris. Again, I don't have a clue how you'd collect the debris, since you'd need hundreds of square kilometers of capture area to be effective. Unless you fly-fish for them... Anyways, if you do figure that part out, finding use for that mass is not hard.

It's been a few years, but from what I remember from the class, the procedure for a single-engine is to try a restart right away. If you fail after a couple of attempts, then you are in situation where you start trading altitude for air speed, so you need to have an idea of where you are going to land if your engine doesn't start. From there on, if you have altitude/range for it, you keep trying to maintain speed and trying to restart the engine. If not, you switch to best glide. On a twin, you just keep flying level, unless second engine quits. But I don't know how much of this is applicable to jet turbines. I also don't know how much of that I'd remember if I was actually flying an airplane that suffered a failure. Our CFI used to say, "Propeller is like a fan. When it stops spinning, you start sweating." And I was sweating enough just landing a 172 with an instructor next to me. I do recall that landing was done with almost no throttle, though. So while I'm sure engine failure would cause me some new gray hairs, it's not something I'd worry about getting into the plane. I really need to get back to that once I have a paying job.

On Earth, pretty much the only material worth recycling is aluminum, because of the amount of energy it takes to recover metallic aluminum from ores. Everything else is cheaper to just dig out and refine afresh. Consideration for recovering space debris are pretty much the same. It's all trash, but there is a lot of energy in it. I don't know if there is any practical way to do so, but if one could cheaply recover debris, it'd be worth just for the mass that's already orbiting. That's a big "if", I know.

Dragon, I suggest you try this with a decent simulator. MS FSX or X-Flight 9 should suffice in this case. I suspect, you'll find that flaring a 747 on final with dead engines is a lot harder than it sounds. But as I've said originally, under ideal situation, it is possible to land. Real world rarely provides these, however.

The computation yielding 5.5 is for clean config. I've made no assumption of configuration or AoA in qualitative explanation, either. Given same on both, higher density, and consequently, lower air speed, results in much higher drag induced by wingtip vortices. How much of a difference that's going to make is going to depend on wing loading, among other things. 747 is an example of an airplane that can only glide in low air density.

Of course. When you know in advance what sort of magnitudes you are going to be dealing with, you can just hard-code fixed point. Most MCU applications are going to be like this. But if you are writing something a little more general, it's not a lot of work to have computations done in fixed-point style, but with ability to adjust the position of the point. It's not true floating-point, because you don't adjust the point for every operation and for each operand. But that's why you don't lose much in terms of performance. You have to do shifts between multiplication operations anyways. Might as well hold the shift amount in one of the registers and shift by that, rather than by fixed amount. (Actually, is there even an immediate shift in SSE/AVX instruction sets?)

That's a bad idea. Because prior to multiplying these together, you'd do a 128 bit shift to the right. If you happen to work with small numbers, you'd be losing most of your precision. You are much better off keeping track of the exponent somewhere in software, rather than having a truly fixed point.

SSE2 can already do 2 double precision operations at a time. But no, performing two double operations at the same time doesn't give you 128 bit precision. However, for many practical applications, where exponent does not change much, 256 bit integer math gives you the same results. AVX2 does, indeed, allow you to perform 256 bit integer operations, which you can use to fake 128 bit floating point math in most cases. True. But I think, the implication is a way to combine data without significant loss of performance, which is not going to be the case.

I don't know why I wrote 777 in place of 747. The 777 is a much better glider, yeah. Pretty much any twin is designed to be landed with dead engines for safety reasons. With four engines, assumption that at least one engine is running is a common one. Sorry about confusion with the numbers. This is not even true for 2D air foil. Glide ratio is a function of a Reynolds number. Here is an analysis of a typical foil used by Boeing. 747 uses several variations of this foil along the length of the wing. Take a look at the Cl/Cd polar. The slope of the tangent will give you best glide ratio. 747 has MAC of 327.8" and VY of around 180KIAS at maximum landing weight. That gives me R = 50x106 at sea level. At cruise altitude, air density is about a third, so under ideal circumstance, the velocity needs to increase by sqrt(3) to maintain lift at the same glide slope. That puts Reynolds number at less than 30x106. I don't have polars for Reynolds numbers that high, but if the trend holds, this would actually reduce glide ratios. However, real wings aren't constant 2D cross-sections of infinite span. The above is only useful to demonstrate the fact that the statement that glide ratio does not depend on altitude is wrong at a very fundamental level. For a real airplane, things are far, far more complicated. I'm going to skip all the factors related to the way the wing profile changes along its length, and the fact that body contribution is very different at different air speeds. What's critical for a 747 is actually the wing loading and finite wing span. That's what kills the 747 performance at low air speeds and, consequently, high air densities. Without getting into all complexities, lift depends on transverse circulation induced by an airfoil. Unfortunately, due to finite span, that tends to induce longitudinal circulation. Id est, tip vortices. While circulation is actually higher at cruise altitude, the low density and high air speed means that it is distributed over much larger volume of air. That results in much gentler wing tip vortices as well. At low speeds, the wing tip vortex is much more compact. It has a smaller radius, and the air flow past the wing is much slower, all while the impulse transfered to the air remains the same. This results in much higher air speed in the wing tip vortex, so more energy has to be transferred to air. More power lost to wing tip vortices at lower air speed means dramatically higher drag. This also brings up another interesting aspect. For a 2D air foil, wing loading isn't even a factor in the performance. You have a fixed glide ratio, and it depends on qualities of the wing only. This is absolutely not the case for a real airplane. A heavily loaded airplane does not glide as well as an empty one. This is pretty intuitive, but another thing you don't seem to account for in your statement. The reason is exactly the same. Higher wing loading means stronger wingtip vortex. That means more energy lost to the vortex, and that means higher drag. But, I believe, people will still ask me for references. Here is the manual. On pages 42-43 you can find the climb rate for maximum takeoff weight to be 2000FPM at 210KIAS with 880,000 lbs of weight and about 250,000lbs of thrust. That's a climb at a slope of +9%. That indicates an excess thrust of 9% of weight, meaning drag of 160,000lbs, indicating a glide ratio of 5.5. That's a bit higher than ratio of 4 I've mentioned earlier, but this is clean configuration. Add a bit of flaps to try to bring the landing speed down, and you're right back to 4 or bellow.

The ability of an airplane to glide is defined by the ratio of its lift coefficient to its drag coefficient, called the glide ratio. Both of these are a function of air speed and air density. The best glide speed is usually equal to the airplane's VY speed, which also gives best rate of climb. So if you are trying to land an airplane unpowered, you shoot for VY. Airplanes with good glide ratio are relatively easy to land with no engines. A typical light airplane will have a glide ratio in the 8-10 range and unpowered landing is often part of standard training. High performance gliders can have glide ratio of over 60. These can stay in the air pretty much indefinitely with a good pilot, thanks to thermals. An airliner, typically, has much lower glide ratio. A 777, as an extreme case, has a glide ratio of only 4. (Actually, a 777 has a glide ratio of almost 10 at cruise altitude, but that's due to it being designed for flight at low density and high speeds. At low altitudes, it's only marginally better than a brick.) A wing suit has glide ratio of about 2.5-3. Helicopter in auto-rotation is also about 3, but helicopter can flare a lot better than an airplane can due to energy stored in the rotor, so they can do a soft landing unpowered, despite being very bad at "gliding". A 777 with all four engines dead is typically considered incapable of unpowered landing. It needs at least 1 engine running. Though, under ideal conditions, a survivable landing is possible with that glide ratio. Technically, so does the brick. If you add a small tale to keep brick's orientation stable, it can do better than 1.

Of course. The question was, why it's recommended for helicopters to autorotate with some minimal forward speed, and if that's going to be a problem for rotor kite approach to capsule landing, which is what I was trying to address. My point was that helicopter is designed for powered flight, which doesn't make it an ideal vehicle for autorotation, and hence the much softer landing with forward speed. If your goal is to build a replacement for parachute, you can design around that.

People spending too much time laying down having serious heart troubles is part of the reason I'm sure there will be problems. It's fine if you sleep 8 hours a day. But if you're horizontal 24/7, it's bad.

It's not that simple, unfortunately. I'm sure it'd help, but evenly distributed weight throughout your body is not exactly the same as weight you wear on your body. One problem that jumps immediately to mind is blood circulation. You heart has to constantly combat gravity to pump blood into your brain and out of your legs. At 40% gravity, heart wouldn't have to work nearly as hard. That can cause all sorts of problems, including serious ones like arrhythmia. It might be possible to offset that with exercises, of course, and it might be possible to deal with all of the effects of 40% gravity, but my point is that it's not going to be as simple as wearing a weighted suit. Though, that may be a start. Nibb31, there are, without a doubt, problems with low gravity. What we don't know is how severe they are, and which ones are reversible. But I completely agree with Winter Man that a variable gravity setup in orbit would be a good way to start looking into it.

Depends on the size of the sphere. The energy output is still going to be exactly the same as that of the star, just at a lower temperature. That puts the sphere at 390K at 1AU. But it's easier to build a smaller sphere. So if you make it, say 0.2AU in diameter, it will be red-hot, at over 1,200K, with substantial glow to it.