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I've received comments that explain to me that the analogy between rocket attitude control during a hovering maneuver and the act of balancing pencil on the end of a finger is a helpful and good one; solve one problem and you've solved the other, and I've just read elsewhere that it's definitely not, and thinking that way can be referred to as "the pendulum rocket fallacy".

Question: Is it possible to outline to what extent the analogy is helpful and in what ways it is inadequate or breaks down completely? If so, is it possible to do this by including some math, and not just paragraphs of prose? That may be most easily doable by quoting some source (a tutorial, some slides, a talk, a textbook, etc.) rather than trying to roll your own.

It might be somewhat related to the notion that having an engines near the top of a rocket helps stabilize it. Then again, it might not. don't let this distract from the question at hand!

Thanks!

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  • $\begingroup$ Answered here space.stackexchange.com/a/9688/6944 $\endgroup$
    – Organic Marble
    Commented Mar 27, 2021 at 12:21
  • $\begingroup$ @OrganicMarble is my question about a normal rocket hovering also answered there, or only to the "might be somewhat related to the notion that " blurb at the bottom? $\endgroup$
    – uhoh
    Commented Mar 27, 2021 at 12:33
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    $\begingroup$ Could there be some confusion here between the pendulum rocket fallacy and the inverted pendulum control problem? Rocket control is not a pendulum problem of any sort due to gravity not producing any torque on the system, but it has some similarities to the inverted pendulum problem in the effect of moment arms and moments of inertia. $\endgroup$
    – Christopher James Huff
    Commented Mar 27, 2021 at 13:28
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    $\begingroup$ @ChristopherJamesHuff it sounds like you've hit the nail on the head; there is some confusion here! I think that a short answer based on your comment is all that's needed. $\endgroup$
    – uhoh
    Commented Mar 27, 2021 at 13:38
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    $\begingroup$ I guess this is why aviation has very strict rules WRT aircraft reference for "lift, weight, thrust, and drag". Since the gravity vector becomes misaligned with the thrust vector, a fixed thrust vector cannot compensate for its effect whether it is on top or bottom. (Do not forget aerodynamic forces too). $\endgroup$
    – Robert DiGiovanni
    Commented Mar 28, 2021 at 2:18

4 Answers 4

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The pendulum fallacy is the belief that rockets would be passively stable with engines at the top, with the rocket "hanging" from them. The error lies in expecting gravity to pull the body of the rocket down while the engines pull it up. In reality, gravity acts on the body of the rocket and the engines equally, exerting no torque (except for negligible tidal forces), and the engines actually pull it forward, regardless of where that direction is with respect to gravity, or where the engines' notion of forward actually is with respect to the vehicle's center of mass. Any pointing error will cause the rocket to spin just as much with a "tractor" configuration as it does with a "pusher" configuration.

The inverted pendulum problem is control of an upside-down rigid pendulum by moving or applying torque to the base, with gravity exerting a toppling force. For example, balancing a broomstick or pencil on one's hand. Rockets are not really inverted pendulums, the disturbing torque from misaligned thrust is independent of the vehicle's orientation and gravity, but their response to such misaligned thrust or outside disturbances is similar and balancing an inverted pendulum is sometimes used as an analogy to rocket control. This analogy may not be accurate in every detail, but is not an instance of the pendulum fallacy.

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    $\begingroup$ A "gravity turn" doesn't involve a torque beyond the initial maneuver to pitch over, the effect of gravity is on the rocket's trajectory, not its attitude. Gravity applies a toppling force to an inverted pendulum, a restoring force to a conventional pendulum, and no torque at all (ignoring tidal forces) to a rocket. Off-axis thrust applies a torque, but one that is independent of orientation, unlike gravity on a pendulum. And aerodynamics is another issue entirely. $\endgroup$
    – Christopher James Huff
    Commented Mar 27, 2021 at 17:08
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    $\begingroup$ Er...no. No torque is needed to continue the pitch maneuver, which will continue until the engines are used to apply a countering torque to end the maneuver, and that torque in no way depends on gravity, they would have the same effect on the vehicle in empty space. No "gravitational torque" is involved in a gravity turn. $\endgroup$
    – Christopher James Huff
    Commented Mar 27, 2021 at 17:53
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    $\begingroup$ I suggest you read it. You have some major misconceptions. $\endgroup$
    – Christopher James Huff
    Commented Mar 27, 2021 at 17:54
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    $\begingroup$ @user39728 Regarding "A gravity turn most decidedly involves a gravitational torque to continue to pitch the rocket over." That is very incorrect. Gravitational torque is rather small. A vehicle launching from the Moon must exert torques with attitude thrusters or gimbaled thrusters to keep the vehicle on the desired trajectory. A vehicle launching from the Earth does not follow a "gravity turn" while it is in the atmosphere. You have asked multiple questions regarding PEG. PEG wouldn't be needed if gravitational torque was sufficient to turn the vehicle. $\endgroup$
    – David Hammen
    Commented Mar 28, 2021 at 1:19
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    $\begingroup$ @user39728 "...by aligning it with its velocity vector?" No, you've got that backward. Gravity does not rotate the rocket. What it does do is adjust its velocity. So, a properly done gravity turn will put the rocket on its intended flight path, which means the vertical axis of the rocket can be aligned with with its velocity vector. But the actual alignment isn't done by gravity, which, as I said, does not produce any torque on the rocket. The alignment is accomplished through thrust vectoring/fins/maneuvering thrusters/reaction wheels/etc. $\endgroup$
    – HiddenWindshield
    Commented Mar 28, 2021 at 1:43
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In the inverted pendulum problem:

  • gravity exerts a vertical force on the pendulum, at the center of gravity
  • the support of the pendulum (like the finger under the pencil) exerts a vertical force on the pendulum, at the bottom of it

In a rocket:

  • gravity is the same
  • engines exert a force along the long axis of the rocket, where the engine is (which doesn't matter much, as you can see if you draw it)

The main effect of that difference is the runaway that happens in the inverted pendulum:

  • if the pendulum tips, the support still exerts the force vertically, creating a moment of forces, which encourages the pendulum to tip further
  • if the rocket tip, it goes sideways, but there is no feedback loop as there is no moment created by the force the engine exerts on the rocket
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  • $\begingroup$ If you did tilt the rocket so that a line between the centres of mass of rocket and planet did not pass through the thrust base I would expect toppling from gravity but not caused by rocket thrust. Is that right? Two forces, one rocket aligned, one observer aligned, causing a turning moment. $\endgroup$
    – Peter Wone
    Commented Mar 29, 2021 at 4:06
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    $\begingroup$ @PeterWone No; with a rocket, two forces are both aligned at center of mass. To generate torque, you need a force that is not aligned at center of mass. The pendulum's gravitational force is aligned with the center of mass of the pendulum, but the support is not, so you get torque; the rocket's thrust is aligned, the gravity is aligned, so you get zero torque. If you made a rocket where the rockets always magically faced downward, then it acts like an inverted pendulum. Do so with the rockets at the top of the rocket, and the rocket dangles like a pendulum. But that is hard, so... $\endgroup$
    – Yakk
    Commented Mar 29, 2021 at 18:20
  • $\begingroup$ oh, right, the gravity acts equally on all points, it's a not a string attached to the centre of mass $\endgroup$
    – Peter Wone
    Commented Mar 29, 2021 at 21:23
  • $\begingroup$ "The main effect of that difference is the runaway that happens in the inverted pendulum" You don't explain what the difference is, let alone why it would result in a different effect. What you say about each situation apply to both situations. $\endgroup$
    – Acccumulation
    Commented Mar 30, 2021 at 4:21
  • $\begingroup$ @Acccumulation I thought I did, sorry if that's not clear: the difference is the moment created by the support exerting a force which remains vertical and is no longer aligned with the center of mass, as soon as the pencil starts tipping. $\endgroup$
    – njzk2
    Commented Mar 30, 2021 at 20:13
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My controls professor, Dr. Carroll Johnson at UAH, taught the rocket problem by first demonstrating the "try to balance a pencil on it's point" device. He then went on to have us attempt to stabilize a rocket system with various control techniques. The first exercise made a lot of unrealistic assumptions (no wind, perfect balance, etc.) just to make it possible to stabilize with a simple controller. Progressive exercises removed these assumptions and added more conditions. The final result was a controller that was unstable and this was his point: The unstable rocket system can only be stabilized by a controller that is itself unstable. He called this a "homeopathic system" or a "homeopathic instability" (I don't recall which). Made sense at the time, got an A in the course, promptly forgot most everything except this lesson. Here’s a link to his paper on the subject https://doi.org/10.1080/00207178108922912

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    $\begingroup$ That last sentence cause me to chuckle. It made me think of a state-space control class I took in the early 80s. The only thing I remembered after the semester ended was the example of a San Francisco BART train oscillating between stations, never coming to a full stop (apparently this did happen when they were ringing out the control system). That was one of two examples that prof used. The other one started "assume a system in N-space...". I got a B and I was damn proud of it. $\endgroup$
    – Flydog57
    Commented Mar 30, 2021 at 2:13
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If the engine is rigidly connected to the payload, then whether it is connected at the top or the bottom doesn't matter; in either case, it provides a constant torque, and as the payload rotates, the engine will rotate along with it. The angle between the engine-payload displacement and the force remains constant. The rocket will continue to rotate, but not in as catastrophic manner as in the pencil situation.

If the engine is attached non-rigidly to the payload, however, then an engine at the top will exhibit a phenomenon opposite to what we see in a bottom engine case. In both cases, the payload rotates with respect to the engine, but with a bottom engine, the payload rotates away from the direction of the force, causing a positive feedback loop and making the rocket unstable. With a top engine, the payload rotates towards the direction of the force, causing a negative feedback loop and making the rocket stable.

So the most stable configuration would be an engine with a payload connected with a rope, but obviously that has serious engineering challenges. The least stable configuration would be an engine with a payload connected with a hinge.

TL; DR
The payload is rigidly attached to the engine. When you hold a pencil with your finger, the pencil is not rigidly attached to your finger. If it were, it would be quite easy to keep it from falling.

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