I think I may have answered my own question.
Correct me if Im wrong but it seems that as long as the center of pressure / center of lift, as measured over the span of each blade, is higher than the fulcrum then the system will be inherently stable. Is this the case with most two bladed helicopters? I suppose it has to be.
If I take, for example, two pieces of string and tie two of the ends together and then tie that knot to the top of a stick with a ball of clay at the bottom then, when suspended by the two free ends of the string pair, the system will always try to distribute the dead load (stick with clay or fuselage) between the two strings equally (in this case the two free ends of the string pair represent the center of pressure as acting on each rotor blade). If the two fixed ends are now tilted then the high side will become more loaded than the low side and will thus try to rectify itself by returning to a state of equal load between the two strings.
If however I replace the string with two solid skewer sticks and fix their free ends on a plane in space lower than the point where the stick with clay attaches then the system will be unstable. As the plane in space is tilted then more of the weight from the stick with clay will be distributed to the low side. The system will still be in balance but the cg would have changed to an aggravating condition of instability rather than a self rectifying condition.
So, if I have an underslug teeter block then I will have the advantage of rotor stability during low rotor rpm. However, a self stabilizing condition will only occur in flight if the triangle of forces is, in effect, upside down forming a cone of which the origin is on the teetering pivot and the base is on the same plane as the center of pressure between the blades (which needs to be higher than the pivot).
Does this sound about right? Critique is welcome.