How to normalize a mesh into -1 to 1, then revert from normalized mesh to original one?

Question

I have a mesh model in X, Y, Z format. Lets say.

Points *P;

In first step, I want to normalize this mesh into (-1, -1, -1) to (1, 1, 1). Here normalize means to fit this mesh into a box of (-1, -1, -1) to (1, 1, 1).

then after that I do some processing to normalized mesh, finally i want to revert the dimensions to similar with the original mesh.

step-1: P = Original Mesh dimensions;

step-2: nP = Normalize(P); // from (-1, -1, -1) to (1, 1, 1)

step-3: cnP = do something with (nP), number of vertices has increased or decreased.

step-4: Original Mesh dimensions = Revert(cnP); // dimension should be same with the original mesh

how can I do that?

Answer 1

I know how easy it can be to get lost in programming and completely miss the simplicity of the underlying math. But trust me, it really is simple.

The most intuitive way to go about your problem is probably this:

1. determine the minimum and maximum value for all three coordinate axes (i.e., x, y and z). This information is contained by the eight corner vertices of your cube. Save these six values in six variables (e.g., min_x, max_x, etc.).

2. For all points p = (x,y,z) in the mesh, compute

   q =  ( 2.0*(x-min_x)/(max_x-min_x) - 1.0
          2.0*(y-min_y)/(max_y-min_y) - 1.0
          2.0*(z-min_z)/(max_z-min_z) - 1.0 )
   

   now q equals p translated to the interval (-1,-1,-1) -- (+1,+1,+1).

3. Do whatever you need to do on this intermediate grid.

4. Convert all coordinates q = (xx, yy, zz) back to the original grid by doing the inverse operation:

   p =  ( (xx+1.0)*(max_x-min_x)/2.0 + min_x
          (yy+1.0)*(max_y-min_y)/2.0 + min_y
          (zz+1.0)*(max_z-min_z)/2.0 + min_z )
   

5. Clean up any mess you've made and continue with the rest of your program.

This is so easy, it's probably a lot more work to find out which library contains these functions than it is to write them yourself.

Answer 2

It's easy - use shape functions. Here's a 1D example for two points:

-1 <= u <= +1
x(u) = x1*(1-u)/2.0 + x2*(1+u)/2.0
x(-1) = x1
x(+1) = x2

You can transform between coordinate systems using the Jacobean.

Let's see what it looks like in 2D:

-1 <= u <= =1
-1 <= v <= =1
x(u, v) = x1*(1-u)*(1-v)/4.0 + x2*(1+u)*(1-v)/4.0 + x3*(1+u)*(1+v)/4.0 + x4*(1-u)*(1+v)/4.0
y(u, v) = y1*(1-u)*(1-v)/4.0 + y2*(1+u)*(1-v)/4.0 + y3*(1+u)*(1+v)/4.0 + y4*(1-u)*(1+v)/4.0