sevenperforce

Yikes, that drippy bit looked terrifying. This is why I'm still squicked about human landings on Starship. The only difference between these tests and prior static fires is that they're feeding the Raptors from the LOX header tank. I wonder if the long flow line from the LOX header tank is resulting in an out-of-envelope temperature condition for the LOX. The LOX flow is used to regeneratively cool the LOX preburner, so if its temperature is a few degrees warmer than the LOX that Raptor is used to accepting, it could have insufficient cooling and thus melt parts of the preburner or turbopump. That would also seem to align with what Elon was saying about the pressure in the header tank rising uncontrollably. There's a GOX line that runs from the heat exchanger to the header tank for autogenous press. If it was coming in too hot and the header tank was already too warm, it could cause boiloff in the tank. Only tangentially related...Elon has COVID and is whining about it.

I knew it!

It’s also a wet dress rehearsal for the second stage.

The more charge you build up, the more energy it will take to fire them back fast enough.

Contingency Decelerative Compressibility.

Motion, electrical current, and magnetism have a very close relationship. If you put motion into a magnet you get electrical current. This is a dynamo. If you put electrical current into a magnet you get motion. This is a motor. If you put electrical current into motion you get magnetism. This is an electromagnet.

It will when the constellation is fully operating.

But you cannot design it to 105%. You are proposing that you can get something over 100% efficiency. You might not recognize it, but you are.

No matter how much money you spend, you cannot overcome the conservation of momentum. F1 = -F2 is not nebulous. Inefficiency means the utilization of a resource is less than 100%. Suppose you have a system with three different components: A, B, and C. Let's suppose that A represents propellant purity, B represents combustion efficiency, and C represents nozzle efficiency. The total efficiency of your system is the product of the efficiency of its components. If E stands for efficiency, then we would represent this by saying E = A x B x C. Let's say that A is currently 99%, B is 95%, and C is 92%. So E = A x B x C = 86.5%. Now let's talk about incremental gains. We can improve B, or we can improve C, or we can even improve A just a little bit. But no matter how much we improve A, B, or C, we can never ever get E to a number greater than 100%. Ever. Even if we make A = 99.99% and B = 99.99% and C = 99.99%, E is still only going to be 99.97%. You cannot get E to be over 100% unless A, B, or C are also over 100%. If you disagree, please do not use words. Show me maths.

4.6 km is not quite bit enough to be gravitationally rounded. I have posited that our classification scheme ought to reflect our need for clear communication and science education.

Batteries do not "have" electrons. Electrons are in everything and around everything. Electric batteries are a way to use chemistry to push the electrons which already exist around in a circle with a certain amount of force, and that force can then be used to do things (like power a lightbulb). An incandescent electric flashlight has a battery and a filament. The battery pushes electrons in a circle, through the filament, and the force of the battery pushing those electrons around in the circle is released as heat in the filament, which causes it to get hot and glow. Batteries are not a bottle of compressed electrons. Solar panels do not "get" electrons from the sun. Rather, solar panels already have electrons in them, and the impact of sunlight pushes those electrons around in a circle, which provides the aforementioned force. You can build an electron gun to fire electrons, and firing those electrons can produce thrust, but you still need to get the electrons from somewhere.

In terms of conservation of momentum, it's perfectly "legal". Any momentum you gain from the bounce has to be offset by negative momentum (in the opposite direction) from the ship you are bouncing the photons off of. Consider when SpaceX separates the first and second stages from the Falcon9. They just use some kind of gas piston or spring or some such thing, right? Well, in this case the photons are just the spring in the system, pushing the two ships apart. It wouldn't even have to be photons. It could be a stream of BBs. But the nice thing about photons is that they always move at the speed of light. However, they wouldn't be able to accelerate you infinitely. I'm sure that some sort of time/space effect would cause them to lose energy compared to you as you sped up. Probably related to "redshift" or something like that. There are proposals to do this with laser-driven light sails. An ordinary laser-driven light sail uses a sail on the probe and laser emitters in LEO. However, you can add a fresnel lensed mirror to the power station in LEO, allowing the photons to bounce back and forth many times. Conservation of momentum is preserved not only by the thrust force on the LEO mirror, but the fact that the light is bouncing off a moving object. It is redshifted by the bounce and thus loses momentum. The faster the probe is moving away, the more dramatic the redshift. You can always improve efficiency, but those gains are incremental. You cannot defeat conservation of momentum. That actually wouldn't work. "Using" the gas means pulling energy out of it; that means less energy to push the bullet with. This, this, a thousand times this. Work is energy. If your gas does work on one application, it has lost energy for ALL other applications. You cannot get that energy back. [snip]

That gives me an idea. If you built a really, really big pusher plate.....

Also electrons are ions.

I agree — I’ve never liked the scooper approach. I am not sure what would be better, but it seems like it has problems. The issue with a standard payload bay door approach, on the other hand, is that you are deploying at 90 degrees from the load axis. With the whale mouth you can at least tilt the payload mount to get the right release vector, but with the 90-degree approach you can’t. So that’s not particularly great either. I’m not sure it’s a problem with a good solution.

I think we all came up with ideas like this. I had a system in which gravity and magnets and electric motors worked together to produce perpetual motion...of course I didn’t realize that the energy required to operate the electric motor was equal to the potential energy gained by the magnetic field. A lot of it has to do with a lack of understanding of force, energy, momentum, and mass. You can do a lot of fun things with different forces in different directions but you’re not going to violate the conservation of energy. You can waste a lot of energy but you’re never going to violate the conservation of momentum. Well, the example proposed by @SOXBLOX was someone throwing a rock, not shooting a bullet. But let’s use your bullet example because it might get you closer to where you are trying to go. Over and over you keep talking about trying to capture the gases or some parts of the particles, only somehow without preventing the thrust. This doesn’t work, of course, because the exhaust gas IS the thrust, in any meaningful sense. However, with the bullet example, you have two different components, right? You have the bullet itself and you also have the escaping gas, and they both contribute to recoil. Is there something you can do there? Let’s find out. A 5.56x45 mm NATO round fired from an AR-15 uses two grams of powder and fires a four-gram bullet. I have a feeling we will get tripped up if we try to talk about force, so instead let’s just focus on momentum. The bullet, escaping from the barrel at 950 m/s, has a momentum of 0.004 kg * 950 m/s = 3.8 kg*m/s. The recoil momentum imparted to the rifle is -3.8 kg*m/s. But, just as implied, we aren’t done yet. We haven’t factored in the exhaust gases from the propellant. The exhaust gases are also escaping from the barrel at 950 m/s and they mass 2 grams, so they have a momentum of 1.9 kg*m/s, making the total recoil momentum imparted to the rifle -5.7 kg*m/s. But let's say we now want to capture that escaping gas and recycle it somehow, so we put some sort of balloon structure around the muzzle that will allow the bullet to escape but will trap the gases. Let’s say it can trap 100% of the gases. The gases exit the barrel at 950 m/s, but they are immediately slowed down and stopped by the balloon structure. That stop imparts -1.9 kg*m/s to the gases, and their equal and opposite force on the balloon imparts +1.9 kg*m/s on the balloon and the entire rifle. So the gas is trapped, but the total impulse on the rifle is -5.7 kg*m/s + 1.9 kg*m/s = -3.8 kg*m/s, just the same as if the bullet had been fired but the exhaust gases had never escaped at all. So yes, you can capture part of the propellant, but that only makes it as if the propellant had never existed in the first place. ”But what if you added a muzzle brake or a silencer that redirected the exhaust gases out at an angle to make them easier to capture?” And sure, you can do that. Let’s say your muzzle brake directs the gases out to the side at a 45 degree angle. That way, the balloon can expand outward radially instead of expanding backward and half of the forces will cancel out. Good, right? The problem is that directing the gas out at a 45 degree angle also reduces the impulse provided to the rifle by half. So in the end, the recoil imparted to the rifle is still just -3.8 kg*m/s, no matter how the gas is captured. If we had a machine that violated the conservation of energy then we would have a machine that violated the conservation of energy.

No, the formula does not have nuances. F1 = -F2 There is nothing to expand. There are no other variables. That is the entire equation. If an exhaust particle produces a force on your vehicle (that’s thrust), it experiences an equal and opposite force propelling it away. It is now moving away (and at a high rate of speed too). If you want to stop it, you have to exert force on it. And if you exert force on it, it will exert an equal and opposite force on you. That is the way reality works. Arguing otherwise is like arguing that you can generate infinite energy by plugging an extension cord into itself.

I’m going to go out on a limb and predict that they will launch fully extended, not folded flat as they are now.

It is a scientific issue, not an engineering issue. It cannot be done. There are no nuances in the formula to play off of. It is not possible. The formula is the simplest formula imaginable. It has no nuances. This is the formula: F1 = -F2 That's it. That's the formula. There's no room in that formula to play around. A particle in the exhaust of a rocket ceases to produce thrust when it is no longer in contact with the engine bell. Once it leaves the engine bell, it is lost forever. If you attempt to use ANY structure to recover that exhaust particle, then it will produce an opposite thrust on that structure, thus canceling its initial thrust. What you are suggesting cannot be done.

Yep. "Payloads are integrated into the Starship fairing vertically in ISO Class 8 (Class 100,000) cleanrooms. Then the integrated payload stack is transferred to the launch pad and lifted onto the Starship vehicle, while maintaining the same vertical orientation throughout the entire process. Conditioned air is delivered into the fairing during encapsulated ground processing while in the processing facility and on the launch pad." The fixed fin-legs on Superheavy definitely help keep the center of pressure back. More importantly, though, a computer-programmed ascent typically flies prograde all the way through the atmosphere, with virtually no deviation beyond the initial pitchover. If the angle of attack stays at zero relative to the airstream, the forward wings never "bite" into the air and cause torque on the vehicle. Finally, the Raptors boast 15 degrees of gimbal which is a LOT of control authority to maintain that heading.

All the cranes hanging around as it comes together reminds me of the early Iron Man suit up scenes...except instead of machine arms putting it together, it's cranes with people, and instead of Tony Stark getting wrapped in armor, it's a spaceship. I wonder how they are attaching the downcomer...I would have imagined someone would be inside to line up the piping?

So damn beautiful.

If you're near a binary black hole, you can use their gravitational waves to slingshot yourself to ridiculous speeds. A binary black hole is the ultimate gravitational assist machine. Conservation of energy is not violated because you are robbing the system of some minute percentage of its angular momentum. Of course the accretion disc will shred you but that's beside the point.

Only once in a bleu moon. Or maybe the Apollo astronauts tracked mold in on their boots and now the moon is slowly turning bleu.

Any propellant you capture from the engine and put back into the system will be propellant that does not contribute to thrust. If you want thrust, you have to let the propellant go. It is that simple. Still not convinced? Imagine, for a moment, that you have a very large spaceship with a very large, empty cargo bay. You mount a small rocket engine on one side of the cargo bay. No matter how much thrust that rocket engine provides, it will not produce any net thrust on the spaceship because it will be counteracted by the equal and opposite force of the exhaust hitting the other side of the cargo bay. You can open up holes on the outside of the cargo bay to get thrust from escaping gases, but the ONLY thrust you will get will be from the gases that escape. You will get NO thrust from gases that remain trapped. Zilch. Zero. Nada. You cannot lift yourself off the ground by tugging on your own bootstraps.