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How to approximately (or exactly, if it's easier) find a function which describes the 'density' of the path of a 2D double pendulum?

Double pendulum Image from the program at http://dllu.net/dp/

Here are some initial step(s) I would take: I would use the solutions to the usual 2D double pendulum to calculate the path for arbitrary m,M,l and L, and then 'compress' the path into a circular probability distribution (the circumference) and then make an equation describing the vertical probability.

  1. To simplify, 'compress' all the points into a circle. The thickness of each infinitesimal arc on the circumference of the circle represents the probability that the path of the pendulum would be on the radial line (the line that intersects the center of the circle and the infinitesimal arc).

enter image description here

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What type of pendulum? is the radius constant? 2D or 3D? single or double? – nbubis Dec 27 '12 at 8:12
A 2D double pendulum. The distance between each pair of pendulums is constant. – Raindrop Dec 27 '12 at 8:13
I wouldn't jump to any conclusions from this applet - it looks like the pendulum in the simulation is losing kinetic energy after some point due to "friction" probably caused by floating point inaccuracies. I would think the "real" distribution would be less dense at the bottom. – nbubis Dec 27 '12 at 8:20
I would use the solutions to the usual 2D double pendulum to calculate the path for arbitrary m,M,l and L, and then 'compress' the path into a circular probability distribution (the circumference) and then make an equation describing the vertical probability – Raindrop Dec 27 '12 at 8:24
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