Z-Man

That is not negative absolute pressure, that is just pressure that is lower than the pressure of the surrounding air.

Glad someone else noticed that, too.

Do they have an I_sp of 1 like the hypothetical example I stated? No? Well, then.But still, where would they be in stupid_chris' graph? Keep in mind that for SRBs, you need to take the maximum TWR to be somewhat consistent. I tell you where they are. Deeply in the region he considers OP. Are they really OP? What's their payload fraction to LKO? With staging allowed in this case.

Yes. You can read the intention from the atmospheric I_sp; the KR-2L has a lower one than the other new engines, and you can attach to it from below. It is meant to be an awesome space engine for larger vessels. Increasing its mass would be the right move to reinforce that. That would be an engine balanced according to your graph. But in practice, it would be useless. You would not be able to build a rocket clearing even 1 km in its first stage with it. I am using this extreme example to demonstrate that your analysis method is wrong there, but it actually is wrong everywhere. That you have to cull outliers also shows that. Amaroq's metric is much better. Specifically, single stage payload fraction to LKO is an excellent measure for lifter engines. I personally think that larger engines should be at least as good at that particular job than smaller engines so players are not forced to decide between part count (use large engine) and efficiency (use cluster of smaller engines).

I said TWR infinite. And I did not mean an engine balanced according to your formula. I meant a hypothetical engine clearly overpowered by your formula, but not really.Mind you, I just disagree with your method. Some of the conclusions are right. The new engines are more powerful as lifters of heavy stuff than anything we had before.

That is a special case of a power law. Because it implies that a hypothetical engine with I_sp of 1, any thrust and zero mass is better than anything we currently have because it has infinite TWR. Because it implies the old Ion engine is useless. It does not take into account that different situations require you to weigh TWR differently, and in none of them should you be looking at TWR*I_sp exclusively. I gave a more correct formula earlier.

I have to redact my attraction and total electron-positron annihilation claim. It doesn't work like that. True, the matter and antimatter penetrate each other. True, in the overlap zone, electrons and positrons annihilate, leaving a plasma of nuclei and anti-nuclei. However, that plasma is overall electrically neutral and the leftover matter and antimatter are, too. So there is no reason for the electrons and positrons to leave their cozy spot. I was assuming they quickly redistribute over the whole former shape of the blob because... of reasons. That is silly, of course. And without that, the rest of the process crumbles. The nuclei are free to react and certainly enough of them are hurled away from the plasma and into the leftover matter and antimatter and leave enough momentum there to blow everything apart.

Yes, that's what I was saying, wasn't I? I may not have been overly clear.It is, of course, entirely possible my entire prediction here is wrong. It's purely based on some rough arguments about energy and time scales and a lot of handwaving.

I don't object the use of graphs in general or the general principle that high TWR should go with low I_sp, at least for engines of similar mass and size. What I object to is the use of a power law for it. That is simply factually wrong.

The electron-positron reaction, which happens quicker, only produces gamma rays. What mechanism do you propose for the heating? The only energy you have available for the heating is the kinetic energy of the electrons. But that energy is only half the negative potential binding energy, and the electric binding energy of the nucleus-antinucleus is about the same initially and can go much, much higher (well, lower since it is negative) as the nuclei can pack themselves more densely now. Sure, individual nuclei can be ejected, but there is not enough energy available to completely scatter them in a way that prevents further annihilation.You are right in one point, though: The state the matter/antimatter conglomerate is in can best be described as a plasma.

a2soup: Yeah, of course for lower mass landers a lower mass engine would win, largely independently of small I_sp and TWR differences. m4v: No. For lower required total TWR, clusters of 48-7S also become less attractive. That's the point. And the whole trouble of the 48-7S was that clusters of it were better than anything else for launches from Kerbin. Well, not any more.

True to some extent, still: Twice the TWR does not make a twice as useful engine. The best lander engine depends on where you want to land. For Tylo, sure, it's the 48-7S. For Mun, no. Purely by stats, it's the LV-N, actually. I think. Don't have my spreadsheet here. But if the total TWR you aim for goes below 1 or so, the 48-7S stops being attractive and the LV family takes over. Do the proper math and you'll see.

If I read the excel file correctly, the curve is an auto-fit power law without significance. The graph and formula used give too much weight to TWR. Twice the TWR does not make a twice as useful engine.Yes, the 48-7s is a bit overpowered in the lifter/lander department. Still, it's a valid comparison point for new engines simply by the fact that it has been that way for a couple of updates. Either it's intended to be this way or the devs intend to make a global rebalance pass sometime in the future, probably only after the economic aspects of the game are in place.

Unfortunately, no. Two lumps of matter can only touch at the surface, true. That is, essentially, because the Pauli exclusion principle forbids two electrons to be in the same state, so the electron shells of lump 1 can't go where the shells of lump 2 are. If Lump 2 is antimatter, however, things are different. The Pauli principle does not apply, the shell positrons already are in an intrinsically different state. Matter and antimatter can penetrate. They still get blown apart once the reaction starts, you will probably say. Well, not small blocks. What happens next is that some electrons and positrons annihilate into gamma rays. Gamma rays are highly penetrating, they quickly leave the immediate area and do not blow apart the lumps. And once only a small fraction of positrons and electrons are gone, what you have is a massively negatively charged lump of antimatter and a massively positively charged astronaut. The lump of antimatter will immediately be pulled entirely into the astronaut's hand where the rest of the positrons are eliminated. That's already 0.05%-0.1% of the total mass converted into gamma rays, again escaping without blowing anything up. I'm not familiar enough with strong interactions and the relevant cross sections to say what happens to the nuclei once they find each other (which takes longer than the electron positron reaction because they are heavier). Maybe some (anti)nucleons annihilate on first contact and the recoil blasts the nuclei apart; in that case, still at least 2% more of the mass is converted into energy. Maybe before the first annihilation takes place, the entire two nuclei first penetrate each other (again, they're allowed to), in that case, energy conversion can be close to complete for identical elements of matter and antimatter.

Kasuha: Using cubic octagonal struts, you can also cluster 24 mainsails around the S3-KS, maybe more if you disable snap. For when you need that little bit of extra thrust. No debug menu part clipping required, though of course there is plenty of clipping. TWR*Isp is not a sensible engine metric. The most sensible short one is Isp*(1-TWRdesired/TWR), where TWRdesired is the desired total TWR of the vehicle. So about 2 for a launch engine and however low you want to go for space. That formula assumes you are building an infinite asparagus with perfect fuel tanks and shed engines as soon as you no longer need their thrust. Just a bit of nitpicking. Personally, I think it is fine if designated huge stage 1 launch engines are strictly (but slightly) better than everything smaller. After all, their weight disqualifies them for most applications once in space. The new parts just highlight how shoddy the Mainsail is at its job The real launch engine efficiency comparison would be a 48-7S cluster, and they're not much better than that. If I were calling the shots, I'd have given them a slightly lower vacuum Isp, though, just to make absolutely clear they are for lifter stages only.

Yep, that's what I observed too with a shorter string, and it's what was predicted. I did not make videos of that because the stable circling radius changes a lot with the gyro's precession frequency and thus rotation speed, and the rate of change there was simply too big to get anything meaningful out of it.

That's actually not a bad idea. I still have one, and as these gimmicks go, it's not getting much use Still, you'd need good bearings and a way to get it up to a decent rotation. I hear ya. Pro tip: Find one that does not explode. At least in my case, a lot of it is probably due to the friction in the bearings, that causes the gyro axis the string is attached to to rotate a little. The force from the string would thus be applied off-axis. And, well, there are those horrible vibrations. In the MIT experiment, the rope is sturdy and short and gets wound up. That causes a considerable amount of torque around the z axis counteracting the precession, which in turn pushes the wheel rotation axis downwards. And yes, even ignoring those practically relevant disturbances, analytical progress is difficult. Maybe switching to a frame co-rotating with the gyro precession helps? Simulation is probably the best approach so one can see what you can try to prove analytically in the first place.

You mean like here? https://www.youtube.com/watch?v=8H98BgRzpOM Well, it depends on the parameters. Check K^2's formula for the É << Ω case. The maximum deviation in your initial conditions is roughly dÉ/Ω. Plug in your own numbers and see what you get. In the video, É is a bit below 1/s, Ω for the ca. 1m string would be 3/s, which would give a maximum deviation of about 0.3 d. Looks about right. And what the formula neglects, as stated, is the torque caused by z and r not being colinear. Well... such a torque is quickly swallowed up by a gyro, which reacts with an inclination change. You can observe it in that video. Watch how quickly the oscillation of the string dies down. That is not normal friction, that is the wheel eating up torque. In the other realm of É >> Ω, the formula predicts z=-d or precession around the gyro's center of mass. I managed to make a video. We have a high, reasonably open stairwell at work where I was able to easily set up a very long gyro pendulum. I did not measure the length, but it were 4 storeys, so at least 10 m and Ω > 1. É for the toy gyro is about 3/s at the best of times. The gyro had a mass of 84g (yes, I used it as the weight previously. I already had it on a rope. Sue me.), mass of the string is about 1g. Video: http://youtu.be/N2trsfBqxPA As you can see, before it gives in to friction, the gyro always precesses pretty much about its own center of mass. Definitely not around the suspension point. Difficulties: Crappy gyroscope with too much friction, winds down too fast. The string was more elastic than anticipated, causing the gyro to bounce up and down. Crappy gyroscope. What can I say, it already was 5 Euros over the budget I usually allow for internet debates. And it was the only one the toy store had. And incidentally, the É >> Ω is also pretty much equivalent to a gyroscope precessing on a very light tower you always demanded. So there.

I merely followed your instructions. You did not specify an upper limit to the force to put into the pendulum, and the example values you gave for r vs L were already well outside the limits of the linear approximation. Does not matter. Your experiment was irrelevant anyway. K^2, about the potentially chaotic behaviour for arbitrary initial conditions in the case of É << Ω: Don't worry about it as long as d << L. Do an energy and angular momentum (z-axis, naturally) conservation analysis. If the gyro tilts up or down, it transfers both energy and angular momentum into the pendulum degrees of freedom. And if it is fast spinning, the pendulum degrees of freedom can't take the large amount of angular momentum they would need to pick up while being constrained by the amount of energy they get. It really works out beautifully and precisely: If you ignore d and start with a horizontal gyro, it is locked into the horizontal plane by the conservation laws precisely if É < Ω. Edit: for finite values of d/L, I suppose you still get narrow enough limits on the gyro's stilt to be able to ignore it. Edit: I realized that the strict result only applies for initial states with zero energy in the pendulum degrees of motion. No matter. For É << Ω, the conservation laws still provide narrow boundaries on the allowed inclination of the gyro.

Reread the analysis. Behaviour is supposed to get complicated near the resonance. Far off the resonance either way, behaviour is expected to be more regular. Which is precisely what I observed (and, of course, others before me) and reported. I've swung around enough weights on strings in my life to know how they behave. But fine, if you insist to waste precious bits on the internet on an irrelevant side point of a side point, I've done it. Rope length was 60 cm, weight was suspended 3 cm off its center of mass, mass was 84g in case it matters to you. Swung it around. Counted how many rounds it did over the course of a minute.Radius of about 25 cm: rotation frequency was 39.5 per minute (̉ۡ = 2.07/s) Radius of somewhere between 40 and 50 cm: rotation frequency was 46 per minute (̉ۡ = 2.41/s) Radius of somewhere close to 55 cm: rotation frequency was 60 per minute (̉ۡ = 3.14/s) That was about as fast as I dared to fling the thing around, the rope was from the other experiment and thus not chosen for durability or painless handling. Anyway, clearly not constant.

Oh, so I misinterpreted your vague instructions there. The more important points still stand. And now, by the way, ̉ۡ varies. With a fixed string length, it increases as the movement radius increases, it has no upper bound. No way around it. It is a mass going in circles, yes, but not in itself a gyroscope. The formula specifically requires a gyroscope that is spinning all by itself, suspended on a string attached to it off-center. The weight is not itself spinning around its center of mass, so the formula does not apply. Not a gyro on a string. And of course it did behave differently. An inert mass, forced onto the same relative motion to the tower, would just have toppled over. The radius can be obtained for the very specific case that the pendulum is in a purely precession driven state. The state you instructed us to consider exclusively ("Then do not push it!"). The magnitude of the linear momentum at any given time in that state is fixed given the fixed specified variables.

The equation was specifically only valid under these conditions: You have a gyro precessing with angular velocity ̉ۡ. There is no such thing in your setup. z is small compared to L. Not the case in your experiment, they are about equal. The formula is for the movement component induced by the precession. In your experiment, you are using the other, regular pendulum motion mode. So yeah, of course the formula would describe the experiment incorrectly. The experiment does not even come close to testing the formula. Furthermore, contrary to your claims, L (the RELEVANT rope length.Total rope length does not matter, what does matter is the length from the weight/gyro to the point it is fixed, in this case your hand) does vary. ̉ۡ, which I assume you interpret as the angular velocity of the mass around your hand, can vary too.

You mean with ̉ۡ as the precession angular velocity? That is impossible in general. If you try and release the gyro, you'll just have it in a mixed state of regular pendulum motion and precession driven motion.I actually went to the toy store and got a moderately suitable gyro and tried that. When suspended on a string, it behaves as calculated within the wide limits of observational accuracy. Its slowest precession is about one revolution in two seconds, that's slow enough to be slower than the oscillation given the 40-50 cm rope length I used. Unfortunately, the bearings are crap and it drops down to resonant precession within about ten seconds, so its movement is nowhere near stable enough to measure the displacement. But the qualitative behavior is as expected, sign change of z included. K^2's formula for the displacement z of the suspension point, where d is the horizontal distance of the CoM from the suspension point, L the rope length, g Earth's surface gravity, ̉ۡ as above. Derived for a semi-steady state where, when seen from above, the gyro rotates around the center (the vertical axis through the rope's upper suspension point. That is an Ansatz. Try calculating it yourself. It gives one mode of movement (edit: technically, it's a special solution to an inhomogeneous differential equation here, I'm calling it a mode because it's actually variable if you allow different inclinations of the gyro), the other is the regular pendulum motion.

Yes. That is precisely the basis for what K^2 calculated. Sorry, this may be a language barrier thing. Every one of Momentus' statements falls into one of three categories: 1. Yeah, that's what mainstream physics says. 2. No, that's absolutely not how this works. 3. Wait, what is he trying to say? So, Momentus: Please, what exactly are you saying? Actually, the only assumption you need for this to hold away from the actual resonance is that L is large compared to d and z. It will correctly describe the semi-steady state where z and d are parallel. Of course, the full dynamics get really messy near the resonance and it requires fine tuning to actually get the system into that state, but hey. And as usual, ̉ۡ needs to be large compared to the spin frequency of the gyro.

Show that it happens precisely so. I say it won't. You'd observe a complicated overlay motion instead, one component looking a bit like you describe, the other being the regular pendulum motion with low amplitude. In the initial conditions you describe (gyro axis horizontal, string exactly vertical) that amplitude may (I have not done a full analysis, but edit: as K^2 said, the regular pendulum motion can be considered driven by the precession) get lower as the rotation speed of the gyro increases, so if you ever did the experiment yourself (holding the string in your hand, I'd assume), you probably did not see it.And of course the precession speed is independent of the string length. Why would it not be? If the bench still moves in that case, your claim would be equally refuted. Point is, the heavy tower moves. Do you understand how F=ma works?