# Re: Biped Stability

I'm not sure how you'd do it in a system with multiple pivot points such as a bipedal robot, but in a system where a rigid body rotates about an axis,
it's stability is calculated using potential energy relative to any fixed point (I usually use the pivot itself).
First, find the centre of gravity of the rigid body, work out an equation of it's vertical height in terms of the angle it rotates through, then use E=mgh to form an equation of potential energy in terms of the angle.
Next, differentiate with respect to the angle. When the first derivative is zero, you have a position of equilibrium. At a point of equilibrium, if the second derivative is positive, the equilibrium is said to be stable (that is, it will take external force to shift it from that point, like a pendulum hanging vertically below it's pivot). If the second derivative is negative, the equilibrium is unstable (like a pendulum balanced vertically above it's pivot - the slightest nudge or wobble will be enough to make it fall down to the stable position)
Hopefully this will help a bit, but you'll probably want to look it up in some mechanical engineering textbooks to find out what to do in more complicated situation.
Tom

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Hi Duncan
This is my best guess because I'm still very much a bipedal beginner. When things go wrong with a biped it gets worse exponentially. If you add extra weight at the top you will buy yourself a little extra time but be prepared for a lot of crab-like motion if you don't have enough articulation in the hip or a fast enough response time.
Am I right? No doubt someone is about to tell me :o)
best regards
Robin G Hewitt
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Yes well I'm definitely beginner who has been thrown in the deep end *chuckle* The only mechanics I've studied is 'statics' where things don't move...
I understand that adding more weight to the top of a biped is effectively raising the COG and would make the robot less stable while standing still. Although, wouldn't a disc weight on top make the biped more stable during 'single support' meaning it would not have to lean over as far to keep the ZMP under the stance foot?
-Duncan
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On Tue, 22 Jul 2003 22:39:10 +1000, "Blog the Haggis"

The actual weight itself doesn't contribute to the location of the ZMP, it's more how you accelerate the mass. If you are moving very slowly, the ZMP will co-incide with the COG. It's only when you start moving at a speed where the moments resulting from the acceleration of the disc dominate your dynamic equations that the ZMP will deviate from the COG.
From Vukobratkovics work, the ZMP is the point where all moments cancel. No acceleration = No moment.
Have a look for the literature from Waseda University, in particular the work done on the WL series bipeds and WABIAN humanoid. The work by Yamaguchi and Takanishi in the early 90's is similar to what you are trying to do.
HTH
damo
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Thanks, I'll look it up.
-Duncan
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Thanks Tom, This gives me a good point to start from, time to brush off the textbooks...
-Duncan
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From having worked on a simple biped i can safely say that the height of the torso from the grade has little effect on bipedal stability.But like as already bin said it will certainly buy you more time for the actuators to re-act to pertubations. Theres a simple way to illustrate this by balancing a broomhandle on the tip of your finger, you can pretty well manage that, but cut the handle in half and you will immediately notice the difference in your reaction time. I cant say much about the mathematics of bipedal motion, most of my work with regards to motion control was with CPGs (central pattern generators). There a couple quadrupeds named Tekken and Patrush that use this idea. Im not sure if this can be accomplished using a biped but research with Tekken and Patrush did prove that motion control using a CPG is possible.
Larry.
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The geometry is actually irrelevant !
It's how you move it that matters. Humans can walk with a 90Kg backpack, a stiff leg, bodgie knee, leg brace, a kid hanging off each leg, carrying a 40kg bag of cement, etc. etc, but have difficulty when drunk, or after a head injury. It's all in the software. If the successful motion of your biped is hypersensitive to teh difference between a disk and a torso, you'll have major problems.
Just a thought.
Joe.

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Joe, To append three additional points....flex range, speed, and power. A kid hanging off of you is managable as long as you can contort quickly to shift the COG (adjusted to the acceration) and lift his/their weight. ;o)
Just as a suggestion (which some will scream, "HOW!" or "NOT POSSIBLE!" ;o), but did you think of using a near-time (hate the phrase "real time") dynamic motion calculator that takes into account multiple parts movement and accelerations and a theortetical "earth" (in other words relative to something that is theoretically unmovable). The calculation time must be very small to be accurate plus you will need an occasional "earth" reset (such visual input...like a horizon detect...or more simply a double gyro).
There will never be an absolute position as there will never be an absolute acceleration, so to simply try to calculate how to stand seems easy until it is done in the real world. It must be dynamic and constantly re-adjusted to be "close enough". Think of a SPICE'ish engine doing the simple position/acceleration vector math on a small time scale. It drifts quickly, so must be reset constantly (otherwise, like closing your eyes and walking around).
Just my 1 1/2 cents worth.
Bruce

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Hi Joe
It's not a matter of correct foot placement either or stepping stones would be an impassable barrier.
OTOH, if you give someone their first pair of roller skates or stilts then they fall over.
But only for a while.
best regards
Robin G Hewitt
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Hi Joe,
My original post was regarding how to mathematically prove the relative dynamic stability of a biped with an upper torso mass to one without. Of course both can achieve successful motion but that was not what I asked.
Much thanks to Tom, Robin, Damo & others who answered my query in July :-)
regards, -Duncan