20

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!

uhoh
  • 148,791
  • 53
  • 476
  • 1,473
  • Answered here https://space.stackexchange.com/a/9688/6944 – Organic Marble Mar 27 '21 at 12:21
  • @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? – uhoh Mar 27 '21 at 12:33
  • 6
    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. – Christopher James Huff Mar 27 '21 at 13:28
  • 1
    @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. – uhoh Mar 27 '21 at 13:38
  • 1
    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). – Robert DiGiovanni Mar 28 '21 at 02:18

4 Answers4

37

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.

Christopher James Huff
  • 10,784
  • 1
  • 29
  • 50
  • Thanks, this is exactly what I needed. +1 – uhoh Mar 27 '21 at 16:03
  • OK I don't know about this. Gravity acts equally on an inverted pendulum too. And the torque you see in inverted pendula comes from the force couple between gravity at the center of mass and the normal force at the base. This is no different from a rocket: there is a force couple with gravity at the center of mass and a thrust force at the base. That couple will produce a torque, which is why you get a gravity turn if you just point your thrust vector at your CoM. Which of course means you'd first need to pitch the rocket over for the torque to act on it, but this is true of pendula too. –  Mar 27 '21 at 16:43
  • But it's true that a rocket is not just an inverted pendulum. If the rocket is aerodynamically stable (center of pressure behind center of mass), then drag will resist any tendency for the rocket to flip over. So a rocket is nothing like an inverted pendulum, which will most decidedly fall the moment a random perturbation causes it to pitch over ever so slightly. But the thrust-gravity force couple is analogous to the normal force-gravity force couple of a pendulum. –  Mar 27 '21 at 16:48
  • 2
    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. – Christopher James Huff Mar 27 '21 at 17:08
  • 1
    A gravity turn most decidedly involves a gravitational torque to continue to pitch the rocket over. Between the end of pitchover and MECO, a Falcon 9 pitches down some 50 deg while maintaining zero angle of attack. –  Mar 27 '21 at 17:32
  • And Christopher, gravity doesn't discern between pendulum and rocket. Gravity alone acts at the center of mass and by itself produces no torque on a body. But gravity in a force couple does produce torque. It's that torque from that force couple that causes the inverted pendulum to topple, and it's that torque from that force couple that causes a rocket to continue to pitch down during a gravity turn. –  Mar 27 '21 at 17:34
  • Chris I suggest you read the wiki on gravity turns: https://en.wikipedia.org/wiki/Gravity_turn#Downrange_acceleration. –  Mar 27 '21 at 17:53
  • 6
    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. – Christopher James Huff Mar 27 '21 at 17:53
  • 2
    I suggest you read it. You have some major misconceptions. – Christopher James Huff Mar 27 '21 at 17:54
  • I can't make drawings, but I'll try one more time. After the pitchover, gravity has a component perpendicular to the velocity vector. That perpendicular component will cause acceleration perpendicular to the velocity vector. That perpendicular acceleration will cause the velocity vector to rotate down. As velocity rotates down, the angle of attack becomes nonzero if not compensated by a rotation of the engine nozzles---which causes the thrust vector to no longer pass directly through the center of mass, giving it a small component perpendicular to the length of the rocket. There's your torque. –  Mar 27 '21 at 18:13
  • 1
    But you're right, I'm wrong. You don't need a force couple to produce rotation about the center of mass. Put the pendulum in deep space far away from any massive bodies, where gravity is near zero, and you can still cause a body to rotate merely by applying a single force not passing through its center of mass. In a gravity turn, that force is thrust, not gravity. So while gravity rotates the velocity vector down, thereby forcing the thrust vector to rotate in order to keep zero angle of attack, it's the thrust vector alone that causes the rocket to rotate. –  Mar 27 '21 at 18:18
  • 1
    Actually, drag by itself would ensure that the rocket pitches down during a gravity turn, even if you were to completely shut down the engines, as it would act to keep the angle of attack at zero. Bottom line is gravity rotates the velocity vector down, causing drag to rotate the rocket down or, in a controlled rocket programmed to follow a zero angle-of-attack gravity turn, causing the thrust vector to rotate so as to track the rotating velocity vector. –  Mar 27 '21 at 18:31
  • But no, the rotation from a pitchover would not just continue through a gravity turn. Even if you paused the rotation at the end of the pitchover, so that the rocket is pitched at an angle but has zero rotational momentum, the rocket would continue to rotate due to drag and/or a controlled thrust vector to track the velocity vector as it is rotated down by gravity. –  Mar 27 '21 at 18:34
  • 2
    @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. – David Hammen Mar 28 '21 at 01:19
  • David, you know a lot and I respect that. But PEG is needed to precisely insert the vehicle in the proper orbital plane with the proper orbital parameters (altitude, velocity, flight path angle). A gravity turn by itself would turn the vehicle roughly horizontal relative to the horizon by SECO time. But it would give you zero control over the necessary orbital parameters. So it's not that a gravity turn wouldn't turn the rocket enough, it's that it wouldn't give you the control needed for orbital insertion. –  Mar 28 '21 at 01:30
  • 1
    And isn't the purpose of a gravity turn partly to minimize aerodynamic stresses on the rocket---by aligning it with its velocity vector? That's useful only while in the atmosphere. It's precisely in the atmosphere that a gravity turn is most valuable. Above the atmosphere, it would still be a fuel efficient maneuver, but since you have orbital parameters to hit, PEG is the maneuver to use, unless you don't care exactly what orbital parameters you hit at insertion. –  Mar 28 '21 at 01:36
  • 6
    @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. – HiddenWindshield Mar 28 '21 at 01:43
  • 1
    OK, I see I'm confusing everybody. Gravity produces acceleration with a component perpendicular to the velocity vector. That acceleration component will change the velocity vector so that it points down slightly relative to its original direction. As the velocity vector points down, the angle of attack of the rocket becomes nonzero, causing drag to rotate the body of the rocket until it's aligned with the velocity vector. If outside the atmosphere, the controlled thrust vector would do the rotating so as to keep zero angle of attack (which by definition means tracking the velocity vector). –  Mar 28 '21 at 01:48
  • So: gravity rotates velocity vector causing drag/thrust vector to rotate proportionally in order to maintain zero angle of attack---which by definition is what a gravity turn maneuver does. So no, the gradual pitching of a gravity turn is not done by gravity directly. It's done by drag/thrust vectoring---but it's done only because gravity is continuously rotating the velocity vector down. –  Mar 28 '21 at 01:49
  • 2
    @user39728 You could say that gravity does, indirectly, rotate the rocket that way. But that's not usually how its phrased. There are any number of other things that can affect the velocity of the rocket, all of which would produce the same rotation, so we usually say the "cause" of the rotation you describe is aerodynamic effects. The Pendulum Rocket Fallacy that the OP asked about is about the direct effects of gravity and thrust. Adding aerodynamics to the problem just complicates things. The Fallacy still holds even in finless rockets, or rockets moving too slowly for the fins to work. – HiddenWindshield Mar 28 '21 at 02:05
  • I honestly don't even know how the comments became all about gravity turns, ha ha. But the effect of gravity is dominant during gravity turns, and without it, there would be no rotating velocity vector, and therefore no body rotation from drag (as angle of attack would never deviate from zero), and no need to rotate the thrust vector while in the thick of the atmosphere to minimize aerodynamic stresses (though you probably would still do it to minimize gravity losses). OK, I'll stop dragging this on :D –  Mar 28 '21 at 02:09
  • @user39728 "Put the pendulum in deep space far away from any massive bodies, where gravity is near zero" According to the equivalence principle, the physics are exactly the same as if the rocket were in deep space with no gravity/massive bodies, but surrounded by air that is accelerating upwards for some reason. – Acccumulation Mar 30 '21 at 04:28
  • Accumulation, I have no idea what your point is. If you didn't get it, I was conceding in that comment that gravity by itself cannot in fact cause a body to rotate. Because gravity always acts through the center of mass. And because a free body always rotates about its center of mass. So that any force acting through that center of mass, like gravity, cannot directly cause it to rotate. I don't know what your point is about drag. –  Mar 30 '21 at 14:30
  • So to recap what I've said so far: gravity adds vertical acceleration to rocket causing velocity vector to gradually pitch down causing drag to naturally rotate rocket to keep it aligned with velocity vector (zero angle-of-attack position) and/or the thrust vector to rotate so that it no longer passes through the rocket's center of mass in order to bring the rocket to zero angle of attack (if the rocket happens to be controlled to maintain zero angle of attack). –  Mar 30 '21 at 14:33
  • There is no drag outside the atmosphere and gravity turns are relevant even on airless bodies or even pure orbital flyby maneuvers, and engine vectoring to keep thrust aligned with the trajectory is not a gravitational torque. Gravity simply does not rotate the vehicle in a gravity turn, even indirectly. – Christopher James Huff Mar 30 '21 at 14:38
24

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
njzk2
  • 368
  • 1
  • 9
  • 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. – Peter Wone Mar 29 '21 at 04:06
  • 1
    @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... – Yakk Mar 29 '21 at 18:20
  • oh, right, the gravity acts equally on all points, it's a not a string attached to the centre of mass – Peter Wone Mar 29 '21 at 21:23
  • "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. – Acccumulation Mar 30 '21 at 04:21
  • @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. – njzk2 Mar 30 '21 at 20:13
1

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

Community
  • 1
John Grant
  • 11
  • 1
  • 1
    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. – Flydog57 Mar 30 '21 at 02:13
0

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.

Acccumulation
  • 695
  • 3
  • 7