sevenperforce

My guess is that the plumbing from the header tank flows into a common entry point in the thrust puck plumbing along with the plumbing from the main tank, so testing one tests all of them. However they may keep all three Raptors if they want to test the flow capabilities of the header tank. For all I know they're just swapping this Raptor out for a different one for other reasons.

Ah, good point. Yeah, so it's more likely they only remove 2 engines and leave the third if they want to do a header-flow static fire.

Who's to say they won't?

Oh that's pretty. I wonder if they are removing two of the Raptors to do a static fire test fed from the LOX header tank, once the nose cone is attached. If something goes wrong they don't want to lose all three Raptors. Or they may even remove all three Raptors just to pressure-test the header tank, for the same reason.

The tug of war value is the ratio of the forces acting on a body: Fp/FS, where Fp is the force of the companion on the body and FSis the force of the star on the body. Newton's law of gravitation says that F = G*m1*m2/d2. So in computing the tug-of-war value for a particular body, the mass of the body being acted upon drops out, and Fp/FS = (Mp*dS2)/(MS*dp2). In considering the tug-of-war value for Charon, we run the calculation and find that Fp/FS = ([mass of Pluto]*[39.5 AU]2)/([mass of Sol]*[19,640 km]2) = 596. So Charon is definitely a moon of Pluto and not a planet. But if we run the tug-of-war value for Pluto, we find that Fp/FS = ([mass of Charon]*[39.5 AU]2)/([mass of Sol]*[19,640 km]2) = 73.7. So Pluto is definitely a moon of Charon and not a planet. (Note that this is not the same for other planet-moon combinations in our solar system. The tug-of-war value for Io with respect to Jupiter is 3244 while the tug-of-war value for Jupiter with respect to Io is 0.1526; the tug-of-war value for Phobos with respect to Mars is 189.8 while the tug-of-war value for Mars with respect to Phobos is a miniscule 0.0000032. However, the tug-of-war value for Neptune with respect to Triton is 1.72, so take that for what it is worth...after all, Triton is one of the few solar system bodies which is almost definitely a captured dwarf planet.) I'm not saying we can't use the tug-of-war value, particularly when talking about exoplanets. The tug-of-war value seems like a great way to use a simple equation based on available data to make educated guesses about anomalous formation. For exoplanets which exert more force on their moons than their star, but which are subject to less force from their moons than their star, we can conclude that the given exomoon likely formed from the same accretion disc. For exomoons where this is not the case, it's likely that something significant happened like capture or collision. After all, the only moons in our system where this is not the case -- Luna, Charon, Triton, and Carpo -- were either collisions or irregular captures. It's fairly easy to call attention to this; you can simply say, "Not only have we discovered an exoplanet with a moon, but the balance of gravitational force in this system makes it almost certain that its moon didn't come from its accretion disc!" But this isn't a reason not to call it a moon. In fact, I think failing to call it a moon makes it more confusing, not less confusing. I think we should choose definitions which make science more accessible to laypeople, not less accessible. Should they be consistent? Yes. Should they have value beyond our own solar system? Yes. But they should make people see space as something they can understand and learn more about. If our definitions make space and the solar system harder for lay people to figure out, I think we're barking up the wrong tree.

Is that a COPV on the outside of the nosecone?

It's not aliens.

I suppose if you want to use the first photograph taken of Earth's surface, you'd use this one:

Photons have mass; they just do not have rest mass. Particles which travel at the speed of light (photons and gluons) do not carry any energy other than their kinetic energy. We, along with all ordinary objects (that is, objects with rest mass) experience the universe as a combination of unbounded dimensions (space) and a fixed dimension (time). Massless particles, on the other hand, experience a universe in which time is unbounded, but space is fixed. An ordinary particle can slow down or speed up or turn; a massless particle can do none of those things. A massless particle experiences no time at all, just a fixed, eternal path between the point where it is created and the point where it is destroyed. Of course, we observe a photon as an electromagnetic fluctuation that takes energy from one atom and carries it through space to another atom. Because it has energy, it therefore has mass. It's just a very, very small amount of mass. I'm not quite sure what you're asking here, but keep in mind that you can't just get power out of heat. You have to move something from hot to cold in order to generate power. The average temperature of a closed system can never decrease. No, that is not possible. The momentum of the engine exhaust striking the boundary of the "large area" is exactly equal to the momentum that the engine exhaust imparts to the engine when it is originally fired. What you are describing is a "reactionless thruster" and it is not possible to create one in this universe. You can't get a push without pushing against something else. One common idea for creating a reactionless thruster is to say, "What if you fire your exhaust out really fast, but then slow it down more gradually?" This is a quick way to expend a lot of energy, but because thrust is proportional to the momentum of the propellant, it doesn't really help you. The total momentum gain will be exactly balanced out by momentum loss, every time.

I doubt that you could have enough actuators to fully simulate the kick-flip gimbal maneuver. You've got the engines pushing against their gimbal mounts at full gimbal, the gimbal mounts torquing the thrust puck, the thrust puck transferring transverse forces to the skirt, and the entire aerodynamic load on the vehicle at the same time.

Well, the Roche limit of Earth is between 1.5 and 2.9 Earth radii, depending on whether you choose the rigid or the fluid computation. It is generally accepted that the Moon formed near the Roche limit but close to it, at around 4 Earth radii. So the tug-of-war value I calculated (103:1) was based on 4 Earth radii, not based on the Roche limit. Also on the topic of the Asimov measure -- by his numbers, Charon is definitely a moon of Pluto, not a double planet system. This seems strange because the Pluto-Charon system seems much closer to being a double planet than the Earth-moon system. Even worse, if you do the calculation in reverse, you find that Pluto is actually a moon of Charon by his definition. I'm not saying that the tug-of-war value isn't a valid or useful measure. But I'm just not convinced it has the greatest utility, overall. There are a bunch of factors to consider, honestly. You've even brought some of them up yourself. You'd almost want to make a "pros and cons" list. Did the satellite form by capture or accretion? Will the satellite escape in the short term? What about in the long term? Is the satellite tidally locked? Will the primary ever become tidally locked to the satellite? What are the ratios of the masses? Does the primary pull on the satellite more than the star? Is the satellite's orbit ever convex? Is the barycentre of the system ever outside the primary? Is the satellite gravitationally rounded? Would the satellite be able to clear/dominate its present orbit in the absence of the primary? Is the angular momentum of the satellite's orbit greater than the angular momentum of the primary's rotation? All of these factors can tell us important things about the system and about the relationship of the two bodies. Some of these factors change over time; others do not. Some could be applied to multi-stellar systems; others cannot. The angular momentum measure is an interesting one, given the relationship between angular momentum and planetary disc formation. It's also a totally conserved value so it won't change as readily over time. The rotational angular momentum of Earth is 7.2e33 kg*m2/s while the orbital angular momentum of the Moon is 2.9e34 kg*m2/s. So the angular momentum of the Moon dominates the total angular momentum of the Earth-Moon system, making it less a moon and more a co-satellite. Likewise, the angular momentum of Pluto is 8.6e29 kg*m2/s, while the angular momentum of Charon is 5.4e30 kg*m2/s, which creates the same conclusion. In contrast, the combined orbital angular momentum of all the moons of Jupiter is less than 1% of the rotational angular momentum of Jupiter, so Jupiter dominates the angular momentum of its system. Then again, Jupiter's orbital angular momentum is more than 60% of the total angular momentum of the entire solar system. So that measure is problematic if we want to apply it to stellar systems.

That’s a flex.

To simplify even further, perhaps past the point of rigorous accuracy.... Antimatter doesn’t produce thrust; it can only produce heat. If you want thrust then you’ll have to combine that heat with some kind of propellant so that the propellant can explode out the back of your engine and produce thrust.

5, thrust puck failure at restart, seems like the most likely failure mode to me because there’s really no way to test for it without a full-up launch. Transverse loading during gimbal is going to put some unique stresses on the thrust puck and the entire airframe.

Right. Those will be the fixed, non-throttleable engines on the periphery of the Super Heavy once it is upgraded. Early Super Heavy boosters will likely use the regular Raptors.

This topic did inspire a TikTok.... https://vm.tiktok.com/ZMJaSJvbg/

Oh, I didn't realize we had seen the stump. If that is the case then yeah, it has to flow all the way to the engines. Which I suppose means a different valve design for LOX than for CH4. I wonder how autogenous pressurization is handled for the LOX header tank. Gaseous CH4 can be piped into the CH4 main tank which in turn pressurizes the header; you only need to fire the engines for a split second before you have ullage and can open the valves to maintain head pressure. But you'd need a separate GOX press line running all the way up to the header tank. I wonder if they have a separate CH4 gas line for the CH4 header tank too, just as a failsafe. A loss seems more likely than not. The big question is where the loss will take place. I think there are a few particularly significant failure modes: Structural/valve/engine failure from a longer-duration burn and associated heat, stress, and vibration (the engines can probably take it, since they've all gone through full-duration test fires, but I don't know about the thrust puck) Aerodynamic control loss due to structural or systems failure for the flap drive mechanism Aerodynamic control loss due to computational challenges Restart failure Thrust puck failure at restart Hard/off-center landing with leg failure and tipover I think the most likely failure modes, in descending order, are 5, 2, and 6. I would give them a 35% chance of success.

It depends on what the matter and antimatter comprise. Matter and antimatter aren't intrinsically attracted to each other by their nature; it all depends on charge. A positron is attracted to an electron, for example, and an antiproton is attracted to a proton. But that attraction is the result of electromagnetic charge, not anything to do with the nature of antimatter. An antineutron is not particularly attracted to a neutron; they both have no net electric charge. Antihydrogen, which is the atom formed when a positron binds to an antiproton, is electrically neutral and is not electrically attracted to anything. However, it does have spin, which can be used to confine it in carefully-shaped magnetic fields. We were first able to synthesize a significant amount of antihydrogen in 2010, where 38 antihydrogen atoms were trapped in a magnetic trap and persisted for a sixth of a second before annihilating. Within a few months, we were able to trap antihydrogen atoms for several minutes at a time. But that's about as far as we've gotten. We can produce individual antideuterium and antihelium atoms but it's really challenging and long-term confinement remains an open problem.

I wonder A LOT about how the plumbing is set up there. Presumably there is a valve setup which can feed either from the header tanks or from the mains. If they did a partial detanking to empty the mains, however, the T/W for three Raptors firing at once would be pretty extreme...might stress the hold-down clamps. If this static fire went off without a hitch, then I am guessing we see two more: the first to simulate the header-fed burn by itself, and then the second where all tanks are filled and they feed first from the mains for a split second, then cut the engines, then switch valves and fire from the headers a few minutes later without any recycle. I would have thought it possible that for the CH4 tanks, the main simply feeds into the header, which in turn feeds the engines, and so they close a flow valve between the main and the header immediately after MECO in order to trap the header tank in the "full" state. This avoids needing an extra flow path near the engines. However they'd definitely need a separate flow path and valve near the engines for the header tank, obviously. In that C-bass animation on the last page, it showed the feed line from the LOX header tank coming down through the main LOX tank, internal to the vehicle. Do we know if that's the design, or will the LOX downcomer from the header tank flow along an external raceway for SN8? Looking more closely at how the angle on the squid-fins/canards works, I wonder if I was wrong to assume that the lateral edge of the new fins doesn't run parallel to the side of Starship. This view, at least, suggests that the LOX downcomer is at least internal to the fairing. Whether it exits the fairing and runs along the outside of the lower section or whether it remains internal...well that remains to be seen.

I wonder if they will do a partial detank and then do a header-tank static fire as well, feeding only from the header tanks to simulate the restart and landing.

Any time you think "this is how explosive antimatter is" without doing the math first, multiply it by a thousand. And then you'll still be too conservative. A classic M-80 holds three grams of pyrotechnic flash powder which I believe has a TNT equivalence of 0.45 or thereabouts. Do the math on TNT tonne equivalence and that gives you 5,648 joules for an M-80 firework. Using e=mc2 (and keeping in mind that you only need half as much antimatter because it will react with anything), we learn that you need 3.1e-11 grams of antimatter to equal the blast of an M-80. For reference, that's slightly more than the mass of a single human sperm cell. (small squibble -- it's not "reaction mass" in this context, but "blast yield") What about something the size of a small pill? Well, I have a bottle of extra strength 500 mg Tylenol in my desk. Each pill is half a gram. A half-gram of antimatter would annihilate a half-gram of normal matter, so that's one gram of mass being converted into pure energy. Dial up e=mc2 again, and we get the answer: 9e13 Joules, or 21 kilotons, the same as the blast yield of the Fat Man nuclear warhead that destroyed Nagasaki. The Fat Man was a 6.4-kg critical mass inside of a 4.7-tonne implosion warhead and it converted almost exactly one gram of its critical mass into pure energy -- an efficiency of 0.016%. Antimatter, in contrast, is 100% efficient. In the Orion Drive, the actual nuclear blast does practically nothing to accelerate the ship. The propulsion comes from the propellant -- a dense tungsten tamper which is placed in the casing. The tungsten propellant is about 30-40% of the total mass of the warhead. When the nuke is triggered, the x-rays vaporize the tungsten and turn it into a jet of plasma that is blasted back toward the ship, delivering the desired impulse. The only reason you need the pusher plate, though, is because there is a minimum size that a nuke can be. If you have antimatter containment, you can simply feed microgram-quantities of antimatter into a stream of propellant continuously. The Raptor engine, for example, has an energy output of 3.52 GW (yes, that's three times the infamous Back To The Future value). To match that with an antimatter engine, you'd only need a flow of 19.6 micrograms of antimatter per second...that's just under five gnats per minute. Of course, to match the thrust of the Raptor engine, you're also going to need to feed about 700 kilograms of cryogenic propellant into the chamber per second. The more liquid propellant you use, the more thrust you get, but the lower your efficiency is.

@K^2, your thoughts?

Well, it wouldn’t be an explosion in the conventional sense. The initial fireball from a nuclear weapon, for example, is primarily surrounding air that was boiled into plasma by the extreme x-ray flux from the nuclear chain reaction. In space, however, there’s no air, so there’s nothing to “explode” in that familiar sense. An antimatter annihilation reaction would produce an extremely bright burst of x-rays and gamma rays, enough to fry the ship, but the only momentum imparted to the ship would be from the photons themselves. As @K^2 noted, we are talking about gigawatt levels of heat for just a single Newton of thrust. If you want more thrust then you’ll need to surround the annihilation reaction with water or liquid methane or liquid hydrogen or something to both absorb the heat and have something with enough mass to actually explode outward and produce meaningful thrust.

The binding energy between the quarks is the relativistic kinetic energy of the gluons which mediate the strong force.

Nope -- the vast majority of mass is relativistic mass from the energetic motion of the particles it contains. The nucleus of an atom has more mass than the constituent neutrons and protons it contains. A proton has a mass of 1.6726e-24 grams and it is made of two up quarks and one down quark. The rest mass of an up quark is 4.02e-27 grams and the rest mass of a down quark is 8.59e-27 grams. Add that up and you find that the constituent quarks in a single photon mass just 1.663e-26 grams, just under one percent of the mass of the proton. 99% of the mass of the proton comes from the kinetic energy of those quarks and the gluons that hold them together. When you thrown in the mass of the neutron and the associated binding energy of the gluons which hold the nucleus together, the total "rest mass" in an atom is less than 0.2% of the total mass. You are quite literally made of energy.