It could be that I've got the math for the handwaving wrong. Let me see... (Update from the future: most of this post is stream-of-consciousness, so I change my mind a bunch)

If, say, an R-R-R arrangement and an R-S-Z arrangement are viewed in terms of symmetries, then R-S-Z has half as many, because it's not mirror-symmetric (a common theme here). However, things get really hairy from this point of view. It may be that I have to consider the initially available symmetries. In addition, this idea, while promising, doesn't address the issue of damping that I brought up... Hopefully, the two simply operate independently.

I think... weighting a state rotation to twice a state reflection makes sense in various intuitive fashions, and provides the behavior I want. So far, this has all been a highly simplified exercise in working out tartonic potentials between zero and infinity. However, I fear that the behaviors I predict could be somewhat... confusing.

So, to summarize: to calculate the tartonic potential of a given charge, from infinity to close to a group of charges... Count how many unlike charges there are, and subtract the number of like charges. Now, from that, subtract four times the product of the two numbers of unlike charges. The result is the amount of energy 'released' by taking the given charge from infinity to zero.

Now, back to that four, because it's the most arbitrary aspect of all that. I want a coefficient that works based on distance, and will keep three different charges on the points of an equilateral triangle precisely balanced. This means... two attractive forces, sixty degrees apart... Each thirty off the horizontal, equates to sqrt(3) attraction, which must be balanced by sqrt(3) repulsion from the midpoint, which is closer... In other words, four thirds as strong, per effective charge. As a result, the effective charge is just three-quarters of root three. Now, let me see... if I make it isosceles, with one vertex twice as charged... Huh. The implication I get is, that's only isosceles if I say it is. Buh? Or maybe not even then. It looks like doing things like that forces any distribution of unlike charges on the vertices of an equilateral triangle to balance. I mean, the charge groups themselves won't be stable, but the overall shape will be, if you ignore them...

(Sanity check: since most cirquons contain two charges, the addition of a charge in one place would repel the other cirqon that shares that charge. I... think that could work.) *scribble scribble scribble* Completely unworkable in that form.

Hold on, let me check if the potentials work at all, here...

Cirquons resist coming together, but that's to be expected... Now, what kind of behavior do I get if... Balance charges in a line, with altered repulsions confined to particles... Hm... let's give the equilateral triangle another shot... No... The real trick is getting a triangle with a really short leg to work nicely. Such an arrangement should cause minimal repulsion along the short side... Which implies that equilateral triangles are only stable at one length scale. I don't think "all of the necessary components and catalysts" are there for my reasoning, but this should have been obvious to me before.

This should mean that the 'ghost charges' in my way of viewing this manifest in proportion to the distribution of all appropriate charges, and that the ghost charge effect has to decrease faster than inverse square. This is because I want the effect to be strong when they're close together, and that means there needs to be very little far-away contribution... But I also want it to be finite at zero. I don't know what kind of distribution could fit those specifications. I mean, I know normal distribution would work for weighting the effect, but I'd like to look into the alternatives.

In any case, what I'd really like is for the ghosting to precisely cancel the attraction at some distance, be twice it at another, and approach zero at infinity.

(Hm... another thought about the symmetries idea is that it should be in terms of perimeter. So, twice as far apart would ghost half as much, which... doesn't work)

Anyway, I'm going to go walk for a bit, and think about this. Sorry for being all rambly.