Surface Identification, Area and Volume Calculations in a Linked Simplex Mesh

Intro

The simplex meshes are point representations with straight lines interpolated between the points. The limitation with this is that the number points often has to be increased significantly to get curves or surfaces which look smooth.

Summing of the simplexes generally seems to underestimate the volume. Adding more points can be taxing on resources (in computational time and memory). A natural step it to look at interpolation.

The boundary or surface identification is related to integration which is volume and area in that they are the same problem. Integration needs the boundary identified and represented.

Area Calculation

Here is the problem. The aim is to calculate the area of the triangle by summing the tessellations triangles.

As a solution I propose to use Cubic Hermite Interpolation in Vector Space to firstly identify the boundary curve of surfaces in 2D linked simplex tessellations.

Then with this curve add the region between the interpolating curve and the straight line segment for each simplex on the boundary.

The calculation is then a sum of the convex hull's area (all the triangles) and a sum of the area between the convex hull and the interpolated curve.

Interpolation and Circles and Spheres

A simple strategy is to use a surface interpolation function to sample the surface on the boundary. Consider a lone simplex. Interpolate and add points along its boundary. Retessellate in a fan, assuming the boundary is on one side only. Then sum the individual pies.

In 2D a pie is calculated with r*cos(theta). Approximate pie with r as the average of the two sides and calculate theta the angle between the two sides.

In 3D a pie is three points on a sphere and the center of the sphere. The length of the three points to the center must be equal so let the three points be on the boundary, the center be the other point in the simplex within the mesh. Let the radius be the average of the three points to the center.

<TODO> Find a formla for the volume calculation. Essentially given a unit sphere and three points on it find the volume of the pie.

A pie could more formally be defined with the pivot at the N-dimensional sphere's center, with n edges to the surface. The edges form halfspaces which partition the space. These halfspaces and the N-dimensional sphere bound the pie.

http://local.wasp.uwa.edu.au/~pbourke/geometry/sphere/
m=(a+b+c)/2
A=area of triangle on sphere surface=
  4*arctan( (tan(m/2)*tan((m-a)/2)*tan((m-b)/2)*tan((m-c)/2) )^(1/2) )
V=4/3*Pi*r^3
S=4*Pi*r^2
volume of pi = 
  A/S*V = r/3*A