Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to create a machine that moves on two points (Wheels or legs). Because of the extremely difficult nature of perfectly balancing the parts, I am wondering is there any way to create a mechanical mechanism to balance it. I know that it can be done with many different electric circuits, but I am wondering is it even theoretically possible to create such a system? And if it's possible, does anyone know of any systems that do that?

P.S. Could someone retag this appropriately? I'm new to this SE, and I'm not sure how to tag it.

share|cite|improve this question
This is probably covered by the segway patent, but this was the major hurdle. – Ron Maimon May 15 '12 at 22:29
Yes, but I don't want to use any type of computer. – J. Walker May 15 '12 at 22:48
You could do it with a mechanical contraption, which has a bunch of gyroscopes connected to mechanical motorized weight-movers, but I don't see why you don't want a computer--- the result will be equivalent. – Ron Maimon May 16 '12 at 1:23
Look here as well:… – IljaBek May 16 '12 at 18:19
Are gyroscopes allowed? – user1631 May 16 '12 at 20:47
up vote 3 down vote accepted

A passive machine on a surface (of gravitationaly attracting sphere = ideal Earth) cannot be in a stable or meta stable state with only two supports.

A di-pod cannot span an area, in a way that it's center of gravity stays within if tilted.

unlike the doll below, which has always a projected area around it's point of support under it's center of gravity - a two-legged system spans only a line, which is easy to cross.

stable doll

Self-balancing system should revert to it's state after a small (and to be defined, in the example below it would be $x_a - x_b$) dicsplacement. In terms of a potential this is the case for stable and meta-stable states. Potential of a meta-stable (x_a) and a stable (x_c) state In the picture the state of your system is related to potential (y-axis), and some abstract position is at the x-axis: e.g. Your system is staying "upright" at $x_a$, it will go back to $x_a$ if pushed with less effort than $\Delta U$, and will be "falling down" untill it reaches next stable position at $x_c$ "lying down"

Addendum: A bag hook could be considered stable virtually having only one support, but it's not an option, for a) moving machine b) on a surface

Upon further reflection - the hook is actually hanging on a table, which has at least three supports itself.

bag hook

share|cite|improve this answer

It's certainly possible. The legged version is called "passive dynamic walking", and you can see a good example of it in action here. The trick is that, unless you go for the cheap trick of having very large feet, you need an energy source to maintain the balance. In the case of the walker in the video the energy comes from gravitational potential, as it walks down an incline, although versions powered by electric motors have also been made. You may notice that the walker's gait is rather more natural-looking than, for example, that of Sony's AIBO robot, which uses a more computational approach.

For a two-wheeled rather than two-legged robot, at least one way to do it is to put one wheel in front of the other and let the front one rotate on a set of forks that are angled forward - you then have, essentially, a bicycle, and a well balanced bicycle will roll quite stably without a rider as long as it has enough forward momentum. If you want the robot to be able to stand still then I imagine it gets a bit harder - I can't think of a simple way of the top of my head, although that doesn't mean there isn't one.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.