3 ms·

Easily. The yardstick is unstable in inclination if it isn't vertical. The rocket is unstable in horizontal position if it isn't vertical. But it doesn't start

by jakub_h 11y ago

Easily. The yardstick is unstable in inclination if it isn't vertical. The rocket is unstable in horizontal position if it isn't vertical. But it doesn't start to rotate as long as the thrust vector goes through the CG. The thrust vector has stable orientation relative to the rocket (if you don't gimbal), whereas the force vector doesn't have stable orientation relative to the yardstick.

gozur88 11y ago
I don't see how that can be true. The rocket will still start to rotate even if the thrust vector goes through the center of gravity, unless that vector is exactly vertical (a point of instability). As I understand it that's the key insight to the pendulum fallacy.
- jakub_h 11y ago
  No, it won't, because neither the thrust vector nor the gravity causes a moment in this situation. The key insight to the pendulum fallacy is actually that this situation doesn't create orientation-dependent momentum. That would make it a pendulum, but it isn't like that, hence the "fallacy" part.
  - gozur88 11y ago
    I see what you're saying. That would imply in the absence of an atmosphere a gravity turn isn't really possible. You'd pick up horizontal velocity, relatively slowly, but assuming you didn't exert some kind of control your rocket would always stay at the attitude it had at the end of the pitchover.
    - jakub_h 11y ago
      Yes, what is called "gravity turn" turns the velocity vector, but if the rocket didn't actively reorient itself (so as to keep in line with the velocity vector), gravity itself wouldn't rotate the rocket (change its attitude), because almost-homogeneous gravity field won't impart (meaningful) momentum. (Satellites can actually use it that way, but the time constants are substantial and not really relevant for an ascending rocket on a timescale of minutes.) However, this is independent of the presence of the atmosphere. In real life, both scenarios need active control of attitude.
zardo 11y ago
Note: This is completely wrong. The yard stick has a normal force vector from the hand passing through its center of mass. It doesn't start to rotate unless an outside force is applied. Of course in the real world it requires active control because the most minute misalignment causes a feedback that moves it further out of alignment. The rocket is exactly the same, if the thrust vector is not directly between the center of mass of the rocket, and the Earth, it starts to tip over. If it were a stable system, misalignment would push the system back into alignment.
- DenisM 11y ago
  The engines are attached to the rocket. Unless the engines are deliberately twisted, the thrust vector is going through the center of mass at all times, regardless of tilt. So if a rocket tilts in flight, it will not tilt further and it will not fall back. The yard stick, if tilted, will tilt further still and eventually fall, unless caught.
- jakub_h 11y ago
  It's not the case with a rocket, unless you're flying forward with a liquid propellant rocket and the fuel tanks are almost empty so that the center of mass is significantly behind the center of drag (and the atmosphere is appreciable). That is an inverted pendulum situation. Outside of atmosphere and/or at low velocities, there's no instability of this kind, and "flying backwards" (=landing) makes the rocket actually auto-stable (like with a dart), even if the time constants may not necessarily be meaningful for useful control here.
dandelany 11y ago
> 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.

by jakub_h 11y ago

gozur88 11y ago

I don't see how that can be true. The rocket will still start to rotate even if the thrust vector goes through the center of gravity, unless that vector is exactly vertical (a point of instability). As I understand it that's the key insight to the pendulum fallacy.

jakub_h 11y ago
No, it won't, because neither the thrust vector nor the gravity causes a moment in this situation. The key insight to the pendulum fallacy is actually that this situation doesn't create orientation-dependent momentum. That would make it a pendulum, but it isn't like that, hence the "fallacy" part.
- gozur88 11y ago
  I see what you're saying. That would imply in the absence of an atmosphere a gravity turn isn't really possible. You'd pick up horizontal velocity, relatively slowly, but assuming you didn't exert some kind of control your rocket would always stay at the attitude it had at the end of the pitchover.
  - jakub_h 11y ago
    Yes, what is called "gravity turn" turns the velocity vector, but if the rocket didn't actively reorient itself (so as to keep in line with the velocity vector), gravity itself wouldn't rotate the rocket (change its attitude), because almost-homogeneous gravity field won't impart (meaningful) momentum. (Satellites can actually use it that way, but the time constants are substantial and not really relevant for an ascending rocket on a timescale of minutes.) However, this is independent of the presence of the atmosphere. In real life, both scenarios need active control of attitude.

zardo 11y ago

Note: This is completely wrong. The yard stick has a normal force vector from the hand passing through its center of mass. It doesn't start to rotate unless an outside force is applied. Of course in the real world it requires active control because the most minute misalignment causes a feedback that moves it further out of alignment. The rocket is exactly the same, if the thrust vector is not directly between the center of mass of the rocket, and the Earth, it starts to tip over. If it were a stable system, misalignment would push the system back into alignment.

DenisM 11y ago
The engines are attached to the rocket. Unless the engines are deliberately twisted, the thrust vector is going through the center of mass at all times, regardless of tilt. So if a rocket tilts in flight, it will not tilt further and it will not fall back. The yard stick, if tilted, will tilt further still and eventually fall, unless caught.
jakub_h 11y ago
It's not the case with a rocket, unless you're flying forward with a liquid propellant rocket and the fuel tanks are almost empty so that the center of mass is significantly behind the center of drag (and the atmosphere is appreciable). That is an inverted pendulum situation. Outside of atmosphere and/or at low velocities, there's no instability of this kind, and "flying backwards" (=landing) makes the rocket actually auto-stable (like with a dart), even if the time constants may not necessarily be meaningful for useful control here.

dandelany 11y ago

> 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.