spiral coordinates

SPIRAL COORDINATES - 260505

IMAGINE THE TRUNK OF A TREE AS A SIMPLIFIED CYLINDER. IMAGINE THIS CYLINDER INTERSECTS WITH A RANDOM ARRANGEMENT OF BRANCHES, THEMSELVES TO BE MODELED AS SIMPLE CYLINDERS, GROW AND EXTEND INTO RECURSIVE CYLINDERS INTERPENETRATING CYLINDERS. THOUGH THEIR ARRANGMENT IN SPACE IS COMPLICATED AND LACKS RECTILINEARITY, SUPPOSE WE WISHED TO APPLY THE SAME RIGORS OF CARTESIAN SPACE TO FIND THE LOCATION OF A LEAF. ANY TERMINAL OR “LEAF” AT THE EDGE OF THIS WEB OF INTERSPEARING CYLINDERS LACKS ANY ELEGANT DESCRIPTION OF ITS POSITION, EITHER RELATIVE TO THE SPACE AROUND IT OR TO THE NETWORK OF PARENT CYLINDERS. IF WE IMAGINE A WALL (WITH A CONVENIENT GRID DRAWN UPON IT, IF SUCH A THING CAN BE IMAGINED) AND A LIGHTSOURCE, WITH THE TREE BETWEEN, THE TREES SHADOW WILL BE CAST UPON THE GRIDWALL BY A PROJECTION FROM 3 SPACE TO 2 SPACE:

P:R^3->R^2.

THE POSITION OF A LEAF MAY BE DESCRIBED BY THE FUNCTION WHICH TAKES IN ITS POSITION ON THE WALL AND COMPUTES THE INVERSE OF P TO GET ITS POSITION IN 3 SPACE. THIS, OF COURSE, REQUIRES THAT WE REPROJECT THE TWO DIMENSIONS INTO THREE, NEGLECTING THE INEVITABLE OVERALAPS BETWEEN LEAVES.

SO WHY NOT JUST RECORD THE LEAVES POSITION IN 3 SPACE? THOUGH EASY TO DO WE LOSE OTHER INFORMATION AND, AT THE RISK OF HYPERBOLE, DO VIOLENCE TO THE BEAUTY AND COMPLEXITY OF FRACTAL BRANCHING. IF WE USE THE CORE OF THE TRUNK AS ITS ZERO VECTOR AND CALCULATE THE POSITION RELATIVE TO ZERO WE GET THE POSITION IN OUR FAMILIAR SPACE. EASY ENOUGH. I HOPE EVEN THE MOST CALOUS ANTINATURALIST — VEINS FLOWING WITH CRUED OIL AND HANDS SHAKEY FROM A DAY OF CLEARCUTTING — WILL AGREE WE LOSE SOMETHING IMPORTANT AND INTERESTING ABOUT THE FRACTALINE SHAPE OF THE TREE, WHICH ONLY TERMINATES IN A LEAF.

SO HOW SHOULD THE POSITION OF A LEAF BE DEFINED?

MY PROPOSAL IS THAT WE CAN USE SPIRAL COORDINATES

SPIRAL COORDINATES, AS I WILL HEREIN COIN THE TERM, DESCRIBES A CYLINDRICALLY BRANCHING STRUCTURE USING A RIGHHAND RULE TO CREATE ORDINAL VECTORS.

STARTTING AT THE BASE OF THE TREE AND FOLLOWING THE CURVE OF OUR RIGHTHAND (WITH OUR THUMB PARALLELLING THE STARTING UPWARD THRUST OF THE TRUNK) EACH NODE IS GIVEN AN ORDINAL EQUAL TO WHEN WE ENCOUNTER IT ALONG THIS SPIRAL. THE FIRST BRANCH ON A TRUNK IS GIVEN THE ORDINAL 1, THE SECOND 2, ETC. EVERY BRANCH WHICH EXTENDS FROM THE TRUNK WILL HAVE A UNIQUE ORDINAL. IF THE TRUNK ITSELF SPLITS, THERE SHOULD ALWAYS BE A MORE OBVIOUS CHOICE BETWEEN ‘TRUNK’ AND ‘BRANCH’. IF NO SUCH CHOICE EXISTS, SELECT THE TALLER OF THE TWO AS THE TRUNK SO THE TERMINUS OF THE TRUNK IS ALSO USUALLY THE TALLEST POINT ON THE TREE. THE COORDINATE OF THIS POINT WILL BE EQUAL TO THE NUMBER OF BRANCHES WHICH EXTEND FROM THE TRUNK PLUS ONE, SINCE EACH OF THOSE LOWER BRANCHES WERE ENCOUNTERED BEFORE THIS FINAL ONE. THIS LEAF, AT THE TOP OF THE TRUNK, WILL BE SAID TO HAVE A DIMENSION OF 1 SINCE THE NUMBER OF ORDINALS NEEDED TO DESCRIBE ITS POSITION IS ALSO 1. IF A LEAF IN QUESTION EXTENDS FROM ONE OF THESE LOWER BRANCHES WE RECORD THE ORDINAL OF THAT BRANCH AND BEGIN COUNTING BRANCHES OF THAT BRANCH UNTIL WE FIND THE BRANCH WHICH CONTAINS SAID LEAF, AND THEN CONTINUE THIS PROCESS UNTIL WE ARE AT THE TERMINUS.

MY CLAIM, AND THE REASON FOR THE PRECEEDING PREAMBLE, IS THAT NO LEAF WILL EVER HAVE A DIMENSION GREATER THAN 8 (OR 7, IF WE DONT COUNT THE STEM OF THE LEAF AS A BRANCH), OR IN OTHERS WORDS, NO LEAF WILL HAVE MORE THAN 8 PARENT BRANCHES INCLUDING THE TRUNK.