
Re: Calculating the area of a closed 3D path or ring
Posted:
Mar 14, 2013 7:21 AM


On Mar 13, 3:25 pm, Narasimham <mathm...@gmail.com> wrote: > 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 coplanar. > > 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 > > coplanar, that Area A would not be equal to Area B. > > I am looking for a numericalmethods 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 nonplanar rigid loop (arc = s single > > parameter) in 3space 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 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. > > Hope it can begin. > > Narasimham
The simplest case is axisymmetric, si= pi/4 and the lines intersect orthogonally for a catenoid of revolution.
r = c cosh(z/c) = c cosh( t ) where ( r,t,z ) are in a cylindrical coordinate system.

