# Divide the 3D surface into equal patches

12 ビュー (過去 30 日間)
Trinh Nam 2021 年 5 月 26 日
コメント済み: darova 2021 年 7 月 24 日
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

サインインしてコメントする。

### 採用された回答

darova 2021 年 5 月 29 日
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 サインインしてコメントする。

### その他の回答 (5 件)

darova 2021 年 5 月 26 日
Try griddata to interpolate
##### 1 件のコメント表示非表示 なし
Trinh Nam 2021 年 5 月 27 日
I already tried, although it can generate the rectangle patch, but the area of these patch are not equal

サインインしてコメントする。

darova 2021 年 5 月 27 日
Try to interpolate in polar system. Find center of a circle clc,clear
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) ##### 0 件のコメント表示非表示 -1 件の古いコメント

サインインしてコメントする。

Trinh Nam 2021 年 5 月 28 日
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)
##### 1 件のコメント表示非表示 なし
darova 2021 年 5 月 28 日
• 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) サインインしてコメントする。

Trinh Nam 2021 年 5 月 29 日
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
##### 0 件のコメント表示非表示 -1 件の古いコメント

サインインしてコメントする。

Trinh Nam 2021 年 5 月 30 日

Hi Darova
thanks so much for your help.
t = [0 cumsum(sqrt(dx+dy+dz))]; % parameter --> is it you calculating the arc lenght ?
I don't know 2 lines below, could you please explain ?
t1 = linspace(0,t(end),size(x,2)); % new parameter (why in herem you used size(x,2) instead of size(x,1) ?
x1(i,:) = interp1(t,x(i,:),t1); --> I don't fully understand this interpretion ..
##### 3 件のコメント表示非表示 2 件の古いコメント
darova 2021 年 7 月 24 日
Try to use simple mean
xc = mean(x(i:i+1,j:j+1)); % x center coordinate of the (i,j) patch

サインインしてコメントする。

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!