proposal of a new crown measurement Robert Leverett May 30, 2007 10:20 PDT
 RE: proposal of a new crown measurement Edward Frank May 30, 2007 15:46 PDT
 Back to Ed Robert Leverett May 31, 2007 04:30 PDT
 Ed,    I think that there is room for both crown area and crown volume determinations to be reasonably pursued. The projected crown area (shadow imprint) is something we can do and area projections of that sort have been done for trees like the Angel Oak. But regardless, of what we choose as a standard, I am of the opinion that our largest trees deserve to be measured, and thus documented, in a wide variety of ways and crown volume is certainly one of those ways, although a computationally chellenging one. Fitting the crown of a tree to a standard form is a good start. Bob
 Crown Measures Edward Frank May 31, 2007 14:05 PDT
 Bob, When you do your crown area calculation, if you have a fat tree you need to be sure you measure to what would be the center of the tree, or else the sides of the triangle may be upwards of 7 feet or more short.   You are certainly aware of this but it should be mentioned in a formal description of the methodology. As an example of the crown volume calculation look at the Audubon Park Oak on http://www.nativetreesociety.org/fieldtrips/louisiana/audubon/audubon_park_live_oaks.htm The Audubon Park Oak- CBH-35’ 2”, Spread-165’, and Height-60’ The branches droop downward so the crown actually extends to the ground, rather from the branching point upward. It could be characterized as half of an ellipsoid: Volume = (1/2)4/3 (pi) a b c where a, b and c are the radii of the major axis. c = height = 60',   a = 1/2 max spread = 1/2(165) = 82.5', b = 1/2 spread at 90 degrees from max (unknown but say it was 80% the maximum spread) = 66'. Then the volume of the crown would be  683, 892 cubic feet. The crown area would be roughly 17,100 square feet using those assumptions, and without your polygonal modeling. It would be interesting to see what the actual b axis length would be, and what number you would derive from your polygonal crown area modeling. The "footprint" of the crown is something I think would be worth measuring on massive trees such as these. Ed
 RE: Crown Measures Robert Leverett Jun 01, 2007 08:15 PDT
 Ed,    Yes, absolutely. That's why I included ray length + 0.5*diameter.    I think someone by some method calculated the cross sectional area covered by the Angel Oak in South Carolina. On occasion those kinds of calculations are performed and reported in newspaper articles, but the methodology is never described. Bob
 Re: Crown Measures Beth Koebel Jun 01, 2007 12:53 PDT
 Bob, I was thinking (look out! the smoke is pouring out of my ears.) about how to measure the crown volume. I don't have everything in place but one idea that maybe you could expand on. I think that you could use right pyramids with the top facing in to the trunk. The edges of the base would meet each other until you have covered the whole tree. How one would do this from the ground I don't know. The formula for the volume of a right pyramid is (area of base) X height/ 3 the base is a polygon so you would need the formula for that as well and it is, according to Pocket Ref, page 478, Area = n ar/2 = nrSQR tan(angle) = nRSQR/sin2(angle) where a = length of a side(all sides being equal) n = number of sides r = distance from the center of the polygon to the edge of a side R = distance from the center of the polygon to the corner where two sides meet. Pocket Ref also simplifies this by the number of sides the polygon has on page 478. sides     area 3.........0.4330aSQR 4.........1.0000aSQR 5.........1.7205aSQR 6.........2.5981aSQR 7.........3.6339aSQR 8.........4.8284aSQR 9.........6.1818aSQR 10........7.6942aSQR 11........9.3656aSQR 12.......11.1962aSQR where a = length of a side when all sides are equal. Beth
 Back to Beth Robert Leverett Jun 04, 2007 07:17 PDT
 Beth,    Thanks for making the contribution. Please keep your thinking cap on. We make progress this way.    Perhaps the biggest drawback to applying the right pyramid shape is its regularity. The base is a regular polygon (equal sides) and that imposes both two and three-dimensional limits that our unruly trees just won't respect. Ellipsoids may be a better solution, but even there, we impose a regular shape that errant limbs and gaps between limbs and multiple limb layers don't respect. However, there is no need to rule any form out. We should keep the regular pyramid in our reperatoire. Who knows, maybe we'll discover tree pyramid power.    The non-regular polyon approach that I proposed a few e-mails back was just to measure projected crown area in a way that allows for non-regular limb projections - anywhere they occur. We seek to respond to actual tree geometry. The question is what do we do when we project the two-dimensional base into 3 dimensions. We might think of enclosing the crown by a series of prisms each with a triangular base representing a slice out of the non-regular basal polygon used to calculate projected crown area. We would then only need to multiply the projected crown area by the height of the tree to compute the multi-prism crown enclosure. This is just a tightening of the reins over the cylinder that Ed has proposed as a starting point.     We could also use the triangles making up the polygon to construct a series of pyramids, but they wouldn't be right pryamids. Nonetheless, we could see where the approach leads. Some of the pyramids will encompass too much space and in others not enough. By contrast, the prisms will always include too much space. I can't think of any tree that I'ver ever seen that would be an exception, although Lombardy Poplars come close.     Back to the encompassing prism model. If we could determine at several randomly chosen points within the multi-prism shape whether we were inside foliage or outside, we could adjust the volume of the multi-prism shape accordingly by a simple percentage calculations. The key to this method is to be able to determine if we are within the crown enclosure or not at each point. That is not easy, but I think I have a partial solution to the problem, which I'll propose in a future e-mail. Bob