K^2

Yes, I was thinking that pendulum oscillations should just precess with gyro precession frequency. But I do not know how to prove it. I will run a full simulation to see what really happens in these cases. Just to make sure. And yes, as usual, Momentus got it completely backwards. I give up on him.

That is an absolutely hilarious statement. Fortunately, we do have Relativistic Quantum Field Theory, which tells us exactly how particles behave near light speed, and we have used it to make computations, such as anomalous magnetic moment of electron, to confirm these equations to 12 orders of magnitude. Experiments with similar precision exist in General Relativity. Between the two, equations of Special Relativity, as a special case of the above, are some of the most precisely tested.

Yet you refuse to do the experiment. Hm. And for the tenth time, there are two separate oscillations here. One due to pendulum and one due to gyro. Not giving the system a "push", id est, having zero pendulum-caused oscillation, was one of your own requirements. For the sake of more educated here, the problem is governed by a second order differential equation. Working with complex numbers to represent 2D displacement of center of mass, call it r, and the point where string supports the gyro, call it z, we can define: r = z + deiÉt Here, d is distance between the point on axis where we attach the string of length L to the center of mass of gyro with mass m. Since string is the only source of force, equation of motion is very simple. mr'' = -mzg/L Note that this equation completely ignores torque on the gyro that may be caused by the string. This is a valid assumption under some conditions I'm going to discuss later. The most general solution, one that includes gyro-pendulum interactions, is extremely complex, chaotic, and is similar to a double-pendulum problem. Fortunately, we don't have to deal with all that. Masses cancel, and we substitute z = r - deiÉt. r'' + rg/L = deiÉt g/L At this point, it should be obvious that this is a driven harmonic oscillator. As with any inhomogeneous linear equation, we can start by solving the homogeneous part. r0'' + r0 Ω² = 0 Here, I substituted Ω² = g/L for simplicity, which leads to a general solution. r0 = aeiΩt + be-iΩt For some pair of complex numbers a and b to be determined from boundary condition or other information. So now we need a particular solution for the inhomogeneous part. Naturally, it has to have eiÉt term, so we go with a trial solution r1 = ceiÉt. Substituting that into differential equation, we have an algebraic equation. cΩ² - cɲ = dΩ² c = dΩ²/(Ω²-ɲ) And since a general solution is just sum of general homogeneous solution and particular solution, r0+r1, we have solution to the system. r = aeiΩt + be-iΩt + deiÉtΩ²/(Ω²-ɲ) This gives us center of mass location as a function of time. It's clear that there are two separate oscillations going on there. One has frequency Ω and is identical to the motion of pendulum. Relative to it, there is a second oscillation with frequency É, which is frequency of gyro's precession. So the only question left is the values for a and b. In general, they can be anything. All you have to do is give the right amount of push to the system from the start. But Momentus wanted to talk about the case where gyro precesses around an almost vertical string. That is the case with minimal energy of the system. Just for sake of completeness, let us write down the energy of the system. It contains a kinetic term and a potential energy term. E = mr'²/2 + z²g/(2L) If we consider r = r0 + r1, since r1 has no parameters, by triangle inequality, the above is minimized when the above equation is minimized for r0. And since it is trivially 0 for a = b = 0 and cannot be negative, that is our minimum. In other words, the absolute minimum amount the pendulum-gyro system can have is solution with a = b = 0 r = deiÉtΩ²/(Ω²-ɲ) Which happens to be identical to our particular solution. In this case, we can look at motion of the suspension point, rather than center of mass. Substitution the above into equation for z, we have: z = deiÉtΩ²/(Ω²-ɲ) - deiÉt = deiÉtɲ/(Ω²-ɲ) Naturally, if you just want absolute displacement, we are looking at the norm of z. |z| = dɲ/(Ω²-ɲ) = dLɲ/(g - ɲL) And, as I have pointed out earlier, this also satisfies the condition on centrifugal force. The above equation is clearly singular when Ω²=ɲ. In fact, there are serious problems when two are even similar. ɲ << Ω² results in positive displacement, which just counters centrifugal force as gyro slowly precesses. However, the case where ɲ >> Ω² is also entirely valid. Here, center of mass stays relatively put near the center, as the suspension point rapidly spins around. That's a case of a weak gyro, one which Momentus doesn't seem to even consider. But it is also a valid solution. One should be able to guess that when ɲ ~ Ω², the z and r are not colinear. That is a serious problem. As I've mentioned near the start, we are ignoring torque on the gyro. That's perfectly fine so long as z and r have the same direction. There really is no torque. Gyro stays horizontal and continues to precess at a constant rate. But if orientation of z and r do not match, which is going to be the case for near-resonance or for a bad choice of parameters a and b, this assumption falls apart. worst part is that this torque will cause gyro to turn towards vertical. As that happens, center of mass lowers or rises, changing potential energy stored in gyro. As a result, you have energy exchange between pendulum and gyro. As I've mentioned, it's similar to interaction in a double-pendulum and is chaotic. So general solution, for general starting conditions or general relation between É and Ω, is not available. But provided for a far-from-resonance condition and initial condition with minimal energy ("no push"), we do have an exact solution which does give us a fixed displacement provided here and in earlier posts. The most important part to take from all of this is that it is impossible to have oscillations with two different "orbits". This is not a simple pendulum. There is one solution with a = b = 0. And any significant deviation from that, because É and Ω do not match, result in a significant angle between z and r, causing gyro to do weird things. So again, there is only one solution of the type Momentus wants to talk about, giving unique displacement, which he'd be able to confirm if he actually bothered to do the experiment rather than be an annoying troll.

You should not think of it as a warp field. It is a bubble. It has walls, but space inside is just ordinary space.

Yeah. If it passes pendulum test for you, I am out of obvious ideas. I would have to replicate it and start excluding variables with more experiments and using sensors to try and figure out where all the forces are.

That's because you don't understand any physics. I actually know the amount of momentum that gyro will have. It follows from equations of motion. So why are YOU refusing to do the actual experiment described? No, I need video of gyro on a string. I KNOW how a pendulum works. YOU don't know how a gyro works. We need you to try the actual experiment described, because you obviously have no idea what you are talking about.

That is like pulling yourself by own hair. It should be noted that it is possible to build space-time geometry where this is the case. This might be one of these. They are not physical, however.

Center of mass is at r = z+d. So centrifugal force is m(z+d)ɲ. The force due to displacement is -mgz/L. I'll let you do the math. Or you can solve differential equation for motion of the gyroscope, which is what I did, and get this as a particular solution for when the whole thing isn't oscillating. General solution includes precession with this displacement about the normal oscillations of the pendulum. Something you really need to know how to derive before you argue with people on such topics. Once again, have fun doing the math. Or better yet, do the actual experiment and take measurements. The equation I've given is absolutely correct.

What does Star Trek have anything to do with it?

And you have completely missed the point in your own experiment. É and Sqrt(g/L) are two distinct frequencies. É is frequency of precession, which has NOTHING to do with the frequency of pendulum oscillation. In fact, É = mgd/L [This would be the angular momentum L, not length of string L. Pardon the confusion.], exactly as I've described before. Again, d is the distance between center of mass and the point where the string is attached to the gyro axis. And yes, a pendulum moves in an ellipse about the equilibrium point. But we aren't interested in that. We are interested in motion of a gyro suspended on a string. Ground state, no push, or whatever you want to call this condition. Just the precession of the gyro. And guess what, there is still deflection. The point where string is attached to the axis of the gyro is going to be shifted from center by distance z that I have specified. The expression can actually be simplified further, and I suggest you use that for an experiment, if we note that Lɲ in denominator is small. So if we take natural frequency of pendulum to be Ω, so that Ω² = g/L, then the displacement is: z = d ɲ/Ω² Does that make it cleaner for you? I suggest you go and test this. Take a string of sufficient length so that É/Ω isn't too small, and actually measure the displacement. Gyro WILL orbit point directly bellow the point of suspension with frequency É, and the point where string is attached to axis WILL be displaced from vertical by z.

Where would the momentum go? You can send a gravitational wave and get recoil from that, but that would carry away same amount of energy that an EM wave would. In other words, you are back to photon drive, and you might as well just hook up a flashlight to your power plant. And we have neither the flashlights nor power plants with sufficient power to make it worthwhile. No, even in the STL mode, this is not going to propel you. It can move you from one location in the system to another, but your ship will still obey energy and momentum conservation, and you'll have to use conventional fuel to adjust your orbit. Still, a drive capable of even 10% of speed of light would change the way we see space travel within our system even if we have to burn conventional rockets as well.

You forgot boosts. Oh, and I've replied to your gyro pendulum "conundrum" above with an actual equation. Get back to me when you manage to verify it.

Momentus, gyro suspended from a string will turn around point ROUGHLY bellow point of suspension, UNLESS gyro precession frequency resonates with pendulum frequency. It's a trivial thing to derive. This is a driven oscillator problem, and driving force out of resonance will not push pendulum far from equilibrium point. This is, seriously, an Undergraduate level mechanics problem. The reason this does not violate conservation of momentum is because small deflections will occur. Point of rotation will not be PERFECTLY under suspension point, resulting in small angle of the string, which is going to provide horizontal component of the force. Again, trivially derived. Edit: In fact, given a gyroscope with precession frequency É suspended from the string of length L making distance d from suspension point to center of mass, the deflection of the suspension point from center is going to be given by: z = d Lɲ/(g - Lɲ) Clearly, if you take a very slowly precessing gyro, with É small, the deflection z will also be very small. But it is not zero. P.S. Almost forgot. This is taken in the limit of É being not very large. Specifically, ɲ << g/L. In other words, precession frequency much lower than pendulum frequency. This is to avoid resonance. A more general expression can be derived, but I'm bored with this problem already.

Again. You have not demonstrated that. You only allude to 3rd party having claimed to have demonstrated it in dubious experiments. I am not claiming that heavy gyro going around light tower would not be proof. I'm saying that this did not happen. Experiment you are describing never took place. Video you have posted is of a heavy tower and light gyro. And the experiment on ice was not performed under frictionless conditions. It's that simple. I'll see if I can get the air track setup at uni to try and run heavy-gyro-light-base experiment for you. I think, it'd look even more impressive with tower being able to move only in 1D, dancing back and forward under the gyro that stays still. The problem is, ironically, finding a suitable gyro. Cue the new age crap. You'll start talking about crystals next, right?

Oh, yes. Here it comes again. This is a conspiracy by the scientific community! Duh! Again, you keep making up stories of possible conspiracy, but you haven't provided any evidence that a machine violating conservation of momentum exists. I'm not saying violating Newtonian Physics, because I personally know plenty of devices that do that. Ever used a GPS? Yeah. That thing takes into account the fact that time on a GPS satellite flows at a different rate than on Earth. Due to both the gravitational field of Earth and relative motion of satellite and receiver. It simply wouldn't work without that correction. Total violation of Newtonian Physics. Except, conservation of momentum is a central result in General Relativity as well. Moreso, momentum is part of the stress-energy tensor, which is a conserved charge of Poincare local symmetry. This isn't even General Relativity, we are talking about general field theory. I understand that these are all consepts way, way over your head. You are still stuck on freshman mechanics. But you seriously need to start getting through your head that you are going after the concept which is the most fundamental principle in all of physics. Conservation of momentum isn't some observational law, like Newton's Laws. It follows from mathematical theorems based on most fundamental symmetries of space-time structure. Other predictions of these theories are tested to 12 decimal places, both from measurements in GR and Quantum Mechannics. It's something that we know holds true for elementary particles and neutron stars in other gallaxies as a matter of fact. And you're trying to prove it wrong with third-hand account of a bad experiment with a gyro? I feel bad for you. In all honesty, and with no offense meant. I've given you equations to work with, and if you feel you need to prove it to yourself, you should have everything you need. But if you are still under an illusion that you understand something that some of us don't, you are simply wrong, and nothing you've brought up so far has shown anything other than your ignorance of the subject.

Wait, OP's nick is ZodiaK, and I'm just now noticing it? Is that a weird coincidence, or are you just a fan of Cryptography, ZodiaK?

You do not compute lift/drag from Reynolds Number. A table of values that is either experimentally obtained or numerically evaluated for a given wing foil/geometry is only valid for a given Reynolds Number. So you need to take your specific wing, compute Reynolds Number for your aircraft, and that will give you a specific table or polar that you should be using to obtain the data you need. So if you go to the link that PUNiSH3R posted, you'll see at the top left of the diagram "NACA 4412 12%" and "Re = 105". This is for a wing of "infinite" span with geometry NACA 4412 and is valid at Reynolds Number 105. To be honest, I'm not sure what the 12% refers to. It might be the maximum thickness as fraction of the chord, but that seems redundant, as "4412" already specifies that.

My main concern is for storms. Corals are pretty brittle. Pieces break off during storms even on parts that are completely submerged. A floating structure would be under all sorts of stress. But maybe if it just barely floats, allowing for larger waves to roll over it, and is stronger than regular coral, maybe? You'd also have to modify how this thing grows. Normal coral grows in tiny corallites, which are usually rings of tiny little calcium disks. (I was unfortunate enough to run into a coral while diving once, and able to study the patterns on my skin for nearly two weeks.) These definitely have the room to hold gas for buoyancy, but you'd have to come up with some very clever modification to prevent gas from escaping. I don't know if our GE knowledge is up to it at the moment.

Pluto travels through a pretty densely populated part of our system. I think, if it weren't for gas giants, inner planets would probably end up with way more tiny moons as well. Somebody with better understanding of Solar system formation might be prepared to correct me, though.

First of all, you have that backwards. The faster the object is moving through space, the faster it moves through time. It ages slower, from perspective of "stationary" observer, but that's the same thing as getting to the future faster. If you understand this better in terms of mathematics, the proper velocity for "stationary" object is u = (c, 0). That's traveling at speed of light c through time, and 0 through space. If an object is moving relative to your frame of reference, it's proper velocity is u = (γc, γv), where γ = 1/Sqrt(1 - v²/c²) is the Lorentz factor. As v approaches c, γ tends towards infinity. At v = 0, γ = 1. None of this has anything to do with dark energy, however. Dark energy is the pressure term in the average stress-energy tensor density of the universe. You should probably just read the article on Stress Energy and ask questions you end up having. But you really need to understand a bit about that, at least qualitatively, before you can understand dark energy.

The discussion we have is on how much the velocity of the ship relative to star changes if the Alcubierre Drive is used within the system. Unlike flat space-time, there is a change in velocity and heading. It is a consequence of the conservation laws. While extra energy might come from warp drive itself, there is nothing you can do about momentum conservation. If you warp from 1AU to 10AU and your velocity does not change, you suddenly have 10x as much angular momentum. I did find geodesic equations for Schwarzschild metric in a nice form that can be taken to Newtonian limit (rs << r) for the most general case, including space-like geodesics that the FTL ship will follow. (And in v << c case, these do match classical trajectories.) I should be able to use these to work out precise trajectories for warp ship under different warp factors and the resulting change in velocity. From conservation laws alone, I would say that, most likely, you will need to perform a burn to Hohmann transfer as if you are planning a normal trip, then engage warp (In what direction? Or can warp drive simply follow Hohmann?) And if you drop out at apoapsis altitude, you should have the same velocity as you would if you followed classical trajectory without warping. Depending on trajectory of the warp bubble, you might end up with a completely different argument of the periapsis, however. As you can see, I still have some questions, but I'm getting close to working it all out.

Larger ceiling also reduces odds of any tangential forces being passed down through supports. But yeah, it doesn't hurt to have two different runs. You can never have too much data. If you'll need help on how to average data from two different runs, I'll be happy to guide you.

There is nothing indeterminate about quantum mechanics.

Which we don't have. We still have a video of two objects moving about a common center of mass, exactly as one expects it to. If you have some other video, or exact numbers for masses of gyro and tower, please, bring them forward. But the video you've linked to demonstrates a light gyro going about a heavy tower.

When dealing with any sort of numerical computation, the big question is always convergence. You can't, in general, get an exact result for any computation. But provided a "nice" dynamic, you can get as close as you need. If this is a simulation, there are three main possibilities. First, we might simply not have probed the universe with sufficient precision to matter. Second, it might actually be impossible to probe with sufficient precision because of how the dynamics workout. Plank scales come to mind. Finally, if simulation exists for our sake, precision could be adjusted on demand. The more sensitive the experiment, the more time the computer spends on making sure all the numbers match up. Attention-sensitive algorithm might sound contrived, but when you get right down to it, it's just like using multiple scales for turbulence simulation. The fact that it's a quantum system only increases the size of the state space. It doesn't actually make the simulation harder. And if we are considering a computer that can run our entire universe, it's already something infinitely more complex, so it hardly makes a difference.