K^2

Technically, yes. I was a bit lazy about doing this strictly analytically, and I'm not sure how hard it'd be to actually do. If you take h to infinity, even thought time required for ascent is finite, fuel requirement diverges because acceleration goes to infinity as h does. So you probably need to set up cutoff height, optimize for v0, and only then take the limit. Alternatively, one can be satisfied with the fact that numerically v0 = vt works. That's the attitude I chose. Call me lazy if you must.

Actually, you could measure tidal forces to figure out quite a few things about your orbit assuming you know the relevant parameters for the planet you are orbiting. But yes, you can't just measure gravity directly, and that means much, much finer measurements to get any useful data.

I'm not too familiar with how things are done these days, but I know that on early flights, all of the computations for orbit transfers were done on the ground. The flight control would have tracking data, they'd use it to figure out the orbit the ship is in, find the necessary burn times and orientations and send these to the pilot. So if you want to think of it in KSP terms, the flight control center would have equivalent of the map view in KSP and would be able to set up transfer nodes. The pilot would receive instructions equivalent to what the little blue marker on the nav ball tells you. That is, set heading for whatever it needs to be, wait for specific time, then burn for that many seconds. So in that respect, KSP is pretty similar to what actually happens, except the player has access to both sides of the mission planning and execution.

They be heavy. So it's all about which one's going to require less additional mass on board. Bringing more RCS monoprop or installing huge gyro wheels. For a Lunar lander, gyros would make zero sense. For the CM such a thing would probably be more viable, but CM didn't need that many precise orientation changes, so RCS was probably still an easier option. And as somebody already mentioned, reaction wheels (or CMGs) aren't a replacement for RCS. It's not like KSP where reaction wheels can absorb any quantity of angular momentum. They can store some, but it's limited. One way or another, you have to have RCS thrusters on board. But on longer missions or for better precision, CMGs might reduce the amount of RCS monoprop you need significantly enough to be of use. For scientific equipment there are additional considerations, of course.

Oh. Uniform? No arms or anything? Then it's uniform in time, as well, and that makes it a 2D problem. That's totally solvable. Still a hell of a mess, of course, since velocity field will depend on metric and vice versa, and that means the equations will have to be solved very precisely. But since the grid is only 2D, there are no hardware limitations. Just complexity of the code. This would be a lot of fun to implement. I can already see how it can be spread over a computing cluster nicely. If you seriously want to try it and don't mind taking the brunt of the analytic work, I can work out all the numerics. There's probably a paper in it if nobody has done this with high enough fidelity before.

Is this even doable numerically? There are no obvious symmetries to exploit, so this would be a full 4D sim. And I don't know if there might be a better way to do this, but the only thing that comes to mind is lattice, and 4D lattice computations are tricky. Even in QFT a considerable amount of trickery goes into it, and that stuff's all 100% linear.

That will either not work at all or be too much of a pain. Has to do with the fact that each spaceship consists of many many meshes each one with its own transform matrices. 3D Ripper won't know which models go together, so it won't be able to rip it all into a single 3D file. You might be able to rip individual components and put them together by hand, but I suspect there is an easier way to get the models from a Unity game.

When you say "Newtonian," do you mean pure Newtonian, or something along the lines of linearized gravity? Because I can entirely believe gravitomagnetic effects to be quite significant on the scale of galaxies. But if they are actually saying that the problem is highly non-linear, that would be rather unexpected. I really would like to see some computations supporting that.

Heh. Yeah, I missed something. Sorry, I'm actually doing all of this in Mathematica, which lets me skip steps. What's missing is the ∂F/∂h term. I'll make an edit. Hm. This is a much messier thing to differentiate than it looked. I'm really glad I wasn't doing this on paper. I'm feeling a bit lazy to fill in the steps, but hopefully, you can figure it out. The equations in the main post should be right now.

Well, the basic idea is that total fuel used (ignoring ISP changes due to pressure) is proportional to trust. Now, there is choice in what we are optimizing with respect to, but if an integral over F(t) from some t1 to t2 can be reduced with minor variations in v(t), then you haven't done the best possible job. Now, given altitude as h(t), velocity is h'(t) and acceleration is h''(t). (Just making note of notation.) Then in units of ship's mass, we can write. F(t) = h''(t) + g + k(t)h'²(t) Here k(t) is the drag coefficient given by k(t) = k0 exp(-h(t)/H). This is proportional to àin your computations. So what we want to optimize is a functional S[h(t)] = ∫F(t)dt. Any function h(t) that satisfies such an optimization will also satisfy the Euler-Lagrange Equation. For a function F(t) which depends on h(t), h'(t), and h''(t), the following is satisfied. ∂F/∂h - (d/dt)∂F/∂h' + (d²/dt²)∂F/∂h'' = 0 In other words, whatever the optimal ascent is going to be, it's going to satisfy the following equation. (d/dt) (∂k(t)/∂h h'²(t) - 2 k(t)h'(t) = 0 Two things should be noted right away. First, if k(t) was constant, we'd have v(t) is constant as solution. Second is that this doesn't depend on g and allows for infinitely many solutions. For instance, v(t)=0 is clearly a solution. This isn't THE optimal one, just a constraint on possible optimal solutions. But it's a very useful constraint. Taking time derivative, we get the actual differential equation for velocity. 2k0 exp(-h(t)/H) h''(t) = k0 exp(-h(t)/H) h'²(t)/H Which, thankfully, simplifies. 2H v'(t) = v²(t) That isn't so bad, and general solution is easy to find. v(t) = -2H/(t+C) Where C can be any constant. On one hand, this is a solution. Give me v(t=0) and I can find C that will make this work. But that's not the interesting part. Consider the velocity that satisfies the following equation. v(t) = v0 exp(h(t)/(2H)) You will realize that if v0 is terminal velocity at the surface, the above is terminal velocity at altitude h(t). This differential equation can also be solved, and with h(t=0) = 0 it yields the following. h(t) = -2H ln((2H-tv0)/(2H)) What's more interesting is that I can find the velocity of the above profile. v(t) = 2H v0/(2H - tv0) And this, in turn, satisfies the earlier equation with C = -1/v0. In other words, ascent at terminal velocity through exponentially thinning atmosphere is locally optimal. And therefore, being the optimal solution for constant density is the overall optimal way to ascend through exponentially thinning atmo. Edit: Fixed ∂F/∂h term.

It's not really so much about how well they are defined. It's more about the fact that when you have boundaries in your abilities to probe a model within some parameter space, it's absolutely useless discussing whether the model applies beyond these boundaries. It's one thing if the boundaries are purely due to practical limitations. Then we can expect things to be nice and continuous. But if there is an actual singularity, like the v=c case, there is just absolutely no reason to assume that mathematical continuation of the model beyond that boundary has anything whatsoever to do with reality. Grandfather paradox is almost trivially resolved in field theory, so it's not a very compelling argument. Me and Stochasty have actually had discussion on possibility of time travel in GR without involving exotic matter. I still need to track down some of these papers, by the way. Point is, time travel might very well be within realms of physical possibility, so even if it remains out of practical reach, we still couldn't discount a possibility simply based on it leading to time travel related paradoxes.

As far as I know, accretion disks are denser closer to the star, so there is more stuff to collect for a gas giant close in. I don't know if it explains all of it, but it might be the big part of the puzzle.

I'd say the more interesting statement of Standard Model is that even if they somehow did exist, none of the known particles could interact with them.

You are looking at mass changes. I basically disregarded that because in KSP drag is proportional to the mass, so everything can be done in units of ship's mass. What I was working on is accounting for dÃÂ/dt terms. Your derivation simply assumes that ÃÂ is constant, but it really isn't, and it would throw a major monkey wrench in your computations, because then you can't just say, "What is optimal v?" You have to say, "Which function, v(t), optimizes total fuel consumption." And that's Variation Calculus problem. Fortunately, some really smart people found some really nice shortcuts there.

Most likely, it means absolutely nothing. A formula like that is only good up to a singularity. In other words, all of the Lorentz boost formulas are only valid for v<c. But if we assume that these formulas are correct and try to ask what it would mean for particles assuming everything else extends to that region nicely as well, we can look at the Shrodinger Equation for the answer. So long as velocity is high enough, say, v > 10c, we should be able to use non-relativistic wave equations. In that case, the wave vector for a particle satisfies ħ²k²/(2m) = iE for some real value E. That gives us solution, k = (1+i) Sqrt(Em)/ħ. Since this quantity has both imaginary and real parts, the wave itself will be losing amplitude. In other words, if a state with v>c can exist and follows algebraic extensions of equations we have for sub-light particles, such a state must immediately decay. But again, it most likely has nothing to do with reality to begin with.

Yes, light does bend space-time. Though, not quite in the same way that a massive point particle would. It has to do with the fact that if light has a lot of energy it also has a lot of momentum, and momentum factors into stress-energy tensor which is ultimately the source of gravity. As a good example, there exists a Vaidya Metric, which is a solution for a star that radiates a lot of energy, and that radiated energy alters the gravitational field around the star.

It was there in .20. I had a flag planted next to it. It's one of the Squad monuments. I can probably get a craft there in a short while to check it over in .21.

Optimal climb rate is, indeed, terminal velocity. For constant density, the proof is trivial. The thrust, in units of ship mass, is equal to F = g + kv². The amount of time needed to rises to some altitude h is t = h/v. In that time you consume some quantity of fuel proportional to Ft, so your goal is to find maximum of (g + kv²)/v. Differentiating that with respect to v and setting derivative to zero yields k - g/v² = 0. Or g = kv². Note that this is the same exact condition as setting F = 0, meaning you just found terminal velocity, except going upwards instead of down. Substituting into original equation we have F = 2g, meaning thrust is twice the weight. Now, the problem is actually more complicated because air density is decreasing exponentially, so the value k is varying as k(h) = k0 exp(-h/H), where H is scale height. Math for this problem is extremely hairy and involves calculus of variation. I have solved this problem, and if anyone here wants to see the proof, I can outline it. But the result is exactly the same. Climb must be performed at terminal velocity for given altitude. In other words, v(h) = v0 exp(h/(2H)), where v0 is terminal velocity at sea level. Now, however, thrust must compensate for drag and acceleration. By the time acceleration is significant, however, you should already be well into gravity turn, so it turns out that TWR of 2 is still a good one to shoot for. Two dimensional problem involving gravity turn is extremely complex, and I have not found a solution. I have verified a number of "rules of thumb" to not be true optimums, but even solving for optimal ascent numerically has proven to be beyond the "simple" methods. Edit: There is an interesting consequence for all of this. Again, I'm not going to get into all the math, but you can estimate the extra delta-V required to climb out of atmosphere as 4gH/v0. Again, H is scale height and v0 is terminal velocity at whatever height you start out from. This means that the total minimum delta-V required to reach stable circular orbit is roughly dV = 4gH/v0 + Sqrt[MG/(h+R)]. Here R is radius of the planet, M is its mass, G is the gravitational constant, and h is the altitude of your final circular orbit. For Kerbin, that works out to 4,128m/s, which is very close to empirically established value.

This requires clarification. Technically speaking, E = mc² is wrong. The correct equation is E² = p²c² + (mc²)². The reason for that is that the symbol 'm' is reserved for rest mass which is invariant for an object. However, inertial (and therefore gravitational) mass remains proportional to total energy. Since inertial mass of an object is equal to γm, it is also entirely fair to write E = γmc². Gamma is, of course, Lorentz Boost factor and it satisfies 1/γ² = 1 - v²/c². Since p = γmv, the two expressions for energy are equivalent. Your point is entirely valid, however, since it is the inertial mass of the object that determines the force required to accelerate it. As for relativistic rocket formula, something very interesting happens. While from perspective of the outside observer the ship accelerates less and less, requiring more and more fuel to get closer to speed of light at alarmingly faster rate than classical rocket formula would suggest, it's not actually as bad as it seems. Due to space contraction, it becomes increasingly faster for the ship to cover distance as measured by ship's clock. If instead of looking at speed relative to various objects you consider proper velocity of the rocket, which is how far you get in certain amount of ship time, eventually the equivalent "rocket formula" becomes linear. That is, to go twice as fast, you need twice as big a fuel fraction. And that actually makes it possible to go really, really far, provided you have good fuel. Unfortunately, this doesn't let you break the light speed barrier. You can get insanely far on reasonable amount of fuel and within life time of the crew, but on Earth many, many years will pass. That isn't actually true. It's a common misconception arising from poor understanding of special relativity, but there is no self-consistent theory that allows for faster than light travel without warping space-time.

Won't work. Basically, when you get really close to the black hole, Kepler's Laws break down. In particular, you no longer follow an elliptic trajectory. Once you cross event horizon, you'll just keep falling in towards the singularity at the center of the BH. There are coordinate systems in which the object bellow the event horizon is traveling faster than the speed of light, but in GR this is a very loose statement. Yes, you are traveling faster than light relative to remote observer if you draw the coordinates a certain way. But it doesn't really have any physical meaning. What's important about the speed of light limit is that it's a true limit locally. That means you can't be moving faster than light past something. And that's still going to be true even after you deep bellow the event horizon.

It might look like equally fast to human eye, but the actual times aren't even in the ball park. The liquid fuel has to mix with oxidizer to ignite. That takes time. On large rockets, quite a bit of time. We are talking order of seconds here. Still almost instant to human eye, but long enough for electronic systems to respond and even do something about it. SRB is already a mix of fuel and oxidizer. Once the shockwave starts, it goes as fast as that shockwave propagates. We are talking small fraction of a second. Even if you have time to register it, you simply can't outrun this explosion. And as Themohawkninja pointed out, an explosion of a liquid rocket's engine at least has a chance to go quietly. With SRB, there is no such chance. The only mode of failure is catastrophic explosion.

There are several parts to it. First, as Jaydee says, you can't shut a solid booster off. If you have a problem with it, you have to either ride it out or jettison it. There are no other options. Then there is the fact that a smallest crack in the fuel can lead to an explosion. It's rather difficult to inspect a solid engine for things like that. Finally, the biggest issue. When the SRB blows, the whole thing does. All of your fuel is sitting within microns of all of your oxidizer. It's a powder keg, almost literally. If your liquid engine explodes, it's bad, but all of your fuel won't go into a fireball all at once. You have a chance to fire up an emergency escape system. With boosters, the whole thing will be over before even automatic systems can react to it. Liquid engines come with their own hazards, of course, but most of these are things where you'd have some sort of control over it. Apollo 13 came home safely. Columbia did not, but at least there was a window of opportunity for problem to be detected. With Challenger there was never even a chance. The shuttle was doomed while standing on the launch pad.

How about an MHD Generator? You can get efficiency comparable to gas turbine. Losses are still considerable, but like Drunkrobot said, there are going to be first world nations that such an engine would still be able to power.

It will melt, of course, but so do most hybrid fuels. The point is that it burns away faster than it melts. So it's not a problem.

One of the bigger reasons is that they are flying relatively small rockets. The smaller the stage, the heavier the tank has to be to keep cryogenic hydrogen as fraction of fuel carried. For something like Dragon, it makes no sense to fly with LH2 because extra weight of the tanks will offset any benefit from higher ISP.