K^2

The amount of nuclear material on GLONASS sats is insignificant. This is just press making noise out of nothing. As for the launch, looks like engine failure to me. Of course, that's a very broad diagnosis. Almost anything that can go wrong with the first stage is some sort of engine failure.

That is an absolutely beautiful shot. I'm assuming reflector telescope? How big was the mirror on this one?

Except there are known solutions involving only positive definite stress-energy tensor that allow for closed time-like loops. In other words, GR allows for time travel without need for exotic matter. So you can roll your causality in a pipe and smoke it. Not to mention that once you realize that you are dealing with a field theory, all causality "paradoxes" are trivially resolved. As far as the Alcubierre Metric itself goes, yes, it requires negative energy densities. But first of all, there is nothing that suggests that positive-definite solution does not exist. In contrast, existence of "time machine" solutions suggests the opposite. And it is still not clear how Casimir Effect factors into all of this. There have been a lot of theoretical work done on possibility of using Casimir Effect to stabilize traversable wormholes, which have the same problem as Warp Drive. Known solutions to both require negative energy densities. So it's entirely possible that we already know how to achieve such a state, if only on microscopic levels. And don't knock the interferometry experiments. What they are doing to test the warp drive goes far beyond any direct, laboratory measurements of space-time curvature we have done to date. Yes, we are all pretty certain what the outcome is going to be. This is more of the test of precision to which we can solve these equations numerically and to which we can conduct measurement. But you have to agree, calling space-time curvature speculative when people are measuring it in the lab is quite misinformed. That is actually wrong. Standard model predicts div(=0. It follows from gauge symmetries of the QCD Lagrangian.

That will cost you a LOT of extra Delta-V. But for practicing landings, it might be a good approach.

If all of your wings are attached straight, then your angle of attack is just the pitch relative to prograde vector. Otherwise, you have to add geometric angle of attack of the particular lifting surface. And that, by the way, is how you can change your effective AoA. Why do you need any of that, though?

On the nav ball, look for the prograde and retrograde markers. Simply point the ship at the retrograde marker and burn. This will slow you down both horizontally and vertically. Bring your speed down to about 5m/s, and you should have no trouble touching down. Edit: LV-909 is a good engine for Mun landing, yes. If it's not enough thrust to get you to the Mun, add an extra stage with a larger engine.

Well, the funding for "Breakthrough Propulsion Program" ended in 2002. So it's outdated by over a decade. There is current experimental research on warp drive technology being done at NASA right now. So it went from "it's purely theoretical," to "we are doing experiments on it," in the meantime. Inaccuracies are in pretty much everything. Alcubierre Warp Drive is not a speculation. It is a prediction of an extremely well-tested theory. It is yet unclear if all the requirements can be met, but this isn't something that people are just making up. Neither is warping of space-time speculative. Again, it's a prediction of General Relativity. Theory without which the GPS satellites would not work. I mean, at this point, you might as well just say that gravity is just a speculation. And Special Relativity has nothing to do with the discussion of the FTL propulsion. All of the limitations with infinite energy, etc., only happen if you try to breach the speed of limit in flat space-time. Newsflash, it's not flat. Speed of light is a local limitation, and so long as you obey it locally, absolutely none of the problems the article raises show up. Global causality is violated, but there is absolutely no reason why it shouldn't be.

Yeah. I punch the equations used by KSP into Mathematica and solve them there. Fortunately, they are very simple equations, so NDSolve method has no trouble with them.

This is both outdated and written by a person who does not understand the subject.

It can't catch up from the rear. Correct. But it wouldn't be encountered at FTL speeds from the front. That's the whole point of the statement that warp drive does not result in local violations of the speed of light limit. Warp bubble is going to frame-drag anything it encounters. That includes "tugging" on the light shining from the front so that it is still traveling at precisely the light speed when encountered by the ship. That will result in severe blue shift, and angles are going to be all distorted. In fact, observer on the ship will be able to look around and see the outside universe in every direction. Except that the light that seems to be coming from behind is also going to be actually from the front. Hm. Maybe I should write up a simulation of what the sky is going to look like when you are traveling under a warp drive.

Not hard at all. Manufacturing of these devices is not illegal. Only export from US is. So as long as all of the components are made in China, it's not a problem. And there are plenty of such units on the market. That said, civilian band GPS is really not good enough for missile guidance without a decent INS to back it up.

Alright, so in simulation, for periapsis of 12,745m after entering Duna SOI on transfer from Kerbin, I'm getting an apoapsis of 240km. Acceleration never exceeds 1G during the maneuver. This is consistent with what Vanamonde suggested, so I say go for it.

That's only true if you want to park low in Jool's orbit. If your goal is transfer to Laythe, it's going to be considerably less. Suppose, you are heading to Jool on Hohmann transfer orbit from Kerbin. You'll hit Jool's SOI at 1,750m/s. If your periapsis is right there at the edge of the system, transfer to Laythe will cost you 1,700m/s and you'll reach Laythe SOI going 1,310m/s. If your goal is parking in Laythe orbit, you'll have to burn these 1,310 and then some, but that's an absolute minimum for parking anywhere in Laythe SOI. So your total Delta-V is 3,060m/s. Now, suppose you are going to pass right above Jool's atmo. You'll buzz the giant at 9,740m/s and it will cost you 1,080m/s to slow down to Laythe transfer. You'll reach Laythe's SOI going 1,270m/s relative to it. So your total Delta-V is going to be 2,450m/s. So you do save 600m/s with extremely low periapsis. Of course, if you are going to drop that low, you might as well aerobrake and save a lot more. A more typical periapsis of a few thousand km over Jool will save a lot less. Finally, if you decide to hit Laythe directly, by carefully adjusting your periapsis from a distance, you'll enter Laythe SOI at 4,860m/s. So this is definitely not the best option if you are planning to put a station there. On the other hand, if you are planning a direct landing on Laythe, there is no reason not to do that. You'll hit Laythe atmo going 5,450m/s, and that's perfectly survivable even in Laythe's thiner atmosphere. (Though, once they enable over-G damage, you'll have to chose re-entry angle very, very carefully. If you come in too steeply, you can easily hit 50G+.)

You don't really have to lower the periapsis before getting there. The difference in fuel will be fairly minimal. If you aren't planning aerobraking, it might make things easier. But otherwise, once you're on transfer orbit, just set up a new node and play with the controls. See which ones make periapsis go down. If inclination is the reason you aren't getting good periapsis, best place to adjust that is at ascending/descending nodes. But yeah, once you set the navigation node, it's just a matter of burning in direction of blue marker. The only reason you might be getting off course this way is that your "control from" part of the ship might not match the direction of your engines. That can happen if your overall ship is a multitude of docked parts. Just find a docking port, capsule, or probe that points the right way, click on it, select, "Control from here," and you should be set.

Sure it can. There is no event horizon in the warp bubble, so in principle, you can communicate with ship under warp. There can be some engineering difficulties associated with extreme red/blue shifts, order reversal, and lensing that signal undergoes, but you can send a message across the bubble.

There is an important difference here. If a spot of light appears to be moving faster than speed of light, it still can't convey information from one point to another faster than light. A ship under warp drive can.

On asparagus with auxiliary engines? Yes, definitely. But it still only really matters the first 20km or so. If your upper stages have the "wrong" TWR it will be a relatively minor setback in terms of fuel, so long as your turn corresponds with TWR your rocket is capable of.

Fuel-optimal TWR for vertical ascent is exactly 2. That follows from the fact that optimal ascent rate is equal to terminal velocity* and except for actual liftoff, acceleration is negligible until you reach thinner atmosphere. Of course, as fuel is used up, TWR at maximum thrust will increase. So you should start at either a little lower than 2, as UmbralRaptor suggests, or you have to cut the throttle a bit as you go up. Of course, once you begin gravity turn, the turn itself will determine optimal thrust, and I haven't been able to compute the optimal solution even for the simple case of Kerbin's atmosphere. A lot of people are recommending different things, and I have been able to confirm at least a few as not true optimums, but for the most part, if you follow general advice, it's pretty close. * Yes, even for the case when atmospheric density decreases with altitude. At least, for exponential case used in the game. So your velocity-altitude profile is going to be v(h) = v0 exp(h/(2H)), where v0 is terminal velocity near Kerbin's surface, which is about 100m/s, and H is the scale height, which is 5000m for Kerbin. In other words, you should be traveling at 100m/s very soon after liftoff and reach 270m/s when 5km up. Again, once you begin the gravity turn, this formula breaks down.

Yes, warp drive violates global causality. But there is absolutely no law stating that causality must be obeyed globally. It is a local property, and warp drive does preserve local causality. As for wormholes, there is no known mechanism for a change in topology of space-time, and one must alter space-time topology to create a new wormhole. Of course, it's possible that there are enough microscopic wormholes around that we don't need to create one from scratch. Only manipulate it to make it traversable. That we can do with known physics, at least, if not with known materials. But we haven't discovered any actual wormholes, so that's a bit of a moot point. Warp drive is currently the only FTL method which we actually understand the physics of. There are still some major theoretical hurdles, like requirement for negative energy densities, but these are a problem for traversable wormholes as well.

Whether it is a fact or not depends on your definition of fact. How would you define it? Can you give me an example of something that is a fact? Defining fact as anything that is supported by well-established theory is not a bad way to define it. That's a good, pragmatic definition that a lot of people hold to. But yes, theory by its very design must be falsifiable. Otherwise, it is useless. The only theory that cannot be falsified is a theory that makes no testable predictions. And if it makes no testable predictions it makes no predictions that we can make use of.

Yes. And it should.

Yes, all angles are in radians. So if your input is in degrees, don't forget to convert them. It's easy to go from eccentric anomaly to mean anomaly. m = t - e * sin(t). But you need to go backwards and get t from m. I'm not sure there is an exact solution for that, but there are some iterative methods you can use. Off the top of my head, this should work. //Given m and e, compute t. t = 0; for(j = 0; j < 100; j++)t = m + e * sin(t); The higher the number in "j < 100" the better the approximation is going to be. There is probably a better way.

Parameter is a general term for the independent variable in the parametric equation. The simplest parametric equation for ellipse is written with eccentric anomaly as parameter. That's the equation you used. But it's possible to use a different parameter, including the mean anomaly. You'll just need a different equation. (Or a map from mean anomaly to eccentric anomaly.) Mean anomaly is defined in such a way that it advances by equal amounts over equal time intervals. In other words, if you want to define the rate at which toe object traverses orbit in terms of degrees/second, you are talking about mean anomaly. So if you still want to use parametric equations based on eccentric anomaly, you need to first convert mean anomaly to eccentric anomaly. But like I said, the correction is small if the eccentricity is small, so I'd focus on getting everything else to work first.

I'd start with making sure that your orbit has the star at the focus rather than center. As I mentioned earlier, your equations set the star in the geometrical center of the ellipse. That's not a problem for a circular orbit, since the foci coincide with center for circle, but the moment you add a bit of eccentricity, it will look wrong. You have equation for ellipse with origin on center: x = a*cos(t); z = b*sin(t); Instead, you want equation where origin is at the focus. This is achieved by simply shifting the ellipse by focal distance. x = a*e + a*cos(t); z = b*sin(t); Inclination (i) is simply a rotation around the Z axis (since your major axis runs along X). In general, a rotation around Z axis looks like this: x_after_rotation = cos(angle) * x_before_rotation - sin(angle) * y_before_rotation; y_after_rotation = sin(angle) * x_before_rotation + cos(angle) * y_before_rotation; But prior to inclination y is zero. So if you apply this formula, the equation for an inclined ellipse looks like this: x = cos(i) * (a*e + a*cos(t)); y = sin(i) * (a*e + a*cos(t)); z = b*sin(t); The z coordinate doesn't change since Z is the axis we are rotating around. In principle, there are two more angles you need to set the orbit. That's the argument of periapsis and longitude of ascending node. But if you are modeling a solar system, where all inclinations are small, you can just add a rotation around the Y axis. If you do so, the full equation takes the following form. x = cos(f) * cos(i) * (a*e + a*cos(t)) + sin(f) * b*sin(t);; y = sin(i) * (a*e + a*cos(t)); z = cos(f) * b*sin(t) - sin(f) * cos(i) * (a*e + a*cos(t)); Here f is the angle by which you rotate position of the major axis. For most notable objects in the Solar system, this is more than adequate. Oh, and maltesh is absolutely correct about the fact that your equations are for eccentric anomaly (variable t in all of the above), not mean anomaly. What that means is that for highly eccentric orbits your objects won't traverse the orbit at quite the right rate. Mean anomaly and eccentric anomaly are closely related, however, so it's not very difficult to recompute one from the other. I can show you how to do that if you'll want to correct for this.

Hm. What about this bit of code in Body.Update() transform.Translate(Coords); Does this call translate object to Coords or by Coords? Because if it's the later, it would explain a lot, including why z doesn't go negative.