How to find projection plane for 3d points

Math and Physics Programming

Started by mmakrzem July 16, 2015 08:38 PM

3 comments, last by mmakrzem 8 years, 8 months ago

1,037

Author

July 16, 2015 08:38 PM

I have a set of points defined in 3D space. I’d like to find the optimal plane for which I could project all of these 3d points onto and have them “nicely” spaced out. I realize that there may not be an optimal solution if all my points are poorly spatially distributed. As a simple example, imagine we have points (illustrated as white dots below) such that they encircle a sphere.

[attachment=28178:points.jpg]

I’d like to find the cross section plane that is illustrated in the image below. Could someone suggest how we could find this plane given only the x,y,z coordinates of the 3d points?

[attachment=28177:plane.jpg]

--- @www.CreateStones.com

Randy Gaul

2,835

July 16, 2015 09:07 PM

You can try using Newell's plane building algorithm:

http://box2d.org/2014/01/troublesome-triangle/

http://cs.haifa.ac.il/~gordon/plane.pdf

mmakrzem

1,037

Author

July 16, 2015 09:14 PM

Thanks, I'll give that a try

--- @www.CreateStones.com

clb

2,152

July 22, 2015 03:24 PM

"Nicely" is a bit informal, but if somehow maximizing the area of the distances between the points is important, then two approaches that come to mind:

- Use PCA to compute the direction of maximum spread and use the axis information to form the plane, e.g. https://en.wikipedia.org/wiki/Principal_component_analysis and http://setosa.io/ev/principal-component-analysis/

- First compute the diameter of the 3D point set, http://www.tcs.fudan.edu.cn/rudolf/Courses/Algorithms/Alg_ss_07w/Webprojects/Qinbo_diameter/intro.htm to find the first direction for one of the axes of the plane, and then form the second axis by finding the diameter of the 2D point set that is formed by projecting along the first chosen axis.

These might give different kinds of planes compared to Newell's method.