NTS,

Zane Moore's approach to evening the playing field in big tree volume competitions may have applicability to the more customary point competitions by showing us the way to reducing the circumference points from fused trunks of coppice forms.

If we have a tree that we definitely believe to be from one root system, but subordinate trunks have developed from the root collar and fused with a main trunk, each up to some height, do we exclude all of subordinate trunks, include those that make contact with the main trunk at 4.5 feet, or roll dice? Remember, here we assume on tree - just a complex form. The following formula discounts subordinate stems and is set up for basic reticle measurement of the diameters of each stem an the point of splitting from the main trunk plus the measurement of the whole structure at 4.5 feet or lower if that is where they all come together. In the formula below, Di represents the ith diameter where i=1 is for the main trunk. si=height where the ith trunk splits from the main trunk, h = total height of main stem to top of tree. s1=h to include the whole contribution of the main stem, n =total number of stems including the main trunk.

Note that this approach follows Zane's lead. It is algebraically equivalent to: computing the cross-sectional areas of each stem at the point of separation and the apportioning the total cross-sectional area at 4.5 feet or lower, if that is where they all are fused, between the n stems. Then the areas of the subordinate stems are discounted by the factors si/h, remembering that s1=h. The areas of the main and discounted subordinate stems are added together and then the equivalent circumference computed from that combined area. The process can be expressed in terms of diameters since my assumption is that they will be measured via the reticle, except possibly for the combined group at 4.5 feet or lower. The composite measurement may be done with a tape in terms of circumference and then converted into equivalent area based on a circle.

When I return to Massachusetts, I'm going to test the formula out on some eastern deciduous trees. We owe Zane a vote of thanks for pointing the way to possibly solving a thorny problem that has heretofore had no satisfactory solution.

Bob