Divide the 3D surface into equal patches

Hi, I am looking for a method or algorithm to divide a 3D surface into equal patches. The detail is explained below

Background:

the initial plate has grids as in picture (1). initial position of A1 (a1,b1,c1) …. A121 (a121, b121, c121). All of this position is known. A1 is the origin A1(0,0,0)

After deformation, the plate becomes 3D curve and grids has new position A’1(u1,v1,w1) …A’100(u100, v100,w100).

Currently, I used 3D scanner to get the point cloud data of deformed plate and successfully get the equation of 3D surface.

My question is: How to divide the 3D surface into 100 patches with equal area to find the grid A’2 …A’100 position. (A’1 is the origin A1(0,0,0) and coincide with A1) I already tried matlab built-in function and other software but the result gives me the mesh with different in area. (example: area of S1 ≠ S2 ≠ S100)

Attachment is the point cloud data after processed

Thanks for your support.

Answer 1

[darova]

Try arc length interpolation. Original link: LINK

function main
clc,clear

    % generate some data
[x,y] = meshgrid(-1:0.4:1);
z = x.^2+y.^2;
surf(x,y,z,'facecolor','none')
    % interpolation
[x1,y1,z1] = myinterp1(x,y,z);
[x2,y2,z2] = myinterp1(x1',y1',z1');
hold on
plot3(x2,y2,z2,'.r')
hold off
axis equal

function [x1,y1,z1] = myinterp1(x,y,z)
x1 = x*0;
y1 = y*0;
z1 = z*0;
for i = 1:size(x,1)
    dx = diff(x(i,:)).^2;
    dy = diff(y(i,:)).^2;
    dz = diff(z(i,:)).^2;
    t = [0 cumsum(sqrt(dx+dy+dz))];     % parameter
    t1 = linspace(0,t(end),size(x,2));  % new parameter
    x1(i,:) = interp1(t,x(i,:),t1);
    y1(i,:) = interp1(t,y(i,:),t1);
    z1(i,:) = interp1(t,z(i,:),t1);
end

end
end

[Trinh Nam]

Hi Darova, thanks for your support. I don't understand some points, could you please explain ?

1. Why the 3D surface is translated along the X axis after running the code ?

2. Why we cannot calculate the "Red point" in the Top and Bottom edge of the 3D surface while it can find the "Red point" along Left edge and Right edge.

3. Why we need to translate the X +15 unit and Z for -100 unit ? , According to the code

plot3(x+15,y,z-100) and [t,r] = cart2pol(x+15,z-100)

Thanks for your explaination

[darova]

• 1. Why the 3D surface is translated along the X axis after running the code ?

I don't understand the question. What translation?

• 2. Why we cannot calculate the "Red point" in the Top and Bottom edge of the 3D surface while it can find the "Red point" along Left edge and Right edge.

We can. Try griddedInterpolant instead. It has extrapolation property

• 3. Why we need to translate the X +15 unit and Z for -100 unit ? , According to the code

It's needed to place your data on the circle (to interpolate it in polar syste)

[Trinh Nam]

Hi Darova

thanks for your explain. But i think this solution is not general. We cannot apply for the saddle shape or the pillow shape.

Saddle shape General shape Pillow shape

I am looking for the general way to do this job, I am appreciate if you have any algorithm

[Akhila]

Hi Trinh

Did you get the solution to your problem?

I also need this from my work, if you have the solution could you share it with me?

Answer 2

[darova]

Try griddata to interpolate

[Trinh Nam]

I already tried, although it can generate the rectangle patch, but the area of these patch are not equal

Answer 3

[darova]

Try to interpolate in polar system. Find center of a circle

clc,clear

data = load('curve.txt');
x = data(:,1);
y = data(:,2);
z = data(:,3);
t = linspace(0,2*pi,20);
[x1,y1] = pol2cart(t,100);
 plot3(x+15,y,z-100)
%line(x1,y1)

[t,r] = cart2pol(x+15,z-100);
t0 = linspace(max(t(:)),min(t(:)),10);
y0 = linspace(min(y(:)),max(y(:)),10);
[T,Y] = meshgrid(t0,y0);
R = griddata(t,y,r,T,Y);
[X,Z] = pol2cart(T,R);
hold on
plot3(X,Y,Z,'.r')
hold off
view(45,45)