4 ms·

> The thrust vector has stable orientation relative to the rocket (if you don't gimbal) This is what makes your argument moot - because they do gimbal, and the

> The thrust vector has stable orientation relative to the rocket (if you don't gimbal) This is what makes your argument moot - because they do gimbal, and the gimbaling is what creates the moment/torque. In a way that is very nearly analogous to twisting your hand around slightly to balance a broomstick. > The yardstick is unstable in inclination if it isn't vertical. The rocket is unstable in horizontal position if it isn't vertical. The rocket is unstable in position and inclination (attitude). So you're right that there's more to it, but it's still the inverted pendulum problem for attitude, plus the rocket formulas for translation and added constraints of landing location. Are you saying Bezos is lying in this post when he explicitly says "When you do a vertical landing, you’re solving the classic inverted pendulum problem"? He would know, and I don't see a reason for him to lie about it.

NamTaf 11y ago
Fair call on the gimbaling at the end, in that instance it does impart a moment around the CoM and will invoke a torque, though that's really only at the end once much of the vertical velocity is cancelled. As a result it does represent the inverted pendulum right before landing. Interestingly, because the CoM is so low due to the low volume of fuel by the time of landing, it's likely not as big a torque as you'd first expect (all the mass is down the base where the engines are). In that case, rocket height isn't going to have as a big an impact because it's not like it's equally dense up the entire rocket structure. I'd argue he's likely overstating the effect of rocket size on how difficult it is to achieve.
jakub_h 11y ago
> This is what makes your argument moot - because they do gimbal, and the gimbaling is what creates the moment/torque. In a way that is very nearly analogous to twisting your hand around slightly to balance a broomstick. The point of the inverted pendulum problem is that the torque on a long, thin, ground-supported object (or even just any ground-supported object with a narrow base or a hinge on its bottom) is a function of its orientation. That gives you a second order differential equation for tilt. This is NOT the case with the free-flying rocket because the engine is rotating along with the rocket. You could simulate the inverted pendulum actively, by rotating the engine so that it has a fixed orientation relative to the ground (up to the gimbaling limit, which is usually <10 degrees), but you actually want to do the opposite with the landing rocket (with significant damping to boot). Basically, you want the active control system to simulate a normal pendulum, just turned upside down (but with the damping so that it doesn't actually oscillate). > The rocket is unstable in position and inclination (attitude). Not the latter unless you make it behave in that way. You could also check on the Apollo LM landing guidance, maybe it will become more clear if you study that. > So you're right that there's more to it, but it's still the inverted pendulum problem for attitude No, it's not. That's the point. > Are you saying Bezos is lying in this post when he explicitly says "When you do a vertical landing, you’re solving the classic inverted pendulum problem"? He would know, and I don't see a reason for him to lie about it. No, I'm saying that this phrasing makes no sense. Remember that Goddard made a similar mistake when reasoning about rockets and pendulums, so arguments to authority are demonstrably to be avoided here. A significant clarification on his part could make it more comprehensible what he actually meant, but making any kind of analogy to a ground-supported object is clearly bad when talking about a free-flying body.
- dandelany 11y ago
  This is an excellent response - thanks.

NamTaf 11y ago

Fair call on the gimbaling at the end, in that instance it does impart a moment around the CoM and will invoke a torque, though that's really only at the end once much of the vertical velocity is cancelled. As a result it does represent the inverted pendulum right before landing. Interestingly, because the CoM is so low due to the low volume of fuel by the time of landing, it's likely not as big a torque as you'd first expect (all the mass is down the base where the engines are). In that case, rocket height isn't going to have as a big an impact because it's not like it's equally dense up the entire rocket structure. I'd argue he's likely overstating the effect of rocket size on how difficult it is to achieve.

jakub_h 11y ago

> This is what makes your argument moot - because they do gimbal, and the gimbaling is what creates the moment/torque. In a way that is very nearly analogous to twisting your hand around slightly to balance a broomstick. The point of the inverted pendulum problem is that the torque on a long, thin, ground-supported object (or even just any ground-supported object with a narrow base or a hinge on its bottom) is a function of its orientation. That gives you a second order differential equation for tilt. This is NOT the case with the free-flying rocket because the engine is rotating along with the rocket. You could simulate the inverted pendulum actively, by rotating the engine so that it has a fixed orientation relative to the ground (up to the gimbaling limit, which is usually <10 degrees), but you actually want to do the opposite with the landing rocket (with significant damping to boot). Basically, you want the active control system to simulate a normal pendulum, just turned upside down (but with the damping so that it doesn't actually oscillate). > The rocket is unstable in position and inclination (attitude). Not the latter unless you make it behave in that way. You could also check on the Apollo LM landing guidance, maybe it will become more clear if you study that. > So you're right that there's more to it, but it's still the inverted pendulum problem for attitude No, it's not. That's the point. > Are you saying Bezos is lying in this post when he explicitly says "When you do a vertical landing, you’re solving the classic inverted pendulum problem"? He would know, and I don't see a reason for him to lie about it. No, I'm saying that this phrasing makes no sense. Remember that Goddard made a similar mistake when reasoning about rockets and pendulums, so arguments to authority are demonstrably to be avoided here. A significant clarification on his part could make it more comprehensible what he actually meant, but making any kind of analogy to a ground-supported object is clearly bad when talking about a free-flying body.

dandelany 11y ago
This is an excellent response - thanks.