On Wednesday, March 13, 2013 6:36:38 AM UTC+5:30, Math Guy wrote: > Looking for some thoughts about how to understand this problem. > A closed loop (an irregular ring) is defined by a set of n points in > space. > Each point has an (x,y,z) coordinate. The points are not co-planar. > Typically, this ring would approximate the perimeter of a horse saddle, > or a potato chip. The number of points (n) is typically from 6 to 12 > (usually 9) but will never be more than 16. ( chip like Pringles brand ?) > The way I see it, there are two ways to understand the concept of the > area of this ring. > a) if a membrane was stretched across the ring, what would the area of > the membrane be? Think of the membrane as a film of soap - which > because of suface tension would conform itself to the smallest possible > surface area. This would be Area A. > b) if the ring represented an aperture through which some material (gas, > fluid) must pass, or the flux of some field (electric, etc). This would > be Area B. > I theorize that because the points that define this ring are not > co-planar, that Area A would not be equal to Area B. > I am looking for a numerical-methods formula or algorythm to calculate > the "area" of such a ring, and because I believe there are two different > areas that can be imagined, there must be two different formulas or > algorithms, and thus I'm looking for both of them. > If I am wrong, and there is only one "area" that can result from such a > ring, then I am looking for that formula.
I liked the problem. The problem of Rado and Plateau are classical, but I found
no easier guide for this problem and gave it up, temporarily at least.
To help you towards its solution, we may still try. But before attempting a numeric solution,scalar invariants are to be first understood.
When curvature and torsion of a closed non-planar rigid loop (arc = s single
parameter) in 3-space are given, you want to find the minimal area.
From differential geometry/surface theory, mean curvature H = (k1 + k2)/2 = 0
ds^2 = E du^2 + 2 F du dv + G dv^2. Two parameters u,v are linked to edge arc
parameter s. u and v should be chosen such that E N + G L = 2 F M , if that
surface should be of minimal area.
Area = Integral sqrt( E G - F^2)du dv
If pressure is introduced across a soap film of rigid boundary,normal curvatures increase, Gauss curvatures also increase.Normal curvature is kn, surface tension = T, then
kn = p/ T.
Normal curvatures are zero along asymptotic directions. In my view they are most natural parameter lines to deal with this problem.
So for each p, there is one minimal sutface that can be defined. p =0 is the minimal area film of _perhaps_ minimal minimal integral curvature. Integ K dA.
Normal curvature changes with direction si. Now kn = k1 cos(si)^2 + k2 sin(si)^2 ; si = 0 or pi/2 for principal directions. This is from Euler's relation. Positive and negative kn areas areas are partitioned by kn = 0 lines.
Gauss curvature K = k1* k2 < 0 when H = 0 invariably at any saddlle point of soap film.