How can we optimize the conversational experience of a group, and once the group gets too large, divide the group in such a way that the conversational quality of the two groups is maintained?
The quality of a conversation is directly proportional to the number of ideas presented and understood therein.
Ideas increase with both the number and diversity of participants.
Understanding requires dialectical inquiry, which in turn requires bandwidth.
I define bandwidth to be the number of minutes in one hour each person has to communicate her ideas to the others in the group in such a manner that each person has an equal opportunity to speak. For example, a conversation between two people shares one channel of communication and each has thirty minutes of bandwidth. A conversation between eight people shares 28 channels of communication and each person has about one minute of bandwidth.
When the group forms, members are familiarized with the ideas of talk dancing, bandwidth, and with the Occupy Wall Street Hand Signals. The group is responsible for the conversational flow, and those who do not respect the bandwidth of others should expect to be called to order by the “wrap it up” hand signal from others in the group.
My own experience suggests the optimal group size to be about 8 people. Eight people can have a lot of ideas. More than 8 people in a group imply less than a minute of bandwidth for each person. It’s hard to express an interesting idea in less than a minute. By the time you get to twelve in a group there is less than thirty seconds of bandwidth available.
I would suggest that a group divide in two when it reaches about ten people, and certainly no more than twelve. Let the group elect a ballot counter. Then each person write down their own name together with four (if there are 10 in the group) or five (if twelve) others they would like in their group. Then divide the groups so that everyone has at least one person they wanted in their group besides themselves.
If there are persons with less than five votes then there are those with more than five. Pair off the ones with the most votes with those with the least.