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
Hello,
>
> If I had a choice between building a biped with just a lower torso or
> building one which had a disc weight simulating the upper torso, how would I
> go about proving which one was most stable?
>
> Intuitively it would seem that the one with the disc weight would be more
> stable but I am searching for a mathematical reason.
>
> Any pointers would be greatly appreciated,
>
> -Duncan (duncan_mcdonald at ieee dot org)
>
>