I do not understand how tightrope walkers return themselves to equilibrium.

I am not concerned with the direction along the rope or wire where their base can be large, and they are able to move their foot forwards or backwards.

I am thinking about the system through the wire. For simplicity, can we assume the wire is perfectly taut so the contact point is fixed? The walker is a mass (or connected articulated masses) balanced above a point (the wire) to which they are connected by a low friction pivot.

The walker is at equilibrium when their centre of mass is directly above the tightrope.

I've read the top few Google hits. They mention a rod or other device may be carried, to provide a larger moment of inertia that will damp the onset of rotations. They also mention that it lowers the centre of gravity, which will reduce the turning moment that begins rotation. This makes sense and I can see that would help. What they don't mention is how a walker returns themselves to equilibrium once they have moved away from it.


This is probably very confused, but whatever a walker does, they as a whole (system) must conserve centre of mass (and momentum), right? Because of this I don't understand how anything that they can do (moving a leg out, say) can actually make a difference! Any change they make just leaves them with their mass differently spread about but the centre of mass in the same place, and still not at equilibrium.

Standing on one leg

I think that a similar system is at play when I balance on one leg. I've noticed that if I stand on my right leg and force myself out of equilibrium by falling to the right, then I will lift up my left leg to somehow balance myself. Why doesn't conservation ensure that lifting my leg like this simply topples me further to the right?

"Practical" use

Given that a tightrope walker can balance, it must surely be possible to build a (compound) inverted pendulum that can balance without moving its ground contact point in the way that these control systems typically do?

  • 1
    $\begingroup$ You may be interested to know this question has been asked in many other forms, from ice skaters to astronauts to falling cats. $\endgroup$
    – user10851
    Jul 13, 2015 at 9:46
  • $\begingroup$ @ChrisWhite Thank you, that does help, particularly this cat/astronaut-righting answer. Another answer mentions a reaction wheel which is not something a person has available, but does demonstrate how it's possible for a system to adjust its attitude. $\endgroup$
    – Benjohn
    Jul 13, 2015 at 10:02
  • $\begingroup$ So, when I'm trying to balance on one leg (right), and I "throw" my other leg up (left), that acceleration gives me a turning moment that's sufficient to pull my whole body centre of mass back over my right foot, at which point I can decelerate the leg. I'm still a bit confused why this doesn't unbalance me again, though. $\endgroup$
    – Benjohn
    Jul 13, 2015 at 10:08
  • $\begingroup$ @Ben I suppose the tight-rope walker is using a sort of negative feedback loop to stabilise themselves. falling to the right -> move to the left. now falling to the left -> move to the right etc Probably they are constantly unbalancing themselves, corecting etc as above $\endgroup$
    – innisfree
    Jul 13, 2015 at 11:17
  • $\begingroup$ it depends how good the tight-rope walker is at making the corrections, but a negative feedback loop could make him return to equilibrium, even if he was unsettled by e.g. a gust of wind or a mistake. $\endgroup$
    – innisfree
    Jul 13, 2015 at 11:21

1 Answer 1


To simplify we can treat the wire as perfectly taut.

This allows the rope to exert a force on the tight rope walker (TRW)

Also for simplicity imagine the TRW as a solid body with a solid body pole that can be rotated about an axis parallel to the rope at a fixed location relative to the TRW.

Now, if the TRW is tipping clockwise their center of gravity will be to the right of the rope and moment created by gravity will add clockwise momentum. Similarly, if the TRW is tipping counter-clockwise gravity will add anti-clockwise momentum.

So the angular position of the TRW determines if clockwise rotational momentum is added or subtracted from the TRW+pole system. This angular position is determined by integrating rotational momentum of the TRW (divided by the moment of inertia) The amount of rotational momentum the TRW has can be modified by transferring momentum to and from the pole. While this doesn't directly affect the TRW+pole system's rotational momentum, it allows the TRW to control their angle, which in turn will affect the momentum of the system.

So if the TRW has too much clockwise momentum, and wants to move back to equilibrium, they can transfer a lot of clockwise momentum to the pole to temporarily tip themselves in the counterclockwise direction. If they tip far enough, then gravity will be eating away at the extra clockwise momentum, and the TRW can then take back some of the momentum given to the pole to bring themselves back to a neutral angle.

  • $\begingroup$ Thanks! I think this makes a lot of sense – but I are you are sometimes saying inertia in place of momentum? The fundamental element seems to be that it's not about moving mass around to adjust the centre of gravity. Instead, it's about temporary transfers of angular momentum from one part of the system to another? $\endgroup$
    – Benjohn
    Jul 13, 2015 at 12:50
  • 1
    $\begingroup$ @Benjohn Wow I really didn't get enough sleep last night. Thanks, for the heads up. $\endgroup$
    – Rick
    Jul 13, 2015 at 13:01
  • $\begingroup$ @Benjohn The angular momentum transfers allow temporary relocation of the center of mass, and the center of mass location allows permanent increase or decrease of angular momentum. So it's about both. $\endgroup$
    – Rick
    Jul 13, 2015 at 13:05
  • $\begingroup$ @Benjohn - At the same time that clockwise momentum is transferred to the pole, the walker is also exerting a right force onto the wire, coexistent with the wire exerting a left force onto the walker. This left linear force from the wire corresponds to moving the center of mass of walker and pole back to the left. $\endgroup$
    – rcgldr
    Jul 11, 2016 at 0:50

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