It's been a while for me. After thinking about the uphill roller, I dug around to see if it had been mulled over. Sadly I could find very little. If anyone knows of more, point me in the right direction.
@SilverTiger,I agree. It will definitely try to settle at some elevation at some off-center location. But this is what (I think) I'm trying to go for. At its starting position, seated at the base of the rims, its CoG is elevated, making it want to roll towards the diverging closed-loop path in front of it. When it does roll, it will do so until its CoG is centered along its axis of rotation. Of course, the point at which it settles will be an off-center elevation along the rims. The question I would love to have answered is that once it settles in some off-center location, will gravity succeed in bringing it back down toward the base, achieving some sort of mechanical driving oscillation, or will it at some point find equilibrium? Which will win out in the end? Its settling point or the imbalance in the rims caused by its need to settle off-center? Or neither one? I'm only hoping that what I have here is two objects constantly fighting each other over equilibrium. Moreover, would friction be the ultimate deciding referee in this case?
Did this vertical uphill roller wheel ever get tested? I am very interested in variations of this concept. So far, I can't find any fundamental reason this wouldn't work, just considering balance. I can see possible problems with z axis forces, but those could be eliminated.
I do not think the COM of anything has to drop in order to cause rotation, it only needs to be allowed to rotate in the direction of torque and maintain that torque during rotation. I know that sounds redundant but it is the key question; What kind of system does not lose its torque during rotation? This system at least doesn't need to trade energy all the time to reset between positive and negative torques.
I beg anyone to share their thoughts on this. This idea is a high candidate project for me.
Respectfully,
Tim[/u]